Lecture - 4 MAK

ENERGY ANALYSIS
OF CLOSED SYSTEM
1
ENERGY ANALYSIS OF
CLOSED SYSTEM
• In LO#2 we studied the various form of energy, and
energy transfer, and we studied the conservation of
energy and energy balance Ein – Eout = ΔE
• In LO#3 we learned how to determine the
thermodynamics properties of substance.
• In this LO#4 we will apply the energy balance relation
in closed systems (no mass flow).
• We will study the moving boundary work or P dV work
used in reciprocating devices such as cars and
compressors engines.
2
MOVING BOUNDARY WORK
• One form of mechanical work is
associated with the expansion or
compression of a gas in a piston-
cylinder device.
• The expansion and compression work
is often called moving boundary
work, or simply boundary work.
• The total boundary work done by a
moving piston is obtained by adding all
the differential works from the initial
state to the final state:
3
Moving boundary work (P dV work): Quasi-equilibrium process:
The expansion and compression work A process during which the system
in a piston-cylinder device. remains nearly in equilibrium at all
times.
Wb is positive → for expansion

Wb is negative → for compression
P is Absolute pressure.
The work associated
with a moving
boundary is called
boundary work.
A gas does a differential

amount of work δWb as it
forces the piston to move
by a differential amount ds.
4
The boundary work done during a

process depends on the path
followed as well as the end states.
The area under the process curve on a P-V

diagram is equal, in magnitude, to the work
done during a quasi-equilibrium expansion
or compression process of a closed system.
5
BOUNDARY WORK = AREA UNDER THE
P-V DIAGRAM PROCESS CURVE
• The area under the process curve on a p-v diagram
is equal, in magnitude, to the work done during a
quasi- equilibrium expansion or compression
process of a closed system.
• The process could go in many different paths as the

gas expand from state1 to state 2.
• The will lead to different area under the process
curve.
6
EXAMPLE 1 (CONSTANT VOLUME PROCESS)
A rigid tank contains air at 500 kPa and 150°C.
As a result of heat transfer to the surroundings, the temperature and
pressure inside the tank drop to g 65°C and 400 kPa, respectively.
Determine the boundary work done during; a this process.
In constant volume process we have no area under the curve

And the work will equal to Zero.
7
EXAMPLE 2 (CONSTANT PRESSURE PROCESS)
A frictionless piston-cylinder device contains 5 kg of steam at 400 kPa
and 200°C. Heat is now transferred to the steam until the temperature
reaches 250°C. Determine the work done by the steam during this
process.
Contact P :
Wb = P (V2-V1)
In constant pressure process we have rectangular area under the curve

8
EXAMPLE 3 (CONSTANT TEMPERATURE PROCESS)
A piston-cylinder device initially contains 0.4 m3 of air (Ideal Gas)
at 100 kPa and 80°C. The air is now compressed to 0.1 -m3 in
such a way that the temperature inside the cylinder remains
constant. Determine the work done during this process.
9
POLYTROPIC PROCESS
• During actual expansion and compression
process of gases, pressure and volume are
often related by:
• Where n and C are constants.

• A process of this kind is called Polytropic
process.
10
POLYTROPIC PROCESS
• A process of this kind is called a polytropic process
11
Summary: Polytropic, Isothermal, and Isobaric processes
Polytropic process: C, n (polytropic exponent) constants
Polytropic
process
Polytropic and for ideal gas
When n = 1
(isothermal process)
Constant pressure process
What is the
boundary work
for a constant- Schematic and
volume P-V diagram for
process? a polytropic
process.
12
ENERGY BALANCE FOR
CLOSED SYSTEMS
13
ENERGY BALANCE FOR
CLOSED SYSTEMS
14
EXAMPLE 4
A piston-cylinder device contains 25 g of saturated
water vapor that is maintained at a constant
pressure of 300 kPa.
A resistance heater within the cylinder is turned on
and passes a current of 0.2 A for 5 min from a
120-V source. At the same time, a heat loss of 3.7
kJ occurs,
(a) Show that for a closed system the boundary
work Wb and the change in internal energy ΔU
in the first-law relation can be combined into
one term, Δ H, for a constant- pressure
process.
(b) Determine the final temperature of the steam.
15
EXAMPLE 5
16
EXAMPLE 5
17
EXAMPLE 5
A rigid tank is divided into two equal parts by a partition. Initially,
one side of the tank contains 5 kg of water at 200 kPa and 25°C,
and the other side is evacuated.
The partition is then removed, and the water expands into the
entire tank. The water is allowed to exchange heat with its
surroundings until the temperature in the tank returns to the initial
value of 25°C. Determine
(a) the volume of the tank,

(b) the final pressure
(c) the heat transfer for this process.
18
EXAMPLE 5
19
EXAMPLE 5
20
EXAMPLE 5
21
SPECIFIC
HEATS
22
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
• In thermodynamics, we are interested in two kinds of specific
heats: specific heat at constant volume cv and specific heat
at constant pressure cp.
Ex. for Helium, Cp is always greater than Cv 23

SPECIFIC HEATS
24
INTERNAL ENERGY, ENTHALPY, AND
SPECIFIC HEATS OF IDEAL GASES
Approximated as follows
25
If the final temperature T2 is not known, the specific heats may be evaluated
at T, or at the anticipated average temperature.
26
Three ways to find cv and cp
27
SPECIFIC HEAT RELATIONS OF
IDEAL GASES
Specific heat ratio
28
EXAMPLE 6
Air at 300 K and 200 kPa is heated at constant pressure to 600 K.
Determine the change in internal energy of air per unit mass,
using:
(a) data from the air table
(b) the average specific heat value .
29
EXAMPLE 6
Difference between the two methods?
30
EXAMPLE 7
An insulated rigid tank initially contains 0.7 kg of helium at 27°C
and 350 kPa. A paddle wheel with a power rating of 0.015 kW is
operated within the tank for 30 min. Determine
(a) the final temperature and
(b) the final pressure of the helium gas.
31
EXAMPLE 7
32
EXAMPLE 7
The cv value of helium is determined from Table A-2a to be
cv = 3.1156 kJ/kg-°C.
Substituting this and other known quantities ,we obtain
Absolute
33
EXAMPLE 8
A piston-cylinder device initially contains 0.5 m3 of nitrogen gas at 400
kPa and 27°C.
An electric heater within the device is turned on and is allowed to pass
a current of 2 A for 5 min from a 120-V source.
Nitrogen expands at constant pressure, and a heat loss of 2800 J
occurs during the process. Determine the final temperature of
nitrogen.
34
EXAMPLE 8
35
SUMMARY
Various forms of boundary work are expressed as follows:
36
SUMMARY
The energy balance for a closed system is expressed as:
37
SUMMARY
For ideal gases u, h, cv and cp are functions of temperature
alone. The Δu and Δh of ideal gases are expressed as
38

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ENERGY ANALYSIS

Wb is positive → for expansion

A gas does a differential

The boundary work done during a

The area under the process curve on a P-V

• The process could go in many different paths as the

In constant volume process we have no area under the curve

In constant pressure process we have rectangular area under the curve

• Where n and C are constants.

Polytropic and for ideal gas

Constant pressure process

(a) the volume of the tank,

Ex. for Helium, Cp is always greater than Cv 23

Three ways to find cv and cp

Specific heat ratio

Difference between the two methods?

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