Thermodynamics 1 Initial Draft 1

INITIAL DRAFT – MANUSCRIPT
FOR THERMODYNAMICS 1
TABLE OF CONTENTS
CHAPTER I. PROPERTIES OF MATTER
1.1 Scope and Definition of Thermodynamics ……………………………………………….. 1

1.2 Laws of Thermodynamics ………………………………………………………………… 3
1.3 Analysis of Motion ……………………………………………………………………….. 3
1.4 Analysis of Forces ………………………………………………………………………… 10
1.5 Plane Mensuration ………………………………………………………………………… 23
1.6 Volume ……………………………………………………………………………………. 26
1.7 Basic Properties of Fluid ………………………………………………………………….. 30
1.8 Pressure …………………………………………………………………………………… 38
1.9 Measurement of Pressure …………………………………………………………………. 38
1.10 Pascal’s Law …………………………………………………………………………….. 39
1.11 Variation in Pressure …………………………………………………………………….. 39
1.12 Stacks of Immiscible Fluids ……………………………………………………………... 40
1.13 Pressure-Measuring Instruments ………………………………………………………… 40
1.14 Hydrostatic Force on Plane Surfaces ……………………………………………………. 46
1.15 Temperature ……………………………………………………………………………… 53
Chapter 1 Supplementary Problems ……………………………………………………….. 57 - 87
CHAPTER II. ENERGY AND POWER CONCEPTS
2.1 Conservation of Mass …………………………………………………………………….. 88

2.2 Work and Energy …………………………………………………………………………. 89
2.3 Power ……………………………………………………………………………………... 89
2.4 Unit Conversions of Energy and Power ………………………………………………….. 90
2.5 Conversion of Mass into Energy …………………………………………………………. 91
2.6 General Forms of Energy ………………………………………………………………… 92
2.7 Types of Stored Energy ………………………………………………………………….. 92
2.8 Heat ………………………………………………………………………………………. 95
2.9 Specific Heat ……………………………………………………………………………… 95
2.10 Types of Heat ……………………………………………………………………………. 96
2.11 Mechanical Work ………………………………………………………………………… 98
2.12 Other Forms of Work/Energy ……………………………………………………………. 99
2.13 Total Energy and Total Mechanical Energy ……………………………………………… 104
2.14 Energy Balance For Thermodynamic Systems ………………………………………….. 104
2.15 Energy Balance For Steady Flow Engineering Devices …………………………………. 107
2.16 Energy Balance in Steam Power Plant …………………………………………………… 112
Chapter 2 Supplementary Problems ……………………………………………………….. 134 - 146
CHAPTER III. IDEAL GAS
3.1 Equation of State ………………………………………………………………………… 147

3.2 Gas Constant …………………………………………………………………………….. 147
3.3 Gas Constant and Specific Heat ………………………………………………………… 148
3.4 Derived Equations of State ……………………………………………………………… 149
3.5 Gas Laws ……………………………………………………………………………….. 150
3.6 Buoyancy in Ideal Gas ………………………………………………………………….. 151
3.7 Analysis of Mixture of Ideal Gases …………………………………………………….. 152
Chapter III Supplementary Problems ………………………………………………………. 172 – 185
CHAPTER IV. IDEAL GAS PROCESSES
4.1 General Equations For Ideal Gas Processes ………………………………………………,, 186

4.2 Entropy …………………………………………………………………………………….. 187
4.3 Second Law and Irreversibility ……………………………………………………………. 189
4.4 Isometric Process …………………………………………………………………………... 191
4.5 Isobaric Process ……………………………………………………………………………. 193
4.6 Isothermal Process …………………………………………………………………………. 196
4.7 Isentropic Process ………………………………………………………………………….. 199
4.8 Polytropic Process …………………………………………………………………………. 203
4.9 Process Curves …………………………………………………………………………….. 206
4.10 Compressors – Application of Ideal Gas Processes ……………………………………… 207
Chapter IV Supplementary Problems ………………………………………………………. 220 - 225
MODULE I: PROPERTIES OF MATTER THERMODYNAMICS I
I. PROPERTIES OF MATTER
Learning Objectives: At the end of this chapter, the student shall
1. Understand thermodynamics, its theory, and its application to real life.
2. Analyze thermodynamic problems using force and motion analysis.
3. Using fluid properties to solve thermodynamic problems.
4. Define pressure and enumerate the six types of pressure.
5. Relate pressure to some real-life applications.
6. Determine the hydrostatic forces on submerged bodies.
7. Define temperature, different scales of temperature, and measurements in temperature
1.1 SCOPE AND DEFINITION OF THERMODYNAMICS
Thermodynamics is a branch of physical science that deals with different concepts, theories, phenomena of
energy and related properties of matter, transformation of heat energy from one form or another, It comes from two
Greek words “therme” which means “heat” and “dynamics” which means “motion”. There are two branches of
thermodynamics:
a. Statistical Thermodynamics - deals with the microscopic point of view of thermodynamics, dealing with the
thermodynamics of molecules, atoms, and particles.
b. Classical Thermodynamics- deals with the macroscopic point of view of thermodynamics, composed of
engineering thermodynamics systems, visible and in the macroscopic level.
The following are operational terms related to thermodynamics:
1. Fluid - it is a substance that exists as a continuum characterized by low resistance to flow and tendency to assume
the shape of a container. A fluid may characterize as a liquid or a gas.
Type of Fluid Applications
Steam Steam Turbine
Air Air Compressor
Air and Fuel Mixture Internal Combustion Engine
Water Hydraulic Turbine
2. Substance - is made up of molecules.
a) Simple substance - is one whose state is defined by two independent variables.
b) Pure Substance - is one that is homogeneous in composition and homogeneous and invariable in chemical
aggregation.
3. Boundary- the hypothetical or actual enclosure of the system, can be fixed, moved, real, or imaginary.
4. System - a portion of the universe, an atom, a galaxy, a certain quality of matter that one wishes to study. There
are six classification of thermodynamic systems,
a) Closed System - there is no exchanged of matter within the surroundings
b) Open System - one across whose boundaries there is a flow of mass and energy.
c) Isolated System - one completely impervious to its surroundings, neither mass nor energy.
d) Steady Flow System - there is no change in mass flowing in the system.
e) Steady State System - there is no change of state and phase in the system.
f) Adiabatic System – there is no exchange of heat from the system to the surroundings.
5. Properties- descriptive characteristics of a system.
a) Intensive Properties - independents of the mass.
b) Extensive Properties - dependent upon the mass and for those characteristics that gives total values.
c) Specific Properties - properties that is for unit mass.
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6. State - it is the condition as identified though the properties of a substance. We can determine for any given
variable if it is a thermodynamic property if
a) The property is a single value for each state
b) The change in value for given two prescribed states is a single value.
c) Any variable whose change is fixed throughout the end states.
7. Phase - it is a quantity of matter that is homogeneous throughout in chemical composition and physical structure.
Three Phases Phase Standard Molecular
Of Matter Temperature Arrangement
(a) Solid Ice 𝑡 ≤ 0°𝐶 Compact
(b) Liquid Water 0°𝐶 ≤ 𝑡 ≤ 100°𝐶 Little free of motion
(c) Gas Steam 𝑡 ≥ 100°𝐶 Free of Motion
8. Homogeneous Substance - a single phase system
9. Heterogeneous Substance - two or more phase system. Example is a mixture of air and water vapor in air
conditioning systems, or liquid and water vapor in turbine exhaust.
10. Phase Change - there is a change in state of a substance.
a. Freezing/Solidification - phase change from liquid to solid.
b. Melting - phase change from solid to liquid.
c. Vaporization - phase change from liquid to gas. The liquid vaporized is called as vapor.
d. Condensation - phase change from gas to liquid. The liquid undergoing condensation is called as
condensate.
e. Sublimation - phase change from solid to gas without passing the liquid state.
11. Thermodynamics Equilibrium - is achieved if the temperature and pressure at all points are same ; there should
be no velocity gradient ; the chemical equilibrium is also necessary. Systems under temperature and pressure
equilibrium but not under chemical equilibrium are sometimes said to be in metastable equilibrium conditions. It is
only under thermodynamic equilibrium conditions that the properties of a system can be fixed. There are three types
of equilibrium:
a) Thermal Equilibrium -the temperature of the system does not vary with time throughout the system.
b) Mechanical Equilibrium - the pressure of the system does not vary with time throughout the system.
c) Chemical Equilibrium - no chemical reaction takes place within the system, and the reaction does not
vary with time throughout the system.
12. Process -occurs when the system undergoes a change in a state or an energy transfer at a steady state. There
are four categories of process.
a. Non-Flow Process - process in which there is a fixed mass in a boundary having a change of state.
Closed systems undergo non-flow process.
b. Steady State Process - also called as flow process. It is a process wherein a fixed mass may enter and
leave the system. Open systems undergo flow/steady state process.
c. Quasi-Static Process - also called as reversible process. The term ‘quasi’ means ‘almost’. This process
is a succession of equilibrium states, allowing the system to change state back and forth at a given time.
d. Irreversible Process - are actual processes, wherein at a particular state, there is actual phase of that
state prior to its ideal state.
e. Cycle - any process or series whose end states are almost identical.
f. Point Functions - are functions whose properties can be located in coordinate axes. Examples are pressure,
temperature, and volume. These can be written as a function between two state points
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g. Path Functions - are functions that pertains to the path or the area beneath the curve in a certain process.
Examples are work and heat. These cannot be written as a function between two state points.
h. State Postulate - states that a state can be always be determine if there atleast two given thermodynamic
properties.
1.2 LAWS OF THERMODYNAMICS

There are four laws of thermodynamics
1. Zeroth Law of Thermodynamics (Law of Thermal Equilibrium) - when two bodies isolated to each other and
has a thermal equilibrium with a third body, then both bodies are thermal equilibrium to each other,
2. First Law of Thermodynamics (Law of Conservation of Energy) - states that energy is neither created nor
destroyed but transferred from one form to another.
3. Second Law of Thermodynamics (Law of Entropy) - states that wherever energy is transformed, its amount cannot
be conserved and permanently reduced to lower level.
4. Third Law of Thermodynamics (Law of Absolute Zero) - states that all physical system cannot exceed below
absolute temperature. At absolute zero, matter cease to exists.
1.3 ANALYSIS OF MOTION

Motion is defined as the change in position. Motion can be classified as absolute or relative in motion. The
point moving in space describes a line is called its path. The direction and sense of motion is defined either right or
left, positive or negative, clockwise or counterclockwise.
Types of Motion according to its path.
1. Rectilinear Motion - it is defined as the motion in a straight line. Consider a body having initial position 𝑠0 ,initial
speed 𝑣0 and time 𝑡 respectively.
Variable Definition Formula

Linear Displacement Change in position with respect to time 𝚫𝒔 = 𝒔 − 𝒔𝟎
Linear Velocity It is the rate of change of displacement with respect to time. 𝚫𝒔 𝒔 − 𝒔𝟎
𝒗= =
𝚫𝒕 𝒕 − 𝒕𝟎
Linear Acceleration It is the rate of change of velocity with respect to time 𝚫𝒗 𝒗 − 𝒗𝟎
𝒂= =
𝚫𝒕 𝒕 − 𝒕𝟎
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2. Instantaneous Motion – the microanalysis view of motion. It measures the infinitesimal change of motion per
unit time.
Instantaneous Velocity Δ𝑠 𝒅𝒔
𝑣 = lim =
∆𝑡→0 Δ𝑡 𝒅𝒕
Instantaneous Acceleration Δv 𝒅𝒗 𝒅𝟐 𝒔
𝑎 = lim = =
∆𝑡→0 Δ𝑡 𝒅𝒕 𝒅𝒕𝟐
Kinematic Differential Equation 𝒂 𝒅𝒔 = 𝒗 𝒅𝒗
3. Rectilinear Motion with Constant Acceleration – we can now derive the three kinematic equations for constant
acceleration.
From the instantaneous acceleration equation:

𝑑𝑣
𝑎=
𝑑𝑡
𝑑𝑣 = 𝑎𝑑𝑡
𝑣 𝑡
∫ 𝑑𝑣 = 𝑎 ∫ 𝑑𝑡
𝑣0 0
𝑣 − 𝑣0 = 𝑎𝑡
𝒗 = 𝒗𝟎 + 𝒂𝒕 (First Kinematic Equation)
Using the instantaneous velocity equation
𝑑𝑠
𝑣=
𝑑𝑡
𝑑𝑠 = 𝑣 𝑑𝑡
𝑑𝑠 = (𝑣0 + 𝑎𝑡)𝑑𝑡
𝑠 𝑡
∫ 𝑑𝑠 = ∫ (𝑣0 + 𝑎𝑡)𝑑𝑡
0 0
𝟏
𝒔 = 𝒗𝟎 𝒕 + 𝒂𝒕𝟐 (Second Kinematic Equation)
𝟐
From the kinematic differential equation
𝑎 𝑑𝑠 = 𝑣 𝑑𝑣
𝑠 𝑣
𝑎 ∫ 𝑑𝑠 = ∫ 𝑣𝑑𝑣
𝑜 𝑣0
1
𝑎(𝑠 − 𝑠𝑜 ) = (𝑣 2 − 𝑣0 2 )
2
𝟐 𝟐
𝒗 = 𝒗𝟎 + 𝟐𝒂(𝒔 − 𝒔𝟎 ) (Third Kinematic Equation)
Sign conventions for the acceleration a
• a is positive if it is moving in the positive direction or decelerating in the negative direction.
• a is negative if it is decelerating in the positive direction, known as deceleration or accelerating in the
negative direction, known as negative acceleration.
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4. Vertical Motion – if a body is thrown in the vertical motion, the acceleration is due to the gravitational force
against the body. Hence the acceleration is due to gravity g. The value of g is :
FPS system CGS system MKS system SI
2 2 2
32.174 ft/s 980.66 cm/s 9.8066 m/s 9.8066 m/s2
Consider a body from rest and positioned from a certain height h. We can describe its motion using the kinematic
equations, replacing the variable s with the variable h.
𝒗 = 𝒗𝟎 ± 𝒈𝒕
𝟏
𝒉 = 𝒗𝟎 𝒕 ± 𝒈𝒕𝟐
𝟐
𝒗𝟐 = 𝒗𝟐𝟎 ± 𝟐𝒈𝒉
Sign conventions for the acceleration due to gravity:

• Positive if the body is moving downward.
• Negative if the body is moving upward.
5. Projectile Motion - it is the motion in two dimensions. The path of the body is curvilinear or parabolic. The
body that is fired is called as the projectile. There are three cases in dealing projectile motion:
Case A: A projectile is fired horizontally from a height 𝑦0 . 𝑦 < 𝑦0
Case B: A projectile is fired at which the initial and final position are equal. 𝑦 = 𝑦0
Case C: A projectile is fired at ground, and to overcome a certain height y. 𝑦 > 𝑦𝑜
Definition of Terms
• Range – it is the horizontal displacement of the projectile upon reaching its final position.
• Maximum Height – it is the highest vertical position of the projectile with respect to a fixed position
• Time of Flight – it is the total time it takes for the projectile to reach its final position from its initial position.
• Apex – it is the highest point in the path of the projectile. The time it takes for the projectile to reach its
maximum height is half its time of flight.
General Steps in Solving Projectile Motion Problems:
1. Use the kinematic equation to determine its horizontal motion. The acceleration in the x-direction is always
constant at any point in the path of the projectile.
2. Use again the kinematic equation to determine its vertical motion. The acceleration in the y-direction is not
constant at any point and due to gravitational forces.
3. Solve for the required /needed variable.
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Case A: A projectile is fired horizontally from a height 𝑦𝑜 – let 𝑦0 = ℎ or the height at which the projectile is fired
horizontally. When the projectile is fired horizontally, the velocity vector in the vertical direction is always equal
to zero.
Horizontal Motion 𝑿 = 𝒗𝒙 𝒕
Vertical Motion 𝟏
𝒀 = 𝒈𝒕𝟐
𝟐
Time of Flight
𝟐𝒀
𝒕=√
𝒈
Case B: A projectile is fired in which its initial position is equal to its final position – consider the projectile that is
fired at a given angle θ, hence
Velocity Vectors 𝒗𝒐𝒙 = 𝒗𝟎 𝒄𝒐𝒔𝛉

𝒗𝒐𝒚 = 𝒗𝟎 𝐬𝐢𝐧𝛉
Horizontal Motion 𝑿 = 𝒗𝒐 𝒕𝒄𝒐𝒔𝛉
Vertical Motion 𝟏
𝒀 = 𝒗𝟎 𝒕𝒔𝒊𝒏𝛉 − 𝐠𝐭 𝟐
𝟐
We will now derive the range, maximum height, and the time of flight of the projectile.
The time of flight can be determine using the horizontal motion:

𝑿
𝒕=
𝒗𝟎 𝒄𝒐𝒔𝛉
The time it takes for the projectile to reach its maximum height is one half the total time of flight
𝑿
𝒕𝟏/𝟐 =
𝟐𝒗𝟎 𝒄𝒐𝒔𝛉
From the first kinematic equation for vertical motion, assuming upward motion
𝑣𝑦 = 𝑣0𝑦 − 𝑔𝑡
Since 𝑣𝑜𝑦 = 𝑣𝑜 𝑠𝑖𝑛θ
𝑣𝑦 = 𝑣𝑜 𝑠𝑖𝑛θ − 𝑔𝑡
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Using the horizontal motion for projectile, wherein R=X

𝑅 = 𝑣0 𝑡1/2 𝑐𝑜𝑠θ
𝑅
𝑡1/2 =
2𝑣0 𝑐𝑜𝑠θ
When the projectile is at its final position or at the ground, 𝑣𝑦 = 0, hence
𝑔𝑅
0 = 𝑣𝑜 𝑠𝑖𝑛θ −
2𝑣0 𝑐𝑜𝑠θ
2𝑣𝑜 2 𝑠𝑖𝑛θ𝑐𝑜𝑠θ
𝑅=
𝑔
From trigonometry: 𝑠𝑖𝑛2θ = 2𝑠𝑖𝑛θ𝑐𝑜𝑠θ, hence the range of the projectile fired at a given angle is
𝒗𝟎 𝟐 𝒔𝒊𝒏𝟐𝛉
𝑹=
𝒈
Using the third kinematic equation for vertical motion, assuming upward motion:
𝑣𝑦 2 = 𝑣𝑜𝑦 2 − 2𝑔ℎ
Since 𝑣𝑜𝑦 = 𝑣𝑜 𝑠𝑖𝑛θ , and at maximum height ℎ = ℎ𝑚𝑎𝑥 , the velocity in the y direction is zero. Hence
0 = 𝑣0 2 𝑠𝑖𝑛2 θ − 2𝑔ℎ𝑚𝑎𝑥
𝒗𝟎 𝟐 𝒔𝒊𝒏𝟐 𝛉
𝒉𝒎𝒂𝒙 =
𝟐𝒈
Case C: If a projectile is initially fired at an angle from the ground and to just pass through a height h. – we can
use the horizontal motion and relating it to the vertical motion of the projectile.
For the horizontal motion:

𝑋 = 𝑣0 𝑡𝑐𝑜𝑠θ
For the vertical motion:
1
𝑌 = 𝑌0 +𝑣0 𝑡𝑠𝑖𝑛θ + 𝑔𝑡 2
2
Since
𝑋
𝑡=
𝑣0 𝑐𝑜𝑠θ
Hence
2
𝑋 1 𝑋
𝑌 − 𝑌0 = 𝑣0 𝑠𝑖𝑛θ ( )+ 𝑔( )
𝑣0 𝑐𝑜𝑠θ 2 𝑣0 𝑐𝑜𝑠θ
Thus:
𝒈𝑿𝟐
𝒀 − 𝒀𝟎 = 𝑿𝒕𝒂𝒏𝛉 −
𝟐𝒗𝟎 𝟐 𝒄𝒐𝒔𝟐 𝛉
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6. Rotation - is defined as that motion of a rigid body in which the particles move in a circular path with their
centers on a fixed straight line that is called the axis of rotation. The planes on the circles in which the particles
move perpendicular to the axis of rotation.
a. Kinematics of Rotation
Consider a pulley free to rotate around an axle O under the action of a weight W
suspended from a cord wound around the pulley. Assume the weight descends at
s as shown in the figure. This will unwind the pulley a length of chord equal to s
so that point B on the rim will occupy the position of A. The angular distance θ
through which the pulley rotates is subtended by radii drawn from point A to B.
The relation between linear displacement and angular displacement of the pulley is described by the equation
𝒔 = 𝒓𝜽
The angular displacement(θ) is defined as the angular distance swept through by any line in a rigid body, measured
in radians, degrees, or revolutions.
Differentiating with respect to time t, assuming the radius of the pulley is constant.
𝑑𝑠 𝑑𝜃
=𝑟
𝑑𝑡 𝑑𝑡
The angular velocity (𝜔) is the rate of change of angular displacement. Hence
𝒅𝜽
𝝎=
𝒅𝒕
Relating linear velocity to angular velocity:
𝒗 = 𝒓𝝎
Differentiating again with respect to time t, assuming again that the radius is constant:
𝑑𝑣 𝑑𝜔
=𝑟
𝑑𝑡 𝑑𝑡
The angular acceleration (𝛼) is the rate of change of angular velocity, hence:
𝒅𝝎
𝜶=
𝒅𝒕
Relating linear acceleration to angular acceleration:
𝒂 = 𝒓𝜶
We can also define the tangential and normal components of acceleration in terms of kinematic equations of rotation.
The direction tangential acceleration is also equal to the direction of the linear acceleration. We can conclude that
the magnitude of linear acceleration and tangential acceleration are equal
𝒂𝒕 = 𝒓𝜶
Since v=rω, the normal acceleration of any point on the rim of the pulley is given by
𝒂𝒏 = 𝒓𝝎𝟐
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The three kinematic differential equations are as follows

Variable Angular Velocity 𝒅𝜽
𝝎=
𝒅𝒕
Variable Angular Acceleration 𝒅𝝎
𝜶=
𝒅𝒕
Kinematic Differential Equation of 𝛚𝒅𝛚 = 𝛂𝒅𝛉
Rotation
Summarizing all the derived equations, together with the kinematics of rectilinear motion, and relating rectilinear
to rotational motion.
Rectilinear Motion Rectilinear to Rotational Motion Rotational Motion
𝒅𝒔 𝒔 = 𝒓𝜽 𝒅𝜽
𝒗= 𝝎=
𝒅𝒕 𝒗 = 𝒓𝝎 𝒅𝒕
𝒅𝒗 𝒅 𝒔𝟐
𝒂𝒕 = 𝒓𝜶 𝒅𝝎
𝒂= = 𝜶=
𝒅𝒕 𝒅𝒕𝟐 𝒅𝒕
𝒂𝒏 = 𝒓𝝎𝟐
𝒂 𝒅𝒔 = 𝒗 𝒅𝒗 𝛚𝒅𝛚 = 𝛂𝒅𝛉
We can also derive the equations of rotation for constant angular acceleration, same as the constant acceleration for
linear motion
Rectilinear Motion Rotational Motion
𝒗 = 𝒗𝒐 ± 𝒂𝒕 𝝎 = 𝝎𝒐 ± 𝜶𝒕
𝟏 𝟏
𝒔 = 𝒔𝒐 + 𝒗𝒐 𝒕 ± 𝒂𝒕𝟐 𝜽 = 𝜽𝒐 + 𝝎𝒐 𝒕 ± 𝜶𝒕𝟐
𝟐 𝟐
𝟐 𝟐 𝟐 𝟐
𝒗 = 𝒗𝟎 ± 𝟐𝒂(𝒔 − 𝒔𝟎 ) 𝝎 = 𝝎𝟎 ± 𝟐𝜶(𝜽 − 𝜽𝟎 )
Rules of Direction: The sense of positive θ determines the sense of positive ω and α. (+) for acceleration and (-)
for deceleration
The period T is the amount of time it takes for a body to complete one revolution. Hence the reciprocal of period is
called as the frequency. Units of frequency are in terms of hertz (Hz)
𝟏
𝒇=
𝑻
The centripetal acceleration 𝑎𝑐 is defined as the acceleration of a particle moving in a circular path. Hence
𝒗𝟐
𝒂𝒄 =
𝒓
Where r is the radius of the circle.
For rotating parts in the machine, it is common for mechanical engineers to describe its speed in terms of revolutions
per unit time. The periphery speed, also called as the surface speed is the linear speed of the circumference of a
revolving body. From the definition of velocity in terms of angular speed
𝑣 = 𝑟ω
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Since one revolution 𝑁̇ is equivalent to 2π radians, and ω is expressed in radians per second, then the periphery
speed of rotating body expressed in revolutions per second is
𝒗 = 𝟐𝛑𝒓𝑵̇
𝒗 = 𝛑𝑫𝑵̇
Where 𝑁̇ is in revolutions per second. If two rotating bodies (say a pulley and a
belt) moving with the same angular velocity, then the ratio of speed of the
pulley and belt is equivalent to the ratio of their respective radius.
𝒗𝒃 𝒓 𝒃
=
𝒗𝒂 𝒓𝒂
To convert angles, the following is the system of units in converting angles.

CONVERSION OF ANGLE MEASUREMENTS
1 revolution = 360° (Degrees)
= 2π rad (Radians)
= 400 grad (gradians)
= 400 gons (gradians)
= 6400 mils (Military Angle)
1.4 ANALYSIS OF FORCES

Isaac Newton defined the three laws of motion.
1. Law of Inertia -a body at rest will remain at rest and a body moving in constant velocity will remain in constant
velocity unless acted by a net unbalanced external force.
2. Law of Acceleration - the acceleration is directly proportional to the force and inversely proportional to its mass.
3. Law of Interaction - forces applied must be balance by an equal opposite force.
From Newtons Second Law of Motion:

𝐹
𝑎∝
𝑚
𝐹
𝑎=𝑘
𝑚
𝒎𝒂
𝑭=
𝒌
Where k is called the proportionality constant and m is the mass which is defined as the absolute quantity of matter,
When k is unity, its value is defined in different systems of units:
English System Metric System SI
FPS CGS MKS 𝑘𝑔𝑚 − 𝑚
1.0
K 𝑠𝑙𝑢𝑔 − 𝑓𝑡 𝑔𝑚 − 𝑐𝑚 𝑘𝑔𝑚 − 𝑚 𝑁 − 𝑠2
1.0 1.0 1.0 𝑘𝑔𝑚 − 𝑚
𝑙𝑏𝑓 − 𝑠2 𝑑𝑦𝑛𝑒 − 𝑠2 𝑁 − 𝑠2 1000
𝑘𝑁 − 𝑠2
10
Interpretation of the value of 𝑘 where k is unity:

1 pound force is needed to accelerate 1 slug mass at 1 ft per sec2
1 dyne is needed to accelerate 1 gram-mass at 1 cm per sec2
1 newton is needed to accelerate 1 kilogram-mass at 1 m per sec2
1 kN is needed to accelerate 1000 kilogram-mass at 1 m per sec2
Weight is defined as the gravitational force exerted to the body

where a = g. Hence.
𝒎𝒈
𝑾=
𝒌
𝒎𝒈
𝒌=
𝑾
For proportionality constant k where its value is not unity in different system of units.
English Metric SI
FPS CGS MKS 𝑘𝑔𝑚 − 𝑚
1.0
K 𝑙𝑏𝑚 − 𝑓𝑡 𝑔𝑚 − 𝑐𝑚 𝑘𝑔𝑚 − 𝑚 𝑁 − 𝑠2
32.174 9806.6 9.8066
𝑙𝑏𝑓 − 𝑠2 𝑔𝑓 − 𝑠2 𝑘𝑔𝑓 − 𝑠2 1000 𝑘𝑔𝑚 − 𝑚
𝑘𝑁 − 𝑠 2
Interpretation of the value of 𝑘 where k is not unity:

1 lbf is required to accelerate 32.174 pound-mass at 1 ft per sec2
1 gram-force is required to accelerate 9806.6 gram-mass at 1 cm per sec2
1 kilogram-force is required to accelerate 9.8066 kilogram-mass at 1 m per sec2
1 N is required to accelerate 1 kilogram-mass at 1 m per sec
Where
N is in newtons kgm is kilogram-mass
lbf is in pound-force gm is gram-mass
gf is in gram-force lbm is pound-mass
kgf is in kilogram-force
Standard conversion of mass and weight are as follows: (Note: All mechanical engineering students must memorize
the said conversions)
MASS AND WEIGHT CONVERSION FACTORS
1 kgf = 1 kilopond 1 stone = 14 lbm
= 2.205 lbf 1 quintal = 1 centner
=9.8066 N = 1 hundredweight
1 lbf = 32.174 poundals = 100 kgm
=4.4482 N 1 slug = 1 english slug
=7000 grains = 1 gee pound
=256 drums = 32.174 lbm
= 16 ounces (oz) = 14.59 kgm
11
1N = 1 x 105 dynes 1 hyl = 1 metric slug

1 kip = 1 kilopound = 1 mug
= 1000 lbf = 1 TME
1 ton = 1 english ton = 1 par
= 1 short ton = 1.055 slug
= 2000 lbm = 9.8066 kgm
= 907.03 kgm 1 blob = 1 slinch
1 tonne = 1 metric ton = 12 slugs
= 1 long ton = 1 𝑙𝑏𝑓 − 𝑠2 /𝑖𝑛2
= 1000 kgm
= 2205 lbm
Since k is equal for both cases, we can relate the force acting on the body against its weight using proportionality
constant. Hence
𝑾 𝑭
=
𝒈 𝒂
The pound -force (lbf) is the unit of force in the English system may be express in terms of pounds(lb) and poundals.
The poundal uses pound-mass and the pound uses slug as its mass. Hence
𝒍𝒃𝒎 − 𝒇𝒕
𝟏 𝒑𝒐𝒖𝒏𝒅𝒂𝒍 = 𝟏. 𝟎
𝒔𝟐
𝒔𝒍𝒖𝒈 − 𝒇𝒕
𝟏𝒑𝒐𝒖𝒏𝒅 = 𝟏. 𝟎
𝒔𝟐
Analysis of forces can be used when thermodynamic problems involves
a. Static Equilibrium
b. Dynamic Equilibrium
c. Friction Forces
d. Centripetal Forces
e. Variation in gravitational forces
f. Newton’s Law of gravitation
a. Static Equilibrium - Consider forces that passes through the same point. These forces are called concurrent
forces, and the point it passes is called as the point of concurrency. If the body is not in uniform motion, then its
velocity and acceleration is equal to zero. It is in static equilibrium. Doing summation of forces with respect to its
components and equate each component to zero.
Doing summation of forces in the x direction
∑ 𝑭𝒙 = 𝟎
𝑭𝟏𝒙 + 𝑭𝟐𝒙 + 𝑭𝟑𝒙 = 𝟎
Doing summation of forces in the y- direction:

∑ 𝑭𝒚 = 𝟎
𝑭𝟏𝒚 + 𝑭𝟐𝒚 + 𝑭𝟑𝒚 = 𝟎
12
b. Dynamic Equilibrium -If a body is moving but in a constant velocity, the acceleration is still zero. This is called
as dynamic equilibrium. Now if a body is moving caused by an external force, then the resultant of the external
forces applied to the body is the vector summation of the effective forces(REF). This is called as D’Alemberts
principle. From the given figure, by doing summation of forces
𝑊
∑ 𝐹𝑥 = 𝑎
𝑔
𝑊
𝐹 − 𝐹𝑓 = 𝑎
𝑔
∑ 𝐹𝑦 = 0
𝑁−𝑊 =0
Where
𝐹 = external net fore
𝑅𝐸𝐹 = reversed effective force
𝐹𝑓 = the force to overcome friction, also called as frictional force which is always parallel to the surface.
𝑁 = normal force, which is the force that is always perpendicular to the surface
𝑊 = weight
𝑔 = acceleration due to gravity
𝑎 = the constant acceleration of the moving body
c. Friction Forces. There are three types of friction.
1. Static friction - it is the friction associated when the body is at rest. For the body to move, it must overcome the
static friction. The value 𝑓𝑠 is the coefficient of static friction.
𝑭𝒇 = 𝒇𝒔 𝑵
2. Impending Friction - To overcome friction, the force must exceed the maximum static friction. At this moment,
the body will as its verge of moving, also known as impending motion.
𝑭𝒇𝒎𝒂𝒙 > 𝒇𝒔 𝑵
3. Kinetic Friction - it is the friction associated when the body is moving continuously. The value fk is the
coefficient of kinetic friction.
𝑭𝒇 = 𝒇𝒌 𝑵
The angle which the resultant of the normal force and the limiting force of friction makes with the normal is
called as the angle of friction θ.
𝐹𝑓
𝑡𝑎𝑛ϕ =
𝑁
𝑓𝑁
𝑡𝑎𝑛𝜙 =
𝑁
𝒕𝒂𝒏𝝓 = 𝒇
13
On the other hand, the angle of repose is the angle that the plane in contact between two bodies makes with the
horizontal when the upper body is just on the point of sliding. We can now relate the angle of repose and friction
angle.
∑ 𝐹𝑥 = 0
𝑁 − 𝑊 𝑐𝑜𝑠α = 0
∑ 𝐹𝑦 = 0
𝐹𝑓 − 𝑊𝑠𝑖𝑛α = 0
Dividing both equation
𝑊𝑠𝑖𝑛α 𝐹𝑓
=
𝑊𝑐𝑜𝑠α 𝑁
𝑡𝑎𝑛α = 𝑡𝑎𝑛ϕ
𝛂=𝛟
Hence at the verge of moving, the angle of repose is at its maximum when it is equal to the angle of friction
For belt friction in pulleys, the equation is represented as

𝑻𝟏
= 𝒆𝒇𝛃
𝑻𝟐
Where 𝑇1 is the tight section of the belt,
𝑇2 is the loose section of the belt,
β is the contact angle of the pulley and the belt
𝑓 is the coefficient of friction
d. Centripetal Force - is the force necessary to keep an object moving in a
curved path and that is directed inward toward the center of rotation.
𝐹𝑐 = 𝑚𝑎𝑐
𝒎𝒗𝟐
𝑭𝒄 =
𝒓
Consider a body, say a pendulum with a bob of weight W and is supported by cord of length L. By force balance,
and creating a free body diagram on the blob of the pendulum:
∑ 𝐹𝑥 = 𝑚𝑎
𝑊
𝐹𝑐 − 𝑇𝑐𝑜𝑠𝜃 = 𝑎
𝑔
∑ 𝐹𝑦 = 0
𝑇 𝑠𝑖𝑛 𝜃 − 𝑊 = 0
The two given equations are useful when dealing problems involving
vertical circles and pendulums.
14
e) Variation in Gravitational Forces - Sometimes it is necessary to determine the gravity acceleration observed
(go) at a certain height h given a variation in gravitational acceleration Δg at a standard gravitational acceleration
(gs) as a datum line
By ratio and proportion:

𝒉𝒐 𝒉
=
𝜟𝒈 𝒈𝟎 − 𝒈𝒔
d. Newton’s law of gravitation - states that the force of attraction between two bodies is directly proportional to the
product of their masses and inversely proportional to the radius of separation squared between the two bodies. In
symbols:
1
𝐹 ∝ 𝑚1 𝑚2 ∝
𝑠2
𝒎𝟏 𝒎𝟐
𝑭=𝑮
𝒔𝟐
Where
𝐹 =force of attraction between two bodies
𝐺 = gravitational constant
𝑠 = distance of separation between two bodies
Where G is called as the Cavendish constant, also known as the gravitational constant. Its units are:
FPS CGS MKS SI
G 𝑙𝑏𝑓 −𝑓𝑡 2 𝑑𝑦𝑛𝑒 − 𝑐𝑚2 𝑘𝑔𝑓− 𝑚2 𝑁 − 𝑚2
3.29 × 10−11 6.67 × 108 6.802 × 10−12 6.67 × 10−11
𝑙𝑏𝑚2 𝑔𝑚2 𝑘𝑔𝑚2 𝑘𝑔𝑚2
As the distance of separation increases, the force of interaction decreases. At the point where the two masses
are on top with each other, s approaches zero. Hence by limits:
1
lim F = lim
𝑠→0 𝑠→0 𝑠 2
1
lim ≈ +∞
𝑠→0 𝑠 2
Hence
𝐹 ≈ +∞
15
Solved Problems
Problem 1-1: Five masses are in a region where g = 30.5 fps are as follows: m1 is 500 gm, m2 weighs 800 gf, m3
weighs 15 poundals, m4 weighs 3 lbf and m5 is 0.1 mass of slug. Determine the total mass express in (a) grams (b)
pounds (c) slugs.(d) What is the total weight in grains?
Solution. The total mass is the sum of each individual mass. For this type of problem, convert each mass in terms
of a single unit, then add
𝑚 𝑇 = ∑ 𝑚𝑖 = 𝑚1 + 𝑚2 + 𝑚3 + 𝑚4 + 𝑚5
𝑖=1
For 𝑚1
𝑚1 = 500𝑔𝑚
For 𝑚2
𝑚2 𝑔
𝑊2 =
𝑘
𝑊2 𝑘
𝑚2 =
𝑔
𝑙𝑏𝑚 − 𝑓𝑡
(800𝑔𝑓) (32.174 )
𝑙𝑏𝑓 − 𝑠2 2.205𝑙𝑏𝑓 1000𝑔𝑚
𝑚2 = ( )( )
𝑓𝑡 1000𝑔𝑓 2.205𝑙𝑏𝑚
30.5 2
𝑠
𝑚2 = 843.91 𝑔𝑚
For 𝑚3
𝑚3 𝑔
𝑊3 =
𝑘
𝑊3 𝑘
𝑚3 =
𝑔
(15 𝑝𝑜𝑢𝑛𝑑𝑎𝑙𝑠) (32.174 )
𝑙𝑏𝑓 − 𝑠2 1𝑙𝑏𝑓 1000𝑔𝑚
𝑚3 = ( )( )
𝑓𝑡 32.174 𝑝𝑜𝑢𝑛𝑑𝑎𝑙𝑠 2.205𝑙𝑏𝑚
30.5
𝑠2
𝑚3 = 223.04 𝑔𝑚
For 𝑚4
𝑚4 𝑔
𝑊4 =
𝑘
𝑊4 𝑘
𝑚4 =
𝑔
(3 𝑙𝑏𝑓) (32.174 ) 1000𝑔𝑚
𝑙𝑏𝑓 − 𝑠2
𝑚4 = ( )
𝑓𝑡 2.205𝑙𝑏𝑚
30.5 2
𝑠
𝑚4 = 1435.22𝑔𝑚
For 𝑚5
32.174 𝑙𝑏𝑚 1000 𝑔𝑚
𝑚5 = 0.1𝑠𝑙𝑢𝑔 ( )( )
1 𝑠𝑙𝑢𝑔 2.205 𝑙𝑏𝑚
𝑚5 = 1459.14 𝑔𝑚
16
a) In terms of gram mass:

𝑚 𝑇 = (500 + 843.91 + 223.04 + 1435.22 + 1459.14)𝑔𝑚
𝒎𝑻 = 𝟒𝟒𝟔𝟏. 𝟑𝟏 𝒈𝒎 ⇒ 𝑨𝒏𝒔.
b) In terms of pound mass
2.205𝑙𝑏𝑚
𝑚 𝑇 = 4461.31𝑔𝑚 ( )
1000𝑔𝑚
𝒎𝑻 = 𝟗. 𝟖𝟑𝟕𝟐 𝒍𝒃𝒎 ⇒ 𝑨𝒏𝒔.
c) In terms of slugs
1𝑠𝑙𝑢𝑔
𝑚 𝑇 = 9.8372𝑙𝑏𝑚 ( )
32.174𝑙𝑏𝑚
𝒎𝑻 = 𝟎. 𝟑𝟎𝟓𝟕 𝒔𝒍𝒖𝒈 ⇒ 𝑨𝒏𝒔.
d) The total weight in grains would be:
𝑚𝑇 𝑔
𝑊𝑇 =
𝑘
𝑓𝑡
(9.8372𝑙𝑏𝑚) (30.5 2 ) 3000𝑔𝑟𝑎𝑖𝑛𝑠
𝑠
𝑊𝑇 = ( )
𝑙𝑏𝑚 − 𝑓𝑡 1𝑙𝑏𝑓
32.174
𝑾𝑻 = 𝟐𝟕𝟗𝟕𝟔. 𝟏𝟐 𝒈𝒓𝒂𝒊𝒏𝒔 ⇒ 𝑨𝒏𝒔.
Problem 1-2: For a ballistics study, a 1.9 gm bullet is fired into soft wood. The bullet strikes the wood surface
with a velocity of 380 m/s and penetrates 0.15 m deep. Find
a) The constant deceleration of the bullet in m/s2
b) The time required to stop the bullet in s.
c) The constant retarding force in N
Solution:
a) The constant deceleration of the bullet would be defined by the third kinematic equation
𝑣 2 = 𝑣0 2 + 2𝑎𝑠
Since the final velocity is zero when it hits the soft wood, then the acceleration is
𝑣 2 − 𝑣0 2
𝑎=
2𝑠
𝑚2
(02 − 3802 ) 2
𝑎= 𝑠
2(0.15𝑚)
𝒎
𝒂 = −𝟒𝟖𝟏𝟑𝟑𝟑. 𝟑𝟑 𝟐 ⇒ 𝑨𝒏𝒔.
𝒔
The negative sign indicates that the bullet is decelerating in the chosen positive direction as shown in the figure.
17
b) The time required to stop the bullet can be determined using the first kinematic equation.
𝑣 = 𝑣0 + 𝑎𝑡
𝑣 − 𝑣0
𝑡=
𝑎
𝑚
(0 − 380) 2
𝑡= 𝑠
𝑚
−481333.33 2
𝑠
𝒕 = 𝟎. 𝟎𝟎𝟎𝟕𝟖𝟗 𝒔 ⇒ 𝑨𝒏𝒔.
c) The constant retarding force F’ is the force that is against the force F that causes motion to be bullet. Assuming
equilibrium condition at the time when the bullet hit the wood surface, by doing summation of force equal to zero
in the horizontal direction
+→ ∑ 𝐹ℎ = 0
𝐹 − 𝐹′ = 0
𝐹′ = 𝐹
𝑚𝑎
𝐹′ =
𝑘
𝑚
(1.9𝑔𝑚) (−481333.33 2 ) 1𝑘𝑔𝑚
𝐹′ = 𝑠 ( )
𝑘𝑔𝑚 − 𝑚 1000𝑔𝑚
1000
𝑘𝑁 − 𝑠2
𝑭′ = −𝟎. 𝟗𝟏𝟓𝒌𝑵 ⇒ 𝑨𝒏𝒔.
The negative sign indicates that the force is acting opposite to the chosen positive direction.
Problem 1-3: A nut is thrown downward from the bottom of an elevator moving upward at 3 meters per second. It
strikes the ground after two seconds. How far above the bottom of the elevator will the nut be after 0.50 seconds?
Solution.
𝑚
The time it takes for the elevator to move upward at 𝑣 = 3 when the nut hits
𝑠
the ground can be determined using the first kinematic equation. The sign is
negative since the elevator is moving upward.
𝑣 = 𝑣0 − 𝑔𝑡𝑢𝑝
𝑣0 − 𝑣
𝑡𝑢𝑝 =
−𝑔
𝑚
(0 − 3)
𝑡𝑢𝑝 = 𝑠
𝑚
−9.8066 2
𝑠
𝑡𝑢𝑝 = 0.306𝑠
18
The total time required for the elevator to reach the ground would be
𝑡 = 𝑡𝑢𝑝 + 𝑡𝑑𝑜𝑤𝑛
𝑡𝑑𝑜𝑤𝑛 = 𝑡 − 𝑡𝑢𝑝
𝑡𝑑𝑜𝑤𝑛 = (2 − 0.306)𝑠
𝑡𝑑𝑜𝑤𝑛 = 1.694𝑠
The height of the elevator from the ground can be determined using the second kinematic equation
1
𝑠 = 𝑠0 + 𝑣𝑜 𝑡 − 𝑔𝑡 2
2
Since the initial velocity and initial displacement are equal to zero, assuming the elevator starts from a fixed
ground, the height s of the elevator would be
1
𝑠 = 𝑔𝑡 2
2
At 𝑡 = 1.694 𝑠
1 𝑚
𝑠 = (9.8066 2 ) (1.694𝑠)2
2 𝑠
𝑠 = 14.071 𝑚
At 𝑡 = 0.50 𝑠, the position of the nut from the position where the elevator assumed to come at rest would be
found by using the second kinematic equation. The sign is positive since the motion of the nut is downward. Since
the nut is downward, the velocity is also in the downward direction, hence negative.
1
𝑠′ = 𝑠0 + 𝑣0 𝑡 + 𝑔𝑡 2
2
𝑚 1 𝑚
𝑠′ = 14.071 𝑚 + (−3 ) (0.50𝑠) + (9.8066 2 ) (0.50𝑠)2
𝑠 2 𝑠
𝒔′ = 𝟏𝟑. 𝟕𝟗𝟕𝒎 ⇒ 𝑨𝒏𝒔.
Problem 1-4: The motion of a particle is represented by the function 𝑠 = 15𝑡 3 + 24𝑡 2 − 76𝑡 + 350 where s is in
inch per second and t is in second. Find
a) the velocity and acceleration at t= 2 seconds
b) the position of the particle when it comes to a rest
Solution.
a) The velocity function is defined as the derivative of displacement. Hence
𝑑𝑠
𝑣=
𝑑𝑡
𝑑
𝑣 = (15𝑡 3 + 24𝑡 2 − 76𝑡 + 4)
𝑑𝑡
𝑣 = 45𝑡 2 + 48𝑡 − 76
The velocity at 𝑡 = 2 𝑠 would be
𝑣 = 45(2)^2 + 48(2) − 76
𝐢𝐧
𝐯 = 𝟐𝟎𝟎 ⇒ 𝑨𝒏𝒔.
𝐬
The acceleration function is the derivative of the velocity function, hence
𝑑𝑣
𝑎=
𝑑𝑡
𝑑
𝑎 = (45𝑡 2 + 48𝑡 − 76)
𝑑𝑡
𝑎 = 90𝑡 + 48
19
The acceleration at 𝑡 = 2 𝑠 would be

𝑎 = 90(2) + 48
𝒊𝒏
𝒂 = 𝟐𝟐𝟖 𝟐 ⇒ 𝑨𝒏𝒔.
𝒔
b) The velocity of the particle is zero when it is at rest. Thus at 𝑣 = 0

𝑣 = 45𝑡 2 + 48𝑡 − 76
0 = 45𝑡 2 + 48𝑡 − 76
Using quadratic formula
−𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑡=
2𝑎
If a =45, b=48, and c = -76
−48 ± √(48)2 − 4(45)(−76)
𝑡=
2(45)
−48 ± 12√111
𝑡=
90
Take the negative sign:
−48 − 12√111
𝑡=
90
𝑡 = −1.38 𝑠
Since time t cannot be lesser than zero, reject this solution.
Take the positive sign:
−48 + 12√111
𝑡=
90
𝑡 = 0.871 𝑠
Since time t is positive, it is the solution to the quadratic equation. Solving for the position at time 𝑡 = 0.871 𝑠
from the displacement function.
𝑠 = 15𝑡 3 + 24𝑡 2 − 76𝑡 + 350
𝑠 = 15(0.871)3 + 24(0.871)2 − 76(0.871) + 350
𝒔 = 𝟑𝟏𝟏. 𝟗𝟐 𝒊𝒏 ⇒ 𝑨𝒏𝒔.
Problem 1-5: Calculate the ratio of gravitational force between two bodies to the total weight, having masses of a
1 ton and 1 tonne,0.5mm apart.
Solution:
For the total weight of the two bodies

(𝑚1 + 𝑚2 )𝑔
𝑊𝑇 =
𝑘
Where
𝑚1 = 1𝑡𝑜𝑛 = 907.03𝑘𝑔𝑚
𝑚2 = 1𝑡𝑜𝑛𝑛𝑒 = 1000𝑘𝑔𝑚
20
Thus
𝑚
(907.03 + 1000)𝑘𝑔𝑚 (9.8066 )
𝑊𝑇 = 𝑠2
𝑘𝑔𝑚 − 𝑚
1000
𝑘𝑁 − 𝑠2
𝑊𝑇 = 18.701𝑘𝑁
For the force of attraction between two bodies

𝑚1 𝑚2
𝐹𝑔 = 𝐺
𝑠2
𝑘𝑁 − 𝑚2
(6.67 × 10−14 ) (907.03𝑘𝑔𝑚)(1000𝑘𝑔𝑚)
𝑘𝑔𝑚2
𝐹𝑔 =
(0.0005𝑚)2
𝐹𝑔 = 0.243𝑘𝑁
Hence the ratio of gravitational force between two bodies to total weight is
𝐹𝑔 0.243𝑘𝑁
=
𝑊𝑇 18.791𝑘𝑁
𝑭𝒈
= 𝟎. 𝟎𝟏𝟑
𝑾𝑻
Analysis: We can see that the gravitational attraction between two bodies is 1.3% the total weight of two bodies,
hence in most thermodynamic problems, we neglect the gravitational attraction unless specified.
Problem 1-6: Note that the gravity acceleration at equatorial seal level is 𝑔 = 32.088𝑓𝑝𝑠2 and that its variation
is−0.003𝑓𝑝𝑠2 per 1000 ft ascent. Find the height in miles above the point for which
a. The gravity acceleration becomes g = 30.504 fps2
b. The weight of the given man is decreased by 5%
c. What is the weight of a 180 lbm man atop the 29 131 ft Mt. Everest in Tibet, relative to this point.
Solution:
a) By Ratio and Proportion:

𝑓𝑡
ℎ (30.504 − 32.088) 2
= 𝑠
1000𝑓𝑡 𝑓𝑡
−0.003 2
𝑠
1𝑚𝑖
ℎ = 528000𝑓𝑡 ( )
5280𝑓𝑡
𝒉 = 𝟏𝟎𝟎 𝒎𝒊 ⇒ 𝑨𝒏𝒔.
b) Since mass is an absolute quantity, then 𝑚0 = 𝑚𝑠
𝑊𝑜 𝑊𝑠
=
𝑔𝑜 𝑔𝑠
21
Since 𝑊0 = (1 − 0.05)𝑊𝑠 = 0.95𝑊𝑠

0.95𝑊𝑠 𝑊𝑠
=
𝑔𝑜 𝑔𝑠
𝑔𝑜 = 0.95𝑔𝑠
𝑓𝑡
𝑔𝑜 = 0.95 (32.088 )
𝑠2
𝑓𝑡
𝑔𝑜 = 30.4836 2
𝑠
By Ratio and Proportion:
𝑓𝑡
ℎ (30.4836 − 32.088) 2
= 𝑠
1000𝑓𝑡 𝑓𝑡
−0.003 2
𝑠
1𝑚𝑖
ℎ = 534800𝑓𝑡 ( )
5280𝑓𝑡
𝒉 = 𝟏𝟎𝟏. 𝟐𝟗 𝒎𝒊 ⇒ 𝑨𝒏𝒔.
c) Since mass is an absolute quantity, we can solve first for the observed gravitational acceleration 𝑔𝑜 of given
height ℎ = 29131 𝑓𝑡. By ratio and proportion
𝑓𝑡
29131 𝑓𝑡 (𝑔𝑜 − 32.088) 𝑠2
=
1000 𝑓𝑡 𝑓𝑡
−0.003 2
𝑠
𝑓𝑡
𝑔𝑜 = 32.000607 2
𝑠
For the weight of the man at 𝑔𝑜 = 32.000607𝑓𝑡/𝑠2
𝑚𝑜 𝑔𝑜
𝑊𝑜 =
𝑘
𝑓𝑡
(180𝑙𝑏𝑚) (32.000607 2 )
𝑠
𝑊𝑜
32.174
𝑾𝒐 = 𝟏𝟕𝟗. 𝟎𝟑𝒍𝒃 ⇒ 𝑨𝒏𝒔.
22
1.5 PLANE MENSURATION

Plane mensuration involves the study of length, area, and surface area of common planar figures. Table below
shows length conversions and area conversions.
CONVERSION OF LENGTH
1 kilometer (km) = 1000 m 1 inch (in) = 2.54 cm
1 meter (m) = 100 cm = 25.4 mm
= 1000 mm 1 foot = 12 in = 3 hands
= 3.281 ft 1 mil = 1 thou
= 39.372 in = 0.001 in
−6
1 micron (μ) = 10 m 1 yard (yd) = 3ft
−9
1 millimicron (𝑚𝜇)= 10 m 1 fathom (ftm) = 6 ft
4
1 cm = 10 1 furlong (fur) = 5.5 yd
−10
1 angstrom (A) = 10 m 1 chain (ch) = 22 yd
= 100 links (li)
1 rod (rd) = 16.5 ft
1 mile (mi) = 1 statute mile
= 5280 ft
= 1760 yd
= 1.609 km
1 nautical mile (nmi) = 6080 ft
= 1.15 mi
1 league = 1 nautical league (NL)
= 3 nmi
1 cable length (cbl) = 100 ftm
= 720 yd
1 rope = 20 ft
1 vara = 1 Spanish yard
= 33.33 in
1 foot rule = 1.23 in
CONVERSION OF AREA
1 square foot(𝑓𝑡 2 ) = 929 𝑐𝑚2
1 acre (ac) = 43560 𝑓𝑡 2
= 4046.8 𝑚2
1 hectare (ha) = 10 000 𝑚2
= 11 960 𝑦𝑑 2
= 100 ac
1 section = 1 statute square mile
= 1 square mile (𝑚𝑖 2 )
= 640 ac
2
1 square meter (𝑚 ) = 10.76 𝑚𝑖 2
23
I. Engineering Formulas – the following formulas are commonly used in dealing thermodynamic problems in
relations to plane geometry.
Geometrical Figure Formula

Rectangle 𝑨 = 𝒂𝒃
𝑷 = 𝟐(𝒂 + 𝒃)
where
b = base
a = height
Square 𝑨 = 𝒂𝟐
𝑷 = 𝟒𝒂
Where
a = edge / side
Triangle 𝑷 =𝒂+𝒃+𝒄
𝟏
𝐀 = 𝐛𝐡
𝟐
Where
a and c are sides of triangle
b = base
h = height or altitude
Circle 𝑪 = 𝟐𝛑𝑹
𝑪 = 𝛑𝑫
𝑨 = 𝛑𝑹𝟐
𝛑
𝑨 = 𝑫𝟐
𝟒
Where
R = radius
D = diameter
Annulus 𝑨 = 𝛑(𝑹𝟐 − 𝒓𝟐 )
𝛑
𝑨 = (𝑫𝟐 − 𝒅𝟐 )
𝟒
Where
R = outside radius
r = inside radius
D = outside diameter
d = inside diameter
Sector of a Circle 𝟏 𝟐
𝑨= 𝑹 𝛉
𝟐
Where
R = radius
θ = angle in radians
24
II. Approximation Formulas - is used to find areas of irregular figures. The figure is divided into vertical strips n
of equal length d and measured its height y. The greater the the number of strips n , the more accurate the
approximation. The following are approximation formulas
Simpson’s Rule 𝟏
𝑨= 𝒅[(𝒚𝟏 + 𝒚𝒏 ) + 𝟒(𝒚𝟏 + 𝒚𝟑 + 𝒚𝟓 + ⋯ + 𝒚𝒏−𝟏 ) + 𝟐(𝒚𝟐 + 𝒚𝟒 + 𝒚𝟔 + 𝒚𝟐𝒏−𝟏 )]
𝟑
Trapezoidal Rule 𝟏
𝑨= 𝒅(𝒚𝟏 + 𝟐𝒚𝟐 + 𝟑𝒚𝟑 + ⋯ + 𝟐𝒚𝒏 + 𝒚𝒏−𝟏 )
𝟐
Durand’s Rule 𝟐 𝟏𝟏 𝟏𝟏 𝟐
𝑨 = 𝒅 ( 𝒚𝟏 + 𝒚𝟐 + 𝒚𝟑 + ⋯ + 𝒚𝒏−𝟐 + 𝒚𝒏−𝟏 + 𝒚𝒏 )
𝟓 𝟏𝟎 𝟏𝟎 𝟓
3/8 Simpson’s Rule 𝟑 𝟑 𝟓
𝑨 = 𝒅(𝒚𝟏 + 𝟑𝒚𝟐 + 𝟑𝒚𝟑 + 𝒚𝟒 ) − 𝒅
𝟖 𝟖𝟎
Hardy’s Rule 𝟏
𝑨= 𝒅(𝟐𝟖𝒚𝟏 + 𝟏𝟔𝟐𝒚𝟐 + 𝟐𝟐𝟎𝒚𝟒 + 𝟏𝟔𝟐𝒚𝟔 + 𝟐𝟖𝒚𝟖 + ⋯ + 𝒚𝒏 )
𝟏𝟎𝟎
Weddie’s Rule 𝟑
𝑨= 𝒅(𝒚𝟏 + 𝟓𝒚𝟐 + 𝒚𝟑 + 𝟔𝒚𝟒 + 𝟓𝒚𝟓 + 𝒚𝒏 + ⋯ + 𝟓𝒚𝒏−𝟏 + 𝒚𝒏 )
𝟏𝟎
Note that the approximation formulas are only usable when dealing with irregular planar area for actual
measurements.
III. Using Integral Calculus – if a given irregular figure is represented by a curve 𝑦 = 𝑓(𝑥) and is bounded by
lines 𝑥 = 𝑎 and 𝑥 = 𝑏, then the area of the irregular figure can be represented by the equation
𝒃
𝑨 = ∫ 𝒇(𝒙)𝒅𝒙
𝒂
Where
𝐴 = area under the curve
𝑎 = lower limit
𝑏 = upper limit
𝑓(𝑥) = the function of the curve
𝑑𝑥 = differential
IV. Using a Planometer – a planometer is a

device use to measure the area of irregular planar
figures.
25
1.6 VOLUME
Volume is defined as a property of matter that occupies space bounded by its lateral surface area. The lateral
surface area (LSA) is defined as the area of the sides of the solid, while the total surface area is defined as the sum
of the lateral area and the base area. The following below shows unit conversions in volume.
VOLUME CONVERSIONS
1L = 1000 cm3 1 ft3 = 7.481 gal

= 1.057 qt = 1728 in3
= 61.025 in3 = 28.317 L
= 0.03532 ft3 1 bbl = 42 gal
1 m3 = 1000 L 1 dr = 55 gal
1 mL =1 cm3 1 bu = 4 pk
= 1 cc = 64 pt
1 gal = 1 U.S gallons = 8 gal
= 3.785 L 1 pk = 2 gal
= 4 qt 1 cavan = 25 gantas
= 8 pt = 75 L
1 U.K gal = 1 imperial gallon 1 ganta = 8 chupas
= 1 British gallon =3L
= 1.201 gal 1 chupa = 4 apatans
= 0.375 L
Where:
L in liters pk in pecks bu in bushel
mL in milliliters cc in cubic centimeter gal in gallons
qt in quarts dr in drums pt in pints
bbl in barrels
Volume can be measured by the following:
I. By Volume Displacement Method - Using a graduated cylinder, pour initially some water. Measure the volume
in mL of the water. Place the irregular solid to the
graduated cylinder and measure again its volume in mL.
The volume of the solid is the difference of the volume
displaced to the volume of the water.
𝑽𝒔𝒐𝒍𝒊𝒅 = 𝑽𝒔𝒐𝒍𝒊𝒅+𝒘𝒂𝒕𝒆𝒓 − 𝑽𝒘𝒂𝒕𝒆𝒓
26
II. By Engineering Formulas - The volume of the solid is subdivided into three categories
A. Solids in which the cross-sectional area b is always the same in any height h of the solid.
In general
𝑽 = 𝒃𝒉
Where V is the volume of the solid

b is the cross-sectional area of the solid
h is the height of the solid
Note: The space diagonal or the diagonal of the solid is the longest line joining the two opposite vertices of the
solid. The face diagonal is the diagonal of one of the faces of the solid.
Cube Rectangular Parallelepiped
Volume: 𝑽 = 𝒍𝒘𝒉
Lateral Surface Area: 𝐿𝑆𝐴 = 2ℎ(𝑙 + 𝑤)
Volume: 𝑽 = 𝒔𝟑 Total Surface Area: 𝑇𝑆𝐴 = 2(𝑤ℎ + 𝑤𝑙 + 𝑙ℎ)
Lateral Surface Area: 𝐿𝑆𝐴 = 4𝑠2 Face Diagonal: 𝑑 = √𝑙2 + ℎ2
Total Surface Area: 𝑇𝑆𝐴 = 6𝑠2 Space Diagonal: 𝐷 = √𝑙2 + 𝑤 2 + ℎ2
Face Diagonal: 𝑑 = 𝑠√2
Space Diagonal: 𝐷 = 𝑠√3
Prism Cylinder
Volume For Right Cylinder:

π
a) For any Prism: 𝑉 = 𝑏ℎ Volume: 𝑉 = 𝑑 2 ℎ
4
b) Given Lateral edge e and area of the right Lateral Surface Area: 𝐿𝑆𝐴 = π𝐷ℎ
section R Total Surface Area: 𝑇𝑆𝐴 = 2π𝑟(𝑟 + ℎ)
𝑉 = 𝑅𝑒
Where For Oblique Cylinder:
ℎ = 𝑒𝑠𝑖𝑛θ
𝑅 = 𝐵𝑠𝑖𝑛θ Volume: 𝑉 = 𝑅𝑒
Where
Lateral Surface Area: 𝐿𝑆𝐴 = 𝑃𝑒 ℎ = 𝑒𝑠𝑖𝑛θ
Total Surface Area: 𝑇𝑆𝐴 = 2𝑏 + 𝑃𝑒 𝑅 = 𝐵𝑠𝑖𝑛θ
27
B. Solids in which the cross-sectional area b varies from a point to a given height h. The general formula
would be:
𝟏
𝑽 = 𝒃𝒉
𝟑
Where
b = base of the solid
h = height of the solid
Pyramid Cone
For any given base: For right circular cone:

Volume: Volume:
1 π 2
𝑉 = 𝐵ℎ 𝑉= 𝑟 ℎ
3 3
Slant Height: Lateral Surface Area:
𝐿𝑆𝐴 = π𝑟𝑙
𝑙 = √ℎ 2 + 𝑎 2
Total Surface Area:
Lateral Surface Area:
1 𝑇𝑆𝐴 = π𝑟(𝑟 + 𝑙)
𝐿𝑆𝐴 = 𝑃𝑙 Slant height:
2
Total Surface Area: 𝑙 = √ℎ 2 + 𝑟 2
1
𝑇𝑆𝐴 = 𝐵 + 𝑃𝑙
2
C. Solids in which it is generated by revolving a certain figure on a fixed axis. The general formula would be:
𝟏
𝑽 = 𝒃𝒉
𝟑
Where
B = cross sectional area at the middle of the solid.
H = height of the solid
Definition of Terms:
1. Zone(Z) – it is the surface of revolution of a sphere and equal to the lateral surface area of a
spherical segment.
2. Spherical Lune – it is the slice of the surface of the sphere cut by two planes through the
azimuthal axis with dihedral angles.
3. Spherical Wedge – it is the volume formed by the spherical lune.
4. Spherical Polygon – it is a polygon on the surface of the sphere whose sides are arcs of great
circle.
28
Sphere Spherical Segment Spherical Sector
Volume
For one base spherical segment 1
𝑉 = 𝑍𝑅
πℎ3 3
𝑉= (3𝑅 − ℎ) Where
Volume: 6
π 3 𝑆 = π𝑅(2ℎ + 𝑅) 𝑍 = 2π𝑅ℎ
𝑉= 𝑑 Surface Area:
6
Surface Area: For two base spherical segment, 𝑆 = π𝑅(2ℎ + 𝑅)
𝑆 = 4π𝑟 2 it is also known as a spherical
frustrum:
πℎ3
𝑉= (3𝑎2 + 3𝑏 2 + ℎ2 )
6
𝑆 = π(2𝑅ℎ + 𝑎2 + 𝐵2 )
Spherical Wedge Spherical Polygon Spherical Pyramid
Volume of the Wedge: Surface Area of the Spherical

2 Volume:
𝑉 = 𝑅3θ Polygon: 1
3 𝑉 = 𝑟3 𝐸
3
𝑆 = 𝑟2 𝐸 Where
Surface Area of the Lune: Where
𝐸 = (∑ θ) − π(𝑛 − 2)
𝑆= 2𝑟 2 θ 𝐸 = (∑ θ) − π(𝑛 − 2)
III. Using Prismatoidal Formula - it is used for irregular solids, with atleast three given base areas
𝒉
𝑽= (𝑨 + 𝟒𝑨𝒎 + 𝑨𝟐 ); 𝐴𝑚 ≠ 0
𝟔 𝟏
Where
𝐴1 = area of the top base

𝐴2 = area of the bottom base
𝐴𝑚 = area of the middle part
29
IV. Number of Plates To Be Used: If initially a solid is to be fabricated by steel plates whose dimensions are in
terms of length of width, the number of steel plates to be used would be:
(Total Surface Area)(Number of Units to be Fabricated)(1 + % material wasted)

Number of Plates =
Area of the steel plate
Where
Total Surface Area (TSA) in units of length per fabricated unit
Number of units to be fabricated in terms of the number of units to be made.
Area of the steel plate in units of area per plate.
1.7 BASIC PROPERTIES OF FLUID

The following are basic properties of fluid applied to incompressible fluids.
A. Density - also called as mass density or specific density, is the mass per unit volume.
𝒎
𝝆=
𝑽
B. Specific Volume - is the volume per unit mass, and the reciprocal of density.
𝑽
𝛎=
𝒎
𝟏
𝛎=
𝛒
C. Specific Weight - is also called as the unit weight or weight density, is the weight per unit volume
𝑾
𝛄=
𝑽
Since 𝑊 = 𝑚𝑔/𝑘 and 𝑉 = 𝑚/ρ, then
𝛒𝒈
𝛄=
𝒌
D. Specific Gravity - also called a relative density, is the ratio of the density of the substance to some density of
relative or standard substance. Since it is a ratio, the units of specific gravity are dimensionless.
a. Specific Gravity for Solids and Liquids - the reference substance is water at 4°C.
𝑾 𝛒 𝛄
𝑺𝑮 = = =
𝑾𝑯𝟐𝑶 𝛒𝑯𝟐𝟎 𝛄𝑯𝟐𝑶
b. Specific Gravity for Gases – the reference substance is air at 1 atm and 20°C.
𝑾 𝛒 𝛄 𝑴𝑾
𝑺𝑮 = = = =
𝑾𝒂𝒊𝒓 𝛒𝒂𝒊𝒓 𝛄𝒂𝒊𝒓 𝑴𝑾𝒂𝒊𝒓
30
The specific gravity of the substance determines how lighter is the
substance with respect to the reference substance. Consider a non-
homogeneous mixture in the container. Mercury has a greater value
of specific gravity since it is heavier than oil or water. Oil as we
know floats above water because water is heavier than oil. Thus
𝑺𝑮𝒐𝒊𝒍 < 𝑺𝑮𝒘𝒂𝒕𝒆𝒓 < 𝑺𝑮𝒎𝒆𝒓𝒄𝒖𝒓𝒚
Standard Values for Specific Gravity of Common Fluids

Incompressible Water and Air (Unless Specified)
𝑘𝑔𝑚 Fluid SG
ρ𝑤 = 1000 3
𝑚 Water at 4°C 1.0
𝑙𝑏𝑚 Blood at 37°C 1.06
= 62.4 3
𝑓𝑡 Ice at 0°C 0.916
𝑘𝑁 Mercury 13.6
γ𝑤 = 9.8066 3
𝑚 Oil 0.8
𝑙𝑏𝑓 Molasses 1.5
= 62.4 3
𝑓𝑡 Carbon Tetrachloride (CCl4) 1.6
𝑘𝑔𝑚 Brine (Salt and Water Solution) 1.15
ρ𝑎𝑖𝑟 @𝑆𝑇𝑃 = 1.29 3 Gasoline 0.68
𝑚
𝑘𝑔𝑚 Alcohol 0.79
ρ𝑎𝑖𝑟 @20℃ = 1.20 3 Glycerin 1.26
𝑚
Air (Atmospheric Air) 1.0
E. Mass Flow Rate – it is defined as the mass flow per unit time.
𝒎
𝒎̇ =
𝒕
F. Volume Flow Rate – it is defined as the amount of volume flowing per unit time.
𝑽
𝑽̇ =
𝒕
Consider a pipe as shown in the figure. The volume flow rate is directly proportional to the constant velocity of
the fluid and the cross-sectional area of the homogeneous pipe. The volume flow rate in this type of problem is
also called as the discharge rate of the fluid in the pipe.
The volume flow rate of the fluid is
𝐐 = 𝑽̇ = 𝑨𝒗
Hence the mass flow rate of the fluid would be
𝒎̇ = 𝛒𝑨𝒗
Hence the mass flow rate of the fluid would be

𝒎̇ = 𝛒𝑨𝒗
Where
ρ = density of the fluid
𝐴 = the cross-sectional area
𝑣 = velocity of flow
31
Common units for Fluid Properties
Fluid Properties FPS MKS SI
𝛒 𝒍𝒃𝒎 𝒌𝒈𝒎 𝒌𝒈𝒎
𝒇𝒕 𝟑 𝑳 𝒎𝟑
𝛎 𝒇𝒕 𝟑 𝑳 𝒎𝟑
𝒍𝒃𝒎 𝒌𝒈𝒎 𝒌𝒈𝒎
𝛄 𝒍𝒃𝒇 𝒌𝒈𝒇 𝑵 𝒌𝑵
𝟑 ,
𝒇𝒕 𝑳 𝒎𝟑 𝒎 𝟑
𝒎̇ 𝒍𝒃𝒎 𝒌𝒈𝒎 𝒌𝒈𝒎
𝒔 𝒔 𝒔
𝑽̇ 𝑓𝑡 3 𝑔𝑎𝑙 𝑳 𝑳 𝒎𝟑
, (𝐺𝑃𝑀) , (𝑳𝑷𝑴) (𝑪𝑴𝑺)
𝑠 𝑚𝑖𝑛 𝒔 𝒎𝒊𝒏 𝒔
Solved Problems
Problem 1.7: 5.04 kgm of glycerin is contained in a 4 liter flask. Calculate its (a)specific density, (b) specific
volume, (c) weight density, and (d) relative density.
Solution
a) For density/specific density:
𝑚
ρ=
𝑉
5.04𝑘𝑔𝑚
ρ=
1𝑚3
4𝐿 ( )
1000𝐿
𝝆 = 𝟏𝟐𝟔𝟎 𝒌𝒈𝒎/𝒎𝟑 ⇒ 𝐴𝑛𝑠.
b) For specific volume:
𝑉 1
ν= =
𝑚 ρ
1 𝑚3
ν=
1260 𝑘𝑔𝑚
𝛎 = 𝟎. 𝟎𝟎𝟎𝟕𝟗𝟑𝟕𝒎𝟑 /𝒌𝒈𝒎 ⇒ 𝐴𝑛𝑠.
c) For specific weight/weight density:

ρ𝑔
γ=
𝑘
𝑘𝑔𝑚 𝑚
(1260 3 ) (9.8066 2 )
𝑚 𝑠
γ=
1000
𝑘𝑁 − 𝑠 2
𝛄 = 𝟏𝟐. 𝟑𝟓𝟔 𝒌𝑵/𝒎𝟑 ⇒ 𝑨𝒏𝒔.
32
Problem 1.8: Two liquids of different densities 1500 kgm/m3 and 500 kgm/m3 are poured together into a 100L
tank, filling it. If the resulting density of the mixture is 800 kgm/m3, find
a. the respective quantities of liquids used in kgm and
b. weight of the mixture in kN at standard gravitational acceleration.
Solution:
For the total mass of the mixture.

𝑚𝑚 = 𝑚1 + 𝑚2
ρ𝑚 𝑉𝑚 = ρ1 𝑉1 + ρ2 𝑉2
𝑘𝑔𝑚 1𝑚3 𝑘𝑔𝑚 𝑘𝑔𝑚
(800 ) (100𝐿) ( ) = (1500 ) 𝑉1 + 500 𝑉
𝑚 3 1000𝐿 𝑚3 𝑚3 2
1500𝑉1 + 500𝑉2 = 80 (Equation 1)
For the total volume of the mixture

𝑉1 + 𝑉2 = 0.10 (Equation 2)
Two equations two unknowns. Solving equations 1 and 2 simultaneously yields:

𝑉1 = 0.03𝑚3
𝑉2 = 0.07𝑚3
a) For the mass of each liquid
𝑚1 = ρ1 𝑉1
𝑘𝑔𝑚
𝑚1 = (1500 3 ) (0.03𝑚3 )
𝑚
𝒎𝟏 = 𝟒𝟓𝒌𝒈𝒎
𝑚2 = 𝑚𝑡𝑜𝑡 − 𝑚1
𝑚2 = (80 − 45)𝑘𝑔𝑚
𝒎𝟐 = 𝟑𝟓𝒌𝒈𝒎
b) For the total weight of the mixture
𝑚𝑇 𝑔
𝑊𝑇 =
𝑘
𝑚
(80𝑘𝑔𝑚) (9.8066 2 )
𝑊𝑇 = 𝑠
1000
𝑘𝑁 − 𝑠2
𝑾𝑻 = 𝟎. 𝟕𝟖𝟒𝟓𝒌𝑵
Problem 1.9: A horizontal cylindrical tank is filled with liquid having a mass of 1 tonne to a depth of 0.75 m. The
diameter and length are 1m and 3m respectively. Find the following.
a) The density of the liquid in kgm/m3, lbm/bbl, g/mL. and slug/ft.
b) The specific volume of the liquid in in3/slug, yd3/hyl, m3/ton, and gal/lbm.
c) The specific weight of the liquid in kN/m3, lbf/ft3, kgf/dr, and dyne/cm3.
d) The specific gravity of the liquid.
e) How long will it take a 150 GPM pump to fill the tank completely with liquid assuming the tank is initially
empty?
f) How long will it take a 150 GPM pump to fill the tank completely with liquid assuming the tank has the same
liquid having a depth of 0.75m?
g) How many 8ft by 4 ft steel plates are needed to fabricate 10 units of these if 25% of the material used is wasted?
33
Solution:
Sketch the cylinder, and the cross-sectional area of the cylinder.
From our general formula for volume, whose cross-sectional area is always equal at any given height.
𝑉 = 𝑏ℎ
From the figure:

𝑏 = 𝐴1 + 𝐴2
𝐴2 = 𝐴𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
By Pythagorean theorem
𝑥 2 + 0.252 = 0.52
𝑥 = 0.433 𝑚
For area of triangle A 2

1
𝐴2 = (2) (0.433)(0.25)
2
𝐴2 = 0.10825𝑚2
For A1
𝐴1 = 𝐴𝑠𝑒𝑐𝑡𝑜𝑟
1
𝐴1 = 𝑟 2 θ
2
Where θ must always be in radians.
For angle α
0.25
α = 𝑎𝑟𝑐𝑐𝑜𝑠 ( )
0.50
α = 60°
34
For angle θ
π
θ = (360 − 2α) ( 𝑟𝑎𝑑)
180
π
θ = [360 − 2(60)] ( 𝑟𝑎𝑑)
180
θ = 4.1888 𝑟𝑎𝑑
Solving for the area of the sector of the circle.

1
𝐴1 = (0.5)2 (4.1888)
2
𝐴1 = 0.5236𝑚2
For b
𝑏 = 𝐴1 + 𝐴2
𝑏 = (0.5236 + 0.10825)𝑚2
𝑏 = 0.63185𝑚2
For the volume of the liquid in the tank

𝑉 = 𝑏ℎ
𝑉 = (0.63185𝑚2 )(3𝑚)
𝑉 = 1.8956𝑚3
Another solution: The total base area b is the area of the segment of given angle θ. Note that the angle units must
be in radians.
𝑏 = 𝐴𝑠𝑒𝑔𝑚𝑒𝑛𝑡
1
𝑏 = 𝑟 2 (θ − 𝑠𝑖𝑛θ)
2
1
𝑏 = (0.5𝑚)2 (4.1888 − 𝑠𝑖𝑛4.1888)
2
𝑏 = 0.63185𝑚2
For the volume of the liquid in the tank

𝑉 = 𝑏ℎ
𝑉 = (0.63185𝑚2 )(3𝑚)
𝑉 = 1.8956𝑚3
(a) Solving for density in units
In terms of 𝑘𝑔𝑚/𝑚3
𝑚
ρ=
𝑉
1000𝑘𝑔𝑚
ρ=
1.8956𝑚3
𝛒 = 𝟓𝟐𝟕. 𝟓𝟑𝟕𝒌𝒈𝒎/𝒎𝟑 ⇒ 𝑨𝒏𝒔.

In terms of 𝑙𝑏𝑚/𝑏𝑏𝑙
𝑘𝑔𝑚 2.205𝑙𝑏𝑚 1𝑚3 3.785𝐿 42𝑔𝑎𝑙

ρ = 527.537 3 ( ) ( )( )( )
𝑚 1𝑘𝑔𝑚 1000𝐿 1𝑔𝑎𝑙 1𝑏𝑏𝑙
𝛒 = 𝟏𝟖𝟒. 𝟗𝟏𝟕 𝒍𝒃𝒎/𝒃𝒃𝒍 ⇒ 𝑨𝒏𝒔.
In terms of 𝑔𝑚/𝑚𝐿
𝑘𝑔𝑚 1000𝑔𝑚 1𝑚3 1𝑐𝑚3
ρ = 527.737 ( ) ( ) ( )
𝑚3 1𝑘𝑔𝑚 1003 𝑐𝑚3 1𝑚𝐿
𝛒 = 𝟎. 𝟓𝟐𝟕𝟕 𝒈𝒎/ 𝒎𝑳 ⇒ 𝑨𝒏𝒔.
35
In terms of 𝑠𝑙𝑢𝑔/𝑓𝑡 3
𝑙𝑏𝑚 1𝑠𝑙𝑢𝑔 1𝑏𝑏𝑙 7.481𝑔𝑎𝑙

ρ = 184.917 ( )( )( )
𝑏𝑏𝑙 32.174𝑙𝑏𝑚 42𝑔𝑎𝑙 1𝑓𝑡 3
𝛒 = 𝟏. 𝟎𝟐𝟑𝟕 𝒔𝒍𝒖𝒈/𝒇𝒕𝟑 ⇒ 𝑨𝒏𝒔.
In terms of 𝑖𝑛3 /𝑠𝑙𝑢𝑔:
𝑉
ν=
𝑚
1.8956𝑚3 1𝑘𝑔𝑚 32.174𝑙𝑏𝑚 39.373 𝑖𝑛3
ν= ( )( )( )
1000𝑘𝑔𝑚 2.205𝑙𝑏𝑚 1𝑠𝑙𝑢𝑔 1𝑚3
𝛎 = 𝟏𝟔𝟖𝟕. 𝟖𝟕𝟐 𝒊𝒏𝟑 /𝒔𝒍𝒖𝒈 ⇒ 𝑨𝒏𝒔.
In terms of 𝑦𝑑 3 /ℎ𝑦𝑙
1𝑓𝑡 3 1𝑦𝑑 3 1.055𝑠𝑙𝑢𝑔
ν = 1687.872𝑖𝑛3 /𝑠𝑙𝑢𝑔 ( ) ( ) ( )
12𝑖𝑛 3𝑓𝑡 1ℎ𝑦𝑙
𝛎 = 𝟎. 𝟎𝟑𝟖𝟏𝟕 𝒚𝒅𝟑 /𝒉𝒚𝒍 ⇒ 𝑨𝒏𝒔.
In terms of 𝑚3 /𝑡𝑜𝑛
1.8956𝑚3 907.03𝑘𝑔𝑚
ν= ( )
1000𝑘𝑔𝑚 1𝑡𝑜𝑛
𝛎 = 𝟏. 𝟕𝟏𝟗𝟒 𝒎𝟑 /𝒕𝒐𝒏 ⇒ 𝑨𝒏𝒔.
In terms of 𝑔𝑎𝑙/𝑙𝑏𝑚
1.8956𝑚3 1𝑘𝑔𝑚 1000𝐿 1𝑔𝑎𝑙
ν= ( )( 3 )( )
1000𝑘𝑔𝑚 2.205𝑙𝑏𝑚 1𝑚 3.785𝐿
𝛎 = 𝟎. 𝟐𝟐𝟕𝟏 𝒈𝒂𝒍/𝒍𝒃𝒎 ⇒ 𝑨𝒏𝒔.
(c) Solving for the specific weight of the liquid
ρ𝑔
γ=
𝑘
In terms of 𝑘𝑁/𝑚3
𝑘𝑔𝑚 𝑚
(527.537 ) (9.8066 2 )
𝑚3 𝑠
γ=
1000
𝑘𝑁 − 𝑠 2
𝛄 = 𝟓. 𝟏𝟕𝟑𝟑 𝒌𝑵/𝒎𝟑 ⇒ 𝑨𝒏𝒔.
In terms of 𝑙𝑏𝑓/𝑓𝑡 3 , we can convert it using our known standard values for water
𝑙𝑏𝑓
𝑘𝑁 62.4 𝑓𝑡 3
γ = 5.1733 3 ( )
𝑚 9.8066 𝑘𝑁
𝑚3
𝟑
𝛄 = 𝟑𝟐. 𝟗𝟏𝟖𝒍𝒃𝒇/𝒇𝒕 ⇒ 𝑨𝒏𝒔.
In terms of 𝑘𝑔𝑓/𝑑𝑟
𝑙𝑏𝑓 1𝑘𝑔𝑓 1𝑓𝑡 3 35𝑔𝑎𝑙
γ = 32.918 3 ( ) ( )( )
𝑓𝑡 2.205𝑙𝑏𝑓 7.481𝑔𝑎𝑙 1𝑑𝑟
𝛄 = 𝟔𝟗. 𝟖𝟒𝟓 𝒌𝒈𝒇/𝒅𝒓 ⇒ 𝑨𝒏𝒔.
36
In terms of 𝑑𝑦𝑛𝑒/𝑐𝑐
𝑘𝑁 1000𝑁 1𝑑𝑦𝑛𝑒 1𝑚 3
γ = 5.1733 ( ) ( ) ( )
𝑚3 1𝑘𝑁 105 𝑁 100𝑐𝑚
𝛄 = 𝟓. 𝟏𝟕𝟑𝟑 × 𝟏𝟎−𝟖 𝒅𝒚𝒏𝒆/𝒄𝒄 ⇒ 𝑨𝒏𝒔.
d) For the specific gravity of the liquid:

γ
𝑆𝐺 =
γ𝑤
5.1733𝑘𝑁/𝑚3
𝑆𝐺 =
9.8066𝑘𝑁/𝑚3
𝑺𝑮 = 𝟎. 𝟓𝟑 ⇒ 𝑨𝒏𝒔.
e) Using the definition of volume flow rate, assuming that the tank is initially empty:
𝑉
𝑉̇ =
𝑡
𝑉
𝑡=
𝑉̇
π
(1𝑚)2 (3𝑚)
𝑡= 4
𝑔𝑎𝑙 3.785𝐿 1𝑚3
150 ( )( )
𝑚𝑖𝑛 1𝑔𝑎𝑙 1000𝐿
𝒕 = 𝟒. 𝟏𝟓 𝒎𝒊𝒏 ⇒ 𝑨𝒏𝒔.
f) Using the definition of volume flow rate, assuming that the tank initially has a liquid at a depth of 0.75 m
𝑉
𝑡=
𝑉̇
𝑉 = 𝑉𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 − 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙
π
(1𝑚)2 (3𝑚) − 1.8956𝑚3
𝑡= 4
𝑔𝑎𝑙 3.785𝐿 1𝑚3
150 ( )( )
𝑚𝑖𝑛 1𝑔𝑎𝑙 1000𝐿
𝒕 = 𝟎. 𝟖𝟏𝟏𝟑 𝒎𝒊𝒏 ⇒ 𝑨𝒏𝒔.
g) For the total number of plates needed, use the formula
(Total Surface Area)(Number of Units to be Fabricated)(1 + % material wasted)

Number of Plates =
Area of the steel plate
For a closed horizontal cylindrical tank, the total surface area would be
𝑇𝑆𝐴 = 2π𝑟(𝑟 + 𝐿)
Hence
𝑚
2π(0.5𝑚)(0.5 + 3𝑚) (10tanks)(1 + 0.25)
Number of Plates = 𝑡𝑎𝑛𝑘
ft 2 1m 2
(8)(4) ( )
plate 3.281ft
𝑁 = 46.237 𝑝𝑙𝑎𝑡𝑒𝑠
𝑵 ≈ 𝟒𝟕 𝒑𝒍𝒂𝒕𝒆 ⇒ 𝑨𝒏𝒔.
37
1.8 PRESSURE
Pressure is defined as the force per unit cross sectional area, such that the single equivalent force is perpendicular
and at the centroid of the cross-sectional area. In actual, pressure is in terms of intensity, that varies in the differential
area. In this topic, we will discuss only the average pressure.
𝒅𝑭
𝑷=
𝒅𝑨
𝑭
𝑷=
𝑨
Consider a column of liquid, say water, in which the column has a cross sectional area A and the depth of the liquid
inside the column has a depth h. The weight of the water W acts as a force that is perpendicular to the cross-sectional
area of the vessel, hence
𝑊
𝑃=
𝐴
𝑉
Since 𝑊 = γ𝑉, and 𝐴 = , then ℎ
ℎ
𝑷 = 𝛄𝒉
in which is called as the hydrostatic pressure. 𝑃𝐵
1.9 MEASUREMENT OF PRESSURE

There are six types of pressure, namely atmospheric, gage, vacuum, absolute, total vacuum, and hydrostatic
pressure.
1. Atmospheric pressure - it is the pressure at any point of the earth’s surface from the weight of the standard
atmospheric air above. The atmospheric pressure can be measured by a barometer, and also called as the
barometric pressure. Under standard conditions and at sea level, the values of atmospheric pressure and pressure
conversions are as follows:
CONVERSION OF PRESSURE
𝑃𝑎𝑡𝑚 = 1 atm 1 bar = 0.10 MPa
= 101.325 kPa = 100 kPa
= 0.101325 MPa 1 mmHg = 1 torr
=1.01325 bar 1 kPa = 0.145 psi
= 14.7 psi =1000𝑁/𝑚2
= 760 mmHg 1psi = 1 𝑙𝑏𝑓/𝑖𝑛2
= 760 torr 1 psf = 1 𝑙𝑏𝑓/𝑓𝑡 2
= 29.92 in Hg
= 34 ft. 𝐻2 𝑂
𝑘𝑔𝑓
= 1.0332 2
𝑚
38
2. Gage Pressure - it is the pressure that is above the atmospheric pressure. Gage pressure is measurable by a
pressure measuring instrument.
𝑃𝑔 > 𝑃𝑎𝑡𝑚
3. Vacuum Pressure - it is the pressure that is below atmospheric pressure, also called as a negative gage
pressure.
𝑃𝑣𝑎𝑐 < 𝑃𝑎𝑡𝑚
4. Total Vacuum Pressure - it has a zero-pressure value. In total vacuum or absolute vacuum, heat and molecules
cease to exist.
𝑃=0
5. Absolute Pressure - it is the algebraic sum of the atmospheric pressure to the measured pressure. When
converting gage pressure to absolute pressure, add the atmospheric pressure to the gage pressure
𝑷𝒂𝒃𝒔 = 𝑷𝒈 + 𝑷𝒂𝒕𝒎
And when converting vacuum pressure to absolute pressure, subtract the atmospheric pressure to the vacuum
pressure.
𝑷𝒂𝒃𝒔 = 𝑷𝒂𝒕𝒎 − 𝑷𝒗𝒂𝒄
6. Hydrostatic Pressure - it is the pressure exerted at the bottom of the column of a static fluid. Hydrostatic
pressure is also a gage pressure.
𝑃 = γℎ
Note: For absolute pressure, units are added by a letter “a” at the end of its name (ex. psia, kPaa) or the word
“abs” to denote absolute pressure (ex. mmHg abs, torr abs). For gage pressure, units are added by a letter “g” to
denote that it is a gage pressure (ex. psig, kPag, torrg, mmHg g). For vacuum pressure, units are added by the
word “vac” to indicate vacuum pressure (ex. psi vac, kPa vac, mmHg vac).
1.10 PASCALS LAW

Pascal’s Law states that the pressure of a static liquid is the same in all directions. The pressure at any type of
configuration, for the same liquid, has equal pressure. This statement is also called as the hydrostatic paradox.
1 2 3 4 5 6
Since the depth ℎ for any configuration are equal, then

𝑃1 = 𝑃2 = 𝑃3 = 𝑃4 = 𝑃5 = 𝑃6 = γℎ
The theorem above only applies to liquids. For gases, the pressure measured at any point in a closed vessel is
always equal to the pressure at any point of the said vessel.
1.11 VARIATIONS IN PRESSURE

For incompressible fluids, the specific weight is assumed to be always constant. If the specific weight is a
function of height h, then 1
𝑃 = γℎ
Differentiating implicitly:
𝑑𝑃 = γ𝑑ℎ + ℎ𝑑γ
Since γ is constant, 𝑑γ = 0 ℎ
𝑑𝑃 = γ𝑑ℎ
2
Putting a negative sign on the equation indicates that as the height increases, the pressure decreases.
𝑑𝑃 = −γ𝑑ℎ
39
Integrating from upper limits to lower limits:

2 1
∫ 𝑑𝑃 = −γ ∫ 𝑑ℎ
1 2
𝑃2 − 𝑃1 = γ(ℎ2 − ℎ1 )
Since ℎ = ℎ2 − ℎ1, then
𝑃2 − 𝑃1 = γℎ
The hydrostatic pressure at the bottom is
𝑷𝟐 = 𝑷𝟏 + 𝛄𝒉
1.12 STACKS OF IMMISCIBLE FLUDIS
Consider the tank shown in the figure, filled with liquids of different densities, and with air on top with a given
gage pressure. The pressure is defined as
𝑛
𝑃𝑏 = ∑ γ𝑖 ℎ𝑖
𝑖=1
Where n is the number of liquids. Consider the classic stacks of fluid that involves air, water, oil, and mercury. The
pressure at the air and oil interface is:
𝑃1 = 𝑃𝑎𝑖𝑟 Air ℎ𝑎
The pressure at the oil – water interface is 1
𝑃2 = 𝑃𝑎𝑖𝑟 + γ𝑜 ℎ𝑜 Oil ℎ𝑜
The pressure at the water-mercury interface is:
𝑃3 = 𝑃𝑎𝑖𝑟 + γ𝑜 ℎ𝑜 + γ𝑊 ℎ𝑊 2
The pressure at the bottom of the tank is Water ℎ𝑤
𝑃𝑏 = 𝑃𝑎𝑖𝑟 + γ𝑜 ℎ𝑜 + γ𝑊 ℎ𝑊 + γℎ𝑔 ℎℎ𝑔
3
Mercury ℎℎ𝑔
B
1.13 PRESSURE-MEASURING INSTRUMENTS
There are several ways of measuring pressure. Pressure measuring instruments may be subdivided into three
categories
a. Mechanical Instruments - consists of manometers and pressure gauges.
b. Electromechanical Instruments - These instruments usually employ a mechanical means for detecting the
pressure and electrical means for indicating or recording the detected pressure.
c. Electronic Instruments- pressure instruments that depends on the physical change that can be detected or
recorded electronically.
The following are mechanical pressure instruments:
1. Barometer - invented by Evangelista Toricelli. It is a device that measures atmospheric pressure. The barometer
consists of a closed end glass tube initially filled with mercury. The tube is submerged into a vessel of mercury and
turning it outside down, creating a small volume in the fluid in the tube. The height of the fluid in the tube is the
barometric measurement, measured in inches of Mercury (inHg) or millimeter of Mercury (mmHg). The
measurement of this pressure is in terms of pressure head.
40
The barometric pressure can be measured from A and B,

assuming the pressure at point A is zero
𝑃𝐵 = 𝑃𝐴 + γ𝐻𝑔 ℎ
𝑷𝒃𝒂𝒓𝒐 = 𝑷𝑩 = 𝛄𝑯𝒈 𝒉
2. Manometer - it is a devise that measures the gage pressure of a liquid. There are two types of manometer:
a. Open Type manometer – a part of the tube is open to the atmosphere. It measures the gage pressure at a
selected point in the manometer.
b. Closed Type manometer - no part is open tp the atmosphere. It measures the difference between two
pressure points.
To solve any manometer problems, the following are key procedures:

✓ Decide the pressure if it is to be expressed in terms of force per unit area or in terms of pressure heads.
For pressure heads, convert all liquids into units of length of water (ex. meters of water, feet of water).
✓ Starting from the end point, number in order the interface of different liquids.
✓ Identify point of equal pressure. Recall Pascal law that states that pressure are always equal of the same
height and liquid at any configuration.
✓ Proceed from level to level, adding if going down and subtracting if going down. To convert pressure
head of any liquid into water.
✓ For gases, the pressure is always equal at any point in it. If a gas is enclosed in a manometer and the
problem does not mentioned its pressure, assume its pressure equal to zero. The pressure of the interface
where the gas is enclosed is always equal.
The following are type of manometer

A. Piezometer - the simplest form of manometer used for measuring moderate pressures of liquid. Consider two
fluids A and B. Using pressure heads
41
B. U-Tube Open Manometer - consists of glass tube bent in U shape, one end of which is connected to a point at
which pressure is to be measured and another end remains open to the atmosphere.
C. Differential Manometer - is used to measure the difference in pressures between two points. There are two
types of differential manometer:
a. U-Tube Differential Manometer - most common type of differential manometer.
b. Inverted U-Tube Manometer - used when measuring difference of two pressure whose accuracy is a
major consideration.
42
Solved Problems
Problem 1.12: For a barometric pressure of 100 kPa. What is the absolute pressure of
a) 20 psig in kPa and torr
b) 3 psi vac in mmHg and kgf/cm2
c) a gage pressure numerically twice its absolute pressure in bar and in Hg.
Solution
From the definition of absolute pressure:
a) 𝑃𝑎𝑏𝑠 = 𝑃𝑏𝑎𝑟𝑜 + 𝑃𝑔
101.325𝑘𝑃𝑎
𝑃𝑎𝑏𝑠 = 100𝑘𝑃𝑎 + 20𝑝𝑠𝑖 ( )
14.7𝑝𝑠𝑖
𝑷𝒂𝒃𝒔 = 𝟐𝟑𝟕. 𝟖𝟔𝒌𝑷𝒂𝒂 ⇒ 𝑨𝒏𝒔.
760𝑡𝑜𝑟𝑟𝑎𝑏𝑠
𝑃𝑎𝑏𝑠 = 237.86𝑘𝑃𝑎𝑎 ( )
101.325𝑘𝑃𝑎𝑎
𝑷𝒂𝒃𝒔 = 𝟏𝟕𝟖𝟒. 𝟏𝟎 𝒕𝒐𝒓𝒓 𝒂𝒃𝒔 ⇒ 𝑨𝒏𝒔.
b) 𝑃𝑎𝑏𝑠 = 𝑃𝑏𝑎𝑟𝑜 − 𝑃𝑣𝑎𝑐

101.325𝑘𝑃𝑎
𝑃𝑎𝑏𝑠 = 100𝑘𝑃𝑎 − 3𝑝𝑠𝑖 ( )
14.7𝑝𝑠𝑖
760𝑚𝑚𝐻𝑔𝑎
𝑃𝑎𝑏𝑠 = 79.32𝑘𝑃𝑎𝑎 ( )
101.325𝑘𝑃𝑎𝑎
𝑷𝒂𝒃𝒔 = 𝟓𝟒𝟓. 𝟗𝟓𝒎𝒎𝑯𝒈 𝒂 ⇒ 𝑨𝒏𝒔.
𝑘𝑔𝑓
1.0332
2 𝑎𝑏𝑠
𝑃𝑎𝑏𝑠 = 545.95𝑚𝑚𝐻𝑔𝑎 ( 𝑐𝑚 )
760𝑚𝑚𝐻𝑔 𝑎
𝒌𝒈𝒇
𝑷𝒂𝒃𝒔 = 𝟎. 𝟕𝟒𝟐𝟐 ⇒ 𝑨𝒏𝒔.
𝒄𝒎𝟐
c. If the gage pressure is twice the barometric pressure:
𝑃𝑔 = 2𝑃𝑏𝑎𝑟𝑜 (Equation 1)
𝑃𝑎𝑏𝑠 = 𝑃𝑏𝑎𝑟𝑜 + 𝑃𝑔 (Equation 2)
Substituting equation 1 to equation 2:

𝑃𝑎𝑏𝑠 = 𝑃𝑏𝑎𝑟𝑜 + 2𝑃𝑏𝑎𝑟𝑜
𝑃𝑎𝑏𝑠 = 3𝑃𝑏𝑎𝑟𝑜
1𝑏𝑎𝑟𝑎
𝑃𝑎𝑏𝑠 = 3(100𝑘𝑃𝑎𝑎) ( )
100𝑘𝑃𝑎𝑎
𝑷𝒂𝒃𝒔 = 𝟑𝒃𝒂𝒓 𝒂 ⇒ 𝑨𝒏𝒔.
29.92𝑖𝑛𝐻𝑔
𝑃𝑎𝑏𝑠 = 3𝑏𝑎𝑟𝑎 ( )
1.01325𝑏𝑎𝑟𝑎
𝑷𝒂𝒃𝒔 = 𝟖𝟖. 𝟓𝟗 𝒊𝒏𝑯𝒈 𝒂𝒃𝒔 ⇒ 𝑨𝒏𝒔.
43
Problem 1.13: Find the gage and absolute pressure in kPag and psia at the middle and at the bottom of the tank
respectively. 𝑆𝐺𝑜𝑖𝑙 = 0.85, 𝑆𝐺𝐻𝑔 = 13.55.
Solution
For the pressure at the middle of the tank, from 1 to 2:
𝑃𝑚 − 𝑃1 = γ𝑜 ℎ1
𝑃𝑚 = 𝑃1 + γ𝑜 ℎ1
𝑃𝑚 = 𝑃1 + 𝑠𝑜 γ𝑤 ℎ1
𝑘𝑁
𝑃𝑚 = 20𝑘𝑃𝑎 + 0.85 (9.8066 3 ) (1𝑚)
𝑚
𝑷𝒎 = 𝟐𝟖. 𝟑𝟒𝒌𝑷𝒂𝒈 ⇒ 𝑨𝒏𝒔.
For the gage pressure at the bottom of the tank, from 2 to 4

𝑃𝐵 = 𝑃𝑚 + γ𝑤 ℎ2 + γ𝐻𝑔 ℎ3
𝑃𝐵 = 𝑃𝑚 + γ𝑤 ℎ2 + 𝑠𝐻𝑔 γ𝑤 ℎ3
𝑘𝑁 𝑘𝑁
𝑃𝐵 = 28.34𝑘𝑃𝑎 + (9.8066 3 ) (1.8𝑚) + 13.55 (9.8066 3 ) (0.2𝑚)
𝑚 𝑚
𝑃𝐵 = 72.57𝑘𝑃𝑎𝑔
For the absolute pressure at the bottom of the tank

𝑃𝐵 = 𝑃𝑔
𝑃𝑎𝑏𝑠 = 𝑃𝑔 + 𝑃𝑎𝑡𝑚
14.7𝑝𝑠𝑖
𝑃𝑎𝑏𝑠 = (72.57 + 101.325)𝑘𝑃𝑎 ( )
101.325𝑘𝑃𝑎
𝑷𝒂𝒃𝒔 = 𝟐𝟓. 𝟐𝟑 𝒑𝒔𝒊𝒂 ⇒ 𝑨𝒏𝒔.
Problem 1.14: Mr. Atienza carries a barometer from the ground floor atop the building. On the ground level the
barometer reads 30.150 inHg abs, while at the top level it reads 28.607 inHg abs. Assume that the average
atmospheric air density was 0.075 lbf /ft3, estimate the height of the building.
Solution:
From the definition of pressure for incompressible air:

𝑃 = γℎ
44
Differentiating the equation implicitly gives:

𝑑𝑃 = γ𝑑ℎ + ℎ 𝑑γ
For incompressible fluids, the specific weight is assumed to be constant, hence

𝑑γ = 0
Thus:
Integrate from its upper limit (2) to its lower limit (1):
2 1
∫ 𝑑𝑃 = γ ∫ 𝑑ℎ
1 2
𝑃2 − 𝑃1 = γ(ℎ1 − ℎ2 )
From the figure, ℎ = ℎ1 − ℎ2

Thus
𝑃2 = 𝑃1 + γℎ
Solving for the approximate height of the building:
(𝑃2 − 𝑃1 )𝑘
ℎ=
ρ𝑔
𝑙𝑏𝑓 144𝑖𝑛2
14.7 2 ( )
𝑖𝑛 1𝑓𝑡 2 𝑙𝑏𝑚 − 𝑓𝑡
(30.150 − 28.607)𝑖𝑛𝐻𝑔 ( ) (32.174 )
29.92 𝑖𝑛𝐻𝑔 𝑙𝑏𝑓 − 𝑠2
ℎ=
𝑙𝑏𝑓 𝑓𝑡
0.075 3 (32.174 2 )
𝑓𝑡 𝑠
𝒉 = 𝟏𝟒𝟓𝟓. 𝟓𝟒 𝒇𝒕 ⇒ 𝑨𝒏𝒔.
Problem 1.15: For the open manometer shown below, what is the pressure at A in kPag? Let ℎ1 =30 cm,
ℎ2 =48cm and relative density of mercury to be 13.59.
Summing up all the pressure from point A to 2 equal

to zero
∑𝑃 = 0 ↓ + ↑ −
𝑃𝐴 + γ𝑤 ℎ1 − γ𝐻𝑔 ℎ2 − 𝑃2 = 0
Since at point 2, the manometer is open to the
atmosphere, thus
𝑃2 = 0
Hence
𝑃𝐴 = γ𝐻𝑔 ℎ2 − γ𝑤 ℎ1
𝑘𝑁 𝑘𝑁
𝑃𝐴 = (13.59) (9.8066 3 ) (0.48𝑚) − (9.8066 3 ) (0.30𝑚)
𝑚 𝑚
𝑷𝑨 = 𝟔𝟏. 𝟎𝟑𝒌𝑷𝒂𝒈 ⇒ 𝐴𝑛𝑠.
45
1.14 HYDROSTATIC FORCES ON PLANE SURFACES

The hydrostatic force F is defined as the product of the area and the intensity of the pressure along its centroid.
It is also the force exerted by liquids to bodies submerged at rest.
𝐅 = 𝛄𝐀𝐡̅
where:
𝐹 = hydrostatic force in kN, lbf

γ = specific weight of liquid in kN/m3, lbf/ft3
ℎ̅ = distance from the center of gravity to the open
surface
ℎ𝑝 = distance from the center of pressure to the open
surface
𝐺 = center of gravity
𝑃 = center of pressure
𝑒 = eccentricty
The value of 𝑒 is determined as
𝑰𝒈
𝒆=
̅
𝑨𝒉
Where
𝐼𝑔 = moment of inertia with respect to the center of gravity
𝐴 = cross-sectional area of the body
Hence the distance from the center of pressure to the open surface is
̅+𝒆
𝒉𝒑 = 𝒉
̅
*Derivation of Formula 𝑭 = 𝛄𝑨𝒉
To derive this formula, consider a body that is submerged in a liquid as shown below. We will now derive the
hydrostatic force for a simple plane surface having a plane area A.
Liquid Surface
θ
̅𝑦 ̅ℎ ℎ𝑝
𝑦𝑝
𝐹 𝑑𝐴
𝐺
P
𝑒
From the definition of pressure

P = γh
Consider a differential force dF acting on a differential area dA. Hence
dF
P=
dA
46
dF
= γh
dA
From the figure,
h = y sinθ
Hence
dF
= yγsinθ
dA
Integrating both sides
∫ dF = γsinθ ∫ y dA
Recall that
y̅ = ∫ y dA
Hence
F = γAy̅sinθ
Since h̅ = y̅sinθ
𝐅 = 𝛄𝐀𝐡̅
Where
F = hydrostatic force exerted by the liquid to the body.
y̅ = is the location of center of gravity G along the incline plane.
h̅ = is the location of center of gravity along the liquid surface.
A = is the plane area that is submerged to the liquid.
γ = is the specific weight of the liquid.
*Derivation of the formula 𝒉𝒑 = 𝒉 ̅+𝒆

The location of the hydrostatic force that acts on the body is called as the center of pressure. The pressure
force is defined as an intensity and increase from the end of the body to the bottom of the body where the pressure
force is at maximum.
From the figure, summing all moments at point O:

θ
yp F = ∫ y dF
Since dF = γy sinθdA, and F = γAy̅sinθ
𝑦𝑝
yp γAy̅sinθ = γsinθ ∫ y 2 dA
∫ 𝑦 𝑑𝐹
yp y̅A = ∫ y 2 dA 𝑭
Recall that for the moment of inertial along the x axis with respect to O:
IO = ∫ y 2 dA
Hence
yp y̅A = Io
Since we are dealing with the centroid, we can determine the centroidal moment of inertia using the parallel axis
theorem, where
IO = IG + Ay̅ 2
Hence:
yp y̅A = Ig + Ay̅ 2
47
The location of center of pressure along the plane would be:

𝐈𝐠
𝐲𝐩 = 𝐲̅ +
𝐀𝐲̅
Ig
The value ̅
is sometimes referred by engineers as the eccentricity which is defined as the difference of the
Ay
length of location of center of pressure to center of gravity.
𝐈𝐠 𝑰𝒈
𝐞= =
𝐀𝐲̅ 𝑨𝒉 ̅
Hence
𝐲𝐩 = 𝐲̅ + 𝐞
𝒉𝒑 = 𝒉 ̅+𝒆
Where
𝐲𝐩 is the location of center of pressure along the liquid surface.
𝐞 is the eccentricity
𝐈𝐠 is the centroidal moment of inertia or the moment of inertia along the centroidal axis
𝐈𝐨 is the moment of inertia at the fluid surface
The following are useful formulas in finding centroid and moment of inertia for plane surfaces.
Centroidal Moment of Inertia
Plane Surface Figure
and Centroid
𝑏ℎ3
𝐼𝑔 =
36
𝑏
𝑥̅ =
Triangle 2
1
𝑦̅ = ℎ
3
𝑏ℎ3
𝐼𝑔 =
12
𝑏
𝑥̅ =
Rectangle 2
ℎ
𝑦̅ =
2
π𝐷 4
𝐼𝑔 =
64
π𝑅 4
𝐼𝑔 =
Circle 4
𝐷
𝑥̅ = 𝑦̅ =
2
48
Solved Problems
Problem 1.16 Find the hydrostatic force in kgf and its location of the following:
a. A 2ft by 3 ft rectangular plate submerged fully into oil, whose base is at the oil surface.
b. A 4ft by 6 ft rectangular plate, in which half of the plate is submerged in water.
c. A 2ft radius circular plate, in which is 1ft below the free surface of molasses.
Solution:
a) A 2ft by 3 ft rectangular plate submerged fully into oil, whose base is at the oil surface.
For the hydrostatic force acting on the rectangular

plate
𝐹 = γ𝐴ℎ̅
𝐹 = 𝑠𝑜 γ𝑤 𝐴ℎ̅
𝑙𝑏𝑓
𝐹 = 0.80 (62.4 3 ) (5𝑓𝑡)(4𝑓𝑡)(1.5𝑓𝑡)
𝑓𝑡
1𝑘𝑔𝑓
𝐹 = 449.28𝑙𝑏𝑓 ( )
2.205𝑙𝑏𝑓
𝑭 = 𝟐𝟎𝟑. 𝟕𝟔 𝒌𝒈𝒇 ⇒ 𝑨𝒏𝒔.
For the eccentricity

𝐼𝑔
𝑒=
𝐴ℎ̅
𝑏ℎ3
𝑒 = 12
𝐴ℎ̅
ℎ2
𝑒=
12ℎ̅
(3𝑓𝑡 2 )
𝑒=
12(1.5𝑓𝑡)
𝑒 = 0.5 𝑓𝑡
For the location of center of pressure

ℎ𝑝 = ℎ̅ + 𝑒
ℎ𝑝 = (1.5 + 0.5)𝑓𝑡
𝒉𝒑 = 𝟐𝒇𝒕 from the liquid surface ⇒ 𝑨𝒏𝒔.
b) A 4ft by 6 ft rectangular plate, in which half of the plate is submerged in water.
49
For the hydrostatic force acting on the plate

𝐹 = γw 𝐴ℎ̅
𝑙𝑏𝑓
𝐹 = (62.4 3 ) (1.5𝑓𝑡)(3𝑓𝑡)(4𝑓𝑡)
𝑓𝑡
1 𝑘𝑔𝑓
𝐹 = 1123.2 𝑙𝑏𝑓 ( )
2.205𝑙𝑏𝑓
𝑭 = 𝟓𝟎𝟗. 𝟑𝟗 𝒌𝒈𝒇 ⇒ 𝑨𝒏𝒔.
For the eccentricity

𝐼𝑔
𝑒=
𝐴ℎ̅
𝑏ℎ3
𝑒 = 12
𝐴ℎ̅
ℎ2
𝑒=
12ℎ̅
(3𝑓𝑡 2 )
𝑒=
12(1.5𝑓𝑡)
𝑒 = 0.5 𝑓𝑡

ℎ𝑝 = ℎ̅ + 𝑒
ℎ𝑝 = (1.5 + 0.5)𝑓𝑡
𝒉𝒑 = 𝟐𝒇𝒕 from the liquid surface ⇒ 𝑨𝒏𝒔.
c. A 2ft radius circular plate, in which is 1ft below the free surface of molasses.
For the hydrostatic force acting on the plate:

𝐹 = γ𝐴ℎ̅
𝐹 = 𝑠𝑚𝑜𝑙 γ𝑤 𝐴ℎ̅
𝑙𝑏𝑓
𝐹 = (1.5) (62.4 3 ) (π)(2𝑓𝑡)2 (3𝑓𝑡)
𝑓𝑡
1𝑘𝑔𝑓
𝐹 = 3528.64 𝑙𝑏𝑓 ( )
2.205𝑙𝑏𝑓
𝑭 = 𝟏𝟔𝟎𝟎. 𝟐𝟗 𝒌𝒈𝒇 ⇒ 𝑨𝒏𝒔.
For eccentricity
𝐼𝑔
𝑒=
𝐴ℎ̅
π𝑟 4
𝑒 = 42
π𝑟 ℎ̅
𝑟2
𝑒=
4ℎ̅
(2𝑓𝑡)2
𝑒=
4(3𝑓𝑡)
𝑒 = 0.33 𝑓𝑡
ℎ𝑝 = ℎ̅ + 𝑒
ℎ𝑝 = (3 + 0.33)𝑓𝑡
𝒉𝒑 = 𝟑. 𝟑𝟑 𝒇𝒕 from the liquid surface⇒ 𝑨𝒏𝒔.
50
Problem 1-17: Compute for the force required to open the gate shown below in kN.
Solution: Consider the figure below showing all the forces acting on the gate.
For 𝐹1 , the hydrostatic force exerted by water on the submerged body.

𝐹1 = γ𝑤 𝐴1 ̅̅̅
ℎ1
𝑘𝑁
𝐹1 = (9.8066 3 ) (4 𝑚)(8 𝑚)(6 𝑚)
𝑚
𝐹1 = 1882.87𝑘𝑁
51
For 𝐹2 , the hydrostatic force exerted by oil on the submerged body.

𝐹2 = γ𝑜 𝐴2 ̅̅̅
ℎ2
𝐹2 = 𝑠𝑜 𝛾𝑤 𝐴2
𝑘𝑁
𝐹2 = (0.8) (9.8066 3 ) (4 𝑚)(4 𝑚)(2 𝑚)
𝑚
𝐹2 = 251.05 𝑘𝑁
For 𝑒1 , the eccentricity of the hydrostatic force exerted by the water:

𝐼𝑔
𝑒1 =
𝐴1 ̅̅̅
ℎ1
h1 2
e1 =
12h̅̅̅1
(8𝑚)2
𝑒1 =
12(6𝑚)
𝑒1 = 0.89𝑚
For 𝑒2 , the eccentricity of the hydrostatic force exerted by the oil:

𝐼𝑔
𝑒2 =
̅̅̅2
𝐴2 ℎ
h2 2
e2 =
12h̅̅̅2
(4𝑚)2
𝑒2 =
12(2𝑚)
𝑒2 = 0.67 𝑚
Taking summation of moments equal to zero:
∑ 𝑀𝐴 = 0
𝐹2 (6 + 𝑒2 ) + (𝐹𝑐𝑜𝑠30)(8) − 𝐹1 (4 + 𝑒1 ) = 0
(251.05𝑘𝑁)(6 + 0.67)𝑚 + 𝐹𝑐𝑜𝑠30(8𝑚) − (1882.87𝑘𝑁)(4 + 0.89𝑚) = 0
(1882.87 𝑘𝑁)(4 + 0.89)𝑚 − (251.05𝑘𝑁)(6 + 0.67)𝑚
𝑭=
(8𝑚)𝑐𝑜𝑠30
𝑭 = 𝟏𝟎𝟖𝟕. 𝟐𝟔 𝒌𝑵 ⇒ 𝑨𝒏𝒔.
Analysis: In order to move the gate, 𝑭 ≥ 𝟏𝟎𝟖𝟕. 𝟐𝟔 𝒌𝑵
52
1.15 TEMPERATURE
Temperature is the measure of hotness or coldness of a body. Temperature measures the amount of heat
that is transferred from one body to another until it reaches thermal equilibrium. Basically, there are two types of
temperature.
1. Arbitrary Temperature - it is the temperature t that varies with the temperature reading. The english system for
arbitrary temperature is the Fahrenheit (°F) that uses mercury thermometers. For the metric system, we use Celsius
scale (°C). Fahrenheit scale is subdivided into 180 divisions, while Celsius are subdivided into 100 division; the
reason why Celsius scale is also called as centigrade scale. Assuming standard atmospheric pressure, for the boiling
point (steam point) and the freezing point (ice point) of water.
We can now derive each temperature scale as a function of another scale. For example, we would wish to convert °F
as function of °C, using a coordinate system, we can derive the conversions
°C °F
0 32
𝑡𝐶 𝑡𝐹
100 212
By ratio and proportion, since the

coordinate system employs similar
triangles:
𝑡𝐶 − 0 𝑡𝐹 − 32
=
100 − 0 212 − 32
𝑡𝐶 𝑡𝐹 − 32
=
100 180
100
𝑡𝐶 = (𝑡 − 32)
180 𝐹
𝟓
𝒕𝒄 = (𝒕𝑭 − 𝟑𝟐)
𝟗
To convert °𝐶 as a function of °𝐹
𝟗
𝒕𝑭 = 𝒕 + 𝟑𝟐
𝟓 𝑪
53
2. Absolute Temperature - it is the temperature T that is based on the fixed or absolute reading. The english
unit of absolute temperature is called as the Rankine (°R) and for the SI/metric units is called as Kelvin (K). The
scale is based on the absolute zero temperature.
Absolute zero for Rankine and Kelvin scales:

𝑇𝑅 = 0°𝑅 = −460℉
𝑇𝐾 = 0𝐾 = −273℃
Note that at temperature below the absolute zero temperature, all molecules and heat cease to exists and thus
validates the third law of thermodynamics.
Converting arbitrary to absolute temperature

𝑻𝑲 = 𝒕𝒄 + 𝟐𝟕𝟑
𝑻𝑹 = 𝒕𝑭 + 𝟒𝟔𝟎
In thermodynamic problems, we are most focused on the change of temperature of the system because it
indicates if heat is lost or gained by the system. The rate of change of absolute temperature is also the rate of
change of arbitrary temperature. In differential terms (microanalysis)
a) Relationship between Kelvin and Celsius scale:

𝑇𝐾 = 𝑡𝑐 + 273
𝑑𝑇𝐾 = 𝑑𝑡𝑐 + 𝑑(273)
𝒅𝑻𝑲 = 𝒅𝒕𝒄
𝚫𝑲 = 𝚫°𝑪
b) Relationship between Rankine and Fahrenheit scale:
𝑇𝑅 = 𝑡𝑓 + 460
𝑑𝑇𝑅 = 𝑑𝑡𝑓 + 𝑑(460)
𝑑𝑇𝑅 = 𝑑𝑡𝑓
𝚫°𝑹 = 𝚫°𝑭
c) Relationship between Fahrenheit and Celsius scale:
9
𝑡𝐹 = 𝑡𝐶 + 32
5
9
𝑑𝑡𝐹 = 𝑑𝑡𝑐 + 𝑑(32)
5
𝑑𝑡𝐹 = 1.8𝑑𝑡𝐶
𝚫°𝑭 = 𝟏. 𝟖𝚫°𝑪
The following are notable properties for change in temperature
a. The change in absolute temperature is also equivalent to the change in arbitrary temperature.
b. For the change in temperature, the following conversions can be used:
𝚫𝑲 = 𝚫°𝑪
𝚫°𝑹 = 𝚫°𝑭
𝚫°𝑭 = 𝟏. 𝟖𝚫°𝑪
54
Solved Problem:
Problem 1.18: Convert the following temperature readings:

a. 122 °F to °C and to K
b. - 30 °C to °F and to °R
c. 942 °R to °C and to K
Solution
a) 122 °F to °C and to K
5
𝑡𝑐 = (𝑡𝑓 − 32)
9
5
𝑡𝑐 = (122 − 32)
9
𝒕𝒄 = 𝟓𝟎℃ ⇒ 𝑨𝒏𝒔.
𝑇𝐾 = 𝑡𝑐 + 273
𝑇𝐾 = 50 + 273
𝑻𝑲 = 𝟑𝟐𝟑𝑲 ⇒ 𝑨𝒏𝒔.
b) - 30 °C to °F and to °R
9
𝑡𝑓 = 𝑡𝑐 + 32
5
9
𝑡𝑓 = (−30℃) + 32
5
𝒕𝒇 = −𝟐𝟐℉ ⇒ 𝑨𝒏𝒔.
𝑇𝑅 = 𝑡𝑓 + 460
𝑇𝑅 = −22 + 460
𝑻𝑹 = 𝟒𝟑𝟖°𝑹 ⇒ 𝑨𝒏𝒔.
c) 942 °R to °C and to K
𝑡𝑓 = 𝑇𝑅 − 460
𝑡𝑓 = 942 − 460
𝑡𝑓 = 482℉
5
𝑡𝑐 = (t f − 32)
9
5
𝑡𝑐 = (428 − 32)
9
𝒕𝒄 = 𝟐𝟐𝟎℃ ⇒ 𝑨𝒏𝒔.
𝑇𝐾 = 𝑡𝑐 + 273
𝑇𝐾 = 220 + 273
𝑻𝑲 = 𝟒𝟗𝟑𝑲 ⇒ 𝑨𝒏𝒔.
55
Problem 1.19: A fahrenheit and celsius thermometer are both immersed in a fluid
a. At what absolute temperature in kelvin and rankine will both thermometer readings are equal?
b. What is the fluid temperature if the fahrenheit reading is numerically twice that of the celcius reading?
Solution:
a) If 𝑡𝑓 = 𝑡𝑐
9
𝑡𝑓 = 𝑡𝑐 + 32
5
9
𝑡𝑐 = 𝑡𝐶 + 32
5
9
𝑡𝑐 (1 − ) = 32
5
𝑡𝑐 = −40℃
𝑡𝑓 = −40℉
𝑇𝐾 = 𝑡𝑐 + 273
𝑇𝐾 = −40 + 273
𝑻𝑲 = 𝟐𝟑𝟑𝑲 ⇒ 𝑨𝒏𝒔.
𝑇𝑅 = 𝑡𝑓 + 460
𝑇𝑅 = −40 + 460
𝑻𝑹 = 𝟒𝟐𝟎℉ ⇒ 𝑨𝒏𝒔.
b) If 𝑡𝑓 = 2𝑡𝑐
9
𝑡𝑓 = 𝑡𝑐 + 32
5
9
2𝑡𝑐 = 𝑡𝑐 + 32
5
9
𝑡𝑐 (2 − ) = 32
5
𝒕𝒄 = 𝟏𝟔𝟎℃ ⇒ 𝑨𝒏𝒔.
𝑡𝑓 = 2(160)
𝒕𝒇 = 𝟑𝟐𝟎℉ ⇒ 𝑨𝒏𝒔.
1.20 Engineer X proposed a new temperature scale °X denoting 10°X and 100°X as the freezing and boiling point
of water respectively.
a. Derive an expression of this temperature in terms of celsius scale.
b. What would be the absolute zero temperature of this temperature scale?
Solution:
a) By ratio and proportion
𝑡𝑥 − 10 𝑡𝑐 − 0
=
100 − 10 100 − 0
𝒕𝒙 = 𝟎. 𝟗𝟎𝒕𝒄 + 𝟏𝟎 ⇒ 𝑨𝒏𝒔.
b) By ratio and proportion

100 − 𝑇𝑥 100 − (−273)
=
100 − 10 100 − 0
𝑻𝒙 = −𝟐𝟑𝟓. 𝟕°𝑿
56
SUPPLEMENTARY PROBLEMS___________________
CHAPTER I. PROPERTIES OF MATTER
Name: _______________________________________________________ Class Number: __________
Course/Year/Section:____________________________________________ Score: _________________
Directions: Solve the following problems and show all complete solutions. Any erasures in the computations will
consider your answer as wrong. Enclosed your final answer by a rectangle. Answers with wrong units are considered
also as wrong.
1. Five masses in a region where g = 30.5 fps are as follows: 𝑚1 is 1500 kgm, 𝑚2 weighs 850 kgf, 𝑚3 weighs
150 slugs, 𝑚4 weighs 3 tons and 𝑚5 is 50 slinches. Determine the total mass express in (a) kilograms (b) pounds
(c) hyl, and (d) tonnes. (e) What is the total weight in pound force?
Ans. (a)
2. For a ballistic study, a 2.0 gm bullet is fired into soft wood. The bullet strikes the wood surface with a velocity
of 400 m/s and penetrates 2 inches. Find
a. The constant retarding force in lbf.
b. The time required to stop the bullet in seconds.
c. The constant deceleration of the bullet in fps.
d. The velocity of the bullet half-way of its penetration in mps.
Ans. (a)
57
3. If 50 kg of mass are placed on the pan of a spring balance located on a freight elevator and if local gravity
acceleration is 9.70 m/s2, then
a. When the elevator is moving with an upward acceleration of 2.5 m/s2 what will the balance read?
b. If the elevator is stopped, what will be the balance read?
c. If the supporting cable breaks, what will the balance read?
d. If the balance reads 350 N, what would be the acceleration of the mass? Is the elevator moving upward
or downward?
Ans. (a)
4. Compute for the gravitational force between a proton and an electron in an atom whose radius of electron orbit
is 5.29 x 10-11 m. Report your answers in units of N and dynes. Mass for proton and electron are 1.66 x 10 -27 kgm
and 9.11x10-31 kgm respectively.
Ans.
58
5. A mass of 0.1 slug in space is subjected to an external vertical force if 4 lb. If the local gravity acceleration is
30.5 fps2 and friction effects are neglected, determine the acceleration of the mass in fps 2 if the vertical force is
acting (a) upward (b) downward
Ans.
6. A system has a mass of 30 lbm. What is the total force necessary to accelerate it 15 fps 2
a. If it is moving on a horizontal frictionless plane
b. If it is moving vertically upward where local g = 31.50 fps2
Ans.
7. A girl weighing 470 N hangs suspended on the end of the rope 8m long. If a friend pushes her laterally to one
side until the rope makes an angle of 35 degrees with the vertical
a. What lateral force is required in N?
b. What is the tension in the rope in N?
c. If local g = 970 cps2, what is her mass in kgm and lbm?
Ans.
59
8. An astronaut located on the surface of the moon (g = 5.36 fps 2) places a crater sample on a spring scale, a
reading of 7.5lbf is noted. (a)What is the sample mass in lbm? (b)If the scale had been a balance type, what
reading in lbf would have been noted?
Ans.
9. What is the mass in kilograms and weight in newtons (g=9.65mps2) of (a) a 4000 lbm automobile (b) 235 lbm
fullback?
Ans.
10. Find the mass in grams and weight in dynes (g=9.65mps2) of (a) 77 grains of moisture (b) 12 oz of salt.
Ans.
11. Note that the gravity acceleration at equatorial seal level is g = 32.0565fps2 and that its variation is -0.002 fps2
per 1500 ft ascent. Find the height in miles above the point for which
a. The gravity acceleration becomes g = 30 fps2
b. The weight of the given man is decreased by 4%
c. What is the weight of a 160 kgm man atop the 5000 m mountain, relative to this point?
60
12. The mass of a given airplane at sea level at local g=32.10fps2 is 10 tons.
a. Find its mass in lbm, slugs, and kgm
b. Find its gravitational weight in lbf and N when it is travelling a 50000 ft elevation. The acceleration
due to gravity g decreased by 3.33 x 10 -6 fps2 for each foot of elevation.
13. Calculate the magnitude of the gravity acceleration on the surface of the moon and also at the point 1000 km
above the surface of the moon. Ignore the gravity effects of the earth. The moon has a mean radius of 1740 km and
a mass of 7.4 x 1022 kgm.
14. How far from the earth must a body be along a line toward the sun so that the gravitational pull of the sun
balances of the earth? Earth to sun distance is 9.3x107 mi; mass of the sun is 3.24 x 10 5 times mass the earth.
61
15. A car moves with uniform acceleration travels 10 m between the 20th second of its motion and 9 m during the
16th second of its motion. What is its initial velocity and its constant acceleration?
16. A bus runs at a velocity of 60mph. A motorcycle is running at the same direction of the bus 300ft apart. If brakes
are applied to the bus, causing a constant deceleration of 5ft/s2, in what time will the motorcycle overtake the bus,
and how far each vehicle had travelled?
17. Two cars A and B is 500 meters apart. Car A runs from rest with a constant acceleration of 10m/s. Car B is
against the direction of car B runs at a constant velocity of 8m/s. When and where will the cars crash?
62
18. A boy throws a stone from an elevator moving at 5 fps until the stone reaches the shaft of the elevator. How
high is the elevator from the shaft?
19. A block is thrown vertically upward and returns to the ground in 15 seconds. What is its initial velocity in fps
and how high did it go in meters?
20. A stone is dropped from a building 50 m high at the same instant another stone is thrown upward from the
ground with an initial velocity of 10mps. When and where do they pass, and what relative velocity?
63
21. The motion of a particle is represented as 𝑠 = 5𝑡 3 + 2𝑡 2 − 𝑡 + √5 , where s is in centimeters and t is in

seconds. Determine
a) The acceleration when t=2s
b) The velocity when t=3s
c) Time elapsed when the particles come to a stop.
d) The time elapsed when the particle is at its 500 mm mark
22. The economical speed for a certain material to be used for belts in pulley system is around 4400 fpm. A pulley
of what diameter in inches should be used on a motor running at 1760 rpm to give the required speed, if there is no
slipping between the belt and the pulley surface.
64
23. In order to call Juliet, Romeo must throw a 5 gram pebble from the ground to reach the window, which is 5 ft
above from the ground. If the motion is assumed curvilinear, at what velocity should he throw the pebble at 45° so
that Juliet will hear the sound of the pebble? Romeo is 4m away from the window.
24. Compute the velocity of an earth satellite whose distance is 4.5 × 107 m. Take mass of earth to be approximately
5.98 × 1024 𝑘𝑔𝑚.
Ans. 10717.9 kph
25. A plane dropped a bomb at an elevation 1000 meters from the ground intended to hit a target which elevation
is 200 m from the ground. If the plane was flying at a velocity of 300 kph, at what distance from the target must the
bomb be dropped to hit the target? Wind velocity and atmospheric pressure to be disregarded.
Ans. 1064.2 m
65
26. It is known that the gravitational acceleration in the moon is 1/6 th of the earth’s gravitational acceleration. If 1
gram of stone is throne straight upward in the earth, determine
a. The mass of the stone when it is thrown straight upward to the moon
b. How high should be able to throw the stone straight upward on the moon? Assume that the throwing
speeds are the same.
*27. What is the maximum angle with respect to the horizontal that can a 0.25in inside diameter water jet can be
fired and filled a 1.4384 gal container for one minute? The water jet is 2.49 m above the ground and shoots the
container 10m above the ground and 10m away from it. Assume motion of water jet is curvilinear.
66
28. The mass of air in a room 3m x 5m x 20m is 350 kgm. Find the density of air in kg/m 3.
29. 100 grams of water is mixed to 150 grams of alcohol (p = 790 kgm/m 3). Calculate the specific gravity of the
mixture and specific volume of the mixture in m 3/kgm.
30. A 40 kN tank weights 64 kN with kerosene and 70 kN when filled with the same volume of water. Find the
specific gravity of kerosene.
31. A 45 cm cube of ice melts, the liquid water is transferred in a spherical container until it becomes full. If the
specific volume of ice at 0 C is 1.094 cm3/gm and of water is 1.002 cm3/g , determine the diameter of the spherical
vessel to be used assuming no loss of mass due to evaporation.
67
32. Two liquids of different densities 1200 lbm/ft3 and 760 lb/ft3 are poured together into a 1gallon tank, filling it.
If the resulting density of the mixture is 900 lbm/ft3 , find the respective quantities of liquids used in lbm and weight
of the mixture in lbf at local g= 32.0 fps2
33. It is estimated that the earth has a mean radius of 6.37 x 106 m and the air comprising the atmosphere above it
is 8.05x104 m high. Atmospheric air is generally composed of Nitrogen, Oxygen, and Argon with volumetric
proportions of 78,21, and 1 percent and densities of 1.165,1.33, and 1.783 in kgm/m 3 respectively. Determine the
mass in kgm and density in kgm/m3 of the atmospheric air.
68
34. A cylindrical drum 3ft high and 2ft diameter is filled with a fluid whose density is 40 pounds per cubic foot.
Local g is 31.90 foot per square second. Determine
a. The volume occupied by the fluid in cubic foot.
b. The total mass of the fluid in pounds and slugs.
c. The specific volume of cubic foot per pound.
d. The specific weight in pound force per cubic foot
35. 0.690 g/cc liquid flows through a 5 cm inside diameter pipe at 8.3 m/s. Determine the mass flow rate in kg/s.
36. The specific gravity of mercury relative to water is 13.55. What is the specific weight of mercury? The specific
weight of water is 62.4 lb per cubic foot.
69
37. A batch of concrete consisted of 200 lbs fine aggregate, 350 lbs coarse aggregate, 94 lbs, cement, and 5
gallons water. The specific gravity of the sand and gravel may be taken as 2.65 and that of the cement as 3.10.
a. What is the weight of the concrete in place per cubic foot?
b. How much by the weight of the cement is required to produce one cubic yard?
38. A spherical tank is full of water that has a total mass of 10,000 kg. If the outside diameter of the tank is 2722
mm, how thick is the wall of the tank in mm?
39. What is the specific gravity of a solid sphere that has a radius of 15 mm and a mass of 0.038 kgm?
70
*40. A hose shoots water straight up a distance of 2.5m. The end opening on the hose has an area of 0.75 cm 2. How
much water comes out in liters per minute?
41. Two pistons A and B of the hydraulic jack are on the same level. Piston A is 100 cm 2 while piston B is 500cm2.
Piston A carries a 500 kg load. Find the required force F on tonne force at piston B to carry the load.
42. A 2kg piston with 0.5 m diameter is acted by a pressurized gas. What should be the pressure of the gas in kPaa
that would move the piston upward?
43. A condenser vacuum gage reads 715 mmHg. When the barometer stands at 757 mmHg, state the absolute
pressure in the condenser in bars and kN/m2.
71
44. A closed piston cylinder set up contains liquid and water vapor at 100°𝐶. The 0.10m diameter piston, exposed
to the atmosphere is externally loaded with a force of 20kN, what pressure of water is needed to push the piston in
kPaa?
45. Given the barometric pressure of 14.7 psia, make these conversions:
a. 80 psig to psia and to atm.
b. 20 inHg vacuum to in.Hg. Absolute and to psia.
c. 10 psia to psi vac and to Pa.
d. in.Hg gage to psia, to torrs, and to Pa.
72
46. Due to unavailability of mercury, a barometer was constructed using water in the closed column with a certain
amount of oil on top to prevent the boiling of water. For standard atmospheric pressure and temperature, what height
of oil will be required in ft and how high will the column of water will be in ft if the minimum pressure at the water-
oil interface is to be 1 psia? Specific gravity of oil is 0.8.
47. A mercury U-Tube manometer is attached to the side of a nearly full water tank. The reading of the mercury
column is 15.5 inHg gage. Although air on the water maintains the pressure, the other 10 in of the manometer is
full of water. For water, p = 62.3 lbm/ft3, for mercury p = 846 lbm/ft3. If the location is standard atmospheric
pressure and temperature, find:
a. The pressure of the tank where the manometer is attached in psia
b. The pressure 10 ft below the level of the tank in psia.
73
48. A 30 m vertical column fluid has a density of 1878 kgm/m3 is located where g= 9.65 mps2. Find the pressure at
the base of the column in kPaa.
49. A weatherman carried an aneroid barometer from the ground floor to his office atop the Sears tower in Chicago.
On the ground level, the barometer reads 30.165 inHg abs; topside it reads 29.609 in Hg abs. Assume that the
average atmospheric air density is 0.07543 lb/ft3, estimate the height of the building in ft.
74
50. Given two compartment vessels shown in the figure. Gage A reads 85 psig, Gage B inside compartment X
reads 25 psig. If the barometer reads 30.61 inHg,
a. Determine the reading of gage C in psia.

b. Determine the reading of gage C in psia if gage A is 15 in Hg vac and gage B is 5.4 psig.
c. If gage C reads 350 psig and B reads 125 psig, what will gage A read in psig?
d. If A reads 0 psig, what is the maximum reading that gage B can have? Reading of Gage C is 350 psig.
e. If gage A and C reads the same reading, what would be its reading if gage B is 0 psig?
f. If gage A is twice gage C, what would be its reading if gage B reads 10 psi vac?
75
*51. The composite piston has a total mass m in lbm and it is supported by pressure at C by a flowing methane gas.
a. If pressure at A is 100 psia and pressure at B is 20 psia with m = 40 lb m, find pressure at C.
b. If pressure at A is 40 psia, pressure at B is 18 psia, and pressure at C is 25 psia, find the mass of the
composite piston.
76
52. A vertical composite fluid column whose upper end is open to the atmosphere is composed of 18 in mercury
(SG=13.45), 26 in of water whose density is 62 lbm/ft3 and 32 in of oil (SG =0.825). Determine the pressure
a. At the base of the column
b. At the oil-water interface
c. At the water mercury interface
77
53. A simple mercury manometer connected into a flow line gives readings as shown in the figure. Local gravity is
standard and the mercury density is 0.48 lb/in3. Find the pressure at points X and Y when the flow line and the left
leg contains
(a) air whose density is 0.072 lb/ft3
(b) water whose density is 62.1 lb/ft3
(c) If the local gravity is 30 fps2, what are the pressure at (a) and (b)?
78
54. A manometer containing water (density = 62.1 lb/ft3) and mercury (specific gravity = 13.55) connects two
pressure regions A and B as shown in the figure. The local gravity acceleration is 𝑔 = 32.00𝑓𝑝𝑠2 .
a) If 𝑃𝐵 = 50𝑝𝑠𝑖𝑔, find 𝑃𝐴
b) Same as (a) , but in lieu of the mercury, the fluid is a special compound with specific gravity of 2.00
79
55. A solid frictionless piston P whose mass is 80 lb is pulled up the inside of a 6-in vertical pipe that has its lower
end in a pool of water ( density = 62.4 lb/ft3) and its upper end open to the 15 psia atmosphere. The water rises in
contact with the piston to a height of 20 ft above the surface of the pool. If the local gravity acceleration is g = 31
fps2, find
(a) the pull F on the piston required at the 20 ft height.
(b) the pressure exerted by the water on the piston at this point
(c) If the vapor pressure of the water is 0.5 psia (point at which water boils and produces vapor), to what
maximum height may the piston be raised and still maintain contact with the liquid water (no evaporation)?
(d) From (c), what will the pull on the piston at this point?
80
56. A rectangular gate 3m high and 1.5 m wide is submerged 2 m deep with its width along the water surface.
Calculate the pressure acting on the gate and the location of center of pressure.
57. The square gate ABCD having edge of 60 cm. Side AB is 1.7 m below the free water surface. Determine the
difference between the hydrostatic forces of subpanels ABC and BCD.
81
*58. A triangular gate is submerged half in oil (SG=0.80) and half in water such that its top edge is flashed with oil
surface. What is the ratio of the force exerted by the water acting on the lower half to that oil acting on the upper
half?
82
59. 3 A vertical circular gate in a tunnel 8 m in diameter has oil (SG=0.8) on one side and air on the other side. If
oil is 12 m above the invert and the air pressure is 40 kPa, where will be a single support be located above the
inverted of the tunnel to hold the gate in position?
Oil surface
Hinge
12m Pg = 40kPa
8m
83
60. Consider a 200ft high, 1200 ft wide dam filled with capacity. Determine (a) the hydrostatic forces on the dam,
and (b) the force per unit area of the dam near the top and near the bottom.
Ans. a) 𝟏. 𝟒𝟗𝟖 × 𝟏𝟎𝟔 𝒍𝒃𝒇 b) 0 psi, 86.670 psi
61. Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1m
high and 0.9m wide, and the top edge of the door is 8m below the water surface. Determine the net force acting on
the door and the location of the pressure center if (a) The car is well sealed and contains air at atmospheric, (b) the
car is filled with water. Ans. a) 83.037 kN, 8.562 m, b) 0 kN
62. A room in a lower level of a cruise ship has a 30 cm diameter circular window. If the midpoint of the window
is 5m below the water surface, determine the hydrostatic force acting on the window, and pressure center. Take
specific gravity of seawater to be 1.025. Ans. 3.467 kN, 5.001 m
84
63. A circular plate is vertically submerged 3ft below the free surface of water. If the eccentricity is 0.225 ft,
determine the radius of the plate in inches. Ans. 25.844 in
64. A triangular plate has a gage pressure of 200 kPa along its centroid when it is submerged fully at a depth with
its base parallel to the water surface. For an eccentricity of 0.25m, determine the height of the triangle.
Ans. 9.578 m
*65. 5. A square plate 2m is immersed vertically into water. How must a dividing line be drawn parallel to the
surface so as to divide the plate into two areas, the total forces upon which shall be equal?
Ans. 1.414 m
85
66. Convert the following:

(a) 122°𝐹 to °𝐶 and °𝐹,
(b) −25°𝐶 to °𝐹 and °𝑅,
(c) 942°𝑅 to °𝐶 and to 𝐾, and
(d) 373 𝐾 to °𝐹 and to °𝑅
*67. (a) If two thermometers, one reading °𝐶 an the other 𝐾, are inserted in the same system, under what
circumstance will they bot have the same numerical reading? (b) What will be the system’s temperature when the
absolute thermometer reads twice the numerical reading of the Celsius thermometer?
86
68. (a) Define a new temperature scale, say °𝑁, in which the boiling and freezing points of water are 1000°𝑁 and
100°𝑁, respectively, and correlate this scale with the Fahrenheit and Celsius scales. (b) The °𝑁 reading on this
scale is a certain number of degrees on a corresponding absolute temperature scale. What is this absolute
temperature at 0°𝑁?
69. What is the change in temperature in °𝐹 for a water temperature rise of 15°𝐶?
*70. For a particular thermocouple, if one junction is maintained at 0°𝐶 (cold junction) and the other junction is
used to probe to measure the desired Celsius temperature 𝑡, the voltage 𝐸 generated in the circuit is related to the
temperature 𝑡 as 𝐸 = 𝑡(𝑎 + 𝑏𝑡). Further, for this thermocouple, when 𝐸 = in in millivolts mV, the two constants
are 𝑎 = 0.25 and 𝑏 = −5.5 × 10−4
(a) What are the units of constants 𝑎 and 𝑏?
(b) What is the value of 𝐸 for a temperature of 100°𝐶?
(c) Find the rate of change of 𝐸 per °𝐶 for a temperature of 100°𝐶.
87
MODULE II. ENERGY AND POWER CONCEPTS THERMODYNAMICS I
II. ENERGY AND POWER CONCEPTS

Learning Objectives: At the end of this chapter, the mechanical engineering students shall:
1. Apply conservation of mass to open and closed systems.
2. Define work, energy, and power.
3. Convert different systems of units of energy and power.
4. Compute the energy associated when mass travels at a speed of light.
5. Enumerate the general forms of energy.
6. Explain each type of energy.
7. Define heat, specific heat, and types of heat.
8. Analyze the other types of work and energy.
9. Differentiate work non-flow from work-steady flow.
10. Apply conservation of energy to open and closed systems.
11. Apply conservation of energy to steady flow steady state engineering devices.
12. Apply conservation of energy to a simple steam power plant
2.1 CONSERVATION OF MASS
Conservation of mass states that the change in mass in a system is equivalent to the difference of mass flowing
out to the system to the mass flowing into the system.
(a) For closed system, the change in mass is constant, wherein it is the difference between the mass leaving
and the mass leaving the system.
𝚫𝒎̇ = 𝒎𝒐𝒖𝒕̇ − 𝐦𝐢𝐧 ̇
(b) For open systems, the change in mass is zero, wherein there is no change in mass through the process. The
mass entering the system is also equal to the mass leaving the system. This is also called as the mass balance for
open systems.
𝚫𝒎̇ = 𝟎
𝒎𝒊𝒏̇ = 𝒎𝒐𝒖𝒕
̇
Consider the steady flow system below. At section 1, mass is entering the system and at section 2 mass is
leaving the system. We can also relate the mass flow rate of the system in terms of the inlet and outlets cross
sectional area, velocity of the fluid, and density/specific volume of the fluid. By mass balance on the system:
𝒎𝒊𝒏̇ = 𝒎𝒐𝒖𝒕 ̇ 𝚫𝒎̇ = 𝟎

𝑚̇ 1 = 𝑚̇ 2
1 2
𝜌1 𝑉1̇ = 𝜌2 𝑉2̇
𝛒𝟏 𝑨𝟏 𝒗𝟏 = 𝛒𝟐 𝑨𝟐 𝒗𝟐
𝑨𝟏 𝐯𝟏 𝑨𝟐 𝐯𝟐
=
𝛎𝟏 𝛎𝟐
Where 𝐴 = cross sectional area
v = constant velocity
ν = specific volume
The mass balance for open system is also called as the continuity equation for fluids.
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2.2 WORK AND ENERGY

Energy is defined as the capacity to do work. The energy is a scalar quantity, and the energy of a system of
bodies is the algebraic sum of the magnitudes of various forms of energy.
Work is a form of energy, is also the product of displacement of a body and the component of the unbalance
constant force in the direction of the displacement. A body never contains work, it has the capacity to do work.
Consider a body of mass m acted by an unbalanced constant force F.
If an infinitesimal displacement ds is acted by an unbalanced constant force F, the work W would be
𝑑𝑊 = 𝐹𝑑𝑠
Summing all ds, by integration

𝑤 𝑠
∫ 𝑑𝑊 = 𝐹 ∫ 𝑑𝑠
0 0
𝑾 = 𝑭𝒔
The differential equation 𝒅𝑾 = 𝑭𝒅𝒔 is called as the general equation of work, and the derivation is still the same
in deriving some other forms of work.
2.3 POWER
Power is defined as the rate of change of work per unit time. In differential symbols
𝒅𝑾
𝑷=
𝒅𝒕
The average power or the power is defined as:
𝑾
𝑷 = 𝑾̇ =
𝒕
Since 𝑊 = 𝐹𝑠, hence
𝐹𝑠
𝑃=
𝑡
𝑷 = 𝑭𝐯
The power is also equal to the product of the unbalance constant force to the velocity of the body. From our
equation of power, we can see that the work done is also equivalent to the product of power and time.
𝑾 = 𝑷𝒕
Power also can be defined as the product of the energy E to the ratio of the mass flow rate to the mass of the
system. It can be also defined as the product of energy E to the ratio of volume flow rate to the volume of the system.
𝑚̇
𝑃 =𝐸( )
𝑚
𝑉̇
𝑃 =𝐸( )
𝑉
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2.4 UNIT CONVERSIONS OF ENERGY AND POWER
It is important for mechanical engineering students to memorize all the unit conversions of energy and power
since we are dealing with thermodynamic systems. Units of power can be also in terms of the product of units of
energy to the units of time. The following are the base units of energy and power in different systems of units.
(Note: Memorize all these conversions!)
FPS CGS KGS/metric SI
Energy BTU Erg kgf-m J, kJ
Power hp metric hp metric hp W, kW
CONVERSION FACTORS FOR ENERGY

1 BTU = 778.16 ft-lbf 1 ev = 1.6025 × 10−12 erg
= 1055 J = 1.6025 × 10−12 dyne -cm
= 1.055 kJ = 1.6025 × 10−22 kJ
= 252 cal
1 CHU = 1.8 BTU 1J = 1 N-m
1 Th = 100 000 BTU = 1 W-s
1 erg = 1 dyne-cm = 1 V-C
= 10-10 kJ = 107 erg
1 Cal = 1 kcal 1 kJ = 1 kW-s
=1000 cal 1 kW-hr = 3412.32 BTU
1 cal = 4.187 J = 3600 kJ
1 kcal = 4.187 kJ
1 kgf-m = 9.8066 J
Where
BTU = British Thermal Units Cal = Calorie
CHU = Calorie Heat Unit cal = calorie
Th =Therm/ Thermal Heat Unit ev = electrovolts
J = Joules W = watts
V = volts C = coloumbs
CONVERSION FACTORS FOR POWER
1W = 1 VA 1 english hp = 1 hp 1 metric hp = 1 Mhp

= 1 VC/s = 746 W = 1 french horse
= 1 J/s = 0.746 kW = 1 pfer-starky (P.S)
= 1 A2Ω = 550 ft-lbf/s =1 cheuval vapeur (CV)
= 1 A2/mho = 33000 ft-lbf/min = 1 power of a horse
= 42.41 BTU/min = 736 W
= 1 mule = 0.736 kW
= 1 Arabian mule = 75 kgf-m/s
= 4500 kgf-m/min
Other non-standard units of power
1 Boiler Mhp = 33480 BTU/hr
= 35322 kJ/hr
4
1ponselet(p) = 𝑀ℎ𝑝
3
= 1.315 hp
Where
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V = volts hp = horsepower Ω = ohms
A = ampere P.S = pfer-starky p = ponselet
J = Joules C.V = cheuva vapeur Mhp = metric horsepower
Notable definitions of units of energy and power are as follows:
1. British Thermal Unit (BTU) - it is the amount of energy required to raise a temperature of a unit pound force
by a unit degree fahrenheit.
2. Calorie Heat Unit (CHU) - it is the thermochemical energy unit alternative for calorie
3. calorie (cal) - it is the amount of energy required to raise the temperature of unit gram mass at unit degree
celsius.
4. Calorie (Cal) - it is the amount of energy required to rause the temperature of unit kilogram mass by a unit
degree celsius.
5. Joule (J) - the amount of energy done by a newton force displaced by a unit meter, named after James Joules.
It is the SI unit of energy.
6. Erg (erg) - it is the amount of energy done by a dyne-force that is displaced by a unit centimeter.
7. Thermal heat (therm/th) - it is the approximate energy equivalent of burning 100 cubic foot of natural gas.
8. English horsepower (hp) -also called as the mechanical horsepower rating. It is also the rating of performance
of certain machines. It is invented by James Watt by using a mule to measure its capacity in carrying a load of 550
pounds of coal covering a unit foot in unit second.
9. Metric horsepower (Mhp) -also called as the electrical horsepower rating, pferstarky (P.S) and cheuva vapeur
(CV). Instead of using mule, french horse was used and it was loaded by 75 kilograms of coal, covering a unit
meter in unit second.
10. Watt (W) - the SI unit of power, named after James Watt. It is the standard electrical unit of power.
2.5 CONVERSION OF MASS INTO ENERGY
A mass m can be converted into energy E if it will travel at a speed of light c. Energy at this point is at
maximum when this mass is fully consumed. Albert Einstein’s famous equation for theory of relativity applies the
conversion of mass into energy.
𝒎𝒄𝟐
𝑬=
𝒌
where E is the energy associated when the mass is converted by means of burning without energy loss.
c is the speed of light, where 𝒄 = 𝟑 × 𝟏𝟎𝟖 𝒎/𝒔
k is a proportionality constant of the equation, for consistency of unit conversions.
If this mass also called as rest mass or the initial mass 𝑚0 moves at a certain velocity v lesser than the speed of
light, then the new mass m of the body as it moves at this speed is defined as:
𝒎𝟎
𝒎=
𝟐
√𝟏 − 𝒗𝟐
𝒄
This concept of mass- energy conversion is applicable in determining the mass of nuclear fuel to be used in nuclear
power plants. The mass of nuclear fuel is defined as the source of heat in the plant in for the nuclear power plant to
produce electricity. Hence
𝒎𝑵𝑭 𝒄𝟐
𝑸=
𝒌
where Q is the heat or energy that is needed to run the nuclear powerplant
𝑚𝑁𝐹 is the mass of nuclear fuel
c is the speed of light
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2.6 GENERAL FORMS OF ENERGY

Energy is subdivided into two general forms:
A. Stored Energy - energy that is within the property of the system and cannot be transferred from another
system. Stored energy are point functions, and associated with the change of energy. It is also dependent on the
mass of the system. The following are kinds of stored energy.
⚫ Potential Energy / Gravitational Potential Energy (PE)
⚫ Kinetic Energy/ Mechanical Kinetic Energy (KE)
⚫ Flow Energy/ Flow Work (𝑊𝑓 )
⚫ Internal Energy (U)
⚫ Enthalpy (H)
B. Transient Energy - it is the energy that can be transferred from one system to another. Transient energy are
path functions, and does not associates with the change of energy. There are two kinds of transient energy.
⚫ Mechanical Work (W)
◆ Work Steady Flow (WSF)
◆ Work Non-Flow (WNF)
⚫ Heat (Q)
Energy can be defined in terms of
(a) Energy units ( Ex.kJ , BTU)
(b) Specific energy or energy per unit mass, denoted as small letters ( Ex: kJ/kg, BTU/lbm)
(c) Energy per unit time or in units of power, denoted as capital letters with dot on top of the letter.
(Ex: kW, hp)
2.7 TYPES OF STORED ENERGY
In this section, we will fully understand each kind of stored energy.
1. Potential Energy - it is also called as gravitational potential energy. It is the energy of a body due to its position
or its elevation at a specified datum line. From the figure below, if a mass m is hold at a certain height z, then that
mass stores potential energy.
In terms of energy units:
𝒎𝒈𝒛
𝑷𝑬 =
𝒌
In terms of energy per unit mass basis:
𝑔𝑧
𝑝𝑒 =
𝑘
In terms of energy per unit time basis
𝑚̇𝑔𝑧
𝑃𝐸̇ =
𝑘
The change of potential energy is defined as
Δ𝑃𝐸 = 𝑃𝐸2 − 𝑃𝐸1
𝒎𝒈(𝒛𝟐 − 𝒛𝟏 )
𝚫𝑷𝑬 =
𝒌
2. Kinetic Energy - also called as mechanical kinetic energy. It is the energy possessed by a moving body by
virtue of its momentum. From the figure below, if a mass m that moves at a velocity v them that mass stores
kinetic energy. v
𝒎𝒗𝟐
𝑲𝑬 =
𝟐𝒌
𝑣2
𝑘𝑒 =
2𝑘
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𝑚̇ 𝑣 2
̇ =
𝐾𝐸
2𝑘
The change of kinetic energy is defined as
Δ𝐾𝐸 = 𝐾𝐸2 − 𝐾𝐸1
𝒎(𝒗𝟐 𝟐 − 𝒗𝟏 𝟐 )
𝚫𝑲𝑬 =
𝟐𝒌
3. Flow Energy - is also called as the flow work or flow resistance energy. It is a stored energy of a body to resists
another body to get into the body of another system. Work must be done to move the body against the resistance
caused by another body. From the given figure, we will derive the equation of flow energy.
From our definition of work

𝑊 = 𝐹𝑠 1
Since 𝑠 = 𝐿; then 2
𝑊 = 𝐹𝐿 1
2
But V = AL , F = PA , hence in terms of energy units:
𝑾𝒇 = 𝑷𝑽
𝑤𝑓 = 𝑃ν
𝑊̇𝑓 = 𝑃𝑉̇
The change of flow energy is
Δ𝑊𝑓 = 𝑊𝑓2 − 𝑊𝑓1
𝚫𝑾𝒇 = 𝑷𝟐 𝑽𝟐 − 𝑷𝟏 𝑽𝟏
4. Internal Energy - also called as inner work, internal work, and intrinsic energy. It is the energy stored within a
body or a substance by virtue of the ability and configuration of its molecules and vibration of the atoms withing
the molecules. The term internal energy first appeared with its symbol U in the works of Rudolf Clausius
𝑼 = 𝒎𝒖
𝑈
𝑢=
𝑚
𝑈̇ = 𝑚̇𝑢
The change of internal energy in terms of energy units
Δ𝑈 = 𝑈2 − 𝑈1
𝚫𝑼 = 𝒎(𝒖𝟐 − 𝒖𝟏 )
The change of internal energy in terms of energy per unit mass basis:
𝚫𝒖 = 𝒖𝟐 − 𝒖𝟏
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The internal energy of the system is a stored energy, which is the sum of all
the microscopic forms of energy in the system. It is related to the molecular
structure, and degree of molecular activity, and can also view as the sum of all the
kinetic and potential energies of the molecules. The following are the portions of
internal energy in detail:
a) Translational Kinetic Energy - energy associated with the translational motion
of molecules.
b) Rotational Kinetic Energy - energy associated with the rotational motion of
molecules.
c) Vibrational Kinetic Energy - energy associated with the vibration of atoms
within molecules
d) Internal Potential Energy - is the energy due to relative position of its
molecules and attraction of molecules to one another.
e) Sensible Energy - the portion of the internal energy associated with the kinetic
energies of molecules.
f) Latent Energy - the internal energy associated with the phase of the system.
g) Chemical Energy -the internal energy associated with the atomic bonds in the
molecule.
h) Nuclear Energy - it is the tremendous amount of energy associated with strong
bonds within the nucleus of the atom itself.
5. Enthalpy - is a stored energy sometimes called as the combination energy. It is

defined as the sum of the internal energy and flow energy. It is the composite property
to all fluids relating flow work and internal energy.
𝐻 = 𝑚ℎ
𝑯 = 𝑷𝑽 + 𝑼
ℎ = 𝑃ν + 𝑢
𝐻̇ = 𝑃𝑉̇ + 𝑈̇
The change of enthalpy in terms of energy units
Δ𝐻 = 𝐻2 − 𝐻1
Δ𝐻 = 𝑚(ℎ2 − ℎ1 )
Δ𝐻 = (𝑃2 𝑉2 + 𝑈2 ) − (𝑃1 𝑉1 + 𝑈1 )
𝚫𝑯 = 𝚫𝑷𝑽 + 𝚫𝑼
The change of enthalpy in terms of energy per unit mass basis:
Δℎ = ℎ2 − ℎ1
Δℎ = (𝑃2 𝑣2 + 𝑢2 ) − (𝑃1 𝑣1 + 𝑢1 )
Δℎ = Δ𝑃𝑣 + Δ𝑢
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2.8 HEAT
Heat (Q) is defined as the energy in transition from one system to another because of differences of
temperature. Heat flows always from higher temperature system to lower temperature system. If two bodies have
the same temperature, then thermal equilibrium is reach.
Sign Conventions of Heat
i. Q is positive if heat is absorbed or received by the system.
ii. Q is negative if heat is released by the system.
The general formula of heat in terms of mass, specific heat c, and

temperature for any state at energy basis is:
𝑸 = 𝒎𝒄𝚫𝑻 Units For Specific Heat

𝑸 = 𝒎𝒄𝚫𝒕 FPS MKS SI
𝐵𝑇𝑈 𝑘𝑐𝑎𝑙 𝑘𝐽
If heat is in terms of energy per unit mass basis:
𝑙𝑏𝑚 − °𝐹 𝑘𝑔𝑚 − 𝐾 𝑘𝑔𝑚 − °𝐶
𝑄
𝑞= 𝑘𝑐𝑎𝑙 𝑘𝐽
𝑚 𝐵𝑇𝑈
𝑞 = 𝑐Δ𝑇 𝑘𝑔𝑚 − °𝐶 𝑘𝑔𝑚 − 𝐾
𝑙𝑏𝑚 − °𝑅 𝑐𝑎𝑙
𝑞 = 𝑐Δ𝑡
In terms of energy per unit time basis 𝑔𝑚 − °𝐶
𝑐𝑎𝑙
𝑄̇ = 𝑚̇𝑐Δ𝑇
𝑄̇ = 𝑚̇𝑐Δ𝑡 𝑔𝑚 − 𝐾
2.9 SPECIFIC HEAT
Another property that relates heat to another state properties is the specific heat. The specific heat c is defined
as the ratio of the change in energy in form of heat to the change in temperature of a given fluid for a particular
process. The specific heat c in energy basis is defined as
𝑸
𝒄=
𝒎𝚫𝑻
𝑸
𝒄=
𝒎𝚫𝒕
and the specific heat at unit mass basis is
𝑞
𝑐=
Δ𝑇
𝑞
𝑐=
Δ𝑡
In differential form
𝒅𝒒
𝒄=
𝒅𝒕
There are two kinds of specific heat
1. Specific Heat at Constant Volume Process (Isometric Process) - the specific heat on a constant volume process
is defined as the change in internal energy per degree change in temperature.
a) In macroanalysis, we defined heat as the change in internal energy for constant volume process.
𝑄 = Δ𝑈
𝑸 = 𝒎𝒄𝒗 𝚫𝑻
𝑸 = 𝒎𝒄𝒗 𝚫𝒕
Where 𝑐𝑣 is the specific heat at constant volume process
95
.
b) In microanalysis, we defined the specific heat at constant volume in differential
elements
𝑑𝑄 = 𝑚𝑐𝑣 𝑑𝑇
𝑑𝑞 = 𝑐𝑣 𝑑𝑇
Where 𝑐𝑣 must be a function of temperature( 𝑐𝑣 (𝑇)). Hence the heat would be
𝑸 𝟐
∫ 𝒅𝑸 = 𝒎 ∫ 𝒄𝒗 (𝑻)𝒅𝑻
𝟎 𝟏
2. Specific Heat at Constant Pressure Process (Isobaric Process) - the specific heat on a constant pressure process
is defined as the change in enthalpy per degree change in temperature.
a) In macroanalysis, we defined heat as the change in enthalpy for constant pressure

process.
𝑄 = Δ𝐻
𝑸 = 𝒎𝒄𝒑 𝚫𝑻
𝑸 = 𝒎𝒄𝒑 𝚫𝒕
Where 𝑐𝑝 is the specific heat at constant pressure process
b) In microanalysis, we define the specific heat at constant pressure in differential elements.

𝑑𝑄 = 𝑚𝑐𝑝 𝑑𝑇
𝑑𝑞 = 𝑐𝑝 𝑑𝑇
3. Where 𝑐𝑝 must be a function of temperature( 𝑐𝑝 (𝑇)). Hence the heat would be
𝑸 𝟐
∫ 𝒅𝑸 = 𝒎 ∫ 𝒄𝒑 (𝑻)𝒅𝑻
𝟎 𝟏
2.10 TYPES OF HEAT

When dealing with heat, there are general two types of heating:
1. Sensible Heat - it is the heat that cause a change in temperature of a system without changing the phase of the
system. Sensible heat is defined as:
𝑄𝑠 = 𝑚𝑐Δ𝑡
𝑄𝑠 = 𝑚𝑐Δ𝑇
If two or more bodies came in contact with different temperatures, then heat will flow from higher temperature
body to lower temperature body. At such time, these bodies in contact will come in thermal equilibrium,as validated
by the zeroth law of thermodynamics. Assuming that there is no change in the phase of the bodies being in contact
and no heat is loss within the system, a “heat balance” is determined:
∑𝑄 = 0
𝑛
∑ 𝑄𝑖 = 0
𝑖=1
𝑛
∑ 𝑚𝑖 𝑐𝑖 (𝑇 − 𝑇𝑖 ) = 0
𝑖=1
where T represents the thermal equilibrium temperature.
2. Latent Heat - it is the amount of heat needed to change the phase of the system without changing its temperature.
The general formula for latent heat of any system is
𝑄𝐿 = 𝑚𝐿
where L is the latent heat of the system, either latent heat of fusion, or latent heat of vaporization.
96
There are basically two types of latent heat.

a) Latent Heat of Fusion (ℎ𝑓 ) - also called as the heat of fusion, is the amount of energy or the amount of
enthalpy that must be added for a system to change phase between a solid and a liquid without changing
its temperature It is also the amount of energy absorbed during melting and also equivalent to the energy
released during freezing
b) Latent Heat of Vaporization (ℎ𝑣 /ℎ𝑓𝑔 )- also called as the heat of vaporization, is the amount of energy or
the amount of enthalpy to be added on a system to change phase from liquid into gas. It is also the energy
absorbed during vaporization and also equivalent to the energy released during condensation.
The following are useful constant of specific heats and latent heat of water. (All of these constants must be memorize
by mechanical engineering students!). Assuming that the following constant pressures are measured at standard
pressure and temperature (STP)
I. Specific Heats of Water (H2O) at Constant Pressure

English/FPS Metric / MKS SI
𝑩𝑻𝑼 𝒌𝒄𝒂𝒍 𝒌𝑱
𝒄𝒑𝒘 = 𝟏 𝒄𝒑𝒘 = 𝟏 𝒄𝒑𝒘 = 𝟒. 𝟏𝟖𝟕
𝒍𝒃𝒎 − °𝑭 𝒌𝒈𝒎 − °𝑪 𝒌𝒈𝒎 − °𝑪
𝒄𝒑𝒘 = 𝟏 𝒄𝒑𝒘 = 𝟏 𝒄𝒑𝒘 = 𝟒. 𝟏𝟖𝟕
𝒍𝒃𝒎 − °𝑹 𝒌𝒈𝒎 − 𝑲 𝒌𝒈𝒎 − 𝑲
𝒄𝒂𝒍 𝑱
𝒄𝒑𝒘 = 𝟏 𝒄𝒑𝒘 = 𝟒. 𝟏𝟖𝟕
𝒈𝒎 − °𝑪 𝒈𝒎 − °𝑪
𝒄𝒂𝒍 𝑱
𝒄𝒑𝒘 = 𝟏 𝒄𝒑𝒘 = 𝟒. 𝟏𝟖𝟕
𝒈𝒎 − 𝑲 𝒈𝒎 − 𝑲
Other specific heats

b) Specific Heat of Ice at Constant Pressure
𝒄𝒑𝒊 = 𝟎. 𝟓𝒄𝒑𝒘
c) Specific Heat of Steam at Constant Pressure
𝒄𝒑𝒔 = 𝟎. 𝟒𝟒𝟓𝟒𝒄𝒑𝒘
II. Latent Heat
a) Latent Heat of Fusion (Ice)
𝒉𝒇 = 𝟏𝟒𝟒 𝒉𝒇 = 𝟖𝟎 𝒉𝒇 = 𝟑𝟑𝟓
b) Laten Heat of Vaporization (Steam)

𝒉𝒇𝒈 = 𝟗𝟕𝟎. 𝟑 𝒉𝒇𝒈 = 𝟓𝟒𝟎 𝒉𝒇𝒈 = 𝟐𝟐𝟓𝟕
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2.11 MECHANICAL WORK

Mechanical Work is a transient energy developed due to shaft rotation or to piston movement. There are
basically two types of mechanical work.
Sign Conventions for Work

i. Positive (+) if the work is done by the system.
ii. Negative (-) if the work is done to the system or on the system
1. Work Non-Flow - also called as the non-flow work. It is the mechanical work developed in closed systems and
non-flow systems. It is also called as the moving boundary work since there is no transfer of mass in a closed
system, the mass remains the same but the energy changes. Most common examples is the movement of pistons in
piston cylinder arrangement. The non-flow work is computed as the area under the PV diagram.
𝟐
𝑾𝑵𝑭 = ∫ 𝑷 𝒅𝑽
𝟏
2. Work Steady-Flow - also called as the steady flow work. It is the mechanical work developed in an open
system. It is also the work done due to shaft rotation in some steady flow engineering devices to produce energy.
Most common example is a turbine. The steady flow work is computed as the area at the “back” of the PV curve,
assuming that ΔKE=0 and t ΔPE=0.
𝟐
𝑾𝑺𝑭 = ∫ 𝑽𝒅𝑷
𝟏
Note: In some textbooks and other

reference books, the steady flow work is
indicated by a negative sign, means that
the work is done by a working substance
to the system. Neglect the negative sign
for easier calculations and let point 2 be
any value (higher limit) that is higher
than the value in point 1 (lower limit) in
the integral.
98
2.12 OTHER FORMS OF WORK/ENERGY

Aside from the general forms of work, there are other types of work applicable to some engineering
thermodynamic problems.
1. Shaft Work - it is the energy transmission by a rotating shaft. It is proportional to the torque applied and the
number of revolutions of the shaft.
𝑊𝑠ℎ̇ = 𝐹𝑣
Since 𝑇 = 𝐹𝑟, and 𝑣 = 2π𝑟𝑁 , then
𝑾𝒔𝒉̇ = 𝟐𝛑𝑻𝛚
where T is the torque applied to the rotating shaft

N is the number of revolutions per unit time.
ω is the angular velocity or angular speed.
Since shaft work are in terms of energy per unit time basis (power), and it is dependent to the torque, we can
use the formula below for easy conversion of shaft work to power,
a) For SI units b) For English units
𝑻𝑵 𝑻𝑵
̇ =
𝑾𝒔𝒉 , 𝒌𝑾 𝑾𝒔𝒉 ̇ = , 𝒉𝒑
𝟗. 𝟓𝟒𝟗 × 𝟏𝟎 𝟔 𝟗𝟔𝟑𝟐𝟎𝟓
Where Where
T is in kN-m T is in lb-in
N is in revolutions per minute (rpm) N in revolutions per minute (rpm)
2. Spring Work - it is also called as elastic potential energy. It is the energy

required for a force F acting on an elastic system, such as springs to either
compressed or stretched the system. Hooke’s Law states that the force acting
on the spring is directly proportional to its free length. In symbols
𝐹∝𝑥
𝑭 = 𝒌𝒙
where
x is called as its free length
k is called as the spring index, spring constant, or stiffness of spring
having units of force per unit length.
If work is done on the spring, then the elastic potential energy is defined as the differential element of work done
per differential element of displacement of the spring. Thus:
𝑑𝑊𝑠𝑝 = 𝐹𝑑𝑠
Also 𝑑𝑠 = 𝑑𝑥
Hence
𝑊𝑠𝑝 2
∫ 𝑑𝑊𝑠𝑝 = ∫ 𝐹 𝑑𝑥
0 1
Since 𝐹 = 𝑘𝑥, then
𝑊𝑠𝑝 2
∫ 𝑑𝑊𝑠𝑝 = 𝑘 ∫ 𝑥 𝑑𝑥
0 1
1
𝑊𝑠𝑝 = 𝑘(𝑥2 2 − 𝑥1 2 )
2
𝟏
𝑾𝒔𝒑 = 𝒌𝚫𝒙𝟐
𝟐
99
3. Strain Energy - it is the energy required for a force F to stretch the system (elastic bars) within an elastic region.
For this purpose, we would like to have a glimpse of the strength of mater (mechanics of deformable bodies) basic
formulas of stress and strain. In strength of materials, we do not assume that the bodies are rigid, but the bodies can
deform.
a) Stress - is the ratio of applied load P in units of force per cross sectional area A. It is
also the pressure definition in solids.
𝐹
𝑠=
𝐴
When the solid is under tension or tensile forces, it is said to acquire tensile stress,
otherwise compressive stress when the solid is under compression or compressive forces.
𝑇 𝐶
𝑠𝑡 = 𝑠𝑐 =
𝐴 𝐴
(Tensile Stress) (Compressive Stress)
b) Strain - is the ratio of the elongation of the material δ or Δx by its original length L. Hence
δ Δ𝑥
ϵ= =
𝐿 𝐿
c) Modulus of Elasticity - it is the ratio of stress to strain at proportional limit, assuming that the stress
is directly proportional to the strain.
𝑠∝ϵ
𝑠 = 𝐸ϵ
𝑠
𝐸=
ϵ
Solving F in terms of E,l, A, and x yields
𝐹𝐿
𝐸=
𝐴𝑥
The force would be
𝑥
𝐹 = 𝐸𝐴
𝐿
Since work is done to the system, work is assume to be negative. For a differential load dF applied and elongates
the rod at a differential element dx,
𝑑𝑊𝑠𝑡 = −𝐹𝑑𝑥
Hence
𝐸𝐴
𝑑𝑊𝑠𝑡 = − 𝑥𝑑𝑥
𝐿
𝑊𝑠𝑡 𝑥
𝐸𝐴
∫ 𝑑𝑊𝑠𝑡 = − ∫ 𝑥𝑑𝑥
0 0 𝐿
100
𝐸𝐴𝑥 2
𝑊𝑠𝑡 = −
2𝐿
From strain equation
𝑥 = ϵ𝐿
𝑥 2 = ϵ2 𝐿2
Hence the work done in stretching elastic bars is
𝟏
𝑾𝒔𝒕 = − 𝑬𝑨𝛜𝟐 𝑳
𝟐
4. Surface Tension Work - it is the work done against a resisting surface tension. It is the work done on stretching
a liquid film. It is also applicable to rotating elements in lubrication. Surface Tension is the property of a fluid,
usually in a liquid-gas interface exerts a net force per unit length. In symbols
𝑭
𝛔=
𝒃
where b is the length of the fluid where the force is applied.

The work done in stretching a liquid film at the given area shown to the
left is:
𝑑𝑊 = 𝐹𝑑𝑠
Also 𝑑𝑠 = 𝑑𝐿, hence
𝑑𝑊 = 𝐹𝑑𝐿
Since 𝐹 = σ𝑏, then
𝑑𝑊 = σ𝑏𝑑𝐿
𝑊 𝐿
∫ 𝑊 = σ𝑏 ∫ 𝑑𝐿
0 0
𝑾 = 𝛔𝒃𝑳
For a soap bubble film, the work done to the circular film is
1
𝑊 = πσ𝑟 2
4
5. Electrical Work - is the work done by the electrons, which is also the product of potential difference and 'the
charge of the electrons. For this purpose, we will review some basic electrical engineering principles in
electrostatics.
a) Coulomb’s Law - states that the electric force is proportional to the product of the magnitude of two
electrical charge and inversely proportional to the square of the distance between them. In symbols
𝑘𝑄1 𝑄2
𝐹𝑒 =
𝑟2
𝑁𝑚2
Where k is called as the coulomb’s constant, wherein 𝑘 = 9 × 109 2
𝐶
Q is the charge of the particle, in units of coulombs (C)
b) Electric Field - it is the electric force per unit charge. Its units are in unit force per unit charge.
𝐹𝑒
𝐸𝑓 =
𝑄
101
c) Electric Potential Energy - it is the work done on the charge to move the charge on a certain
displacement.
𝑊 = ∫ 𝐹𝑑𝑥
𝑥
𝑊 = 𝐸𝑓 𝑄 ∫ 𝑑𝑥
0
𝑊 = 𝐸𝑓 𝑄𝑥
d) Potential difference - it is the product of the electric field intensity E to the displacement
𝑉 = 𝐸𝑓 𝑥
Using the equations for electric potential and potential difference, we can now define the electrical work
𝑾𝒆 = 𝑽𝑸
e) Ohms Law - states that the resistance is directly proportional to the potential difference and inversely
proportional to the current.
𝑽
𝑹=
𝑰
The resistance is defined as the ability of the material to resist electric flow.
f) Current - is the flow of electrons per unit time, in units of coulombs per unit time. The standard unit of
current is called as the ampere (amp) or (A) which is the flow of 1 coulomb in a unit second.
𝑸
𝑰=
𝒕
g) Conductance - it is the reciprocal of resistance. Its units are in terms of mho, siemens, and inverted
omega symbol. These units are equivalent to each other.
𝟏
𝑮=
𝑹
We can now use ohm’s law to relate electrical work to voltage,
current, and resistance
𝑾 = 𝑽𝑰𝒕
𝑾 = 𝑰𝟐 𝑹𝒕
𝑽𝟐 𝒕
𝑾=
𝑹
h) Electrical Power - the electrical power is somehow identical to mechanical power. The electrical power
P is also defined as the electrical work per unit time. Eliminating t from the previous equations yields the
fundamental electrical power equations that are very useful in mechanical engineering.
Fundamental Electrical Power Description Units
Formulas
𝑷 = 𝑰𝑽 Power as a function of current and 𝑉−𝐴
voltage
𝑽𝟐 Power as a function of voltage and 𝑉2
𝑷= resistance
𝑹 Ω
𝑷=𝑰 𝑹 𝟐 Power as a function of current and 𝐴2 − Ω
resistance
𝑷=𝑽 𝑮 𝟐 Power as a function of voltage and 𝑉 2 − 𝑚ℎ𝑜
conductance
𝑰 𝟐 Power as a function of current and 𝐴2 − 𝑚ℎ𝑜
𝑷= conductance
𝑮
102
6. Paddle Work - also called as stirring work. If a weight is placed on a pulley, and then the paddle is rotating and
supplying work to the system as the weight is falling. Hence potential energy is converted into paddle work.
𝑊𝑝𝑎𝑑𝑑𝑙𝑒 = ∫ 𝑊𝑑𝑧 = ∫ 𝑇𝑑 θ
7. Magnetic Work - the work done per unit volume on a magnetic material through which the magnetic and
magnetization fields are uniform is
2
𝑊𝑚𝑎𝑔𝑛𝑒𝑡 = ∫ 𝐻𝑑𝐼
1
Where H is defined as the magnetic field strength
I is the component of magnetization in the direction of the field
8. Work done to accelerate or raise the body - the work done to raise the body is
equivalent to the change in potential energy, while the work done to accelerate the
body is the change in kinetic energy.
9. Electrical Polarization Work - the generalized force is the electric field

strength and the generalized displacement is the polarization of the medium (the
sum of all electric dipole moments of the molecules)
103
2.13 TOTAL ENERGY AND TOTAL MECHANICAL ENERGY

In this section, we will generalize different forms of energy into a single magnitude. The conservation of
energy or the first law of thermodynamics states that energy is neither created nor destroyed but it is converted
from one energy into another.
A. Total Energy - the total energy of a system, assuming that other types of work are negligible in the system is
the sum of the internal, potential, and kinetic energy.
𝐸𝑡𝑜𝑡 = 𝑚𝑒𝑡𝑜𝑡
𝑬𝒕𝒐𝒕 = 𝑼 + 𝑷𝑬 + 𝑲𝑬
𝑔𝑧 𝑣 2
𝐸𝑡𝑜𝑡 = 𝑚 (𝑢 + + )
𝑘 2𝑘
Where 𝑒𝑡𝑜𝑡 is the total energy in the unit mass basis
B. Total Mechanical Energy - the mechanical energy of a system is defined as the energy that can be converted
into mechanical work completely and directly into a mechanical device such as an ideal turbine.
𝐸𝑚𝑒𝑐ℎ = 𝑚𝑒𝑚𝑒𝑐ℎ
𝑬𝒎𝒆𝒄𝒉 = 𝑾𝒇 + 𝑷𝑬 + 𝑲𝑬
𝑔𝑧 𝑣 2
𝐸𝑚𝑒𝑐ℎ = 𝑚 (𝑃ν + + )
𝑘 2𝑘
In unit mass basis
𝑃 𝑔𝑧 𝑣 2
𝑒𝑚𝑒𝑐ℎ = + +
ρ 𝑘 2𝑘
2.14 ENERGY BALANCE FOR THERMODYNAMIC SYSTEMS

In general, the energy balance at any system is defined as
Δ𝐸𝑠𝑦𝑠𝑡𝑒𝑚 = 𝐸𝑜𝑢𝑡 − 𝐸𝑖𝑛
a) Energy Change in a System - in absence of other effects, the change in the total energy of the system is
defined as the sum of the change in internal, potential, and kinetic energy. Hence:
Δ𝐸𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑚Δ𝑒𝑠𝑦𝑠𝑡𝑒𝑚
𝚫𝑬𝒔𝒚𝒔𝒕𝒆𝒎 = 𝚫𝑼 + 𝚫𝑷𝑬 + 𝚫𝑲𝑬
𝑔Δ𝑧 Δ𝑣 2
Δ𝐸𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑚 (Δ𝑢 + + )
𝑘 2𝑘
104
b) Energy Change for Open Systems - the energy change in open systems or flow systems is zero, in which the
energy entering the system is equal to the energy leaving the system. It is the conservation of energy for open
systems. The equation below is the steady flow energy equation (SFEE) for open system. The work done for open
systems is the steady flow work. By energy balance on the open system below.
𝑬𝒊𝒏 = 𝑬𝒐𝒖𝒕 𝚫𝑬 = 𝟎
In terms of energy units
𝑷𝑬𝟏 + 𝑲𝑬𝟏 + 𝑾𝒇𝟏 + 𝑼𝟏 + 𝑸 = 𝑷𝑬𝟐 + 𝑲𝑬𝟐 + 𝑾𝒇𝟐 + 𝑼𝟐 + 𝑾𝑺𝑭
The steady flow work would be:
𝑾𝑺𝑭 = 𝑸 − 𝚫𝑷𝑬 − 𝚫𝑲𝑬 − 𝚫𝑾𝒇 − 𝚫𝑼
In terms of each individual energy equations

𝑔(𝑧2 − 𝑧1 ) 𝑣2 2 − 𝑣1 2
𝑊𝑆𝐹 = 𝑚 [𝑞 − − − (𝑃2 𝑣2 − 𝑃1 𝑣1 ) − (𝑢2 − 𝑢1 )]
𝑘 2𝑘
In terms of specific steady flow work, in which all energies are in per unit mass basis
𝑒𝑖𝑛 = 𝑒𝑜𝑢𝑡 Δ𝑒 = 0
𝑝𝑒1 + 𝑘𝑒1 + 𝑤𝑓1 + 𝑢1 + 𝑞 = 𝑝𝑒2 + 𝑘𝑒2 + 𝑤𝑓2 + 𝑢2 + 𝑤𝑆𝐹
The steady flow work in energy per unit mass basis would be
𝑤𝑆𝐹 = 𝑞 − Δ𝑝𝑒 − Δ𝑘𝑒 − Δ𝑤𝑓
𝑔(𝑧2 − 𝑧1 ) 𝑣2 2 − 𝑣1 2
𝑤𝑆𝐹 = [𝑞 − − − (𝑃2 𝑣2 − 𝑃1 𝑣1 ) − (𝑢2 − 𝑢1 )]
𝑘 2𝑘
If the open system is given in terms of enthalpies, with negligible kinetic and potential energy, the steady flow
energy equation reduces to
𝑄 + 𝐻1 = 𝐻2 + 𝑊𝑆𝐹
𝑾𝑺𝑭 = 𝑸 − 𝚫𝑯, 𝚫𝑲𝑬 = 𝚫𝑷𝑬 = 𝟎

𝑊𝑠𝑓 = 𝑄 − (𝐻2 − 𝐻1 )
In terms of specific steady flow work
𝑤𝑠𝑓 = 𝑞 − (ℎ2 − ℎ1 )
Note: You do not need to memorize all these energy equations. Knowing the concept of energy balance is a must
because most thermodynamic systems are ideally open systems.
105
c) Energy Change for Closed Systems - for closed systems, the energy balance is still the same such that the
energy entering the system is equal to the energy leaving the system. For an ideal closed system, the volume of
the system is assumed to be constant unless specified. The enthalpy for closed systems is equivalent to its internal
energy at constant volume, and the work done by the piston shown below is also the non-flow work. The equation
below is also called as the non-flow energy equation (NFEE).
By Energy balance on the system

𝑃𝐸1 + 𝐾𝐸1 + 𝑈1 + 𝑄 = 𝑃𝐸2 + 𝐾𝐸2 + 𝑈2 + 𝑊𝑁𝐹
The non-flow work would be
𝑊𝑁𝐹 = 𝑄 − Δ𝑈 − Δ𝑃𝐸 − Δ𝐾𝐸

𝑔(𝑧2 − 𝑧1 ) 𝑣2 2 − 𝑣1 2
𝑊𝑁𝐹 = 𝑚 [𝑞 − (𝑢2 − 𝑢1 ) − − ]
𝑘 2𝑘
In terms of specific non flow work or per unit mass basis
𝑔(𝑧2 − 𝑧1 ) 𝑣2 2 − 𝑣1 2
𝑤𝑁𝐹 = [𝑞 − (𝑢2 − 𝑢1 ) − − ]
𝑘 2𝑘
If change in potential and kinetic energy are negligible, in most cases for applying energy balances for close
system, the equation reduces to
𝑾𝑵𝑭 = 𝑸 − 𝚫𝑼, 𝚫𝑲𝑬 = 𝚫𝑷𝑬 = 𝟎
Note:
a) When work is done by the piston, W is positive, hence heat must be rejected to the system. As the
heat is rejected, there must be a stored energy (internal energy to balance both energy).
b) Conversely if heat is added to the system, Q is positive, and work is done to the system in which
W is negative. There must be a store energy again (internal energy) to balance both work and heat.
The change in internal energy places a vital role in energy balance for closed systems
106
2.15 ENERGY BALANCE FOR STEADY FLOW ENGINEERING DEVICES.

Most mechanical engineering devices in thermodynamics are assumed to be ideal, hence these devices are
assumed to be open systems, and conservation of mass (mass balance) and conservation of energy (energy
balance) is applicable. Steady flow engineering devices are classified into three.
A) Energy Producing Devices - energy in which converts heat and other forms of energy to produce
work, or electricity. The work in these devices is done by the system.
B) Energy Consuming Devices - energy in which work, or electricity must be done to the device in
order to operate. The work in these devices is done to the system
C) Heat Exchangers – are steady flow devices whose solely purpose is to transfer and redirect heat
from a designated storage or application. These devices do not produce work.
The following are some of the common steady flow engineering devices.
1. Turbine - is a steady flow device that generates work and electricity. There are two common types of turbine.
a) Water Turbine - used in hydroelectric
powerplants and usually installed
beneath dams. It harnesses potential
energy from the water and the flow of
water will move the blades in the
turbine. The turbine is connected to a
generator that produces power.
From the given figure, by energy balance,

assuming that there is no change in internal
energy and no heat is loss or gain into the
system:
𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡 Δ𝐸 = 0
𝑃𝐸1 + 𝐾𝐸1 + 𝑊𝑓1 + 𝑄 = 𝑃𝐸2 + 𝐾𝐸2 + 𝑊𝑓2 + 𝑊𝑆𝐹
The work done by the water turbine would be:
𝑊𝑆𝐹 = −Δ𝑃𝐸 − Δ𝐾𝐸 − Δ𝑊𝑓
𝒈(𝒛𝟐 − 𝒛𝟏 ) 𝒗𝟐 𝟐 − 𝒗𝟏 𝟐
𝑾𝑺𝑭 = −𝒎𝒘𝒂𝒕𝒆𝒓 [ + + (𝑷𝟐 𝒗𝟐 − 𝑷𝟏 𝒗𝟏 )]
𝒌 𝟐𝒌
b) Gas/Steam Turbines - used in most powerplants. Steam or gas passes through the turbine and part of its
energy converted to work. The turbine output runs the generator to produce power.
Neglect change in potential and internal energy, the

energy balance for gas or steam turbines is:
𝐾𝐸1 + 𝐻1 = 𝑄 + 𝐾𝐸2 + 𝐻2 + 𝑊𝑆𝐹
The work done by the steam/gas turbine would be:
𝑊𝑆𝐹 = −𝑄 − Δ𝐻 − Δ𝐾𝐸
𝒗𝟐 𝟐 − 𝒗𝟏 𝟐
(𝒉
𝑾𝑺𝑭 = −𝒎𝒔𝒕𝒆𝒂𝒎 [𝒒 + 𝟐 − 𝒉𝟏 + ) ]
𝟐𝒌
107
2. Pump - is a steady flow device that draws liquid from lower level to higher level by increasing the fluid
pressure. Work is required to run the pump and this may be supplied by an electric motor or a diesel engine.
From the given figure, by energy balance, assuming that there is no change in internal energy and no heat is loss
or gain into the system:
𝑃𝐸1 + 𝐾𝐸1 + 𝑊𝑓1 + 𝑊𝑆𝐹 = 𝑃𝐸2 + 𝐾𝐸2 + 𝑊𝑓2
The work done to the pump is equivalent to

𝑊𝑆𝐹 = Δ𝑃𝐸 + Δ𝐾𝐸 + Δ𝑊𝑓
𝒈(𝒛𝟐 − 𝒛𝟏 ) 𝒗𝟐 𝟐 − 𝒗𝟏 𝟐
𝑾𝑺𝑭 = 𝒎𝒘𝒂𝒕𝒆𝒓 [ + + (𝑷𝟐 𝒗𝟐 − 𝑷𝟏 𝒗𝟏 )]
𝒌 𝟐𝒌
3. Centrifugal Compressor - is a steady flow device that increases the pressure of air and supplies the same at
moderate pressure and in large quantity. Work is required to run the centrifugal compressor, and requires an
electric motor coupled to the compressor.
From the given figure, by energy balance, assuming that there is no change in potential and kinetic energy.
𝐾𝐸1 + 𝐻1 + 𝑊𝑆𝐹 = 𝑄 + 𝐾𝐸2 + 𝐻2
The work done to the centrifugal compressor would

be
𝑊𝑆𝐹 = 𝑄 + Δ𝐻 + Δ𝐾𝐸
𝒗𝟐 𝟐 − 𝒗𝟏 𝟐
𝑾𝑺𝑭 = 𝒎𝒂𝒊𝒓 [𝒒 + (𝒉𝟐 − 𝒉𝟏 ) + ]
𝟐𝒌
4. Reciprocating Compressor - is a steady flow device that draws in air from the atmosphere and supplies at a
considerable higher pressure in small quantities. It considers a steady flow system if the compressor includes a
receiver in which it reduces the fluctuations of flow.
Applying energy balance, neglecting change in potential and internal energy.

𝐾𝐸1 + 𝐻1 + 𝑊𝑆𝐹 = 𝑄 + 𝐾𝐸2 + 𝐻2
The work done to the reciprocating compressor is

𝑊𝑆𝐹 = 𝑄 + Δ𝐻 + Δ𝐾𝐸
𝒗𝟐 𝟐 − 𝒗𝟏 𝟐
𝑾𝑺𝑭 = 𝒎𝒂𝒊𝒓 [𝒒 + (𝒉𝟐 − 𝒉𝟏 ) + ]
𝟐𝒌
108
5. Boiler -is a steady flow heat exchanger device, part of the steam generating unit. Its main purpose is to transfer
heat from a heat source to an incoming high pressure liquid in order to increase its temperature and change phase
to become steam.
Applying energy balance, since no work is made to the system and

neglect change in potential and kinetic energy.
𝐻1 + 𝑄 = 𝐻2
The heat added to the steam is defined as

𝑄 = 𝐻2 − 𝐻1
𝑸𝑨 = 𝒎𝒔𝒕𝒆𝒂𝒎 (𝒉𝟐 − 𝒉𝟏 )
where QA is the heat added to the steam.
For coal steam powerplants, coals are used as a fuel to heat steam into the boiler. Coals have heating values
(HV), also called as calorific value (CV) wherein it is the energy required to burn per unit mass of fuel. For ideal
steam generating unit, the heat that is added to the boiler can be defined as
𝑸𝑨 = 𝒎𝒄𝒐𝒂𝒍 𝑯𝒗
where Hv is the heating value of fuel whose units are in energy per unit mass. Equationg both equations will give
us the amount of coal required to heat the steam ideally.
𝑚𝑠𝑡𝑒𝑎𝑚 (ℎ2 − ℎ1 ) = 𝑚𝑐𝑜𝑎𝑙 𝐻𝑣
𝒎𝒔𝒕𝒆𝒂𝒎 (𝒉𝟐 − 𝒉𝟏 )
𝒎𝒄𝒐𝒂𝒍 =
𝑯𝒗
For nuclear steam powerplants, nuclear fuel are used in order to heat steam. The heat added must be
𝑚𝑁𝐹 𝑐 2
𝑄𝐴 = , then the mass of nuclear fuel needed would be:
𝑘
𝑚𝑁𝐹 𝑐 2
= 𝑚𝑠𝑡𝑒𝑎𝑚 (ℎ2 − ℎ1 )
𝑘
𝒌
𝒎𝑵𝑭 = 𝒎𝒔𝒕𝒆𝒂𝒎 (𝒉𝟐 − 𝒉𝟏 ) 𝟐
𝒄
6. Condenser - is a steady flow device whose main purpose is to reject heat in order for the steam to condenser,
for a change phase from steam vapor to liquid water at constant pressure. Applying energy balance, since no work
is made to the system and neglect change in potential and kinetic energy:
𝐻1 = 𝐻2 + 𝑄𝑅
The heat rejected by the condenser from the steam is
𝑄𝑅 = 𝐻1 − 𝐻2
𝑸𝑹 = 𝒎𝒔𝒕𝒆𝒂𝒎 (𝒉𝟏 − 𝒉𝟐 )
109
To condense the steam, a circulating cooling fluid, usually a refrigerant or cooling water absorbs the heat rejected
by the condenser. By energy balance, assuming the system is the cooling water:
𝐻𝑎 + 𝑄𝑅 = 𝐻𝑏
The heat rejected by the condenser in terms of cooling water
𝑄𝑅 = 𝑚𝑤 𝑐𝑝𝑤 (𝑡𝑏 − 𝑡𝑎 )
𝑄𝑅 = 𝑚𝑤 𝑐𝑝𝑤 Δ𝑡
𝑸𝑹 = 𝛒𝒘 𝑽𝒘 𝒄𝒑𝒘𝚫𝒕
Sometimes in designing condensers, we are required to determine the amount of cooling water in terms of
volume that must circulate. Equating the equations:
ρ𝑤 𝑉𝑤 𝑐𝑝𝑤 Δ𝑡 = 𝑚𝑠𝑡𝑒𝑎𝑚 (ℎ1 − ℎ2 )
𝒎𝒔𝒕𝒆𝒂𝒎 (𝒉𝟏 − 𝒉𝟐 )
𝑽𝒘 =
𝛒𝒘 𝒄𝒑𝒘 𝚫𝒕
7. Evaporator - is a steady flow device used in refrigeration plant to carry heat from the refrigerator to maintain the
low temperature. A Refrigerant liquid passed through the evaporator and absorbs heat from the refrigerating space,
decreasing the temperature of the system. The “freezer” in your domestic refrigerator is technically called as an
evaporator because heat that is stored from different food and product that is inside the freezer is absorbed by the
evaporator, allowing the temperature of all the things inside the freezer below its room temperature.
Applying energy balance, since no work is made to the

system and neglect change in potential and kinetic
energy.
𝐻1 + 𝑄𝐴 = 𝐻2
The heat absorbed by the evaporator is

𝑸𝑨 = 𝑯𝟐 − 𝑯𝟏
where QA is the heat absorbed by the refrigerant, also
called as the refrigeration effect.
8. Nozzle -it is a steady flow engineering device that increases the velocity of the
fluid at the expense of its pressure. Hence the nozzle inlet cross sectional area is
lesser than its discharge cross sectional area. As velocity increases, the pressure
decreases. This is generally used to convert the part of the energy of the steam
into kinetic energy of steam supplied to the turbine.
9. Diffusers - is a steady flow engineering device that increase the pressure at the
expense of its velocity. Decreasing or slowing down the motion of fluid allows
an increase in pressure. Hence the diffuser inlet is larger than the discharge cross
sectional area. As the pressure increases, the velocity decreases
110
For a steam nozzle, then no work is done and neglect change in potential and internal energy. Assume no heat is
lost or gained by the nozzle, the energy balance would be
𝐾𝐸1 + 𝐻1 = 𝐾𝐸2 + 𝐻2
Hence the velocity leaving the nozzle is
Δ𝐾𝐸 = −Δ𝐻
𝒗𝟐 𝟐 − 𝒗𝟏 𝟐
= −(𝒉𝟐 − 𝒉𝟏 )
𝟐𝒌
The final velocity would be
𝑣2 = √𝑣1 2 − 2𝑘Δℎ
If initial velocity is so small that it is very negligible:

𝑣2 = √−2𝑘Δℎ
𝒗𝟐 = √𝟐𝒌(𝐡𝟏 − 𝐡𝟐 )
10. Mixing Chambers - it is a steady flow device where mixing process between two or more liquids occur. The
mixing chamber does not to be a “chamber” itself. Most mixing chambers are well insulated, and no heat is loss
or gain by this system.
Applying mass balance from the given figure.
𝑚𝑖𝑛̇ = 𝑚𝑜𝑢𝑡 ̇ Δ𝑚̇ = 0 𝒎𝟑
𝑚̇ 1 + 𝑚̇ 2 = 𝑚3
Applying energy balance to the system, neglecting
change in potential and kinetic energy, and assume
that there is no shaft work made by the system.
𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡 Δ𝐸 = 0 𝒎𝟐
𝐻1 + H2 = 𝐻3
𝑚1 ℎ1 + 𝑚2 ℎ2 = 𝑚3 ℎ3
The enthalpy after two liquids are mixed would be: 𝒎𝟏

𝑚1 ℎ1 + 𝑚2 ℎ2
ℎ3 =
𝑚3
From mass balance
𝒎𝟏 𝒉 𝟏 + 𝒎𝟐 𝒉 𝟐
𝒉𝟑 =
𝒎𝟏 + 𝒎𝟐
111
2.16 ENERGY BALANCE IN STEAM POWER PLANT

A steam powerplant is an industrial facility that uses steam as a working substance to produce electricity.
A simple steam powerplant basically has four components: The boiler, turbine, condenser, and pump.
For any system, it is possible for a system to have its efficiency. The efficiency is defined as the ratio of output to
input.
𝒐𝒖𝒕𝒑𝒖𝒕
𝒏=𝒆=
𝒊𝒏𝒑𝒖𝒕
In terms of energy producing devices, the output represents the actual energy produced by the system, and
the input represent the theoretical energy produced by the system.
𝐚𝐜𝐭𝐮𝐚𝐥 𝐞𝐧𝐞𝐫𝐠𝐲 𝐩𝐫𝐨𝐝𝐮𝐜𝐞𝐝
𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐜𝐲 =
𝐭𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐞𝐧𝐞𝐫𝐠𝐲 𝐩𝐫𝐨𝐝𝐮𝐜𝐞𝐝
We will first investigate the efficiencies in a simple steam turbine. For a turbine, the turbine efficiency , also
called as the indicated efficiency, would be
𝑾𝑻 ′
𝒏𝒕 =
𝑾𝑻
Where 𝑊𝑇 ′ is the actual work or the indicated work of the turbine.
The mechanical efficiency is the ratio of the brake work and the indicated work. The brake work is also called as
the available work, in which it is the work available at the turbine shaft.
𝑾𝑩
𝒏𝒎𝒆 = ′
𝑾
The electrical efficiency, also called as the generator efficiency or alternator efficiency, is the ratio of break work
to combined work. The combined work is the electrical energy available to consumers.
𝑾𝒌
𝒏𝒆𝒆 =
𝑾𝑩
The overall efficiency or also called as the plant efficiency is the ratio of electrical energy charge to the plant
to the heat added due to the burning of fuel.
𝑾𝒌
𝒏𝒐𝒗𝒆𝒓 =
𝑸𝑨 ′
For the heat added when coal is used:
𝑸𝑨 ′ = 𝒎𝒄𝒐𝒂𝒍 𝑯𝒗
112
For the heat added when nuclear fuel is used:
𝒎𝑵𝑭 𝒄𝟐
𝑸𝑨 ′ =
𝒌
For energy consuming devices, the output represents the theoretical input energy to the system, while the input
represents the actual energy input to the system.
𝐭𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐢𝐧𝐩𝐮𝐭 𝐞𝐧𝐞𝐫𝐠𝐲
𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐜𝐲 =
𝐚𝐜𝐭𝐮𝐚𝐥 𝐢𝐧𝐩𝐮𝐭 𝐞𝐧𝐞𝐫𝐠𝐲
For a boiler, the boiler efficiency would be the ratio of actual heat due to burning of fuel to the heat added to
steam.
𝑸𝑨
𝒏𝑩 =
𝑸𝑨 ′
Where Q’ is the actual heat added due to the burning of fuel.
For a pump, the pump efficiency is the ratio of theoretical work input to the pump to the actual work input to the
pump.
𝑾𝒑
𝒏𝒑 =
𝑾𝒑 ′
Where 𝑊𝑝 ′ is the actual pump work.
The diagram below shows a simpler version of a

steam power plant. By energy balance on the
whole plant:
𝑄𝐴 + 𝑊𝑝 = 𝑊𝑇 + 𝑄𝑅
𝑊𝑇 − 𝑊𝑝 = 𝑄𝐴 − 𝑄𝑅
The net work would be
𝑾𝒏𝒆𝒕 = 𝑾𝑻 − 𝑾𝒑
Assuming that pump work is negligible compare
to the value of the turbine work
𝑊𝑛𝑒𝑡 = 𝑊𝑇
The thermal efficiency of the powerplant, is defined as the ratio of the net work to heat added.
𝑊𝑛𝑒𝑡
𝑒=
𝑄𝐴
Hence if pump work is negligible (𝑊𝑝 = 0), then
𝑾𝑻
𝒆=
𝑸𝑨
113
Solved Problems:
1. A fluid moves in a steady flow manner between two sections in a fluid flow line. At section 1: 𝐴1 = 1𝑓𝑡 2 , 𝑣1 =
1000𝑓𝑝𝑚, 𝑣1 = 4𝑓𝑡 3 /𝑙𝑏 . At section 2:𝐴2 = 2𝑓𝑡 2 , ρ2 = 0.20𝑙𝑏/𝑓𝑡 3 .Calculate
a) the flow in lb/hr
b) The velocity at section 2 in fps.
Solution:
a) The mass flow rate at section 1 is:
𝑚̇ 1 = ρ1 𝐴1 v1
𝐴1 v1
𝑚̇ 1 =
ν1
𝑓𝑡 60𝑚𝑖𝑛
(1𝑓𝑡 2 ) (1000 )( )
𝑚𝑖𝑛 1ℎ𝑟
𝑚̇ 1 =
𝑓𝑡 3
4
𝑙𝑏𝑚
𝒍𝒃𝒎
𝒎̇ 𝟏 = 𝟏𝟓𝟎𝟎
𝒉𝒓
b) By mass balance:
𝑚𝑖𝑛̇ = 𝑚𝑜𝑢𝑡 ̇ Δ𝑚 = 0
𝑚̇ 1 = 𝑚̇ 2
ρ1 𝐴1 v1 = ρ2 𝐴2 v2
𝐴1 v1
= 𝜌2 𝐴2 v2
𝜈1
𝐴1
v2 = ρ2 𝑣1 v1 ( )
𝐴2
𝑙𝑏𝑚 𝑓𝑡 3 𝑓𝑡 1𝑚𝑖𝑛 1𝑓𝑡 2
𝑣2 = (0.20 3 ) (4 ) (1000 )( )( )
𝑓𝑡 𝑙𝑏𝑚 𝑚𝑖𝑛 60𝑠 2𝑓𝑡 2
𝒇𝒕
𝒗𝟐 = 𝟔. 𝟔𝟔
𝒔
2. Two gaseous streams enter a mixing chamber through two sections and leaves through one section. Entrance
conditions are 𝐴1 = 450𝑐𝑚2 , v1 = 150 𝑚𝑝𝑠, 𝑣1 = 0.625𝑚3 /𝑘𝑔 and 𝐴2 = 380𝑐𝑚2 , 𝑣2 = 550𝑚𝑝𝑠, ρ2 = 2𝑘𝑔/
𝑚3 . At the exit, 𝑣3 = 245𝑚𝑝𝑠 and 𝑣3 = 0.58𝑚3 /𝑘𝑔. Determine
a) The mass flow rate at section 2 in kg/s
b) area of the exit section in sq. cm.
Solution:
a) The mass flow rate at section 2 is
𝑚̇ 2 = ρ2 𝐴2 v2
𝑘𝑔 1𝑚 2 𝑚
𝑚̇ 2 = (2 3 ) (380𝑐𝑚2 ) ( ) (55 )
𝑚 100𝑐𝑚 𝑠
𝒌𝒈
𝒎̇ 𝟐 = 𝟒. 𝟏𝟖
𝒔
b) By mass balance
𝑚𝑖𝑛̇ = 𝑚𝑜𝑢𝑡 ̇ Δ𝑚 = 0
𝑚̇ 1 + 𝑚̇ 2 = 𝑚3
ρ1 𝐴1 v1 + ρ2 𝐴2 v2 = ρ3 𝐴3 v3
𝐴1 v1 𝐴3 v3
+ 𝜌2 𝐴2 v2 =
𝑣1 𝑣3
𝑣3 𝐴1 v1
𝐴3 = ( ) ( + 𝜌2 𝐴2 v2 )
v3 𝑣1
114
𝑚3 𝑚 1𝑚 2
0.58 2
𝑘𝑔𝑚 (450𝑐𝑚 ) (150 𝑠 ) (100𝑐𝑚 ) 𝑘𝑔 2) (
1𝑚 2 𝑚
𝐴3 = ( 𝑚 ) [ 3 + (2 3 ) (380𝑐𝑚 ) (55 )]
245 𝑚 𝑚 100𝑐𝑚 𝑠
𝑠 0.625
𝑘𝑔𝑚
100𝑐𝑚 2
𝐴3 = 0.035463𝑚2 ( )
1𝑚
𝑨𝟑 = 𝟑𝟓𝟒. 𝟔𝟑𝒄𝒎𝟐
3. The 600 kg hammer or a pile driver is lifted 2m above a piling head. Local g=9.65 m/s 2.
a) What is the change of potential energy in kJ?
b) If the hammer is released, what will be its velocity in m/s at the instant it strikes the piling?
c) Compare the values of kinetic energy and potential energy, then make a conclusion.
Solution.
a) The change in potential energy of the hammer is

𝑚𝑔𝑧
𝑃𝐸 =
𝑘
𝑚
(600𝑘𝑔𝑚) (9.65 ) (2𝑚)
𝑃𝐸 = 𝑠
1000
𝑘𝑁 − 𝑠2
𝑷𝑬 = 𝟏𝟏. 𝟓𝟖 𝒌𝑱
b) Using the 3rd kinematic equation of motion

𝑣 2 = 𝑣𝑜 2 + 2𝑔ℎ
The initial velocity at that point is zero, thus
𝑣 = √2𝑔ℎ
𝑚
𝑣 = √2 (9.65 ) (2𝑚)
𝑠
𝒎
𝒗 = 𝟔. 𝟐𝟏𝟑
𝒔
c) The kinetic energy of the system would be:
𝑚𝑣 2
𝐾𝐸 =
2𝑘
𝑚 2
(600𝑘𝑔𝑚) (6.213 )
𝐾𝐸 = 𝑠
2 (1000 )
𝑘𝑁 − 𝑠 2
𝑲𝑬 = 𝟏𝟏. 𝟓𝟖 𝒌𝑱
Hence we can conclude that the potential energy is converted into kinetic energy as it strikes the ground.
115
4. The flow energy of a moving fluid at 100 GPM is 200 BTU/min, What is the gage pressure in kPag at this point?
Solution
From the definition of flow energy, in energy per unit time basis
𝑊̇𝑓 = 𝑃𝑉̇
The absolute pressure at this point is defined as
𝑊̇𝑓
𝑃=
𝑉̇
𝐵𝑇𝑈 778.16𝑓𝑡 − 𝑙𝑏𝑓 12𝑖𝑛
200 ( )( )
𝑚𝑖𝑛 1𝐵𝑇𝑈 1𝑓𝑡
𝑃=
𝑔𝑎𝑙 231𝑖𝑛3
100 ( )
𝑚𝑖𝑛 1𝑔𝑎𝑙
2
𝑃 = 80.85𝑙𝑏𝑓/𝑖𝑛
𝑃 = 80.85 𝑝𝑠𝑖𝑎
Hence the gage pressure at that point would be
𝑃𝑎𝑏𝑠 = 𝑃𝑔 + 𝑃𝑎𝑡𝑚
𝑃𝑔 = 𝑃𝑎𝑏𝑠 − 𝑃𝑎𝑡𝑚
101.325𝑘𝑃𝑎𝑔
𝑃𝑔 = (80.25 − 14.7)𝑝𝑠𝑖𝑔 ( )
14.7𝑝𝑠𝑖𝑔
𝑷𝒈 = 𝟒𝟓𝟏. 𝟖𝟑𝒌𝑷𝒂𝒈
5. 1 kgm/s of ice at -25C is converted into steam at 120C. Determine the following
a) The total amount of heat in kW needed in the conversion.
b) If the heater efficiency is 80%, what is the power input of the heater in CHU/hr?
c) If the heating vent has a resistance of 2kΩ, what is the voltage, the electrical charge per minute, and the
conductance in mho?
Solution:
−20℃ 0℃ 0℃ 100℃ 100℃ 120℃
a) Recall the definition of sensible and latent heat. The heat required to convert ice to steam is:
̇ = 𝑄𝑠1
𝑄𝑡𝑜𝑡 ̇ + 𝑄𝐿1̇ + 𝑄𝑠2
̇ + 𝑄𝐿2
̇ + 𝑄𝑠3̇
𝑄𝑡𝑜𝑡̇ = 𝑚̇𝑐𝑝𝑖 Δ𝑡𝑖 + 𝑚̇ℎ𝑓 + 𝑚̇𝑐𝑝𝑤 Δ𝑡𝑤 + 𝑚̇ℎ𝑣 + 𝑚̇𝑐𝑝𝑠 Δ𝑡𝑠
𝑄𝑡𝑜𝑡 ̇ = 𝑚̇[𝑐𝑝𝑖 Δ𝑡𝑖 + ℎ𝑓 + 𝑐𝑝𝑤 Δ𝑡𝑤 + ℎ𝑣 + 𝑐𝑝𝑠 Δ𝑡𝑠 ]
𝑘𝑔𝑚 𝑘𝐽 𝑘𝐽 𝑘𝐽
𝑄𝑡𝑜𝑡 ̇ =1 [(0.5) (4.187 ) (0 − (−25))°𝐶 + 335 + (4.187 ) (100 − 0)°𝐶
𝑠 𝑘𝑔𝑚 − °𝐶 𝑘𝑔𝑚 𝑘𝑔𝑚 − °𝐶
𝑘𝐽 𝑘𝐽
+ 2257 + 0.4454 (4.187 ) (120 − 100)°𝐶]
𝑘𝑔𝑚 𝑘𝑔𝑚 − °𝐶
𝑸𝒕𝒐𝒕 ̇ = 𝟑𝟏𝟎𝟎. 𝟐𝟑 𝒌𝑾
b) From efficiency
𝑜𝑢𝑡𝑝𝑢𝑡
𝑒=
𝑖𝑛𝑝𝑢𝑡
Since electrical input is needed to produce the heat required
̇
𝑄𝑡𝑜𝑡
𝑒=
𝑃
𝑄𝑡𝑜𝑡̇
𝑃=
𝑒
116
𝐵𝑇𝑈
3100.23𝑘𝑊 42.41 𝑚𝑖𝑛 60𝑚𝑖𝑛 1𝐶𝐻𝑈
𝑃= ( )( )( )
0.80 0.746𝑘𝑊 1ℎ𝑟 1.8𝐵𝑇𝑈
𝑪𝑯𝑼
𝑷 = 𝟕𝟑𝟒𝟑𝟔𝟓𝟐. 𝟓𝟎
𝒉𝒓
c) From electrical power

𝑉0 2
𝑃=
𝑅
𝑉0 = √𝑃𝑅
3100.23𝑘𝑊 1000𝑊
𝑉0 = √ ( ) (2000Ω)
0.80 1𝑘𝑊
𝑽𝟎 = 𝟖𝟖𝟗𝟑𝟕. 𝟓𝟓𝑽
From ohm’s law, we can determine the electric charge per minute:
𝑉
𝑅=
𝐼
𝑉
𝐼=
𝑅
880375𝑉
𝐼=
2000Ω
1𝐶/𝑠 60𝑠
𝐼 = 44.02𝐴 ( )( )
1𝐴 1𝑚𝑖𝑛
𝑪
𝑰 = 𝟐𝟔𝟒𝟏. 𝟐
𝒎𝒊𝒏
For the conductance
1
𝐺=
𝑅
1
𝐺=
2000Ω
𝑮 = 𝟎. 𝟎𝟎𝟎𝟓 𝒎𝒉𝒐
6. What is the change of internal energy in BTU, kW-hr, PS-min, kcal, erg, hp-hr, ft-lbf, and kgf-m in heating 2
kgm of oxygen gas from 540R to 5000R at constant atmospheric pressure while its volume changes by 5 ft3 if
5.375 47.8
𝑐𝑝 = 0.36 − + 𝑖𝑛 𝐵𝑇𝑈/𝑙𝑏𝑚 − °𝐹.
√𝑇 𝑇
Solution:
From the microscopic definition of enthalpy
Δ𝐻 = 𝑚 ∫ 𝑐𝑝 𝑑𝑇
5000
5.375 47.8
Δ𝐻 = 𝑚 ∫ (0.36 − + ) 𝑑𝑇
540 √𝑇 𝑇
5000
Δ𝐻 = 2𝑘𝑔𝑚[0.36𝑇 − 2√5.375𝑇 + 47.8𝑙𝑛𝑇]540
2.205𝑙𝑏𝑚
Δ𝐻 = 2𝑘𝑔𝑚 ( ) [(0.36(5000) − 2√5.375(5000) + 47.8𝑙𝑛5000)
1𝐵𝑇𝑈
𝐵𝑇𝑈
− (0.36(540) − 2√5.375(540) + 47.8𝑙𝑛540)]
𝑙𝑏𝑚 − °𝑅
𝚫𝑯 = 𝟓𝟐𝟗𝟗. 𝟐𝟗 𝑩𝑻𝑼
117
From our macroscopic definition of enthalpy:
Δ𝐻 = Δ𝑈 + Δ𝑊𝑓
Δ𝐻 = Δ𝑈 + Δ𝑃𝑉
Since Pressure is constant, 𝑃1 = 𝑃2 = 𝑃, then the internal energy would be:
Δ𝑈 = Δ𝐻 − 𝑃Δ𝑉
𝑙𝑏𝑓 144𝑖𝑛2 1𝐵𝑇𝑈
Δ𝑈 = 5299.29𝐵𝑇𝑈 − (14.7 2 ) (5𝑓𝑡 3 ) ( 2 )( )
𝑖𝑛 1𝑓𝑡 778.16𝑓𝑡 − 𝑙𝑏𝑓
Δ𝑈 = 5285.69 𝐵𝑇𝑈
In terms in kW-hr
1.055𝑘𝐽 1𝑘𝑊 − ℎ𝑟
Δ𝑈 = 5285.69𝐵𝑇𝑈 ( )( )
1𝐵𝑇𝑈 3600𝑘𝐽
𝚫𝑼 = 𝟏. 𝟓𝟒𝟗 𝒌𝑾 − 𝒉𝒓
In terms of PS-min
1𝑃. 𝑆 60𝑚𝑖𝑛
Δ𝑈 = 1.549𝑘𝑊 − ℎ𝑟 ( )( )
0.736𝑘𝑊 1ℎ𝑟
𝚫𝑼 = 𝟏𝟐𝟔. 𝟐𝟖 𝑷𝑺 − 𝒎𝒊𝒏
In terms of erg
0.736𝑘𝑊 60𝑠 1𝑘𝐽 1010 𝑒𝑟𝑔
Δ𝑈 = 126.28𝑃𝑆 − 𝑚𝑖𝑛 ( )( )( )( )
1𝑃𝑆 1𝑚𝑖𝑛 1𝑘𝑊 − 𝑠 1 𝑘𝐽
𝚫𝑼 = 𝟓. 𝟓𝟖 × 𝟏𝟎𝟏𝟑 𝒆𝒓𝒈
In terms of hp-hr
10−10 𝑘𝐽 1𝐵𝑇𝑈 1ℎ𝑝 1ℎ𝑟
Δ𝑈 = 5.58 × 1013 𝑒𝑟𝑔 ( )( )( )( )
1𝑒𝑟𝑔 1.055𝑘𝐽 42.41 𝐵𝑇𝑈 60𝑚𝑖𝑛
𝑚𝑖𝑛
𝚫𝑼 = 𝟎. 𝟐𝟎𝟕𝟗 𝒉𝒑 − 𝒉𝒓
In terms of ft-lbf
𝑓𝑡 − 𝑙𝑏𝑓
550 3600𝑠
Δ𝑈 = 0.2079 ℎ𝑝 − ℎ𝑟 ( 𝑠 )( )
1ℎ𝑝 1ℎ𝑟
𝚫𝑼 = 𝟒𝟏𝟏𝟔𝟏𝟐 𝒇𝒕 − 𝒍𝒃𝒇
In terms of kgf-m
1𝐵𝑇𝑈 1.055𝑘𝐽 1𝑘𝑔𝑓 − 𝑚 1000𝐽
Δ𝑈 = (411612𝑓𝑡 − 𝑙𝑏𝑓) ( )( )( )( )
778.16𝑓𝑡 − 𝑙𝑏𝑓 1𝐵𝑇𝑈 9.8066𝐽 1𝑘𝐽
𝚫𝑼 = 𝟓𝟔𝟗𝟎𝟖. 𝟖𝟒 𝒌𝒈𝒇 − 𝒎
118
7. A powerplant is to produce 1 × 1015 kW-hr of energy in one year. If plant conversion efficiency is 35%,
compute
a) The mass of coal needed by the plant in kgm if heating value is 20000 kcal/kgm.
b) The mass of nuclear fuel in kgm.
c) The energy wasted in the plant in CHU, PS-min, and hp-hr.
d) Recommend dimensions of a spherical tank to serve as a container of fuel in (a) and (b) if specific
gravity of coal and nuclear fuel are 2 and 10 respectively.
e) The gallons of water that must circulate in the condenser if its temperature rise is 20°C.
f) The electrical output in kW-hr and hp-min if mechanical and electrical efficiencies are 85% and 93%
respectively.
Solution: The schematic diagram below shows a simple operation of a steam powerplant, either using coal or
nuclear fuel. A simple steam powerplant composed of four main components, the steam generating unit, turbine,
condenser, and pump. In this energy analysis, we assume that the pump work is zero.
a) For the mass of coal needed,
𝑒=
𝑊𝑇
𝑒=
𝑄𝐴
𝑊𝑇
𝑒=
𝑚𝑐𝑜𝑎𝑙 𝐻𝑣
𝑊𝑇
𝑚𝑐𝑜𝑎𝑙 =
𝑒𝐻𝑣
1𝑘𝐽 3600𝑠
(1 × 1015 𝑘𝑊 − ℎ𝑟) ( )( )
𝑚𝑐𝑜𝑎𝑙 = 1𝑘𝑊 − 𝑠 1ℎ𝑟
𝑘𝑐𝑎𝑙 𝑘𝐽
0.35 (20000 ) (4.187 )
𝑘𝑔𝑚 𝑘𝑔𝑚
𝒎𝒄𝒐𝒂𝒍 = 𝟏. 𝟐𝟑 × 𝟏𝟎𝟏𝟒 𝒌𝒈𝒎
b) If nuclear fuel is used instead of coal

𝑒=
𝑊𝑇
𝑒=
𝑄𝐴
𝑊𝑇 𝑘
𝑒=
𝑚𝑁𝐹 𝑐 2
𝑊𝑡 𝑘
𝑚𝑁𝐹 = 2
𝑒𝑐
𝑘𝑔𝑚 − 𝑚 1𝑘𝐽 3600𝑠
(1 × 1015 𝑘𝑊 − ℎ𝑟) (1000 )( )( )
𝑘𝑁 − 𝑠2 1𝑘𝑊 − 𝑠 1ℎ𝑟
𝑚𝑁𝐹 =
𝑚 2
0.35 (3 × 108 )
𝑠
𝒎𝑵𝑭 = 𝟏𝟏𝟒𝟐𝟖𝟓. 𝟕𝟏𝒌𝒈𝒎
119
c) By energy balance on the powerplant.
𝑄𝐴 + 𝑊𝑝 = 𝑊𝑇 + 𝑄𝑅
Since 𝑊𝑝 = 0, then the heat rejected would be
𝑄𝑅 = 𝑄𝐴 − 𝑊𝑇
𝑄𝑅 = 𝑚𝑐𝑜𝑎𝑙 𝐻𝑣 − 𝑊𝑇
𝑘𝑐𝑎𝑙 4.187𝑘𝐽 3600𝑘𝐽 1𝐵𝑇𝑈 1𝐶𝐻𝑈
𝑄𝑅 = [(1.23 × 1014 𝑘𝑔𝑚) (20000 )( ) − (1 × 1015 𝑘𝑊 − ℎ𝑟) ( )] [ ][ ]
𝑘𝑔𝑚 1𝑘𝑐𝑎𝑙 1𝑘𝑊 − ℎ𝑟 1.055𝑘𝐽 1.8𝐵𝑇𝑈
𝑸𝑹 = 𝟑. 𝟓𝟐 × 𝟏𝟎𝟏𝟖 𝑪𝑯𝑼
In terms of PS-min
1𝐵𝑇𝑈 1.055𝑘𝐽 1𝑘𝑊 − 𝑠 1𝑃𝑆 1𝑚𝑖𝑛
𝑄𝑅 = 3.52 × 1018 𝐶𝐻𝑈 ( )( )( )( )( )
1.8𝐶𝐻𝑈 1𝐵𝑇𝑈 1𝑘𝐽 0.736𝑘𝑊 60𝑠
𝑸𝑹 = 𝟏. 𝟓𝟏 × 𝟏𝟎𝟏𝟕 𝑷𝑺 − 𝒎𝒊𝒏
In terms of hp-hr
0.736𝑘𝑊 1ℎ𝑝 1ℎ𝑟
𝑄𝑅 = 1.51 × 1017 𝑃𝑆 − 𝑚𝑖𝑛 ( )( )( )
1𝑃𝑆 0.746𝑘𝑊 60𝑚𝑖𝑛
𝑸𝑹 = 𝟐. 𝟒𝟖 × 𝟏𝟎𝟏𝟓 𝒉𝒑 − 𝒉𝒓
d) From the definition of fluid properties:

𝑚
ρ=
𝑉
ρ
𝑆𝐺 =
ρ𝑤
𝑚
𝑉=
𝑆𝐺ρ𝑤
π
For sphere, 𝑉 = 𝑑 3
6
π 3 𝑚
𝑑 =
6 𝑆𝐺ρ𝑤
Solving for the diameter d:

3 6𝑚
𝑑=√
π𝑆𝐺ρ𝑤
For the diameter of spherical tank to contain nuclear fuel (SG=10)
3 6(114285.71𝑘𝑔𝑚)
𝑑=√
𝑘𝑔
π(10) (1000 3 )
𝑚
𝒅 = 𝟐. 𝟕𝟗 𝒎
For the diameter of spherical tank to contain coal (SG=5)

1
3 6(1.23 × 10 4𝑘𝑔𝑚)
𝑑=√
𝑘𝑔
π(5) (1000 3 )
𝑚
𝒅 = 𝟒𝟖𝟗𝟕. 𝟑𝟐 𝒎
120
e. Doing energy balance on the condenser, taking the cooling water as working substance:
𝐻𝑎 + 𝑄𝑅 = 𝐻𝑏
𝑄𝑅 = 𝑚𝑤 𝑐𝑝𝑤 (𝑡𝑏 − 𝑡𝑎 )
𝑄𝑅 = 𝑚𝑤 𝑐𝑝𝑤 Δ𝑡
𝑄𝑅 = ρ𝑤 𝑉𝑤 𝑐𝑝𝑤 Δ𝑡
𝑄𝑅
𝑉𝑤 =
ρ𝑤 𝑐𝑝𝑤 Δ𝑡
60𝑚𝑖𝑛 42.41𝐵𝑇𝑈 1.055𝑘𝐽
(2.48 × 1015 ℎ𝑝 − ℎ𝑟) ( )( )( )
1ℎ𝑟 1ℎ𝑝 − 𝑚𝑖𝑛 1𝐵𝑇𝑈
𝑉𝑤 =
𝑘𝑔𝑚 𝑘𝐽 1𝑚3 3.785𝐿
(1000 3 ) (4.187 ) (20℃) ( )( )
𝑚 𝑘𝑔𝑚 − ℃ 1000𝐿 1𝑔𝑎𝑙
𝑽𝒘 = 𝟐. 𝟏𝟎 × 𝟏𝟎𝟏𝟔 𝒈𝒂𝒍
f) From the turbine below, the turbine is coupled to the generator. Hence this is called a turbogenerator. The
brake work is the work available at the shaft of the turbine. The power output is the power produced by the
generator due to shaft rotation.
From the given figure

𝑒=
𝑃𝑒
𝑛𝑒𝑒 =
𝑊𝐵
𝑊𝐵
𝑛𝑚𝑒 =
𝑊𝑇
𝑃𝑒 = 𝑛𝑒𝑒 𝑛𝑚𝑒 𝑊𝑇
𝑃𝑒 = (0.85)(0.90)(1 × 1015 𝑘𝑊 − ℎ𝑟)
𝑷𝒆 = 𝟕. 𝟔𝟓 × 𝟏𝟎𝟏𝟒 𝒌𝑾 − 𝒉𝒓
In terms of hp-min
1ℎ𝑝 60𝑚𝑖𝑛
𝑃𝑒 = 7.65 × 1014 𝑘𝑊 − ℎ𝑟 ( )( )
0.746𝑘𝑊 1ℎ𝑟
𝑷𝒆 = 𝟔. 𝟏𝟓𝟑 × 𝟏𝟎𝟏𝟓 𝒉𝒑 − 𝒎𝒊𝒏
121
8. A shaft is applied by a torque of 200 N-m and rotates at a rate of 4000rpm. Find the power transferred by the
shaft in hp and kW.
Solution: From the definition of power

𝑃 = 𝐹𝑣
𝑇
𝑃= 𝑣
𝑟
𝑃 = 𝑇ω
𝑟𝑒𝑣 2π 1𝑚𝑖𝑛
𝑃 = (0.2𝑘𝑁 − 𝑚) (4000 )( )( )
𝑚𝑖𝑛 1𝑟𝑒𝑣 60𝑠
𝑷 = 𝟖𝟑. 𝟕𝟖 𝒌𝑾
In terms of hp
1ℎ𝑝
𝑃 = 83.78𝑘𝑊 ( )
0.746𝑘𝑊
𝑷 = 𝟏𝟏𝟐. 𝟑𝟏 𝒉𝒑
9. A 60 hyl, 53 cm radius flywheel is slowed from 250 to 235 rpm. How much kinetic energy is released?
Solution: From the definition of kinetic energy

𝑚(𝑣2 2 − 𝑣1 2 )
Δ𝐾𝐸 =
2𝑘
But for angular velocity:
𝑣 = 𝑟ω
Hence
𝑚𝑟 2 (ω2 2 − ω1 2 )
Δ𝐾𝐸 =
2𝑘
9.8066𝑘𝑔𝑚 𝑟𝑒𝑣 2 2π 2 1𝑚𝑖𝑛 2
(60ℎ𝑦𝑙) ( ) (0.53𝑚)2 (2352 − 2502 ) ( ) ( )
1ℎ𝑦𝑙 𝑚𝑖𝑛2 1𝑟𝑒𝑣 60𝑠
Δ𝐾𝐸 =
2 (1000 )
𝑘𝑁 − 𝑠2
𝚫𝑲𝑬 = −𝟔. 𝟓𝟎𝟑 𝒌𝑱
Note: A negative sign indicates that the kinetic energy of the flywheel is decreasing. It is due to the fact that as
speed decreases, the kinetic energy also decreases as well.
10. A spring has a natural length of 20cm and it has a length of 40cm when a constant 40N force is applied to it.
How much work is done in stretching the spring from 35 cm to 38 cm?
Solution:
From Hooke’s Law
𝐹 = 𝑘𝑥
𝐹
𝑘=
𝑥
40𝑁
𝑘=
0.4𝑚
𝑘 = 100 𝑁/𝑚
122
Solving for the work done in stretching the spring
1
𝑊 = 𝑘(𝑥2 2 − 𝑥1 2 )
2
1 𝑁
𝑊 = (100 ) (0.382 − 0.352 )𝑚2
2 𝑚
𝑾 = 𝟏. 𝟎𝟗𝟓 𝑱
11. A train weighing 1450 tonne is pulled up a 2% grade by 5 MW engine. Train resistance is 8710kg. What is the
speed of the train in kph?
Solution. The grade of the road is determined as a slope with a rise of 2 and a run of 100. Since a resisting force is
acting on the train parallel to the surface of the road, hence doing summation of forces in the x’ direction to solve
the force required in pulling the train.
∑ 𝐹𝑥′ = 0
𝑅 + 𝑊𝑠𝑖𝑛θ − 𝐹 = 0
𝐹 = 𝑅 + 𝑊𝑠𝑖𝑛θ
The power needed to pull the train is

𝑃 = 𝐹𝑣
The velocity made by the engine in pulling the train is
𝑃
𝑣=
𝐹
𝑃
𝑣=
𝑅 + 𝑊𝑠𝑖𝑛θ
𝑃𝑘
𝑣=
𝑔(𝑚𝑅 + 𝑚 𝑇 𝑠𝑖𝑛θ)
𝑘𝑁 − 𝑚 𝑘𝑔𝑚 − 𝑚 1𝑘𝑚 3600𝑠
(5000 ) (1000 )( )( )
𝑠 𝑘𝑁 − 𝑠2 1000𝑚 1ℎ𝑟
𝑣=
𝑚 2
(9.8066 2 ) [8710 + 1450 × 103 ( 2 )] 𝑘𝑔𝑚
𝑠 √2 + 1002
𝒌𝒎
𝒗 = 𝟒𝟖. 𝟔𝟖
𝒉𝒓
12. Calculate the pump horsepower rating in order to lift water from a 5m deep well to fill 5000 L tank in a
minute
Solution: The power input to the pump is converted into potential energy of the water.
𝑊
𝑃=
𝑇
𝑃𝐸
𝑃=
𝑡
𝑚𝑔ℎ
𝑃=
𝑡
𝑃 = 𝑚̇𝑔ℎ
𝑃 = 𝑉̇ρ𝑔ℎ
𝑃 = 𝑉̇γℎ
𝐿 1𝑚3 𝑘𝑁 1ℎ𝑝 1𝑚𝑖𝑛
𝑃 = (5000 )( ) (9.8066 3 ) (5𝑚) ( )( )
𝑚𝑖𝑛 1000𝐿 𝑚 0.746𝑘𝑊 60𝑠
𝑷 = 𝟓. 𝟒𝟖 𝒉𝒑
123
13. Calculate the power of a steam jet 15mm in diameter moving at 750 m/s. Take density of steam to be 0.794
kg/m3.
Solution: From the steam jet, power input is fully converted into kinetic energy of steam.
𝑊
𝑃=
𝑡
𝐾𝐸
𝑃=
𝑡
𝑚𝑣 2
𝑃=
2𝑘𝑡
𝑚̇𝑣 2
𝑃=
2𝑘
ρ𝑉̇ 𝑣 2
𝑃=
2𝑘
Since 𝑉̇ = 𝐴𝑣
ρAv 3
𝑃=
2𝑘
π
Assuming the cross-sectional area is a circle, then 𝐴 = 𝑑 2, hence
4
πρ𝑑 2 𝑣 3
𝑃=
8𝑘
kgm 𝑚 3
π (0.794 3 ) (0.015𝑚)2 (750 )
m 𝑠
𝑃=
8 (1000 )
𝑘𝑁 − 𝑠2
𝑷 = 𝟐𝟗. 𝟔𝟎 𝒌𝑾
14. The sun generates 1 kW per square meter when used as a source of solar collector. 80% of heat collected is
used to heat 3 liters per min of water. Temperature rises to 4.78°C. Recommend dimensions of a rectangular solar
collector to be used in the process, assuming that length is thrice its width.
Solution: Assuming constant pressure process, the heat added to water is

𝑄𝐴̇ = 𝑚𝑤 ̇ 𝑐𝑝𝑤 Δ𝑡
𝑄𝐴 = ρw V̇𝑐𝑝𝑤 Δ𝑡
̇ (Equation 1)
Since only 80% of the heat collected is used to heat water, the heat transfer of the collector to the water is
̇
𝑄𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑜𝑟 = 𝑞̇ 𝐴
̇
𝑄𝐴 = 0.80𝑄𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑜𝑟
𝑄𝐴̇ = 0.80𝑞̇ 𝐴
For a rectangular solar panel, 𝐴 = 𝐿𝑊. For the given constraints, 𝐿 = 3𝑊, hence the area of the solar panel as a
function of width would be
𝐴 = 3𝑊 2
Thus
𝑄𝐴̇ = (0.80)(3)𝑞̇ 𝑊 2
𝑄𝐴̇ = 2.4𝑞̇ 𝑊 2 (Equation 2)
Equating equations 1 and 2 and solving for the width W:

ρw V̇𝑐𝑝𝑤 Δ𝑡 = 2.4𝑞̇ 𝑊 2
ρw V̇𝑐𝑝𝑤 Δ𝑡
𝑊=√
2.4𝑞̇
124
𝑘𝑔𝑚 𝐿 𝑘𝐽 1𝑚3 1𝑚𝑖𝑛
(1000 3 ) (5 ) (4.187 ) (4.78℃) ( )( )
𝑚 𝑚𝑖𝑛 𝑘𝑔 − ℃ 1000𝐿 60𝑠
𝑊=√
𝑘𝐽
2.4 (1 2 )
𝑚 −𝑠
𝑊 =0.834m
Hence
𝐿 = 3𝑊
𝐿 = 3(0.834𝑚)
𝐿 = 2.502 𝑚
The recommended dimensions for the solar panel would be 2.502m x 0.834 m.
15. A river flowing towards the lake at 3 m/s at a rate of 500 CMS at a location 90 m below the lake surface.
Calculate
a) The total mechanical energy of the river per unit mass
b) The total energy of the river per unit mass.
c) The power generation potential of the entire river at that location.
Solution:
a) The hydrostatic pressure P below the given height is
𝑃𝑔 = γℎ
𝑘𝑁
𝑃𝑔 = (9.8066 3 ) (90𝑚)
𝑚
𝑃𝑔 = 882.594𝑘𝑃𝑎𝑔
Since hydrostatic pressure is gage pressure, we shall need to find the absolute pressure
𝑃 = 𝑃𝑔 + 𝑃𝑎𝑡𝑚
𝑃 = (882.584 + 101.325)𝑘𝑃𝑎𝑎
𝑃 = 983.909 𝑘𝑃𝑎𝑎
a) The total mechanical energy of the river per unit mass is defined as
𝑃 𝑔𝑧 𝑣 2
𝑒𝑚𝑒𝑐ℎ = + +
ρ 𝑘 2𝑘
𝑘𝑁 𝑚 𝑚 2
882.594 2 (9.8066 2 ) (90𝑚) (3 )
𝑒𝑚𝑒𝑐ℎ = 𝑚 + 𝑠 + 𝑠
𝑘𝑔𝑚 𝑘𝑔𝑚 − 𝑚 𝑘𝑔𝑚 − 𝑚
1000 3 1000 2 (1000 )
𝑚 𝑘𝑁 − 𝑠2 𝑘𝑁 − 𝑠 2
𝒌𝑱
𝒆𝒎𝒆𝒄𝒉 = 𝟏. 𝟕𝟕
𝒌𝒈𝒎
b) The total energy of the river per unit mass is

𝑔𝑧 𝑣 2
𝑒𝑡𝑜𝑡𝑎𝑙 = + +𝑢
𝑘 2𝑘
Neglecting specific internal energy, the total energy per unit mass basis is:
𝑔𝑧 𝑣 2
𝑒𝑡𝑜𝑡𝑎𝑙 = +
𝑘 2𝑘
125
𝑚 𝑚 2
(9.8066 2 ) (90𝑚) (3 )
𝑒𝑡𝑜𝑡𝑎𝑙 = 𝑠 + 𝑠
𝑘𝑔𝑚 − 𝑚 𝑘𝑔𝑚 − 𝑚
1000 2 (1000 )
𝑘𝑁 − 𝑠2 𝑘𝑁 − 𝑠2
𝒌𝑱
𝒆𝒕𝒐𝒕𝒂𝒍 = 𝟎. 𝟖𝟗
𝒌𝒈𝒎
c) The potential energy of the river is fully converted into power. Hence
𝑊
𝑃=
𝑇
𝑃𝐸
𝑃 =
𝑡
𝑚𝑔ℎ
𝑃 =
𝑡
𝑃 = 𝑚̇𝑔ℎ
𝑃 = 𝑉̇ρ𝑔ℎ
𝑃 = 𝑉̇γℎ
𝑚3 𝑘𝑁
𝑃 = (500 ) (9.8066 3 ) (90𝑚)
𝑠 𝑚
𝑷 = 𝟒𝟒𝟏𝟐𝟗𝟕 𝒌𝑾
16. A 20 kg mass is initially 15m above the free surface of the water tank. Initially the stone and water are at the
same temperature. If the stone falls unto the water, determine ΔU, ΔPE, ΔKE, Q, and W when
a) The stone is about to enter the water
b) The stone has come to rest in the tank
c) The heat is transferred to the surroundings in such an amount that the stone and water comes from their
initial temperature
Solution:
a) Applying energy balance for closed system.
𝑃𝐸1 + 𝐾𝐸1 + 𝑈1 + 𝑄 = 𝑃𝐸2 + 𝐾𝐸2 + 𝑈2 + 𝑊𝑁𝐹
The non-flow work would be
Since the stone is about to enter the water, there is no heat, no work done, and no change in internal energy. Thus:
𝑸 = 𝑾𝑵𝑭 = 𝚫𝑼 = 𝟎
From the energy balance,
Δ𝑃𝐸 = −Δ𝐾𝐸
The change in potential energy of the stone would be
𝑚𝑔Δ𝑧
Δ𝑃𝐸 =
𝑘
𝑚
(20𝑘𝑔𝑚) (9.8066 2 ) (0 − 15)𝑚
Δ𝑃𝐸 = 𝑠
1
𝑁 − 𝑠2
𝚫𝑷𝑬 = −𝟐𝟗𝟒𝟏. 𝟗𝟖 𝑱
126
The change in kinetic energy is:
𝚫𝑲𝑬 = 𝟐𝟗𝟒𝟏. 𝟗𝟖 𝑱
b) When the stone has come to rest in the tank , there is no heat, no work done, no change in kinetic energy but as
the stone comes in contact with water, the internal energy of the water increases by the magnitude of the potential
energy of the stone.
From the energy balance
𝑃𝐸1 + 𝐾𝐸1 + 𝑈1 + 𝑄 = 𝑃𝐸2 + 𝐾𝐸2 + 𝑈2 + 𝑊𝑁𝐹
Since
𝑸 = 𝑾𝑵𝑭 = 𝚫𝑲𝑬 = 𝟎
Then
Δ𝑈 = −Δ𝑃𝐸
The change in potential energy of the stone would be
𝑚𝑔Δ𝑧
Δ𝑃𝐸 =
𝑘
𝑚
(20𝑘𝑔𝑚) (9.8066 2 ) (0 − 15)𝑚
Δ𝑃𝐸 = 𝑠
1
𝑁 − 𝑠2
𝚫𝑷𝑬 = −𝟐𝟗𝟒𝟏. 𝟗𝟖 𝑱
The change in internal energy would be:
𝚫𝑼 = 𝟐𝟗𝟒𝟏. 𝟗𝟖 𝑱
c) When the water and stone come from their initial temperature, there is no work done, and zero change in
kinetic energy. The heat is equivalent to the internal energy, which is equal also to the potential energy as the
stone comes to rest.
By energy balance
𝑃𝐸1 + 𝐾𝐸1 + 𝑈1 + 𝑄 = 𝑃𝐸2 + 𝐾𝐸2 + 𝑈2 + 𝑊𝑁𝐹
Since
𝑾𝑵𝑭 = 𝚫𝑲𝑬 = 𝟎
Since the potential energy is converted to internal energy:
𝚫𝑼 = −𝚫𝑷𝑬 = 𝟐𝟗𝟒𝟏. 𝟗𝟖 𝑱
Hence
𝑄 = −Δ𝑈
𝑸 = −𝟐𝟗𝟒𝟏. 𝟗𝟖 𝑱
127
17. A fluid enters an apparatus at 450 fps, initially, the pressure of the fluid is 120 psia, the specific volume of 5
cu. ft. per lb., and the internal energy is 390 BTU/lb. The fluid leaves the apparatus at 25 psia, specific volume of
20 cu. ft. per lb., an exit velocity of 1000 fps, and internal energy of 120 BTU/lb. The heat radiation loss is 10
BTU/lb. Determine the work steady flow.
By energy balance on per unit mass basis:
𝑒𝑖𝑛 = 𝑒𝑜𝑢𝑡 Δ𝑒 = 0
𝑝𝑒1 + 𝑘𝑒1 + 𝑤𝑓1 + 𝑢1 = 𝑞𝑟 + 𝑝𝑒2 + 𝑘𝑒2 + 𝑤𝑓2 + 𝑢2 + 𝑤𝑆𝐹
The steady flow work in energy per unit mass basis would be
𝑤𝑆𝐹 = −𝑞𝑟 − Δ𝑝𝑒 − Δ𝑘𝑒 − Δ𝑤𝑓
𝑔(𝑧2 − 𝑧1 ) 𝑣2 2 − 𝑣1 2
𝑤𝑆𝐹 = [−𝑞𝑟 − − − (𝑃2 𝑣2 − 𝑃1 𝑣1 ) − (𝑢2 − 𝑢1 )]
𝑘 2𝑘
Neglecting change in potential energy, Δ𝑝𝑒 = 0

𝑣2 2 − 𝑣1 2
𝑤𝑆𝐹 = [𝑞 − − (𝑃2 𝑣2 − 𝑃1 𝑣1 ) − (𝑢2 − 𝑢1 )]
2𝑘
2
2 2 𝑓𝑡
𝐵𝑇𝑈 (1000 − 450 ) 𝑠2 𝑙𝑏𝑓 𝑓𝑡 3 144𝑖𝑛2 1𝐵𝑇𝑈
𝑤𝑠𝑓 = −10 − − [25(20) − 120(5)] 2 ( )( )
𝑙𝑏𝑚 𝑙𝑏𝑚 − 𝑓𝑡 𝑖𝑛 𝑙𝑏𝑚 𝑓𝑡 2 778.16𝑓𝑡 − 𝑙𝑏𝑓
2 (32.174 )
𝑩𝑻𝑼
𝒘𝑺𝑭 = 𝟐𝟔𝟐. 𝟓𝟖
𝒍𝒃𝒎
18. A turbine operates under steady flow conditions, receiving steam at the following state: pressure 1000 kPa,
temperature 180 deg. C, enthalpy 2700 kJ/kg, speed 50 m/s, and elevation 2 m. The steam leaves the turbine at the
following state: pressure 50 kPa, enthalpy 2500 kJ/kg, speed 100 m/s, and elevation 0 m. Heat is lost to the
surroundings at the rate of 0.50 kJ/s. If the rate of steam flow through turbine is 0.50 kg/s, what is the power
output of the turbine in kW.
Solution: By energy balance in the turbine.
𝑃𝐸1 + 𝐾𝐸1 + 𝑊𝑓1 + 𝑈1 = 𝑄𝑅 + 𝑃𝐸2 + 𝐾𝐸2 + 𝑊𝑓2 + 𝑈2 + 𝑊𝑆𝐹
𝑃𝐸1 + 𝐾𝐸1 + 𝐻1 = 𝑄𝑅 + 𝑃𝐸2 + 𝐾𝐸2 + 𝐻2 + 𝑊𝑆𝐹
𝑊𝑆𝐹 = −𝑄𝑅 − Δ𝑃𝐸 − Δ𝐾𝐸 − Δ𝐻

𝑔(𝑧2 − 𝑧1 ) 𝑣2 2 − 𝑣1 2
𝑊𝑆𝐹 = −𝑄𝑅̇ − 𝑚̇ [ + + (ℎ2 − ℎ1 )]
𝑘 2𝑘
𝑚 𝑚𝑠
kJ 𝑘𝑔𝑚 (9.8066 𝑠2 ) (0 − 2)𝑚 (502 − 1002 ) 2 𝑘𝐽
𝑊𝑆𝐹 = −0.50 − (0.50 )[ + 𝑠 + (2500 − 2700) ]
s 𝑠 𝑘𝑔𝑚 − 𝑚 𝑘𝑔𝑚 − 𝑚 𝑘𝑔𝑚
10000 2 (10000 )
𝑘𝑁 − 𝑠2 𝑘𝑁 − 𝑠2
̇ = 𝟏𝟎𝟏. 𝟑𝟖𝟓𝒌𝑾
𝑾𝑺𝑭
128
19. A steady state steady flow compressor draws 500CFM of air whose density is 0.079 lb/cu.ft and discharge it
with a density of 0.304 lb/cu.ft. At suction, pressure is at 15 psia, at discharge, pressure is at 80 psia. The increase
in specific internal energy 33.8 BTU/lb. The heat comes from the air cooling is 13 BTU/lb. Neglect changes in
potential and kinetic energy., determine the work done on the air in BTU/min.
Solution:
The mass of air entering the compressor is
𝑚̇ = 𝑚̇ 1 = 𝜌1 𝑉1̇ (Equation 1)
By energy balance
𝑃𝐸1 + 𝐾𝐸1 + 𝑊𝑓1 + 𝑈1 += 𝑄𝑅 + 𝑃𝐸2 + 𝐾𝐸2 + 𝑊𝑓2 + 𝑈2 + 𝑊𝑆𝐹
𝑊𝑆𝐹 = −𝑄𝑅 − Δ𝑃𝐸 − Δ𝐾𝐸 − Δ𝑊𝑓 − Δ𝑈
Neglecting change in kinetic and potential energy, then the steady flow work would be:
𝑊𝑆𝐹 = −𝑚̇[𝑞 + (𝑃2 𝑣2 − 𝑃1 𝑣1 ) + Δ𝑢 ] (Equation2)
Substituting equation 1 to equation 2, then

𝑊𝑆𝐹 = −𝜌1 𝑉1̇ [𝑞 + (𝑃2 𝑣2 − 𝑃1 𝑣1 ) + Δ𝑢 ]
𝑃2 𝑃1
𝑊𝑆𝐹 = −𝜌1 𝑉1̇ [𝑞 + ( − ) + Δ𝑢 ]
ρ2 ρ1
𝑙𝑏𝑚 ft 3 𝐵𝑇𝑈 80 15 𝑙𝑏𝑓 𝑓𝑡 3 144𝑖𝑛2 1𝐵𝑇𝑈
𝑊𝑆𝐹 = − (0.079 3 ) (500 ) [13 +( − ) ( )( )
𝑓𝑡 min 𝑙𝑏𝑚 0.304 0.079 𝑖𝑛2 𝑙𝑏𝑚 1𝑓𝑡 2 778.16𝑓𝑡 − 𝑙𝑏𝑓
BTU
+ 33.8 ]
lbm
𝑩𝑻𝑼
𝑾𝑺𝑭 = −𝟐𝟑𝟖𝟒. 𝟐𝟖
𝒍𝒃𝒎
20. 15 lbm/min of steam passes through a turbine at a specific enthalpy of 15000 BTU/lbm, 30 fps velocity and a
pressure of 10 kip/ft2. It leaves at the exit section 10 ft from the entrance section with a specific enthalpy of 10000
BTU/lbm and 20 fps velocity. If the heat lost from the turbine is 15% of the turbine work, and the mechanical and
electrical efficiencies are 80% and 90% respectively, determine the power developed in the turbine in SI units.
Solution. By energy balance:

𝑃𝐸1 + 𝐾𝐸1 + 𝐻1 = 𝑄𝑅 + 𝑃𝐸2 + 𝐾𝐸2 + 𝐻2 + 𝑊𝑆𝐹
The steady flow work is

Since 𝑄𝑅 = 0.15𝑊𝑆𝐹 , then

𝑊𝑆𝐹 = −0.15𝑊𝑆𝐹 − Δ𝑃𝐸 − Δ𝐾𝐸 − Δ𝐻
1.15𝑊𝑆𝐹 = −Δ𝑃𝐸 − Δ𝐾𝐸 − Δ𝐻
−Δ𝑃𝐸 − Δ𝐾𝐸 − Δ𝐻
𝑊𝑆𝐹 =
1.15
𝑚 𝑔(𝑧2 − 𝑧1 ) 𝑣2 2 − 𝑣1 2
𝑊𝑆𝐹 = − [ + + (ℎ2 − ℎ1 )] (Equation1)
1.15 𝑘 2𝑘
129
Since the turbine is coupled to a generator having mechanical and electrical efficiencies, the power output would
be
𝑒=
𝑃
𝑛𝑒𝑒 =
𝑊𝐵
𝑊𝐵
𝑛𝑚𝑒 =
𝑊𝑇
𝑃 = 𝑛𝑒𝑒 𝑛𝑚𝑒 𝑊𝑇 (Equation 2)
Since 𝑊𝑇 = 𝑊𝑆𝐹 , then substitute equation 1 to 2:
−𝑛𝑒𝑒 𝑛𝑚𝑒 𝑚 𝑔(𝑧2 − 𝑧1 ) 𝑣2 2 − 𝑣1 2

𝑃= [ + + (ℎ2 − ℎ1 )]
1.15 𝑘 2𝑘
𝑙𝑏𝑚 (32.174 𝑓𝑡 ) (0 − 10)𝑓𝑡 𝑓𝑡 2
−(0.80)(0.90) (15 ) 𝑠 2 (202 − 302 ) 2 1𝐵𝑇𝑈
𝑃= 𝑚𝑖𝑛 [ + 𝑠 ][ ]
1.15 𝑙𝑏𝑚 − 𝑓𝑡 𝑙𝑏𝑚 − 𝑓𝑡 778.16𝑓𝑡 − 𝑙𝑏𝑓
32.174 2 (32.174 )
[ 𝑙𝑏𝑓 − 𝑠2 𝑙𝑏𝑓 − 𝑠2
𝐵𝑇𝑈 0.746𝑘𝑊
+ (10000 − 15000) ( )
𝑙𝑏𝑚 42.41 𝐵𝑇𝑈
𝑚𝑖𝑛
]
The electrical power available would be:

𝑷 = 𝟖𝟐𝟓. 𝟗𝟖𝒌𝑾
21. A steady flow steady state thermodynamic system receives 100 lb/min of a fluid at 30 psia and 20°F and
discharges at a point 80 ft above the entrance section at 150 psia and 600°F. The fluid enters with a velocity of 7200
fpm and leaves with a velocity of 2400 fpm. During this process, there are supplied 25000BTU/hr of heat from an
external source, and the increase in enthalpy is 2 BTU/lb. Determine the work done in hp and kgf-m/min.
Solution: By energy balance on the turbine:
𝑃𝐸1 + 𝐾𝐸1 + 𝐻1 + 𝑄𝐴 = +𝑃𝐸2 + 𝐾𝐸2 + 𝐻2 + 𝑊𝑆𝐹
𝑊𝑆𝐹 = 𝑄𝐴 − Δ𝐻 − Δ𝑃𝐸 − Δ𝐾𝐸
𝑔(𝑧2 − 𝑧1 ) 𝑣2 2 − 𝑣1 2
𝑊𝑆𝐹 = 𝑄𝐴 − 𝑚̇ [Δℎ + + ]
𝑘 2𝑘
𝐵𝑇𝑈 1ℎ𝑟
𝑊𝑆𝐹 = 25000 ( )
ℎ𝑟 60𝑚𝑖𝑛𝑠
𝑓𝑡 2 2
lbm BTU (32.174 2 ) (80 − 0)𝑓𝑡 (24002 − 72002 ) 𝑓𝑡 2 (1𝑚𝑖𝑛) 1𝐵𝑇𝑈
𝑠 𝑚𝑖𝑛 60𝑠 ] [
− (100 ) 2 +[ + ]
min lbm 𝑙𝑏𝑚 − 𝑓𝑡 𝑙𝑏𝑚 − 𝑓𝑡 778.16𝑓𝑡 − 𝑙𝑏𝑓
32.174 2 2 (32.174 2 )
[ 𝑙𝑏𝑓 − 𝑠 𝑙𝑏𝑓 − 𝑠 ]
𝐵𝑇𝑈
𝑊𝑆𝐹 = 92231.96
𝑚𝑖𝑛
In terms of hp:
130
𝐵𝑇𝑈 1ℎ𝑝
𝑊𝑆𝐹 = 92231.96 ( )
𝑚𝑖𝑛 42.41 𝐵𝑇𝑈
𝑚𝑖𝑛
𝑾𝑺𝑭 = 𝟐𝟏𝟕𝟒. 𝟕𝟕𝒉𝒑
In terms of kgf-m/min
𝑘𝑔𝑓 − 𝑚
0.746𝑘𝑊 1𝑃𝑆 75 60𝑠
𝑊𝑆𝐹 = 2174.77ℎ𝑝 ( )( )( 𝑠 )( )
1ℎ𝑝 0.736𝑘𝑊 1𝑃𝑆 1𝑚𝑖𝑛
𝒌𝒈𝒇 − 𝒎
𝑾𝑺𝑭 = 𝟗. 𝟗𝟐 × 𝟏𝟎𝟔
𝒎𝒊𝒏
22. What is the temperature change in dropping water into a 15m waterfalls in kelvin?
Solution: By energy balance, assuming that no heat is added or rejected by the water, and there is no shaft work
present in the system, Neglect also change in kinetic energy.
𝑃𝐸1 + 𝐻1 = 𝑃𝐸2 + 𝑃𝐸2
Δ𝐻 = Δ𝑃𝐸
Expressing enthalpy in terms of specific heat at constant pressure (microanalysis), then

Δ𝐻 = Δ𝑃𝐸
𝑚𝑤 𝑔(𝑧2 − 𝑧1 )
𝑚𝑤 𝑐𝑝𝑤 Δ𝑇 =
𝑘
Solving for Δ𝑇
𝑔(𝑧1 − 𝑧2 )
Δ𝑇 =
𝑘𝑐𝑝𝑤
𝑚
(9.8066
𝑠 2 ) (15 − 0)𝑚
Δ𝑇 =
𝑘𝑔𝑚 − 𝑚 𝑘𝑁 − 𝑚
(1000 ) (4.187 )
𝑘𝑁 − 𝑠2 𝑘𝑔𝑚 − 𝐾
𝚫𝑻 = 𝟎. 𝟎𝟑𝟓𝟏𝟑 𝑲
23. A closed gaseous system undegoes a reversible process during which 25 BTU of heat are rejected. The
volume changes from 5 to 2 cubic foot, and the pressure remains at 50psia, find the change in internal energy.
131
Solution: By energy balance
𝑊𝑁𝐹 = 𝑄𝑅 + Δ𝑈
Since the work non flow can be determine as the area under the PV diagram:
2
𝑊𝑁𝐹 = ∫ 𝑃𝑑𝑉
1
Since the pressure is constant, then
2
𝑊𝑁𝐹 = 𝑃 ∫ 𝑑𝑉
1
Integrating yields:
𝑊𝑁𝐹 = 𝑃(𝑉2 − 𝑉1 )
Hence the change in internal energy would be
Δ𝑈 = 𝑊𝑁𝐹 − 𝑄𝑅
Δ𝑈 = 𝑃(𝑉2 − 𝑉1 ) − 𝑄𝑅
𝑙𝑏𝑓 3
144𝑖𝑛2 1𝐵𝑇𝑈
(5
Δ𝑈 = (50 2 ) − 2)𝑓𝑡 ( ) ( ) − 25𝐵𝑇𝑈
𝑖𝑛 1𝑓𝑡 2 778.16𝑓𝑡 − 𝑙𝑏𝑓
𝚫𝑼 = 𝟐. 𝟕𝟓𝟖 𝑩𝑻𝑼
24. During a reversible process executed by a non flow system, the pressure increases from 344.7 kPa to 1378.96
kPa in accordance to PV=C. If the initial volume is 85 liters, and the internal energy increases by 22577J, Find the
heat transfer in kJ.
Solution: By energy balance, since an increase in pressure means work is done to compressed the fluid, and is
need to reject heat.
𝑊𝑁𝐹 = 𝑄𝑅 + Δ𝑈
Since the work non flow can be determine as the area under the PV diagram:
2
1
From the given law, 𝑃𝑉 = 𝐶, also
𝑃1 𝑉1 = 𝑃2 𝑉2 = 𝐶
𝑉2 𝑃1
=
𝑉1 𝑃2
𝐶
𝑃=
𝑉
Integrating
2
𝑑𝑉
𝑊𝑁𝐹 = 𝐶 ∫
1 𝑉
𝑊𝑁𝐹 = 𝐶[𝑙𝑛𝑉]12
𝑊𝑁𝐹 = 𝐶(𝑙𝑛𝑉2 − 𝑙𝑛𝑉1 )
𝑉2
𝑊𝑁𝐹 = 𝐶𝑙𝑛
𝑉1
Since 𝐶 = 𝑃1 𝑉1 = 𝑃2 𝑉2 , then
𝑉2
𝑊𝑁𝐹 = 𝑃1 𝑉1 𝑙𝑛
𝑉1
132
𝑉 𝑃
Since 2 = 1, then the non flow work would be:
𝑉1 𝑃2
𝑃1
𝑃2
This the heat transfer would be
𝑄𝑅 = 𝑊𝑁𝐹 − 𝛥𝑈
𝑃1
𝑄𝑅 = 𝑃1 𝑉1 𝑙𝑛 − Δ𝑈
𝑃2
𝑘𝑁 1𝑚3 344.7𝑘𝑃𝑎𝑎
𝑄𝑅 = (344.7 2 ) (85𝐿) ( ) 𝑙𝑛 − 22.577𝑘𝐽
𝑚 1000𝐿 1378.98𝑘𝑃𝑎𝑎
𝑸𝑹 = −𝟔𝟑. 𝟏𝟐𝟏𝒌𝑱
25. A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal
energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does
100 kJ of work on the fluid. Determine the final internal energy of the fluid. Neglect the energy stored in the
paddle wheel.
By energy balance, assuming that the system considering is a closed system:

𝑊𝑃𝑎𝑑𝑑𝑙𝑒 + 𝑈1 = 𝑄𝑅 + U2
Solving for the final internal energy:

𝑈2 = 𝑊𝑃𝑎𝑑𝑑𝑙𝑒 + 𝑈1 − 𝑄𝑅
𝑈2 = (100 + 800 − 500)𝑘𝐽
𝑼𝟐 = 𝟒𝟎𝟎𝒌𝑱
133
SUPPLEMENTARY PROBLEMS___________________
CHAPTER II. ENERGY AND POWER CONCEPTS
Directions: Solve the following problems and show all complete solutions. Any form of erasures in your solutions
will consider your whole answer as wrong. Always enclosed your final answer. Answers with wrong units are
considered as wrong. Take you final answer atleast two or three decimal places.
Conservation of Mass
1. A 0.70 g/cc liquid flows through a 5 cm inside 9. A 9.95 ft diameter by 14.96 ft height is receiving
diameter pipe at 8 m/s. Determine the mass flow rate water at a rate of 299 GPM and discharging through
in kg/s. a 5.49 in I.D. line with a constant velocity of 5.1 fps.
Ans: 11 kg/s At a given instant, the tank is half full. Find the water
2. How many gallons of water per minute is flowing level in cm and the mass change in the tank in lbm
through a 10 in inside diameter tube at a speed of 10 13 mins later.
m/s? Ans: 165.85 cm, 9883.874 lbm
Ans:8032.35 gpm 10. Fluid flows in a steady flow manner through a
3. If it takes 12 seconds to empty a one-liter oil converging tube. At inlet, the pressure is 690 kPaa
container, then what is the mass flow rate of oil and a constant density of 0.838 kgm/m 3. If 125
leaving the container? Take the density of oil to be m3/min of air enters at the rate of 94 meters per
850 kg/m3. minute, and the exit section has an inside diameter of
Ans: 0.0708 kg/s 350 mm, determine:
4. A pump discharges 20 kg/min of diesel of relative a) The mass flow rate in kg/min.
density of 0.8 to a spherical tank. It takes 1 hour, 20 b) The diameter of the entrance section in
minutes, and 36 seconds to fill the said tank. What is mm.
the diameter of the tank in meters? c) The exit velocity in meters per minute
Ans: 1.567 m Ans: a) 104.75 kg/min, b)1301.21 mm, c) 1299.32
5. A horizontal pipe of cross section 8 cm 2 has a kg/min
constriction of cross section of 2 cm2. Gasoline flows
in the larger pipe with a speed at 10 cm/s. What is the Energy and Mass Relation
speed in the constriction in cm/s?
Ans: 40 cm/s 11. It is estimated that the annual electrical energy
6. Air flows through the compressor under steady consumption in the country is 1.58 x 10 17 J.
flow conditions. At suction, specific volume is 0.75 Determine how many kilograms of matter would
cubic meter per kg, velocity of 30 m/s. At discharge, have to be converted to produce this much energy?
0.25 cubic meter per kg of specific volume, and Ans: 1.76 kgm
velocity of 40 m/s. Determine the diameter ratio of 12. It is estimated that the United States consumes
the discharge pipe to the suction pipe. annually about 1.75 x 1015 W-hr of electrical energy.
Ans:𝒅𝟐 /𝒅𝟏 = 𝟎. 𝟓𝟎 How many kilograms of matter would have to be
7. Water is being drained through a hole at the destroyed to yield this energy?
bottom of a tank at a velocity of 10m/s. For a tank Ans: 1.17 kgm
diameter of 1.18m and a hole diameter of 5cm, find 13. Scientist have recently developed a powerful
the rate of change of water level in the tank in meters pulse laser for research in materials. Find its energy
per minute. output in each of the following conditions:
Ans: 1.08 m/min a) 20 J in 10 psec in hp.
8. Determine the volume flow rate of oil as it flows b) 200J in 35 nsec in BTU/min and kgf-m/min.
at 1.75 m/s through 20 tubes, each tube having 20mm c)800J in 1msec in CV
inside diameter. d)At its peak pulse, it will produce 10 terawatts for
Ans:0.011 m3/s a period of 10 psec, find its discharge in hp-hr and
CHU.
Ans: a) 2.68x109 hp, b) 3.25x108 BTU/min, 3.49 x
1010 kgf-m/min, c) 1086.96 CV, d) 0.0396 hp-hr
134
14. A system contains a unit gram mass of matter. 24. A 10-kg body falls from rest, with negligible
How much energy in kJ may be derived from this interaction with its surroundings (no friction).
system if all of the mass could be converted into Determine its velocity after it falls 5 m.
energy? What will be its mass in grams when its
moves at one half-light speed? 25. Compute the kinetic energy of a 3lb mass, when
Ans: 300kJ, 1.1547 gm it falls in air from rest through a height of 15500in in
15. A cryogenic picovoltmeter using liquid helium BTU.
reads 4 pV when used in a circuit upon which 80mA
has been impressed. Find the resistance of the circuit 26. A 50 lb mass has a potential energy of -4 BTU
in pΩ and power in fW, BTU/min, and metric hp. with respect to a given datum within the earth’s
Ans: 50 p𝛀, 320 fW, standard gravitational field.
16. An electron has a rest mass of 9.11 x 10-28 gm. a) Find its height in ft relative to the datum.
What is its mass when moving with a speed of 0.90c? b) If the gravitational field is suddenly disturbed
Ans: 2.09x10-27 gm such that the local gravity becomes 25 fps2, what will
17. A nuclear bomb containing plutonium releases be the potential energy of the mass in BTU?
1.62 x 1014 J of energy during a 1 microsecond Ans:
explosion. If the rest mass of the products is less than 27. A 300 lbm hammer or a pile driver is lifted 3ft
the original mass by 1/10000, determine: above a piling head. Local g=32.0 ft/s2.
a) The power produced by the explosion in a) What is the change of potential energy in kJ,
BTU/min, kW, and hp. BTU, and CHU?
b) The original mass of the plutonium in lbm. b) If the hammer is released, what will be its
Ans: a) 9.2 x 1018 BTU/min, b) 39.69 lbm velocity in m/s at the instant it strikes the piling?
Potential and Kinetic Energy
28. A girl weighing 470N holds suspended on the end
18. A rocket with mass of 3500 lb travels at 27860 of a rope 8m long.
fps. Determine its kinetic energy in CHU a) What will be her gain in potential energy in
Ans: 1081.86 CHU BTU when a friend swings her to one side so that
19. A diver hits the water 5 meters deeps from a the rope makes an angle of 35 degrees with the
certain height 0.79 seconds later. At what height did vertical?
the diver jumped? b) If local g=9.65 m/s2, what is her mass in kg
and lb?
20. A ball is thrown vertically upward from the Ans: a) 0.65 BTU, b) 48.45 kgm, 106.83 lbm
ground and a student gazing out the window sees it 29. There are 400 kg/min of water being handled by
moving upward pass him at 5m/s. The window is a pump. The lift is from a 20 m deep well and the
10m above the ground. How high does the ball go delivery velocity is 15m/s. Local g=9.70 m/s2.Find:
above the ground? a) The change in potential energy in kJ/kgm,
Ans: 11.27 m BTU/lbm, and kgf-m/kgm.
21. A 200 gram apple is thrown from the edge of a b) The kinetic energy in CHU/lbm and erg/gm?
tall building with an initial speed of 20m/s. What is c) The required power of the pumping unit in hp,
the change in kinetic energy in hp-sec if it strikes the BTU/hr, and kgf-m/s
ground at 50 m/s? What is the height of the building Ans. a) 0.194 kJ/kgm, b)0.02687 CHU/lbm, c)
in ft? 30. When an automobile is travelling at 60 km/hr, its
Ans: 0.2815 hp-s, 351.30 ft engine is developing 25 hp.
22 The combined mass of a car and its passenger a) Find the total resisting force in N.
travelling at 72 kph is 1500kgm. Find the kinetic b) Find the total resisting force in N if the road is
energy of the combined mass in kW-hr. 2% graded.
Ans. 5 kW-hr c) Assuming that the resisting force is
23. A car of mass 2000 kg travels with a velocity of proportional to the speed, what horsepower must
75 km/h. Find the kinetic energy. How high should it the engine develop to drive the automobile at
be lifted in the standard gravitational field to have a 100km/hr.
potential energy that equals the kinetic energy?
135
31. A 2000 kg elevator accelerated upward uniformly c) The change in flow energy in ft-lbf/min, hp-
at 1m/s2 from a stop position. Local g=9.70 m/s2. hr/min, PS-min/hr, kcal/hr, kW, kgf-m/hr, and
a) What is the tension of the lifting cable in N? atm-ft3/min.
b) At the end of 4s operation, what will be the
kinetic and change in potential energy , both in Internal Energy
kJ and also in BTU?
38. An insulated tank initially contains 0.25 kg of gas
32. A system is composed of a 10000 lb elevator with an internal energy of 200 kJ/kg. Additional gas
moving downward with 5fps and a 6000lb with an internal energy of 300 kJ/kg and enthalpy of
counterweight moving upward at 5 fps, and a braking 400 kJ/kg enters the tank until the total mass of gas
pulley with connecting cables. Assume the kinetic contained is 1 kg. Compute for the final internal
energy of the cable and rotating parts to be energy in kJ/kg of the gas in the tank
negligible, determine the frictional energy absorbed
by the break when the elevator is uniformly stopped 39. In an internal combustion engine, during the
in 4ft in ft-lbf and BTU. compression stroke the heat rejected to the cooling
water is 50 kJ/kg and the work input is 100 kJ/kg.
33. A 64000 lbm airplane is traveling at 1000fps. Compute the change in the internal energy of the
a) How much is the kinetic energy in hp-hr? working fluid.
b) If it suddenly noses vertically upward at this
speed at g=32 ft/s2, through what vertical distance 40. A solar collector receiving solar radiation at the
will it moves in miles? rate of 0.6 kW/m2 transforms it to the internal energy
of a fluid. The fluid is heated from 313K to 350 K to
34. There are required 33.76kJ of gravitational work run a heat engine. If the heat engine has efficiency of
to elevate a mass 76.22m in the earth’s gravitational 50% and is to deliver 2.5 kW power, what is the
field where the local gravity is 9.75 m/s2 minimum area of the solar collector?
a) Find its mass in kg.
b) If the initial potential energy of the mass was 41. A body containing 0.5 kg of gas whose cv=0.98
10551Nm with respect to the earth’s surface, kJ/kg-K falls from a balloon 4 km above the earth’s
determine the final elevation in meters above surface. Find the temperature rise of the gas when the
the surface. box hits the ground
Flow Energy 42. The internal energy of a system is a function of

temperature given by the formula E=30+0.27t in kJ.
35. The flow energy of a moving fluid at 10 barrels If this system executes a process for which the work
per minute is 5hp.What is the vacuum pressure in done by it per degree temperature increases is 0.75
psia at this point? kN-m, what is the heat interaction per degree
temperature increases in kJ?
36. A pump forces 1 GPM of water horizontally from
an open well to a closed tank where the pressure is 43. The specific internal energy of a certain system
0.01 MPa. Compute the work the pump must do upon is given by the equation 𝑢 = 66.77 + 0.30(𝑡 −
the water in an hour just to force the water into the 100) + 0.0078(𝑣 − 0.0161) where t is in °F and v
tank against the pressure. is in cu.ft / lbm. If pressure is constant at 100 psig
and volume changes from 0.01608 to 0.01658 cu.ft /
37. A centrifugal air compressor compresses 200
lbm, then the temperature changes from 100°F to
CFM from 12 psia to 90 psia. The initial and final
200°F. Find the values of specific heat at constant
specific volumes are 12.6 and 3.25 ft3/lbm
respectively. If the inlet suction line is 4 in inside pressure and specific heat at constant volume
diameter and the discharge line is 25 in inside
diameter. Determine
a) The flow rate in lb/min and bbl/min.
b) The change in velocity in fps.
136
44. Thermally insulated battery is being discharged Heat

at atmospheric pressure and constant volume. During
a 1 hour test it is found that a current of 50 A and 2 49. If 2BTU of heat is added to 15 lb of water,
V flows while the temperature increases from 20°C determine the temperature rise in C
to 32.5°C. Find the change in internal energy of the
cell during the period of operation. 50. Water initially at 35°C flowing through a pipe
losses 14 kW of heat to the surrounding so that its
temperature is reduced by 28°C. Determine the mass
45. A 0.8-lbm object traveling at 200 ft/sec enters a
of water flow in kg/s.
viscous liquid and is essentially brought to rest
before it strikes the bottom. What is the increase in
51. A 4 kW 20L water heater is switched on for 10
internal energy, taking the object and the liquid as the
minutes. The heat capacity of water is 4.187 kJ/kg-
system? Neglect the potential energy change.
K. Assuming total electrical energy has gone into
heating the water, find the increase in water
46. A vertical piston cylinder arrangement has a
temperature.
piston mass of 50 kg with a face area of 300cm2,
containing 10 g of air. The initial volume occupied
52. An electric storage battery which can exchange
by the air is 5L. A decrease in internal energy
heat goes through a complete cycle of two processes.
amounting to 4kJ occurs because of the temperature
In process 1 to2, a 2.8 kW-h of electrical work-flow
difference between the surroundings and the volume
into the battery while 732 kJ of heat flow out to the
inside the cylinder that caused the volume to
atmosphere. During process 2-1, 2.4 kW-h of work
decrease by 2L. The atmospheric pressure at the top
flow out the battery. What is the heat transfer from
of the piston is 101.325 kPa. Neglect friction effects
process 2 to 1?
between the surface of the piston and the cylinder.
Determine the change in internal energy of air in
53. 2 lbm/s of ice at -30F is converted into steam at
kJ/kg.
300F. Determine the following
a) The total amount of heat in kW needed in the
47. A piston cylinder contains 0.50 kg of air and is
conversion.
fitted with electrical resistor. The mass of the piston
b) If the heated efficiency is 90%, what is the
is 60 kg and with a face area of 1500 cm2. The heat
power input of the heater in CHU/hr and kgf-m/s?
transfer produced by the electric current passes
through the resistor is 2 kJ that caused the volume to
54. A 12-gram piece of aluminum (cp = 0.215 cal/g-
increase by 50 L, while maintaining a constant
K) at 70°C is placed in a beaker that contains 35
pressure. At initial and final state, the air and the
grams of water at 15°C. At what temperature will
piston are at rest. Neglect the friction effects between
they come to thermal equilibrium?
the surface of the piston and the cylinder and take
local acceleration due to gravity at standard
55. 30 grams of gas inside a cylinder fitted with a
conditions. Determine the change in internal energy
piston has a temperature of 150 ℃. The piston is
in kJ.
moved with a mean force of 200 N so that it moves
60 mm and compressed the gas. The temperature
48. There are 2 kg of fluid mass in a closed container
rises to 21°C as a result. Calculate the heat transfer
at rest on a given datum line at local g=9.75 m/s2 .
given cv = 718 J/kg-K.
The container is now raised vertically up to 1000 m.
Initially the internal energy of the fluid was 20 kJ.
56. A 100 g copper vessel contains 150 g of water at
Determine the final internal energy if swirling
velocity of the fluid is 50 m/s. 25°C. A 60 g block at 70°C is dropped to the water.
The temperature equilibrium is 32°C. If the specific
heat of copper is 0.093 kcal/kg-°C, what is the
specific heat of the block?
137
57. A 1/4 cubic meter of water contained in a 1kg 64. Determine the change of enthalpy in kJ needed to
aluminum and heated from 20℃ to 25℃. Specific heat 1 lbm of hydrogen gas from 540 R to 5400 R if
heat of aluminum is 20% that of water. What is the 1.87𝑇 9.95
𝑐𝑝 = 0.364 + + ; 𝐵𝑇𝑈/𝑙𝑏𝑚 − °𝑅
total heat gained by the water and container in cal? 104 √𝑇
58. 0.5 lbs of water and 0.5 lbs of ice are in thermal 65. A boiler contains 50 gallons of water. Heat
equilibrium at 0°C. If 0.75lb of steam at 120°C are transfer rate in water is given by 𝑄̇ = 80(450 − 𝑇)
added, , where 𝑄̇ is in BTU/hr and T is the temperature of
a) Find the final temperature of the mixture water at any instant in °F. How many minutes does it
b) How many kilograms of steam condensed? take to heat water from its ice point to its steam
point?
59. A closed system consists of 1 gram of water.
When vaporized at 1 atm, it occupies 1671 cm3; the 66. In an oil cooler, oil flows steadily through a tube
heat of vaporization is 539 cal/gram. Find the submerged in a steady stream of cooling water.
external work and the increase in enthalpy during Under the steady flow conditions, the oil enters at
vaporization in calories. 90℃ and leaves a 30℃, while the water enters at 25℃
and leaves at 70℃. The specific heat of oil at a given
60. A 1.2kΩ 30A heater converts m kg of ice at -10C
temperature t in ℃ is given by 𝑐𝑝 = 1.68𝑡 +
to steam at 130C for 10 mins. Determine the amount
m of ice that is converted to steam. (10.5 × 10−4 )𝑡 2 kJ/kg-K. Find the amount of
cooling water needed in kg/s to cool 2.75 kg/s of oil.
61. Water at 30°C is to be cooled by putting ice cubes
Power and Energy Conversion
at 0°C, each ice cube has a mass of 5g. Water is
stored in a glass having a water equivalent of 5g.
67. For each of the following, identify whether the
How many ice cubes should be added to water to unit is a form of energy or power. If energy, convert
reach -3°C? it to BTU, and for power, convert it to ft-lb/s
a) 400 kW-hr
62. Determine the amount of saturated steam needed b) 25 ev
to convert 10 kgm of ice to water at 40°C in kgm at c) 30W
constant atmospheric pressure if the final d) 100 hp
temperature of the water from steam is the same as e) 32.2 slug-ft2/s2
the temperature of water from the ice.
68. Convert 10 CHU to
63. A classroom has a seating capacity of 50, and a) kW-hr
each student dissipates an average heat rate of 18 b) Therm
BTU/hr. The classroom also contains 5 light bulbs, c) hp-min
when opened each dissipates 30 W of heat. On an d) kJ
average 8 hr teaching operations, walls from the e) Erg
classroom absorbs 100 kJ of heat.
69. How much energy in BTU is produced by a
a) If 50hp air conditioner is bought to cool down 60MW engine operating for 30 mins?
the classroom and to be used for meeting purposes
after teaching hours, how many air conditioners 70. An engine has an efficiency of 35%. It uses 5
should be bought? gallons of gasoline per hour. Gasoline has a heating
b) If electricity charge amounts to ₱8.25 per kW- value if 20000 BTU/lb, and SG of 0.80. What is the
hr, how much should be charged to the school for power output of the engine in kW?
using the classroom, assuming that the aircon and 5
light bulbs are on, and there are 2kW allowance
electricity usage for personal used.
138
71. Find the power in watts expended by a 50 kgm 80. A motor driven pump transfers 5000L of oil
worker to carry a 3 kg crate in a 25 step stair case, through an elevation of 16 m. If delivery velocity is
each step is 30 cm high in 1 minute, 10 m/s, what is the input power to the pump in hp and
kW-hr?
72. A pump is lifting water through a 5.5m to fill a
570m3 tank. The overall energy efficiency is 80%. 81. If 85% of the electrical power input to the motor
Calculate the length of time that 7.5 kW applied to is available to lift a 1500 kgm elevator for 30 seconds
the pump will require to complete the job in minutes. at 20 meters, determine the horsepower rating.
73. A body is being dragged uniformly along a 82. The sun generates 1 kW/m2 when used as a
horizontal surface by a force of 45 kg acting at an source of solar collectors. A collector with an area of
angle of 20 degrees to the horizontal. Find the work 1 m2 heat water. The flow rate is 3 L/min. What is the
done in moving the body in kW temperature rise in the water in C?
74. A 50 inches diameter diamond saw blade is 83. A water brake consists of a metal cylinder
mounted on a pulley driven steel shaft, required a containing water and a paddle wheel to dissipate,
blade peripheral linear speed of 150ft/s. Motor drive through fluid friction, the work output of a test
is 125 hp at 1200 rpm with 6 inches diameter engine. The brake is full of 100 gallons of water, at
pulley. Determine the shaft rpm to attain the blade the time of the pump, which circulates the water
peripheral speed required through a cooler and back, falls leaving the water
trapped in the brake. At the time of failure, the
75. Forty percent of the electrical input to a motor
temperature of the brake is 120°F. The engine is
driven pump is converted into a hydraulic jet 12 mm
in diameter for the purpose of washing down ashes. continuously delivering 120hp to the brake. The
Find the jet velocity in m/s. Motor rating descriptions brake will automatically shut if the water reaches its
are 3 phase, 220V, 7.5 A with a power factor of 85%. steam point. How long will it take for the brake to
(Note: Power Factor is the ratio of the theoretical automatically shut down, from the time where the
power to actual power. For a 3phase motor, multiply brake fails?
the actual power by √3)
84. A steam powerplant has an output of 10 MW with
76. A turbogenerator rotating mass has a moment of 75% waste heat. It runs for one-week continuous
inertia of 555 hyl-m2. It is delivering 2500 kW at operation.
1800 rpm. The load suddenly increases to 2550 kW, a) Determine the energy output in CHU, kW-hr,
the developed steam power remaining unchanged. and ft-lbf
What is the resulting speed in rpm after 10 seconds? b) Determine the energy chargeable to the plant in
MJ
77. A hoist is to raise a 1135 kg mine cafe at a rate of c) Determine the energy wasted in the plant in
4.6 m/s. Mechanical efficiency of the hoist is 92%. BTU/min.
What is the power in kW to drive at this speed? d) Recommend dimensions for an inverted conical
tank whose diameter is one half its depth, as a
78. How much power is there in the kinetic energy of storage of coal having a heating value of 40000
the atmosphere whose density is 1.217 kg/m 3 at 56 kJ/kg for one year continuous operation
kph wind velocity in kW? Consider that the
atmosphere passes through a 3m diameter circular Other Forms of Work
area normal to the velocity
85. Given two small electrical charges Q1 and Q2
79. A dam holds 200000 m3/day of water at a height positioned on the x-axis as follows: Q1= +4 μC at x=
of 492 ft above the valley floor. If the hydraulic -3m, Q2 = +1 μC at x=+2m. Find the position on the
turbine is situated in the valley floor, what is the x-axis of a third charge Q3, that experiences no net
maximum power that can be generated in kW- force from these two charges.
hr/year?
139
86. A triangle ABC has these dimensions AB=4m 95. A spherical soap bubble, having a radius of 3
BC=5m AC=3m. Electrical charges are located at the inches is formed by means of blowing through a
vertices as follows, +30 μC at A, -160 μC at B, and soapy blow pipe. If surface tension is 15 dynes/cm ,
+90 μC at C. Find the magnitude and direction of the find the work done in dyne-cm to overcome the
net force on charge A in newtons and the direction surface tension of the bubble.
respect of the axis of AB.
96. A solid disk flywheel has a moment of inertia of
87. Determine the atmospheric work done in ft-lbf as 200 kgm-m2 is rotating at a speed of 900rpm. What
a 2 in cube ice melts in a region of 1 atm. At 32 F, is its rotational kinetic energy in therm?
these densities obtain for water: liquid 62.42 and
solid 57.15 in lb/ft3. 97. A 1200 hp boiler feed pump connected to electric
motor has a drive shaft rotating at 2000 RPM. How
88. A 1 kg of ice at 0℃ is completely melted into much torque is on the shaft?
water at 0℃ at 1 bar pressure. The latent heat of
fusion of water is 333 kJ/kg and the densities of water 98. A 12 V battery is receiving a constant charge
and ice at 0 ℃ are 999.0 kg/m3 and 916 kg/m3, from a generator. The voltage across the terminal is
respectively. What is the approximately value of the 12.5V and the current is 10A. Determine the input
work done? power in hp.
89. A balloon initially collapsed and flat is slowly 99. A 5W battery supplies electricity to two resistors.
filled with a gas at 100 kPa, so as to form into a When the resistors are arranged in series, the current
sphere of 1m radius. What is the work done by the flowing through the resistors are measured to be 2 A.
gas in the balloon during the filling process. When the resistors are arrange in parallel, the total
current flowing through the resistors is measured to
90. Heat is transferred to an elastic sphere containing be 2.5 A. Determine the ohmic values of these
gas at atmospheric; the diameter of the sphere is 2m. resistors.
Because of heating the sphere diameter increases by
2.4 m and the gas pressure increases in proportion to 100. A constant force moves an 18 in conductor with
the diameter. Find the work of the gas during the a velocity of 25fps orthogonally across a magnetic
heating process. field whose flux density is 1 N-s-m/C. The conductor
carries a current of 20A. Find the force and the rate
91. There are required 124 ft lbf or work to compress of work produced.
a spring so that the final length of spring after
compression is 2.5 in. If the spring has an index of 101. A shock absorber pumps oil internally through
100 lbf/in, find the free length of the spring. a flow resistance to dissipate mechanical energy. The
force acting on the absorber is given by
92. Two springs are designed to absorb the kinetic 𝑑𝑥
energy of a 2000-kgm vehicle. Determine the spring 𝐹=2
𝑑𝑡
constant necessary if the maximum compression is to where F is in lbf-s/in and dx/dt is in inches per
be 100 mm for a vehicle speed of 10 m/s. second. The displacement is given by
𝑥 = 6𝑠𝑖𝑛2π𝑡
93. A 10 ft wire having a modulus of elasticity of 30
where x is in inches and t is in seconds. The absorber
x 106 psia is stretched by an applied force of 1200
lbf. What is the work done to the wire in in-lbf? and the oil has a combine mass of 5 lbm, and the
average specific heat of the absorber and oil is 0.25
94. An aerial soap film is formed by wetting a wire BTU/lb-°F. How long will it take the absorber to
frame then moving the slide wire away from a leg If change temperature from 70 to 160°F?
the wire is enclosed in such a way it has the
dimensions of 6cm x 10 cm. Find the work done
against the resisting surface tension of 25 dyne per
centimeter in kgf.
140
Mechanical Work
102. Evaluate the non-flow work in terms of 108. During reversible process executed by a non-
𝑃1 , 𝑉1 , 𝑃2 , and 𝑉2 of a fluid undergoing a reversible flow system the pressure increases from 50 psia to
state in accordance of each of the following relations: 200 psig in accordance to PV=C and the internal
a) 𝑃 = 𝐶 energy increases 21.4 BTU; the initial volume is 3
b) 𝑉 = 𝐶 cubic foot.
c) 𝑃𝑉 = 𝐶 a) Find the heat is calories.
d) 𝑃𝑉 2 = 𝐶 b) If the change of internal energy is neglected,
e) 𝑃𝑉 3 = 𝐶 what is the change of enthalpy in kJ?
f) 𝑃𝑉(𝑙𝑛𝑉) = 𝐶
200 109. A gas undergoes a thermodynamic cycle
𝑔) 𝑃 = 2 + 2 consisting of three processes beginning at an initial
𝑉
state where 𝑃1 = 1𝑏𝑎𝑟, 𝑉1 = 1.5𝑚3 𝑎𝑛𝑑𝑈1 =
103. If 6L of a gas at a pressure of 100 kPaa are 512𝑘𝐽. The processes are as follows: For process 1-
compressed reversibly according to PV2 = C until the 2, compression with PV=C to 𝑃2 = 2𝑏𝑎𝑟 and 𝑈2 =
volume becomes 2L, find the final pressure in kPaa 960𝑘𝐽. For process 203, W=0 and Q=-150 kJ. For
and work non-flow in J. process 301 W=+50 kJ neglecting change in PE and
KE. What is the heat transfer from process 1 to 2?
104. During a reversible process exerted by a non-
flow system, the pressure varies from 345 kPaa to 110. An imaginary engine receives heat and does
1400 kPaa in accordance with PV = C. The internal work on a slowly moving piston at such rates that the
energy increases to 22500 J, and the initial volume cycle operation on the PV diagram can be
is 85 L. Find the heat in Joules. represented as a circle 10cm in diameter , wherein
1cm=300kPa and 1cm=0.10 m3/kg. What is the work
105. A gasesous substance whose properties are done by each kg of working fluid for each cycle of
unknown, undergoes a internally reversible process operation?
during which V = -0.10P + 300 where V is in cubic
food at P is pounds per square foot abs. Neglect Energy Balance For Open Systems
change in potential energy.
a) Find the nonflow work in BTU if the pressure 111. A 14 300 kg airplane is flying at an altitude of
changes from 1000psfa to 100 psfa . 497 m at a speed of 214 km/h. Determine the
b) Find the steady flow work in BTU airplane's total mechanical energy.
c) If the process is a steady flow with an
increasing kinetic energy of 25BTU and 112. A 4 kg/s of water at 40°𝐶 is mixed with 6 kg/s
enthalpy decreasing of 300 BTU, then of water at 100℃ in a steady flow process. Find the
determine the net work and heat, both in BTU. temperature of the resulting mixture in ℃.
d) If the process in non-flow, determine the
work and change in internal energy, both in 113. A tank contains 25L of oil of relative density of
BTU 0.9 and temperature of 10C. The oil is heated for 15
min by a 2.16kW heater. Determine the final
106. If 6 L/s at 100kPaa of fluid is compressed temperature of oil assuming heat loss through the
reversibly to 2 L/s according to the law PV 2=C, tank as 160 kW. Specific heat of oil is 48% of
determine the specific heat of water at constant pressure.
a) work steady flow in hp and
b) work non flow in PS. 114. Consider an 80-gallon hot water heater. Over a
fifteen-minute time period hot water flows out of the
107. If a gas of volume 6000cm3 and at a pressure of hot water heater at 0.75 kg/s and cold water at 0.5
100 kPa is compressed quasistatically in a piston kg/s flows into the hot water heater. How full in
cylinder according to 𝑃𝑉 2 = 𝐶 until the volume percent is the hot water heater at the end of fifteen
becomes 2000cm3. What is the work done? minutes? You may take the water temperature to be
85ºC. Note at density of water at 85°C = 968.2 kg/m3.
141
115. In a steady flow adiabatic turbine, the changes 121. Steam turbine receives 70 pounds of steam per
in enthalpy, kinetic energy, and potential energy of minute with an enthalpy of 1600 Btu per pound and
the working fluid from inlet to exit are 1150 kJ/kg, velocity of 100 ft/sec. It leaves the turbine at 900
10kJ/kg, and zero respectively. What is the turbine ft/sec and 1320 Btu/lb enthalpy. The radiation loss is
work? 84,000 Btu/hr. Find the horsepower output.
116. Steam flows through an adiabatic turbine at a 122. Steam enters a turbine stage with enthalpy of
rate of 100 lb/min. At suction, the pressure is at 175 3628 kJ/kg 70 m/sec and leaves the same stage with
psia , volume of 3.16 ft3/lb, and internal energy of an enthalpy of 2846 kJ/kg and a velocity of 124
1170 BTU/lb. At exhaust, the pressure is 0.813 psia, m/sec. Calculate the work done by the steam.
the volume increases 100 times and the internal
energy is 900 BTU/lb. Assume that there is no 123. Steam enters in a steady flow manner in an
change in velocity and elevation between the suction adiabatic turbine at 80 m/s, 2.5kPaa, 500C with a
and exhaust. Find the turbine horsepower rating specific enthalpy of 3462.10 kJ/kgm. It leaves the
required. turbine at 40 m/s,10kPaa, with a specific enthalpy of
2393.21 kJ/kgm. At suction, the specific volume is
117. A turbine operates under steady flow 1.2 L/kgm. If the flow is 25 kgm/s, determine:
conditions, receiving steam at the following state: a) The change in kinetic energy in kW
pressure 1000 kPa, temperature 180 deg. C, enthalpy b) The work done by the turbine in hp.
2700 kJ/kg, speed 50 m/s, and elevation 2 m. The c) The cross-sectional area of the suction in m2
steam leaves the turbine at the following state:
pressure 50 kPa, enthalpy 2500 kJ/kg, speed 100 m/s, 124. Steam enters a turbine at steady state with a
and elevation 0 m. Heat is lost to the surroundings at mass flow rate of 4600 kg/h. The turbine develops a
the rate of 0.50 kJ/s. If the rate of steam flow through power output of 1000 kW. At the inlet the pressure is
turbine is 0.50 kg/s, what is the power output of the 0.05 MPa, the temperature is 400 °C, and the velocity
turbine in hp? is 10 m/s. At the exit, the pressure is 10 kPa, the
quality is 0.9, and the velocity is 50 m/s. Calculate
118. In a gas turbine the gases enter the turbine at rate the rate of heat transfer between the turbine and
of 5 kg/s with a velocity of 50 m/s and enthalpy of surroundings, in kW. Take h 1 = 3178.9 kJ/kg ; h2=
900 kJ/s and leave the turbine with 150 m/s and 2345 kJ/kg.
enthalpy of 400 kJ/kg. The loss of heat from the
gases to the surrounding is 25 kJ/kg. Assume 125. In a gas turbine unit, the gases flow through the
R=0.285 kJ/kg-K and cp=1.004 kJ/kg-K. The inlet turbine is 15 kg/s and the power developed by the
condition to be at 100 kPa and 27℃.Find the work turbine is 12000 kW. The enthalpies of gases at the
transfer in kW. inlet and outlet are 1260 kJ/kg and 400 kJ/kg
respectively, and the velocity of gases at the inlet and
119. In a turbine unit, the gases flow through the outlet are 50 m/s and 110 m/s respectively.
turbine is 15 kg/s and the power developed by the Calculate :
turbine is 12000 kW. The enthalpies of gases at the a) The rate at which heat is rejected to the turbine
inlet and outlet are 1260 kJ/kg and 400 kJ/kg b)The diameter of the inlet pipe given that the
respectively and the velocity of gases at the inlet and specific volume of the gases at the inlet is 0.45 m3/kg.
outlet are 100 m/s and 50 m/s respectively. What is
the rate at which heat is rejected to the turbine? 126. Stream of gases at 7.5 bar, 750°C and 140 m/s
is passed through a turbine of a jet engine. The
120. 10000 kg/hr of steam with an enthalpy of 2778 stream comes out of the turbine at 2.0 bar, 550°C and
kJ/kg enters the turbine and leaves with an enthalpy 280 m/s. The process may be assumed adiabatic. The
of 2168 kJ/kg. Determine the turbine power output if enthalpies of gas at the entry and exit of the turbine
thermal efficiency is 35%. are 950 kJ/kg and 650 kJ/kg of gas respectively.
Determine the capacity of the turbine if the gas flow
is 5 kg/s.
142
127. A centrifugal air compressor compresses 200 132. There are 24000 lb/hr of steam supplied to a
CFM from 12 psia to 90 psia. The initial specific nozzle at an abosulte pressure of 200 psia. The speed
volume is 12.6 ft3/lbm and the final specific volume of the steam entering the nozzle is 5000 fpm having
is 3.25 ft3/lbm. If the inlet suction line is 4 in ID and a specific volume of 2.29 cu. ft/lb and internal energy
the discharge line is 2.5 in ID line, determine of 1114 BTU/lb. Exit conditions of steam has a
a) The change in flow energy in PS specific volume of 26.77 cu. ft/lb and internal energy
b) The mass flow rate in kg/s of 1080 BTU/lb. Find the exit velocity.
c)the change in velocity in mi/hr
133. Steam at 1000 lb/ft3 pressure and 330°R has a
128. A compressor draws 6000 kg/hr of atmospheric specific volume of 6.5 ft3/lb and a specific enthalpy
air and delivers it at a higher pressure. At inlet and of 9800 ft-lb/lb. Find the internal energy per pound
exit, the enthalpies for air are 300 kJ/kgm and 509 mass of steam.
kJ/kg respectively. The heat rejected by the water
coolant is 5 kW. Neglecting change in potential and 134. Air enters air compressor with a mass flow rate
kinetic energy, determine the power required to drive of 0.70 kg/s with a specific enthalpy of 290 kJ/kg and
the compressor in hp. leaves it with 450 kJ/kg of specific enthalpy.
Velocities at inlet and exit are 6 m/s and 2 m/s
129. In an air compressor air flows steadily at the rate respectively. Assuming adiabatic process, what is the
of 0.5 kg/s through an air compressor. It enters the power input to the compressor in kW?
compressor at 6 m/s with a pressure of 1 bar and a
specific volume of 0.85 m3/kg and leaves at 5 m/s 135. Steam with an enthalpy of 2168 kJ/kg enters the
with a pressure of 7 bar and a specific volume of 0.16 condenser at a rate of 22050 lbm/hr. If the condensate
m3/kg. The internal energy of the air leaving is 90 has an enthalpy of 251 kJ/kg, determine the heat
kJ/kg greater than that of the air entering. Cooling rejection rate in BTU/min.
water in a jacket surrounding the cylinder absorbs
heat from the air at the rate of 60 kJ/s. Calculate : 136. Air and fuel enter a furnace used for home
(i) The power required to drive the compressor ; heating. The air has an enthalpy of 302 kJ/kg and the
(ii) The suction and discharge pipe cross-sectional fuel has an enthalpy of 43 207 KJ/kg. The gases
areas. leaving the furnace have an enthalpy of 616 kJ/kg. of
fuel. The house requires 17.6 KW of heat. What is
130. 12 kg of air per minute is delivered by a the fuel consumption per day?
centrifugal air compressor. The inlet and outlet
conditions of air are v1 = 12 m/s, P1 = 1 bar, ν1 = 0.5 137. The power plant furnace burns coal at the rate
m3/kg and v2 = 90 m/s,P2 = 8 bar,ν2= 0.14 m3/kg. The of 108,200 kg/hr. Air at 100.8 kPa, 28°C is supplied
increase in enthalpy of air passing through the at the rate of 13.8 kg/kg coal. Determine the volume
compressor is150 kJ/kg and heat loss to the flow rate of air flow in m3/min.
surroundings is 700 kJ/min. Find
a) Motor power required to drive the compressor 138. The enthalpy of air is increased by 139.586
b)Ratio of inlet to outlet pipe diameter assuming kJ/kg in a compressor. The rate of air flow is 16.42
that inlet and discharge lines are at the same level. kg/min. The power output is 48.2 KW. Find the heat
loss in the compressor in kW.
131. In a test of water cooled air compressor, it is
found that the shaft work required to drive the 139. The air with enthalpy of 100 kJ/kg is
compressor is 175 kJ/kg of air delivered and the compressed by an air compressor to a pressure and
enthalpy of air leaving is70 kJ/kg greater than that temperature at which its enthalpy becomes 200
entering and that the increase in enthalpy of kJ/kg. The loss of heat is 40 kJ/kg from the
circulating water is 92 kJ/kg. Compute the amount of compressor as the air presses through it. Neglecting
heat transfer to the atmosphere from the compressor kinetic and potential energies, what is the power
per kg of air. required for an air mass flow of 0.5 kg/s?
143
140. A steam condenser receives 10 kg/s of steam 147. At the inlet to a certain nozzle the enthalpy of
with an enthalpy of 2770 kJ/kg. Steam condenses and fluid passing is 2800 kJ/kg, and the velocity is 50
leaves with an enthalpy of 160 kJ/kg. Cooling water m/s. At the discharge end the enthalpy is 2600 kJ/kg.
passes through the condenser with temperature The nozzle is horizontal and there is negligible heat
increase from 13C to 24C. Determine the inside loss from it.
diameter of the tube where cooling water flows if a) Find the velocity at exit of the nozzle.
water passes and leaves at the same rate of 5 m/s. b) If the inlet area is 900 cm2 and the specific
volume at inlet is 0.187 m3/kg, find the mass flow
141. Steam with an enthalpy of 800 kcal/kg enters a rate.
nozzle at a velocity of 80 m/sec. Find the velocity of c) If the specific volume at the nozzle exit is 0.498
the steam at the exit of the nozzle if its enthalpy is m3/kg, find the exit area of nozzle.
reduced tp 750 kcal/kg. assuming the nozzle is
horizontal and disregarding heat losses. Take g=9.81 148. One of the sections of the heating plant in which
m/s2. there are no pumps enters a steady flow of water at a
temperature of 50°C and a pressure of 3 bar (h = 240
142. Liquid water with a constant density of 1000 kJ/kg). The water leaves the section at a temperature
kg/m3 enters a nozzle at the rate 10 L/min. The inlet of 35°C and at a pressure of 2.5 bar (h = 192 kJ/kg).
of the nozzle has a diameter of 1.50 cm, the diameter The exit pipe is 20 m above the entry pipe. Assuming
of the exit is 0.75 cm. Find the velocities of the water change in kinetic energy to be negligible, evaluate
at the inlet and exit. the heat transfer from the water per kg of water
flowing
143. Steam enters a converging‐diverging nozzle
operating at steady state a t0.05 MPa and 400 °C and 149. A centrifugal pump delivers 50 kg of water per
a velocity of 10 m/s. The steam flows through the second. The inlet and outlet pressures are 1 bar and
nozzle with negligible heat transfer and no 4.2 bar respectively. The suction is 2.2 m below the
significant change in potential energy. At the exit, centre of the pump and delivery is 8.5 m above the
conditions are 0.01 MPa, and the velocity is 665 m/s. centre of the pump. The suction and delivery pipe
The mass flow rate is 2 kg/s. Determine the exit diameters are 20 cm and 10 cm respectively.
enthalpy and area of the nozzle if the enthalpy and Determine the capacity of the electric motor to run
specific volume of steam at the inlet conditions is the pump.
3278.9 kJ/kg and 26.445 cu.m /kg respectively.
150. A steady-state steady flow thermodynamic
144. At the inlet to a certain nozzle, the enthalpy of system receives 100 lb/min of a fluid at 30 psia and
the fluid passing is 3000 kJ/kg and the velocity is 60 2000°F, and discharges it from a point 80 ft above
m/s and the specific volume of 0.187 m 3/kg. At the the entrance section at 150 psia and 6000°F. The
discharge, the enthalpy is 2762 kJ/kg. The nozzle is fluid enters with a velocity of 7200 fpm and leaves
horizontal and there is negligible heat loss from it. with a velocity of 2400 fps. During this process, there
Determine the velocity at the exit of the nozzle. are supplied 25 000 BTU/hr of heat from an external
source and the increase in enthalpy is 20 BTU/lb.
145. Water enters a cooling tower at 35C at the rate Determine the work done in hp.
of 180000 kg/hr and is cooled to 25C. Atmospheric
air is used for cooling enters the tower with enthalpy 151. A boiler produce 600 kg of steam per hour from
of 24 kJ/kgm and leaves with an enthalpy of 88 feedwater at 40C. Assuming the boiler to be a steady
kJ/kgm. Find the mass flow rate in lbm/s. flow system and neglecting potential and kinetic
changes, find the rate at which the heat is
146. A blower handles 1 kg/s of air at 20°C and transformed in kW. The enthalpy of steam is 660
consumes a power of 15 kW. The inlet and outlet kcal/kgm
velocities of air are 100 m/s and 150 m/s
respectively. Find the exit air temperature, assuming
adiabatic conditions. Take cp of air to be 1.005 kJ/kg-
K.
144
152. During flight, the air speed of a turbojet engine 158. A hydraulic pump is handling 5CFS of 60F
is 250 m/s. Ambient air temperature is – 14°C. Gas water through an 8 in inside diameter suction line and
temperature at outlet of nozzle is 610°C. a 6 in inside diameter discharge line. The suction
Corresponding enthalpy values for air and gas are gauge is on the pump center line and reads 10 in Hg
respectively 250 and 900 kJ/kg. Fuel air ratio is vac; the discharge gauge is 20 ft above the center
0.0180. Chemical energy of fuel is 45 MJ/kg. Owing
line. If the energy input to the water is 100 hp, find:
to incomplete combustion 6% of chemical energy is
a) The reading of the discharge gauge in psi.
not released in the reaction. Heat loss from the engine
b) If the discharge gauge reads 100 psig, what
is 21 kJ/kg of air. Calculate the velocity of the
would be the pump horsepower input to the water?
exhaust jet.
Energy Balance For Closed Systems
153. Air at a temperature of 20°C passes through a
heat exchanger at a velocity of 40 m/s where its
159. An inventor claims a closed system that does 3
temperature is raised to 820°C. It then enters a
BTU of work and rejects 2 BTU of heat. Change in
turbine with same velocity of 40 m/s and expands till
internal energy is zero. Validate the claim. If it is
the temperature falls to 620°C. On leaving the
wrong, what is the actual change of internal energy?
turbine, the air is taken at a velocity of 55 m/s to a
nozzle where it expands until the temperature has
160. A closed system undergoes a process 1-2 for
fallen to 510°C. If the air flow rate is 2.5 kg/s.Take
which Q and W are 20kJ and 50kJ respectively. If the
the approximate enthalpy of air as h = cpt, where cp
system is returned to state 1 and Q=-10kJ, what is the
is the specific heat equal to 1.005 kJ/kg°C and t the
value of W?
temperature.calculate :
a) Rate of heat transfer to the air in the heat
161. A closed system executes a series of process for
exchanger
which the work, heat, and internal energy are related
b)The power output from the turbine assuming no
to each other. Complete the table below, assuming
heat loss ;
that energy change is zero.
c) The velocity at exit from the nozzle, assuming no
heat loss.
W Q ΔU
154. Water falls from a height 500m. What is the rise 10hp 500 BTU/min _____kW
in temperature of water in ℃ at the bottom if all of
the energy remains in the water? 65 BTU _____BTU -25 BTU
____ hp 25 kW 0
155. Compute the temperature rise of water falling
continuously from an elevation of 30 m in kelvin. -390 ft-lbf -2 BTU ___ hp-hr
200 000 gm-cm 50000 gm-cm ___kgm-m
156. A centrifugal air compressor compresses 200
CFM from 12 psia to 90 psia. The initial specific
volume is 12.5 ft3/lb, and the final specific volume is 162. A 12 V battery receives a quick 20 min charge
3.25 ft3/lb. If the inlet suction line is 4 in ID and the during time it receives a steady current supply of 50
discharge line is 2.5 in ID, determine the work in ft- amp. In this period, it receives a heat loss of
lbf/min. 150BTU. Find the change in internal energy during
this period
157. In a test of water jacketed compressor, the shaft
work required is 90 kJ/kg of air compressed. During 163. 5000 BTU/hr of heat is leaking through an
compression, increase in enthalpy of air is 30 kJ/kg imperfect insulation in a cooled system. A
of air and increase in enthalpy of circulation cooling thermoelectric system maintains the region and
water is 40 kJ/kg of air. The change in velocity is operates at 40amps and 24V. Find the heat rejected
negligible. Find the amount of heat lost to the by the thermoelectric system.
atmosphere from the compressor per kg of air.
164. A system receives 75kJ of heat while it does 45
kJ of work. How much energy is stored in the
system?
145
165. Five persons occupying the same room gives

400 BTU/hr after a power failure in an air
conditioning system. Find the change in internal
energy of the room air at the end of 10 min following
the power failure.
166. For a certain system executing a cyclic process

250J of heat are absorbed by the system and 100J of
heat are rejected. The system also receives 30 W-s of
electrical power whole it moves a 3 kg mass
vertically by means of a pulley arrangement. How far
does the mass move in meters? Local g=9.65 m/s2.
167. The contents of a well- insulated tank are heated

by a resistor of 23 Ω in which 10 A current is
flowing. Consider the tank with its contents as a
thermodynamic system. The work done by the
system and the heat transfer to the system are
positive. Find the rates of heat (Q), work (W) and
change in (ΔU) during the process in kW.
168. Gas contained in a closed system consisting of

piston cylinder arrangement is expanded. Work done
by the gas during expansion is 50 kJ. Decrease in
internal energy of the gas during expansion is 50 kJ.
Decrease in internal energy of the gas during
expansion is 80 kJ. Find the heat transfer during the
process
169. A paddle wheel is used to create work in an

inflexible tank containing an unknown fluid. The
fluid rejects 12BTU of heat while the initial and final
internal energies are 6.25 and 0.50 BTU,
respectively. Determine the work created by the
paddle wheel to the fluid in cal. Neglect the energy
stored in the paddle wheel
170. 8kg of gas is contained in a vertical piston

cylinder assembly. The area and the mass of the
piston are 300cm2 and 15 kg, respectively. A spring
is loaded in the piston, which the force exerted varies
linearly with displacement, x. At initial position, the
piston is at x=0, the spring exerts no force. Heat is
added to the air causing the gas to expand until the
piston reached the stops at x=0.08m, and the heat
transfer ceases. The atmosphere is 1bar, the observed
gravitational acceleration is 9.81 m/s2, consider the
spring constant to be 8500 N/m and neglect friction
between the piston and the cylinder. Determine the
work in J.
146
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147
CHAPTER III. IDEAL GAS THERMODYNAMICS I
III. IDEAL GAS___________________________________
Learning Objectives: At the end of this chapter, the student shall
1. Enumerate the characteristics of an Ideal Gas
2. Define the Equation of State and the Gas Constant.
3. Relate gas constant, specific heat at constant pressure and at constant volume.
4. Derive the different equations of state in terms of fluid properties.
5. Derive the different ideal gas laws.
6. Solve problems involving mixtures of ideal gas.
7. Differentiate volumetric analysis from gravimetric analysis
8. Apply gas laws to solve some engineering word problems
3.1 EQUATION OF STATE

A gas is said to be ideal if it follows the ideal gas equation, also known as the equation of state or the ideal
gas law. It relates the pressure, volume, mass, and temperature of gas. Hence
𝑷𝑽 = 𝒎𝑹𝑻
Where
Properties English / FPS MKS SI
𝑷 = 𝑨𝒃𝒔𝒐𝒍𝒖𝒕𝒆 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆
Volume (V) ft3 L m3
𝑽 = 𝑽𝒐𝒍𝒖𝒎𝒆
Mass (m) lbm kgm kgm
𝒎 = 𝒎𝒂𝒔𝒔
Temperature (T) °R K K
𝑹 = 𝑮𝒂𝒔 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
Gas Constant (R) 𝑓𝑡 − 𝑙𝑏𝑓 𝑘𝑐𝑎𝑙 𝐽
𝑻 = 𝑨𝒃𝒔𝒐𝒍𝒖𝒕𝒆 𝑻𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆
𝑙𝑏 − °𝑅 𝑘𝑔𝑚 − 𝐾 𝑘𝑔 − 𝐾
3.2 GAS CONSTANT

The gas constant of any given gas R can be found at Item B1 (Problems in Thermodynamics by Faires and
Simmang). If the molecular weight of the gas is only given, we can use the formula:
̅
𝑹 Molecular Weight Units
𝑹= English/FPS MKS/SI
𝑴𝑾
Where 𝑙𝑏𝑚 𝑘𝑔𝑚
̅
𝑹 = 𝑼𝒏𝒊𝒗𝒆𝒓𝒔𝒂𝒍 𝑮𝒂𝒔 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝑙𝑏𝑚 − 𝑚𝑜𝑙 𝑘𝑔𝑚 − 𝑚𝑜𝑙
𝑴𝑾 = 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒓 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒘𝒆𝒊𝒈𝒉𝒕
Standard values of universal gas constant are as follows
Standard Values of Universal Gas Constant

FPS SI
𝑓𝑡 − 𝑙𝑏𝑓 𝑘𝐽
𝑅̅ = 1545.32 𝑅̅ = 8.3143
𝑙𝑏𝑚 − 𝑚𝑜𝑙 − °𝑅 𝑘𝑔𝑚 − 𝑚𝑜𝑙 − 𝐾
𝐵𝑇𝑈 𝐿 − 𝑎𝑡𝑚
𝑅̅ = 1.986 𝑅̅ = 0.0821
𝑚𝑜𝑙 − °𝑅 𝑚𝑜𝑙 − 𝐾
147
The molecular weight MW is defined as the summation of the products of the atomic mass/weight and the
number of atoms each element in a compound.
𝑴𝑾 = ∑ 𝑵𝑨𝑨𝒘
If the number of moles is given as n in lbm-mol or kgm-mol, then
𝑚
𝑛=
𝑀𝑊
Then the molecular weight would be the mass per number of moles.
𝒎
𝑴𝑾 =
𝒏
Where
𝑵𝑨 = 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒂𝒕𝒐𝒎𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒃𝒆𝒊𝒏𝒈 𝒄𝒐𝒏𝒔𝒊𝒅𝒆𝒓𝒆𝒅
𝑨𝑾 = 𝑨𝒕𝒐𝒎𝒊𝒄 𝒘𝒆𝒊𝒈𝒉𝒕
𝒏 = 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔
𝒎 = 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅
3.3 GAS CONSTANT AND SPECIFIC HEAT
We will relate the specific heat at constant pressure, specific heat at constant volume, and gas constant. From
the definition of enthalpy:
𝐻 = 𝑈 + 𝑊𝑓
Since 𝐻 = 𝑚𝑐𝑝 𝑡 , 𝑈 = 𝑚𝑐𝑣 𝑡 , and 𝑊𝑓 = 𝑃𝑉 , then
𝑚𝑐𝑝 𝑡 = 𝑚𝑐𝑣 𝑡 + 𝑃𝑉
Also if 𝑃𝑉 = 𝑚𝑅𝑇, 𝑡ℎ𝑒𝑛
𝑚𝑐𝑝 𝑡 = 𝑚𝑐𝑣 𝑡 + 𝑚𝑅𝑇
𝑐𝑝 𝑡 = 𝑐𝑣 𝑡 + 𝑅𝑇
Differentiating implicitly in terms of temperature:
𝑐𝑝 𝑑𝑡 = 𝑐𝑣 𝑑𝑡 + 𝑅𝑑𝑇
Since we know that 𝑑𝑡 = 𝑑𝑇 for change in temperature, then
𝒄𝒑 = 𝒄𝒗 + 𝑹
The equation above is sometimes called as the Meyer’s Equation. Hence
𝒄𝒑 = 𝒄𝒗 + 𝑹
𝒄𝒗 = 𝒄𝒑 − 𝑹
𝑹 = 𝒄𝒑 − 𝒄𝒗
Hence the gas constant R is defined also as the difference of specific heats of gases at constant volume to
constant pressure. Another parameter for properties of gases is called the ratio of specific heats. It is the ratio of
constant pressure to constant volume specific heats.
𝒄𝒑
𝒌= 𝑤ℎ𝑒𝑟𝑒 𝒌 > 𝟏
𝒄𝒗
We can also relate the Meyer’s Equation from the ratio of specific heats. Since 𝑐𝑝 = 𝑘𝑐𝑣 , then
𝑘𝑐𝑣 = 𝑐𝑣 + 𝑅
The specific heat at constant volume would be:
𝑹
𝒄𝒗 =
𝒌−𝟏
And the specific heat at constant pressure would be:
𝒌𝑹
𝒄𝒑 =
𝒌−𝟏
148
For most cases, we are dealing with air as an ideal gas. We also note that the air is the reference substance for
specific gravity of gases. From Item B1, the following are properties of air (specifically dry air):
Properties of Dry Air English SI

Specific Heat at Constant Pressure 𝐵𝑇𝑈 𝑘𝐽
𝑐𝑝 = 0.24 𝑐𝑝 = 1.0062
𝑙𝑏𝑚 − °𝑅 𝑘𝑔𝑚 − 𝐾
Specific Heat at Constant Volume 𝐵𝑇𝑈 𝑘𝐽
𝑐𝑣 = 0.1714 𝑐𝑣 = 0.7186
Gas Constant 𝑓𝑡 − 𝑙𝑏𝑓 𝐽
𝑅𝑎𝑖𝑟 = 53.342 𝑅𝑎𝑖𝑟 = 287.08
Ratio of Specific Heats 𝑘𝑎𝑖𝑟 = 1.4
3.4 DERIVED EQUATIONS OF STATE.
Like liquids, gases are also fluids and characterized fluid properties. Using fluids properties and equation of
state, we can derive the following
From the equation of state

𝑃𝑉 = 𝑚𝑅𝑇
𝑚
𝑃 = 𝑅𝑇
𝑉
𝑷
𝛒= (Density of Ideal Gas)
𝑹𝑻
1
Since 𝑣 =
ρ
𝑹𝑻
𝒗= (Specific Volume of Ideal Gas)
𝑷
From equation of state:
𝑚𝑔
If 𝑊 = , then
𝑘
𝑃𝑉𝑔
𝑊= (Weight of Ideal Gas)
𝑘𝑅𝑇
𝑊
Since γ = , then
𝑉
𝑷𝒈
𝛄= (Specific weight of Ideal Gas)
𝒌𝑹𝑻
If the ideal gas is in terms of mass and volume flow rates:
𝑷𝑽̇ = 𝒎̇𝑹𝑻
(Power Gas Equation)
From the equation of state:
𝑅̅
If 𝑚 = 𝑛(𝑀𝑊) and 𝑅 = , then:
𝑀𝑊
𝑅̅
𝑃𝑉 = 𝑛(𝑀𝑊) ( )𝑇
𝑀𝑊
̅𝑻
𝑷𝑽 = 𝒏𝑹 (Molar Gas Equation)
149
If the equation of state is in terms of differential, assuming mass and gas constant
𝑑(𝑃𝑉) = 𝑚𝑑(𝑅𝑇)
𝑷𝒅𝑽 + 𝑽𝒅𝑷 = 𝒎𝑹𝒅𝑻 (Micro-Analysis Gas Equation)
In the equation of state is in terms of time t, assuming that mass and gas constant does not change with respect
to time. Hence
𝑑
(𝑃𝑉) = 𝑚𝑅𝑑𝑇
𝑑𝑡
𝑑𝑉 𝑑𝑃 𝑑𝑇
𝑃 +𝑉 = 𝑚𝑅
𝑑𝑡 𝑑𝑇 𝑑𝑡
𝑷𝑽̇ + 𝑽𝑷̇ = 𝒎𝑹𝑻̇ (Time Rate Gas Equation)
3.5 GAS LAWS
The following laws below are also derived from the ideal gas law, assuming that one of its working variable,
either pressure, volume, or temperature are assume to be constant. The term “iso” means constant. These gas laws
can only be used for gases assumed to be “ideal”,
1. Combined Gas Law -states that the pressure is inversely proportional to volume and directly proportional to
temperature. Consider two points 1 and 2 where there is a change on one of the state variables. From the ideal gas
law
If mass and gas constant does not change, then it is assumed to be constant. Hence mR=C. The equation of state
will become: 1 2
𝑷𝑽
=𝑪
𝑻 𝑃1 𝑃2
From point 1 to point 2 𝑉2
𝑉1
𝑇1 𝑇2
𝑷𝟏 𝑽 𝟏 𝑷𝟐 𝑽 𝟐
=
𝑻𝟏 𝑻𝟐
2. Boyle’s Law - states that from a constant temperature or isothermal process, the pressure is directly proportional
to volume or specific volume. Thus
𝟏 𝑷
𝑷∝ 2
𝑽
𝑷𝟐
𝑷𝑽 = 𝑪
𝑷𝒗 = 𝑪
From point to point 2: 𝑷𝑽 = 𝑪
𝑷𝟏 𝑽 𝟏 = 𝑷𝟐 𝑽 𝟐
𝑃1 𝑣1 = 𝑃2 𝑣2
𝑷𝟏 𝑽 𝟐 1
= 𝑷𝟏
𝑷𝟐 𝑽 𝟏
𝑃1 𝑣2
= 𝑽𝟐 𝑽𝟏
𝑃2 𝑣1 𝑽
150
3. Charles Law - states that from a constant pressure or isobaric/isopiestic process, the volume is directly
proportional to its temperature. T
2
𝑽∝𝑻 𝑻𝟐
𝑽
=𝑪
𝑻
From point 1 to point 2
𝑽𝟏 𝑽𝟐
=
𝑻𝟏 𝑻𝟐
𝑉2 𝑇2
= 1
𝑉1 𝑇1 𝑻 𝟏
𝑽𝟐 V
𝑽𝟏
4. Gay Lusaac’s Law - states that for a constant volume or isometric/isochoric/isovolumic process, the pressure is
directly proportional to its temperature. P 2
𝑃∝𝑇 𝑷𝟐
𝑷
=𝑪
𝑻
From point 1 to point 2
𝑷𝟏 𝑷𝟐
=
𝑻𝟏 𝑻𝟐
𝑃1 𝑇1 1
= 𝑷𝟏
𝑃2 𝑇2
𝑻𝟏 𝑻𝟐 T
3.6 BUOYANCY IN IDEAL GAS
Archimedes principle states that “a body floats when submerges to a fluid because of the buoyant force
supporting the body whose magnitude is equal to the displaced fluid.”. The principle behind the Archimedes
principle is called as the principle of buoyancy. The buoyant force acting on the body is defined as:
𝑾𝑩
𝑭𝑩 = 𝛄𝑩 𝑽𝒔
where
𝑭𝒃 is the buoyant force

𝛄𝒃 is the specific weight of the body 𝑽𝒔
𝑽𝒔 is the volume of the body that is fully submerged to the liquid. 𝑭𝑩
The volume that is submerged is defined as the volume that is displaced in the liquid. We can also relate the
specific gravity of liquid to the specific gravity of the fluid
a. If the specific gravity of the body is less than the specific gravity of the fluid (𝑺𝑮𝑩 < 𝑺𝑮𝒇 ), then the
body will just float and some part of it will submerged into the fluid.
b. If the specific gravity of the body is greater than the specific gravity of the fluid (𝑺𝑮𝑩 > 𝑺𝑮𝒇 ), then the
body will fully sink into the fluid.
151
If it is wished to determine the portion of a body submerged, we can apply equilibrium analysis:
𝑾𝑩
+↓ ∑ 𝐹𝑣 = 0
𝑊𝐵 − 𝐹𝐵 = 0
𝑊𝐵 = 𝐹𝐵
γ𝐹 𝑉𝑠 = γ𝐵 𝑉𝐵
𝑽𝒔
𝑭𝑩
Where γ𝐹 is the specific weight of the fluid
γ𝐵 is the specific weight of the body
𝑉𝑠 is the volume of the body submerged to the fluid, or the volume of the fluid displaced by the body
𝑉𝐵 is the volume of the body
3.7 ANALYSIS OF MIXTURE OF IDEAL GASES
For a mixture of ideal gas, it follows the Dalton’s Law of Partial Pressures which is also called the law partial
pressure. It states that for a mixture of gasses in a closed container, the following holds true:
1. The pressure of the mixture is the sum of the absolute pressure of each individual gas exerted
2. The volume of the mixture is equal to the volume of the container. Since the fluid is a gas, each volume of the
gas is equivalent to the volume of the container. In other words, the volume of each gases is equal to each other in
a closed container.
3. The temperature of the mixture is equal to the temperature of the container. In other words, the temperature of
each gas is the same for the temperature of other gases.
4. The total mass in the mixture is the sum of the individual mass of each gas.
5. The molar mass in the mixture in the sum of the individual molar mass of each gas.
Consider ideal gases A,B, and C are inside in the container. Gas A Gas B Gas C
Using Dalton’s Law of Partial Pressure: 𝑷𝑨 𝑷𝑩 𝑷𝑪
𝑷𝒎 = 𝑷𝑨 + 𝑷𝑩 + 𝑷𝑪 𝑽𝑨 𝑽𝑩 𝑽𝑪
𝑽𝒎 = 𝑽𝑨 = 𝑽𝑩 = 𝑽𝑪 𝑻𝑨 𝑻𝑩 𝑻𝑪
𝑻𝒎 = 𝑻𝑨 = 𝑻𝑩 = 𝑻𝑪 𝒎𝑨 𝒎𝑩 𝒎𝑪
𝒎𝑻 = 𝒎𝑨 + 𝒎𝑩 + 𝒎𝑪 𝒏𝑨 𝒏𝑩 𝒏𝑪
𝒏𝒎 = 𝒏𝑨 + 𝒏𝑩 + 𝒏𝑪
There are two ways of mixture analysis: the gravimetric analysis and the volumetric analysis.
1. Gravimetric analysis - an analysis of mass percentage, also called as the mass fraction.
𝒎
Gravimetric analysis of gas A %𝑮𝑨 = 𝑨 × 𝟏𝟎𝟎
𝒎𝒎
𝒎
Gravimetric analysis of gas B %𝑮𝑩 = 𝑩 × 𝟏𝟎𝟎
𝒎𝒎
𝒎
Gravimetric analysis of gas C %𝑮𝑪 = 𝑪 × 𝟏𝟎𝟎
𝒎𝒎
Total gravimetric analysis %𝑮𝑨 + %𝑮𝑩 + %𝑮𝑪 = 𝟏𝟎𝟎%
2. Volumetric analysis - is an analysis of mole percentage, also called as the mole fraction.
𝒏
Volumetric analysis of gas A %𝑽𝑨 = 𝑨 × 𝟏𝟎𝟎
𝒏𝒎
𝒏𝑩
Volumetric analysis of gas B %𝑽𝑩 = × 𝟏𝟎𝟎
𝒏𝒎
𝒏𝑪
Volumetric analysis of gas C %𝑽𝑪 = × 𝟏𝟎𝟎
𝒏𝒎
Total volumetric analysis %𝑽𝑨 + %𝑽𝑩 + %𝑽𝑪 = 𝟏𝟎𝟎%
152
To express gravimetric analysis as a function of its volumetric analysis:
𝑚𝐴 𝑛𝐴(𝑀𝑊𝐴 )
%𝐺𝐴 = =
𝑚𝑚 𝑛𝑚 (𝑀𝑊𝑚 )
𝑴𝑾𝑨
%𝑮𝑨 = %𝑽𝑨 ( )
𝑴𝑾𝒎
To express volumetric analysis as a function of its gravimetric analysis:
𝑴𝑾𝒎
%𝑽𝑨 = %𝑮𝑨 ( )
𝑴𝑾𝑨
Other properties of mixture of gases are as follows
1. The molecular weight of the mixture is

̅
𝑹
𝑴𝑾𝒎 =
𝑹𝒎
2. The gas constant R of the mixture is

𝑹𝒎 = %𝑮𝑨𝑹𝑨 + %𝑮𝑩 𝑹𝑩 + %𝑮𝑪 𝑹𝑪
For nth ideal gases
𝑛
𝑅𝑚 = ∑ %𝐺𝑖 𝑅𝑖 = 𝐺1 𝑅1 + 𝐺2 𝑅2 + 𝐺3 𝑅3 + ⋯ + 𝐺𝑛 𝑅𝑛
𝑖=1
3. For the specific heat of the mixture
a) For specific heat of mixture at constant pressure

𝒄𝒑𝒎 = %𝑮𝑨 𝒄𝒑𝑨 + %𝑮𝑩 𝒄𝒑𝑩 + %𝑮𝑪 𝒄𝒑𝑪
For nth ideal gases
𝑛
𝑐𝑝𝑚 = ∑ %𝐺𝑖 𝑐𝑝𝑖 = 𝐺1 𝑐𝑝1 + 𝐺2 𝑐𝑝2 + 𝐺3 𝑐𝑝3 + ⋯ + 𝐺𝑛 𝑐𝑝𝑛

𝑖=1
b) For the specific heat of mixture at constant volume.

𝒄𝒗𝒎 = %𝑮𝑨 𝒄𝒗𝑨 + %𝑮𝑩 𝒄𝒗𝑩 + %𝑮𝑪 𝒄𝒗𝑪
For nth ideal gases
𝑛
𝑐𝑣𝑚 = ∑ %𝐺𝑖 𝑐𝑣𝑖 = 𝐺1 𝑐𝑣1 + 𝐺2 𝑐𝑣2 + 𝐺3 𝑐𝑣3 + ⋯ + 𝐺𝑛 𝑐𝑣𝑛

𝑖=1
4. The ratio of specific heat of the mixture would be:

𝒄𝒑𝒎
𝒌𝒎 =
𝒄𝒗𝒎
5. The density and specific volume of the mixture would be

𝑷𝒎
𝛒𝒎 =
𝑹𝒎 𝑻𝒎
𝟏 𝑹𝒎 𝑻𝒎
𝒗𝒎 = =
𝛒𝒎 𝑷𝒎
6. The specific weight of the mixture would be
𝑷𝒎 𝒈
𝛄𝒎 =
𝒌𝑹𝒎 𝑻𝒎
Where k = proportionality constant
153
SOLVED PROBLEMS:
1. Determine the weight of the methane gas (CH 4) in a 2m diameter spherical container at 0°C and 20kPag in kgf.
Solution
Using ideal gas law: 𝐶𝐻4
𝑃𝑉 = 𝑚𝑅𝑇 𝑑 = 2𝑚 𝑃𝑔 = 20𝑘𝑃𝑎𝑔
the gas constant for methane would be 𝑡 = 0°𝐶
𝑅̅
𝑅=
𝑀𝑊
The molecular weight of the methane is the summation of the products of the atomic mass and number of atoms of
its respective element components. Average atomic number of carbon is 12, and atomic number of hydrogen is 1,
hence
𝑀𝑊 = 12(1) + 1(4)
𝑘𝑔
𝑀𝑊 = 16
𝑘𝑔 − 𝑚𝑜𝑙
The gas constant would be
𝑘𝐽
8.3143
𝑘𝑔 − 𝑚𝑜𝑙 − 𝐾
𝑅=
𝑘𝑔
16
𝑘𝑔 − 𝑚𝑜𝑙
𝑘𝐽
𝑅 = 0.51964
𝑘𝑔𝑚 − 𝐾
Using equation of state, taking note that the temperature and pressure must be always in their respective absolute
values.
𝑚𝑔
Since 𝑊 = , then
𝑘
𝑊𝑘𝑅𝑇
𝑃𝑉 =
𝑔
𝑃𝑉𝑔
𝑊=
𝑘𝑅𝑇
π
For the volume of the sphere: 𝑉 = 𝑑 3 , hence
6
π𝑃𝑔𝑑 3
𝑊=
6𝑘𝑅𝑇
𝑘𝑁 𝑚
(20 + 101.325) 2 (9.8066 2 ) (2𝑚)3 1000𝑁 1𝑘𝑔𝑓
𝑊= 𝑚 𝑠 [ ][ ]
𝑘𝑁 − 𝑚 𝑘𝑁 − 𝑚 1𝑘𝑁 9.806𝑁
6 (1000 ) (0.51964 ) (0 + 273)𝐾
𝑘𝑁 − 𝑠 2 𝑘𝑔𝑚 − 𝐾
𝑾 = 𝟑. 𝟓𝟖𝟐 𝒌𝒈𝒇
Another solution is using the gas constant of methane that can be found on Item B-1. For methane, the gas constant
would be
𝑘𝐽
𝑅 = 0.51845
154
Substituting the value of R from the table
𝑘𝑁 𝑚
(20 + 101.325) 2 (9.8066 2 ) (2𝑚)3 1000𝑁 1𝑘𝑔𝑓
𝑊= 𝑚 𝑠 [ ][ ]
𝑘𝑁 − 𝑚 𝑘𝑁 − 𝑚 1𝑘𝑁 9.806𝑁
6 (1000 ) (0.51845 ) (0 + 273)𝐾
𝑘𝑁 − 𝑠 2 𝑘𝑔𝑚 − 𝐾
𝑾 = 𝟑. 𝟓𝟗𝟏 𝒌𝒈𝒇
For additional information, the percent weight difference would be
3.591 − 3.582
Percent Difference = × 100
3.582
Percent Difference = 0.25%
There is atleast 0.25% difference between the value when R used is from the table and the R value from the
molecular weight. Hence we can use the values of gas constant in item B-1 with minimum errors to ease the burden
of computing molecular weight.
2. Provide 4 kg of gaseous substance 300 kJ of heat at constant volume, so that the temperature rise is 80K.
Determine
a) The specific heat at constant volume in kJ/kg-K and BTU/lbm-°R
b) If k = 1.55, compute the value of cp in kJ/kg-K and R in kcal/kgm-K
c) Identify the gas among the following:𝐶𝑂2 , 𝑂2 , 𝑁2 , 𝑆𝑂2 , and 𝐶𝐻4
d) What is the density of the gas in kgm/m3 ?
Solution
(a) For the specific heat at constant volume:
𝑄 = 𝑚𝑐𝑣 Δ𝑇
𝑄
𝑐𝑣 =
𝑚Δ𝑇
300𝑘𝐽
𝑐𝑣 =
4𝑘𝑔(80𝐾)
𝒌𝑱
𝒄𝒗 = 𝟎. 𝟗𝟑𝟕𝟓
𝒌𝒈 − 𝑲
In terms of English units, we can convert it using our definition of specific heat of water at constant pressure as a
conversion factor.
𝑘𝐽 1.0𝐵𝑇𝑈/𝑙𝑏𝑚 − °𝑅
𝑐𝑣 = 0.9375 ( )
𝑘𝑔 − 𝐾 4.187𝑘𝐽/𝑘𝑔 − 𝐾
𝑩𝑻𝑼
𝒄𝒗 = 𝟎. 𝟐𝟐𝟑𝟗
𝒍𝒃𝒎 − °𝑹
(b) For the value of cp if k = 1.55
𝑐𝑝
𝑘=
𝑐𝑣
𝑐𝑝 = 𝑘𝑐𝑣
𝑘𝐽
𝑐𝑝 = 1.55 (0.9375 )
𝑘𝑔 − 𝐾
𝑘𝐽
𝑐𝑝 = 1.4531
𝑘𝑔 − 𝐾
For Gas constant R:
𝑅 = 𝑐𝑝 − 𝑐𝑣
𝑘𝐽 1𝑘𝑐𝑎𝑙
𝑅 = (1.4531 − 0.9375) ( )
𝑘𝑔 − 𝐾 4.187𝑘𝐽
𝒌𝒄𝒂𝒍
𝑹 = 𝟎. 𝟏𝟐𝟑𝟏𝟒
𝒌𝒈 − 𝑲
155
(c) We can identify the gas by its molecular weight. Solving for the molecular weight:
𝑅̅
𝑀𝑊 =
𝑅
8.3143
𝑀𝑊 =
1.4531 − 0.9375
𝒌𝒈𝒎
𝑴𝑾 = 𝟏𝟔. 𝟏𝟐𝟓
𝒌𝒈𝒎 − 𝒎𝒐𝒍
Compute first the molecular weight of each ideal gas:
For 𝐶𝑂2 : MW = 12(1) + 16(2) = 44 kg/kg-mol
For 𝑂2 : MW = 16(2) = 32 kg/kg-mol
For 𝑁2 : MW = 14(2) = 28 kg/kg-mol
For 𝑆𝑂2 : MW = 32(1) + 16(2) = 64 kg/kg-mol
For 𝐶𝐻4 : MW = 12(1) + 1(4) = 16 kg/kg-mol
We can conclude that the gas is most likely to be 𝑪𝑯𝟒 .
(d) The density of the gas can be determined from our properties of fluid:
ρ𝑔 𝑀𝑊𝑔
𝑆𝐺 = =
ρ𝑎𝑖𝑟 𝑀𝑊𝑎𝑖𝑟
𝑀𝑊𝑔
ρ𝑔 = ρ𝑎𝑖𝑟 ( )
𝑀𝑊𝑎𝑖𝑟
𝑘𝑔 16.125
ρ𝑔 = 1.20 3 ( )
𝑚 28
𝛒𝒈 = 𝟎. 𝟔𝟗𝟏𝟏𝒌𝒈/𝒎𝟑
3. A gas initially at 15 psia and 2 ft3/min undergoes process to 90 psia and 0.6ft3/min, during which the enthalpy
𝐵𝑇𝑈
increases by 15.5 BTU/min and 𝑐𝑣 = 2.44 . Determine:
𝑙𝑏𝑚−°𝑅
a) Change in internal energy in hp.
𝐵𝑇𝑈
b) Specific heat at constant pressure in .
𝐵𝑇𝑈
c) Gas constant in .
Solution: 1 2
Δ𝐻 = 15.5 𝐵𝑇𝑈/𝑚𝑖𝑛
𝐵𝑇𝑈
𝑐𝑣 = 2.44
𝑃1 = 15𝑝𝑠𝑖𝑎 𝑃2 = 90𝑝𝑠𝑖𝑎
ft3 𝑓𝑡 3
𝑉1̇ = 2 𝑉2̇ = 0.6
min 𝑚𝑖𝑛
(a) For the change in internal energy:
Δ𝐻 = Δ𝑈 + Δ𝑃𝑉
Δ𝑈 = Δ𝐻 − Δ𝑃𝑉
Δ𝑈 = Δ𝐻 − (𝑃2 𝑉2 − 𝑃1 𝑉1 )
𝐵𝑇𝑈 𝑙𝑏𝑓 𝑓𝑡 3 144𝑖𝑛2 1𝐵𝑇𝑈 1ℎ𝑝
Δ𝑈 = [15.5 − [90(0.6) − 15(2)] 2 ∙ ( 2 )( )] [ ]
𝑚𝑖𝑛 𝑖𝑛 𝑚𝑖𝑛 1𝑓𝑡 778.16𝑓𝑡 − 𝑙𝑏𝑓 42.41𝐵𝑇𝑈/𝑚𝑖𝑛
𝚫𝑼 = 𝟎. 𝟐𝟔𝟏 𝒉𝒑
(b)
For specific heat at constant volume:
156
Δ𝑈 = 𝑚𝑐𝑣 Δ𝑇 (1)
For the specific heat at constant pressure:
Δ𝐻 = 𝑚𝑐𝑝 Δ𝑇 (2)
Since for both specific heats, the mass and change in temperature are equal, we divide equation (1) to (2):
Δ𝑈 𝑚𝑐𝑣 Δ𝑇
=
Δ𝐻 𝑚𝑐𝑝 Δ𝑇
Δ𝑈 𝑐𝑣
=
Δ𝐻 𝑐𝑝
Solving for 𝑐𝑝
𝛥𝐻
𝑐𝑝 = 𝑐𝑣 ( )
𝛥𝑈
𝐵𝑇𝑈 1ℎ𝑝
𝐵𝑇𝑈 15.5 ( )
𝑚𝑖𝑛 42.41𝐵𝑇𝑈/𝑚𝑖𝑛
𝑐𝑝 = 2.44 [ ]
𝑙𝑏𝑚 − °𝑅 . 261ℎ𝑝
𝑩𝑻𝑼
𝒄𝒑 = 𝟑. 𝟒𝟏𝟔𝟕
(c) For gas constant R:

𝑅 = 𝑐𝑝 − 𝑐𝑣
𝑅 = 3.4167 − 2.44
𝑩𝑻𝑼
𝑹 = 𝟎. 𝟗𝟕𝟔𝟕
2. A 50cm steel cube of relative density 8.3 is submerge to the mercury of relative density of 13.6
a) What portion of the cube is submerged to the mercury?
b) What percentage of the body that is not submerged?
Solution
(a) For the height of the body that is submerged to mercury:
By equilibrium
∑ 𝐹𝑣 = 0
𝐹𝐵𝐵 = 𝑊 Mercury
γ𝐻𝑔 𝑉𝑠 = γ𝐵 𝑉𝐵 ℎ𝐵 = 50 𝑐𝑚
γ𝑤 (𝑆𝐺𝐻𝑔 )𝐴ℎ𝑠 = γ𝑤 (𝑆𝐺𝐵 )𝐴ℎ𝐵
𝑆𝐺𝐻𝑔 ℎ𝑠 = 𝑆𝐺{𝐵) ℎ𝐵 ℎ𝑠
𝑆𝐺𝐵
ℎ𝑠 = ℎ𝑏 ( )
𝑆𝐺𝐻𝑔
8.3
ℎ𝑠 = 50𝑐𝑚 ( )
13.6
𝒉𝒔 = 𝟑𝟎. 𝟓𝟐𝒄𝒎
(b)
The percentage of body that is submerged to mercury would be
157
ℎ𝑠
𝑞= × 100
ℎ𝑏
Hence the percentage of the body that is not submerged to mercury would be:
𝑝= 1−𝑞
ℎ𝑠
𝑝 = (1 − ) 100
ℎ𝑏
30.52
𝑝 =1−
50
𝒑 = 𝟑𝟖. 𝟗𝟔%
3. A spherical body is half submerged to water. If the diameter of the body is 3 ft, calculate:
a) The buoyant force supporting the body in lbf.
b) What volume of an anchor in gallons is needed to completely submerged the body to water if the specific
gravity of the anchor is 6 in?
c) What is the relative density of the body?
𝑊𝐵
Solution
(a) For the buoyant force:
𝐹𝐵𝐵 = 𝛾𝑤 𝑉𝑠 Water
1 π
For volume of the body submerged: 𝑉𝑠 = ( 𝑑 3 ), hence: 3 ft
2 6
𝜋 3
𝐹𝐵𝐵 = 𝛾𝑤 ( 𝑑 )
12
𝑙𝑏𝑓 𝜋
𝐹𝐵𝐵 = 64 3 ( ) (3𝑓𝑡)3
𝑓𝑡 12
𝑭𝑩𝑩 = 𝟒𝟓𝟐. 𝟑𝟗𝒍𝒃𝒇 𝑭𝑩𝑩
By summation of forces:
∑ 𝐹𝑣 = 0 𝑊𝐵
𝐹𝐵𝐵 = 𝑊𝐵 Water
𝑊𝐵 = 452.39𝑙𝑏𝑓
(b) By summation of forces:

∑ 𝐹𝑣 = 0
𝐹𝐵𝐵 + 𝐹𝐵𝐴 = 𝑊𝐵 + 𝑊𝐴
γ𝑤 𝑉𝑠 + 𝐹𝐵𝐴 = 𝑊𝐵 + γ𝐴 𝑉𝐴
π
γ𝑤 ( 𝑑 3 ) + γ𝑤 𝑉𝐴 = 𝑊𝐵 + (𝑆𝐺𝐴 )γ𝑤 𝑉𝐴 𝐹𝐵𝐵
6
π 3 π
γ𝑤 ( 𝑑 ) + γ𝑤 𝑉𝐴 = γ𝑤 ( 𝑑 3 ) + (𝑆𝐺𝐴 )γ𝑤 𝑉𝐴
6 12
π 3 π 3
𝑑 + 𝑉𝐴 = 𝑑 + 𝑆𝐺𝐴 𝑉𝐴
6 12
Solving for 𝑉𝐴 𝑊𝐴
πd3
𝑉𝐴 =
12(𝑆𝐺𝐴 − 1)
𝐹𝐵𝐴
158
7.481𝑔𝑎𝑙
π(3𝑓𝑡)3 ( )
1𝑓𝑡 3
𝑉𝐴 =
12(6 − 1)
𝑽𝑨 = 𝟏𝟎. 𝟓𝟕𝟔𝒈𝒂𝒍
c) For relative density of the body:
𝑊𝐵 = γ𝑏 𝑉𝑏
𝑊𝐵 = 𝑆𝐺𝐵 γ𝑤 𝑉𝐵
𝑊𝐵
𝑆𝐺𝐵 =
γ𝑤 𝑉𝐵
For weight of the body:
∑ 𝐹𝑣 = 0
𝐹𝐵𝐵 = 𝑊𝐵
𝑊𝐵 = 452.39𝑙𝑏𝑓
π
Hence if 𝑉𝐵 = 𝑑 3
6
6𝑊𝐵
𝑆𝐺𝐵 =
πγ𝑤 𝑑 3
6(452.39𝑙𝑏𝑓)
𝑆𝐺𝐵 =
𝑙𝑏𝑓
π (64 3 ) (3𝑓𝑡)3
𝑓𝑡
𝑺𝑮𝑩 = 𝟎. 𝟓𝟎
4. It is planned to lift logs from inaccessible virgin forest by means of balloons. Helium at atmospheric pressure
and temperature of 21.1 °C is to be used in the balloon. What minimum balloon diameter will be required for a
gross lifting force of 20 metric tons?
𝑊𝐵
Solution: Air
By summation of forces 𝑃 = 101.325 𝑘𝑃𝑎𝑎 𝐻𝑒2
∑ 𝐹𝑣 = 0 𝑡 = 21.1°C 𝑃 = 101.325 𝑘𝑃𝑎𝑎 𝑑 =?
𝑡 = 21.1°C
𝐹𝐵𝐵 + 𝐹𝐵𝐿 = 𝑊𝐵 + 𝑊𝐿
Since the volume of the log is not mentioned, we assume that the buoyant
force in the log is negligible: 𝐹𝐵𝐵
𝐹𝐵𝐵 = 𝑊𝐵 + 𝑊𝐿
𝐹𝐵𝐵 − 𝑊𝐵 = 𝑊𝐿
γ𝑎𝑖𝑟 𝑉𝐵 − γ𝐻𝑒 𝑉𝐵 = 𝑊𝐿 𝑊𝐿
𝑊𝐿
𝑉𝐵 =
γ𝑎𝑖𝑟 − γ𝐻𝑒
For the volume of balloon submerged in “air”:
π 3
𝑉𝐵 = 𝑑
6
Hence: 𝐹𝐵𝐿
π 3 𝑊𝐿
𝑑 =
6 γ𝑎𝑖𝑟 − γ𝐻𝑒
159
Solving for diameter:

3 6𝑊𝐿
𝑑=√
π(γ𝑎𝑖𝑟 − γ𝐻𝑒 )
From Item B1: The gas constants for air and helium are:
𝐽
𝑅𝑎𝑖𝑟 = 287.08
𝐽
𝑅𝐻𝑒 = 2077.67
Hence
𝑃𝑔 𝑃𝑔
γ𝑎𝑖𝑟 − γ𝐻𝑒 = −
𝑘𝑅𝑎𝑖𝑟 𝑇 𝑘𝑅𝐻𝑒 𝑇
𝑃𝑔 1 1
γ𝑎𝑖𝑟 − γ𝐻𝑒 = [ − ]
𝑘𝑇 𝑅𝑎𝑖𝑟 𝑅𝐻𝑒
𝑘𝑁 𝑚
(101.325 2 ) (9.8066 2 ) 1 1 𝑘𝑔𝑚 − 𝐾
γ𝑎𝑖𝑟 − γ𝐻𝑒 = 𝑚 𝑠 [ − ]
(1000 ) (21.1 + 273)𝐾 0.28708 2.07767 𝑘𝑁 − 𝑚
𝑘𝑁 − 𝑠 2
γ𝑎𝑖𝑟 − γ𝐻𝑒 = 0.0101428𝑘𝑁/𝑚3
The diameter of the balloon would be:
3 6𝑊𝐿
𝑑=√
π(γ𝑎𝑖𝑟 − γ𝐻𝑒 )
3 6𝑚𝐿 𝑔
𝑑=√
πk(γ𝑎𝑖𝑟 − γ𝐻𝑒 )
1000𝑘𝑔𝑚 𝑚
6(20𝑡𝑜𝑛𝑛𝑒) (
3 ) (9.8066 2 )
1𝑡𝑜𝑛𝑛𝑒 𝑠
𝑑=√
𝑘𝑔𝑚 − 𝑚 𝑘𝑁
π (1000 2 ) (0.0101428 3 )
𝑘𝑁 − 𝑠 𝑚
𝒅 = 𝟑𝟑. 𝟑𝟎𝟐 𝒎
160
5. An acetylene gas at 350 kPaa and 30°C is stored into a 10 m3 container. If some of the acetylene gas was used
for welding purposed, the pressure and temperature of the remaining gas inside the tank drops to 250 kPaa and 20°C
respectively.
a) What proportion of the acetylene was used?
b) What volume would the used acetylene gas occupy at 101.325 kPaa and 25°C?
Solution:
1 2 𝑚𝑢𝑠𝑒𝑑 = 𝑚3 =?
𝑃1 = 350𝑘𝑃𝑎𝑎 𝑃2 = 250𝑘𝑃𝑎𝑎
𝑡1 = 30°𝐶 𝑡2 = 20°𝐶
𝑚1 𝑚2
𝑉1 = 𝑉2 = 10𝑚3
Let 𝑚1 = initial mass of the acetylene tank
𝑚2 = mass of acetylene after used.
𝑚3 = 𝑚𝑢𝑠𝑒𝑑 = mass of acetylene used
a) For proportion of acetylene used
At point 1:
𝑃1 𝑉1 = 𝑚1 𝑅𝑇1 (1)
At point 2:
𝑃2 𝑉2 = 𝑚2 𝑅𝑇2 (2)
Since 𝑉1 = 𝑉2 , then dividing equation (1) to (2)
𝑃1 𝑚1 𝑇1
=
𝑃2 𝑚2 𝑇2
Hence
𝑚1 𝑃1 𝑇2
=
𝑚2 𝑃2 𝑇1
𝑚1 (350𝑘𝑃𝑎𝑎)(20 + 273)𝐾
=
𝑚2 (250𝑘𝑃𝑎𝑎)(30 + 273)𝐾
𝑚1
= 1.3538
𝑚2
By conservation of mass:
𝑚𝑖𝑛 = 𝑚𝑜𝑢𝑡 Δ𝑚 = 0
𝑚1 = 𝑚2 + 𝑚3
𝑚3 = 𝑚1 − 𝑚2
Since the percentage of used gas per original mass of gas is needed, then
𝑚3 𝑚2
=1−
𝑚1 𝑚1
Hence
𝑚3 1
= 1−
𝑚1 1.3538
𝑚3
= 0.2613
𝑚1
% 𝒖𝒔𝒆𝒅 = 𝟐𝟔. 𝟏𝟑%
161
(b) For the volume of used acetylene gas:

𝑃3 𝑉3 = 𝑚3 𝑅𝑇3
𝑃3 𝑉3
𝑚3 =
𝑅𝑇3
From 𝑚1
𝑃1 𝑉1
𝑚1 =
𝑅𝑇1
From our answer at (a)

𝑚3
= 0.2613
𝑚1
𝑚3 = 0.2613𝑚1
𝑃3 𝑉3 0.2613𝑃1 𝑉1
=
𝑅𝑇3 𝑅𝑇1
𝑃3 𝑉3 0.2613𝑃1 𝑉1
=
𝑇3 𝑇1
Solving for the volume 𝑉3 :
𝑃1 𝑇3
𝑉3 = 0.2613𝑉1 ( ) ( )
𝑃3 𝑇1
350 25 + 273
𝑉3 = 0.2613(10𝑚3 ) ( )( )
101.325 30 + 273
𝑽𝟑 = 𝟖. 𝟖𝟕𝟖𝒎𝟑
6. A cylindrical tank 2m diameter by 4m height contains oxygen gas at 2MPa and 27°C.
a) How many drums 1m diameter by 2m long which are initially devoid of any gas, can be filled to a pressure
of 10 bar abs and 20°C, assuming that the temperature of the oxygen left in the tank remains 27°C?
b) How many drums 1m diameter by 2m long which contains oxygen gas at 101.325kPaa and 25°C, can be
filled to a pressure of 10 bar abs and 20°C, assuming that the temperature of the oxygen left in the tank
remains 27°C?
Solution:
(1) (2) (3)
d = 2m
d = 1m
a)
𝑃1 = 2𝑀𝑃𝑎 h = 4m 𝑃2 = 10𝑏𝑎𝑟 𝑎 h = 2m
𝑡1 = 27°𝐶 𝑡2 = 27°𝐶
𝑃3 = 10𝑏𝑎𝑟 𝑎
𝑡3 = 20°𝐶
𝑚1 𝑚2 𝑚3 = 𝑚𝑢𝑠𝑒𝑑
𝑉1 𝑉2 𝑉3
By conservation of mass
𝑚𝑖𝑛 = 𝑚𝑜𝑢𝑡
𝑚1 = 𝑚2 + 𝑚3
𝑚3 = 𝑚1 − 𝑚2
𝑃1 𝑉1 𝑃2 𝑉2
𝑚3 = −
𝑅𝑇1 𝑅𝑇2
162
π
Since 𝑉1 = 𝑉2 = 𝑑 2 ℎ, 𝑇1 = 𝑇2 , then
4
𝑉1 (𝑃1 − 𝑃2 )
𝑚3 =
𝑅𝑇1
π𝑑 2 ℎ(𝑃1 − 𝑃2 )
𝑚3 =
4𝑅𝑇1
From Item B1, the gas constant for oxygen would be:
𝐽
𝑅 = 259.90
𝑘𝑔 − 𝐾
Hence
π(2𝑚)2 (4𝑚)(2000 − 1000)𝑘𝑃𝑎𝑎
𝑚3 =
𝑘𝑁 − 𝑚
4 (0.25990 ) (27 + 273)𝐾
𝑘𝑔 − 𝐾
𝑚3 = 161.17𝑘𝑔
For the mass available per drum:

𝑃3 𝑉3
𝑚𝑎𝑣𝑎𝑖𝑙 =
𝑅𝑇3
π
1000 𝑘𝑃𝑎𝑎 ( ) (1𝑚)2 (2𝑚)
𝑚𝑎𝑣𝑎𝑖𝑙 = 4
𝑘𝑁 − 𝑚
(0.25990 ) (20 + 273)𝐾
𝑘𝑔 − 𝐾
𝑘𝑔
𝑚𝑎𝑣𝑎𝑖𝑙 = 20.63
𝑑𝑟𝑢𝑚
Let n = number of drums needed, then
𝑚3
𝑛=
𝑚𝑎𝑣𝑎𝑖𝑙
161.17𝑘𝑔
𝑛=
𝑘𝑔
20.63
𝑑𝑟𝑢𝑚
𝑛 = 7.812 𝑑𝑟𝑢𝑚𝑠
Say
𝒏 = 𝟖 𝒅𝒓𝒖𝒎𝒔
(b) For same used gas, if 𝑃4 = 101.325𝑘𝑃𝑎𝑎, and 𝑡3 = 25℃ are properties of oxygen initially in the drum, then
the new mass available would be
𝑚𝑎𝑣𝑎𝑖𝑙 ′ = 𝑚𝑎𝑣𝑎𝑖𝑙 − 𝑚4
Where
𝑃4 𝑉4
𝑚4 =
𝑅𝑇4
π
101.325 𝑘𝑃𝑎𝑎 ( ) (1𝑚)2 (2𝑚)
𝑚4 = 4
𝑘𝑁 − 𝑚
(0.25990 ) (25 + 273)𝐾
𝑘𝑔 − 𝐾
𝑘𝑔
𝑚4 = 2.055
𝑑𝑟𝑢𝑚
Hence
𝑚𝑎𝑣𝑎𝑖𝑙 ′ = 20.63 − 2.055
𝑘𝑔
𝑚𝑎𝑣𝑎𝑖𝑙 ′ = 18.575
𝑑𝑟𝑢𝑚
163
Then let n ‘ = the number of drums to be used if each drum initially contains oxygen:
𝑚3
𝑛′ =
𝑚𝑎𝑣𝑎𝑖𝑙 ′
161.17𝑘𝑔
𝑛′ =
𝑘𝑔
18.575
𝑑𝑟𝑢𝑚
𝑛′ = 8.678𝑑𝑟𝑢𝑚𝑠
Say
𝒏′ = 𝟗 𝒅𝒓𝒖𝒎𝒔
7. A tire contains a certain volume of air at 30 psig amd 70°F. Due to running conditions, the temperature of the air
in the tire rises to 160°F, what will be the gage pressure? Assume that the tire does not stretch . P Baro=29.50 in Hg.
Solution: 1 2
𝑃𝑏𝑎𝑟𝑜 = 29.50𝑖𝑛𝐻𝑔
𝑃1𝑔 = 30𝑝𝑠𝑖𝑔 𝑃2𝑔 =?

𝑡1 = 70°𝐹 𝑡2 = 160°𝐹
Since volume is constant, we can use Gay Lusaac’s Law:
𝑃1 𝑇1
=
𝑃2 𝑇2
For 𝑃𝑏𝑎𝑟𝑜
14.7𝑝𝑠𝑖
𝑃𝑏𝑎𝑟𝑜 = 29.50𝑖𝑛𝐻𝑔 ( )
29.92𝑖𝑛𝐻𝑔
𝑃𝑏𝑎𝑟𝑜 = 14.494𝑝𝑠𝑖
Hence
30 + 14.494 70 + 460
=
𝑃2𝑔 + 14.494 160 + 460
𝑷𝟐𝒈 = 𝟑𝟕. 𝟓𝟔𝒑𝒔𝒊𝒈
164
10. The atmospheric pressure at the base of a mountain is 730 mmHg and at its top is 365 mmHg. The atmospheric
temperature is 15.55°C, constant from base to top. Local g = 9.75 m/s2. Find the height of the mountain.
Solution: Consider the figure below:

2
ℎ2 𝑃2 = 365𝑚𝑚𝐻𝑔
𝑡 = 15.55℃
𝑚
ℎ=? 𝑔 = 9.75 2
𝑠
ℎ1 1 𝑃1 = 730𝑚𝑚𝐻𝑔
From definition of pressure:

𝑃 = γℎ
Differentiating implicitly:
𝑑𝑃 = γ𝑑ℎ + ℎ𝑑γ
Since γ is assume constant, 𝑑γ = 0, then
We consider air as our working substance, then the specific weight of this ideal gas is
𝑃𝑔
γ=
𝑘𝑅𝑇
Where k = proportionality constant
Thus:
𝑃𝑔
𝑑𝑃 = 𝑑ℎ
𝑘𝑅𝑇
By separation of variables:
𝑑𝑃 𝑔
= 𝑑ℎ
𝑃 𝑘𝑅𝑇
Integrating both sides, from upper limit to its lower limits
1 2
𝑑𝑃 𝑔
∫ = ∫ 𝑑ℎ
2 𝑃 𝑘𝑅𝑇 1
𝑃1 𝑔
𝑙𝑛 = (ℎ − ℎ1 )
𝑃2 𝑘𝑅𝑇 2
But ℎ = ℎ2 − ℎ1, then
𝑃1 𝑔ℎ
𝑙𝑛 =
𝑃2 𝑘𝑅𝑇
Solving for the height of the mountain:
𝑘𝑅𝑇 𝑃1
ℎ= 𝑙𝑛
𝑔 𝑃2
𝑘𝐽
From item B1: 𝑅𝑎𝑖𝑟 = 0.28708
𝑘𝑔−𝐾
𝑘𝑔𝑚 − 𝑚 𝑘𝐽
(1000 ) (0.28708 ) (15.5 + 273)𝐾 730 𝑚𝑚𝐻𝑔
𝑘𝑁 − 𝑠2 𝑘𝑔 − 𝐾
ℎ= 𝑚 𝑙𝑛
9.75 2 365 𝑚𝑚𝐻𝑔
𝑠
𝒉 = 𝟓𝟖𝟖𝟖. 𝟎𝟐 𝒎
165
11. A submarine releases bubbles from the bottom of the sea. At what depth must the bubble be released in meters
so that its diameter upon reaching the seawater surface will be thrice the diameter when it was released. Specific
gravity of the seawater is 1.13 and thermal equilibrium was observed at all times.
Solution. Consider the figure below:
Seawater
(SG = 1.13)
2
h=?
The air bubbles are assumed came from the exhaust of a submarine at a certain depth h , and will rise at the free
surface of seawater. Since temperature is constant, then we can apply Boyle’s Law:
𝑃1 𝑉1 = 𝑃2 𝑉2
π 3
Since 𝑉 = 𝑑 , then
6
π π
𝑃1 𝑑1 3 = 𝑃2 𝑑2 3
6 6
𝑃1 𝑑1 3 = 𝑃2 𝑑2 3
Since 𝑑2 = 3𝑑1 , then
𝑃1 𝑑1 3 = 𝑃2 (3𝑑1 )3
𝑃1 = 27𝑃2 (1)
For 𝑃2
𝑃2 = 𝑃𝑎𝑡𝑚 = 101.325𝑘𝑃𝑎𝑎.
For 𝑃1 ,
𝑃1 = 𝑃1𝑔 + 𝑃𝑎𝑡𝑚
Since 𝑃2𝑔 is a hydrostatic pressure that varies with depth of the bubble
𝑃1𝑔 = 𝛾𝑠𝑤 ℎ
𝑃1𝑔 = (𝑆𝐺𝑠𝑤 )𝛾𝑤 ℎ
𝑃1 = (𝑆𝐺𝑠𝑤 )𝛾𝑤 ℎ + 𝑃𝑎𝑡𝑚
From equation (1)
𝑃1 = 27𝑃2
𝑃1𝑔 + 𝑃𝑎𝑡𝑚 = 27𝑃𝑎𝑡𝑚
(𝑆𝐺𝑠𝑤 )𝛾𝑤 ℎ = 26𝑃𝑎𝑡𝑚
26𝑃𝑎𝑡𝑚
ℎ=
(𝑆𝐺𝑠𝑤 )𝛾𝑤
𝑘𝑁
26 (101.325 2 )
ℎ= 𝑚
𝑘𝑁
1.13 (9.8066 3 )
𝑚
𝒉 = 𝟐𝟑𝟕. 𝟕𝟑 𝒎
166
12. An empty tin can with height of 6 in and a diameter of 3 in is open at one end and closed on the other end. If
the vertical can, with the open end down, be slowly immersed in water, how must will the water rise inside the can
in inches? Thermal equilibrium is observed at all times.
Solution: At point 1, the cylindrical can is open to the atmosphere, and at point (2), air is trapped in the cylindrical
can as its closed end is vertically immersed in the fluid surface of water. Using the figure:
0.25 ft
0.5 ft Closed End

Water
1
0.5 − 𝑥
2
𝑥
Assume air trapped in the cylinder as our working substance, and also if temperature is constant, then we can apply
Boyle’s Law:
𝑃1 𝑉1 = 𝑃2 𝑉2
For 𝑃1 :
𝑃1 = 𝑃𝑎𝑡𝑚 = 14.7 2 ( )
𝑖𝑛 1𝑓𝑡 2
𝑙𝑏𝑓
𝑃1 = 2116.9 2
𝑓𝑡
For 𝑉1 :
π
𝑉1 = 𝑑 2 ℎ
4
π
𝑉1 = (0.25𝑓𝑡)2 (0.5𝑓𝑡)
4
𝑉1 = 0.024545𝑓𝑡 3
For 𝑃2 , it is caused by the hydrostatic pressure at the air-water interface inside the can.
𝑃2𝑔 = γ𝑊 ℎ
Since ℎ = 0.5 − 𝑥
𝑃2𝑔 = γ𝑊 (0.5 − 𝑥)
For absolute pressure:
𝑃2 = 𝑃2𝑔 + 𝑃𝑎𝑡𝑚
𝑙𝑏𝑓 𝑙𝑏𝑓 144𝑖𝑛2
𝑃2 = 62.4 3 (0.5 − 𝑥)𝑓𝑡 + 14.7 2 ( )
𝑓𝑡 𝑖𝑛 1𝑓𝑡 2
𝑙𝑏𝑓
𝑃2 = (2148 − 62.4𝑥) 2
𝑓𝑡
For 𝑉2 , it is the volume of air trapped in the cylinder.
π
𝑉2 = 𝑑 2 ℎ
4
Also if ℎ = 0.5 − 𝑥 :
π
𝑉2 = (0.25𝑓𝑡)2 (0.5 − 𝑥)𝑓𝑡
4
𝑉2 = (0.024544 − 0.049087𝑥)𝑓𝑡 3
167
Hence:
𝑃1 𝑉1 = 𝑃2 𝑉2
𝑙𝑏𝑓 𝑙𝑏𝑓
(2116.9 2 ) (0.024545𝑓𝑡 3 ) = (2148 − 62.4𝑥) 2 (0.024544 − 0.049087𝑥)𝑓𝑡 3
𝑓𝑡 𝑓𝑡
Dropping all the units
(2116.9)(0.024545) = (2148 − 62.4𝑥)(0.024544 − 0.049087𝑥)
Simplify:
51.959 = 52.721 − 105.439𝑥 − 1.532𝑥 + 3.063𝑥 2

Hence:
3.063𝑥 2 − 103.907𝑥 + 0.762 = 0
By Quadratic Formula:
−𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
If 𝑎 = 3.063, 𝑏 = −103.907, and 𝑐 = 0.762:
−(−103.907) ± √(−103.907)2 − 4(3.063)(0.762)
𝑥=
2(3.063)
103.907 ± 103.862
𝑥=
2(3.063)
Take the positive sign:
103.907 + 103.862
𝑥=
2(3.063)
12𝑖𝑛
𝑥 = 33.916𝑓𝑡 ( )
1𝑓𝑡
𝑥 = 406.992 𝑖𝑛
If 𝑥 ≫ ℎ, then we will reject this solution.
Consider the negative sign:

103.907 − 103.862
𝑥=
2(3.063)
12𝑖𝑛
𝑥 = 0.0073457 𝑓𝑡 ( )
1𝑓𝑡
𝒙 = 𝟎. 𝟎𝟖𝟖 𝒊𝒏
Since 𝑥 < ℎ, then this is the answer.
168
13. The gravimetric analysis of dry air is approximately 23.1% oxygen gas and 76.9% nitrogen gas. Calculate:
a) Volumetric analysis
b) The gas constant
c) The respective partial pressures
d) The specific volume and density at 1 atm, 15.6°C
e) How many kilograms of oxygen gas must be added to 2.27 kg of air to produce a mixture which is 50%
oxygen gas by volume?
Solution:
Oxygen Nitrogen
𝑷𝑶𝟐 𝑷𝑵𝟐
𝑽𝑶𝟐 𝑽𝑵𝟐
𝑻𝑶𝟐 𝑻𝑵𝟐
𝒎𝑶𝟐 𝒎𝑵𝟐
𝒏𝑶𝟐 𝒏𝑵𝟐
%𝑮𝑶𝟐 = 𝟐𝟑. 𝟏% %𝑮𝑵𝟐 = 𝟕𝟔. 𝟗%
From Item B1, the following properties of oxygen gas and nitrogen gas are as follows:
𝑘𝑔 𝑘𝑔
𝑀𝑊𝑂2 = 32 𝑀𝑊𝑁2 = 28.016
𝑘𝑔𝑚𝑜𝑙 𝑘𝑔𝑚𝑜𝑙
𝑘𝐽 𝑘𝐽
𝑅𝑂2 = 0.25990 𝑅𝑁2 = 0.29686
𝑘𝑔-K 𝑘𝑔-K
b) For the gas constant of the mixture
𝑅𝑚 = %𝐺𝑂2 𝑅𝑂2 + %𝐺𝑁2 𝑅𝑁2
𝑅𝑚 = 0.231(0.25990) + 0.769(0.296886)
𝒌𝑱
𝑹𝒎 = 𝟎. 𝟐𝟖𝟖𝟏
𝒌𝒈-K
For the molecular weight of the mixture

𝑅̄
𝑀𝑊𝑚 =
𝑅𝑚
8.3143
𝑀𝑊𝑚 =
0.2881
𝑘𝑔
𝑀𝑊𝑚 = 28.859
𝑘𝑔𝑚𝑜𝑙
a) The respective volumetric analysis of each gas would be
𝑀𝑊𝑚
%𝑉𝑂2 = %𝐺𝑂2
𝑀𝑊𝑂2
28.859
%𝑉𝑂2 = 0.231 ( )
32
%𝑽𝑶𝟐 = 𝟐𝟎. 𝟖𝟑%
𝑀𝑊𝑚
%𝑉𝑁2 = %𝐺𝑁2
𝑀𝑊𝑁2
28.859
%𝑉𝑁2 = 0.769( )
28.016
%𝑽𝑶𝟐 = 𝟕𝟗. 𝟐𝟏%
169
Another solution for the volumetric analysis of nitrogen gas
%𝑉𝑁2 = 1-%VN2
%𝑉𝑁2 = 1-0.2083
%VN2 = 79.17%
c) For the pressure of each respective components. Using law of partial pressures:
𝑃𝑚 = 𝑃𝑂2 + 𝑃𝑁2
From gravimetric analysis
𝑚𝑂2
%𝐺𝑂2 =
𝑚𝑚
Using ideal gas law, 𝑃𝑉 = 𝑚𝑅𝑇
𝑃𝑂2 𝑉𝑂2
𝑅𝑂2 𝑇𝑂2
%𝐺𝑂2 =
𝑃𝑚 𝑉𝑚
𝑅𝑚 𝑇𝑚
Since the temperature of the mixture is also equal to the temperature of each components, and the volume of the
mixture is equal also to the volume of the gas, then
𝑃𝑂2 𝑅𝑚
%𝐺𝑂2 =
𝑃𝑚 𝑅𝑂2
At standard conditions, the pressure of air is 101.325 kPaa. Hence the pressure of the mixture is assumed to be at
standard conditions. Solving for the pressure of oxygen gas.
𝑅𝑂2
𝑃𝑂2 = %𝐺𝑂2 𝑃𝑚 ( )
𝑅𝑚
0.2589
𝑃𝑂2 = (0.231)(101.325𝑘𝑃𝑎𝑎) ( )
0.2881
𝑷𝑶𝟐 = 𝟐𝟏. 𝟎𝟑𝒌𝑷𝒂𝒂
Then the pressure exerted by the nitrogen gas would be

𝑃𝑁2 = 𝑃𝑚 -PO2
𝑃𝑁2 = 101.325-21.03
𝑷N2 = 𝟖𝟎. 𝟐𝟗𝟓𝒌𝑷𝒂𝒂
d. The density of the mixture is determined by ideal gas law. At 101.325 kPaa and 15.6°C
𝑃𝑚
𝜌𝑚 =
𝑅𝑚 𝑇𝑚
101.325𝑘𝑃𝑎𝑎
𝜌𝑚 =
𝑘𝐽
(0.2881 ) (15.6 + 273)𝐾
𝑘𝑔-K
𝒌𝒈𝒎
𝝆𝒎 = 𝟏. 𝟐𝟏𝟖𝟔 𝟑
𝒎
The specific volume of the mixture would be:
1
𝜈𝑚 =
𝜌𝑚
1 𝑚3
𝜈𝑚 =
1.2186 𝑘𝑔𝑚
𝒎𝟑
𝝂𝒎 = 𝟎. 𝟖𝟐𝟏
𝒌𝒈𝒎
170
e) For the initial mass of O2 and N2 given that the mass of air is 2.27 kgm.
𝑚𝑂2 = %𝐺𝑂2 𝑚𝑚
𝑚𝑂2 = (0.231)(2.27𝑘𝑔)
𝑚𝑂2 = 0.52𝑘𝑔𝑚
𝑚𝑁2 = 𝑚𝑚 - mO2
𝑚𝑁2 = 2.27-0.52
mN2 = 1.75𝑘𝑔𝑚
For the new mixture, let mO2’ be the mass of oxygen added to the initial mass of the mixture
𝑚𝑚 ′ = 𝑚𝑂2 + 𝑚𝑁2 + 𝑚𝑂2 ′
𝑚𝑚 ′ = 𝑚𝑚 + 𝑚𝑂2 ′
𝑚𝑚 ′ = 2.27 + 𝑚𝑂2 ′
The gravimetric analysis of the added oxygen gas as a function of its volumetric analysis would be:
𝑀𝑊𝑂2
%𝐺𝑂2 ′ = %𝑉𝑂2 ′
𝑀𝑊𝑚 ′
𝑚𝑂2 + 𝑚𝑂2 ′ 𝑀𝑊𝑂2
= %𝑉𝑂2 ′
𝑚𝑚 ′ 𝑀𝑊𝑚 ′
The new molecular weight of the mixture in terms of its desired volumetric analysis:
𝑀𝑊𝑚 ′ = %𝑉𝑂2 ′𝑀𝑊𝑂2 + %𝑉𝑁2 ′𝑀𝑊𝑁2
𝑀𝑊𝑚 ′ = (0.50)(32) + (0.50)(28)
𝑘𝑔
𝑀𝑊𝑚 ′ = 30
𝑘𝑔𝑚𝑜𝑙
= %𝑉𝑂2 ′
𝑚𝑚 ′ 𝑀𝑊𝑚 ′
= %𝑉𝑂2 ′
𝑚𝑚 + 𝑚𝑂2 ′ 𝑀𝑊𝑚 ′
0.52 + 𝑚𝑂2 ′ 32
= (0.50)( )
2.27 + 𝑚𝑂2 ′ 30
0.52 + 𝑚𝑂2 ′ = 0.5333(2.27 + 𝑚𝑂2 ′)
0.52 + 𝑚𝑂2 ′ = 1.2106 + 0.5333𝑚𝑂2 ′
1.2106-0.52
𝑚𝑂2 ′ =
1-0.5333
𝒎𝑶𝟐 ′ = 𝟏. 𝟒𝟕𝟗𝒌𝒈𝒎
171
CHAPTER III: IDEAL GAS __________________
SUPPLEMENTARY PROBLEMS
Directions: Solve the following problems and show all complete solutions. Any erasures in the computations will
consider your answer as wrong. Enclosed your final answer by a rectangle. Answers with wrong units are considered
also as wrong. For gas constants, refer to Item B1.
Buoyancy in Ideal Gas
1. A solid cube material is 0.75 cm on each side. If it is floats in oil of density 800 kgm/m 3 with one third of the
block out of the oil, what is the density of the material of the cube in kgm/m 3?
2. An object weighs 1.5 gram force in air and 1.250 gram force in water. Determine the specific gravity of the
object.
3. A piece of metal weighs 350 N in air and 240 N in water.

a) Determine the volume of the piece of metal in cubic meter.
b) What is the unit weight of the metal in kN/m3?
c) What is the specific gravity of the metal?
172
4. A solid block having a specific gravity of 3.47 is placed in a cylinder containing mercury having a specific
gravity of 13.6.
a) What part of the block is floating above the mercury level in terms of percentage?
b) If the volume of the block is 0.018m3, what is the weight of the block in N?
c) What minimum downward force is required in (b) to make it completely submerged in mercury?
5. An open top cylindrical tank 1 meter in diameter is 2.80 meters tall. The tank floats in water with a draft of 2.30
meters. Unit weight of lead is 110 kN/m3.
a) What is the weight of the tank in kN?
b) How much lead must be fastened to the outside bottom of the tank in order to make it completely submerge
in water in kN?
c) How much lead must be placed inside the tank in order to make it completely submerged in water in kN?
173
6. (a)A spherical balloon containing helium at atmospheric pressure and temperature of 21°C is used to carry a load
of 10 metric tons, what minimum balloon diameter in meters is required for this purpose? (b) Assume that the
balloon is filled with hydrogen at the same pressure and temperature, what total mass can the balloon support in
tonnes?
7. A 1ft steel cube of relative density 8.3 is submerge to the mercury of relative density of 13.45.
a) What portion of the cube in cm is submerged to the mercury?
b) What percentage of the body that is not submerge to the mercury?
174
8. A spherical body is 1/5 submerged to water. If the diameter of the body is 3ft, determine:
a) The buoyant force supporting the body in kgf.
b) What volume of an anchor is needed to completely submerged the body in gallons if the specific gravity of
the anchor is 6.
c) What is the relative density of the body?
9. It is planned to lift and move logs from almost inaccessible forest areas by means of balloons. What minimum
balloon diameter in meters, assume to be spherical will be required for a gross lifting forces of 15 metric tons.
Helium at atmospheric pressure and temperature of 21.1°C is to be used in the balloon.
175
10. A balloon considered spherical, is 30 ft in diameter. The surrounding air is at 60°F and the barometer reads
29.60 in Hg. What gross load may the balloon lift it is filled with
a) Hydrogen at 70°F at atmospheric pressure
b) Helium at 70°F at atmospheric pressure
11. A spherical hot-air balloon is initially filled with air at 120 kPa and 20°C with an initial diameter of 5 m. Air
enters this balloon at 120 kPa and 20°C with a velocity of 3 m/s through a 1-m-diameter opening. How many
minutes will it take to inflate this balloon to a 17-m diameter when the pressure and temperature of the air in the
balloon remain the same as the air entering the balloon?
12. A wooden spherical ball of specific gravity of 0.42 and diameter of 30 cm is dropped from a height of 4.2 m
above the water surface of the pool. If the ball barely touch the bottom of the pool before it begun to float, how
deep is the water pool at that point in m?
.
13. A 30 cm cube steel of relative density 8.5 is placed in mercury. What is the volume of the cube submerged in
mL?
14. A balloon, assumed to be an ellipsoid vertically oriented is filled with helium at standard atmospheric pressure
and temperature. It is needed to carry a 2ft edge cubical container of unknown fluid. Specific gravity of the fluid is
1.5. The surrounding atmospheric air at a certain elevation where the balloon is at atmospheric pressure and 21.1°C.
Recommend dimensions for the ellipsoid in feet, such that the length of the vertical axis is 10% greater than the
horizontal axis and is 20% greater than the axis that is outside the plane.
15. When a solid cylinder is submerged to water, the draft is 5m, and if it is submerged to mercury, the draft is
decreased by 10%. What is the height of the cylinder in inches?
16. An object is dropped from a height of 5m above the water surface of the water pool. If the ball barely touches
the bottom of the pool before it begun to float at, what would be the average depth of the pool in meters?
17. A spherical body is 1/4 submerged to water and a spherical solid is used as an anchor so that the body is fully
submerged into an unknown liquid. What is the ratio of the diameter of the spherical body to the spherical solid?
Assume specific gravity of the liquid is 30% lesser than the specific gravity of the body and 30% greater than the
specific gravity of the solid.
Equation of State and Ideal Gas Laws
18. A bicycle tire has a volume of 600 cm3. It is inflated with carbon dioxide to a pressure of 80 psia at 20°C. How
many grams of carbon dioxide are contained in the tire?
19. A volume of 400cc of air is measured to a pressure of 740 mmHg abs and a temperature of 18°C. What will be
the volume in cc at 760mmHg and 0°C?
20. If the initial volume of an ideal gas is compressed to one half its original volume and twice its original
temperature the pressure is?
176
21. A transportation company specializes in the shipment of pressurized gaseous materials. An order is received for
100L of a particular gas at 32°F and 1 atm. What minimum volume tank in liters as necessary to transport the gas
at 80°F and maximum pressure of 8atm?
22. At STP the density of chlorine is 3.22 kg/m3. What is the weight of the gas in kN if contained in a flask of 100cc
at 24C and 100 kPa?
23. The density of helium is 0.178 kg/m3 at STP. What is the density at 25°C and 98 kPaa in kg/m3.
24. A rigid vessel initially contains helium at 105 kPaa and 15°C. Two kilograms of helium are then added to the
contents so that the final pressure and temperature are 200kPaa and 20C. Determine:
a) The volume of the vessel in gal
b) The final mass of the helium in kgm.
25. A rigid vessel initially contains hydrogen at 200 kPaa and 20°C. Two kilograms of hydrogen are then pumped
out so that the final pressure and temperature is at 105 kPaa and 15°C. Determine
a) The final mass of hydrogen.
b) The volume of the vessel
26. A steel tank initially contains oxygen gas at 450 kPag and 20°C. After 2.5 kg of oxygen had been used, the
pressure and temperature of the gas in the tank was found to be 180 kPag and 20°C respectively. What was the
initial mass of oxygen in the tank in kgm?
27. A 1500 L tank contains nitrogen at a pressure of 400 kPag and a temperature of 320K. Later because of leak, it
was found that the gage pressure has dropped to 320 kPag and the temperature has decreased to 300K. Determine
a) Initial mass of nitrogen in kgm.
b) Remaining nitrogen in kgm.
c) Amount of nitrogen leak out in kgm.
28. A 0.5 m3 carbon dioxide tank at atmospheric pressure and 27°C is placed near a furnace with the exhaust valve
open. The temperature inside the tank reaches 87°C. The exhaust valve is then closed, and the tank is cooled to its
initial temperature. Determine
a) The initial mass of the tank in kgm.
b) The remaining mass of carbon dioxide in the tank n kgm.
c) The final pressure of carbon dioxide in the tank in kPaa.
29. A bicycle pump is full of air at atmospheric pressure. The length of stroke of the pump is 45cm. At what part of
the stroke in cm does the air begin to enter the tire in which the gage pressure is 310 kPag? Assume compression to
be isothermal.
30. A kilomole of methane is 16 kg. Compute the density in kgm/m3 at 20°C and 5 atm.
31. An air bubble rises from the bottom of a well when the temperature is 27°C. Find the percentage increase in
volume of the air bubble if the depth of the well is 5m. Atmospheric pressure is 101.528 kPaa.
32. Determine the mass of the air in a room whose dimensions are 5m x 6m x 7m at 105 kPaa and 27 °C.
33. Ammonia at a temperature if 42°C is contained in a 400 L steel tank.

a) If the tank is heated such that pressure increases by 140%, what is the final temperature in C?
b) What is the change in density in kgm/m3?
177
34. An automobile tire initially contains 0.12 kg of air at 250 kPag and 22°C. Assume that the tire does not stretch
and is equipped with a release valve such that the air pressure in the tire will not exceed 290 kPag. Due to running
conditions, the temperature in the time reaches 80°C. Determine
a) The amount of air that escape in kgm
b) The tire pressure when the remaining air has returned to 22C in kPag
35. A 15 cm diameter by 100 cm length cylindrical steel vessel contained nitrogen at 1315 kPag and 28°C. After
some of the nitrogen was used the pressure and temperature dropped to 480 kPag and 22°C.
a) What proportion of nitrogen used?
b) What volume would the used nitrogen in cc would occupy at 1 atm and 15.5 C?
36. 0.8 m diameter sphere contains helium at 20 bar and 28°C. How many cylindrical drums 12cm in diameter and
70 cm long, which are initially devoid of any gas, can be filled to a pressure of 3 bar at 18°C. Assume the helium
in the sphere remains at 28°C.
37. At an atmospheric pressure and temperature of 760 mmHg and 24°C, how much dry air is more heavier than
moist air in percentage?
38. A 700 m3 blimp is to be filled with helim at 1 atm and 27°C, with the surrounding air also at 27°C.
a) If the helium is stored in 1.2m3 tanks at a pressure of 1500 kPa, how many tanks are required?
b) What maximum weight will the blimp support?
39. Determine the weight of acetylene gas in barrel with base diameters of 2ft and 4 ft respectively, having a depth
of 4ft at 80°F and 60psig in lbf.
40. A submarine releases bubbles from the bottom of the sea. At what depth must the bubbles be released so that
its diameter upon reaching the sea water surface will be thrice the diameter when it was released? Specific gravity
of the sea water is 1.13 and thermal equilibrium was observed at all times.
41. An empty tin can with height of 1ft in and a diameter of 0.25 ft is open at one end and closed at the other end.
If the vertical can, with the open end down, be slowly immersed in water, how far will the water rise in inches inside
the can? Thermal equilibrium is observed at all times.
42. The temperature of an ideal gas remains constant while the absolute pressure changes from 103.4 kPaa to 827.2
kPaa.
a) If the initial volume is 80L, what its final volume in L?
b) For 160 grams of ideal gas, determine the change of density expressed as a percentage of the initial
density.
43. The atmospheric pressure at the base of a mountain is 760 mmHg and at its top is 375 mmHg. Local g=9.75
m/s2. Determine the height of the mountain in meters if
(a) the average air pressure and temperature is 103 kPa and 20℃ ,
(b) The atmospheric temperature is 18C, constant from base to top
(c) If the working substance follows the law 𝑃𝑉1.4 = 𝐶 , having base temperature of 25°𝐶 and peak
temperature of 16°𝐶.
44. An empty tin can with height of 6 in and a diameter of 3 in is open at one end and closed at the other end. If the
vertical can, with the open end down, be slowly immersed in water, how far will the water rise in inches inside the
can when the closed end is 10ft? Temperature at the fluid surface is 22°𝐶 and 16°𝐶 at the depth where the closed
end is located.
178
45. Sulfur dioxide at a temperature of 204°C occupies a volume of 0.3m3.

a) If the volume is increased to 0.9m3 while the pressure is maintained constant, what is the final
temperature in °C?
b) If the initial volume is maintained constant, and the pressure is tripled, what is the final temperature in
°C?
46. The temperature of 4.82 lb of oxygen gas occupying 8 ft3 from 110°F to 200°F while the pressure remains
constant at 115 psia. Determine
a) The final volume in cubic foot.
b) The percentage change in density
c) If volume is constant and absolute temperature is quadrupled, find the final temperature in psia.
47. An automobile tire contains a certain volume if air at 30 psig at 70°F at 29.50 inHg barometric pressure.
a) If due to running conditions, the temperature of the air in the tire rises to 160F what would be the gage
pressure in psig? Assume the tire does not stretch.
b) If the tire initially contains 0.25 lb of air and have an automatic valve that releases air wherever the
pressure exceeds 34 psig, then what mass of air escapes in lbm if in a certain running condition, the
temperature becomes 180°F?
c) From (b), determine the pressure in psig when the remaining air has returned to 70°F.
48. What is the specific volume of gas in ft3/lbm at 180 psia and 90°F when its density is 0.0446 lb/ft3 at 14.7 psia
and 32°F? Calculate its gas constant in ft-lbf/lbm-R and molecular weight in lbm/lbm-mol.
49. A certain gas at 15 inHg abs and 90°F occupies 10 ft 3.

a) If the state is changed until the volume is 30ft3 and the temperature is 540°F, determine the final pressure
in psia.
b) If the state is changed until the initial pressure is trebled and the temperature increases to 400°F,
determine the final volume in cubic foot.
50. A steel company plans on using 17 cubic foot of oxygen gas in processing 1 metric ton of steel. If this volume
is measured at 101.325 kPaa and 21°C, what mass of oxygen gas is needed for a 20000 metric ton/month furnace.
51. A drum 6 in in diameter and 40 in long contained acetylene at 250 psia and 80°F. After some of the acetylene
was used, the pressure was 100 psia and the temperature was 70°F.
a) What proportion of the acetylene was used?
b) What volume in cubic foot would the used acetylene occupy at 14.7 psia and 60°F?
52. A sphere 6 ft in diameter containing oxygen at 300 psia and 80°F.

a) How many drums 6 in in diameter and 2 ft long, which are initially devoid of any gas can be filled to a
pressure of 50 psia and 65°F, assume that the temperature of the oxygen left in the sphere remains at 80F?
b) Same as (a) , except that the drum initially contains oxygen gas at 20 psia and 65°F.
53. Air is pumped into a 20ft3 tank until the pressure is 135 psig. When the pump is stopped, the temperature is
200°F.
a) What is the mass in the tank in lbm and density in lbm/ft3?
b) If the air is allowed to cool to 70°F, what is the pressure of the cooled air in psia and its density after
cooling in lbm/ft3
54. The height of a cylindrical tank containing 15 lbm of gas at 70 psia and115°F is triple its diameter; the gas
constant is R=45 ft-lbf/lbm-R. Find the tank dimensions in feet.
179
55. A 60 ft3 tank containing helium at 250°F is evacuated from atmospheric pressure until the vacuum is 28.9 in
Hg.
a) What is the mass of 250°F of helium left in the tank?
b) What mass of helium was pumped out?
c) If the helium gas left in the tank is cooled to a temperature of 36°F, what would will be its pressure in psia
and inHg abs?
56. A 12 ft3 tank contained hydrogen sulfide gas at 150 psia and 60°F after 5lb of the gas has been drawn put. Before
any gas left in the tank the temperature was 70°F. What mass of gas was in the tank originally in lbm and what was
its pressure in psia?
57. A 50 ft3 tank is being filled with oxygen. At a particular instant the temperature is 240°F, the pressure is 200
psia, and each is increasing at rates of 50°F/sec and 20 psia/sec. Find the flow rate of oxygen in CFS into the tank
at this instant.
58. There are withdrawn 200 ft3 of air measured at 15 psia at 90°F from a 50ft3 tank containing air initially at 100
psia and 140°F. What is the pressure of the air in psia remaining in the tank if the temperature is 130°F.
59. Two spheres, each 6 ft in diameter are connected by a pipe in which there is a valve. Each sphere contains
helium at a temperature of 80°F. With the valve closed, one sphere contains 3.75 lb and the other 1.25 lb of helium.
After the valve has been open long enough for equilibrium to obtain, what is the common pressure in psia in the
spheres if there is no lost or gain of energy?
60. Compute for the wind power produced by atmospheric air at 30°C at a velocity of 10 m/s on a 2m blade diameter
wind mill in watts.
61. Air enters a 16-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. Air is heated as it flows,
and it leaves the pipe at 180 kPa and 40°C. Determine
a) the volume flow rate of air at the inlet,
b) the mass flow rate of air, and
c) the velocity and volume flow rate at the exit.
62. Air enters a nozzle steadily at 50 psia, 140°F, and 150 ft/s and leaves at 14.7 psia and 900 ft/s. The heat loss
from the nozzle is estimated to be 6.5 Btu/lbm of air flowing. The inlet area of the nozzle is 0.1 ft2. Determine
a) the exit temperature of air and
b) the exit area of the nozzle.
63. An insulated rigid tank initially contains 1.5 lbm of helium at 80°F and 50 psia. A paddle wheel with a power
rating of 0.02 hp is operated within the tank for 30 min. Determine
a) the final temperature and
b) the final pressure of the helium gas.
64. A piston–cylinder device initially contains 0.5 m3 of nitrogen gas at 400 kPa and 27°C. An electric heater within
the device is turned on and is allowed to pass a current of 2 A for 5 min from a 120-V source. Nitrogen expands at
constant pressure, and a heat loss of 2800 J occurs during the process. Determine the final temperature of nitrogen.
65. Carbon steel balls (ρ = 7833 kg/m3 and cp = 0.465 kJ/kg°C) 8 mm in diameter are annealed by heating them first
to 900°C in a furnace, and then allowing them to cool slowly to 100°C in ambient air at 35°C. If 2500 balls are to
be annealed per hour, determine the total rate of heat transfer from the balls to the ambient air.
180
66. The volume of a high-altitude chamber is 40 m3. It is put into operation by reducing pressure from 1 bar to 0.4
bar and temperature from 25°C to 5°C. How many kg of air must be removed from the chamber during the process?
Express this mass as a volume measured at 1 bar and 25°C.
67. A steel flask of 0.04 m3 capacity is to be used to store nitrogen at 120 bar,20°C. The flask is to be protected
against excessive pressure by a fusible plug which will melt and allow the gas to escape if the temperature rises too
high.
a) How many kg of nitrogen will the flask hold at the designed conditions ?
b) At what temperature must the fusible plug melt in order to limit the pressure of a full flask to a maximum
of 150 bar?
68. A balloon of spherical shape 6 m in diameter is filled with hydrogen gas at a pressure of 1 bar abs. and 20°C.
At a later time, the pressure of gas is 94 percent of its original pressure at the same temperature
a) What mass of original gas must have escaped if the dimensions of the balloon is not changed?
b) Find the amount of heat to be removed to cause the same drop in pressure at constant volume.
69. Two spheres each 2.5 m in diameter are connected to each other by a pipe with a valve as shown below. One
sphere contains 16 kg of air and other 8 kg of air when the valve is closed. The temperature of air in both sphere is
25°C. The valve is opened and the whole system is allowed to come to equilibrium conditions. Assuming there is
no loss or gain of energy, determine the pressure in the spheres when the system attains equilibrium.
70. A natural gas has a specific gravity of 1.46 relative to the density of air at 100 mbar and 25°𝐶.
a) What is the specific weight of the natural gas?
b) What is the gas constant of natural gas?
c) What is the molecular weight of the natural gas?
71. Oxygen is supplied to a 250mm outside diameter, 10 mm thick, and 1.20m long cylindrical tank. The tank has
a rating maximum temperature capacity of 45°𝐶. How many minutes this tank can supply 0.225 liter per second to
an oxy-acetylene welding set oxygen in the tank has a pressure of 4 MPa and temperature of 45°𝐶 and not to fill
below 500 kPa at 25°𝐶? The welding set is at atmospheric pressure and a temperature of 28°𝐶.
72. The volume of the air in the tire is 30L at 10°C and 180 kPag. If the volume due to running conditions are
decreased by 300 cc and an increase of temperature of 20C, determine the final pressure in psi vac.
73. Determine the ratio of maximum to minimum above atmospheric pressures if the maximum diameter and
minimum radius of air bubbles formed by the exhaust in submarines are 10 cm and 4 cm respectively. Observe
thermal equilibrium at all times.
74. An average human breathe 0.5L of air and can do 25 breathe /min. How many kilograms of air does the person
may intake at atmospheric pressure and temperature of 26°C?
75. Recommend dimensions for the diameter of a fan blade for generating 100kW of electricity by air at 32°C
moving at the speed of 60 kph if power efficiency is 85%.
76. A certain gas at 101.325 kPa and 10C whose volume is 2.83m3 capacity. Before admission, the storage vessel
contained gas at a pressure and temperature of 137.8 kPa and 26°C; after admission the pressure increased to
1171.8 kPa. What should be the final temperature of the gas in the vessel in kelvin?
181
77. A rigid vessel contains an unknown ideal gas initially at 100 kPaa and 15°C. What is the total weight of the
vessel in kgf if 3kg of this ideal gas are then added to the contents so that the final pressure and temperature
becomes 200kPaa and 20°C. Assume that the vessel without this unknown gas is weightless.
78. A container is filled with an unknown ideal gas at 1300kPag and 30°C. After some of this ideal gas was used,
the pressure dropped to 500 kPag and 25°C. What proportion of these ideal gas was used?
79. An empty cylindrical can is open at one end. If the vertical can with the open end down, be slowly immersed in
seawater (s=1.20), then water inside the can will rise up to 2 inches when the closed end is 9.95ft below and
temperature is 15°C.Recommend dimensions of the can such that its diameter is 30% larger than its height in
inches. Temperature above sea surface is assumed to be 24°𝐶
80. The atmospheric pressure at the base of the 6000m mountain at 15°C reads 730mmHg. What would be the
pressure reading at the top of the mountain in psi?
81. A steel factory utilizes 452.6 metric tons of oxygen per month in processing 20000 metric tons per month
furnace. Oxygen is contained in a spherical tank maintained at 101.325kPaa and 25°C. If the oxygen in the tank
can process 2 metric tons of steel, what would be the recommended dimensions of the tank in feet?
82. A submarine 800ft deep releases air bubbles from the bottom of the sea. Assuming that the bubble released will
reach the seawater (SG=1.13) surface, what is the ratio of the diameter when the bubble is released to the
diameter when the bubble reaches the seawater surface?
83. A 40 cubic feet tank is being filled with an unknown ideal gas. At a particular instant, the temperature is 235°F
and the pressure is 195 psia and each is increasing at a rate of 40°F per second and 21 psia per second. Find the
flow rate of these unknown ideal gas in the tank at that instant in GPM.
Specific Heat and Ideal Gas
84. Complete the following ideal gas table and identify each of the following gas. Units for pressure, density,
specific volume, gas constant R and temperature are kPaa, kg/m 3, m3/kgm, kJ/kgm-K, and K respectively.
Gas MW P Ρ ν R T
17 125 295
2.22 0.189 370
48 105 0.35
459 1.68 0.277
85. Complete the ideal gas table and identify the gas. Units for R , cp, and cv are in kJ/kgm-K.
Gas MW R cp cv K
28.97 1.4
1.7549 1.4782
2.078 3.1233
182
86. A gas whose cv = 1.386 kJ/kgm-K and R= 0.485 kJ/ kgm-K is heated from 200 kPaa and 1.8 m3 to a pressure
and volume of 280 kPaa and 2.5 m3 respectively. Determine ΔU and ΔH in kJ.
87. 8 kg of a gas, R = 280 J/kgm-K and cv = 0.78 kJ/kgm-K undergoes a reversible non flow constant pressure
process from 1.2 m3 and 700 kPaa to a state of 480C. Find
a) ΔU, ΔH, Q, WNF in kJ.
b) If the process had been a reversible steady flow type with ΔPE=0 and ΔKE=0, find W SF in kJ.
88. A gas whose k=1.35 is heated from 200 kPaa and 1.8m 3 to a pressure and volume of 285 kPaa and 2.4m3,
respectively. Determine ΔU and ΔH in kJ.
89. Provide 8 kg of gaseous substance 600 kJ of heat at constant volume, so that temperature rises by 100K.
Determine
a) The specific heat at constant volume in kJ/kgm-K and BTU/lbm-R
b) If k=1.55, compute for cp and R in kJ/kgm-K and kcal/kgm-K.
c) What is the density of the gas in kgm/m3?
90. A gas initially at 20 psia and 6ft3/hr undergoes process to 60 psia and 0.8 ft3/hr, during which the enthalpy
increases by 20 BTU/min. cv = 3.25 BTU/lbm-R. Determine
a) ΔU in Mhp.
b) cp in BTU/lbm-R
c) R in BTU/ lbm-R
91. The pressure in 142L/min of air is boosted reversibly from 2068.44 kPaa to 6295.32 kPaa while temperature
remains constant at 24C. ΔU =0
a) Find the work done in non-flow and steady flow processes in hp.
b) For a non-flow process, find Q, ΔH, in hp.
c) For steady flow process during which ΔKE = 5 kJ/min and ΔPE=0, find ΔWf, ΔH,Q, in hp.
92. Assume 10 CFM and 0.988 lbm/min of hydrogen initially at 400 psia and 300°F. Determine the final pressure,
change in internal energy, change in enthalpy, heat, and change in entropy for a process occurs at 120°F , neglecting
change in potential and kinetic energy at
a) Non-flow constant pressure process.
b) Non-flow constant volume process.
c) Steady flow constant pressure process.
d) Steady flow constant volume process.
93. A vessel of capacity 3 m3 contains 1 kg mole of N2 at 90°C.

a) Calculate pressure and the specific volume of the gas.
b) If the ratio of specific heats is 1.4, evaluate the values of cp and cv.
c) Subsequently, the gas cools to the atmospheric temperature of 20°C ; evaluate the final pressure of gas.
d) Evaluate the increase in specific internal energy, the increase in specific enthalpy, increase in specific
entropy and magnitude and sign of heat transfer.
94. Find the work possess for a helium gas at 20C.
95. Two kilograms of gas is confined in a 1 m3 tank at 200 kPaa and 88C. What type of gas is in the tank?
96. Find the enthalpy of helium in kJ/kgm if its internal energy is 200 kJ/kgm.
97. Two kilograms of air in a rigid tank changes its temperature from 32 C to 150C. Find the work done in kJ during
the process.
98. An ideal gas at 0.80 atmospheres and 97C occupies 0.450L. How many moles are in the sample?
183
Analysis of Ideal Gas Mixture
99. The gravimetric analysis of dry air is approximately O2=22.5% and excess nitrogen gas. Calculate
a) Volumetric analysis
b) The gas constant
c) The respective partial pressures
d) The specific volume and density at 0.95 atm, 20°C
e) How many kilograms of N2 must be added to 3 kg of air to produce a mixture which is 60% O2 by volume.
100. How many kilograms of nitrogen must be mixed with 3.60 kg of carbon dioxide in order to produce a
gaseous mixture that is 50% by volume of each constituent? Also determine the molecular weight, gas constant of
the mixture, and the partial pressure of N2 if pressure of CO2 is 138 kPaa.
101. One mol of a gaseous mixture has the following gravimetric analysis: O2=16%, CO2= 44% and N2=40%.
Find
a) Molecular weight of the mixture
b) Mass of each constituent
c) Moles of each constituent in the mixture
e) Rm
f) Partial pressures of each constituent if partial pressure of the mixture is 30 psia.
102. A gravimetric analysis of dry air is approximately 25% and the rest are nitrogen. How many kilograms of
oxygen must be added to 5 kilograms of air to produce a mixture which contains 45% oxygen gas by volume?
103. An 8 ft3 steel drum contains a mixture of 𝐶𝐻4 and 𝐶𝑂2, each 50% by mass at 𝑃𝑚 = 100𝑝𝑠𝑖𝑎, 100℉; 2 lb of
𝑂2 are added to the drum. For the final mixture, find:
a) Gravimetric analysis
b) Volumetric analysis
c) specific heat of mixture at constant pressure
d) new pressure of the mixture.
104. A 283 L drum contains a gaseous mixture of 30% 𝑂2 by volume and the rest are 𝐶𝐻4 at 689.48 kPaa and
37.8℃. How many kilograms of mixture must be bled and what mass of 𝑂2 added in order to produce at the original
pressure and temperature a mixture whose new volumetric composition is 30% 𝐶𝐻4 and the rest are 𝑂2 ?
105. The following are partial pressures of different gaseous species in a vessel:
Ideal gas Partial Pressure
𝐶𝐻4 140 kPaa
𝑁2 55 kPaa
𝐶𝑂 70 kPaa
𝑂2 15 kPaa
Find
a) The respective volumetric analysis
b) The respective gravimetric analysis
c) Molecular weight of the mixture
d) Gas constant of the mixture
e) 𝑐𝑝𝑚
f) 𝑐𝑣𝑚
g) the volume occupied by 45 kg at the pressure of the mixture and temperature of 32℃.
184
Bibliography

Burghardt, D. M. (1986). Engineering Thermodynamics with Applications 3rd Edition. New York City: Harper &
Row, Publishers Inc.
Cengel, Y. A., Boles, M. A., & Kanoglu, M. (2019). Thermodynamics - An Engineering Approach Ninth Edition.
New York City: McGraw-Hill Education.
Dionisio, D. (2018). Machine Design & Allied Subjects- Dynamic Reviewer. Uyao St. Bagumbayan, Roxas
Oriental Mindoro: Educspace Learning Services.
Doughtie, V., & James, W. H. (1954). Elements of Mechanism. New York City: John Wiley & Sons, Inc.
Faires, V. M., Simmang, C. M., & Brewer, A. V. (1978). Problems on Thermodynamics 6th Edition. New York
City: Macmillan Publishing Co., Inc.
185
Chapter IV. IDEAL GAS PROCESSES THERMODYNAMICS 1
IV. IDEAL GAS PROCESSES

At the end of this chapter, the mechanical engineering students shall:
1. Apply concepts of ideal gas law when solving processes.
2. Define entropy and its significance.
3. Determine the PVT relationships for each process.
4. Apply energy equations to ideal gas processes.
5. Solve problems involving two or three continuous process.
4.1 GENERAL EQUATIONS FOR IDEAL GAS PROCESSES

The basic properties of matter, energy and power concepts, and ideal gas laws will be sum up in the table as
shown below, showing PVT relationships, specific heat, and energy equations for ideal gases.
Questions to be Needed Equations Other Tools
Asked
1. PVT Relationships Isometric Process 𝑉 = 𝐶 𝑃2 𝑇2 Energy Balance
=
𝑃1 𝑇1 (Δ𝑃𝐸 = 0, ΔK𝐸 = 0)
Isobaric Process 𝑃 = 𝐶 𝑉2 𝑇2 a) Open Systems
=
𝑉1 𝑇1 𝑊𝑆𝐹 = 𝑄 − Δ𝐻
Isothermal Process 𝑇 = 𝐶 𝑃1 𝑉1 = 𝑃2 𝑉2
Isentropic process 𝑆 = 𝐶 𝑃1 𝑉1𝑘 = 𝑃2 𝑉2𝑘 b) Closed Systems
𝑘−1 𝑊𝑁𝐹 = 𝑄 − Δ𝑈
𝑇2 𝑉1 𝑘−1 𝑃2 𝑘
=( ) =( )
𝑇1 𝑉2 𝑃1 Specific Heat (𝑐)
Polytropic Process 𝑃1 𝑉1𝑛 = 𝑃2 𝑉2𝑛 𝑐 = 𝑐𝑉 𝑉 = 𝐶

𝑛−1
𝑇2 𝑉1 𝑛−1
𝑃2 𝑛 𝑐 = 𝑐𝑝 𝑇 = 𝐶
=( ) =( ) 𝑐 = 𝑐𝑛 (𝑃𝑜𝑙𝑦𝑡𝑟𝑜𝑝𝑖𝑐)
𝑇1 𝑉2 𝑃1
2. Non-Flow Work if 2
Δ𝑃𝐸 = 0, Δ𝐾𝐸 = 0 𝑊𝑁𝐹 = ∫ 𝑃𝑑𝑉 Specific Heat and

1 Energy Relationships
3. Heat added / Heat 2
Rejected 𝑄 𝑄 = 𝑚𝑐Δ𝑡 = 𝑚 ∫ 𝑐𝑑𝑡 𝑇=𝐶

1 𝑐𝑝 = 𝑐𝑣 + 𝑅
2
𝑄 = 𝑚𝑐Δ𝑇 = 𝑚 ∫ 𝑐𝑑𝑇 𝑇=𝐶 𝑅 = 𝑐𝑝 − 𝑐𝑣
1 𝑐𝑝
2 𝑘=
𝑐𝑣
𝑄 = ∫ 𝑇𝑑𝑆 𝑇≠𝐶 𝑅
1 𝑐𝑣 =
4. Change in Internal Δ𝑈 = 𝑚𝑐𝑣 Δ𝑡 𝑘−1
𝑘𝑅
Energy Δ𝑈 = 𝑚𝑐𝑣 Δ𝑇 𝑐𝑝 =
(For Any Process) 𝑘−1
5. Change in Enthalpy Δ𝐻 = 𝑚𝑐𝑝 Δ𝑡 Specific heat for

Δ𝐻 = 𝑚𝑐𝑝 Δ𝑇 Polytropic Process
(For Any Process)
𝑘−𝑛
6. Change in Entropy Δ𝑆 = 𝑚Δ𝑠 𝑐𝑛 = 𝑐𝑣 ( )
1−𝑛
𝑄
𝑄 𝑑𝑄
Δ𝑆 = = ∫ 𝑇=𝐶
𝑇 0 𝑇
186
𝑇2
Δ𝑆 = 𝑚𝑐𝑙𝑛 𝑇≠𝐶
𝑇1
7. Work Steady Flow if 2
Δ𝑃𝐸 = 0, Δ𝐾𝐸 = 0 𝑊𝑆𝐹 = ∫ 𝑉𝑑𝑃

1
8. Work Steady Flow if 𝑊𝑆𝐹 = 𝑄 − Δ𝐻 − Δ𝑃𝐸 − Δ𝐾𝐸

Δ𝑃𝐸 ≠ 0, Δ𝐾𝐸 ≠ 0
4.2 ENTROPY
Entropy is defined as the measure of degree of disorder of molecules or the randomness of movement of
molecules. It is defined as
𝒅𝑸
𝒅𝑺 =
𝑻
T-s (Temperature - Entropy) diagrams are useful in describing
the heat transfer in the system. The area under the T-s diagram
represents the heat transfer (heat added/heat rejected) by the
system. The following are useful formulas for entropy when
dealing with open and closed systems
Assuming 𝑇 ≠ 𝐶 𝑇2
Δ𝑆 = 𝑚𝑐𝑙𝑛
𝑇1
Assuming 𝑇 = 𝐶 𝑄
Δ𝑆 =
𝑇
Closed Systems 𝑇2 𝑉2
Δ𝑆 = 𝑚𝑐𝑣 + 𝑚𝑅𝑙𝑛
𝑇1 𝑉1
𝑉2
Δ𝑆 = 𝑚𝑐𝑝 𝑙𝑛
𝑉1
Open Systems 𝑇2 𝑃2
Δ𝑆 = 𝑚𝑐𝑝 𝑙𝑛 + 𝑚𝑅𝑙𝑛
𝑇1 𝑃1
𝑇2
Δ𝑆 = 𝑚(2𝑐𝑝 − 𝑐𝑣 )𝑙𝑛
𝑇1
From state 1 to state 2,

𝑑𝑄
𝑑𝑆 =
𝑇
187
Since 𝑑𝑄 = 𝑚𝑐𝑑𝑇
𝑑𝑇
𝑑𝑆 = 𝑚𝑐
𝑇
2 2
𝑑𝑇
∫ 𝑑𝑆 = 𝑚𝑐 ∫
1 1 𝑇
𝑇2
𝑆2 − 𝑆1 = 𝑚𝑐𝑙𝑛
𝑇1
𝑻𝟐
𝚫𝑺 = 𝒎𝒄𝒍𝒏
𝑻𝟏
The change in specific entropy would be:
𝚫𝑺
𝚫𝒔 = = 𝒔𝟐 − 𝒔𝟏
𝒎
The derived equation is the general equation of entropy for any process. We can also relate entropy to other
energy equations.
A. For Closed Systems Consider a piston-cylinder arrangement that contains piston. Heat is added to the system
containing ideal gas, and the piston moves.
By Energy Balance:
𝑄 = Δ𝑈 + 𝑊𝑁𝐹
In differential terms:
𝑑𝑄 = 𝑑𝑈 + 𝑑𝑊𝑁𝐹
Since 𝑑𝑄 = 𝑇𝑑𝑆 , 𝑑𝑈 = 𝑚𝑐𝑣 𝑑𝑇m and 𝑑𝑊𝑁𝐹 = 𝑃𝑑𝑉, hence:
𝑇𝑑𝑆 = 𝑑𝑈 + 𝑃𝑑𝑉
𝑚𝑅𝑇
𝑃=
𝑉
Hence:
𝑑𝑉
𝑇𝑑𝑆 = 𝑚𝑐𝑣 𝑑𝑇 + 𝑚𝑅𝑇
𝑉
𝑑𝑇 𝑑𝑉
𝑑𝑆 = 𝑚𝑐𝑣 + 𝑚𝑅
𝑇 𝑉
Dividing both sides by T and integrate from state 1 to state 2:
2 2 2
𝑑𝑇 𝑑𝑉
∫ 𝑑𝑆 = 𝑚𝑐𝑣 ∫ + 𝑚𝑅 ∫
1 1 𝑇 1 𝑉
𝑇2 𝑉2
𝑆2 − 𝑆1 = 𝑚𝑐𝑣 𝑙𝑛 + 𝑚𝑅𝑙𝑛
𝑇1 𝑉1
The change in entropy would be:
𝑻𝟐 𝑽𝟐
𝚫𝑺 = 𝒎𝒄𝒗 𝒍𝒏 + 𝒎𝑹𝒍𝒏
𝑻𝟏 𝑽𝟏
Applying PVT relationship:
𝑉2 𝑇2
=
𝑉1 𝑇1
𝑇2 𝑇2
Δ𝑆 = 𝑚𝑐𝑣 𝑙𝑛 + 𝑚𝑅𝑙𝑛
𝑇1 𝑇1
𝑇2
Δ𝑆 = 𝑚(𝑐𝑣 + 𝑅)𝑙𝑛
𝑇1
Since 𝑐𝑝 = 𝑐𝑣 + 𝑅; hence:
188
𝑻𝟐 𝑽𝟐
𝚫𝑺 = 𝒎𝒄𝒑 𝒍𝒏 = 𝒎𝒄𝒑 𝒍𝒏
𝑻𝟏 𝑽𝟏
B. For Open Systems – consider a steady flow steady system. From the enthalpy equation:
𝐻 = 𝑈 + 𝑃𝑉
In differential terms:
𝑑𝐻 = 𝑑𝑈 + 𝑑(𝑃𝑉)
𝑑𝐻 = 𝑑𝑈 + 𝑃𝑑𝑉 + 𝑉𝑑𝑃
From the closed system:
𝑇𝑑𝑆 = 𝑑𝑈 + 𝑃𝑑𝑉
Eliminating 𝑃𝑑𝑉 gives:
𝑑𝐻 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃
𝑇𝑑𝑆 = 𝑑𝐻 − 𝑉𝑑𝑃
Since:
𝑑𝐻 = 𝑚𝑐𝑝 𝑑𝑇
𝑚𝑅𝑇
𝑉=
𝑃
𝑑𝑃
𝑇𝑑𝑆 = 𝑚𝑐𝑝 𝑑𝑇 − 𝑚𝑅𝑇
𝑃
𝑑𝑇 𝑑𝑃
𝑑𝑆 = 𝑚𝑐𝑝 − 𝑚𝑅
𝑇 𝑃
Integrating both sides from their respective lower and upper limits:
2 2 1
𝑑𝑇 𝑑𝑃
∫ 𝑑𝑆 = 𝑚𝑐𝑝 ∫ − 𝑚𝑅 ∫
1 1 𝑇 2 𝑃
𝑇2 𝑃2
𝑆2 − 𝑆1 = 𝑚𝑐𝑝 𝑙𝑛 + 𝑚𝑅
𝑇1 𝑃1
𝑻𝟐 𝑷𝟐
𝚫𝑺 = 𝒎𝒄𝒑 𝒍𝒏 + 𝒎𝑹
𝑻𝟏 𝑷𝟏
Applying PVT relationships:

𝑃2 𝑇2
=
𝑃1 𝑇1
𝑇2 𝑇2
Δ𝑆 = 𝑚𝑐𝑝 𝑙𝑛 + 𝑚𝑅
𝑇1 𝑇1
𝑇2
Δ𝑆 = 𝑚(𝑐𝑝 + 𝑅)𝑙𝑛
𝑇1
Since 𝑅 = 𝑐𝑝 − 𝑐𝑣
𝑻𝟐
𝚫𝑺 = 𝒎(𝟐𝒄𝒑 − 𝒄𝒗 )𝒍𝒏
𝑻𝟏
4.3 SECOND LAW AND IRREVERSIBILITY
Second law of Thermodynamics simply states that the sum of entropy in the universe (system plus the
surroundings) is constant. The following are properties of entropy:
1. For isolated system, entropy never decreases.

2. For within the system boundaries, entropy is constant or increases.
3. Entropy is produced, can be produced, but cannot destroyed.
4. As entropy is produced, the ability to do work decreases.
The second law of thermodynamics applies to reversible and irreversible processes.

189
A. Reversible Process - a process in which the system undergoes a change state and can return to its original state
by performing its process in reverse. For reversible process, the change of entropy is equal to zero. Reversible
Process is only a theoretical and ideal process. This is also called as ther reversible adiabatic process. Adiabatic
mean “no heat transfer” is made by the system.
Examples:
a. Frictionless movement of piston cylinder arrangement.
b. No deformation
c. Heat transfer occurs between the temperature differences.
d. Compression and expansion of springs
e. Electric current through a zero internal resistance.
f. No Chemical Reactions
g. Mixing two chemicals at the same state.
B. Irreversible Process -it is the actual and real-life process, where if a system undergoes a change state, it cannot
return to its original state. Hence the change of entropy is not constant, and the difference between the final and
initial specific entropy is called the irreversibility of the process. This is also called as the irreversible isentropic
process.
𝐈𝐫𝐫𝐞𝐯𝐞𝐫𝐬𝐢𝐯𝐢𝐥𝐢𝐭𝐲 = 𝑰 = 𝚫𝒔′ = 𝒔′𝟐 − 𝒔𝟏
If the irreversibility is in terms of a given surrounding temperature, then the irreversibility given the heat transfer
is:
𝑸
𝚫𝒔′ =
𝑻𝒐
Examples of irreversible process:

a. Movement of piston with friction.
b. With deformations
c. There is an actual heat transfer.
d. There is a temperature difference between two systems.
e. Electric current through a non-zero internal resistance
f. Spontaneous chemical reactions (ex. Combustion process)
g. Mixing two chemicals of the same state produces another state.
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4.4 ISOMETRIC PROCESS

Isometric Process, also called as isochoric, isovolumic, or isovolumetric process. It is a constant volume
process or specific volume process. Considered a gas in a rigid container. For a rigid (non-deformable) container,
the volume of the gas is equivalent to the volume of the container. Hence 𝑉1 = 𝑉2 = 𝑣1 = 𝑣2 = 𝐶
𝑉1 = 𝑉2 = 𝐶
𝑣1 = 𝑣2 = 𝐶
1) PVT relationships:
𝑷𝟐 𝑻𝟐
=
𝑷𝟏 𝑻𝟏
2) Work Non-Flow if ΔPE = 0 and ΔKE = 0
Using the area under the PV diagram, since V=C, then dV = 0, thus
2
1
𝑾𝑵𝑭 = 𝟎
Another is that at constant volume process, the graph on the PV diagram is a straight line. Since the area
under the line is zero, by inspection on the graph we can say that the non-flow work is zero.
3) Change in internal energy
𝚫𝑼 = 𝒎𝒄𝒗 𝚫𝑻
4) Change in Enthalpy
𝚫𝑯 = 𝒎𝒄𝒑 𝚫𝑻
5) Heat - the magnitude of heat is defined as the area under the T-s diagram, Since V=C, there is a change in
temperature. Hence
2
𝑄 = ∫ 𝑇𝑑𝑆
1
𝑸 = 𝒎𝒄𝒗 𝚫𝑻
191
𝑸 = 𝚫𝑼
We can conclude, that at constant volume process, the heat is also equivalent to the change in internal energy.
6) Change in Entropy
𝒅𝑸
𝒅𝑺 =
𝑻
Since 𝑑𝑄 = 𝑚𝑐𝑣 𝑑𝑇
𝑑𝑇
𝑑𝑆 = 𝑚𝑐𝑣
𝑇
2 2
𝑑𝑇
∫ 𝑑𝑆 = 𝑚𝑐𝑣 ∫
1 1 𝑇
𝑻𝟐
𝚫𝑺 = 𝒎𝒄𝒗 𝒍𝒏
𝑻𝟏
By its PVT relationships:
𝑷𝟐
𝚫𝑺 = 𝒎𝒄𝒗 𝒍𝒏
𝑷𝟏
7) Work Steady Flow if ΔPE = 0 and ΔKE = 0 - the steady flow work is the area “behind” the PV diagram.
2
𝑊𝑆𝐹 = ∫ 𝑉𝑑𝑃
1
Since volume is constant, 𝑉 = 𝐶
2
𝑊𝑆𝐹 = 𝑉 ∫ 𝑑𝑃
1
𝑾𝑺𝑭 = 𝑽(𝑷𝟐 − 𝑷𝟏 )
Since 𝑉1 = 𝑉2 = 𝑉
𝑊𝑆𝐹 = 𝑃2 𝑉 − 𝑃1 𝑉
𝑊𝑆𝐹 = 𝑃2 𝑉2 − 𝑃1 𝑉1
𝑊𝑆𝐹 = 𝑚𝑅𝑇2 − 𝑚𝑅𝑇1
𝑾𝑺𝑭 = 𝒎𝑹(𝑻𝟐 − 𝑻𝟏 )
Factoring out 𝑇1 gives:
𝑇2
𝑊𝑆𝐹 = 𝑚𝑅𝑇1 ( − 1)
𝑇1
From PVT relationships:
𝑷𝟐
𝑾𝑺𝑭 = 𝒎𝑹𝑻𝟏 ( − 𝟏)
𝑷𝟏
8) Work Steady Flow if Work Steady Flow if ΔPE ≠ 0 , ΔKE ≠ 0 - from energy balance:
𝑾𝑺𝑭 = 𝑸 − 𝚫𝑯 − 𝚫𝑷𝑬 − 𝚫𝑲𝑬
192
4.5 ISOBARIC PROCESS

Isobaric process or also called as isopiestic process, is a constant pressure process, that is P1 = P2 = P = C .
Consider a gas inside a piston cylinder arrangement operated by a spring of constant spring index.
From Hooke’s Law:

𝐹 ∝𝑥
𝐹 = 𝑘𝑥
𝐹1 𝐹2
=
𝑥1 𝑥2
Where k is called as the spring constant or spring index.
For heat addition process, if Q is added, then the piston will expand and V increases. Hence
𝐹1 𝐹1
𝑃1 = =π
𝐴1 𝐷2
4 1
𝐹2 𝐹2
𝑃2 = =π
𝐴2 𝐷2
4 2
Since the diameter is constant throughout the heating process, and the force exerted to the piston is caused by the
pressure due to heat addition. Thus
𝐹1 = 𝐹2
𝑃1 = 𝑃2
P-V diagram T-s diagram
193
1) PVT Relationships
𝑽𝟐 𝑻𝟐
=
𝑽𝟏 𝑻𝟏
2. Work Non-Flow if ΔPE = 0 and ΔKE = 0
It is the area under the PV diagram. Since P = C,
then
2
1
2
𝑊𝑁𝐹 = 𝑃 ∫ 𝑑𝑉
1
𝑾𝑵𝑭 = 𝑷(𝑽𝟐 − 𝑽𝟏 )
If 𝑃1 = 𝑃2 = 𝑃, then
𝑊𝑁𝐹 = 𝑃𝑉2 − 𝑃𝑉1
𝑊𝑁𝐹 = 𝑃2 𝑉2 − 𝑃1 𝑉1
𝑊𝑁𝐹 = 𝑚𝑅𝑇2 − 𝑚𝑅𝑇1
𝑾𝑵𝑭 = 𝒎𝑹(𝑻𝟐 − 𝑻𝟏 )
If 𝑇1 is factored out, then

𝑇2
𝑊𝑁𝐹 = 𝑚𝑅𝑇1 ( − 1)
𝑇1
By PVT relationships:
𝑽𝟐
𝑾𝑵𝑭 = 𝒎𝑹𝑻𝟏 ( − 𝟏)
𝑽𝟏
3. Change in internal energy
4. Change in Enthalpy
5. Heat - the magnitude of heat is defined as the area under the T-S diagram, Since P = C, there is a change in
temperature. Hence
2
1
𝑄 = 𝑚𝑐𝑝 Δ𝑇
𝑸 = 𝚫𝑯
We can conclude, that at constant pressure process, the heat is also equivalent to the change in enthalpy.
6. Change in Entropy
𝒅𝑸
𝒅𝑺 =
𝑻
Since 𝑑𝑄 = 𝑚𝑐𝑝 𝑑𝑇
𝑑𝑇
𝑑𝑆 = 𝑚𝑐𝑝
𝑇
2 2
𝑑𝑇
∫ 𝑑𝑆 = 𝑚𝑐𝑝 ∫
1 1 𝑇
𝑻𝟐
𝚫𝑺 = 𝒎𝒄𝒑 𝒍𝒏
𝑻𝟏
From PVT relationships:
194
𝑽𝟐
𝚫𝑺 = 𝒎𝒄𝒑 𝒍𝒏
𝑽𝟏
7. Work Steady Flow if ΔPE = 0 and ΔKE = 0 - the steady flow work is the area “behind” the PV diagram.
Hence
2
1
Since 𝑃 = 𝐶, then 𝑑𝑃 = 0
𝑾𝑺𝑭 = 𝟎
Another thing is that the graph of the PV diagram at constant pressure process is a horizontal line. Hence the
area “behind” the PV diagram is zero.
8. Work Steady Flow if Work Steady Flow if ΔPE ≠ 0, ΔKE ≠ 0 - by energy balance:
𝑊𝑆𝐹 = 𝑄 − Δ𝐻 − Δ𝑃𝐸 − Δ𝐾𝐸

Since 𝑄 = Δ𝐻
𝑾𝑺𝑭 = −𝚫𝑷𝑬 − 𝚫𝑲𝑬
195
4.6 ISOTHERMAL PROCESS

Isothermal Process is called as the constant temperature process, wherein𝑇1 = 𝑇2 = 𝑇 = 𝐶. Considered a gas
that is contained in a piston-cylinder arrangement with perfect coolant.
State 1 State 2
A coolant surrounds the piston-cylinder arrangement and absorbs a temperature equivalent to the temperature of
the heat rejected by he gas in the cylinder. Hence 𝑇1 = 𝑇2 = 𝑇 = 𝐶 and also 𝑡1 = 𝑡2 = 𝑡
1. PVT Relationships
𝑷𝟐 𝑽 𝟏
=
𝑷𝟏 𝑽 𝟐
2. Work Non Flow if ΔPE = 0 , ΔKE = 0 – it is the area under the PV diagram.
2
1
Since 𝑃𝑉 = 𝐶, then
𝐶
𝑃=
𝑉
Also
𝑃1 𝑉1 = 𝑃2 𝑉1 = 𝐶
Hence
196
2
𝑑𝑉
𝑊𝑁𝐹 = 𝐶 ∫
1 𝑉
𝑉2
𝑊𝑁𝐹 = 𝐶𝑙𝑛
𝑉1
𝑽𝟐 𝑽𝟐
𝑾𝑵𝑭 = 𝑷𝟏 𝑽𝟏 𝒍𝒏 = 𝑷𝟐 𝑽𝟐
𝑽𝟏 𝑽𝟏
Using PVT relationships
𝑷𝟏 𝑷𝟏
𝑾𝑵𝑭 = 𝑷𝟏 𝑽𝟏 𝒍𝒏 = 𝑷𝟐 𝑽𝟐
𝑷𝟐 𝑷𝟐
Using ideal gas law [𝑃𝑉 = 𝑚𝑅𝑇]
𝑽𝟐
𝑾𝑵𝑭 = 𝒎𝑹𝑻𝟏 𝒍𝒏
𝑽𝟏
𝑷𝟏
𝑾𝑵𝑭 = 𝒎𝑹𝑻𝟏 𝒍𝒏
𝑷𝟐
3. Change in internal energy - since T = C, then dT = 0. The change in internal energy during isothermal process
is equal to zero.
𝚫𝑼 = 𝟎
4. Change in Enthalpy - since T = C, then dT = 0. The change in enthalpy during isothermal process is equal to
zero.
𝚫𝑯 = 𝟎
5. Heat - the magnitude of heat is defined as the area under the T-s diagram, Since T =C,
2
1
𝑸 = 𝑻(𝑺𝟐 − 𝑺𝟏 )
𝒅𝑸
𝒅𝑺 =
𝑻
𝟐
𝟏 𝑸
∫ 𝒅𝑺 = ∫ 𝒅𝑸
𝟏 𝑻 𝟎
𝑸
𝚫𝑺 =
𝑻
197
7. Work Steady Flow if ΔPE = 0 , ΔKE = 0 - the steady flow work is the area “behind” the PV diagram.
1
2
Since 𝑃𝑉 = 𝐶, then
𝐶
𝑉=
𝑃
Also:
𝑃1 𝑉1 = 𝑃2 𝑉2 = 𝐶
1
𝑑𝑃
𝑊𝑆𝐹 = 𝐶 ∫
2 𝑃
𝑃1
𝑊𝑆𝐹 = 𝐶𝑙𝑛
𝑃2
𝑷𝟏 𝑷𝟏
𝑾𝑺𝑭 = 𝑷𝟏 𝑽𝟏 𝒍𝒏 = 𝑷𝟐 𝑽𝟐 𝒍𝒏
𝑷𝟐 𝑷𝟐
This equation is identical to work non flow. We can conclude that for isothermal process, the work steady flow is
equivalent to work non flow assuming Δ𝑃𝐸 = Δ𝐾𝐸 = 0.
8. Work Steady Flow if Work Steady Flow if ΔPE ≠ 0 , ΔKE ≠ 0 - by energy balance:

Since Δ𝐻 = 0, and 𝑄 = 𝑇Δ𝑆
𝑾𝑺𝑭 = 𝑻𝚫𝑺 − 𝚫𝑷𝑬 − 𝚫𝑲𝑬
198
4.7 ISENTROPIC PROCESS

Isentropic Process is also called as “Reversible Adiabatic Process”, or adiabatic process which means constant
entropy, and there is no heat transfer. Hence 𝑆1 = 𝑆2 = 𝑆 = 𝐶 and 𝑄 = 0. Consider a gas inside a piston cylinder
arrangement with perfect insulator. Heat will not enter nor leave the system because of heat transfer. If gas expands,
the degree of disorderness will increase, and will move the piston causing work.
State 1 State 2
1. PVT Relationships
The isentropic process for ideal gas follows the curve 𝑃𝑉 𝑘 = 𝐶, where k is the ratio of specific heats. Hence
𝑃𝑉 𝑘 = 𝐶
𝑃1 𝑉1 = 𝑃2 𝑉2𝑘 = 𝐶
𝑘
Hence
𝑷𝟐 𝑽𝟏 𝒌
=( )
𝑷𝟏 𝑽𝟐
𝑚𝑅𝑇2
𝑉2 𝑉1 𝑘
=( )
𝑚𝑅𝑇1 𝑉2
𝑉1
𝑇2 𝑉2 𝑉1 𝑘
= ( )
𝑇1 𝑉1 𝑉2
199
𝒌−𝟏
𝑻𝟐 𝑽𝟏
=( )
𝑻𝟏 𝑽𝟐
Using again the ideal gas law for the curve 𝑃𝑉 𝑘 = 𝐶

𝑃1 𝑉1𝑘 = 𝑃2 𝑉2𝑘
𝑚𝑅𝑇1 𝑘 𝑚𝑅𝑇2 𝑘
𝑃1 ( ) = 𝑃2 ( )
𝑃1 𝑃2
𝑃11−𝑘 𝑇1𝑘 = 𝑃21−𝑘 𝑇2𝑘
𝑇2 𝑘 𝑃1 1−𝑘
( ) =( )
𝑇1 𝑃2
𝑇2 𝑘 𝑃2 𝑘−1
( ) =( )
𝑇1 𝑃1
𝒌−𝟏
𝑻𝟐 𝑷𝟐 𝒌
=( )
𝑻𝟏 𝑷𝟏
2. Work Non-Flow if ΔPE = 0 ΔKE = 0 – it is the area under the 𝑃𝑉 𝑘 diagram.

2
1
Since 𝑃𝑉 𝑘 = 𝐶
𝐶
𝑃=
𝑉𝑘
𝑃 = 𝐶𝑉 −𝑘
Hence
2
𝑊𝑁𝐹 = 𝐶 ∫ 𝑉 −𝑘 𝑑𝑉
1
2
𝑉 −𝑘+1
𝑊𝑁𝐹 = 𝐶 [ ]
−𝑘 + 1 1
𝑉2−𝑘+1 − 𝑉1−𝑘+1
𝑊𝑁𝐹 = 𝐶 [ ]
−𝑘 + 1
1
𝑊𝑁𝐹 = [𝐶𝑉21−𝑘 − 𝐶𝑉11−𝑘 ]
1−𝑘
Also 𝑃1 𝑉1𝑘 = 𝑃2 𝑉2𝑘 = 𝐶
1
𝑊𝑁𝐹 = [𝑃 𝑉 𝑘 𝑉1−𝑘 − 𝑃1 𝑉1𝑘 𝑉11−𝑘 ]
1−𝑘 2 2 2
𝑃2 𝑉2 − 𝑃1 𝑉1
𝑊𝑁𝐹 =
1−𝑘
𝐏𝟏 𝐕𝟏 − 𝐏𝟐 𝐕𝟐
𝐖𝐍𝐅 =
𝐤−𝟏
𝑚𝑅𝑇1 − 𝑚𝑅𝑇2
𝑊𝑁𝐹 =
𝑘−1
𝒎𝑹(𝑻𝟏 − 𝑻𝟐 )
𝑾𝑵𝑭 =
𝒌−𝟏
Factor out T1 from the equation
𝑚𝑅𝑇1 𝑇2
𝑊𝑁𝐹 = ( − 1)
1 − 𝑘 𝑇1
Using PVT relationships for isentropic process:
200
𝑚𝑅𝑇1 𝑉1 𝑘−1
𝑊𝑁𝐹 = [( ) − 1]
1 − 𝑘 𝑉2
𝑘−1
𝑚𝑅𝑇1 𝑃2 𝑘
𝑊𝑁𝐹 = [( ) − 1]
1 − 𝑘 𝑃1
If 𝑃1 , 𝑉1 and 𝑘 are given:
𝑷𝟏 𝑽 𝟏 𝑽𝟏 𝒌−𝟏
𝑾𝑵𝑭 = [𝟏 − ( ) ]
𝒌−𝟏 𝑽𝟐
𝒌−𝟏
𝑷𝟏 𝑽 𝟏 𝑷𝟐 𝒌
𝑾𝑵𝑭 = [𝟏 − ( ) ]
𝒌−𝟏 𝑷𝟏
5. Heat - the heat transfer in an isentropic/adiabatic process is zero since the system is insulated. The magnitude of
heat is defined as the area under the T-s diagram. If 𝑆 = 𝐶 then 𝑑𝑆 = 0. Hence mathematically speaking the
magnitude of heat transfer in an isentropic process is zero based on the equation and based also on our assumption
of adiabatic conditions.
𝑸=𝟎
6. Change in Entropy - the change in entropy is equal to zero since 𝑆1 = 𝑆2 . Also, the area under the T-S
diagram has a vertical slope, which means an area of zero. Also, if 𝑄 = 0, then 𝑑𝑄 = 0
𝑄
𝑑𝑄
Δ𝑆 = ∫
0 𝑇
𝚫𝑺 = 𝟎
7. Work Steady Flow if ΔPE = 0 ΔKE = 0 - the steady flow work is the area “behind” the PV diagram.
2
1
Since 𝑃𝑉 𝑘 = 𝐶
𝐶 1/𝑘
𝑉=
𝑃1/𝑘
1/𝑘
Also 𝐶 ≈ 𝐶 (Since a constant raise to a constant
is also a constant)
𝑉 = 𝐶𝑃 −1/𝑘
Hence:
2
𝑊𝑆𝐹 = 𝐶 ∫ 𝑃 −1/𝑘 𝑑𝑃
1
1 2
𝑃−𝑘+1
𝑊𝑆𝐹 = 𝐶 [ ]
1
− +1
𝑘 1
1 1
𝑘 − +1 − +1
𝑊𝑆𝐹 = 𝐶 [𝑃2 𝑘 − 𝑃1 𝑘 ]
1−𝑘
Also
1 1
𝑃𝑘 𝑉 = 𝐶 𝑘 = 𝐶
201
1/𝑘 1/𝑘
𝑃2 𝑉2 = 𝑃1 𝑉1 = 𝐶
1 1
𝑘 − +1 − +1
𝑊𝑆𝐹 = [𝐶𝑃2 𝑘 − 𝐶𝑃1 𝑘 ]
1− 𝑘
1 1
𝑘 1/𝑘 −𝑘 +1 1/𝑘 −𝑘 +1
𝑊𝑆𝐹 = [𝑃 𝑉2 𝑃2 − 𝑃1 𝑉1 𝑃1 ]
1−𝑘 2
𝑘
𝑊𝑆𝐹 = [𝑃 𝑉 − 𝑃1 𝑉1 ]
1−𝑘 2 2
𝒌(𝑷𝟏 𝑽𝟏 − 𝑷𝟐 𝑽𝟐 )
𝑾𝑺𝑭 =
𝒌−𝟏
This equation is almost identical to non-flow work for isentropic process. The steady flow work is also equivalent
to
𝑾𝑺𝑭 = 𝒌𝑾𝑵𝑭
Equations for steady flow work in terms of 𝑃1 , 𝑉1 , 𝑇1 , and 𝑘 has the same derivation as for non-flow work.
𝒎𝑹𝒌(𝑻𝟏 − 𝑻𝟐 )
𝑾𝑺𝑭 =
𝒌−𝟏
𝒌𝑷𝟏 𝑽𝟏 𝑽𝟏 𝒌−𝟏
𝑾𝑺𝑭 = [𝟏 − ( ) ]
𝒌−𝟏 𝑽𝟐
𝒌−𝟏
𝒌𝑷𝟏 𝑽𝟏 𝑷𝟐 𝒌
𝑾𝑺𝑭 = [𝟏 − ( ) ]
𝒌−𝟏 𝑷𝟏
𝑘𝑅
If 𝑐𝑝 = , then
𝑘−1
𝑚𝑅𝑘(𝑇1 − 𝑇2 )
𝑊𝑆𝐹 =
𝑘−1
𝑊𝑆𝐹 = 𝑚𝑐𝑝 (𝑇1 − 𝑇2 )
𝑾𝑺𝑭 = −𝚫𝑯
8. Work Steady Flow if Work Steady Flow if ΔPE ≠ 0 ΔKE ≠ 0 - by energy balance:
Since 𝑄 = 0 and 𝑊𝑆𝐹 = −Δ𝐻
𝑊𝑆𝐹 = 0 + 𝑊𝑆𝐹 − Δ𝑃𝐸 − Δ𝐾𝐸
𝚫𝑷𝑬 = −𝚫𝑲𝑬
202
4.8 POLYTROPIC PROCESS

Polytropic process is a “many type” processes and follows the law 𝑃𝑉 𝑛 = 𝐶 where n is called as the polytropic
index or polytropic constant. Considered a gas contained in a piston-cylinder arrangement with actual coolant. The
derivation for equations for polytropic process is the same as the isentropic process, but k = n.
State 1 State 2
1. PVT relationships
𝑷𝟏 𝑽𝒏𝟏 = 𝑷𝟐 𝑽𝒏𝟐
𝒏−𝟏
𝑻𝟐 𝑽𝟏 𝒏−𝟏 𝑷𝟐 𝒏
=( ) =( )
𝑻𝟏 𝑽𝟐 𝑷𝟏
2. Work Non Flow if ΔPE = 0 , ΔKE = 0

𝑷𝟏 𝑽 𝟏 − 𝑷𝟐 𝑽 𝟐
𝑾𝑵𝑭 =
𝒏−𝟏
𝒎𝑹(𝑻𝟏 − 𝑻𝟐 )
𝑾𝑵𝑭 =
𝒏−𝟏
𝑷𝟏 𝑽 𝟏 𝑽𝟏 𝒏−𝟏
𝑾𝑵𝑭 = [𝟏 − ( ) ]
𝒏−𝟏 𝑽𝟐
203
𝒏−𝟏
𝑷𝟏 𝑽 𝟏 𝑷𝟐 𝒏
𝑾𝑵𝑭 = [𝟏 − ( ) ]
𝒏−𝟏 𝑷𝟏

5. Heat - From the definition of non-flow work if ΔPE = 0 and ΔKE = 0.

𝑊𝑁𝐹 = 𝑄 − Δ𝑈
𝑄 = 𝑊𝑁𝐹 + Δ𝑈
𝑚𝑅(𝑇1 − 𝑇2 )
𝑄= + 𝑚𝑐𝑣 (T2 − T1 )
𝑛−1
𝑚𝑅(𝑇2 − 𝑇1 )
𝑄= + 𝑚𝑐𝑣 (T2 − T1 )
1−𝑛
𝑅
𝑄 = 𝑚Δ𝑇 ( + 𝑐𝑣 )
1−𝑛
Since 𝑅 = 𝑐𝑝 − 𝑐𝑣 , then
𝑐𝑝 − 𝑐𝑣
𝑄 = 𝑚Δ𝑇 ( + 𝑐𝑣 )
1−𝑛
𝑚Δ𝑇
Q= [c − cv + (1 − n)cv ]
1−n p
𝑚Δ𝑇
𝑄= (𝑐 − 𝑛𝑐𝑣 )
1−n 𝑝
Since
𝑐𝑝
𝑘=
𝑐𝑣
And
𝑐𝑝 = 𝑘𝑐𝑣
𝑚Δ𝑇
𝑄= (𝑘𝑐𝑣 − 𝑛𝑐𝑣 )
1−n
𝒄𝒗 (𝒌 − 𝒏)
𝑸 = 𝒎𝚫𝑻 [ ]
𝟏−𝒏
Hence we defined the specific heat at polytropic process
𝒄𝒗 (𝒌 − 𝒏)
𝒄𝒏 =
𝟏−𝒏
In which
𝑸 = 𝒎𝒄𝒏 𝚫𝑻
𝟐
𝒅𝑸
𝚫𝑺 = ∫
𝟏 𝑻
Since 𝑑𝑄 = 𝑚𝑐𝑛 𝑑𝑇
𝟐
𝑑𝑇
𝚫𝑺 = 𝑚𝑐𝑛 ∫
𝟏 𝑻
𝑻𝟐
𝚫𝑺 = 𝒎𝒄𝒏 𝒍𝒏
𝑻𝟏
204
7. Work Steady Flow if ΔPE = 0 ΔKE = 0

𝑾𝑺𝑭 = 𝒏𝑾𝑵𝑭
𝒏(𝑷𝟏 𝑽𝟏 − 𝑷𝟐 𝑽𝟐 )
𝑾𝑺𝑭 =
𝒏−𝟏
𝒎𝑹𝒏(𝑻𝟏 − 𝑻𝟐 )
𝑾𝑺𝑭 =
𝒏−𝟏
𝒏𝑷𝟏 𝑽𝟏 𝑽𝟏 𝒏−𝟏
𝑾𝑺𝑭 = [𝟏 − ( ) ]
𝒏−𝟏 𝑽𝟐
𝒏−𝟏
𝒏𝑷𝟏 𝑽𝟏 𝑷𝟐 𝒏
𝑾𝑺𝑭 = [𝟏 − ( ) ]
𝒏−𝟏 𝑷𝟏
8. Work Steady Flow if Work Steady Flow if ΔPE ≠
0 ΔKE ≠ 0 - by energy balance:
𝑾𝑺𝑭 = 𝑸 − 𝚫𝑯 − 𝚫𝑷𝑬 − 𝚫𝑲𝑬
205
4.9 PROCESS CURVES
Consider the polytropic equation 𝑃𝑉 𝑛 = 𝐶, where n is called as the polytropic index. The range of n can be zero
up to infinity.
• If n = 0 P=C Isobaric process

• If n is at infinity V=C Isometric process
• If n = 1 PV = C Isothermal process
• If n=k 𝑃𝑉 𝑘 = 𝐶 Isentropic process
• If n is some constant 𝑃𝑉 𝑛 = 𝐶 Polytropic process
Proof that if 𝒏 = ∞, V=C:
Assuming 𝑛 = ∞, from the PV diagram,

𝑃𝑉 𝑛 = 𝐶
𝑃𝑉 ∞ = 𝐶
𝑃1 𝑉1∞ = 𝑃2 𝑉2∞
𝑉1 ∞ 𝑃2
( ) =
𝑉2 𝑃1
1
𝑉1 𝑃2 ∞
=( )
𝑉2 𝑃1
1
𝑃 𝑥
Taking the limit of the value ( 2 ) as x approaches infinity
𝑃1
1
𝑃2 𝑥
lim ( ) = 1
𝑥→∞ 𝑃1
Hence
𝑉1
=1
𝑉2
Or simply
𝑉1 = 𝑉2 = 𝑉
206
4.10 COMPRESSORS
An application of isothermal, isentropic, and polytropic process is the compressor work. A compressor is a
device that increases the pressure of a system. Power is required to run the compressor.
Where 𝑾𝒄 is the work of the compressor, or the theoretical / isentropic work of compression.
𝑾𝒊 is the indicated work or the actual work/polytropic work of compression
𝑾𝒃 is the brake work or the work available at the motor shaft
𝑷𝒎 is the power input of the motor to run the compressor
Compressor Equations: (Note: If compressor is not stated whether an open or closed system, the assumption is
always an open system)
For steady flow theoretical work of compression, use For steady flow actual work of compression, use the
the isentropic steady flow work equations: polytropic steady flow work equation:
𝒌(𝑷𝟏 𝑽𝟏 − 𝑷𝟐 𝑽𝟐 ) 𝒏(𝑷𝟏 𝑽𝟏 − 𝑷𝟐 𝑽𝟐 )
𝑾𝑺𝑭 = 𝑾𝑺𝑭 =
𝒌−𝟏 𝒏−𝟏
𝒎𝑹𝒌(𝑻𝟏 − 𝑻𝟐 ) 𝒎𝑹𝒏(𝑻𝟏 − 𝑻𝟐 )
𝑾𝑺𝑭 = 𝑾𝑺𝑭 =
𝒌−𝟏 𝒏−𝟏
𝒌𝑷𝟏 𝑽𝟏 𝑽𝟏 𝒌−𝟏 𝒏𝑷𝟏 𝑽𝟏 𝑽𝟏 𝒏−𝟏
𝑾𝑺𝑭 = [𝟏 − ( ) ] 𝑾𝑺𝑭 = [𝟏 − ( ) ]
𝒌−𝟏 𝑽𝟐 𝒏−𝟏 𝑽𝟐
𝒌−𝟏 𝒏−𝟏
𝒌𝑷𝟏 𝑽𝟏 𝑷𝟐 𝒌 𝒏𝑷𝟏 𝑽𝟏 𝑷𝟐 𝒏
𝑾𝑺𝑭 = [𝟏 − ( ) ] 𝑾𝑺𝑭 = [𝟏 − ( ) ]
𝒌−𝟏 𝑷𝟏 𝒏−𝟏 𝑷𝟏
Efficiencies Applicable to Compressors

1. Adiabatic Compression Efficiency - also called as the isentropic compression efficiency. It is the ratio of the
theoretical compressor work to actual compressor work.
𝑾𝒄
𝛈𝒂 =
𝑾𝒊
2. Mechanical Efficiency
𝑾𝒊
𝛈𝒎𝒆 =
𝑾𝒃
3. Electrical/Motor Efficiency
𝑾𝒃
𝛈𝒆𝒆 = 𝛈𝒎𝒐 =
𝑷𝒎
207
4. Overall Efficiency
𝛈𝒐 = 𝛈𝒆𝒆 𝛈𝒎𝒆 𝛈𝒂
𝑾𝒄
𝛈𝒐 =
𝑷𝒎
SOLVED PROBLEMS
Problem 4-1Ten cubic foot of air at 200 psia and 300°F is cooled to 200°F at constant volume. Determine
a) Final pressure
b) Work
c) Change in internal energy
d) Transferred heat
e) Change in enthalpy
f) Change in entropy
Given:
𝑉1 = 𝑉2 = 10𝑓𝑡 3
𝑃1 = 200𝑝𝑠𝑖𝑎
𝑡1 = 300°𝐹
𝑡2 = 200°𝐹
Solution:
a) 𝑃2 =?
Using PVT relationship for isometric process:
𝑃2 𝑇2
=
𝑃1 𝑇1
𝑇2
𝑃2 = 𝑃1 ( )
𝑇1
300 + 460
𝑃2 = (200𝑝𝑠𝑖𝑎) ( )
200 + 460
𝑷𝟐 = 𝟐𝟑𝟎. 𝟑𝟎𝟑𝒑𝒔𝒊𝒂 ⇒ 𝐴𝑛𝑠.
b) WNF =?
𝑾𝑵𝑭 = 𝟎 ⇒ 𝐴𝑛𝑠.
c) Δ𝑈
Δ𝑈 = 𝑚𝑐𝑣 Δ𝑇
For 𝑚
𝑃1 𝑉1
𝑚=
𝑅𝑇1
From Item B1:
𝑅𝑎𝑖𝑟 = 53.342
𝐵𝑇𝑈
𝑐𝑣𝑎 = 0.1714
𝐵𝑇𝑈
𝑐𝑝𝑎 = 0.24
(200 )( ) (10𝑓𝑡 3 )
𝑖𝑛2 1𝑓𝑡 2
𝑚=
(53.342 ) (300 + 460)°𝑅
𝑚 = 7.104𝑙𝑏𝑚
208
Hence
𝐵𝑇𝑈
Δ𝑈 = 7.104𝑙𝑏𝑚 (0.1714 ) (200 − 300)°𝑅
𝚫𝑼 = −𝟏𝟐𝟏. 𝟕𝟔𝟑 𝑩𝑻𝑼 ⇒ 𝐴𝑛𝑠.
Note: A negative sign indicates that there is a decrease in internal energy:
d) 𝑄 =?
For isometric process
𝑄 = Δ𝑈
𝑸 = −𝟏𝟐𝟏. 𝟕𝟔𝟑 𝑩𝑻𝑼 ⇒ 𝐴𝑛𝑠.
e) Δ𝐻 =?
Δ𝐻 = 𝑚𝑐𝑝 (𝑇2 − 𝑇1 )
𝐵𝑇𝑈
𝛥𝐻 = 7.104𝑙𝑏𝑚 (0.24 ) (200 − 300)°𝐹
𝜟𝑯 = −𝟏𝟕𝟎. 𝟒𝟗𝟔 𝑩𝑻𝑼 ⇒ 𝐴𝑛𝑠.
Note: A negative sign indicates that there is a decrease in enthalpy.
f) Δ𝑆 =?
𝑇2
Δ𝑆 = 𝑚𝑐𝑣 𝑙𝑛
𝑇1
𝐵𝑇𝑈 200 + 460
Δ𝑆 = (7.104𝑙𝑏𝑚) (0.1714 ) 𝑙𝑛 ( )
𝑙𝑏𝑚 − °𝑅 300 + 460
𝑩𝑻𝑼
𝚫𝑺 = −𝟎. 𝟏𝟕𝟏𝟖 ⇒ 𝐴𝑛𝑠.
°𝑹
Note: A negative sign indicates that there is a decrease in entropy.
Problem 4-2: There are 1.36 kg of gas for which R = 377 J/kg-K and k=1.25 that undergoes a non-flow constant
volume process from 551.6 kPaa and 60°C to 1655 kPaa. During the process the gas is internally stirred, and there
are also added 105.5 kJ of heat. Determine
a) Final temperature in °C.
b) Work input in kJ
c) Q in hp-min
d) Change in internal energy in kW-hr
e) Change in entropy in BTU/°R
Given:
𝑚 = 1.36𝑘𝑔
𝐽
𝑅 = 377
𝑘𝑔 − 𝐾
𝑘 = 1.25
𝑃1 = 551.6𝑘𝑃𝑎𝑎
𝑡1 = 60°𝐶
𝑃2 = 1655𝑘𝑃𝑎𝑎
𝑄𝐴 = 105.5𝑘𝐽
209
Solution:
a) 𝑡2 =?
By PVT relationship for isometric process:
𝑃2 𝑇2
=
𝑃1 𝑇1
𝑃2
𝑇2 = 𝑇1
𝑃1
1655𝑘𝑃𝑎𝑎
𝑇2 = (60 + 273)𝐾 ( )
551.6𝑘𝑃𝑎𝑎
𝑇2 = 999.121𝐾
𝑡2 = 999.121 − 273
𝒕𝟐 = 𝟕𝟐𝟔. 𝟏𝟐𝟏°𝑪 ⇒ 𝐴𝑛𝑠.
b) 𝑊𝑝𝑎𝑑𝑑𝑙𝑒
By energy balance
𝑊𝑝𝑎𝑑𝑑𝑙𝑒 + 𝑄𝐴 = Δ𝑈
𝑊𝑝𝑎𝑑𝑑𝑙𝑒 = Δ𝑈 − 𝑄𝐴
For Δ𝑈:
Δ𝑈 = 𝑚𝑐𝑣 Δ𝑇
For 𝑐𝑉
𝑅
𝑐𝑣 =
𝑘−1
𝐽 1𝑘𝐽
377 ( )
𝑘𝑔 − 𝐾 1000𝐽
𝑐𝑣 =
1.25 − 1
𝑘𝐽
𝑐𝑣 = 1.508
𝑘𝑔 − 𝐾
Hence
𝑘𝐽
Δ𝑈 = (1.36𝑘𝑔) (1.508 ) (726.121 − 60)𝐾
𝑘𝑔 − 𝐾
Δ𝑈 = 1366.134𝑘𝐽
Hence
𝑊𝑝𝑎𝑑𝑑𝑙𝑒 = 1366.134𝑘𝐽 − 105.5𝑘𝐽
𝑾𝒑𝒂𝒅𝒅𝒍𝒆 = 𝟏𝟐𝟔𝟎. 𝟔𝟑𝟒𝒌𝑱 ⇒ 𝐴𝑛𝑠.
c) 𝑄 =?
1𝑘𝑊 − 𝑠 1ℎ𝑝 1𝑚𝑖𝑛
𝑄 = 105.5𝑘𝐽 ( )( )( )
1𝑘𝐽 0.746𝑘𝑊 60𝑠
𝑸 = 𝟐. 𝟑𝟓𝟕 𝒉𝒑 − 𝒎𝒊𝒏 ⇒ 𝐴𝑛𝑠.
d) Δ𝑈 =?
1𝑘𝑊 − 𝑠 1ℎ𝑟
Δ𝑈 = 1366.134𝑘𝐽 ( )( )
1𝑘𝐽 3600𝑠
𝚫𝑼 = 𝟎. 𝟑𝟕𝟗𝟓 𝒌𝑾 − 𝒉𝒓 ⇒ 𝐴𝑛𝑠.
e) Δ𝑆 =?
𝑇2
Δ𝑆 = 𝑚𝑐𝑣 𝑙𝑛
𝑇1
𝑘𝐽 726.121 + 273
Δ𝑆 = (1.36𝑘𝑔) (1.508 ) 𝑙𝑛
𝑘𝑔 − 𝐾 60 + 273
𝒌𝑱
𝚫𝑺 = 𝟐. 𝟐𝟓𝟑 ⇒ 𝐴𝑛𝑠.
𝑲
210
Problem 4-3: Three pounds of perfect gas with R=38.8 ft-lbf/lbm-°R and k=1.67 have 350 BTU of heat added
during a reversible non flow constant pressure process. The initial temperature is 100°F. Determine
a) Final temperature
b) Change in enthalpy
c) Work
d) Change in internal energy
e) Change in entropy
Given:
𝑚 = 3 𝑙𝑏𝑚
𝑅 = 38.8
𝑘 = 1.67
𝑄𝐴 = 350𝐵𝑇𝑈
𝑡1 = 100°𝐹
Solution:
a) 𝑡2 =?
From heat added:
𝑄𝐴 = 𝑚𝑐𝑝 Δ𝑇
For 𝑐𝑝
𝑘𝑅
𝑐𝑝 =
𝑘−1
𝑓𝑡 − 𝑙𝑏𝑓 1𝐵𝑇𝑈
1.67 (38.8 )( )
𝑙𝑏𝑚 − °𝑅 778.16𝑓𝑡 − 𝑙𝑏𝑓
𝑐𝑝 =
1.67 − 1
𝐵𝑇𝑈
𝑐𝑝 = 0.1243
Hence
𝑄𝐴 = 𝑚𝑐𝑝 (t 2 − t1 )
𝑄𝐴
𝑡2 = 𝑡1 +
𝑚𝑐𝑝
350𝐵𝑇𝑈
𝑡2 = 100°𝐹 +
𝐵𝑇𝑈
(3𝑙𝑏𝑚) (0.1243 )
𝑙𝑏𝑚 − °𝐹
𝒕𝟐 = 𝟏𝟎𝟑𝟖. 𝟓𝟖𝟗°𝑭 ⇒ 𝐴𝑛𝑠.
b) Δ𝐻 =?
For isobaric process, Δ𝐻 = 𝑄
𝚫𝑯 = 𝟑𝟓𝟎𝑩𝑻𝑼 ⇒ 𝐴𝑛𝑠.
c) 𝑊𝑁𝐹 =?
For non-flow work, by energy balance
𝑄𝐴 = Δ𝑈 + 𝑊𝑁𝐹
𝑊𝑁𝐹 = 𝑄𝐴 − Δ𝑈
For Δ𝑈
Δ𝑈 = 𝑚𝑐𝑣 (𝑡2 − 𝑡1 )
For 𝑐𝑣
𝑅
𝑐𝑣 =
𝑘−1
𝑓𝑡 − 𝑙𝑏𝑓 1𝐵𝑇𝑈
(38.8 )( )
𝑙𝑏𝑚 − °𝑅 778.16𝑓𝑡 − 𝑙𝑏𝑓
𝑐𝑣 =
1.67 − 1
𝐵𝑇𝑈
𝑐𝑣 = 0.07442
211
Hence
𝐵𝑇𝑈
Δ𝑈 = (3𝑙𝑏𝑚) (0.07442 ) (1038.589 − 100)°𝑅
Δ𝑈 = 209.549 𝐵𝑇𝑈
Hence
𝑊𝑁𝐹 = 350𝐵𝑇𝑈 − 209.649𝐵𝑇𝑈
𝑾𝑵𝑭 = 𝟏𝟒𝟎. 𝟑𝟓𝟏𝑩𝑻𝑼 ⇒ 𝑨𝒏𝒔.
d) Δ𝑈 =?
𝚫𝑼 = 𝟐𝟎𝟗. 𝟓𝟒𝟗 𝑩𝑻𝑼 ⇒ 𝐴𝑛𝑠.
e) Δ𝑆 =?
𝑇2
𝑇1
𝐵𝑇𝑈 1038.789 + 460
Δ𝑆 = (3𝑙𝑏𝑚) (0.1243 ) 𝑙𝑛
𝑙𝑏𝑚 − °𝑅 100 + 460
𝑩𝑻𝑼
𝚫𝑺 = 𝟎. 𝟑𝟔𝟕𝟏 ⇒ 𝐴𝑛𝑠.
°𝑹
Problem 4-4 The volume of a system changes from 0.50 m3 to 0.30 m3 at a constant pressure of 700 kPaa. For air,
determine
a) Change in internal energy
b) Change in enthalpy
c) Heat
d) Change in specific entropy
e) If the process is nonflow and internally reversible, what is the work?
Given:
𝑉1 = 0.50𝑚3
𝑉2 = 0.30𝑚3
𝑃2 = 𝑃1 = 𝑃 = 700𝑘𝑃𝑎𝑎
Solution:
a) Δ𝑈 =?
Δ𝑈 = 𝑚𝑐𝑣 (𝑇2 − 𝑇1 )
Δ𝑈 = 𝑐𝑣 (𝑚𝑇2 − 𝑚𝑇1 )
From ideal gas law [𝑃𝑉 = 𝑚𝑅𝑇]
𝑃𝑉
= 𝑚𝑇
𝑅
Hence
𝑃2 𝑉2 𝑃1 𝑉1
Δ𝑈 = 𝑐𝑉 ( − )
𝑅 𝑅
Since 𝑃1 = 𝑃2 = 𝑃
𝑐𝑣 𝑃
Δ𝑈 = (𝑉2 − 𝑉1 )
𝑅
From Item B1
𝑘𝐽
𝑅𝑎𝑖𝑟 = 0.28708
𝑘𝑔 − 𝐾
𝑘𝐽
𝑐𝑣 = 0.7186
𝑘𝑔 − 𝐾
Hence
𝑘𝐽 𝑘𝑁
(0.7186 ) (700 2 ) (0.30 − 0.50)𝑚3
𝑘𝑔 − 𝐾 𝑚
Δ𝑈 =
𝑘𝐽
0.28708
𝑘𝑔 − 𝐾
212
𝚫𝑼 = −𝟑𝟓𝟎. 𝟒𝟑𝟗 𝒌𝑱 ⇒ 𝑨𝒏𝒔.
b) Δ𝐻 =?
Δ𝐻 = 𝑚𝑐𝑝 (T2 − T1 )
Δ𝐻 = 𝑐𝑝 (𝑚𝑇2 − 𝑚𝑇1 )
From ideal gas law
𝑃𝑉
= 𝑚𝑇
𝑅
Hence
𝑃2 𝑉2 𝑃1 𝑉1
Δ𝐻 = 𝑐𝑝 ( − )
𝑅 𝑅
𝑐𝑝 𝑃
Δ𝐻 = (𝑉2 − 𝑉1 )
𝑅
From Item B1:
𝑘𝐽
𝑐𝑝 = 1.0062
𝑘𝑔 − 𝐾
Hence:
𝑘𝐽 𝑘𝑁
(1.0062 ) (700 2 ) (0.30 − 0.50)𝑚3
𝑘𝑔 − 𝐾 𝑚
𝛥𝐻 =
𝑘𝐽
0.28708
𝑘𝑔 − 𝐾
𝜟𝑯 = −𝟒𝟗𝟎. 𝟔𝟗𝟐 𝒌𝑱 ⇒ 𝐴𝑛𝑠.
c) 𝑄 =?
For isobaric process, 𝑄 = Δ𝐻
𝑸 = −𝟒𝟗𝟎. 𝟔𝟗𝟐 𝒌𝑱 ⇒ 𝐴𝑛𝑠.
d) Δ𝑆 =?
𝑇2
𝑇1
By PVT relationship:
𝑉2 𝑇2
=
𝑉1 𝑇1
Hence
𝑉2
𝑉1
Δ𝑆 𝑉2
Δ𝑠 = = 𝑐𝑝 𝑙𝑛
𝑚 𝑉1
𝑘𝐽 0.30𝑚3
Δ𝑠 = (1.0062 ) 𝑙𝑛
𝑘𝑔 − 𝐾 0.50𝑚3
𝒌𝑱
𝚫𝒔 = −𝟎. 𝟓𝟏𝟒 ⇒ 𝐴𝑛𝑠.
𝒌𝒈 − 𝑲
213
Problem 4-5: Four pounds of air gain 0.495 BTU/°R of entropy during a non-flow isothermal process. Initially
the pressure is at 120 psia, and when heat is transferred to the system, the volume expands to 42.5 ft 3. Find
a) Initial volume
b) Initial temperature
c) Work
d) Heat
e) Change in internal energy
Given:
𝑚 = 4 𝑙𝑏𝑚
𝐵𝑇𝑈
Δ𝑆 = 0.495
°𝑅
𝑃1 = 120𝑝𝑠𝑖𝑎
𝑉2 = 42.5𝑓𝑡 3
Solution:
a) 𝑉1 =?
By energy balance, assuming a piston cylinder arrangement as the system. When heat is transferred to the
system, work is done to the system,
𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡
𝑄𝐴 = 𝑊𝑁𝐹 + Δ𝑈
Since Δ𝑈 = 0 for isothermal process:
𝑄𝐴 = 𝑊𝑁𝐹
𝑉2
𝑇Δ𝑆 = 𝑃1 𝑉1 𝑙𝑛
𝑉1
Since 𝑃1 𝑉1 = 𝑚𝑅𝑇
𝑉2
𝑇Δ𝑆 = 𝑚𝑅𝑇𝑙𝑛
𝑉1
𝑉2
Δ𝑆 = 𝑚𝑅𝑙𝑛
𝑉1
Δ𝑆 𝑉2
= 𝑙𝑛
𝑚𝑅 𝑉1
From Item B1:
𝑅𝑎𝑖𝑟 = 53.342
𝐵𝑇𝑈
𝑐𝑣𝑎 = 0.1714
𝐵𝑇𝑈
𝑐𝑝𝑎 = 0.24
𝐵𝑇𝑈 778.16𝑓𝑡 − 𝑙𝑏𝑓

(0.495 )( ) 42.5𝑓𝑡 3
°𝑅 1𝐵𝑇𝑈
= 𝑙𝑛
𝑓𝑡 − 𝑙𝑏𝑓 𝑉1
(4𝑙𝑏𝑚) (53.342 )
42.5𝑓𝑡 3
𝑙𝑛 = 1.8053
𝑉1
42.5𝑓𝑡 3
= 𝑒 1.8053
𝑉1
𝑉1 = (42.5𝑓𝑡 3 )𝑒 −1.8053
𝑽𝟏 = 𝟔. 𝟗𝟖𝟖𝒇𝒕𝟑 ⇒ 𝐴𝑛𝑠.
214
b) For temperature 𝑡1 , from ideal gas law:

𝑃1 𝑉1 = 𝑚𝑅𝑇1
𝑃1 𝑉1
𝑇1 =
𝑚𝑅
(120 2 ) ( ) (6.988𝑓𝑡 3 )
𝑖𝑛 1𝑓𝑡 2
𝑇1 =
(4𝑙𝑏𝑚) (53.342 )
𝑇1 = 565.936°𝑅
𝑡1 = 564.936 − 460
𝒕𝟏 = 𝟏𝟎𝟓. 𝟗𝟑𝟔°𝑭 ⇒ 𝐴𝑛𝑠.
c) 𝑊𝑁𝐹 =?
𝑉2
𝑉1
2
𝑙𝑏𝑓 144𝑖𝑛 3 )𝑙𝑛
42.5𝑓𝑡 3 1𝐵𝑇𝑈
𝑊𝑁𝐹 = (120 2 ) ( 2 ) (6.988𝑓𝑡 3 ( )
𝑖𝑛 1𝑓𝑡 6.988𝑓𝑡 778.16𝑓𝑡 − 𝑙𝑏𝑓
𝑾𝑵𝑭 = 𝟐𝟖𝟎. 𝟏𝟒𝟑𝑩𝑻𝑼 ⇒ 𝐴𝑛𝑠.
d) Q =?
𝑄 = 𝑇Δ𝑆
𝐵𝑇𝑈
Q = (565.936°R) (0.495 )
°𝑅
𝑸 = 𝟐𝟖𝟎. 𝟏𝟑𝟖 𝑩𝑻𝑼 ⇒ 𝐴𝑛𝑠.
e) Δ𝑈 =?
For isothermal process:
𝚫𝑼 = 𝟎 ⇒ 𝐴𝑛𝑠.
Problem 4-6 Three kilograms per second of air at 30°C is compressed isothermally until the pressure increases by
90% of its initial pressure. What is the work done due to compression in kW?
Given:
𝑚 = 3𝑘𝑔/𝑠
𝑡 = 30°𝐶
Solution:
For 𝑊𝑆𝐹
𝑃1
̇ = 𝑃1 𝑉1̇ 𝑙𝑛
𝑊𝑆𝐹
𝑃2
Since 𝑃1 𝑉1̇ = 𝑚̇𝑅𝑇
𝑃1
̇ = 𝑚̇𝑅𝑇𝑙𝑛
𝑊𝑆𝐹
𝑃2
Since 𝑃2 = (1 + 0.90)𝑃1 = 1.90𝑃1
𝑃1
̇ = 𝑚̇ 𝑅𝑇𝑙𝑛
𝑊𝑆𝐹
1.90𝑃1
𝑘𝑔𝑚 𝑘𝐽 1
̇ = (3
𝑊𝑆𝐹 ) (0.28708 ) (30 + 273)𝐾𝑙𝑛
𝑠 𝑘𝑔 − 𝐾 1.90
̇
𝑾𝑺𝑭 = −𝟏𝟔𝟕. 𝟒𝟗𝟓𝒌𝑾 ⇒ 𝐴𝑛𝑠.
Negative sign indicates that work is done to the system, which makes sense since work is done so that air is to be
compressed, increasing its pressure while decreasing its volume in a piston-cylinder arrangement.
215
Problem 4-7 Determine the isentropic horsepower required for 1 kg/s of air at 100kPaa and 30°C to be compressed
to 200kPaa.
Given:
𝑃1 = 100𝑘𝑃𝑎𝑎
𝑇1 = 30°𝐶
For air, 𝑘 = 1.4
Solution: For steady flow work
𝑘(𝑃1 𝑉1 − 𝑃2 𝑉2 )
𝑊𝑆𝐹 =
𝑘−1
Since 𝑃𝑉 = 𝑚𝑅𝑇
𝑘𝑚𝑅(𝑇2 − 𝑇1 )
𝑊𝑆𝐹 =
1−𝑘
Factoring 𝑇1
𝑘𝑚𝑅𝑇1 𝑇2
𝑊𝑆𝐹 = [1 − ]
𝑘−1 𝑇1
By PVT relationships:
𝑘−1
𝑇2 𝑃2 𝑘
=( )
𝑇1 𝑃1
𝑘−1
𝑘𝑚̇𝑅𝑇1 𝑃2 𝑘
̇ =
𝑊𝑆𝐹 [1 − ( ) ]
𝑘−1 𝑃1
kg 𝑘𝐽 1.4−1
(1.4) (1 ) (0.28708 ) (30 + 273)𝐾 200 𝑘𝑃𝑎𝑎 1.4 1ℎ𝑝
̇ = s 𝑘𝑔 − 𝐾
𝑊𝑆𝐹 [1 − ( ) ]( )
1.4 − 1 100 𝑘𝑃𝑎𝑎 0.746𝑘𝑊
̇ = −𝟖𝟗. 𝟑𝟖𝟏𝒉𝒑 ⇒ 𝐴𝑛𝑠.
𝑾𝑺𝑭
Problem 4-8 A perfect gas is expanded polytropically with an initial volume and temperature of 60L and 147°C.
If the final volume and temperature are 210L and 21°C respectively, what is the polytropic index?
Solution:
𝑇2 𝑉1 𝑛−1
=( )
𝑇1 𝑉2
𝑇2 𝑉1
𝑙𝑛 = (𝑛 − 1)𝑙𝑛
𝑇1 𝑉2
𝑇2
𝑙𝑛
𝑇1
𝑛 =1+
𝑉
𝑙𝑛 1
𝑉2
(21 + 273)𝐾
𝑙𝑛
(147 + 273)𝐾
𝑛 = 1+
60𝐿
𝑙𝑛
210𝐿
𝒏 = 𝟏. 𝟐𝟖𝟓 ⇒ 𝐴𝑛𝑠.
216
Problem 4-9 1 kgm/s of air at 10 kPaa and 27°C is compressed isentropically to 100kPaa. If mechanical, electrical,
and adiabatic compression efficiency to be 80%, 70%, and 85%. Determine
a) Final temperature in °C
b) Work of the compressor in hp
c) Power input to the motor in kW
d) The cost of operating 24hr continuous operation if electricity cost ₱6.25/kW-hr.
e) The actual temperature at the exit in °C
f) The equivalent polytropic index
Solution: Consider the compressor shown below. We assume steady flow work.
Given:
𝑚 = 1𝑘𝑔𝑚/𝑠
𝑡1 = 27°𝐶
η𝑚𝑒 = 0.80
η𝑒𝑒 = 0.70
η𝑎 = 0.85
𝑘𝑎𝑖𝑟 = 1.4
a) 𝑡2 =?
By PVT relationships
𝑘−1
𝑇2 𝑃2 𝑘
=( )
𝑇1 𝑃1
𝑘−1
𝑃2 𝑘
𝑇2 = 𝑇1 ( )
𝑃1
1.4−1
100 𝑘𝑃𝑎𝑎 1.4
𝑇2 = (27 + 273)𝐾 ( )
10 𝑘𝑃𝑎𝑎
𝑇2 = 579.209𝐾
𝑡2 = 579.209 − 273
𝒕𝟐 = 𝟑𝟎𝟔. 𝟐𝟎𝟗°𝑪 ⇒ 𝐴𝑛𝑠.
217
b) 𝑊𝑐 =?
For steady flow work of the compressor:
𝑘𝑚𝑅(𝑡2 − 𝑡1 )
̇ =
𝑊𝑆𝐹
1−𝑘
𝑘𝑔𝑚 𝑘𝐽
(1.4) (1 ) (0.28708 ) (306.209 − 27)𝐾 1ℎ𝑝
̇ 𝑠 𝑘𝑔𝑚 − 𝐾
𝑊𝑆𝐹 = ( )
1 − 1.4 0.746𝑘𝑊
𝑾̇ 𝑪 = −𝟑𝟕𝟔. 𝟎𝟔𝟒𝒉𝒑 ⇒ 𝐴𝑛𝑠.
c) 𝑃𝑚 =?
𝑊𝑐
η𝑜 = = η𝑎 η𝑚𝑒 η𝑒𝑒
𝑃𝑚
𝑊𝑐
𝑃𝑚 =
η𝑎 η𝑚𝑒 η𝑒𝑒
𝟎. 𝟕𝟒𝟔𝒌𝑾
−376.064ℎ𝑝 ( )
𝟏𝒉𝒑
𝑃𝑚 =
(0.85)(0.80)(0.70)
𝑷𝒎 = −𝟓𝟖𝟗. 𝟑𝟕𝟖𝒌𝑾 ⇒ 𝐴𝑛𝑠.
d) 𝐶𝑜𝑠𝑡 =?
Operating Cost = (Power input to the motor)(Time)(Selling Price of Electricity)
₱6.25
𝐶𝑜𝑠𝑡 = (589.378𝑘𝑊)(24ℎ𝑟) ( )
𝑘𝑊 − ℎ𝑟
𝑪𝒐𝒔𝒕 = ₱𝟖𝟖𝟒𝟎𝟔. 𝟕 ⇒ 𝐴𝑛𝑠.
e) 𝑡2′ , the actual temperature at exhaust
From the adiabatic efficiency

𝑊𝑐
η𝑎 =
𝑊𝑖
Since
𝑘𝑚𝑅(𝑡2 ′ − 𝑡1 )
𝑊𝑖 =
1−𝑘
And
𝑘𝑚𝑅(𝑡2 − 𝑡1 )
𝑊𝑐 =
1−𝑘
Then
𝑘𝑚𝑅(𝑡2 ′ − 𝑡1 )
η𝑎 = 1−𝑘
𝑘𝑚𝑅(𝑡2 − 𝑡1 )
1−𝑘
𝑡2 − 𝑡1
η𝑎 =
𝑡2 ′ − 𝑡1
′
𝑡2 − 𝑡1
𝑡2 = 𝑡1 +
η𝑎
306.209 − 27
𝑡2′ = 27 +
0.85
𝒕′𝟐 = 𝟑𝟓𝟓. 𝟒𝟖𝟏°𝑪 ⇒ 𝐴𝑛𝑠.
f) For the equivalent polytropic index, from PVT relation

𝑛
𝑇2′ 𝑃2 𝑛−1
=( )
𝑇1 𝑃1
218
𝑇2′ 𝑛 𝑃2
𝑙𝑛 = 𝑙𝑛
𝑇1 𝑛 − 1 𝑃1
355.481 + 273 𝑛 − 1 100

𝑙𝑛 = 𝑙𝑛
27 + 273 𝑛 10
𝑛−1
= 0.3212
𝑛
𝑛 − 1 = 0.3212𝑛
0.6788𝑛 = 1
𝒏 = 𝟏. 𝟒𝟕𝟑 ⇒ 𝐴𝑛𝑠.
219
SUPPLEMENTARY PROBLEMS
CHAPTER IV: IDEAL GAS PROCESSES
Directions: Solve the following problems. Any erasures found in your paper are considered as wrong. Final answers
must be round off to three decimal places. Use Item B1 for gas constants
Isometric Process
1. There are 1.36 kg of air at 137.9 kPaa stirred with 6. Four pounds of NH3 at 200 psia and 360°𝐹 are
internal paddles in an insulated rigid container whose cooled at constant volume to 100°𝐹. Determine
volume is 0.142 m3 until the pressure becomes 689.5 a) Final pressure
kPaa. Determine the following b) Change in internal energy
a) The work input c) Work
b) Change in Flow Energy
c) Heat
7. A gas whose composition is now known has 42.2
2. A reversible nonflow constant volume process kJ of paddle work input at constant volume of 567 L.
decreases the internal energy by 316.6 kJ for every Initially the pressure and temperature are 140 kPaa
2.268 kg of a gas for which R=430 K/kg-K and and 27°𝐶 respectively. For a final temperature of 82.2
k=1.35. For an initial temperature of 204.4°𝐶 °𝐶, and the ratio of specific heat to be 1.21, calculate
determine: a) The change in internal energy
a) Work b) Heat transfer
b) The heat
c) The change of entropy Isobaric Process
8. A constant pressure system contains 80 lb of ideal
3. The internal energy of perfectly insulated rigid 283 gas. 1 hp-min is required to raise its temperature by a
L nitrogen system is increased 2930 kH by means of unit degree Fahrenheit. What is the specific heat of
internal stirrers. If the initial pressure is 1379 kPaa, the ideal gas?
determine
a) The final pressure 9.Two pounds of hydrogen gas simultaneously rejects
b) The entropy generated per kilogram. heat and receives paddle work input in a nonflow
charge of state at constant pressure from an initial
4. During a cryogenic experiment, 2 gram-mole of temperature of 250°𝐹 to a final temperature of 90°𝐹.
oxygen gas are placed in a 1 L Dewer flask at 120 K. If the heat rejected is thrice the paddle work,
After sometime, the temperature is noted to be 215 K. determine
Find a) Change in internal energy
a) The increase in pressure b) Change in enthalpy
b) The heat c) Change in entropy
d) Heat rejected
5. There are 2 kg of gas, for which R=379 J/kg-K and e) Net work
k =1.25 , undergoing a non-flow constant volume
process from 551.6 kPaa and 60°𝐶 to 1655 kPaa. 10. One pound of air has a decrease of 20.58 BTU on
During the process the gas is internally stirred, and its internal energy when its temperature is reduced to
there are also added 106 kJ of heat. Determine one third of its initial temperature during a nonflow
a) temperature after stirring constant pressure process. Determine
b) the work input a) The initial and final temperature
c) heat transfer b) Heat
d) Change in internal energy c) Work
e) Change in entropy e) Change in entropy
220
11. Oxygen gas at a rate of 3 lb/min undergoes a 16. A closed gaseous system contains 14L of methane
reversible isobaric process during which its entropy gas at 414 kPaa and 277 K. A constant paddle work is
changes by -0.35 BTU/lb-°𝑅. The initial volume and supplied while the frictionless piston moves to a point
temperature are 17.75ft3 and 400°𝐹 respectively. where the pressure reaches 138 kPaa. The
a) For non-flow case, calculate Δ𝑈, Δ𝐻,𝑊, and 𝑄 temperature remains constant throughout the process.
b) For steady-flow case, calculate Δ𝑈,Δ𝐻,𝑊. And 𝑄 a) Estimate the work done of the piston
b) What is the work done by the paddle?
12. Air at 690 kPaa in a frictionless piston cylinder
arrangement remains at constant pressure while its 17. The gain in entropy during an isothermal nonflow
temperature changes from 24°𝐶 to 94°𝐶. A heat loss process of 5 lb of air at 60°𝐹 is 0.425 BTU/°𝑅. What
of 14 kJ/kg is experienced, and the input energy is a is the ratio
form of paddle work. For a unit mass of 1kg, a) 𝑉2 /𝑉1
determine b) 𝑃2 /𝑃1
a) The change in enthalpy c) 𝑄/𝑊
b) The change in internal energy
c) The change in entropy 18. 10 kg/min of air at 96 kPaa amd 7.65m 3 is
d) The net work compressed isothermally to 620 kPaa. Air enters at
15 m/s and leaves at 60 m/s/ Determine
13.Assume 5 lb of ideal gas with R=38.7 ft-lbf/lbm- a) Work done by compression
°𝑅 and k=1.668 have 300 BTU of heat added during b) Change in entropy
a reversible constant pressure change of state. For an
initial temperature of 80°𝐹, calculate 19. 800CFM of air at 80°𝐹 and 200 psia is
a) Change in internal energy compressed isothermally to 600 psi. Air enters at
b) Change in enthalpy 75fps and leaves at 150 fps. Determine
c) Change in entropy a) Non-flow work
d) Net Work b) Change in entropy
c) Heat
Isothermal Process
Isentropic Process
14. Two kilograms of helium gas at 100atm, 165 K
expands isothermally to 1 atm. Determine the 20. The internal energy of an ideal gas is given by the
a) The non-flow work expression 𝑢 = 850 + 0.529𝑃𝑣 in BTU/lb where 𝑃
b) The steady flow work is in psia. Determine the isentropic index.
c) Change in internal energy
d) Heat 21. During an isentropic process of 1.36 kg/s of air,
e) Change in entropy the temperature increases from 4.44°𝐶 to 115.6°𝐶.
Determine
15. There are abstracted 317 kW from 1.135 kg/s of a) Non-flow work
gas while the temperature remains constant at 27°𝐶. b) Change in internal energy
For this gas, 𝑐𝑝 = 2.232 and 𝑐𝑣 = 1.713 in kJ/-kg-K. c) Change in enthalpy
For an initial pressure of 590 kPaa, determine d) Change in entropy
a) Initial and Final Volume e) Heat
b) Final pressure
c) Work done
d) Heat
e) Change in entropy
f) Change in enthalpy
221
22. Hydrogen gas is compressed isentropically from 28. Assume 5 lb/s of NH3 gas are compressed
108 psia, 3 CFS, and 40°𝐹 to 256 psia. For a non-flow adiabatically from 100 psia, 170°𝐹 up to 220 psia in
work process, determine a steady flow process. Due to irreversibility, the
a) Temperature after compression entropy increases by 0.195 BTU/s-°𝑅. Determine
b) Volume after compression a) The actual temperature
c) Steady flow work b) The indicated work
d) Change in enthalpy
e) Change in internal energy 29. Nitrogen at 90 psia and 110°𝐹 undergoes an
f) Change in entropy irreversible adiabatic expansion to 15 psia. The ratio
g) Heat of actual work to ideal work is 80%. For a rate being
2 lb/s determine:
23. One cubic foot per minute of air at 160 psia and a) The actual temperature
260°𝐹 expands isentropically to 20 psia. What is b) The actual change in internal energy
(a) the non flow work c) The actual change in enthalpy
(b) the steady flow work d) The actual non-flow work.
24. Oxygen passes through an adiabatic frictionless 30. Assume 5ft3/s of ideal gas at 15 psia and 80°𝐹 is
horizontal nozzle in a steady flow manner. The compressed during an irreversible adiabatic process
isentropic decrease in temperature from entrance to to 90 psia and 522°𝐹. Determine the isentropic index.
exit is −40°𝐶.
a) For an initial velocity of 30 m/s, determine its exit
velocity. Polytropic Process
b) If the entrance state is 345 kPaa, 333K, determine
the exit pressure. 31. The work required to compress air according to
the law 𝑃𝑉1.30 = 𝐶 is 67.79 kJ if there is no flow.
25. Ten pounds per second of air that are compressed Determine the change in internal energy.
from 15 psia, 80°𝐹 to 75 psia in an isentropic manner.
Find 32. Two pounds of air at 90°𝐹 expands with a
a) The final temperature reversible polytropic process with 𝑛 = 1.25 until the
b) Non-flow work pressure is halved. Air enters at 10 fps and leaves at
c) Steady flow work 400 fps. There is no change in elevation for both
suction and discharge. Compute
26. 1 lb/s of an ideal gas undergoes an isentropic a) The change in internal energy
process through an engine at 95.3 psig and volume of b) The change in enthalpy
0.6 ft3/lbm to a final volume of 3.6 ft3/lbm. If 𝑐𝑝 = c) The change in entropy
0.124 and 𝑐𝑣 = 0.093 in BTU/lb-°𝐹, what is d) Non-flow work
a) the final temperature e) Steady-flow work
b) non-flow work f) Heat
c) steady flow work
d) If the actual steady flow work for an adiabatic 33. Compressed 4 kg/s of 𝐶𝑂2 gas polytropically
expansion from its initial state to the same final having an index of 2, from 104 kPaa and 60°𝐶 to
pressure is 18 BTU/sec, and has a change of kinetic 227°𝐶. Assuming ideal gas action, find
energy to be -2 BTU/lb, what is the change in a) Final pressure
entropy? b) Steady Flow work
c) Non-flow work
27. During an isentropic process, the temperature d) Heat
increases from 40°𝐹 up to 340°𝐹. The actual e) Change in entropy
temperature during the adiabatic process is 400°𝐹.
For 4 lb/s of air, determine the change in entropy. 34. The work required to compressed a gas reversibly
according to 𝑃𝑉1.30 = 𝐶 is 50 000 ft-lb if there is no
flow. If the gas is methane, determine the heat
required.
222
e) Determine the amount of water in bbl that is
35. An ideal gas at 2ft3 and 14.4 psia is compressed cooled if temperature rise is 10°𝐶 for one-year
polytropically to 14.4 psia and 100 psia. 9.54 BTU of continuous operation.
non-flow work is done in the process. What is the f) Recommend dimensions for conical frustrum
polytropic index? whose larger base is twice its smaller base, and its
height is 30% larger of the average of the bases.
Applications of Ideal Gas Processes
Additional Problems
36. 0.5 kg/s of atmospheric air at 64°F is compressed Note: Process of the Ideal Gas are not stated for
to a final pressure of 100 psia. What is the motor students to analyze thermodynamic systems.
rating driving the compressor having an efficiency of
70%, if the compression process is 41. Three kilograms of gas is heated from 200°𝐶 to
a) Isothermal 500°𝐶 at constant pressure in a heat exchanger. What
b) Polytropic process with n=1.3 is the entropy change if the gas is ammonia?
c) Reversible adiabatic process
d) Irreversible adiabatic process with a 42. What is the pressure ratio (𝑃2 /𝑃1 ) that allows a
compressor efficiency of 80%. temperature ratio (𝑇2 /𝑇1 ) of 1.5 when carbon dioxide
is subjected to isentropic compression?
37. There are required 2000 kW of compressor power
to handle air adiabatically from 1 atm, 27°𝐶 to 305 43.Air is compressed in a reversible adiabatic manner
kPaa. The initial air velocity is 21 m/s and final air from 25°𝐶 and 105 kPaa to 500 kPaa. What is the
velocity is 85 m/s. work done by compression?
a) If the process is isentropic, find the volume of air
in m3/min handled, measured from the inlet. 44. When 5 kg/s of air undergoes constant enthalpy,
b) If the compression is an irreversible adiabatic to a the entropy increases by 0.14 kW/K. What is the
temperature of 157°𝐶, with the capacity found in (a), pressure ratio?
find the kilowatt input.
45. A system at 200°𝐶 undergoes constant volume
38. The adiabatic power input required to compress process where the internal energy increases by 316.5
50 lb/min of air from 14 psia and 80°𝐹 to a higher kJ/kg, for 2 kg/min of gas which R = 0.43 kJ/kg-K
pressure is 64 hp. Find the discharge pressure. and k = 1.35. Determine for 𝛥𝐾𝐸 = 4 kJ/kg
a) The heat
39. Helium gas expands polytropically through a b) the change in entropy
turbine accordint to the process 𝑃𝑉1.3 = 𝐶. The inlet c) the nonflow and steady flow work
temperature and pressure are measured to be 1200 K
and 800 kPa respectively. The helium exits the turbine 46. There are 1.85 kg of gas for which R = 425 J/kg-
at a pressure of 160 kPa. If the turbine produces 30 K and k =1.35 undergo a non-constant volume
MW, determine the heat transfer in kW. process from 580 kPaa and 50°𝐶 to 1750 kPaa.
During the process the gas is internally stirred and
40. 0.5 kgm/s of air at 400 Pa vac and 250K is there are added 110 kJ of heat. Determine
compressed polytropically according to the law a) Final temperature
PV1.32=C. It passes through an aftercooler at constant b) The work input
pressure and rejects 50 kW of heat so that the water is c) Change in internal energy
cooled when it passes the cooling tower and then enter d) Change in entropy
into a suitable storage tank.
a) Find the final temperature of the air in the 47. 2 kg of air has a decrease of internal energy of 25
compressor. kJ while its centigrade temperature is reduced to one
b) Determine the power input needed to compress third of its initial temperature during a reversible
the air in hp. nonflow isobaric process. Determine
c) Determine the temperature of the air in the a) Initial and final temperature
aftercooler. b) Heat
d) Determine the change in internal energy of the c) Work
air in the aftercooler. d) Change in entropy
223
48. 1.85 kg/min of nitrogen undergoes reversible 54. 3 kg of hydrogen gas undergoes a non-flow
constant pressure process during which its entropy process until its volume is doubled and its pressure is
changes -1.54 kJ/kg-K, 𝑡1 = 215°𝐶. For steady flow halved. What is the change in entropy?
process, determine
a) Change in internal energy 55. 0.26 moles of ammonia gas undergo a state
b) Change in enthalpy change from 500 kPaa and 0.24m3 to 9.5 MPaa and
c) Work 0.012m3. Determine
d) Heat a) The change in internal energy
b) The change in enthalpy
49. (ME BD APR 2007) Find the isothermal work in c) The change in entropy
kJ done by a piston which is acted upon by a working
substance which expands from 0.04 m3 and 400 kPaa 56. 5 kg of helium initially at 30°𝐶 occupies 170 L. A
to 0.10m3. state change to 75°𝐶 causes an entropy increase of 5.2
kJ/K. Determine
50. During reversible process, 1.2 kg/s of a gas rejects a) The final volume
320 kW of heat at constant temperature of 27°𝐶. For b) The final pressure
this gas, 𝑐𝑝 =2.14 and 𝑐𝑉 = 1.62 kJ/kg-K. The initial
pressure is 595 kPaa. For Δ𝑃𝐸 = 0 and Δ𝐾𝐸 = 57. Argon is heated at constant volume from 350 kPaa
4 𝑘𝑊, determine and 38°𝐶 to 285°𝐶. What is the change in specific
a) Initial volume entropy?
b) Final volume
c) Final pressure 58. (ME BD OCT 2005) A 10m3 rigid closed cylinder
d) Change in entropy contains a gas with a temperature of 100°𝐶 and a
e) Change in enthalpy pressure of 350 kPa. 50 kJ of heat is added to the
f) Work non-flow system. Determine the change of internal energy of
g) Work steady flow the system
51. Helium is compressed isentropically from 745 59. An insulated rigid container holds helium at 150
kPaa and 5.1 CMM and 5°𝐶 to 1765 kPaa. Determine kPaa and 30°𝐶. If 250 kJ/kg of paddle work are
a) Initial volume flow rate introduced to the system, determine
b) Final temperature a) the final temperature
c) Steady flow work b) the final pressure
d) Non-flow work c) the change in specific entropy
52. CO2 gas flows through a horizontal duct according 60. 40 kg of air at 345 kPa is contained in a rigid
to PV1.4=C. At suction, 𝑃1 =250 kPaa, 𝑡1 = 150°𝐶, vessel at a temperature of 20°𝐶. Heat is added until
and 𝑣1 = 47 m/s. At discharge, 𝑃2 = 35 kPaa and the temperature reaches 320°𝐶. Determine the heat
𝑡2 = 80°𝐶. Determine
a) The heat transferred 61. An ideal gas with a molecular mass of 36 is heated
b) The change in entropy at constant volume from 250°𝐶 to 485°𝐶. The change
c) The velocity at discharge of internal energy is 180 kJ/kg. What is the change is
specific enthalpy?
53. 2.5 kg of air occupies 300 liters at 1700 kPaa
expands according to PV1.25=C until the pressure 62.8 kg of air at 28°𝐶 is contained in an insulated
becomes 70 kPaa. Determine container. Determine the change in entropy if a paddle
a) The final volume wheel fitted in the container transfers224 kJ of work
b) The change in entropy as it stirs up the air.
c) The non-flow work
d) Heat transferred 63. 0.033 kN-m of torque is made by paddle work for
120 revolutions. If 11.5 kJ of heat are rejected on a
40L air tank at 25°𝐶 and 2 bar, calculate the change
of entropy in the air.
224
64. Air is cooled at constant atmospheric pressure

with an initial specific volume of 1.34 m3/kg to a final
density of 1.205 kg/m3. For 2.5 kg of air, determine
the change in enthalpy.
65. (ME BD APR 2007) A device arrangement

maintaining a pressure of 400 kPa is heated from
27°𝐶 to 227°𝐶. Initially, the volume is 1L. What is
the resulting volume?
66. 5 kg of methane gas enters a constant pressure

heat exchanger. The gas is cooled from 480 to 180 °𝐶.
What is the change in entropy?
67. An unknown ideal gas mixture is cooled at

isopiestic process from 300°𝐶 to 80°𝐶. Assuming R
= 0.524 kJ/kg-K and k=1.39, determine the
a) The change of entropy for 2.3 moles of this gas
b) The final to initial volume ratio
225
References
Burghardt, D. M. (1986). Engineering Thermodynamics with Applications 3rd Edition. New York City: Harper &
Row, Publishers Inc.
Cengel, Y. A., Boles, M. A., & Kanoglu, M. (2019). Thermodynamics - An Engineering Approach Ninth Edition.
New York City: McGraw-Hill Education.
Faires, V. M., Simmang, C. M., & Brewer, A. V. (1978). Problems on Thermodynamics 6th Edition. New York City:
Macmillan Publishing Co., Inc.
226

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Thermodynamics 1 Initial Draft 1

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INITIAL DRAFT – MANUSCRIPT

CHAPTER I. PROPERTIES OF MATTER

1.1 Scope and Definition of Thermodynamics ……………………………………………….. 1

CHAPTER II. ENERGY AND POWER CONCEPTS

2.1 Conservation of Mass …………………………………………………………………….. 88

3.1 Equation of State ………………………………………………………………………… 147

CHAPTER IV. IDEAL GAS PROCESSES

4.1 General Equations For Ideal Gas Processes ………………………………………………,, 186

1.2 LAWS OF THERMODYNAMICS

1.3 ANALYSIS OF MOTION

Variable Definition Formula

From the instantaneous acceleration equation:

Sign conventions for the acceleration due to gravity:

Velocity Vectors 𝒗𝒐𝒙 = 𝒗𝟎 𝒄𝒐𝒔𝛉

The time of flight can be determine using the horizontal motion:

Using the horizontal motion for projectile, wherein R=X

For the horizontal motion:

The three kinematic differential equations are as follows

To convert angles, the following is the system of units in converting angles.

1.4 ANALYSIS OF FORCES

From Newtons Second Law of Motion:

Interpretation of the value of 𝑘 where k is unity:

Weight is defined as the gravitational force exerted to the body

Interpretation of the value of 𝑘 where k is not unity:

1N = 1 x 105 dynes 1 hyl = 1 metric slug

Doing summation of forces in the y- direction:

c. Friction Forces. There are three types of friction.

For belt friction in pulleys, the equation is represented as

By ratio and proportion:

a) In terms of gram mass:

The acceleration at 𝑡 = 2 𝑠 would be

b) The velocity of the particle is zero when it is at rest. Thus at 𝑣 = 0

For the total weight of the two bodies

For the force of attraction between two bodies

a) By Ratio and Proportion:

Since 𝑊0 = (1 − 0.05)𝑊𝑠 = 0.95𝑊𝑠

1.5 PLANE MENSURATION

Geometrical Figure Formula

IV. Using a Planometer – a planometer is a

1L = 1000 cm3 1 ft3 = 7.481 gal

𝑽𝒔𝒐𝒍𝒊𝒅 = 𝑽𝒔𝒐𝒍𝒊𝒅+𝒘𝒂𝒕𝒆𝒓 − 𝑽𝒘𝒂𝒕𝒆𝒓

Where V is the volume of the solid

Cube Rectangular Parallelepiped

Volume For Right Cylinder:

For any given base: For right circular cone:

Spherical Wedge Spherical Polygon Spherical Pyramid

Volume of the Wedge: Surface Area of the Spherical

𝐴1 = area of the top base

𝐴𝑚 = area of the middle part

(Total Surface Area)(Number of Units to be Fabricated)(1 + % material wasted)

1.7 BASIC PROPERTIES OF FLUID

𝑺𝑮𝒐𝒊𝒍 < 𝑺𝑮𝒘𝒂𝒕𝒆𝒓 < 𝑺𝑮𝒎𝒆𝒓𝒄𝒖𝒓𝒚

Standard Values for Specific Gravity of Common Fluids

Hence the mass flow rate of the fluid would be

c) For specific weight/weight density:

For the total mass of the mixture.

1500𝑉1 + 500𝑉2 = 80 (Equation 1)

For the total volume of the mixture