THERMODYNAMICS

Prof. Dr. Ali Kodal

LECTURE NOTES 2023

IGU Aeronautical Engineering Department

UCK208E THERMODYNAMICS
CONCEPTS AND DEFINITIONS
Thermodynamics can be defined as the science of energy and entropy.

Greek therme → heat dynamis → power

Thermodynamics studies power, heat and the properties of matter interactions and
relationships. It depends on observations and has zeroth, 1st, 2nd and 3rd laws.

The most fundamental law : the conservation of energy principle.

It simply states that during an interaction, energy can change from one form to another but the
total amount of energy remains constant.

The first law of thermodynamics is simply an expression of the conservation of energy

principle.

Ein Eout

Ein - Eout = E

The second law of thermodynamics asserts that energy has quality as well as quantity, and
actual processes occur in the direction of decreasing quality of energy. (There has to be losses)

Thermodynamic system: A system is defined as a quantity of matter or a region in space

chosen for study. The mass or region outside the system is called the surroundings.

The real or imaginary surface that separates the system from its surroundings is called the
boundary. The boundary of a system can be fixed or movable.

The system boundary can move with heat or work.

Isolated system: The system does not affected from the surroundings. There will be no mass,
heat or work crossing through the system’s boundaries.

Isolated system:

A closed system:  (also known as a control mass) consists of a fixed amount of mass, and
no mass can cross its boundary.

An open system:  or a control volume, as it is often called, is a properly selected region in

space. It usually encloses a device that involves mass flow such as a compressor, turbine, or
nozzle. Both mass and energy can cross the boundary of a control volume. The boundaries of
a control volume are called a control surface, and they can be real or imaginary.

PROPERTIES OF A SYSTEM
Any characteristic of a system is called a property. Some familiar properties are pressure P,
temperature T, volume V, and mass m.

Some properties are defined in terms of others.

𝑚
Density: 𝜌= (kg/m3)
𝑉

𝜌
Specific gravity: 𝜌𝑠 = 𝜌
𝐻2𝑂
 𝑉 1
Specific volume: 𝑣=𝑚=𝜌 (m3/kg)

Properties are considered to be either intensive or extensive.

Intensive properties are those that are independent of the mass of a system, such as
Temperature, pressure, and density.

Extensive properties are those whose values depend on the size—or extent—of the system.
Total mass, total volume, and total momentum are some examples.

Specific properties: Extensive properties per unit mass are called specific properties.
𝑉 𝐸
e.g. specific volume ( 𝑣 = 𝑚 ), specific energy (𝑒 = 𝑚).

Matter is made up of atoms that are widely spaced in the gas phase. Yet it is very convenient
to disregard the atomic nature of a substance and view it as a continuous, homogeneous matter
with no holes, that is, a continuum.

The continuum idealization allows us to treat properties as point functions and to assume the
properties vary continually in space with no jump discontinuities. This idealization is valid as
long as the size of the system we deal with is large relative to the space between the molecules.

Consider a container filled with oxygen at atmospheric conditions, at 1 atm pressure and 20°C.
The diameter of the oxygen molecule is about 3x10-10 m and its mass is 5.3x10-26 kg. The mean
free path of oxygen is 6.3x10-8 m. There are about 3x1016 molecules of oxygen in the tiny
volume of 1 mm3.

STATE AND EQUILIBRIUM

If we consider a given mass of water, it can exist in various forms. If it is a liquid initially, it
may become a vapor when it is heated or a solid when it is cooled. Thus, we speak of the
different phases of a substance.

A phase is defined as a quantity of matter that is homogeneous throughout.

When more than one phase is present, the phases are separated from each other by the phase
boundaries.

The state may be identified or described by certain observable, macroscopic properties; some
familiar ones are temperature, pressure, and density.
Each of the properties of a substance in a given state has only one definite value, and these
properties always have the same value for a given state, regardless of how the substance arrived
at the state.
Frequently we will refer not only to the properties of a substance but also to the properties of a
system. When we do so, we necessarily imply that the value of the property has significance
for the entire system, and this implies equilibrium.

Thermodynamics deals with equilibrium states. The word equilibrium implies a state of
balance.

In an equilibrium state there are no unbalanced potentials (or driving forces) within the system.
A system in equilibrium experiences no changes when it is isolated from its surroundings.

There are many types of equilibrium, and a system is not in thermodynamic equilibrium unless
the conditions of all the relevant types of equilibrium are satisfied.

Thermal equilibrium: temperature is the same throughout the entire system.

Mechanical equilibrium: no change in pressure at any point of the system with time.

Also Phase equilibrium, chemical equilibrium etc.

The State Postulate

A system is called a simple compressible system in the absence of electrical, magnetic,
gravitational, motion, and surface tension effects.

The state of a simple compressible system is completely specified by two

independent, intensive properties.

Otherwise, an additional property needs to be specified for each effect

If we know two properties, the others can be determined. For example:

Pv = RT  T = f(p,v)
PROCESSES AND CYCLES
Any change that a system undergoes from one equilibrium state to another is called a process,
and the series of states through which a system passes during a process is called the path of
the process

Gas

When we take away the some very small amount of the weight on the piston, the piston will go
up, pressure drops and the specific volume increases.

When a process proceeds in such a manner that the system remains infinitesimally close to an
equilibrium state at all times, it is called a quasistatic, or quasi-equilibrium, process.

If we remove all the weight very suddenly, then the state changes will be non-equilibrium.

Several processes are described by the fact that one property remains constant. The prefix iso-
is used to describe such a process.

isothermal  constant-temperature process

isobaric  constant-pressure process
isochoric  constant-volume process

A system is said to have undergone a cycle, if it returns to its initial state at the end of the
process.

A A

2
2

Process Cycle
1
1 path

B
B
The term steady implies no change with time.
The opposite of steady is unsteady, or transient.
The term uniform, however, implies no change with location over a specified region.

The steady-flow process, which can be defined as a process during which a fluid flows through
a control volume steadily

TEMPERATURE AND THE ZEROTH LAW OF THERMODYNAMICS

When a body is brought into contact with another body that is at a different temperature, heat
is transferred from the body at higher temperature to the one at lower temperature until both
bodies attain the same temperature

The zeroth law of thermodynamics states that if two bodies are in thermal equilibrium with
a third body, they are also in thermal equilibrium with each other.

A B

General temperature scale: Celsius oC, Fahrenheit oF

Absolute temperature scale: Kelvin K, Rankine R

𝑇(𝐾) = 𝑇(℃) + 273.15

𝑇(𝑅) = 𝑇(℉) + 459.67

𝑇(𝑅) = 1.8 𝑇(𝐾)

𝑇(℉) = 1.8 𝑇(℃) + 32

PRESSURE
Pressure is defined as a normal force exerted by a fluid per unit area.

𝐹
𝑃= (𝑘𝑃𝑎)
𝐴
1 Pa = 1 N/m2

1 kPa = 103 Pa

1 MPa = 103 kPa = 106 Pa

1 bar = 105 Pa = 0.1 MPa = 100 kPa

1 atm = 101325 Pa = 101.325 kPa = 1.01325 bar

Example:
1 atm

2 atm

mP = ? 10 cm
Force balance:

𝑃𝐴 = 𝑃𝑎𝑡𝑚 𝐴 + 𝑚𝑝 𝑔
𝑚
𝑚𝑝 (9.81
)
2(101.325 𝑘𝑃𝑎) = (101.325 𝑘𝑃𝑎) + 𝑠2
𝐴 ∙ 1000
101.325 ∙ 103 (𝜋 ∙ 0.052 )
𝑚𝑝 = = 81.1217 𝑘𝑔
9.81

FORMS OF ENERGY
E = total energy (kJ)
𝐸
𝑒 = 𝑚 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 (kJ/kg)
The macroscopic forms of energy are those a system possesses as a whole with respect to some
outside reference frame, such as kinetic and potential energies.

The microscopic forms of energy are those related to the molecular structure of a system and
the degree of the molecular activity, and they are independent of outside reference frames.

The sum of all the microscopic forms of energy is called the internal energy of a system and
is denoted by U.

Total energy → Macroscobic (kinetic, potential)

→ Microscobic (system molecular structures and activities)

The sum of all the microscopic forms of energy = internal energy = U

Kinetic energy:

𝑚𝑉 2
𝐾𝐸 = (𝑘𝐽)
2
On a unit mass basis,

𝐾𝐸 𝑉 2
𝑘𝑒 = = (𝑘𝐽/𝑘𝑔)
𝑚 2
Potential energy:

𝑃𝐸 = 𝑚𝑔𝑧 (𝑘𝐽)

On a unit mass basis,

𝑝𝑒 = 𝑔𝑧 (𝑘𝐽/𝑘𝑔)

𝑚𝑉 2
𝐸 = 𝑈 + 𝐾𝐸 + 𝑃𝐸 = 𝑈 + + 𝑚𝑔𝑧
2
𝑉2
𝑒 = 𝑢 + 𝑘𝑒 + 𝑝𝑒 = 𝑢 + + 𝑔𝑧
2

Most closed systems remain stationary during a process and thus experience no change in their
kinetic and potential energies.
Closed systems whose velocity and elevation of the center of gravity remain constant during a
process are frequently referred to as stationary systems.
The change in the total energy E of a stationary system is identical to the change in its internal
energy U.

The mass and the volume flow rates

The mass flow rate 𝑚̇, which is the amount of mass flowing through a cross section per unit
time.

It is related to the volume flow rate 𝑉̇ , which is the volume of a fluid flowing through a
cross section per unit time,

Mass flow rate:

𝑚̇ = 𝜌𝑉𝑎𝑣𝑔 𝐴𝑐 (𝑘𝑔/𝑠)

 : Fluid density (kg/m3), Vavg: the average flow velocity normal to Ac (m/s)

A: the cross sectional area of the flow (m2)

Volume flow rate: 𝑉̇ = ∫ 𝑉𝑁 𝑑𝐴 = 𝑉𝑎𝑣𝑔 𝐴

𝑉̇
𝑚̇ = 𝜌𝑉̇ =
𝑣
The dot over a symbol is used to indicate time rate.

Energy flow rate:

𝑘𝐽
𝐸̇ = 𝑚̇𝑒 ( 𝑜𝑟 𝑘𝑊)
𝑠

PURE SUBSTANCE
A substance that has a fixed chemical composition throughout is called a pure substance.

Water, nitrogen, helium, and carbon dioxide, for example, are all pure substances.
A mixture of various chemical elements or compounds also qualifies as a pure substance as
long as the mixture is homogeneous. Air, for example.

A mixture of two or more phases of a pure substance is still a pure substance as long as the
chemical composition of all phases is the same. A mixture of ice and liquid water, for example.

PHASES OF A PURE SUBSTANCE

Consider a system, 1 kg of water contained in the piston/cylinder arrangement shown in Fig.
a.

Suppose that the piston and weight maintain a pressure of 1 atm in the cylinder and the initial
temperature is 20◦C.

As heat is transferred to the water, the temperature increases appreciably, the specific volume
increases slightly, and the pressure remains constant.

When the temperature reaches 100◦C, additional heat transfer results in a change of phase, as
indicated in Fig. b. That is, some of the liquid becomes vapor, and during this process both the
temperature and pressure remain constant, but the specific volume increases considerably.

When the last drop of liquid has vaporized, further transfer of heat results in an increase in both
the temperature and specific volume of the vapor, as shown in Fig. c.

The term saturation temperature designates the temperature at which vaporization takes place
at a given pressure. This pressure is called the saturation pressure for the given temperature.
Tsat = f (Psat)

A liquid that is about to vaporize is called a saturated liquid.

If the temperature of the liquid is lower than the saturation temperature for the existing pressure,
it is called either a subcooled liquid (implying that the temperature is lower than the saturation
temperature for the given pressure) or a compressed liquid (implying that the pressure is greater
than the saturation pressure for the given temperature).

When a substance exists as part liquid and part vapor at the saturation temperature, its quality
is defined as the ratio of the mass of vapor to the total mass.

A vapor that is about to condense is called a saturated vapor.

When the vapor is at a temperature greater than the saturation temperature, it is said to exist
as superheated vapor.
A substance at states between 2 and 4 is referred to as a saturated liquid–vapor mixture since
the liquid and vapor phases coexist in equilibrium at these states.

The heat given between 2 and 4 is the latent heat of vaporization.

At a pressure of 22.09 MPa, there is no constant-temperature vaporization process. This point

is called the critical point. At the critical point the saturated-liquid and saturated-vapor states
are identical. The temperature, pressure, and specific volume at the critical point are called the
critical temperature, critical pressure, and critical volume.

For water: Tcr = 374.14oC, Pcr = 22.09 MPa, vcr = 0.003155 m3/kg
THE IDEAL-GAS EQUATION OF STATE

It has been observed experimentally that, to a close approximation, a very-low-density gas

behaves according to the ideal-gas equation of state:

𝑃𝑣̅ = 𝑅̅ 𝑇

𝑘𝑁𝑚
𝑅̅ = 𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙 𝑔𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 8.3144
𝑘𝑚𝑜𝑙 𝐾
𝑃𝑣̅ 𝑅̅ 𝑇
= → 𝑃𝑣 = 𝑅𝑇
𝑀 𝑀

𝑅̅
M = molar mass 𝑅 = 𝑀 = 𝑔𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑃𝑉 = 𝑛𝑅̅ 𝑇 𝑃𝑉 = 𝑚𝑅𝑇

V = total volume, n = mole number, m = mass

The deviation from ideal-gas behavior at a given temperature and pressure can accurately be
accounted for by the introduction of a correction factor called the compressibility factor Z
defined as:

𝑃𝑣
𝑍=
𝑅𝑇
Gases behave differently at a given temperature and pressure, but they behave very much the
same at temperatures and pressures normalized with respect to their critical temperatures and
pressures.

Reduced pressure = Pr = P/Pc , Pc = critical pressure

Reduced temperature = Tr = T/Tc , Tc = critical temperature

Example

Air is at 25oC ve 101.325 kPa conditions. If the gas constant is R = 287 J/kgK, the find the
specific volume and the molar mass?

Solution:

Assume ideal gas: 𝑃𝑣 = 𝑅𝑇

𝑘𝑗
(101.325 𝑘𝑃𝑎)𝑣 = (0.287 )(25℃ + 273.15)
𝑘𝑔𝐾

→ v = 0.8445 m3/kg

𝑅̅ 8314.4 𝑘𝑔
𝑀= = = 28.97
𝑅 287 𝑘𝑚𝑜𝑙

PROPERTY TABLES
For most substances, the relationships among thermodynamic properties are too complex to be
expressed by simple equations. Therefore, properties are frequently presented in the form of
tables.

Some thermodynamic properties can be measured easily, the other properties are calculated by
using the relations between them and measurable properties. The results of these measurements
and calculations are presented in tables in a convenient format.

Enthalpy
𝐻 = 𝑈 + 𝑃𝑉 (𝑘𝐽)

On a unit mass basis, specific enthalpy:

ℎ = 𝑢 + 𝑃𝑣 (𝑘𝐽/𝑘𝑔)

Greek enthalpien  heat

Mollier referred as u + Pv = heat content, or, total heat.

Pv = flow work

u = internal energy

SATURATED LIQUID-VAPOR TABLE:

Total volume: V = Vliq + Vvap  mv = mliq vf + mvap vg

vf = specific volume of saturated liquid

vg = specific volume of saturated vapor

𝑚𝑙𝑖𝑞 𝑚𝑣𝑎𝑝
𝑣= 𝑣𝑓 + 𝑣
𝑚 𝑚 𝑔
𝑚𝑣𝑎𝑝
𝑥= = 𝑞𝑢𝑎𝑙𝑖𝑡𝑦
𝑚
Total mass: m = mliq + mvap

𝑣 = (1 − 𝑥)𝑣𝑓 + 𝑥𝑣𝑔

By using 𝑣𝑓𝑔 = 𝑣𝑔 − 𝑣𝑓 

𝑣 = 𝑣𝑓 + 𝑥𝑣𝑓𝑔
In the absence of compressed liquid data, a general approximation is to treat compressed liquid
as saturated liquid at the given temperature

Similarly:

𝑢 = 𝑢𝑓 + 𝑥𝑢𝑓𝑔

𝑢𝑓𝑔 = 𝑢𝑔 − 𝑢𝑓

ℎ = ℎ𝑓 + 𝑥ℎ𝑓𝑔

ℎ𝑓𝑔 = ℎ𝑔 − ℎ𝑓 : Enthalpy of vaporization (or latent heat of vaporization)

Example

1 kg of liquid-vapor mixture of water is initially at the pressure of 500 kPa and the quality of
0.2 conditions. The mixture is then heated up at constant pressure. If the final state is
superheated vapor at 300oC, then find the followings:

a) vf , vg at 500 kPa?
b) v1?
c) v2?

Solution:
T

P=500 kPa
2
300oC

1
151.86oC

v1 vg v2 v
vf
a) from the saturated liquid-vapor table, for 500 kPa pressure:
vf = 0.001093 m3/kg and vg = 0.3749 m3/kg

b) at the saturated liquid-vapor region

v1 = vf + xvfg = 0.001093 + 0.2(0.3749-0.001093) = 0.0759 m3/kg

c) from superheated vapor table:

 500 kpa
 v2 = 0.5226 m3/kg
300oC

Example

Complete the following table for H2O?

Solution:

T, C P, kPa h, kJ / kg x Phase description

130 275 2069.26 0.7 Saturated liquid-vapor

135 313.22 2159.1 0.737 Saturated liquid-vapor
162 650 684.08 0.0 Saturated liquid
80 600 334.91 ----- Compressed liquid
300 1800 3030 ------ Superheated vapor

Example

A rigid tank has a volume of 3 m3 and contains saturated vapor of water at the pressure of 350
kPa. If the pressure drops to 250 kPa by means of heat transfer, determine the followings:

a) The quality at the final state?

b) Vapor and liquid masses at the final state?
c) The heat transfer from the tank?
350 kPa
Solution T

P(kPa) T (oC) vf (m3/kg) vg (m3/kg)

250 kPa
250 127.44 0.001067 0.7187
350 138.88 0.001079 0.5243
2

𝑉 3 v
a) 𝑚 = = = 5.722 𝑘𝑔
𝑣𝑔 0.5243

v2 = v1 = vf + xvfg  0.5243 = 0.001067 + x2 (0.7187 – 0.001067)

 x2 = 0.729
 b) mv2 = 0.729 x 5.722 = 4.171 kg

mf2 = m- mv2 = 5.722 – 4.171 = 1.551 kg

c) Q1-2 = U2 – U1 = m (u2-u1) = -3171.5 kJ

Example

A B

Consider two tanks, A and B, connected by a valve, as shown in Fig. Each has a volume of 200
L, and tank A has R-12 at 25◦C, 10% liquid and 90% vapor by volume, while tank B is
evacuated. The valve is now opened, and saturated vapor flows from A to B until the pressure
in B has reached that in A, at which point the valve is closed. This process occurs slowly such
that all temperatures stay at 25◦C throughout the process. How much has the quality changed
in tank A during the process?

Solution

Vliquid = 0.1 x 200x10-3 = 0.02 m3

Vvapor = 0.9 x200x10-3 = 0.18 m3

25oC → table

P1 = 651 kPa

vf = 0.000763 m3/kg

vg = 0.026854 m3/kg

vfg = 0.026091 m3/kg

𝑉𝑙𝑖𝑞𝑢𝑖𝑑 0.02
𝑚1𝑙𝑖𝑞𝑢𝑖𝑑 = = = 26.2123 𝑘𝑔
𝑣𝑓 0.000763
𝑉𝑣𝑎𝑝𝑜𝑟 0.18
𝑚1𝑣𝑎𝑝𝑜𝑟 = = 0.026854 = 6.7029 𝑘𝑔
𝑣𝑔

m = mliquid + mvapor = 32.9152 kg

 𝑚1𝑣𝑎𝑝𝑜𝑟 6.7029
𝑥1 = = = 0.2036
𝑚 32.9152
Final state:

0.2 0.2
𝑚𝐵 = = = 7.44768 𝑘𝑔
𝑣𝑔 0.026854

mA2 = m- mB = 32.9152 – 7.44768 = 25.46752 kg

0.2 0.2
𝑣2𝐴 = = = 0.007853 𝑚3 /𝑘𝑔
𝑚𝐴2 25.46752

𝑣2𝐴 − 𝑣𝑓
𝑥2 = = 0.2717
𝑣𝑓𝑔

x2-x1 = 0.0681

Superheated vapor table

Compared to saturated vapor, superheated vapor is characterized by

Lower pressures ( P < Psat at a given T )
Higher tempreatures ( T > Tsat at a given P )
Higher specific volumes ( v > vg at a given P or T )
Higher internal energies ( u > ug at a given P or T )
Higher enthalpies ( h > hg at a given P or T )
Compressed liquid table

The format of table is very much like the format of the superheated vapor tables.

In the absence of compressed liquid data, a general approximation is to treat compressed liquid
as saturated liquid at the given temperature

In general, a compressed liquid is characterized by:

Higher pressures ( P > Psat at a given T)

Lower tempreatures ( T < Tsat at a given P )
Lower specific volumes ( v < vf at a given P or T )
Lower internal energies ( u < uf at a given P or T )
Lower enthalpies ( h < hf at a given P or T )

SPECIFIC HEATS
The specific heat is defined as the energy required to raise the temperature of a unit mass of
a substance by one degree

Specific heat at constant volume : CV

Specific heat at constant pressure : CP

𝜕𝑢
𝐶𝑉 = ( )
𝜕𝑇 𝑉

𝜕ℎ
𝐶𝑃 = ( )
𝜕𝑇 𝑃

Ideal gas equation: Pv = RT

For ideal gas: u = u(T), h = h(T)

𝑑𝑢 = 𝐶𝑉 𝑑𝑇

𝑑ℎ = 𝐶𝑃 𝑑𝑇
2
∆𝑢 = 𝑢2 − 𝑢1 = ∫ 𝐶𝑉 (𝑇) 𝑑𝑇 (𝑘𝐽/𝑘𝑔)
1
 2
∆ℎ = ℎ2 − ℎ1 = ∫ 𝐶𝑃 (𝑇) 𝑑𝑇 (𝑘𝐽/𝑘𝑔)
1

Assume constant CV, CP :

∆𝑢 = 𝑢2 − 𝑢1 = 𝐶𝑉 (𝑇2 − 𝑇1 )

∆ℎ = ℎ2 − ℎ1 = 𝐶𝑃 (𝑇2 − 𝑇1 )

Usually average CV, CP values are used.

h = u + Pv  ideal gas  h = u + RT,  dh = du + R dT  CP dT

= CV dT + R dT

 CP = CV + R

On a molar basis  ̅̅̅

𝐶 ̅̅̅
𝑃 = 𝐶𝑉 + 𝑅𝑢

Specific heat ratio:

𝐶𝑃
𝑘=
𝐶𝑉

(for air k = 1.4, R = 287 J/kgK, CP = 1.005 kJ/kgK )

SPECIFIC HEATS OF SOLIDS AND LIQUIDS

A substance whose specific volume (or density) is constant is called an incompressible
substance. The specific volumes of solids and liquids essentially remain constant during a
process. Therefore, liquids and solids can be approximated as incompressible substances.

 CP = CV = C

Internal energy change

du = C dT 
2
∆𝑢 = 𝑢2 − 𝑢1 = ∫ 𝐶(𝑇) 𝑑𝑇
1

Assume constant, average C = CAV  ∆𝑢 ≅ 𝐶𝐴𝑉 ∆𝑇

Enthalpy Change

dh = du + v dP + P dv = du + v dP (incompressible dv = 0) 

∆ℎ = ∆𝑢 + 𝑣∆𝑃 ≅ 𝐶𝐴𝑉 ∆𝑇 + 𝑣∆𝑃

ENERGY TRANSFER BY HEAT

Heat is defined as the form of energy that is transferred across the boundary of a system at a
given temperature to another system (or the surroundings) at a lower temperature by virtue of
the temperature difference between the two systems.

Several phrases in common use today—such as heat flow, heat addition, heat rejection, heat
absorption, heat removal, heat gain, heat loss, heat storage, heat generation, electrical heating,
resistance heating, frictional heating, gas heating, heat of reaction, liberation of heat, specific
heat, sensible heat, latent heat, waste heat, body heat, process heat, heat sink, and heat source

Heat inlet System

Heat exit

Heat Source

Qin

System

Qout

Heat Sink
Adiabatic Process: No heat transfer Surroundings
System

Insulation

Heat  Q (kJ)

On a unit mass basis:

𝑄
𝑞= (𝑘𝐽/𝑘𝑔)
𝑚

Heat transfer rate: 𝑄̇ (kJ/s =kW)

𝑡2
𝑄 = ∫ 𝑄̇ 𝑑𝑡 (𝑘𝐽)
𝑡1

If 𝑄̇ = 𝑐𝑜𝑛𝑠𝑡  𝑄 = 𝑄̇ ∆𝑡

If the system is taking heat input (+) (its energy level will increase),

If the system is giving heat output (-) (its energy level will decrease)

ENERGY TRANSFER BY WORK

Work is the energy transfer associated with a force acting through a distance. Work is usually
defined as a force F acting through a displacement x, where the displacement is in the direction
of the force.
2
𝑊 = ∫ 𝐹 𝑑𝑥
1

Thermodynamic definition: Work is done by a system if the sole effect on the surroundings
(everything external to the system) could be the raising of a weight.
Consider as a system the battery and motor of Fig (a), and let the motor drive a fan. Does work
cross the boundary of the system? To answer this question using the definition of work given,
replace the fan with the pulley and weight arrangement shown in Fig (b). As the motor turns,
the weight is raised, and the sole effect external to the system is the raising of a weight. Thus,
we conclude that work is crossing the boundary of the system.
Work done by a system is considered positive and work done on a system is considered
negative.

Work  W (kJ)

The work per unit mass of the system, specific work:

𝑊
𝑤= (𝑘𝐽/𝑘𝑔)
𝑚
The work done per unit time is called power:

𝑘𝐽
𝑊̇ = 𝑝𝑜𝑤𝑒𝑟 ( = 𝑘𝑊)
𝑠
𝛿𝑊
𝑊̇ =
𝛿𝑡
2
𝑊1−2 = ∫ 𝛿𝑊
1

Path functions have inexact differentials designated by the symbol, . Therefore, a

differential amount of heat or work is represented by Q or W, instead of dQ or dW.

Properties, however, are point functions (i.e., they depend on the state only, and not on how a
system reaches that state), and they have exact differentials designated by the symbol d.

For example, a change in volume is represented by dV:

2
∆𝑉 = ∫ 𝑑𝑉 = 𝑉2 − 𝑉1 𝑒𝑥𝑎𝑐𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙, 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑝𝑎𝑡ℎ
1
 𝑊1−2 ≠ ∆𝑊 ≠ 𝑊2 − 𝑊1 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛!

WORK DONE AT THE MOVING BOUNDARY

Consider as a system the gas contained in a cylinder and piston:

Total force on the piston = PA

P = Gas pressure

A = piston (cylinder) cross section area

𝛿𝑊 = 𝑃𝐴 𝑑𝐿

Since A dL = dV volume change 

𝛿𝑊 = 𝑃 𝑑𝑉

2
𝑊1−2 = ∫ 𝑃 𝑑𝑉
1

1→2 work done on the system

2→1 work done by the system

The assumption of a quasi-equilibrium process is essential here because each point on line 1–
2 represents a definite state, and these states correspond to the actual state of the system only
if the deviation from equilibrium is infinitesimal.

2
𝑊1−2 = ∫1 𝑃 𝑑𝑉

From state 1 to state 2, we may go with different paths. Therefore the area underneath the
curves (i.e. work) will be different. Therefore, work is a path function. Work can not be written
as W2-W1.

In order to evaluate the integral, we have to know about P-V relation.

The relationship between P and V is given in terms of experimental data, in graphical or by an
analytical relationship.

For example, a polytropic processes:

𝑃𝑉 𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃𝑉 𝑛 = 𝑃1 𝑉1𝑛 = 𝑃2 𝑉2𝑛  𝑃= 𝑉𝑛

2 2
𝑑𝑉 𝑃2 𝑉2 − 𝑃1 𝑉1
𝑊1−2 = ∫ 𝑃 𝑑𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ∫ = 𝑛≠1
1 1 𝑉𝑛 1−𝑛

For n = 1  PV = constant = P1V1 = P2V2

2 2
𝑑𝑉 𝑉2
𝑊1−2 = ∫ 𝑃 𝑑𝑉 = 𝑃1 𝑉1 ∫ = 𝑃1 𝑉1 𝑙𝑛
1 1 𝑉 𝑉1

Example

Consider the system of piston-cylinder containing helium gas, which expands in reversible
manner.

PV1.5 = constant

V1 = 0.1 m3

Gas P2 = 200 kPa P1 = 450 kPa T1 = 250 K

W1-2 = ?

1⁄
𝑃2 𝑉2 −𝑃1 𝑉1 𝑉2 𝑃 𝑛
𝑊1−2 = , 𝑃1 𝑉1𝑛 = 𝑃2 𝑉2𝑛  = (𝑃1 ) 
1−𝑛 𝑉1 2

450 0.667
𝑉2 = 0.1 ( ) = 0.1717 𝑚3
200
200 ∙ 0.1717 − 450 ∙ 0.1
𝑊1−2 = = 21.32 𝑘𝐽
−0.5
Example

2 kg of air is expanding in a reversible isothermal process at 300 K, from 2 m3 to 4 m3. W1-2 =?

Solution:

Ideal gaz PV = mRT, T = const  PV = const (n = 1, polytropic process)

2
𝑑𝑉 𝑉2 4
𝑊1−2 = 𝑚𝑅𝑇 ∫ = 𝑚𝑅𝑇 𝑙𝑛 = 2(0.287)(300)𝑙𝑛 = 119. 36 𝑘𝐽
1 𝑉 𝑉1 2

Electrical Work
Electrons crossing the system boundary do electrical work on the system.

When N coulombs of electrical charge move through a potential difference V, the electrical
work done is

We = VN

Electrical power:

𝑊̇ = 𝑉𝐼 (𝑘𝑊)

I: current (the number of electrical charges flowing per unit time)

2
𝑊𝑒 = ∫ 𝑉𝐼 𝑑𝑡
1

If VI = const 

𝑊𝑒 = 𝑉𝐼 ∆𝑡 (𝑘𝐽)

Shaft Work
Torque: T = Fr

Wsh = Fs, s = (2r) n

 Wsh = 2nT (kJ)

𝑊̇𝑠ℎ = 2𝜋𝑛̇ 𝑇 (𝑘𝑊)

𝑛̇ : The number of revolutions per unit time

Spring Work

F=kx

k: spring constant
1
𝑊𝑠𝑝𝑟𝑖𝑛𝑔 = 2 (𝑥22 − 𝑥12 ) (𝑘𝐽)

Work Done on Elastic Solid Bars

n: normal stress

The Stretching of a Liquid Film

s: surface tension

CONSERVATION OF MASS
We know from relativistic considerations that mass and energy are related by the well-known
equation:

E = m C2

E: Energy, m: mass, C: velocity of light

Energy and mass are conserved quantities.

Conservation of Mass Principle
𝑚𝑖𝑛 − 𝑚𝑜𝑢𝑡 = ∆𝑚𝑠𝑦𝑠𝑡

𝑚𝑖𝑛

Control volume

𝑚𝑜𝑢𝑡

∆𝑚𝑠𝑦𝑠𝑡 = 𝑚𝑓𝑖𝑛𝑎𝑙 − 𝑚𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑚2 − 𝑚1

On a per unit time basis, i.e. in rate form:

𝑑𝑚𝑠𝑦𝑠
𝑚̇𝑖𝑛 − 𝑚̇𝑜𝑢𝑡 = (𝑘𝑔/𝑠)
𝑑𝑡
(The amount of mass flowing per unit time, that is, the mass flow rate, 𝑚̇)

For multiple inlets and outlets:

Σ𝑚𝑖 − Σ𝑚𝑒 = (𝑚2 − 𝑚1 )𝑠𝑦𝑠𝑡

𝑑𝑚𝑠𝑦𝑠
Σ𝑚̇𝑖 − Σ𝑚̇𝑒 =
𝑑𝑡

For closed systems 𝑚𝑖𝑛 = 𝑚𝑜𝑢𝑡 = 0  𝑚𝑠𝑦𝑠𝑡 = 𝑐𝑜𝑛𝑠𝑡

For steady-state processes, there will be no change with time:

Σ𝑚̇𝑖 = Σ𝑚̇𝑒

FLOW WORK AND THE ENERGY OF A FLOWING FLUID

The force applied on the fluid element by the imaginary piston is: F = PA

To push the entire fluid element into the control volume, this force must act through a distance
L.
Thus, the work done in pushing the fluid element across the boundary (i.e., the flow work) is:

𝑊𝑓𝑙𝑜𝑤 = 𝐹𝐿 = 𝑃𝐴𝐿 = 𝑃𝑉 (𝑘𝐽)

The flow work per unit mass is:

𝑤𝑓𝑙𝑜𝑤 = 𝑃𝑣 (𝑘𝐽/𝑘𝑔)

Enthalpy: h = u + Pv ,

u: internal energy

Pv: flow work (due to movement of the fluid)

The total energy of stationary fluid:

𝑉2
𝑒=𝑢+ + 𝑔𝑧 (𝑘𝐽/𝑘𝑔)
2

The total energy of a flowing fluid:

 = e + Pv = u + Pv + ke +pe

𝑉2
𝜃=ℎ+ + 𝑔𝑧 (𝑘𝐽/𝑘𝑔)
2

Energy Transport by Mass

Amount of energy transport:

𝑉2
𝐸𝑚𝑎𝑠𝑠 = 𝑚𝜃 = 𝑚 (ℎ + + 𝑔𝑧) (𝑘𝐽)
2

Rate of energy transport:

𝑉2
𝐸̇𝑚𝑎𝑠𝑠 = 𝑚̇𝜃 = 𝑚̇ (ℎ + + 𝑔𝑧) (𝑘𝑊)
2

If we neglect ke and pe:

 𝐸𝑚𝑎𝑠𝑠 = 𝑚ℎ 𝑎𝑛𝑑 𝐸̇𝑚𝑎𝑠𝑠 = 𝑚̇ℎ

When the mass and properties are changing with time, have to integrate:

𝑉𝑖2
𝐸𝑖𝑛,𝑚𝑎𝑠𝑠 = ∫ 𝜃𝑖 𝛿𝑚𝑖 = ∫ (ℎ𝑖 + + 𝑔𝑧𝑖 ) 𝛿𝑚𝑖
𝑚𝑖 𝑚𝑖 2

WORK IN NONEQUILIBRIUM PROCESS

When the pin is released, the external force per unit area acting on the
system (gas) boundary is comprised of two parts:

𝐹𝑒𝑥𝑡 𝑚𝑝 𝑔
𝑃𝑒𝑥𝑡 = = 𝑃𝑜 +
𝐴 𝐴

The release of pin will lead to nonequilibrium process due high pressure gas inside the cylinder.
The work for nonequilibrium process is:

𝛿𝑊 = 𝐹𝑒𝑥𝑡 𝑑𝐿 = 𝑃𝑒𝑥𝑡 𝑑𝑉

Let the membrane rupture and the gas fill the entire volume. The process is not quasi-
equilibrium. There will be no work, since there is no resistance at the boundary. Fext =0.

Example

Consider a piston/cylinder arrangement, as shown in Fig. The piston is loaded with a mass mp,
the outside atmosphere P0, a linear spring, and a single point force F1. The piston traps the gas
inside with a pressure P.

Σ𝐹 = 𝑚𝑝 𝑔 + 𝑃𝑜 𝐴 + 𝑘𝑠 (𝑥 − 𝑥0 ) + 𝐹1

From the force balance on the piston, gas pressure:

𝑚𝑝 𝑔+𝑘𝑠 (𝑥−𝑥0 )+𝐹1
𝑃 = 𝑃𝑜 + 𝐴


 𝑚𝑝 𝑔 𝐹1 𝑘𝑠
𝑃 = 𝑃𝑜 + + + 2 (𝑉 − 𝑉𝑜 )
𝐴 𝐴 𝐴

 P = C1 +C2 V
2
𝑊1−2 = ∫1 𝑃𝑑𝑉 = area under the process curve

1
𝑊1−2 = 2 (𝑃1 + 𝑃2 )(𝑉2 − 𝑉1 )

THE FIRST LAW OF THERMODYNAMICS

The first law of thermodynamics, also known as the conservation of energy principle.

Energy Balance
𝑡𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙
( )−( )=( )
𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚

𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = ∆𝐸𝑠𝑦𝑠𝑡

∆𝐸𝑠𝑦𝑠𝑡 = 𝐸𝑓𝑖𝑛𝑎𝑙 − 𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐸2 − 𝐸1

1
∆𝐸 = ∆𝑈 + ∆𝐾𝐸 + ∆𝑃𝐸 = 𝑚(𝑢2 − 𝑢1 ) + 𝑚(𝑉22 − 𝑉12 ) + 𝑚𝑔(𝑧2 − 𝑧1 )
2

For stationary systems: KE  0, PE  0  E  U

Mechanisms of Energy Transfer, Ein and Eout
1) Heat transfer, Q
2) Work, W
3) Mass flow, m

𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = (𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡 ) + (𝑊𝑖𝑛 − 𝑊𝑜𝑢𝑡 ) + (𝐸𝑚𝑎𝑠𝑠,𝑖𝑛 − 𝐸𝑚𝑎𝑠𝑠,𝑜𝑢𝑡 ) = ∆𝐸𝑠𝑦𝑠𝑡

𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = ∆𝐸𝑠𝑦𝑠𝑡

𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 ∶ Net energy transfer by heat, work and mass

∆𝐸𝑠𝑦𝑠𝑡 ∶ Change in internal, kinetic and potential energies

In the rate form:

𝐸̇𝑖𝑛 − 𝐸̇𝑜𝑢𝑡 = ∆𝐸̇𝑠𝑦𝑠𝑡

On a per unit mass basis:

𝑒𝑖𝑛 − 𝑒𝑜𝑢𝑡 = ∆𝑒𝑠𝑦𝑠𝑡

In the differential form:

𝛿𝐸𝑖𝑛 − 𝛿𝐸𝑜𝑢𝑡 = 𝑑𝐸𝑠𝑦𝑠𝑡

𝛿𝑒𝑖𝑛 − 𝛿𝑒𝑜𝑢𝑡 = 𝑑𝑒𝑠𝑦𝑠𝑡

For a closed system undergoing a cycle, the initial and final states are identical:

∆𝐸𝑠𝑦𝑠𝑡 = 𝐸2 − 𝐸1 = 0

 𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡

𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = 0 

(𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡 ) + (𝑊𝑖𝑛 − 𝑊𝑜𝑢𝑡 ) + (𝐸𝑚𝑎𝑠𝑠,𝑖𝑛 − 𝐸𝑚𝑎𝑠𝑠,𝑜𝑢𝑡 ) = 0

For a closed system, no mass inlet or exit:

(𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡 ) + (𝑊𝑖𝑛 − 𝑊𝑜𝑢𝑡 ) = 0

For a cycle:

∮ 𝛿𝑄 = ∮ 𝛿𝑊
 Consider the cycles A-B and C-B:

2 1 2 1
∫ 𝛿𝑄𝐴 + ∫ 𝛿𝑄𝐵 = ∫ 𝛿𝑊𝐴 + ∫ 𝛿𝑊𝐵 (1)
1 2 1 2

2 1 2 1
∫ 𝛿𝑄𝐶 + ∫ 𝛿𝑄𝐵 = ∫ 𝛿𝑊𝐶 + ∫ 𝛿𝑊𝐵 (2)
1 2 1 2

Subtracting (1) and (2):

1 1
∫ (𝛿𝑄 − 𝛿𝑊)𝐴 = ∫ (𝛿𝑄 − 𝛿𝑊)𝐶
2 2

Since A and C represent arbitrary processes between states 1 and 2, the quantity δQ – δW is the
same for all processes between states 1 and 2. Therefore, δQ − δW depends only on the initial
and final states and not on the path followed between the two states. Therefore, it is the
differential of a property of the mass. This property is the energy of the mass and is given the
symbol E.

𝛿𝑄 − 𝛿𝑊 = 𝑑𝐸

dE: property of mass, exact differential  energy of the system

𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = ∆𝐸𝑠𝑦𝑠𝑡

For a closed system:

(𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡 ) + (𝑊𝑖𝑛 − 𝑊𝑜𝑢𝑡 ) = ∆𝐸𝑠𝑦𝑠𝑡 = 𝐸2 − 𝐸1

(𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡 ) = 𝑄1−2 (Net heat transfer to the system)

(𝑊𝑖𝑛 − 𝑊𝑜𝑢𝑡 ) = −𝑊1−2 (Net work done by the system)

1
∆𝐸𝑠𝑦𝑠𝑡 = 𝐸2 − 𝐸1 = ∆𝑈 + ∆𝐾𝐸 + ∆𝑃𝐸 = 𝑚(𝑢2 − 𝑢1 ) + 𝑚(𝑉22 − 𝑉12 ) + 𝑚𝑔(𝑧2 − 𝑧1 )
2
 The first law of thermodynamics for closed sytems:

1
𝑄1−2 = 𝑚(𝑢2 − 𝑢1 ) + 𝑚(𝑉22 − 𝑉12 ) + 𝑚𝑔(𝑧2 − 𝑧1 ) + 𝑊1−2
2
Example

An insulated tank contains 0.2 kg of air at 300 K, 100 kPa conditions. The tank is stirred
by a paddle wheel and thus 8 kJ of energy is given to the air. Determine the change in
internal energy, final temperature and pressure of the air?

Solution

𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = ∆𝐸𝑠𝑦𝑠𝑡 = ∆𝑈 + ∆𝐾𝐸 + ∆𝑃𝐸 = ∆𝑈

 ∆𝑈 = 8 𝑘𝐽

Air can be assumed ideal gas:

∆𝑈 = 𝑚𝐶𝑉 (𝑇2 − 𝑇1 )

8 = 0.2 (0.7165) (T2 – 300)

 T2 = 355.8269 K

For constant volume:

𝑇2 355.8269
𝑃2 = 𝑃1 ( ) = 100 ( ) = 118.6089 𝑘𝑃𝑎
𝑇1 300

PROBLEM ANALYSIS AND SOLUTION TECHNIQUE

1. What is the control mass or control volume? It may be helpful to draw a sketch of the
system, illustrating all heat and work flows, and indicating forces.
2. What do we know about the initial state (that is, which properties are known)?
3. What do we know about the final state?
4. What do we know about the process that takes place? Is anything constant or zero? Is
there some known functional relation between two properties?
5. It may be helpful to draw a diagram, a T–v, P–v or T-s diagram.
6. What is our thermodynamic model for the behavior of the substance (for example,
steam tables, ideal gas, and so on)?
7. What is our analysis of the problem (use the first law or conservation of mass, boundary
conditions, etc.)?
8. What is our solution technique? How do we proceed to find what is desired? Is a trial-
and-error solution necessary?
Example

A rigid tank has a volume of 5 m3 and it contains 0.05 m3 saturated liquid water and 4.95 m3
saturated vapor at the pressure of 0.1 MPa. By means of heat transfer,
the final state reaches full saturated vapor. Determine the necessary
heat transfer?

Solution

V = 5 m3, V = const, W1-2 = 0

State 1: Vliq = 0.05 m3, Vvap = 4.95 m3, P1 = 0.1 MPa

State 2: saturated vapor, x2=1, vg, ug, hg

1st law:

𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = ∆𝐸𝑠𝑦𝑠𝑡

𝑄1−2 = ∆𝑈 = 𝑈2 − 𝑈1

(For constant volume processes: Q1-2 = U)

T

𝑉𝑙𝑖𝑞 0.05 2
𝑚1,𝑙𝑖𝑞 = = = 47.94 𝑘𝑔
𝑣𝑓 0.001043
P1
𝑉𝑣𝑎𝑝 4.95
𝑚1,𝑣𝑎𝑝 = = = 2.92 𝑘𝑔
𝑣𝑔 1.6940 1

𝑈1 = 𝑚1,𝑙𝑖𝑞 𝑢𝑓 + 𝑚1,𝑣𝑎𝑝 𝑢𝑔 = 47.94(47.36) + 2.92(2506.1) = 27326 𝑘𝐽

𝑚𝑡𝑜𝑡 = 𝑚1,𝑙𝑖𝑞 + 𝑚1,𝑣𝑎𝑝 = 50.86 𝑘𝑔

𝑚1,𝑣𝑎𝑝
𝑥1 = = 0.0574
𝑚𝑡𝑜𝑡

𝑉 5
𝑣2 = 𝑣1 = = = 0.09831 𝑚3 /𝑘𝑔
𝑚 50.86

𝑣2 = 𝑣𝑔 = 0.09831 𝑚3 /𝑘𝑔  table P2 = 2.03 MPa, u2 = ug = 2600.5 kJ/kg

U2 = m u2 = 50.86 (2600.5) = 132261 kJ

Q1-2 = U2 – U1 = 132261- 27326 = 104935 kJ

Example

A piston-cylinder system contains 0.5 Kg of saturated liquid-vapor mixture at a volume of 0.1

m3 at 0.4 MPa pressure. By applying heat transfer, the final temperature will reach 300oC. Find
the heat transfer and the work for this process?

Solution

Sat. Liq-vap P2 = P1
2
H2O 300oC
Q1-2 1

v1 v2 v
1st law:

𝑄1−2 = 𝑚(𝑢2 − 𝑢1 ) + 𝑊1−2

Work:
2 2
𝑊1−2 = ∫ 𝑃𝑑𝑉 = 𝑃 ∫ 𝑑𝑉 = 𝑃(𝑉2 − 𝑉1 ) = 𝑚𝑃(𝑣2 − 𝑣1 ) = 𝑚(𝑃2 𝑣2 − 𝑃1 𝑣1 )
1 1

 1st law:

𝑄1−2 = 𝑚(𝑢2 − 𝑢1 ) + 𝑊1−2 = 𝑚(𝑢2 − 𝑢1 ) + 𝑚(𝑃2 𝑣2 − 𝑃1 𝑣1 ) = 𝑚(ℎ2 − ℎ1 )

(For constant pressure processes: Q1-2 = H)

𝑉1 0.1
𝑣1 = = = 0.2 = 0.001084 + 𝑥1 0.4614
𝑚 0.5

→ x1 = 0.4311

h1 = hf +x1 hfg = 604.74 + 0.4311(2133.8) = 1524.6 kJ/kg

P2 = P1 = 0.4 MPa
h2 = 3066.8 kJ/kg
T2 = 300 oC Superheated
v2 = 0.6548 m3/kg
Steam tables

𝑄1−2 = 𝑚(ℎ2 − ℎ1 ) = 0.5(3066.8 − 1524.6) = 771.1 𝑘𝐽

𝑊1−2 = 𝑚𝑃(𝑣2 − 𝑣1 ) = 0.5 ∙ 400(0.6548 − 0.2) = 91 𝑘𝐽

ENERGY ANALYSIS OF STEADY-FLOW SYSTEMS

A large number of engineering devices such as turbines, compressors, and nozzles operate for
long periods of time under the same conditions once the transient start-up period is completed
and steady operation is established, and they are classified as steady-flow devices.

Assumptions:

1. The control volume does not move relative to the coordinate frame.
2. The state of the mass at each point in the control volume does not vary with time.
(mCV = const, ECV = const)
3. As for the mass that flows across the control surface, the mass flux and the state of this
mass at each discrete area of flow on the control surface do not vary with time. The
rates at which heat and work cross the control surface remain constant.

Mass balance:

𝑚̇𝑖𝑛 − 𝑚̇𝑜𝑢𝑡 = ∆𝑚̇𝑠𝑦𝑠 = 𝑚̇2 − 𝑚̇1 = 0

𝑚̇𝑖𝑛 = 𝑚̇𝑜𝑢𝑡 (𝑘𝑔/𝑠)

𝑚̇2 = 𝑚̇1 = 𝑚̇𝐶𝑉

For multiple inlets and outlets:

Σ𝑚̇𝑖 = Σ𝑚̇𝑒

Energy Balance: 𝐸̇𝑖𝑛 − 𝐸̇𝑜𝑢𝑡 = ∆𝐸̇𝑠𝑦𝑠𝑡 = 0

 𝐸̇𝑖𝑛 = 𝐸̇𝑜𝑢𝑡

𝑄̇𝑖𝑛 + 𝑊̇𝑖𝑛 + Σ𝑚̇𝑖 𝜃𝑖 = 𝑄̇𝑜𝑢𝑡 + 𝑊̇𝑜𝑢𝑡 + Σ𝑚̇𝑒 𝜃𝑒

𝑉𝑖2 𝑉𝑒2
𝑄̇𝑖𝑛 + 𝑊̇𝑖𝑛 + Σ𝑚̇𝑖 (ℎ𝑖 + + 𝑔𝑧𝑖 ) = 𝑄̇𝑜𝑢𝑡 + 𝑊̇𝑜𝑢𝑡 + Σ𝑚̇𝑒 (ℎ𝑒 + + 𝑔𝑧𝑒 )
2 2

𝑉𝑒2 𝑉𝑖2
𝑄̇1−2 = Σ𝑚̇𝑒 (ℎ𝑒 + + 𝑔𝑧𝑒 ) − Σ𝑚̇𝑖 (ℎ𝑖 + + 𝑔𝑧𝑖 ) + 𝑊̇1−2
2 2

On a unit mass basis, for single inlet-outlet:

(𝑉22 − 𝑉12 )
𝑞 − 𝑤 = (ℎ2 − ℎ1 ) + + 𝑔(𝑧2 − 𝑧1 )
2

If ke  0, pe  0  q – w = h2 - h1

Adiabatic  q = 0,  -w = h2 - h1

Example

The mass rate of flow into a steam turbine is 1.5 kg/s, and the heat transfer from the turbine is
8.5 kW. The following data are known for the steam entering and leaving the turbine.

Inlet Conditions Exit Conditions

Pressure 2.0 MPa 0.1 MPa
Temperature 350◦C
Quality 100%
Velocity 50 m/s 100 m/s
Elevation above reference plane 6m 3m
g = 9.8066 m/s2

Determine the power output of the turbine.

𝑚̇𝑖 = 1.5 kg/s
Solution Pi = 2 MPa
Ti = 350oC
Vi = 50 m/s
Zi = 6 m
𝑊̇1−2

Turbine

𝑄̇1−2
𝑚̇𝑒 = 1.5 kg/s
Pe = 0.1 MPa
xe = 100%
Ve = 100 m/s
Ze = 3 m
From the first law:

𝑉𝑒2 𝑉𝑖2
𝑄̇1−2 = Σ𝑚̇𝑒 (ℎ𝑒 + + 𝑔𝑧𝑒 ) − Σ𝑚̇𝑖 (ℎ𝑖 + + 𝑔𝑧𝑖 ) + 𝑊̇1−2
2 2

With
𝑄̇1−2 = −8.5 𝑘𝑤

From the steam tables, hi = 3137.0 kJ/kg. Substituting inlet conditions gives

𝑉𝑖2 50𝑥50
= = 1.25 𝑘𝑗/𝑘𝑔
2 2𝑥1000

6 × 9.8066
𝑔𝑧𝑖 = = 0.059 kJ/kg
1000

Similarly, for the exit he = 2675.5 kJ/kg and

𝑉𝑒2 100𝑥100
= = 5 𝑘𝑗/𝑘𝑔
2 2𝑥1000

3 × 9.8066
𝑔𝑧𝑒 = = 0.029 kJ/kg
1000

Therefore, substituting into energy balance equation, we obtain

−8.5 = 1.5(2675.5 + 5.0 + 0.029) − 1.5(3137 + 1.25 + 0.059) + 𝑊̇1−2.

𝑊̇1−2 = 678.2 kW

Two observations can be made by referring to this example:

First, the potential energy change did not affect any of the significant figures. In most problems
where the change in elevation is small, the potential energy terms may be neglected.

Second, if velocities are small—say, under 20 m/s—in many cases the kinetic energy is
insignificant compared with other energy quantities. The kinetic energy terms can be neglected.

In many thermodynamic problems, one must make judgments as to which quantities may be
negligible for a given analysis.
Example
Steam flows steadily into a turbine with a mass flow rate of 26 kg/s and a negligible velocity
at 6 MPa and 600°C. The steam leaves the turbine at 0.5 MPa and 200°C with a velocity of 180
m/s. The rate of work done by the steam in the turbine is measured to be 20 MW. If the elevation
change between the turbine inlet and exit is negligible, determine the rate of heat transfer
associated with this process.

Solution
From steam tables (Table A-4, A5, A-6) 26 kg/s

6 MPa
P1 = 6 MPa 
 h1 = 3658.8 kJ/kg
T1 = 600C 
Turbine
P2 = 0.5 MPa 
 h2 = 2855.8 kJ/kg
T2 = 200C 

0.5 MPa
1st law:
200°C
E − E = E system0 (steady) =0
inout
   
Rate of net energy transfer Rate of change in internal, kinetic,
by heat, work,and mass potential, etc. energies

E in = E out
 V2   V2 
m  h1 + 1  = m  h2 + 2  + W out + Q out (since pe  0)
 2   2 
 
 V 2 − V22 
Q out = −W out + m  h1 − h2 + 1 
 2 
 

 V 2 − V22 
Q out = −W out + m  h1 − h2 + 1 
 2 
 
 (0 − 180 m/s) 2  1 kJ/kg 
= 20,000 kW + (26 kg/s) (3658.8 − 2855.8)kJ/kg +  
 2  1000 m 2 /s 2 
= 455 kW

Throttling Valves (expansion valves)

Throttling valves are any kind of flow-restricting devices that cause a significant pressure drop
in the fluid. Some familiar examples are ordinary adjustable valves, capillary tubes, and porous
plugs.
The pressure drop in the fluid is often accompanied by a large drop in temperature, and for
that reason throttling devices are commonly used in refrigeration and air-conditioning
applications.

W1-2 = 0, pe = 0  1st law:

𝑉𝑖2 𝑉𝑒2
ℎ𝑖 + = ℎ𝑒 +
2 2
If we neglect kinetic energy:

ℎ𝑖 = ℎ𝑒

The Joule-Thomson coefficient:

𝜕𝑇
𝜇𝐽 = ( )
𝜕𝑃 ℎ

J > 0  T < 0; J < 0  T > 0

Example

Air enters adiabatic diffuser at 80 kPa, 127 oC conditions in a steady state flow and exits at 100
kPa. The mass flow rate is 6000 kg/h, and the velocity reduces from 230 m/s to 30 m/s.
Determine the temperature and the cross sectional area at the exit?

Solution

Working fluid is air:

R = 0.287 kJ/kg.K

T1 = 400 K  h1 = 400.98 kJ/kg (Table A-17).

Mass balance: 1 = m
m 2 = m


1st law:
E − E out = E system0 (steady) =0
in  
Rate of net energy transfer Rate of change in internal, kinetic,
by heat, work,and mass potential, etc. energies

E in = E out 1 AIR 2
m (h1 + V12 / 2) = m (h2 + V22 /2) (since Q  W
  pe  0)

V 22 − V12 ,
0 = h2 − h1 +
2
 h2 = h1 −
V22 − V12
= 400.98 kJ/kg −
(30 m/s )2 − (230 m/s )2  1 kJ/kg


 = 426.98 kJ/kg
2 2  1000 m 2 /s 2 
 

Table A-17 
T2 = 425.6 K

(b)

v2 =
RT2
=
(
0.287 kPa  m 3 /kg  K (425.6 K ) )
= 1.221 m 3 /kg
P2 (100 kPa )

1 m v 2 (6000 3600 kg/s )(1.221 m 3 /kg )

m = A2V2 ⎯
⎯→ A2 = = = 0.0678 m 2
v2 V2 30 m/s

Example
Refrigerant R-134a enters a compressor at 100 KPa, -24 oC conditions with a volume flow rate
of 1.35 m3/min and exits at 80.0 kPa, 60 oC conditions. Determine the mass flow rate and
power input?

Solution
E − E = E system0 (steady) =0 800 kPa
inout
   
Rate of net energy transfer Rate of change in internal, kinetic,
by heat, work,and mass potential, etc. energies 60°C
E in = E out Compressor
W in + m h1 = m h2 (since ke  pe  0)
W in = m (h2 − h2 ) 100 kPa

-24°C
From R134a tables (Tables A-11, A-12, A-13)
P1 = 100 kPa  h1 = 236.33 kJ/kg

T1 = −24C  v1 = 0.1947 m 3 /kg

P2 = 800 kPa 
h2 = 296.81 kJ/kg
T2 = 60C 

The mass flow rate is

V1 (1.35 / 60) m 3 /s
=
m = = 0.1155 kg/s
v1 0.1947 m 3 /kg
Substituting,
W in = m (h2 − h1 ) = (0.1155 kg/s)(296. 81 − 236 .33)kJ/kg = 6.99 kW
ENERGY ANALYSIS OF UNSTEADY-FLOW PROCESSES
(The Transient Process, Uniform State, Uniform Flow Processes)
Transient model assumptions are as follows:

1. The control volume remains constant relative to the coordinate frame.

2. The state of the mass within the control volume may change with time, but at any instant of
time the state is uniform throughout the entire control volume (or over several identifiable
regions that make up the entire control volume).
3. The state of the mass crossing each of the areas of flow on the control surface is constant
with time, although the mass flow rates may vary with time.

Mass Balance:

𝑚𝑖𝑛 − 𝑚𝑜𝑢𝑡 = ∆𝑚𝑠𝑦𝑠𝑡

∆𝑚𝑠𝑦𝑠𝑡 = 𝑚𝑠𝑜𝑛 − 𝑚𝑖𝑙𝑘 = 𝑚2 − 𝑚1

For multiple inlet and outlets:

Σ𝑚𝑖 − Σ𝑚𝑒 = (𝑚2 − 𝑚1 )𝑠𝑦𝑠𝑡

Energy Balance:

𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = ∆𝐸𝑠𝑦𝑠𝑡

Q1−2 = Σ𝑚𝑒 θ𝑒 − Σ𝑚𝑖 𝜃𝑖 + (𝑚2 𝑒2 − 𝑚1 𝑒1 )𝑠𝑦𝑠𝑡 + W1−2

Or,

(𝑄𝑖𝑛 + 𝑊𝑖𝑛 + Σ𝑚𝑖 𝜃𝑖 ) − (𝑄𝑜𝑢𝑡 + 𝑊𝑜𝑢𝑡 + Σ𝑚𝑒 𝜃𝑒 ) = (𝑚2 𝑒2 − 𝑚1 𝑒1 )𝑠𝑦𝑠𝑡

Example

A rigid, insulated tank that is initially evacuated is connected through a valve to a supply line
that carries steam at 1.4 MPa and 300°C. Now the valve is opened, and steam is allowed to
flow slowly into the tank until the pressure reaches 1.4 MPa, at which point the valve is closed.
Determine the final temperature of the steam in the tank.

Solution

QCV = 0, WCV = 0, me = 0, (m1)CV = 0

Mass balance: m2 = mi

1st Law : mi hi = m2 u2  hi = u2

From steam tables:

1.4 MPa and 300 oC  hi = 3040.4 kJ/kg

u2 = 3040.4 kJ/kg ve P2 = 1.4 MPa  T2 = 452 oC

Example

Steam at 800 kPa, 300 oC conditions is throttling to the pressure of 200 kPa. In this process,
the kinetic energy can be neglected. Determine the final temperature of the steam and the
average Joule-Thomson coefficient?

Solution

1st law: hi = he = 3056.5 kJ/kg

he = 3056.5 kJ/kg ve Pe = 200 kPa  superheated steam tables Te = 292.4 oC

𝜕𝑇
Joule- Thomson coefficient: 𝜇𝐽 = (𝜕𝑃)
ℎ

∆𝑇 −7.6
(𝜇𝐽 )𝑎𝑣𝑔 = ( ) = = 0.0127 𝐾/𝑘𝑃𝑎
∆𝑃 ℎ −600
Example

An adiabatic capillary tube is used in some refrigeration systems to drop the pressure of the
refrigerant from the condenser level to the evaporator level. The R-134a enters the capillary
tube as a saturated liquid at 50oC, and leaves at -12oC. Determine the quality of the refrigerant
at the inlet of the evaporator?

Solution

Mass balance: m 1 = m 2 = m .

E in − E out = E system0 (steady) = 0

E in = E out 50°C
m h1 = m h2
h1 = h2 Sat. liquid

Since Q  W = ke  Δpe  0 . R-134a

The inlet enthalpy of R-134a is, from the refrigerant tables (Table A-11),
T1 = 50C  -12°C
 h1 = h f = 123.49 kJ/kg
sat. liquid 

The exit quality is

T2 = −12C  h2 − h f 123.49 − 35.92
 x2 = = = 0.422
h2 = h1  h fg 207.38

Example

A rigid tank that is initially evacuated is connected through a valve to a supply line that carries
steam at 350 kPa MPa and 200°C (h=2863.1 kJ/kg). Now the valve is opened, and steam is
allowed to flow slowly into the tank until the pressure reaches to the supply line pressure, at
which point the valve is closed. In this process, the heat at the amount of 700 kJ/kg is rejected
from the tank. The ke and pe changes are neglibible. Determine the final internal inergy of the
steam in the tank?

Solution

1st law: mq1-2 = mu2 – 0 – m (hp + 0)

q1-2 = u2 - hp

-700 kJ/kg = u2 – 2863.1 kJ/kg

 u2 = 2163.1 kJ/kg
Example

Steam enters a nozzle at 400°C and 800 kPa with a velocity of 10 m/s, and leaves at 300°C and
200 kPa while losing heat at a rate of 25 kW. For an inlet area of 800 cm2, determine the
velocity and the volume flow rate of the steam at the nozzle exit?

Solution

1st law:
E − E out = E system0 (steady) =0
in  
Rate of net energy transfer Rate of change in internal, kinetic,
by heat, work,and mass potential, etc. energies 400C STEAM 300C
E in = E out
800 kPa 200 kPa
 V2   V2  Q
m  h1 + 1  = m  h2 + 2  + Q out since W  pe  0)
 2   2 
 
V12 V 2 Q
h1 + = h2 + 2 + out
2 2 m
From the steam tables (Tablo A-6):
P1 = 800 kPa v1 = 0.38429 m3/kg

T1 = 400C  h1 = 3267.7 kJ/kg

P2 = 200 kPa v 2 = 1.31623 m3/kg


T1 = 300C  h2 = 3072.1 kJ/kg

Steam mass flow rate:

1 1
 =
m A1V1 = (0.08 m 2 )(10 m/s) = 2.082 kg/s
v1 0.38429 m3/s
Substitute in the 1st law eq.:
(10 m/s) 2  1 kJ/kg  V22  1 kJ/kg  25 kJ/s
3267.7 kJ/kg +   = 3072. 1 kJ/kg +  +
 1000 m /s  2  1000 m /s  2.082 kg/s
2 2 2 2
2
⎯
⎯→V2 = 606 m/s
The volume flow rate at the nozzle exit:

V2 = m
 v 2 = (2.082 kg/s)(1.31 623 m3 /kg) = 2.74 m3 /s

Example

In some certain cases, for specific applications we may need saturated vapor if we have already
superheated steam. By adjusting the mass flow rate of the liquid water, which was sprayed
upon the superheated steam in a mixing chamber, we can able to get saturated vapor conditions.
The steam enters at 3.5 MPa, 400oC, 0.5 kg/s conditions and the liquid water enters at 3.5 Mpa,
40oC conditions. If the saturated vapor leaves the mixing chamber at pressure of 3 MPa,
determine the mass flow rate of the liquid water?
Solution

2
3

(1) 3.5 MPa, 400 oC, 0.5 kg/s  h1 = 3222.3 kJ/kg

(2) 3.5 MPa, 40 oC (hf @ 40 oC)  h2 = 167.57 kJ/kg
(3) Saturated vapor 3 MPa  h3 = hg @ 3 MPa = 2804.2 kJ/kg
Mass balance:

𝑚̇3 = 𝑚̇1 + 𝑚̇2

1st law:
𝑚̇1 ℎ1 + 𝑚̇2 ℎ2 = 𝑚̇3 ℎ3


𝑚̇2 = 0.0793 𝑘𝑔/𝑠

Example

Consider the piston/cylinder arrangement shown in Fig. A frictionless piston is free to move
between two sets of stops. When the piston rests on the lower stops, the enclosed volume is
400 L. When the piston reaches the upper stops, the volume is 600 L. The cylinder initially
contains water at 100 kPa, with 20% quality. It is heated until the water eventually exists as
saturated vapor. The mass of the piston requires 340 kPa pressure to move it against the outside
ambient pressure. Determine the final pressure in the cylinder, the heat transfer, and the work
for the overall process?

Solution

Vmin = 400 L, Vmax = 600 L, P1 = 100 kPa

x1 =0.2, x2 = 1, Pa + mpg/A = 340 kPa H2O

1st law : Q1-2 = m(u2-u1) + W1-2

Saturated
P1 = 100 kPa mixture
u1 = 417.36 + 0.2 (2088.7) = 835.1 kJ/kg

x1 = 0.2 Table v1 = 0.001043 + 0.2 (1.694 – 0.001043) = 0.3396 m3/kg

P1 < 340 kPa  the piston is at the bottom stop.
0.4
V1 = Vmin = 0.4 m3  𝑚 = 0.3396 = 1.177856 𝑘𝑔

0.6
𝑣2 = = 0.5094 𝑚3 /𝑘𝑔
1.177856

If: 𝑣2 < 𝑣𝑔@340 𝑘𝑝𝑎  the piston may have reached to the top stop.

T
P2

2
340 kPa

100 kPa

v1 0.5 0.52 v

There will work done when the piston starts until touching to the top stop.

W1-2 = P (V2 – V1) = 340 x 0.2 = 68 kJ

v2 = vg@P2 table  P2 = 361.32 kPa, u2 = 2549.9 kJ /kg

 Q1-2 = 1.1778 (2549.9 – 835.1) + 68 = 2087.78 kJ

THE SECOND LAW OF THERMODYNAMİCS

The first law places no restriction on the direction of a process, but satisfying the first law does
not ensure that the process can actually occur. This inadequacy of the first law to identify
whether a process can take place is remedied by introducing another general principle, the
second law of thermodynamics.

Consider the system:

Let this system undergo a cycle in which work is first done on the system by the paddle wheel
as the weight is lowered. Then let the cycle be completed by transferring heat to the
surroundings. We know from our experience that we cannot reverse this cycle. That is, if we
transfer heat to the gas, as shown by the dotted arrow, the temperature of the gas will increase
but the paddle wheel will not turn and raise the weight. Even though this would not violate the
first law.

In its broader significance, the second law acknowledges that processes proceed in a certain
direction but not in the opposite direction. The second law also asserts that energy has quality
as well as quantity.

THERMAL ENERGY RESERVOIRS

A relatively large thermal energy capacity (mass × specific heat) that can supply or absorb
finite amounts of heat without undergoing any change in temperature. Such a body is called a
thermal energy reservoir, such as oceans, lakes, atmosphere,
A reservoir that supplies energy in the form of heat is called a source, and one that absorbs
energy in the form of heat is called a sink. Thermal energy reservoirs are often referred to as
heat reservoirs since they supply or absorb energy in the form of heat.

 heat reservoirs
Thermal energy
source

heat

Thermal energy sink

HEAT ENGINES

A heat engine may be defined as a device that operates in a thermodynamic cycle and does a
certain amount of net positive work through the transfer of heat from a high-temperature body
to a low-temperature body.
For example: Steam power plant, the internal combustion engine and the gas türbine.

Heat source, TH

QH

Heat
Wnet
Engine

QL

Heat sink, TL

Simple Heat Engine Cycle = Steam Power Plant

Wnet = Wturb – Wpump

1st law:

𝑊𝑛𝑒𝑡 = 𝑄𝐻 − 𝑄𝐿
Thermal Efficiency:

𝑁𝑒𝑡 𝑤𝑜𝑟𝑘 𝑜𝑢𝑡𝑝𝑢𝑡

𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑡 𝑖𝑛𝑝𝑢𝑡

𝑊𝑛𝑒𝑡
𝜂𝑡ℎ =
𝑄𝐻

By using the 1st law:

𝑄𝐿
𝜂𝑡ℎ = 1 −
𝑄𝐻

REFRIGERATORS AND HEAT PUMPS

A refrigerator or heat pump is a device that operates in a cycle, that requires work, and that
transfers heat from a low-temperature body to a high-temperature body.

The working fluid used in the refrigeration cycle is called a refrigerant.

Basic components of a refrigeration system and typical operating conditions.

Coefficient of Performance

Coefficient of Performance  COP

For refrigerator:

𝐷𝑒𝑠𝑖𝑟𝑒𝑑 𝑜𝑢𝑡𝑝𝑢𝑡 𝑄𝐿
𝐶𝑂𝑃𝑅 = =
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛𝑝𝑢𝑡 𝑊𝑛𝑒𝑡
1st law: 𝑊𝑛𝑒𝑡 = 𝑄𝐻 − 𝑄𝐿

𝑄𝐿 1
𝐶𝑂𝑃𝑅 = =
𝑄𝐻 − 𝑄𝐿 𝑄𝐻 − 1
𝑄𝐿

For heat pump:

𝐷𝑒𝑠𝑖𝑟𝑒𝑑 𝑜𝑢𝑡𝑝𝑢𝑡 𝑄𝐻
𝐶𝑂𝑃𝐻𝑃 = =
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛𝑝𝑢𝑡 𝑊𝑛𝑒𝑡

𝑄𝐻 1
𝐶𝑂𝑃𝐻𝑃 = =
𝑄𝐻 − 𝑄𝐿 1 − 𝑄𝐿
𝑄 𝐻

𝐶𝑂𝑃𝐻𝑃 = 𝐶𝑂𝑃𝑅 + 1

The Second Law of Thermodynamics:

Kelvin–Planck Statement
It is impossible for any device that operates on a cycle to receive heat from a single
reservoir and produce a net amount of work.

Heat Source, TH

QH

Heat
Wnet
Engine

W ≠ QH

In effect, it states that it is impossible to construct a heat engine that operates in a cycle, receives
a given amount of heat from a high-temperature body, and does an equal amount of work.
It must reject heat. i.e. the thermal efficiency cannot be 100%.

The Second Law of Thermodynamics:

Clausius Statement

It is impossible to construct a device that operates in a cycle and produces no effect other
than the transfer of heat from a lower-temperature body to a higher-temperature body.
 Heat Sink, TH

QH

Refrigerator

QL

Heat Source, TL

In effect, it states that it is impossible to construct a refrigerator that operates without an input
of work.

The Kelvin–Planck and the Clausius statements are equivalent in their consequences, and either
statement can be used as the expression of the second law of thermodynamics. The second law
of thermodynamics is based on experimental observations.

REVERSIBLE AND IRREVERSIBLE PROCESSES

The second law of thermodynamics states that no heat engine can have an efficiency of 100
percent. Then one may ask, What is the highest efficiency that a heat engine can possibly have?
Before we can answer this question, we need to define an idealized process first, which is called
the reversible process.

A reversible process is defined as a process that can be reversed without leaving any trace on
the surroundings.
Processes that are not reversible are called irreversible processes.

Some Irreversibilities:
1. Friction
2. Unrestrained Expansion
3. Heat Transfer Through a Finite Temperature Difference
4. Mixing of Two Different Substances
5. Combustion

A process is called internally reversible if no irreversibilities ocur within the boundaries of

the system during the process.
A process is called externally reversible if no irreversibilities occur outside the system
boundaries during the process.
THE CARNOT CYCLE

If the efficiency of all heat engines is less than 100%, what is the most efficient cycle we can
have?
Let us assume that this heat engine, which operates between the given hightemperature and
low-temperature reservoirs, does so in a cycle in which every process is reversible. If every
process is reversible, the cycle is also reversible; and if the cycle is reversed, the heat engine
becomes a refrigerator. In the next section we will show that this is the most efficient cycle that
can operate between two constant-temperature reservoirs. It is called the Carnot cycle and is
named after a French engineer, Nicolas Leonard Sadi Carnot (1796–1832), who expressed the
foundations of the second law of thermodynamics in 1824.

Being a reversible cycle, the Carnot cycle is the most efficient cycle operating between two
specified temperature limits. Even though the Carnot cycle cannot be achieved in reality, the
efficiency of actual cycles can be improved by attempting to approximate the Carnot cycle
more closely.

Consider the working fluid to be a pure substance, such as steam.

The first process occurs in the boiler. Heat is transferred from the high-temperature reservoir
to the water. For this process to be a reversible heat transfer, the temperature of the water must
be only infinitesimally lower than the temperature of the reservoir. This result implies a
reversible isothermal process.

The next process occurs in the turbine without heat transfer and is therefore adiabatic. Since all
processes in the Carnot cycle are reversible, this must be a reversible adiabatic process.
In the next process, heat is rejected at the condenser, from the working fluid to the low-
temperature reservoir. This must be a reversible isothermal process in which the temperature
of the working fluid is infinitesimally higher than that of the low-temperature reservoir.

The final process occurs in the pump. It is a reversible adiabatic process.

If every process could be reversed in which case it would become a refrigerator. The
refrigerator is shown by the dotted arrows and text in parentheses.

Carnot Efficiency:

𝑄𝐿
𝜂𝐶 = 1 − = 𝜓(𝑇𝐿 , 𝑇𝐻 )
𝑄𝐻

Carnot principles

1. The efficiency of an irreversible heat engine is always less than the efficiency of a
reversible one operating between the same two reservoirs.
2. The efficiencies of all reversible heat engines operating between the same two
reservoirs are the same.
Lord Kelvin first proposed:
𝑄𝐿 𝑇𝐿
=
𝑄𝐻 𝑇𝐻

This temperature scale is called the Kelvin scale, and the temperatures on this scale are called
absolute temperatures.

Thus, Carnot efficiency:

𝑇𝐿
𝜂𝐶 = 1 −
𝑇𝐻

The thermal efficiencies of actual and reversible heat engines operating between the same
temperature limits compare as follows:

< 𝜂𝐶 𝑖𝑟𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒
𝜂𝑡ℎ {= 𝜂𝐶 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒
> 𝜂𝐶 𝑖𝑚𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒

Carnot refrigerator and heat pump:

1
𝐶𝑂𝑃𝑅,𝐶𝑎𝑟𝑛𝑜𝑡 =
𝑇𝐻
𝑇𝐿 − 1
 1
𝐶𝑂𝑃𝐻𝑃,𝐶𝑎𝑟𝑛𝑜𝑡 =
𝑇
1 − 𝑇𝐿
𝐻

These are the highest coefficients of performance that a refrigerator or a heat pump operating
between the temperature limits of TL and TH can have.
< 𝐶𝑂𝑃𝑅,𝐶𝑎𝑟𝑛𝑜𝑡 𝑖𝑟𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒
𝐶𝑂𝑃𝑅 {= 𝐶𝑂𝑃𝑅,𝐶𝑎𝑟𝑛𝑜𝑡 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒
> 𝐶𝑂𝑃𝑅,𝐶𝑎𝑟𝑛𝑜𝑡 𝑖𝑚𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒

(Similarly, same for COPHP )

Example

A heat engine has a heat rate input of 1 MW from the heat source at 550oC and rejecting some
heat to surrounding at 300K while producing power of 450 kW. Determine the heat rate
rejected, thermal efficiency and also compare with the Carnot heat engine that’s works between
the same heat source and sink?

Solution:

𝑊̇𝑛𝑒𝑡 = 𝑄̇𝐻 − 𝑄̇𝐿

450 = 1000 − 𝑄̇𝐿

 𝑄̇𝐿 = 550 𝑘𝑊
𝑊̇𝑛𝑒𝑡 450
𝜂𝑡ℎ = = = 0.45
𝑄̇𝐻 1000

Carnot heat engine for the same heat source and sink:

𝑇𝐿 300
𝜂𝐶 = 1 − =1− = 0.635
𝑇𝐻 550 + 273

Power produced for 1 MW energy input:

𝑊̇𝑛𝑒𝑡 = 𝜂𝐶 𝑄̇𝐻 = 0.635 ∙ 1000 = 635 𝑘𝑊

𝑄̇𝐿 = 𝑄̇𝐻 − 𝑊̇𝑛𝑒𝑡 = 1000 − 635 = 365 𝑘𝑊

Actual Carnot
𝜂𝑡ℎ 0.45 0.635
𝑄̇𝐿 550 kW 365 kW
𝑄̇𝐻 1000 kW 1000 kW
𝑊̇𝑛𝑒𝑡 450 kW 635 kW
Example
An air conditioner removes 4 kW amount of heat rate from a room. The temperatures of the
room and the surrounding are 24oC and 35oC, respectively. Determine the necessary power
input?

Solution

Room 𝑄̇𝐿 𝑄̇𝐻

Air Surrounding

24oC Conditioner 35oC

𝑊̇

The minimum power input will be possible for Carnot refrigeration basis.

1 1
𝐶𝑂𝑃𝑅,𝐶𝑎𝑟𝑛𝑜𝑡 = = = 27
𝑇𝐻 35 + 273
𝑇𝐿 − 1 24 + 273 − 1

𝑄̇𝐿 4
𝑊̇ = = = 0.15 𝑘𝑊
𝐶𝑂𝑃𝑅 27

Example
An inventor claims to have developed a refrigeration system that removes heat from the closed
region at -5°C and transfers it to the surrounding air at 25°C while maintaining a COP of 6.5.
Is this claim reasonable? Why?

Solution
Atmosphere, TH

𝑄̇𝐻

Refrigerator 𝑊̇
system

𝑄̇𝐿
Cooled region, TL
TH = 25 oC = 298.15 K, TL = -5 oC = 268.15 K

COP = 6.5?

For Carnot refrigerator:

1 1
𝐶𝑂𝑃𝑅,𝐶𝑎𝑟𝑛𝑜𝑡 = = = 8.938
𝑇𝐻 298.15
𝑇𝐿 − 1 268.15
−1

Since COP < COPC it is possible.

ENTROPY
The second law of thermodynamics leads to the Clausius inequality:

𝛿𝑄
∮ ≤0
𝑇

It was first stated by the German physicist R. J. E. Clausius (1822–1888). This inequality is
valid for all cycles, reversible or irreversible.

Consider Carnot Heat Engine:

TH

QH
Wreversible

QL
TL

𝛿𝑄 𝑄𝐻 𝑄𝐿
∮( ) = − =0
𝑇 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑇𝐻 𝑇𝐿

𝛿𝑄
For all reversible heat engines ∮ 𝛿𝑄 ≥ 0  ∮ =0
𝑇
 𝛿𝑄
For irreversible heat engines ∮ 𝛿𝑄 > 0  ∮ <0
𝑇

𝛿𝑄
For reversible refrigeration cycles ∮ 𝛿𝑄 ≤ 0  ∮ =0
𝑇

𝛿𝑄
For irreversible refrigerator ∮ 𝛿𝑄 < 0  ∮ <0
𝑇

𝐹𝑜𝑟 𝑎𝑙𝑙 𝑐𝑦𝑐𝑙𝑒𝑠:

𝛿𝑄
∮ ≤0
𝑇

ENTROPY—A PROPERTY OF A SYSTEM

(A, B, C paths are reversible)

For the cycle A-B:

2 1
𝛿𝑄 𝛿𝑄 𝛿𝑄
∮ =0=∫ ( ) +∫ ( )
𝑇 1 𝑇 𝐴 2 𝑇 𝐵

For the cycle C-B:

2 1
𝛿𝑄 𝛿𝑄 𝛿𝑄
∮ =0=∫ ( ) +∫ ( )
𝑇 1 𝑇 𝐶 2 𝑇 𝐵

Subtracting:
1 1
𝛿𝑄 𝛿𝑄
∫ ( ) =∫ ( )
2 𝑇 𝐴 2 𝑇 𝐶
 𝛿𝑄
Therefore ∫ 𝑇
will be same for all reversible paths, and depends only on initial and final
states. This quantity is independent of the path, it is therefore a property. This property is called
entropy and is designated S.

𝛿𝑄
𝑑𝑆 ≡ ( ) 𝐸𝑛𝑡𝑟𝑜𝑝𝑦
𝑇 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒

2
𝛿𝑄
𝑆2 − 𝑆1 = ∫ ( )
1 𝑇 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒

Specific entropy:

𝑆
𝑠=
𝑚
In the saturated region:

𝑠 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔

For reversible isothermal process:

2
𝛿𝑄 1 2 𝑄1−2
∆𝑆 = ∫ ( ) = ∫ 𝛿𝑄 =
1 𝑇𝑜 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑇𝑜 1 𝑇𝑜
THE INCREASE OF ENTROPY PRINCIPLE
Consider a cycle that is made up of two processes: process 1-2, which is arbitrary (reversible
or irreversible), and process 2-1, which is internally reversible, as shown in Figure.

From the Clausius inequality:

𝛿𝑄
∮ ≤0
𝑇

2 1
𝛿𝑄 𝛿𝑄
∫ +∫ ( ) ≤0
1 𝑇 2 𝑇 𝑡𝑒𝑟𝑠𝑖𝑛𝑖𝑟
2
𝛿𝑄
∫ + 𝑆1 − 𝑆2 ≤ 0
1 𝑇
2
𝛿𝑄
𝑆2 − 𝑆1 ≥ ∫
1 𝑇

𝛿𝑄
𝑑𝑆 ≥
𝑇
2
𝛿𝑄
∆𝑆𝑠𝑦𝑠𝑡 = 𝑆2 − 𝑆1 = ∫ + 𝑆𝑔𝑒𝑛
1 𝑇

Sgen = entropy generation, it is always positive, or zero for only reversible process.

𝑆𝑔𝑒𝑛 ≥ 0
 ∆𝑆𝑖𝑠𝑜𝑙𝑎𝑡𝑒𝑑 = ∆𝑆𝑡𝑜𝑡𝑎𝑙 ≥ 0

The entropy of an isolated system during a process always increases or, in the limiting case of
a reversible process, remains constant. In other words, it never decreases. This is known as the
increase of entropy principle.

Surroundings Q

System

𝑆𝑔𝑒𝑛 = ∆𝑆𝑡𝑜𝑡 = ∆𝑆𝑠𝑦𝑠𝑡 + ∆𝑆𝑠𝑢𝑟𝑟 ≥ 0

>0 𝑖𝑟𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒
𝑆𝑔𝑒𝑛 {= 0 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒
<0 𝑖𝑚𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒

ISENTROPIC PROCESSES
Reversible + adiabatic  isentropic process

S = 0  S2 = S1

𝛿𝑄
𝑑𝑆 ≡ ( )
𝑇 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒

(𝛿𝑄)𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 = 𝑇𝑑𝑆
THE THERMODYNAMIC PROPERTY RELATION
The differential form of the conservation of energy equation for a closed stationary system
containing a simple compressible substance can be expressed for an internally reversible
process as

𝛿𝑄 = 𝑑𝑈 + 𝛿𝑊

Also,

𝛿𝑄 = 𝑇𝑑𝑆 𝑎𝑛𝑑 𝛿𝑊 = 𝑃𝑑𝑉

Combine:

𝑇𝑑𝑆 = 𝑑𝑈 + 𝑃𝑑𝑉

Enthalpy: H = U + PV  dH = dU + PdV + VdP

 TdS = dH - VdP

ENTROPY CHANGE OF LIQUIDS AND SOLIDS

Recall that liquids and solids can be approximated as incompressible substances since their
specific volumes remain nearly constant during a process.

dV  0  TdS = dU veya Tds = du 

𝑑𝑢 𝐶𝑑𝑇
𝑑𝑠 = =
𝑇 𝑇
2
𝑑𝑇 𝑇2
𝑠2 − 𝑠1 = ∫ 𝐶(𝑇) ≅ 𝐶𝑜𝑟𝑡 𝑙𝑛
1 𝑇 𝑇1

THE ENTROPY CHANGE OF IDEAL GASES

𝑑𝑣
𝑇𝑑𝑠 = 𝑑𝑢 + 𝑃𝑑𝑣 and Pv = RT  𝑇𝑑𝑠 = 𝐶𝑉 𝑑𝑇 + 𝑅 
𝑣

2
𝑑𝑇 𝑣2
𝑠2 − 𝑠1 = ∫ 𝐶𝑉 (𝑇) + 𝑅𝑙𝑛
1 𝑇 𝑣1

Tds = dh - vdP ve Pv = RT 
2
𝑑𝑇 𝑃2
𝑠2 − 𝑠1 = ∫ 𝐶𝑃 (𝑇) − 𝑅𝑙𝑛
1 𝑇 𝑃1
For constant specific heats:

𝑇2 𝑣2
𝑠2 − 𝑠1 = 𝐶𝑉,𝑎𝑣𝑔 𝑙𝑛 + 𝑅𝑙𝑛 (𝑘𝐽⁄𝐾𝑔𝐾)
𝑇1 𝑣1

𝑇2 𝑃2
𝑠2 − 𝑠1 = 𝐶𝑃,𝑎𝑣𝑔 𝑙𝑛 − 𝑅𝑙𝑛 (𝑘𝐽⁄𝐾𝑔𝐾)
𝑇1 𝑃1

Variable Specific Heats (Exact Analysis)

When the temperature change during a process is large and the specific heats of the ideal gas
vary nonlinearly within the temperature range, the assumption of constant specific heats may
lead to considerable errors in entropy-change calculations.

Instead of performing CV(T) and CP(T) integrals each time, it is convenient to perform these
integrals once and tabulate the results. For this purpose, we choose absolute zero as the
reference temperature and define a function s° as

𝑇
𝑜
𝑑𝑇
𝑠 = ∫ 𝐶𝑃 (𝑇)
0 𝑇

Thus
2
𝑑𝑇
∫ 𝐶𝑃 (𝑇) = 𝑠2𝑜 − 𝑠10
1 𝑇
As a result:

𝑃2
𝑠2 − 𝑠1 = 𝑠2𝑜 − 𝑠10 − 𝑅𝑙𝑛 (𝑘𝐽⁄𝐾𝑔𝐾)
𝑃1

Isentropic Processes of Ideal Gases

Specific heat ratio:

𝐶𝑃
𝑘=
𝐶𝑉

𝑅 𝑘𝑅
𝐶𝑃 − 𝐶𝑉 = 𝑅, 𝐶𝑉 = , 𝐶𝑃 =
𝑘−1 𝑘−1
Assume constant CP and CV, for isentropic process:

𝑇2 𝑣2
𝑠2 − 𝑠1 = 𝐶𝑉 𝑙𝑛 + 𝑅𝑙𝑛 = 0
𝑇1 𝑣1

Which can be rearranged as:

𝑇2 𝑣1 𝑘−1
=( )
𝑇1 𝑣2
Similarly:

𝑇2 𝑃2
𝑠2 − 𝑠1 = 𝐶𝑃 𝑙𝑛 − 𝑅𝑙𝑛 = 0
𝑇1 𝑃1

𝑇2 𝑃2 (𝑘−1)⁄𝑘
=( )
𝑇1 𝑃1

Therefore:

𝑇2 𝑣1 𝑘−1 𝑃2 (𝑘−1)⁄𝑘
=( ) =( )
𝑇1 𝑣2 𝑃1

or

𝑃2 𝑣1 𝑘
( )=( )
𝑃1 𝑣2

This can be expressed in more compact form as:

𝑃𝑣 𝑘 = 𝑐𝑜𝑛𝑠𝑡

Isentropic Process-Variable Specific Heats

𝑃2
𝑠2 − 𝑠1 = 𝑠2𝑜 − 𝑠10 − 𝑅𝑙𝑛 =0
𝑃1

Or

𝑃2
𝑠2𝑜 = 𝑠10 − 𝑅𝑙𝑛
𝑃1
Can be rearranged as:

𝑃2 𝑠2𝑜 − 𝑠10 exp(𝑠2𝑜 ⁄𝑅 ) 𝑃𝑟2

= 𝑒𝑥𝑝 = =
𝑃1 𝑅 exp(𝑠1𝑜 ⁄𝑅 ) 𝑃𝑟1

𝑃𝑟 = exp(𝑠 𝑜 ⁄𝑅) = 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

Pr is a dimensionless quantity that is a function of temperature only and it can be tabulated

against temperature.

Utilizing the ideal-gas relation:

𝑃1 𝑣1 𝑃2 𝑣2
=
𝑇1 𝑇2

𝑣2 𝑇2 𝑃1 𝑇2 𝑃𝑟1 𝑇2 ⁄𝑃𝑟2
= = =
𝑣1 𝑇1 𝑃2 𝑇1 𝑃𝑟2 𝑇1 ⁄𝑃𝑟2
The quantity T/Pr is a function of temperature only and is defined as relative specific volume
vr. Thus,
𝑣2 𝑣𝑟2
( ) =
𝑣1 𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑣𝑟1

The values of Pr and vr are listed for air in Table A–17.

REVERSIBLE STEADY-FLOW WORK

The energy balance for a steady-flow device undergoing an internally reversible process can
be expressed in differential form as:

𝛿𝑞𝑟𝑒𝑣 − 𝛿𝑤𝑟𝑒𝑣 = 𝑑ℎ + 𝑑𝑘𝑒 + 𝑑𝑝𝑒

Also, by using the relations 𝛿𝑞𝑟𝑒𝑣 = 𝑇𝑑𝑠 and Tds = dh –vdP  𝛿𝑞𝑟𝑒𝑣 = 𝑑ℎ − 𝑣𝑑𝑃

The energy balance eq. becomes:

𝑑ℎ − 𝑣𝑑𝑃 − 𝛿𝑤𝑟𝑒𝑣 = 𝑑ℎ + 𝑑𝑘𝑒 + 𝑑𝑝𝑒

2
𝑤𝑟𝑒𝑣 = − ∫ 𝑣𝑑𝑃 − ∆𝑘𝑒 − ∆𝑝𝑒
1

In case we neclect ke and pe, for the reversible work output associated with an internally
reversible process in a steady-flow device:
2
𝑤𝑟𝑒𝑣 = − ∫ 𝑣𝑑𝑃
1

For an incompressible fluid v = const, thus:

𝑤𝑟𝑒𝑣 = −𝑣(𝑃2 − 𝑃1 ) 𝑝𝑢𝑚𝑝 𝑤𝑜𝑟𝑘

If we do not have neglected ke and pe for incompressible flow:

𝑤𝑟𝑒𝑣 = −𝑣(𝑃2 − 𝑃1 ) − ∆𝑘𝑒 − ∆𝑝𝑒

For the steady flow of a liquid through a device that involves no work interactions (such as a
nozzle or a pipe section), the work term is zero, then,

𝑉22 − 𝑉12
𝑣(𝑃2 − 𝑃1 ) + + 𝑔(𝑧2 − 𝑧1 ) = 0
2

Which is known as the Bernoulli equation in fluid mechanics.

ISENTROPIC EFFICIENCIES OF STEADY-FLOW DEVICES

Isentropic Efficiency of Turbines

𝑎𝑐𝑡𝑢𝑎𝑙 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑤𝑜𝑟𝑘 𝑤𝑎

𝜂𝑇 = =
𝑖𝑑𝑒𝑎𝑙 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑤𝑜𝑟𝑘 𝑤𝑠

1st law: actual turbine: wa = h1 –h2a and ideal turbine: ws = h1 – h2s

ℎ1 − ℎ2𝑎
𝜂𝑇 =
ℎ1 − ℎ2𝑠

Isentropic Efficiencies of Compressors

Isentropic compressor work 𝑤𝑠
𝜂𝐶 = =
Actual compressor work 𝑤𝑎

𝑤𝑠 ℎ2𝑠 − ℎ1
𝜂𝐶 = =
𝑤𝑎 ℎ2𝑎 − ℎ1
Isentropic Efficiencies of Pumps
𝑤𝑠 ℎ2𝑠 − ℎ1 𝑣(𝑃2 − 𝑃1 )
𝜂𝑃 = = =
𝑤𝑎 ℎ2𝑎 − ℎ1 ℎ2𝑎 − ℎ1

Ideal pump work: 𝑤𝑠 = ℎ2𝑠 − ℎ1 = 𝑣(𝑃2 − 𝑃1 ) (incompressible fluid)

Isentropic Efficiency of Nozzles

2
Actual KE at nozzle exit 𝑉2𝑎
𝜂𝑁 = = 2
Isentropic KE at nozzle exit 𝑉2𝑠

1st law (at the inlet ke  0) for actual and isentropic cases:
2
𝑉2𝑎
ℎ1 = ℎ2𝑎 +
2
2
𝑉2𝑠
ℎ1 = ℎ2𝑠 +
2

ℎ1 − ℎ2𝑎
𝜂𝑁 =
ℎ1 − ℎ2𝑠
 ENTROPY BALANCE

Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑆𝑓𝑖𝑛𝑎𝑙 − 𝑆𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑆2 − 𝑆1

Mechanisms of Entropy Transfer, Sin and Sout

1. Heat Transfer
2
𝛿𝑄 𝑄𝑘
𝑆ℎ𝑒𝑎𝑡 = ∫ ≅∑
1 𝑇 𝑇𝑘

If T = const.  Sheat = Q/T

Entropy transfer by work: Swork = 0

2. Mass Flow

𝑆𝑚𝑎𝑠𝑠 = 𝑚𝑠

̇
𝑆𝑚𝑎𝑠𝑠 = 𝑚̇𝑠

Entropy Generation, Sgen

𝑆𝑖𝑛 − 𝑆𝑜𝑢𝑡 + 𝑆𝑔𝑒𝑛 = ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚

̇ − 𝑆𝑜𝑢𝑡
𝑆𝑖𝑛 ̇ + 𝑆𝑔𝑒𝑛
̇ ̇
= ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚

𝑠𝑖𝑛 − 𝑠𝑜𝑢𝑡 + 𝑠𝑔𝑒𝑛 = ∆𝑠𝑠𝑦𝑠𝑡𝑒𝑚

For reversible process: Sgen = 0

Closed Systems
A closed system involves no mass flow across its boundaries

𝑄𝑘
∑ + 𝑆𝑔𝑒𝑛 = ∆𝑆𝑠𝑦𝑠𝑡 = 𝑆2 − 𝑆1
𝑇𝑘

If it is adiabatic, then Q = 0 

𝑆𝑔𝑒𝑛 = ∆𝑆𝑠𝑦𝑠𝑡 = 𝑆2 − 𝑆1

Reversible  Sgen = 0 S2 = S1 (isentropic, S = const)

System + Surroundings

𝑆𝑔𝑒𝑛 = ∆𝑆𝑠𝑦𝑠𝑡 + ∆𝑆𝑠𝑢𝑟𝑟

∆𝑆𝑠𝑦𝑠𝑡 = 𝑚(𝑠2 − 𝑠1 )

𝑄𝑠𝑢𝑟𝑟
∆𝑆𝑠𝑢𝑟𝑟 =
𝑇𝑠𝑢𝑟𝑟

Control Volumes

𝑄𝑘
∑ + ∑ 𝑚𝑖 𝑠𝑖 − ∑ 𝑚𝑒 𝑠𝑒 + 𝑆𝑔𝑒𝑛 = (𝑆2 − 𝑆2 )𝐶𝑉
𝑇𝑘

in the rate form,

𝑄̇𝑘
∑ ̇
+ ∑ 𝑚̇𝑖 𝑠𝑖 − ∑ 𝑚̇𝑒 𝑠𝑒 + 𝑆𝑔𝑒𝑛 ̇
= ∆𝑆𝐶𝑉
𝑇𝑘

Steady-flow process: ̇ =0
∆𝑆𝐶𝑉

𝑄̇𝑘
̇
𝑆𝑔𝑒𝑛 = ∑ 𝑚̇𝑒 𝑠𝑒 − ∑ 𝑚̇𝑖 𝑠𝑖 − ∑
𝑇𝑘

Steady-flow, single-stream:

𝑄̇𝑘
̇
𝑆𝑔𝑒𝑛 = 𝑚̇(𝑠𝑒 − 𝑠𝑖 ) −
𝑇𝑘

Steady-flow, single-stream, adiabatic:

 ̇
𝑆𝑔𝑒𝑛 = 𝑚̇(𝑠𝑒 − 𝑠𝑖 )

T-S diagram of Carnot Cycle

1-4 and 2-3: reversible isothermal

1-2 and 3-4: reversible adiabatic = isentropic

𝑊𝑛𝑒𝑡 = 𝑄𝐻 − 𝑄𝐿

𝑇𝐿
𝜂𝐶 = 1 −
𝑇𝐻

Example

Water vapor enters a turbine at 6 MPa and 400°C, and leaves the turbine at 100 kPa with the
same specific entropy as that at the inlet. Calculate the difference between the specific enthalpy
of the water at the turbine inlet and exit?

Solution

From steam tables:

T
P1 = 6 MPa  h1 = 3178.3 kJ/kg
 (Table A - 6) 1
T1 = 400C  s1 = 6.5432 kJ/kg  K

The final state is at saturated region, for 100 kPa, sf and sg

from (Table A-5) 2
s
s2 − s f (6.5432 − 1.3028) kJ/kg  K
x2 = = = 0.8653
s fg 6.0562 kJ/kg  K
h2 = h f + x 2 h fg = 417.51 + (0.8653)(2257.5) = 2370.9 kJ/kg

Enthalpy change:
h = h2 − h1 = 2370 .9 − 3178 .3 = −807.4 kJ/kg

Example

A heavily insulated piston–cylinder device contains 0.02 m3 of steam at 300 kPa and 200°C.
Steam is now compressed in a reversible manner to a pressure of 1.2 MPa. Determine the work
done on the steam during this process?
Solution

Isentropic process: s2 = s1 (Tables A-4 ---- A-6),

v 1 = 0.71643 m 3 /kg
P1 = 300 kPa 
 u1 = 2651.0 kJ/kg
T1 = 200C 
s1 = 7.3132 kJ/kg  K
P2 = 1.2 MPa 
 u 2 = 2921.6 kJ/kg
s 2 = s1 
H2O

300 kPa
V 0.02 m 3
m= = = 0.02792 kg
v 1 0.71643 m 3 /kg
Closed adiabatic system, 1st law:
E − Eout = Esystem
in
  
Net energy transfer Change in internal, kinetic,
by heat, work,and mass potential, etc. energies

Wb,in = U = m(u2 − u1 )

Wb,in = m(u 2 − u1 ) = (0.02792 kg )(2921.6 − 2651 .0 ) kJ/kg = 7.55 kJ

Example

Steam enters a steady-flow adiabatic nozzle with a low inlet velocity as a saturated vapor at 6
MPa and expands to 1.2 MPa.
(a) Under the conditions that the exit velocity is to be the maximum possible value, sketch the
T-s diagram with respect to the saturation lines for this process.
(b) Determine the maximum exit velocity of the steam, in m/s.

Solution

P1 = 6000 kPa  h1 = 2784.6 kJ/kg 6000

T
 (Table A - 5)
x1 = 1  s1 = 5.8902 kJ/kg  K kPa
1200
1 kPa
For the maximum exit velocity, the process should
2
be isentropic:
(Table A-6) s

s2 − s f
5.8902 − 2.2159
P2 = 1200 kPa x2 =
= = 0.8533
 s fg 4.3058

s 2 = s1 = 5.8902 kJ/kg  K  h2 = h f + xh fg = 798.33 + 0.8533  1985.4 = 2492.5 kJ/kg

1st law:
 E − E = E system0 (steady) =0
inout
   
Rate of net energy transfer Rate of change in internal, kinetic,
by heat, work,and mass potential, etc. energies

E in = E out
 V2   V2 
m  h1 + 1  = m  h2 + 2  (since W  Q  pe  0)
 2   2 
 
 V 2 − V12 
h1 − h2 =  2 
 2 
 

 V 2 − V12 
h1 − h2 =  2 
 2 
 
0.5

V2 = 
V12 + 2(h1 − h2 ) 0.5   1000 m 2 /s 2
= (0 m/s) 2 + 2(2784.6 − 2492.5) kJ/kg 
 1 kJ/kg



  
= 764.3 m/s

Example

Steam at 3 MPa and 400°C is expanded to 30 kPa in an adiabatic turbine with an isentropic
efficiency of 92 percent. Determine the power produced by this turbine, in kW, when the mass
flow rate is 2 kg/s?

Solution

Mass balance: 1 = m
m 2 = m
 .

1st law:

E − E = E system0 (steady) =0

inout
   
Rate of net energy transfer
by heat, work,and mass
Rate of change in internal, kinetic, P1 = 3 MPa
potential, etc. energies

E in = E out T1 = 400C
m h1 = W a,out + m h2 (since Q  Δke  Δpe  0) Steam
W a,out = m (h1 − h2 ) turbine

T =92%

Using steam tables (Tables A-4 ---- A-6) : P2 = 30

kPa
P1 = 3 MPa  h1 = 3231.7 kJ/kg

T1 = 400C  s1 = 6.9235 kJ/kg  K

s 2s − s f 6.9235 − 0.9441
P2 s = 30 kPa  x 2 s = = = 0.8763
 s 6.8234
s 2 s = s1 fg
 h = h + x h = 289.27 + (0.8763)(2335.3) = 2335.7 kJ/kg
2s f 2 s fg
From the isentropic efficiency definition:

W a ,out =  T W s ,out
=  T m (h1 − h2 s )
= (0.92)(2 kg/s) (3231.7 − 2335.7)kJ/kg
= 1649 kW

Example

The exhaust nozzle of a jet engine expands air at 300 kPa and 180°C adiabatically to 100 kPa.
Determine the air velocity at the exit when the inlet velocity is low and the nozzle isentropic
efficiency is 96 percent?

Solution
Air properties: cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).
Isentropic process:
( k −1) / k
P 
0.4 / 1.4
 100 kPa 
T2 s = T1  2  = (180 + 273 K)  = 331.0 K 300 kPa
 P1   300 kPa 
Air 100 kPa
Mass balance: 1 = m
m 2 = m
 180C

1st law:
E − E = E system0 (steady) =0
inout
  
Rate of net energy transfer Rate of change in internal, kinetic,
by heat, work,and mass potential, etc. energies

E in = E out

 V2   2 
m  h1 + 1  = m  h2 + V 2 
 2   2 
  
V12 V2
h1 + = h2 + 2
2 2
V 22 − V12
h1 − h2 =
2
V − V12
2
c p (T1 − T2 ) = 2 = ke
2
The kinetic energy change for isentropic case:
ke s = c p (T1 − T2 s ) = (1.005 kJ/kg  K)(453 − 331)K = 122.6 kJ/kg

The kinetic energy change for actual case:

ke a =  N ke s = (0.96 )(122 .6 kJ/kg) = 117.7 kJ/kg

Use the energy balance, solve for the velocity:

0.5
  1000 m 2 /s 2 
V2 = (2ke a ) 0.5 = 2(117.7 kJ/kg)   = 485 m/s
  1 kJ/kg 
 
Example
A hot-water stream at 70°C enters an adiabatic mixing chamber with a mass flow rate of 3.6
kg/s, where it is mixed with a stream of cold water at 20°C. If the mixture leaves the chamber
at 42°C, determine
(a) The mass flow rate of the cold water and
(b) The rate of entropy generation during this adiabatic mixing process. Assume all the streams
are at a pressure of 200 kPa.

Solution
Note: T < Tsat @ 200 kPa = 120.21C, all three states are in liquid form, therefore use saturated
values at the same temperature:
Table A-4,
P1 = 200 kPa  h1  h f @70 C = 293.07 kJ/kg
s  s = 0.9551 kJ/kg  K
T1 = 70C  1 f @ 70 C
70C
1
P2 = 200 kPa  h2  h f @ 20 C = 83.91 kJ/kg H2O
s  s = 0.2965 kJ/kg  K 3.6 42C
T2 = 20C  2 f @ 20 C 3
P3 = 200 kPa  h3  h f @ 42 C = 175.90 kJ/kg 20C
s  s 2
T3 = 42C  3 f @ 42 C
= 0.5990 kJ/kg  K

(a)
Mass Balance:
𝑚̇1 + 𝑚̇2 = 𝑚̇3
Energy balance:
E − E out = E system0 (steady) =0
in  
Rate of net energy transfer Rate of change in internal, kinetic,
by heat, work,and mass potential, etc. energies

E in = E out
m h1 + m 2 h2 = m 3h3 (since Q = W = ke  pe  0)

Combine:
 2 h2 = (m
 1h1 + m
m  2 )h3
1 + m


h1 − h3 (293.07 − 175.90)kJ/kg
2 =
m 1 =
m (3.6 kg/s ) = 4.586 kg/s
h3 − h2 (175.90 − 83.91)kJ/kg
3 = m
m 1 + m
 2 = 3.6 + 4.586 = 8.186 kg/s

(b) Entropy Balance:

Sin − Sout + Sgen = Ssystem0 = 0
   
Rate of net entropy transfer Rate of entropy Rate of change
by heat and mass generation of entropy

m 1s1 + m 2 s2 − m 3s3 + Sgen = 0

Sgen = m 3 s3 − m 2 s2 − m 1s1
= (8.186 kg/s )(0.5990 kJ/kg  K ) − (4.586 kg/s )(0.2965 kJ/kg  K ) − (3.6 kg/s )(0.9551 kJ/kg  K )
= 0.1054 kW/K

Example
A 0.18-m3 rigid tank is filled with saturated liquid water at 120°C. A valve at the bottom of
the tank is now opened, and one-half of the total mass is withdrawn from the tank in the liquid
form. Heat is transferred to water from a source at 230°C so that the temperature in the tank
remains constant. Determine
(a) The amount of heat transfer and
(b) The total entropy generation for this process?

Solution

Water properties (Tables A-4 ---- A-6)

v 1 = v f @120 C = 0.001060 m 3 /kg
T1 = 120C 
 u1 = u f @120 C = 503.60 kJ/kg
sat. liquid 
s1 = s f @120 C = 1.5279 kJ/kg  K

Te = 120C  he = h f @120 C = 503.81 kJ/kg


sat. liquid  s e = s f @120 C = 1.5279 kJ/kg  K

Mass Balance:
min − mout = msystem → me = m1 − m2

Energy Balance:
E − Eout = Esystem
in
  
Net energy transfer Change in internal, kinetic,
by heat, work,and mass potential, etc. energies

Qin = me he + m2u2 − m1u1 (since W  ke  pe  0)

The initial and the final state masses inside the tank:
V 0.18 m 3
m1 = = = 169.76 kg
v 1 0.001060 m 3 /kg

m1 = (169.76 kg ) = 84.88 kg = me
1 1
m2 =
2 2
The internal energy and the entropy at the final state:
V 0.18 m 3
v2 = = = 0.002121 m 3 /kg
m2 84.88 kg
v2 −v f 0.002121 − 0.001060
x2 = = = 0.001191
v fg 0.8913 − 0.001060
T2 = 120C  u 2 = u f + x 2 u fg = 503.60 + (0.001191)(2025.3) = 506.01 kJ/kg

x 2 = 0.001191  s 2 = s f + x 2 s fg = 1.5279 + (0.001191)(5.6013) = 1.5346 kJ/kg  K
By using the energy balance equation:
Qin = me he + m2 u 2 − m1u1
= (84.88 kg )(503.81 kJ/kg ) + (84.88 kg )(506.01 kJ/kg ) − (169.76 kg )(503.60 kJ/kg )
= 222.6 kJ
(b) Entropy Balance:
Sin − Sout + Sgen = Ssystem ⎯
⎯→ −me se + Sgen = S tank + Ssource
   
 
Net entropy transfer Entropy Change
by heat and mass generation in entropy

Qsource,out
S gen = me s e + S tank + S source = me s e + (m 2 s 2 − m1 s1 ) −
Tsource
= (84.88 kg )(1.5279 kJ/kg  K ) + (84.88 kg )(1.5346 kJ/kg  K )
− (169.76 kg )(1.5279 kJ/kg  K ) −
222.6 kJ
(230 + 273) K
= 0.1237 kJ/K

EXERGY: WORK POTENTIAL OF ENERGY

Exergy: work potential, which is also called the availability or available energy.

Work = f (initial state, process path, final state)

All the irreversibilities are disregarded in determining the work potential. The system must be
in the dead state at the end of the process to maximize the work output.

A system is said to be in the dead state when it is in thermodynamic equilibrium with the
environment

A system delivers the maximum possible work as it undergoes a reversible process

from the specified initial state to the dead state.

This represents the useful work potential of the system at the specified state and is called
exergy.

Exergy of kinetic energy:

𝑉2
𝑥𝑘𝑒 = 𝑘𝑒 = 𝑘𝑗/𝑘𝑔
2
Exergy of potential energy:

𝑥𝑝𝑒 = 𝑝𝑒 = 𝑔𝑧 𝑘𝑗/𝑘𝑔
REVERSIBLE WORK AND IRREVERSIBILITY

The property exergy serves as a valuable tool in determining the quality of energy and
comparing the work potentials of different energy sources or systems.

Surroundings work:

𝑊𝑠𝑢𝑟𝑟 = 𝑃𝑜 (𝑉2 − 𝑉1 )

The difference between the actual work W and the surroundings work Wsurr is called the
useful work Wu:

𝑊𝑢 = 𝑊 − 𝑊𝑠𝑢𝑟𝑟 = 𝑊 − 𝑃𝑜 (𝑉2 − 𝑉1 )

When a system is expanding and doing work, part of the work done is used to overcome the
atmospheric pressure, and thus Wsurr represents a loss. When a system is compressed,
however, the atmospheric pressure helps the compression process, and thus Wsurr represents a
gain.
Reversible work: Wrev is defined as the maximum amount of useful work that can be produced
(or the minimum work that needs to be supplied) as a system undergoes a process between the
specified initial and final states.

Any difference between the reversible work Wrev and the useful work Wu is due to the
irreversibilities present during the process, and this difference is called irreversibility I. It is
expressed as

Work producing devices:

𝐼 = 𝑊𝑟𝑒𝑣 − 𝑊𝑢
Work consuming devices:
𝐼 = 𝑊𝑢 − 𝑊𝑟𝑒𝑣

Irreversibility can be viewed as the wasted work potential or the lost opportunity to do work.

Reversible process: I=0

SECOND-LAW EFFICIENCY, II

The thermal efficiency and the coefficient of performance for devices as a measure of their
performance. They are defined on the basis of the first law only, and they are sometimes
referred to as the first-law efficiencies. The first law efficiency, however, makes no reference
to the best possible performance, and thus it may be misleading.

For heat engines, define a second-law efficiency, as the ratio of the actual thermal efficiency
to the maximum possible (reversible) thermal efficiency under the same conditions

𝜂𝑡ℎ
𝜂𝐼𝐼 =
𝜂𝑡ℎ,𝑟𝑒𝑣

𝜂𝑡ℎ,𝑟𝑒𝑣 = 𝐶𝑎𝑟𝑛𝑜𝑡 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 (𝜂𝐶 )

The second-law efficiency can also be expressed as the ratio of the useful work output and the
maximum possible (reversible) work output:

Work-producing devices (turbine, etc):

𝑢𝑠𝑒𝑓𝑢𝑙 𝑤𝑜𝑟𝑘 𝑜𝑢𝑡𝑝𝑢𝑡 𝑊𝑢
𝜂𝐼𝐼 = =
𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑤𝑜𝑟𝑘 𝑜𝑢𝑡𝑝𝑢𝑡 𝑊𝑟𝑒𝑣

Work-consuming devices (compressor, etc):

 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑤𝑜𝑟𝑘 𝑖𝑛𝑝𝑢𝑡 𝑊𝑟𝑒𝑣
𝜂𝐼𝐼 = =
𝑎𝑐𝑡𝑢𝑎𝑙 𝑤𝑜𝑟𝑘 𝑖𝑛𝑝𝑢𝑡 𝑊𝑢

For cyclic devices such as refrigerators and heat pumps:

𝐶𝑂𝑃
𝜂𝐼𝐼 =
𝐶𝑂𝑃𝑟𝑒𝑣

EXERGY CHANGE OF A SYSTEM

Closed System Exergy
The second law of thermodynamics states that heat cannot be converted to work entirely, and
thus the work potential of internal energy must be less than the internal energy itself. But how
much less?

To answer that question, we need to consider a stationary closed system at a specified state (P,
T), that undergoes a reversible process to the state of the environment (Po, To). The useful work
delivered during this process is the exergy of the system at its initial state

Consider a piston–cylinder device that contains a fluid of mass m at temperature T and pressure
P.

𝛿𝐸𝑖𝑛 − 𝛿𝐸𝑜𝑢𝑡 = 𝑑𝐸𝑠𝑦𝑠𝑡𝑒𝑚

−𝛿𝑄 − 𝛿𝑊 = 𝑑𝑈

𝛿𝑊 = 𝑃𝑑𝑉 = (𝑃 − 𝑃𝑜 )𝑑𝑉 + 𝑃𝑜 𝑑𝑉 = 𝛿𝑊𝑏,𝑢𝑠𝑒𝑓𝑢𝑙 + 𝑃𝑜 𝑑𝑉

Carnot efficiency:
 𝛿𝑊𝐻𝐸 𝑇𝑜
𝜂𝐶 = = 1−
𝛿𝑄 𝑇

𝑇𝑜 𝑇𝑜
𝛿𝑊𝐻𝐸 = (1 − ) 𝛿𝑄 = 𝛿𝑄 − 𝛿𝑄 = 𝛿𝑄 − (−𝑇𝑜 𝑑𝑆) = 𝛿𝑄 + 𝑇𝑜 𝑑𝑆
𝑇 𝑇

𝛿𝑄 = 𝛿𝑊𝐻𝐸 − 𝑇𝑜 𝑑𝑆

𝛿𝑊𝑡𝑜𝑡𝑎𝑙,𝑢𝑠𝑒𝑓𝑢𝑙 = 𝛿𝑊𝐻𝐸 + 𝛿𝑊𝑏,𝑢𝑠𝑒𝑓𝑢𝑙

𝛿𝑊𝐻𝐸 : due to temperature, 𝛿𝑊𝑏,𝑢𝑠𝑒𝑓𝑢𝑙 : due to pressure

𝛿𝑊𝑡𝑜𝑡𝑎𝑙,𝑢𝑠𝑒𝑓𝑢𝑙 = −𝑑𝑈 − 𝑃𝑜 𝑑𝑉 + 𝑇𝑜 𝑑𝑆

𝛿𝑊𝑡𝑜𝑡𝑎𝑙,𝑢𝑠𝑒𝑓𝑢𝑙 = (𝑈 − 𝑈𝑜 ) + 𝑃𝑜 (𝑉 − 𝑉𝑜 ) − 𝑇𝑜 (𝑆 − 𝑆𝑜 )

Where Wtotal,useful is the total useful work delivered as the system undergoes a reversible process
from the given state to the dead state, which is exergy by definition.
A closed system, in general, may possess kinetic and potential energies, and the total energy
of a closed system is equal to the sum of its internal, kinetic, and potential energies. Noting that
kinetic and potential energies themselves are forms of exergy, the exergy of a closed system of
mass m is
𝑉2
𝑋 = (𝑈 − 𝑈𝑜 ) + 𝑃𝑜 (𝑉 − 𝑉𝑜 ) − 𝑇𝑜 (𝑆 − 𝑆𝑜 ) + 𝑚 + 𝑚𝑔𝑧
2
On a unit mass basis, the closed system (or nonflow) exergy  is expressed as
𝑉2
𝜙 = (𝑢 − 𝑢𝑜 ) + 𝑃𝑜 (𝑣 − 𝑣𝑜 ) − 𝑇𝑜 (𝑠 − 𝑠𝑜 ) + + 𝑔𝑧
2

𝑉22 − 𝑉12
Δ𝑋 = 𝑋2 − 𝑋1 = (𝑈2 − 𝑈1 ) + 𝑃𝑜 (𝑉2 − 𝑉1 ) − 𝑇𝑜 (𝑆2 − 𝑆1 ) + 𝑚 + 𝑚𝑔(𝑧2 − 𝑧1 )
2

𝑉22 − 𝑉12
Δ𝜙 = 𝜙2 − 𝜙1 = (𝑢2 − 𝑢1 ) + 𝑃𝑜 (𝑣2 − 𝑣1 ) − 𝑇𝑜 (𝑠2 − 𝑠1 ) + + 𝑔(𝑧2 − 𝑧1 )
2

Exergy of a Flowing Fluid

Flow work: wflow = Pv

If the atmospheric pressure is Po 
 𝑥𝑓𝑙𝑜𝑤 = 𝑃𝑣 − 𝑃𝑜 𝑣 = (𝑃 − 𝑃𝑜 )𝑣

𝑥𝑓𝑙𝑜𝑤𝑖𝑛𝑔 = 𝑥𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 + 𝑥𝑓𝑙𝑜𝑤

𝑓𝑙𝑢𝑖𝑑 𝑓𝑙𝑢𝑖𝑑

𝑉2
𝑥𝑓𝑙𝑜𝑤𝑖𝑛𝑔 = (𝑢 − 𝑢𝑜 ) + 𝑃𝑜 (𝑣 − 𝑣𝑜 ) − 𝑇𝑜 (𝑠 − 𝑠𝑜 ) + + 𝑔𝑧 + (𝑃 − 𝑃𝑜 )𝑣
𝑓𝑙𝑢𝑖𝑑 2
𝑉2
𝑥𝑓𝑙𝑜𝑤𝑖𝑛𝑔 = (𝑢 + 𝑃𝑣) − (𝑢𝑜 + 𝑃𝑜 𝑣𝑜 ) − 𝑇𝑜 (𝑠 − 𝑠𝑜 ) + + 𝑔𝑧
𝑓𝑙𝑢𝑖𝑑 2
𝑉2
𝑥𝑓𝑙𝑜𝑤𝑖𝑛𝑔 = (ℎ − ℎ𝑜 ) − 𝑇𝑜 (𝑠 − 𝑠𝑜 ) + + 𝑔𝑧
𝑓𝑙𝑢𝑖𝑑 2
Flow exergy:
𝑉2
𝜓 = (ℎ − ℎ𝑜 ) − 𝑇𝑜 (𝑠 − 𝑠𝑜 ) + + 𝑔𝑧
2

𝑉22 − 𝑉12
Δ𝜓 = 𝜓2 − 𝜓1 = (ℎ2 − ℎ1 ) − 𝑇𝑜 (𝑠2 − 𝑠1 ) + + 𝑔(𝑧2 − 𝑧1 )
2

Note that the exergy change of a closed system or a fluid stream represents the maximum
amount of useful work that can be done (or the minimum amount of useful work that needs to
be supplied if it is negative) as the system changes from state 1 to state 2 in a specified
environment, and represents the reversible work Wrev.

EXERGY TRANSFER BY HEAT, WORK, AND MASS

Exergy by Heat Transfer

𝑇𝑜
𝑋ℎ𝑒𝑎𝑡 = (1 − )𝑄
𝑇

Exergy Transfer by Work

𝑊 − 𝑊𝑠𝑢𝑟𝑟 (𝑓𝑜𝑟 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑤𝑜𝑟𝑘𝑠)
𝑋𝑖ş = {
𝑊 (𝑓𝑜𝑟 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑚𝑠 𝑜𝑓 𝑤𝑜𝑟𝑘𝑠)

Exergy Transfer by Mass

𝑋𝑘ü𝑡𝑙𝑒 = 𝑚𝜓
THE DECREASE OF EXERGY PRINCIPLE
AND EXERGY DESTRUCTION

Consider an isolated system shown in Figure:

Energy balance:
𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = ∆𝐸𝑠𝑦𝑠𝑡
 E2 = E1
Entropy balance:
𝑆𝑖𝑛 − 𝑆𝑜𝑢𝑡 + 𝑆𝑔𝑒𝑛 = ∆𝑆𝑠𝑦𝑠𝑡

 Sgen = S2 – S1
Multiplying the second relation by T0 and subtracting it from
the first one gives:
−𝑇𝑜 𝑆𝑔𝑒𝑛 = 𝐸2 − 𝐸1 − 𝑇𝑜 (𝑆2 − 𝑆1 )

Exergy balance:
𝑋2 − 𝑋1 = (𝐸2 − 𝐸1 ) + ⏟
𝑃𝑜 (𝑉2 − 𝑉1 ) − 𝑇𝑜 (𝑆2 − 𝑆1 )
0, 𝑉2 =𝑉1


−𝑇𝑜 𝑆𝑔𝑒𝑛 = 𝑋2 − 𝑋1 ≤ 0

∆𝑋𝑖𝑠𝑜𝑙𝑎𝑡𝑒𝑑 = (𝑋2 − 𝑋1 )𝑖𝑠𝑜𝑙𝑎𝑡𝑒𝑑 ≤ 0

The exergy of an isolated system during a process always decreases (for only a reversible
process, remains constant). This is known as the decrease of exergy principle.

Exergy Destruction

𝑋𝑑𝑒𝑠𝑡𝑟𝑜𝑦𝑒𝑑 = 𝑇𝑜 𝑆𝑔𝑒𝑛 ≥ 0

>0 𝑖𝑟𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒
𝑋𝑑𝑒𝑠𝑡𝑟𝑜𝑦𝑒𝑑 { 0
= 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒
<0 𝑖𝑚𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒
EXERGY BALANCE

𝑋𝑖𝑛 − 𝑋𝑜𝑢𝑡 − 𝑋𝑑𝑒𝑠𝑡𝑟𝑜𝑦𝑒𝑑 = ∆𝑋𝑠𝑦𝑠𝑡𝑒𝑚

In the rate form
𝑋̇𝑖𝑛 − 𝑋̇𝑜𝑢𝑡 − 𝑋̇𝑑𝑒𝑠𝑡𝑟𝑜𝑦𝑒𝑑 = ∆𝑋̇𝑠𝑦𝑠𝑡𝑒𝑚
Unit-mass basis:
𝑥̇ 𝑖𝑛 − 𝑥̇ 𝑜𝑢𝑡 − 𝑥̇ 𝑑𝑒𝑠𝑡𝑟𝑜𝑦𝑒𝑑 = ∆𝑥̇ 𝑠𝑦𝑠𝑡𝑒𝑚

𝑋𝑑𝑒𝑠𝑡𝑟𝑜𝑦𝑒𝑑 = 𝑇𝑜 𝑆𝑔𝑒𝑛 , ̇
𝑋̇𝑑𝑒𝑠𝑡 = 𝑇𝑜 𝑆𝑔𝑒𝑛

For closed system:

𝑋ℎ𝑒𝑎𝑡 − 𝑋𝑤𝑜𝑟𝑘 − 𝑋𝑑𝑒𝑠𝑡 = ∆𝑋𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑋2 − 𝑋1

𝑇𝑜
∑ (1 − ) 𝑄 − [𝑊 − 𝑃𝑜 (𝑉2 − 𝑉1 )] − 𝑇𝑜 𝑆𝑔𝑒𝑛 = 𝑋2 − 𝑋1
𝑇𝑘 𝑘

For control volume:

𝑋𝑖𝑛 − 𝑋𝑜𝑢𝑡 − 𝑋𝑑𝑒𝑠𝑡𝑟𝑜𝑦𝑒𝑑 = ∆𝑋𝑠𝑦𝑠𝑡𝑒𝑚

𝑋ℎ𝑒𝑎𝑡 − 𝑋𝑤𝑜𝑟𝑘 + 𝑋𝑚𝑎𝑠𝑠,𝑖𝑛 − 𝑋𝑚𝑎𝑠𝑠,𝑜𝑢𝑡 − 𝑋𝑑𝑒𝑠𝑡 = (𝑋2 − 𝑋1 )𝐶𝑉

𝑇𝑜
∑ (1 − ) 𝑄 − [𝑊 − 𝑃𝑜 (𝑉2 − 𝑉1 )] + ∑ 𝑚𝑖 𝜓𝑖 − ∑ 𝑚𝑒 𝜓𝑒 − 𝑋𝑑𝑒𝑠𝑡 = (𝑋2 − 𝑋1 )𝐶𝑉
𝑇𝑘 𝑘

Steady-state:

𝑇𝑜
∑ (1 − ) 𝑄̇ − 𝑊̇ + ∑ 𝑚̇𝑖 𝜓𝑖 − ∑ 𝑚̇𝑒 𝜓𝑒 − 𝑋̇𝑑𝑒𝑠𝑡 = 0
𝑇𝑘 𝑘

Single-stream 
𝑚̇𝑖 = 𝑚̇𝑒 = 𝑚̇

Reversible: 𝑋̇𝑑𝑒𝑠𝑡 = 0  𝑊̇ = 𝑊̇𝑟𝑒𝑣 reversible work 

 𝑇𝑜
𝑊̇𝑟𝑒𝑣 = 𝑚̇(𝜓1 − 𝜓2 ) + ∑ (1 − ) 𝑄̇
𝑇𝑘 𝑘
Adiabatic 

𝑊̇𝑟𝑒𝑣 = 𝑚̇(𝜓1 − 𝜓2 )

Second-Law Efficiency of Steady-Flow Devices

Turbine:
𝑊 ℎ1 − ℎ2 𝑇𝑜 𝑠𝑔𝑒𝑛
𝜂𝐼𝐼,𝑡𝑢𝑟𝑏 = = =1−
𝑊𝑟𝑒𝑣 𝜓1 − 𝜓2 𝜓1 − 𝜓2

Compressor:
𝑊𝑟𝑒𝑣 𝜓2 − 𝜓1 𝑇𝑜 𝑠𝑔𝑒𝑛
𝜂𝐼𝐼,𝑐𝑜𝑚𝑝 = = =1−
𝑊𝑖𝑛 ℎ2 − ℎ1 ℎ2 − ℎ1

Heat exchanger:

𝑚̇𝑐𝑜𝑙𝑑 (𝜓4 − 𝜓3 ) ̇
𝑇𝑜 𝑆𝑔𝑒𝑛
𝜂𝐼𝐼,𝐻𝑋 = =1−
𝑚̇ℎ𝑜𝑡 (𝜓1 − 𝜓2 ) 𝑚̇ℎ𝑜𝑡 (𝜓1 − 𝜓2 )

̇
𝑆𝑔𝑒𝑛 = 𝑚̇ℎ𝑜𝑡 (𝑠2 − 𝑠1 ) + 𝑚̇𝑐𝑜𝑙𝑑 (𝑠4 − 𝑠3 )

Example
Which has the capability to produce the most work in a closed system: 1 kg of steam at 800
kPa and 180°C or 1 kg of R-134a at 800 kPa and 180°C? (T0 = 25°C, P0 = 100 kPa).
Solution
Read values from the tables for the given state and the dead state.
For steam:
u = 2594.7 kJ/kg
P = 800 kPa 
 v = 0.24720 m /kg Steam
3
(Table A - 6)
T = 180C 
s = 6.7155 kJ/kg  K
1 kg
u 0  u f @ 25C = 104.83 kJ/kg
T0 = 25C 
 v 0  v f @ 25C = 0.001003 m /kg
3
(Table A - 4)
P0 = 100 kPa 
s 0  s f @ 25C = 0.3672 kJ/kg  K

Steam exergy:
 = mu − u 0 + P0 (v − v 0 ) − T0 ( s − s 0 )
  1 kJ 
(2594.7 − 104.83)kJ/kg + (100 kPa)(0.24720 − 0.001003)m /kg  
3
= (1 kg)   1 kPa  m 3 
− (298 K)(6.7155 − 0.3672)kJ/kg  K 
= 622.7 kJ

For R-134a:
u = 386.99 kJ/kg
P = 800 kPa 
 v = 0.044554 m /kg
3
(Table A - 13)
T = 180C 
s = 1.3327 kJ/kg  K R-134a
u 0  u f @ 25C = 85.85 kJ/kg
T0 = 25C  1 kg
 v 0  v f @ 25C = 0.0008286 m /kg
3
(Table A - 11)
P0 = 100 kPa 
s 0  s f @ 25C = 0.32432 kJ/kg  K

 = mu − u 0 + P0 (v − v 0 ) − T0 ( s − s 0 )
  1 kJ 
(386.99 − 85.85)kJ/kg + (100 kPa)(0.044554 − 0.0008286)m /kg  
3
= (1 kg)   1 kPa  m 3 
− (298 K)(1.3327 − 0.32432)kJ/kg  K 
= 5.02 kJ
Steam has more work potential than R-134a.

Example
An insulated piston-cylinder device contains 0.05 m3 of saturated refrigerant-134a vapor at 0.8
MPa pressure. The refrigerant is now allowed to expand in a reversible manner until the
pressure drops to 0.2 MPa. Determine the change in (he exergy of the refrigerant during this
process and the reversible work. Assume the surroundings to be at 25°C and 100 kPa.

Solution
Reversible adiabatic (isentropic) process: s2 = s1. Using refrigerant R-134a tables (Tables A-
11 ---- A-13),
 v 1 = v g @ 0.8 MPa = 0.02562 m 3 / kg
P1 = 0.8 MPa 
 u1 = u g @ 0.8 MPa = 246.79 kJ/kg
sat. vapor s =s
1 g @ 0.8 MPa = 0.9183 kJ/kg  K

R-134a
V 0.05 m 3
m= = = 1.952 kg 0.8 MPa
v 1 0.02562 m 3 / kg
s2 − s f 0.9183 − 0.15457
x2 = = = 0.9753
s fg 0.78316
P2 = 0.2 MPa 
 v 2 = v f + x 2v fg = 0.0007533 + 0.099867  (0.099867 − 0.0007533) = 0.09741 m /kg
3
s 2 = s1  u = u + x u = 38.28 + 0.9753  186.21 = 219.88 kJ/kg
2 f 2 fg

For isentropic process, the destroyed exergy is zero. We can determine Wrev,out from the
exergy balance,
X − X out − X destroyed0 (reversible) = X system
in    
Net exergy transfer Exergy Change
by heat, work,and mass destruction in exergy

- Wrev,out = X 2 − X1
Wrev,out = X1 − X 2
= 1 −  2

In this case, the exergy change and the reversible work are the same:

 
W rev,out =  1 −  2 = m (u1 − u 2 ) − T0 ( s1 − s 2 )  + P0 ( v1 − v 2 ) = m(u1 − u 2 ) + P0 (v 1 − v 2 )
0

= (1.952 kg)[(246.79 − 219.88) kJ/kg + (100 kPa)(0.02562 − 0.09741)m 3 / kg[kJ/kPa  m 3 ]

= 38.5 kJ

Example
Air is compressed steadily by an 8-kW compressor from 100 kPa and 17°C to 600 kPa and
167°C at a rate of 2.1 kg/min. Neglecting the changes in kinetic and potential energies,
determine
(a) The increase in the exergy of the air and
(b) The rate of exergy destroyed during this process. Assume the surroundings to be at 17°C.

Solution
Air gas constant: R = 0.287 kJ/kg.K 600 kPa
Air tables (Table A-17)
167C
T1 = 290 K ⎯
⎯→ h1 = 290.16 kJ / kg
s1o = 1.66802 kJ / kg  K
AIR
T2 = 440 K ⎯
⎯→ h2 = 441.61 kJ / kg
8 kW
s2o = 2.0887 kJ / kg  K

100
kPa
 Increase in exergy =  2 − 1
0 0
= [(h2 − h1 ) + ke  + pe  − T0 ( s 2 − s1 )]
= (h2 − h1 ) − T0 ( s 2 − s1 )

P2
s 2 − s1 = ( s 2o − s1o ) − R ln
P1
600 kPa
= (2.0887 − 1.66802)kJ/kg  K - (0.287 kJ/kg  K) ln
100 kPa
= −0.09356 kJ/kg  K
Increase in exergy =  2 − 1
= (441.61− 290.16)kJ/kg - (290 K)(−0.09356 kJ/kg  K) 
= 178.6 kJ/kg

Reversible power:
W rev,in = m ( 2 − 1 ) = (2.1/ 60 )(178.60 ) = 6.25 kW
(b) Exergy destroyed:
X destroyed = W in − W rev,in = 8 − 6.25 = 1.75 kW

Example
Air enters a nozzle steadily at 200 kPa and 65°C with a velocity of 35 m/s and exits at 95 kPa
and 240 m/s. The heat loss from the nozzle to the surrounding medium at 17°C is estimated to
be 3 kJ/kg. Determine
(a) The exit temperature and
(b) The exergy destroyed during this process.

Solution
Air gas constant: R = 0.287 kJ/kg.K (Table A-1). Air tables (Table A-17):
T1 = 338 K ⎯
⎯→ h1 = 338.40 kJ/kg
s1o = 1.8219 kJ/kg  K
Energy balance:
E − E out = E system0 (steady) =0
in  
Rate of net energy transfer
by heat, work,and mass
Rate of change in internal, kinetic,
potential, etc. energies
3 kJ/kg
Ein = E out
m (h1 + V12 / 2) = m (h2 + V22 /2) + Q out 35 m/s AIR 240 m/s

V 22 − V12
0 = q out + h2 − h1 +
2

V22 − V12 (240 m/s) 2 − (35 m/s) 2  1 kJ/kg 

h2 = h1 − q out − = 338.40 − 3 −   = 307.21 kJ/kg
2 2  1000 m 2 / s 2 

For h2 value look in Table A-17  T2 = 307.0 K = 34.0 C and s 2o = 1.7251 kJ/kg  K
(a) Exergy destroyed: X destroyed = T0 S gen

Sgen is determined from entropy balance:

S in − S out + S gen = S system0 = 0

   
Rate of net entropy transfer Rate of entropy Rate of change
by heat and mass generation of entropy

Q out
ms1 − m s 2 − + S gen = 0
Tb,surr
Q
S gen = m (s 2 − s1 ) + out
Tsurr
P2 95 kPa
s air = s 2o − s1o − R ln = (1.7251 − 1.8219)kJ/kg  K − (0.287 kJ/kg  K) ln = 0.1169 kJ/kg  K
P1 200 kPa

Thus exergy destroyed:

x destroyed = T0 s gen = Tsurr s gen

 q   3 kJ/kg 
= T0  s 2 − s1 + surr  = (290 K) 0.1169 kJ/kg  K +
  = 36.9 kJ/kg
 Tsurr   290 K 

GAS POWER CYCLES

The study of power cycles is an important part of thermodynamics. The cycles encountered in
actual devices are difficult to analyze because of the presence of complicating effects, such as
friction. To make an analytical study of a cycle feasible, we have to keep the complexities at a
manageable level and utilize some idealizations. When the actual cycle is stripped of all the
internal irreversibilities and complexities, we end up with a cycle that resembles the actual
cycle closely but is made up totally of internally reversible processes. Such a cycle is called an
ideal cycle.

Thermal efficiency:
𝑊𝑛𝑒𝑡
𝜂𝑡ℎ =
𝑄𝑖𝑛

The idealizations and simplifications commonly employed:

1. The cycle does not involve any friction. Therefore, the working fluid does not
experience any pressure drop as it flows in pipes or devices such as heat exchangers.
2. All expansion and compression processes take place in a quasiequilibrium manner.
3. The pipes connecting the various components of a system are well insulated, and heat
transfer through them is negligible.
4. Neglecting the changes in kinetic and potential energies of the working fluid.
THE CARNOT CYCLE

AIR-STANDARD ASSUMPTIONS

1. The working fluid is air, which continuously circulates in a closed loop and always
behaves as an ideal gas.
2. All the processes that make up the cycle are internally reversible.
3. The combustion process is replaced by a heat-addition process from an external source.
4. The exhaust process is replaced by a heat-rejection process that restores the working
fluid to its initial state.
5. Air has constant specific heats whose values are determined at room temperature
(25°C, or 77°F).
AN OVERVIEW OF RECIPROCATING ENGINES

The reciprocating engine is basically a piston–cylinder device. The piston reciprocates in the
cylinder between two fixed positions called the top dead center (TDC)—the position of the
piston when it forms the smallest volume in the cylinder—and the bottom dead center
(BDC)—the position of the piston when it forms the largest volume in the cylinder. The
distance between the TDC and the BDC is the largest distance that the piston can travel in one
direction, and it is called the stroke of the engine. The diameter of the piston is called the bore.
The air or air–fuel mixture is drawn into the cylinder through the intake valve, and the
combustion products are expelled from the cylinder through the exhaust valve. The minimum
volume formed in the cylinder when the piston is at TDC is called the clearance volume. The
volume displaced by the piston as it moves between TDC and BDC is called the displacement
volume.

The ratio of the maximum volume formed in the cylinder to the minimum (clearance) volume
is called the compression ratio r of the engine:

𝑉𝑚𝑎𝑥 𝑉𝐵𝐷𝐶
𝑟= =
𝑉𝑚𝑖𝑛 𝑉𝑇𝐷𝐶

Mean Effective Pressure  MEP

𝑊𝑛𝑒𝑡
𝑀𝐸𝑃 = (𝑘𝑃𝑎)
𝑉𝐵𝐷𝐶 − 𝑉𝑇𝐷𝐶

Reciprocating engines are classified as spark-ignition (SI) engines or compression-ignition

(CI) engines, depending on how the combustion process in the cylinder is initiated. In SI
engines, the combustion of the air–fuel mixture is initiated by a spark plug. In CI engines, the
air–fuel mixture is self-ignited as a result of compressing the mixture above its self-ignition
temperature. In the next two sections, we discuss the Otto and Diesel cycles, which are the ideal
cycles for the SI and CI reciprocating engines, respectively.

OTTO CYCLE: THE IDEAL CYCLE

FOR SPARK-IGNITION ENGINES

It is named after Nikolaus A. Otto, who built a successful four-stroke engine in 1876 in
Germany using the cycle proposed by Frenchman Beau de Rochas in 1862.

The piston executes four complete strokes within the cylinder, and the crankshaft completes
two revolutions for each thermodynamic cycle. These engines are called four-stroke internal
combustion engines.
It consists of four internally reversible processes:
1-2 Isentropic compression
2-3 Constant-volume heat addition
3-4 Isentropic expansion
4-1 Constant-volume heat rejection

𝑄𝐻 − 𝑄𝐿 𝑄𝐿 𝑚𝐶𝑉 (𝑇4 − 𝑇1 ) 𝑇1 (𝑇4 ⁄𝑇1 − 1)

𝜂𝑡ℎ = =1− =1− = 1−
𝑄𝐻 𝑄𝐻 𝑚𝐶𝑉 (𝑇3 − 𝑇2 ) 𝑇2 (𝑇3 ⁄𝑇2 − 1)

𝑇2 𝑉1 𝑘−1 𝑉4 𝑘−1 𝑇3
=( ) =( ) =
𝑇1 𝑉2 𝑉3 𝑇4

𝑇3 𝑇4
=
𝑇2 𝑇1

𝑇1 1
𝜂𝑡ℎ = 1 − = 1 − 𝑘−1
𝑇2 𝑟

𝑉1 𝑉4
𝑟= = = 𝐜𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐨𝐧 𝐫𝐚𝐭𝐢𝐨
𝑉2 𝑉3

When high compression ratios are used, the temperature of the air–fuel mixture rises above the
autoignition temperature of the fuel, causing an early and rapid burn of the fuel at some point,
followed by almost instantaneous inflammation of the end gas. This premature ignition of the
fuel, called autoignition, produces an audible noise, which is called engine knock.
Autoignition in spark-ignition engines cannot be tolerated because it hurts performance and
can cause engine damage.

Example
The compression ratio of an air-standard Otto cycle is 9.5. Prior to the isentropic compression
process, the air is at 100 kPa, 35°C, and 600 cm3. The temperature at the end of the isentropic
expansion process is 800 K. Using specific heat values at room temperature, determine
(a) The highest temperature and pressure in the cycle;
(b) The amount of heat transferred in, in kJ;
(c) The thermal efficiency; and
(d) The mean effective pressure.
Solution
Specific heat values at room temperature: cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R =
0.287 kJ/kg·K, and k = 1.4 (Table A-2).
(a) Process 1-2: isentropic compression
k −1
P
v 
T2 = T1  1  = (308 K )(9.5)0.4 = 757.9 K 3
v 2 
P2v 2 Pv v T  757.9 K 
⎯→ P2 = 1 2 P1 = (9.5)
= 1 1 ⎯ (100 kPa ) = 2338 kPa Qin
Qou 4
T2 T1 v 2 T1  308 K  2
t 1
Process 3-4: isentropic expansion.
v
k −1
v 
T3 = T4  4 
 = (800 K )(9.5)0.4 = 1969 K
v3 
Process 2-3: v = const. heat addition.
P3v 3 P2v 2 T  1969 K 
= ⎯⎯→ P3 = 3 P2 =  (2338 kPa ) = 6072 kPa
T3 T2 T2  757.9 K 

m=
P1V1
=
(100 kPa ) 0.0006 m 3 (
= 6.788 10 − 4 kg
)
(b)
RT1 (
0.287 kPa  m /kg  K (308 K )
3
)
( )
Qin = m(u 3 − u 2 ) = mcv (T3 − T2 ) = 6.788  10 −4 kg (0.718 kJ/kg  K )(1969 − 757.9 )K = 0.590 kJ

(c) Process 4-1: v = const. heat rejection.

( )
Qout = m(u 4 − u1 ) = mcv (T4 − T1 ) = − 6.788  10 −4 kg (0.718 kJ/kg  K )(800 − 308 )K = 0.240 kJ

Wnet = Qin − Qout = 0.590 − 0.240 = 0.350 kJ

Wnet,out 0.350 kJ
 th = = = 59.4%
Qin 0.590 kJ

V max
(d) V min = V 2 =
r
Wnet,out Wnet,out  kPa  m 3
0.350 kJ 
MEP = = =   = 652 kPa
V 1 −V 2 V1 (1 − 1 / r ) ( )
0.0006 m 3 (1 − 1/9.5)  kJ 


DIESEL CYCLE: THE IDEAL CYCLE

FOR COMPRESSION-IGNITION ENGINES

The Diesel cycle is the ideal cycle for CI reciprocating engines. The CI engine, first proposed
by Rudolph Diesel in the 1890s, is very similar to the SI engine, differing mainly in the method
of initiating combustion. In spark-ignition engines (also known as gasoline engines), the air–
fuel mixture is compressed to a temperature that is below the autoignition temperature of the
fuel, and the combustion process is initiated by firing a spark plug. In CI engines (also known
as diesel engines), the air is compressed to a temperature that is above the autoignition
temperature of the fuel, and combustion starts on contact as the fuel is injected into this hot air.
Therefore, the spark plug and carburetor are replaced by a fuel injector in diesel engines
In diesel engines, only air is compressed during the compression stroke, eliminating the
possibility of autoignition (engine knock). Therefore, diesel engines can be designed to operate
at much higher compression ratios than SI engines, typically between 12 and 24.

• 1-2 isentropic compression

• 2-3 constant-pressure heat addition
• 3-4 isentropic expansion
• 4-1 constant-volume heat rejection.

Cutoff ratio:
𝑉3
𝑟𝑐 =
𝑉2
Thermal efficiency:

𝑄𝐿 𝑚𝐶𝑉 (𝑇4 − 𝑇1 ) 𝑇1 (𝑇4 ⁄𝑇1 − 1)

𝜂𝑡ℎ = 1 − =1− = 1−
𝑄𝐻 𝑚𝐶𝑃 (𝑇3 − 𝑇2 ) 𝑘𝑇2 (𝑇3 ⁄𝑇2 − 1)
 𝑟𝑐𝑘 − 1
1
𝜂𝑡ℎ−𝐷𝑖𝑒𝑠𝑒𝑙 = 1 − 𝑘−1 [ ]
𝑟 𝑘(𝑟𝑐 − 1)
For the same compression ratio:
𝜂𝑡ℎ−𝑂𝑡𝑡𝑜 > 𝜂𝑡ℎ−𝐷𝑖𝑒𝑠𝑒𝑙
(In real applications, Diesel engines have much higher compression ratio than Otto engines,
so that: 𝜂𝑡ℎ−𝑂𝑡𝑡𝑜 < 𝜂𝑡ℎ−𝐷𝑖𝑒𝑠𝑒𝑙 )

Dual Cycle
This cycle may better represents the compression ignition, (CI) engines in real applications.

Example
An ideal diesel engine has a compression ratio of 20 and uses air as the working fluid. The state
of air at the beginning of the compression process is 95 kPa and 20°C. If the maximum
temperature in the cycle is not to exceed 2200 K, determine
(a) The thermal efficiency and
(b) The mean effective pressure. Assume constant specific heats for air at room temperature.

Solution
Specific heats for air at room temperature: cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R =
0.287 kJ/kg·K, k = 1.4 (Table A-2).
(a) Process 1-2: isentropic compression. P
k −1
qin
V 
2 3
T2 = T1  1  = (293 K )(20)0.4 = 971.1 K
V 2 
Process 2-3: P = const. heat addition. 4
qout
P3V 3 P2V 2 V T 2200K 1
= ⎯
⎯→ 3 = 3 = = 2.265
T3 T2 V 2 T2 971.1K
v
Process 3-4: isentropic expansion.
 k −1 k −1 k −1
V   2.265V 2 
0.4
 2.265   2.265 
T4 = T3  3  = T3   = T3   = (2200 K )  = 920.6 K
V 4   V4   r   20 
q in = h3 − h2 = c p (T3 − T2 ) = (1.005 kJ/kg  K )(2200 − 971.1)K = 1235 kJ/kg
q out = u 4 − u1 = cv (T4 − T1 ) = (0.718 kJ/kg  K )(920.6 − 293)K = 450.6 kJ/kg
w net,out = q in − q out = 1235 − 450.6 = 784.4 kJ/kg
wnet,out 784.4 kJ/kg
 th = = = 63.5%
q in 1235 kJ/kg

(b) v1 =
RT1
=
(
0.287 kPa  m 3 /kg  K (293 K ) )
= 0.885 m 3 /kg = v max
P1 95 kPa
v max
v min = v 2 =
r
wnet,out wnet,out 784.4 kJ/kg  kPa  m 3 
MEP = = =   = 933 kPa
v1 −v 2 v 1 (1 − 1 / r ) ( )
0.885 m 3 /kg (1 − 1/20)  kJ 


Stirling cycle

Stirling cycle is made up of four totally reversible processes:

1-2 T = constant expansion (heat addition from the external source)

2-3 v = constant regeneration (internal heat transfer from the working fluid to the regenerator)
3-4 T = constant compression (heat rejection to the external sink)
4-1 v = constant regeneration (internal heat transfer from the regenerator back to the working
fluid)
Ericsson Cycle

The Ericsson cycle is very much like the Stirling cycle, except that the two constant-volume
processes are replaced by two constant-pressure processes.

BRAYTON CYCLE: THE IDEAL CYCLE

FOR GAS-TURBINE ENGINES
The Brayton cycle was first proposed by George Brayton for use in the reciprocating oil-
burning engine that he developed around 1870. Today, it is used for gas turbines only where
both the compression and expansion processes take place in rotating machinery.
1-2 isentropic compression (in a compressor)

2-3 Constant-pressure heat addition (combustion chamber)

3-4 isentropic expansion (in a turbine)

4-1 Constant-pressure heat rejection (heat exchanger)

𝑄𝐿 𝐶𝑃 (𝑇4 − 𝑇1 ) 𝑇1 (𝑇4 ⁄𝑇1 − 1)

𝜂𝑡ℎ = 1 − =1− =1−
𝑄𝐻 𝐶𝑃 (𝑇3 − 𝑇2 ) 𝑇2 (𝑇3 ⁄𝑇2 − 1)

𝑃3 𝑃 𝑃2 𝑇 𝑘⁄𝑘−1 𝑃3 𝑇 𝑘⁄𝑘−1 𝑇3 𝑇
= 𝑃2 , = (𝑇2 ) , = (𝑇3 )  = 𝑇2
𝑃4 1 𝑃1 1 𝑃4 4 𝑇4 1

𝑇1 1 1
𝜂𝑡ℎ = 1 − =1− = 1 −
𝑇2 𝑃 (𝑘−1)⁄𝑘 (𝑟𝑝 )
(𝑘−1)⁄𝑘
(𝑃2 )
1
rp= P2/P1 = Pressure ratio

Deviation of Actual Gas-Turbine Cycles from Idealized Ones

ℎ2𝑠 − ℎ1
𝜂𝐶 = = 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
ℎ2 − ℎ1
 ℎ3 − ℎ4
𝜂𝑇 = = 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
ℎ3 − ℎ4𝑠

Example
A gas-turbine power plant operates on the simple Brayton cycle between the pressure limits of
100 and 800 kPa. Air enters the compressor at 30°C and leaves at 330°C at a mass flow rate of
200 kg/s. The maximum cycle temperature is 1400 K. During operation of the cycle, the net
power output is measured experimentally to be 60 MW. Assume constant properties for air at
300 K with cv = 0.718 kJ/kg-K, cp = 1.005 kJ/kg-K, R = 0.287 kJ/kg-K, k = 1.4.

(a) Sketch the T-s diagram for the cycle.

(b) Determine the isentropic efficiency of the turbine for these operating conditions.
(c) Determine the cycle thermal efficiency.

Solution T
873 K 3
(a) The properties of air are given as cv =
0.718 kJ/kg∙K, cp = 1.005 kJ/kg∙K, R = 2s 2
0.287 kJ/kg∙K, k = 1.4.
4
(b) For the compression process, 303 K
1 4s
W Comp = m c p (T2 − T1 ) s
= (200 kg/s) (1.005 kJ/kg  K )(330 − 30)K
= 60,300 kW
For the turbine during the isentropic process,
( k −1) / k
P 
0.4/1.4
 100 kPa 
T4 s = T3  4 
 = (1400 K)  = 772.9 K
 P3   800 kPa 

W Turb,s = m c p (T3 − T4 s ) = (200 kg/s) (1.005 kJ/kg  K )(1400 − 772 .9)K = 126 ,050 kW
The actual power output from the turbine is
W net = W Turb − W Comp
W Turb = W net + W Turb = 60,000 + 60,300 = 120,300 kW

The isentropic efficiency of the turbine is then

W Turb 120,300 kW
 Turb = = = 0.954 = 95.4%
W Turb,s 126,050 kW

(c) The rate of heat input is

Q in = m c p (T3 − T2 ) = (200 kg/s) (1.005 kJ/kg  K )[(1400 − (330 + 273 )]K = 160 ,200 kW

The thermal efficiency is then

W net 60,000 kW
 th = = = 0.375 = 37.5%

Qin 160 ,200 kW

THE BRAYTON CYCLE WITH REGENERATION

In gas-turbine engines, the temperature of the exhaust gas leaving the turbine is often
considerably higher than the temperature of the air leaving the compressor. Therefore, the high-
pressure air leaving the compressor can be heated by transferring heat to it from the hot exhaust
gases in a counter-flow heat exchanger, which is also known as a regenerator or a recuperator.
The thermal efficiency of the Brayton cycle increases as a result of regeneration.

THE BRAYTON CYCLE WITH INTERCOOLING, REHEATING,

AND REGENERATION
VAPOR POWER CYCLES

RANKINE CYCLE
The ideal Rankine cycle and its T_S diagram:

1-2 Isentropic compression in a pump

2-3 Constant pressure heat addition in a boiler
3-4 Isentropic expansion in a turbine
4-1 Constant pressure heat rejection in a condenser

Thermal efficiency:
𝑤𝑛𝑒𝑡
𝜂𝑡ℎ =
𝑞𝐻

Example
A simple Rankine cycle uses water as the working fluid. The boiler operates at 6000 kPa and
the condenser at 50 kPa. At the entrance to the turbine, the temperature is 450°C. The isentropic
efficiency of the turbine is 94 percent, pressure and pump losses are negligible, and the water
leaving the condenser is subcooled by 6.3°C. The boiler is sized for a mass flow rate of 20 kg/s.
Determine the rate at which heat is added in the boiler, the power required to operate the pumps,
the net power produced by the cycle, and the thermal efficiency?

Solution

From the steam tables (Tables A-4, A-5, and A-6),

P1 = 50 kPa  h1  h f @ 75C = 314.03 kJ/kg


T1 = Tsat @ 50 kPa − 6.3 = 81.3 − 6.3 = 75C  v 1 = v f @ 75C = 0.001026 m 3 /kg
 wp,in = v 1 ( P2 − P1 )
 1 kJ  T
= (0.001026 m 3 /kg )(6000 − 50)kPa  
 1 kPa  m 
3
= 6.10 kJ/kg 3
6 MPa
h2 = h1 + wp,in = 314.03 + 6.10 = 320.13 kJ/kg
qin
2
P3 = 6000 kPa  h3 = 3302.9 kJ/kg 50 kPa

T3 = 450C  s 3 = 6.7219 kJ/kg  K 1 qout 4 4
s4 − s f 6.7219 − 1.0912
P4 = 50 kPa  x 4 s = = = 0.8660
 s fg 6.5019 s
s 4 = s3  h
4s = h f + x 4 s h fg = 340.54 + (0.8660)(2304.7) = 2336.4 kJ/kg

h3 − h4
T = ⎯→ h4 = h3 −  T (h3 − h4s ) = 3302.9 − (0.94)(3302.9 − 2336.4) = 2394.4 kJ/kg
⎯
h3 − h4 s

Thus,
Q in = m (h3 − h2 ) = (20 kg/s)(3302.9 − 320.13)kJ/kg = 59,660 kW
W T,out = m (h3 − h4 ) = (20 kg/s)(3302.9 − 2394.4)kJ/kg = 18,170 kW
W P,in = m wP,in = (20 kg/s)(6.10 kJ/kg) = 122 kW
W = W
net − W
T,out = 18,170 − 122 = 18,050 kW
P,in

and
W net 18,050
 th = = = 0.3025
Q in 59,660

THE IDEAL REHEAT RANKINE CYCLE

Increasing the boiler pressure increases the thermal efficiency of the Rankine cycle, but it also
increases the moisture content of the steam to unacceptable levels.

How can we take advantage of the increased efficiencies at higher boiler pressures
without facing the problem of excessive moisture at the final stages of the turbine?

Two possibilities come to mind:

1. Superheat the steam to very high temperatures before it enters the turbine. This would
be the desirable solution since the average temperature at which heat is added would
also increase, thus increasing the cycle efficiency. This is not a viable solution,
however, since it requires raising the steam temperature to metallurgically unsafe
levels.

2. Expand the steam in the turbine in two stages, and reheat it in between. In other words,
modify the simple ideal Rankine cycle with a reheat process. Reheating is a practical
solution to the excessive moisture problem in turbines, and it is commonly used in
modern steam power plants.
The T-s diagram of the ideal reheat Rankine cycle and the schematic of the power plant
operating on this cycle are shown in Fig.

Remember that the sole purpose of the reheat cycle is to reduce the moisture content of the
steam at the final stages of the expansion process. If we had materials that could withstand
sufficiently high temperatures, there would be no need for the reheat cycle.

Example
Consider a steam power plant that operates on the ideal reheat Rankine cycle. The plant
maintains the boiler at 7000 kPa, the reheat section at 800 kPa, and the condenser at 10 kPa.
The mixture quality at the exit of both turbines is 93 percent. Determine the temperature at the
inlet of each turbine and the cycle’s thermal efficiency?

Solution

From steam tables (Tables A-4, A-5, and A-6),

h1 = h f @ 10 kPa = 191.81 kJ/kg
T
v 1 = v f @ 10 kPa = 0.001010 m 3 /kg
3 5
wp,in = v 1 ( P2 − P1 ) 7 MPa
 1 kJ 
= (0.001010 m 3 /kg )(7000 − 10)kPa  
 1 kPa  m 
3
= 7.06 kJ/kg 800 kPa
h2 = h1 + wp,in = 191.81 + 7.06 = 198.87 kJ/kg 2
4
10 kPa
1 6
P4 = 800 kPa  h4 = h f + x 4 h fg = 720.87 + (0.93)(2047.5) = 2625.0 kJ/kg
 s
x 4 = 0.93  s 4 = s f + x 4 s fg = 2.0457 + (0.93)(4.6160) = 6.3385 kJ/kg  K
P3 = 7000 kPa  h3 = 3085.5 kJ/kg

s3 = s 4  T3 = 373.3C
 P6 = 10 kPa  h6 = h f + x 6 h fg = 191.81 + (0.93)(2392.1) = 2416.4 kJ/kg

x 6 = 0.90  s 6 = s f + x 6 s fg = 0.6492 + (0.93)(7.4996) = 7.6239 kJ/kg  K
P5 = 800 kPa  h5 = 3302.0 kJ/kg

s5 = s6  T5 = 416.2C


q in = (h3 − h2 ) + (h5 − h4 ) = 3085.5 − 198.87 + 3302.0 − 2625.0 = 3563.6 kJ/kg
q out = h6 − h1 = 2416.4 − 191.81 = 2224.6 kJ/kg


q out 2224.6
 th = 1 − =1− = 0.3757 = 37.6%
qin 3563.6

THE IDEAL REGENERATIVE RANKINE CYCLE

After leaving the pump, the liquid circulates around the turbine casing, counter flow to the
direction of vapor flow in the turbine. Thus, it is possible to transfer to the liquid flowing around
the turbine the heat from the vapor as it flows through the turbine. the efficiency of this
idealized regenerative cycle is exactly equal to the efficiency of the Carnot cycle with the same
heat supply and heat rejection temperatures.
The idealized regenerative cycle is impractical for two reasons:

1. First, it would be impossible to effect the necessary heat transfer from the vapor in the
turbine to the liquid feedwater.
2. The moisture content of the vapor leaving the turbine increases considerably as a result
of the heat transfer.

The practical regenerative cycle extracts some of the vapor after it has partially expanded in
the turbine and uses feedwater heaters, as shown in Fig.
Example
A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwater
heaters. Steam enters the turbine at 10 MPa and 600°C and exhausts to the condenser at 5 kPa.
Steam is extracted from the turbine at 0.6 and 0.2 MPa. Water leaves both feedwater heaters as
a saturated liquid. The mass flow rate of steam through the boiler is 22 kg/s. Show the cycle
on a T-s diagram, and determine
(a) the net power output of the power plant and
(b) the thermal efficiency of the cycle.

Solution

T
7 7
Turbine
Boiler 6 10 MPa
8 8
5 0.6
4
10 MPa y 1-y
6 3 0.2 MPa
9 2 9
1-y-z
fwh II fwh I Condenser 5 kPa
1 10
4 2
1 s
5 3
P III P II P I
(a) From the steam tables (Tables A-4, A-5, and A-6),
h1 = h f @ 5 kPa = 137.75 kJ/kg
v 1 = v f @ 5 kPa = 0.001005 m 3 /kg
( )  1 kJ
w pI ,in = v 1 (P2 − P1 ) = 0.001005 m 3 /kg (200 − 5 kPa )
 1 kPa  m 3

 = 0.20 kJ/kg

h =h +w = 137.75 + 0.20 = 137.95 kJ/kg  
2 1 pI ,in
 P3 = 0.2 MPa  h3 = h f @ 0.2 MPa = 504.71 kJ/kg

 v 3 = v f @ 0.2 MPa = 0.001061 m /kg
3
sat.liquid

( )  1 kJ
w pII ,in = v 3 (P4 − P3 ) = 0.001061 m 3 /kg (600 − 200 kPa )
 1 kPa  m 3



= 0.42 kJ/kg  
h4 = h3 + w pII ,in = 504.71 + 0.42 = 505.13 kJ/kg

P5 = 0.6 MPa  h5 = h f @ 0.6 MPa = 670.38 kJ/kg


 v 5 = v f @ 0.6 MPa = 0.001101 m /kg
3
sat.liquid

( )  1 kJ
w pIII ,in = v 5 (P6 − P5 ) = 0.001101 m 3 /kg (10,000 − 600 kPa )
 1 kPa  m 3



= 10.35 kJ/kg  

h6 = h5 + w pIII ,in = 670.38 + 10.35 = 680.73 kJ/kg

P7 = 10 MPa  h7 = 3625.8 kJ/kg


T7 = 600C  s 7 = 6.9045 kJ/kg  K
P8 = 0.6 MPa 
 h8 = 2821.8 kJ/kg
s8 = s 7 
s9 − s f 6.9045 − 1.5302
P9 = 0.2 MPa  x 9 = = = 0.9602
 s fg 5.5968
s9 = s7 
h9 = h f + x 9 h fg = 504.71 + (0.9602)(2201.6 ) = 2618.7 kJ/kg

s10 − s f 6.9045 − 0.4762

P10 = 5 kPa  x10 = = = 0.8119
 s fg 7.9176
s10 = s 7 
h10 = h f + x10 h fg = 137.75 + (0.8119)(2423.0) = 2105.0 kJ/kg

The fraction of steam extracted is determined from the steady-flow energy balance equation applied
to the feedwater heaters. Noting that Q  W  Δke  Δpe  0 ,
FWH-2:
E in − E out = E system 0 (steady) = 0
E in = E out

 m h =  m h
i i e e ⎯ ⎯→ yh8 + (1 − y )h4 = 1(h5 )
⎯→ m 8 h8 + m 4 h4 = m 5 h5 ⎯

where y is the fraction of steam extracted from the turbine ( = m 8 / m 5 ). Solving for y,
h5 − h4 670.38 − 505.13
y= = = 0.07133
h8 − h4 2821.8 − 505.13

FWH-1:

 m h =  m h
i i e e ⎯ ⎯→ zh9 + (1 − y − z )h2 = (1 − y )h3
⎯→ m 9 h9 + m 2 h2 = m 3 h3 ⎯

where z is the fraction of steam extracted from the turbine ( = m 9 / m 5 ) at the second stage. Solving
for z,
h3 − h2
z= (1 − y ) = 504.71 − 137.95 (1 − 0.07136) = 0.1373
h9 − h2 2618.7 − 137.95

Then,
 qin = h7 − h6 = 3625.8 − 680.73 = 2945.0 kJ/kg
qout = (1 − y − z )(h10 − h1 ) = (1 − 0.07133 − 0.1373)(2105.0 − 137.75) = 1556.8 kJ/kg
wnet = qin − qout = 2945.0 − 1556.8 = 1388.2 kJ/kg

and
W net = m wnet = (22 kg/s )(1388.2 kJ/kg ) = 30,540 kW  30.5 MW

(b) The thermal efficiency is

q out 1556.8 kJ/kg
 th = 1 − = 1− = 47.1%
q in 2945.0 kJ/kg

COGENERATION

There are many occasions in industrial settings where the need arises for a specific source or
supply of energy within the environment in which a steam power plant is being used to generate
electricity. In such cases, it is appropriate to consider supplying this source of energy in the
form of steam that has already been expanded through the high-pressure section of the turbine
in the power plant cycle.

The turbine is tapped at some intermediate pressure to furnish the necessary amount of process
steam required for the particular energy need—perhaps to operate a special process in the plant,
or in many cases simply for the purpose of space heating the facilities. This type of application
is termed cogeneration.

Cogeneration: the production of more than one useful form of energy (such as process heat
and electric power) from the same energy source.
Energy utilization factor:

𝑊̇𝑛𝑒𝑡 + 𝑄̇𝑃 𝑄̇𝑜𝑢𝑡

𝜀𝑢 = = 1−
𝑄̇𝑖𝑛 𝑄̇𝑖𝑛

𝑊̇𝑛𝑒𝑡 = 𝑄̇𝑖𝑛 − (𝑄̇𝑜𝑢𝑡 + 𝑄̇𝑃 )

COMBINED GAS–VAPOR POWER CYCLES

REFRIGERATORS AND HEAT PUMPS

THE REVERSED CARNOT CYCLE

Since it is a reversible cycle, all four processes that comprise the Carnot cycle can be reversed.
Reversing the cycle does also reverse the directions of any heat and work interactions. The
result is a cycle that operates in the counterclockwise direction on a T-s diagram, which is
called the reversed Carnot cycle. A refrigerator or heat pump that operates on the reversed
Carnot cycle is called a Carnot refrigerator or a Carnot heat pump.
The coefficients of performance of Carnot refrigerators and heat pumps:

1
𝐶𝑂𝑃𝑅,𝐶𝑎𝑟𝑛𝑜𝑡 =
𝑇𝐻
𝑇𝐿 − 1

1
𝐶𝑂𝑃𝐻𝑃,𝐶𝑎𝑟𝑛𝑜𝑡 =
𝑇
1 − 𝑇𝐿
𝐻

THE IDEAL VAPOR-COMPRESSION REFRIGERATION CYCLE

Many of the impracticalities associated with the reversed Carnot cycle can be eliminated by
vaporizing the refrigerant completely before it is compressed and by replacing the turbine
with a throttling device, such as an expansion valve or capillary tube. The cycle that results is
called the ideal vapor-compression refrigeration cycle, and it is shown schematically and
on a T-s diagram in Fig.

The vapor-compression refrigeration cycle is the most widely used cycle for refrigerators, air-
conditioning systems, and heat pumps. It consists of four processes:

1-2 Isentropic compression in a compressor

2-3 Constant-pressure heat rejection in a condenser
3-4 Throttling in an expansion device
4-1 Constant-pressure heat absorption in an evaporator
Coefficient of Performance:

Refrigerator:

𝑄𝐿 1
𝐶𝑂𝑃𝑅 = =
𝑊𝑛𝑒𝑡 𝑄𝐻 − 1
𝑄𝐿

Heat Pump:

𝑄𝐻 1
𝐶𝑂𝑃𝐻𝑃 = =
𝑊𝑛𝑒𝑡 1 − 𝑄𝐿
𝑄 𝐻

SELECTING THE RIGHT REFRIGERANT

When designing a refrigeration system, there are several refrigerants from which to choose,
such as chlorofluorocarbons (CFCs), ammonia, hydrocarbons (propane, ethane, ethylene, etc.),
carbon dioxide, air (in the air-conditioning of aircraft), and even water (in applications above
the freezing point).

Of these, refrigerants such as R-11, R-12, R-22, R-134a, and R-502 account for over 90 percent
of the market in the United States.
Ethyl ether was the first commercially used refrigerant in vapor-compression systems in 1850,
followed by ammonia, carbon dioxide, methyl chloride, sulphur dioxide, butane, ethane,
propane, isobutane, gasoline, and chlorofluorocarbons, among others.
The industrial and heavy-commercial sectors were very satisfied with ammonia, and still are,
although ammonia is toxic.

It is remarkable that the early refrigerants used in the light-commercial and household sectors
such as sulfur dioxide, ethyl chloride, and methyl chloride were highly toxic.

Example

An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid
maintains a condenser at 1000 kPa and the evaporator at 4°C. Determine this system’s COP
and the amount of power required to service a 400 kW cooling load.

Solution
From refrigerant R-134a tables (Tables A-11, A-12, and A-13),
T1 = 4C  h1 = h g @ 4C = 252.77 kJ/kg
 T
sat. vapor  s1 = s g @ 4C = 0.92927 kJ/kg  K
·
P2 = 1 MPa  QH 2
 h2 = 275.29 kJ/kg
s 2 = s1  3 1 MPa ·
Win
P3 = 1 MPa 
 h = hf = 107.32 kJ/kg
sat. liquid  3 @ 1 MPa

4C
h4  h3 = 107.32 kJ/kg ( throttling ) 1
4s 4 ·
QL

Refrigerant mass flow rate: s

Q L 400 kJ/s
Q L = m
 (h1 − h4 ) ⎯
⎯→ m
 = = = 2.750 kg/s
h1 − h4 (252.77 − 107.32) kJ/kg

Required power input:

W in = m
 (h2 − h1 ) = (2.750 kg/s)(275. 29 − 252.77) kJ/kg = 61.93 kW

Using the definition of the Coefficient of Performance:

Q L 400 kW
COPR = = = 6.46

Win 61.93 kW