Thermodynamics: An Engineering Approach, 5th Edition

Yunus A. Cengel, Michael A. Boles

McGraw-Hill, 2008

Chapter 9
GAS POWER CYCLES

Prof. Dr. Ali PINARBAŞI

Yildiz Technical University
Mechanical Engineering Department
Yildiz, ISTANBUL

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1
 GAS POWER CYCLES

9–1 Basic Considerations in the Analysis of Power Cycles 
9–2 The Carnot Cycle and Its Value in Engineering 
9–3 Air‐Standard Assumptions 
9–4 An Overview of Reciprocating Engines 
9–5 Otto Cycle: The Ideal Cycle for Spark‐Ignition Engines 
9–6 Diesel Cycle: The Ideal Cycle for Compression‐Ignition Engines
9–7 Stirling and Ericsson Cycles 
9–8 Brayton Cycle: The Ideal Cycle for Gas‐Turbine Engines 
Development of Gas Turbines 
Deviation of Actual Gas‐Turbine Cycles from Idealized Ones
9–9 The Brayton Cycle with Regeneration 
9–10 The Brayton Cycle with Intercooling, Reheating,and Regeneration 
9–11 Ideal Jet‐Propulsion Cycles 
Modifications to Turbojet Engines 
9–12 Second‐Law Analysis of Gas Power Cycles 

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Objectives

• Evaluate the performance of gas power cycles for which the working fluid
remains a gas throughout the entire cycle.
• Develop simplifying assumptions applicable to gas power cycles.
• Review the operation of reciprocating engines.
• Analyze both closed and open gas power cycles.
• Solve problems based on the Otto, Diesel, Stirling, and Ericsson cycles.
• Solve problems based on the Brayton cycle; the Brayton cycle with
regeneration; and the Brayton cycle with intercooling, reheating, and
regeneration.
• Analyze jet-propulsion cycles.
• Identify simplifying assumptions for second-law analysis of gas power cycles.
• Perform second-law analysis of gas power cycles.

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BASIC CONSIDERATIONS IN THE ANALYSIS OF POWER CYCLES

Most power-producing devices operate on cycles.

Ideal cycle: A cycle that resembles the actual cycle closely but is made up totally
of internally reversible processes is called on.

Reversible cycles such as Carnot cycle have the highest thermal efficiency of
all heat engines operating between the same temperature levels. Unlike ideal
cycles, they are totally reversible, and unsuitable as a realistic model.

Thermal efficiency of heat engines

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Modeling is a powerful  Care should be exercised in the  The analysis of many complex 
engineering tool that provides  interpretation of the results  processes can be reduced to a 
great insight and simplicity at  from ideal cycles. manageable level by utilizing some 
the expense of some loss in  idealizations.
accuracy.

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The idealizations and simplifications in the analysis of power cycles
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 quasi-equilibrium
manner.
3. The pipes connecting the various components of a system are well insulated,
and heat transfer through them is negligible.
On a T-s diagram, the ratio of the area enclosed by the cyclic curve to the area under
the heat-addition process curve represents the thermal efficiency of the cycle. Any
modification that increases the ratio of these two areas will also increase the thermal
efficiency of the cycle.

P‐v and T‐s diagrams, the area enclosed by the process curve represents the net work of the cycle.
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THE CARNOT CYCLE AND ITS VALUE IN
ENGINEERING
The Carnot cycle is composed of four totally reversible processes: isothermal heat
addition, isentropic expansion, isothermal heat rejection, and isentropic
compression.

For both ideal and actual cycles: Thermal efficiency increases with an increase in
the average temperature at which heat is supplied to the system or with a decrease
in the average temperature at which heat is rejected from the system.

A steady‐flow Carnot engine.
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 P‐v and T‐s diagrams of a Carnot cycle.

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EXAMPLE 9–1
Show that the thermal efficiency of a Carnot cycle operating between the temperature limits of TH
and TL is solely a function of these two temperatures and is given by

Solution It is to be shown that the efficiency of a Carnot cycle

depends on the source and sink temperatures alone.
Analysis All four processes that comprise the Carnot cycle are
reversible, and thus the area under each process curve represents
the heat transfer for that process. Heat is transferred to the
system during process 1-2 and rejected during process 3-4.
Therefore, the amount of heat input and heat output for the cycle
can be expressed as

Discussion Notice that the thermal efficiency of a Carnot cycle is independent of the type of
the working fluid used (an ideal gas, steam, etc.) or whether the cycle is executed in a
closed or steady-flow system.

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AIR-STANDARD ASSUMPTIONS
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
The combustion process is replaced  working fluid to its initial state.
by a heat‐addition process in ideal 
cycles.

Cold-air-standard assumptions: When the working fluid is considered to be

air with constant specific heats at room temperature (25°C).
Air-standard cycle: A cycle for which the air-standard assumptions are
applicable.

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AN OVERVIEW OF RECIPROCATING ENGINES

Compression ratio

• Spark-ignition (SI) engines

• Compression-ignition (CI) engines

Nomenclature for reciprocating engines.
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 Mean effective pressure

The mean effective pressure can be used as a 
parameter to compare the performances of 
reciprocating engines of equal size. The engine with a
larger value of MEP will deliver more net work per 
cycle and thus will perform better.

The net work output of a cycle is
equivalent to the product of the 
mean effective pressure and the
displacement volume.

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 Motor Animasyonu

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otto cycle: the ideal cycle for spark-ignition engines

Actual and ideal cycles in spark‐ignition engines and their P‐v diagrams.
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Four-stroke cycle The two-stroke engines are
1 cycle = 4 stroke = 2 revolution generally less efficient than their
four-stroke counterparts but
Two-stroke cycle they are relatively simple and
1 cycle = 2 stroke = 1 revolution inexpensive, and they have high
power-to-weight and power-to-
volume ratios.

Schematic of a two-stroke
T‐s diagram of the ideal Otto cycle. reciprocating engine.
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 4‐ Stroke Cycles

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 2‐ Stroke Cycles

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 2 Stroke Cycles

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 The thermal efficiency of the Otto cycle increases with the specific heat 
ratio k of the working fluid.

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 In SI engines, the compression ratio is limited
by auto ignition or engine knock.

Thermal efficiency of the ideal Otto cycle as a function 
of compression ratio (k = 1.4).

The working fluid in actual engines contains larger molecules such as carbon 

dioxide and the specific heat ratio decreases with temperature, which is one of 
the reasons that the actual cycles have lower thermal efficiencies than the ideal
Otto cycle. The thermal efficiencies of actual spark‐ignition engines range from 
about 25 to 30 %.
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EXAMPLE 9–2
An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is
at 100 kPa and 17°C, and 800 kJ/kg of heat is transferred to air during the constant-volume heat-
addition process. Accounting for the variation of specific heats of air with temperature, determine (a)
the maximum temperature and pressure that occur during the cycle,(b) the net work output, (c) the
thermal efficiency, and (d ) the mean effective
pressure for the cycle.
Solution An ideal Otto cycle with specified compression ratio and
heat input is considered.
Assumptions 1 The air-standard assumptions are applicable. 2
Kinetic and potential energy changes are negligible. 3 The
variation of specific heats with temperature is to be accounted for.
(a) The maximum temperature and pressure in an Otto cycle
occur at the end of the constant-volume heat-addition process
(state 3). To determine the temperature and pressure of air at the
end of the isentropic compression process (state 2), using data
from Table A–17:

The properties Pr (relative pressure) and vr (relative specific volume) are dimensionless
quantities used in the analysis of isentropic processes, should not be confused with the
properties pressure and specific volume.

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Process 1-2 (isentropic compression of an ideal gas):

Process 2-3 (constant-volume heat addition):

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(b) The net work output for the cycle is determined either by finding the boundary (P dV)
work involved in each process by integration and adding them or by finding the net heat
transfer that is equivalent to the net work done during the cycle. We take the latter
approach. However, first we need to find the internal energy of the air at state 4:

Process 3-4 (isentropic expansion of an ideal gas):

Process 4-1 (constant-volume heat rejection):

(c) The thermal efficiency of the cycle is determined from its definition,

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(d ) The mean effective pressure is determined from its definition,

Therefore, a constant pressure of 574.4 kPa during the power stroke would produce the
same net work output as the entire cycle.

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 DIESEL CYCLE: THE IDEAL CYCLE FOR COMPRESSION-
IGNITION ENGINES
In diesel engines, only air is compressed during the compression stroke,
eliminating the possibility of auto ignition (engine knock). Therefore, diesel
engines can be designed to operate at much higher compression ratios
than SI engines, typically between 12 and 24.

In diesel engines, the spark plug is replaced by a fuel injector, and 
only air is compressed during the compression process.

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

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Cutoff ratio

for the same compression ratio

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 Cutoff ratio

Thermal efficiency of the ideal Diesel cycle as a function of 
compression and cutoff ratios (k=1.4).

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 Dual cycle: A more realistic QUESTIONS
ideal cycle model for modern,
Diesel engines operate at
high-speed compression ignition
higher air-fuel ratios than
engine.
gasoline engines. Why?
Despite higher power to
weight ratios, two-stroke
engines are not used in
automobiles. Why?
The stationary diesel engines
are among the most efficient
power producing devices
(about 50%). Why?
What is a turbocharger? Why
are they mostly used in diesel
engines compared to gasoline
P-v diagram of an ideal dual cycle. engines.

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STIRLING AND ERICSSON CYCLES
Stirling cycle
• 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)

A regenerator is a device that borrows energy from the working fluid 
during one part of the cycle and pays it back during another part.

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 T‐s and P‐v diagrams of Carnot, Stirling, and Ericsson cycles.

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The Stirling and Ericsson cycles give a  Both the Stirling and Ericsson cycles are totally
message: Regeneration can increase  reversible, as is the Carnot cycle, and thus:
efficiency.

The Ericsson cycle is very much like the Stirling

cycle, except that the two constant-volume
processes are replaced by two constant-pressure
processes.

The execution of the Stirling cycle. A steady‐flow Ericsson engine.

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Stirling and Ericsson cycles are difficult to achieve in practice because they involve 
heat transfer through a differential temperature difference in all components 
including the regenerator. 

This would require providing infinitely large surface areas for heat transfer or 
allowing an infinitely long time for the process. Neither is practical. In reality, all 
heat transfer processes will take place through a finite temperature difference, 
the regenerator will not have an efficiency of 100 percent, and the pressure 
losses in the regenerator will be considerable. 

More research and development are needed before these engines can compete 
with the gasoline or diesel engines. Both the Stirling and the Ericsson engines are 
external combustion engines.

Despite the physical limitations and impracticalities associated with them, both 
the Stirling and Ericsson cycles give a strong message to design engineers:
Regeneration can increase efficiency. It is no coincidence that modern gas‐turbine 
and steam power plants make extensive use of regeneration. 

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 External combustion offers several advantages. 
 A variety of fuels can be used as a source of thermal energy.

 There is more time for combustion, and thus the combustion

process is more complete, which means less air pollution and
more energy extraction from the fuel.

 These engines operate on closed cycles, and thus a working

fluid that has the most desirable characteristics can be
utilized as the working fluid.

 Hydrogen and helium are two gases commonly employed in these

engines.

 In fact, the Brayton cycle with intercooling, reheating, and

regeneration, which is utilized in large gas-turbine power plants
and discussed later in this chapter, closely resembles the
Ericsson cycle.
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EXAMPLE 9–4
Using an ideal gas as the working fluid, show that the thermal efficiency of an Ericsson cycle is
identical to the efficiency of a Carnot cycle operating between the same temperature limits.

Solution It is to be shown that the thermal efficiencies of Carnot and Ericsson cycles are
identical.
Analysis Heat is transferred to the working fluid isothermally from an external source at
temperature TH during process 1-2, and it is rejected again isothermally to an external sink
at temperature TL during process 3-4.

since P1=P4 and P3=P2.

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 Stirling Cycles

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 Rejenerative Stirling Cycles

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BRAYTON CYCLE: THE IDEAL CYCLE FOR GAS-TURBINE ENGINES
The combustion process is replaced by a constant-pressure heat-addition
process from an external source, and the exhaust process is replaced by a
constant-pressure heat-rejection process to the ambient air.
1-2 Isentropic compression (in a compressor)
2-3 Constant-pressure heat addition
3-4 Isentropic expansion (in a turbine)
4-1 Constant-pressure heat rejection

An open‐cycle gas‐turbine engine. A closed‐cycle gas‐turbine engine.

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 Thermal efficiency

Pressure ratio

T‐s and P‐v diagrams for the ideal Brayton cycle.
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 Pressure ratio

Thermal efficiency of the ideal Brayton cycle as a 
function of the pressure ratio.

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The two major application areas The highest temperature in the cycle is limited
of gas-turbine engines are by the maximum temperature that the turbine
aircraft propulsion and electric blades can withstand. This also limits the
pressure ratios that can be used in the cycle.
power generation.
The air in gas turbines supplies the necessary
oxidant for the combustion of the fuel, and it
serves as a coolant to keep the temperature of
various components within safe limits. An air–
fuel ratio of 50 or above is not uncommon.

For fixed values of Tmin and Tmax, the net work of 
the Brayton cycle first increases with the 
pressure ratio, then reaches a maximum at rp =  The fraction of the turbine work used to drive 
(Tmax/Tmin)k/[2(k ‐ 1)], and finally decreases. the compressor is called the back work ratio.

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Development of Gas Turbines
1. Increasing the turbine inlet (or firing) temperatures
2. Increasing the efficiencies of turbomachinery components (turbines compressors)
3. Adding modifications to the basic cycle (intercooling, regeneration or
recuperation, and reheating).

Deviation of Actual Gas‐Turbine 
Cycles from Idealized Ones

Reasons: Irreversibilities in turbine and

compressors, pressure drops, heat losses

Isentropic efficiencies of the 
compressor and turbine

The deviation of an actual gas‐turbine cycle 
from the ideal Brayton cycle as a result of 
irreversibilities.

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EXAMPLE 9–5
A gas-turbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The gas
temperature is 300 K at the compressor inlet and 1300 K at the turbine inlet. Utilizing the air-
standard assumptions, determine (a) the gas temperature at the exits of the compressor and the
turbine, (b) the back work ratio, and (c) the thermal efficiency.
Solution A power plant operating on the ideal Brayton cycle with a specified pressure ratio
is considered. The compressor and turbine exit temperatures, back work ratio, and the
thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are
applicable. 3 Kinetic and potential energy changes are negligible. 4 The variation of specific
heats with temperature is to be considered.
(a) The air temperatures at the compressor and turbine
exits are determined from isentropic relations.

Process 1-2 (isentropic compression of an ideal gas):

at compressor exit

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 Process 3-4 (isentropic expansion of an ideal gas):

at turbine exit

(b) To find the back work ratio, we need to find the work input to the compressor and the
work output of the turbine:

(c) The thermal efficiency of the cycle is the ratio of the net power output to the total heat input:

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 The thermal efficiency could also be determined from

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EXAMPLE 9–6
Assuming a compressor efficiency of 80 percent and a turbine
efficiency of 85 percent, determine (a) the back work ratio, (b)
the thermal efficiency, and (c) the turbine exit temperature of the
gas-turbine cycle discussed in Example 9–5.

Solution The Brayton cycle discussed in Example 9–5 is

reconsidered. For specified turbine and compressor
efficiencies, the back work ratio, the thermal efficiency,
and the turbine exit temperature are to be determined.

(a) The actual compressor work and turbine work are determined by using the definitions of
compressor and turbine efficiencies,

That is, the compressor is now consuming 59.2 % of the work produced by the turbine (up
from 40.2 %). This increase is due to the irreversibilities that occur within the compressor
and the turbine.
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(b) In this case, air will leave the compressor at a higher temperature and enthalpy, which
are determined to be

(c) The air temperature at the turbine exit is determined from an energy balance on the
turbine:

Then, from Table A–17,

This value is considerably higher than the air temperature at the compressor exit (T2a=598
K), which suggests the use of regeneration to reduce fuel cost.
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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 the hot exhaust gases in a counter-flow
heat exchanger (a regenerator or a recuperator).
The thermal efficiency of the Brayton cycle increases
as a result of regeneration since less fuel is used for
the same work output.

T‐s diagram of a Brayton cycle 
with regeneration.

A gas‐turbine engine with regenerator.
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 Effectiveness of regenerator
The thermal efficiency depends on the ratio of the minimum to maximum
temperatures as well as the pressure ratio.
Regeneration is most effective at lower pressure ratios and low minimum-to-
maximum temperature ratios.

Effectiveness of regenerator

Effectiveness under cold‐air 
standard assumptions

T‐s diagram of a Brayton cycle  Under cold‐air standard 

with regeneration. assumptions

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 Can regeneration be used at high pressure ratios?

Pressure ratio

Thermal efficiency of the ideal Brayton
cycle with and without regeneration.

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EXAMPLE 9–7
Determine the thermal efficiency of the gas-turbine power plant described in Example 9–6 if a
regenerator having an effectiveness of 80 percent is installed.

Solution The gas-turbine power plant discussed in

Example 9–6 is equipped with a regenerator. For a
specified effectiveness, the thermal efficiency of the cycle
is to be determined.
Analysis We first determine the enthalpy of the air at the
exit of the regenerator, using the definition of effectiveness:

Discussion Note that the thermal efficiency of the power plant has gone up from 26.6 to
36.9 percent as a result of installing a regenerator that helps to recuperate some of the
thermal energy of the exhaust gases.

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THE BRAYTON CYCLE WITH INTERCOOLING, REHEATING,
AND REGENERATION
For minimizing work input to compressor and
maximizing work output from turbine:

A gas‐turbine engine with two‐stage compression with intercooling, two‐stage expansion 
with reheating, and regeneration and its T‐s diagram.

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Multistage compression with intercooling: The work required to compress a gas
between two specified pressures can be decreased by carrying out the compression
process in stages and cooling the gas in between. This keeps the specific volume as
low as possible.
Multistage expansion with reheating keeps the specific volume of the working fluid
as high as possible during an expansion process, thus maximizing work output.
Intercooling and reheating always decreases the thermal efficiency unless they are
accompanied by regeneration. Why?

Comparison of work inputs to a single‐stage  Compression and expansion stages increases, the 

compressor (1AC) and a two‐stage compressor  gas‐turbine cycle with intercooling, reheating, 
with intercooling (1ABD). and regeneration approaches the Ericsson cycle.

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EXAMPLE 9–8
An ideal gas-turbine cycle with two stages of compression and two stages of expansion has an
overall pressure ratio of 8. Air enters each stage of the compressor at 300 K and each stage of the
turbine at 1300 K. Determine the back work ratio and the thermal efficiency of this gas-turbine cycle,
assuming (a) no regenerators and (b) an ideal regenerator with 100 percent effectiveness. Compare
the results with those obtained in Example 9–5.

Solution An ideal gas-turbine cycle with two stages of

compression and two stages of expansion is considered. The
back work ratio and the thermal efficiency of the cycle are to
be determined for the cases of no regeneration and maximum
regeneration.
Assumptions 1 Steady operating conditions exist. 2 The air-
standard assumptions are applicable. 3 Kinetic and potential
energy changes are negligible.

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(a) In the absence of any regeneration, the back work ratio and the thermal efficiency are
determined by using data from Table A–17 as follows:

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A comparison of these results with those obtained in Example 9–5 reveals that multistage
compression with intercooling and multistage expansion with reheating improve the back work
ratio (it drops from 40.2 to 30.4 %) but hurt the thermal efficiency (it drops from 42.6 to 35.7 %).

(b) an ideal regenerator with 100 percent effectiveness.

The heat input and the thermal efficiency in this case are

That is, the thermal efficiency almost doubles as a result of regeneration compared to the
no-regeneration case. The overall effect of two-stage compression and expansion with
intercooling, reheating, and regeneration on the thermal efficiency is an increase of 63 %.
and the thermal efficiency will approach.

Adding a second stage increases the thermal efficiency from 42.6 to 69.6 %, an increase
of 27 % points. This is a significant increase in efficiency. Adding more stages, however
can increase the efficiency an additional 7.3 % points at most, and usually cannot be
justified economically.
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IDEAL JET-PROPULSION CYCLES
Gas-turbine engines are widely used to power aircraft because they are light
and compact and have a high power-to-weight ratio.

Aircraft gas turbines operate on an open cycle called a jet-propulsion cycle.

The ideal jet-propulsion cycle differs from the simple ideal Brayton cycle in
that the gases are not expanded to the ambient pressure in the turbine.
Instead, they are expanded to a pressure such that the power produced by
the turbine is just sufficient to drive the compressor and the auxiliary
equipment.

The net work output of a jet-propulsion cycle is zero. The gases that exit the
turbine at a relatively high pressure are subsequently accelerated in a nozzle
to provide the thrust to propel the aircraft.

Aircraft are propelled by accelerating a fluid in the opposite direction to

motion.

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 In jet engines, the high‐temperature and 
high‐pressure gases leaving the turbine 
are accelerated in a nozzle to provide 
thrust.

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Thrust (propulsive force)

Propulsive power

Propulsive efficiency

Propulsive power is the thrust acting 
on the aircraft through a distance per 
unit time.

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 Basic components of a turbojet engine and the T‐s diagram for the ideal turbojet cycle.

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Modifications to Turbojet Engines
The first airplanes built were all propeller-
driven, with propellers powered by engines
essentially identical to automobile engines.
Both propeller-driven engines and jet-
propulsion-driven engines have their own
strengths and limitations, and several
attempts have been made to combine the
desirable characteristics of both in one
engine.
Two such modifications are the propjet Energy supplied to an aircraft
engine and the turbofan engine. (from the burning of a fuel)
manifests itself in various forms.
A turbofan engine.
The most widely used engine
in aircraft propulsion is the
turbofan (or fanjet) engine
wherein a large fan driven by
the turbine forces a
considerable amount of air
through a duct (cowl)
surrounding the engine.
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 A modern jet engine used to power Boeing 777 aircraft. This is a Pratt & 
Whitney PW4084 turbofan capable of producing 374 kN of thrust. It is 4.87 
m long, has a 2.84 m diameter fan, and it weighs 6800 kg.

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 Various engine types:
Turbofan, Propjet, Ramjet, Sacramjet, Rocket

A turboprop engine.

A ramjet engine.

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SECOND-LAW ANALYSIS OF GAS POWER CYCLES

Exergy destruction for a closed system

For a steady‐flow system

Steady‐flow, one‐inlet, one‐exit

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 Exergy destruction of a cycle

For a cycle with heat transfer only with a source and a sink

Closed system exergy

Stream exergy

A second-law analysis of these cycles reveals where the largest

irreversibilities occur and where to start improvements.

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 SUMMARY

 Basic considerations in the analysis of power cycles

 The Carnot cycle and its value in engineering
 Air-standard asssumptions
 An overview of reciprocating engines
 Otto cycle: The ideal cycle for spark-ignition engines
 Diesel cycle: The ideal cycle for compression-ignition engines
 Stirling and Ericsson cycles
 Brayton cycle: The ideal cycle for gas-turbine engines
 The Brayton cycle with regeneration
 The Brayton cycle with intercooling, reheating, and regeneration
 Ideal jet-propulsion cycles
 Second-law analysis of gas power cycles

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The world’s largest turbine, with an output of 340 MW that a new
combined cycle power plant achieves a record-breaking efficiency of
more than 60 % when it goes into operation in 2011.

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 After assembly at Siemens’ gas turbine plant in Berlin, the world’s largest gas
turbine hits the road. The turbine arrives on a flat bed trailer at its destination

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EXAMPLE 9–16
A gas-turbine power plant operates on the simple Brayton cycle between the
pressure limits of 100 and 1200kPa. The working fluid is air, which enters the
compressor at 30°C at a rate of 150 m3/min and leaves the turbine at 500°C.
Using variable specific heats for air and assuming a compressor isentropic
efficiency of 82 percent and a turbine isentropic efficiency of 88 percent,
determine (a) the net power output, (b) the back work ratio, and (c) the
thermal efficiency.
Assumptions 1 The air-standard assumptions are
applicable. 2 Kinetic and potential energy changes
are negligible. 3 Air is an ideal gas with variable
specific heats.

Properties The gas constant of air is R = 0.287 kJ/kg·K

(Table A-1).

70
Analysis (a) For this problem, we use the properties from EES software. Remember that for
an ideal gas, enthalpy is a function of temperature only whereas entropy is a function of both
temperature and pressure.

71
The mass flow rate is determined from

The net power output is

72
(b)
(b)The
Theback
backwork
workratio
ratioisis

(c)
(c)The
Therate
rateofofheat
heatinput
inputand
andthe
thethermal
thermalefficiency
efficiencyare
are

73
EXAMPLE 9-25
A gas-turbine engine with regeneration operates with two stages of
compression and two stages of expansion. The pressure ratio across each
stage of the compressor and turbine is 3.5. The air enters each stage of the
compressor at 300 K and each stage of the turbine at 1200 K. The compressor
and turbine efficiencies are 78 and 86 percent, respectively, and the
effectiveness of the regenerator is 72 percent. Determine the back work
ratio and the thermal efficiency of the cycle, assuming constant specific
heats for air at room temperature.
Assumptions 1 The air-standard assumptions are applicable.
2 Kinetic and potential energy changes are negligible.
3 Air is an ideal gas with constant specific heats.

74
Analysis The work inputs of each stage of compressor are identical, so are the work
outputs of each stage of the turbine.

75
76
 Thermodynamics: An Engineering Approach, 5th Edition
Yunus A. Cengel, Michael A. Boles
McGraw-Hill, 2008

Chapter 10
VAPOR AND COMBINED POWER
CYCLES

Prof. Dr. Ali PINARBAŞI

Yildiz Technical University
Mechanical Engineering Department
Yildiz, ISTANBUL

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1
 VAPOR AND COMBINED POWER CYCLES

10–1 The Carnot Vapor Cycle

10–2 Rankine Cycle: The Ideal Cycle for Vapor Power Cycles
Energy Analysis of the Ideal Rankine Cycle
10–3 Deviation of Actual Vapor Power Cycles from Idealized Ones
10–4 How Can We Increase the Efficiency of the Rankine Cycle?
Lowering the Condenser Pressure (Lowers Tlow,avg)
Superheating the Steam to High Temperatures (Increases Thigh,avg)
Increasing the Boiler Pressure (Increases Thigh,avg)
10–5 The Ideal Reheat Rankine Cycle
10–6 The Ideal Regenerative Rankine Cycle
Open Feedwater Heaters
Closed Feedwater Heaters
10–7 Second-Law Analysis of Vapor Power Cycles
10–8 Cogeneration
10–9 Combined Gas–Vapor Power Cycles

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 Objectives
• Evaluate the performance of gas power cycles for which the
working fluid remains a gas throughout the entire cycle.
• Analyze vapor power cycles in which the working fluid is
alternately vaporized and condensed.
• Analyze power generation coupled with process heating called
cogeneration.
• Investigate ways to modify the basic Rankine vapor power cycle to
increase the cycle thermal efficiency.
• Analyze the reheat and regenerative vapor power cycles.
• Analyze power cycles that consist of two separate cycles known
as combined cycles and binary cycles.

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 THE CARNOT VAPOR CYCLE

The Carnot cycle is the most efficient cycle operating between two specified
temperature limits but it is not a suitable model for power cycles. Because:

Process 1-2 Limiting the heat transfer processes to two-phase systems

severely limits the maximum temperature that can be used in the cycle (374°C
for water)

Process 2-3 The turbine cannot handle steam with a high moisture content
because of the impingement of liquid droplets on the turbine blades causing
erosion and wear.

Process 4-1 It is not practical to design a compressor that handles two phases.
The cycle in (b) is not suitable since it requires isentropic compression to
extremely high pressures and isothermal heat transfer at variable pressures.

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 Carnot Cycles

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 T‐s diagram of two Carnot vapor cycles.

1-2 isothermal heat addition in a boiler

2-3 isentropic expansion in a turbine
3-4 isothermal heat rejection in a condenser
4-1 isentropic compression in a compressor

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RANKINE CYCLE: THE IDEAL CYCLE FOR VAPOR POWER CYCLES

Many of the impracticalities associated with the Carnot cycle can be eliminated by
superheating the steam in the boiler and condensing it completely in the condenser.

The cycle that results is the Rankine cycle, which is the ideal cycle for vapor power
plants. The ideal Rankine cycle does not involve any internal irreversibilities.

The simple ideal Rankine cycle.
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 Rankine Cycles

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Energy Analysis of the Ideal Rankine Cycle
Steady-flow energy equation

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 The efficiency of power plants in the U.S. is
often expressed in terms of heat rate,

The thermal efficiency can be interpreted as the ratio of the area enclosed by the
cycle on a T-s diagram to the area under the heat-addition process.

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EXAMPLE 10–1
Consider a steam power plant operating on the simple ideal Rankine cycle. The steam enters the
turbine at 3 MPa and 350°C and is condensed in the condenser at a pressure of 75 kPa. Determine
the thermal efficiency of this cycle.

Solution A steam power plant operating on the simple ideal Rankine cycle is considered.
The thermal efficiency of the cycle is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes
are negligible.

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 That is, this power plant converts 26 % of the heat it receives in the boiler to net work. An
actual power plant operating between the same temperature and pressure limits will have
a lower efficiency because of the irreversibilities such as friction.
Discussion Notice that the back work ratio (rpw =win/wout) of this power plant is 0.004, and thus
only 0.4 % of the turbine work output is required to operate the pump. Having such low back
work ratios is characteristic of vapor power cycles. This is in contrast to the gas power cycles,
which typically have very high back work ratios (about 40 to 80 %). It is also interesting to note
the thermal efficiency of a Carnot cycle operating between the same temperature limits

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DEVIATION OF ACTUAL VAPOR POWER CYCLES FROM IDEALIZED ONES

The actual vapor power cycle differs from the ideal Rankine cycle as a result of
irreversibilities in various components.
Fluid friction and heat loss to the surroundings are the two common sources of
irreversibilities.

Isentropic efficiencies

(a) Deviation of actual vapor power cycle from the ideal Rankine cycle.       

(b) The effect of pump and turbine irreversibilities on the ideal Rankine cycle.

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EXAMPLE 10–2
A steam power plant operates on the cycle shown in Figure. If the isentropic efficiency of the turbine
is 87 % and the isentropic efficiency of the pump is 85 %, determine (a) the thermal efficiency of the
cycle and (b) the net power output of the plant for a mass flow rate of 15 kg/s.

Solution A steam power cycle with specified turbine and pump efficiencies is considered.
The thermal efficiency and the net power output are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes
are negligible.

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 (a) The thermal efficiency of a cycle;

Pump work input:

Turbine work output:

Boiler heat input:

(b) The power produced by this power plant is

Resim şu anda görüntülenemiy or.

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HOW CAN WE INCREASE THE EFFICIENCY OF THE RANKINE
CYCLE?
The basic idea behind all the modifications to increase the thermal efficiency
of a power cycle is the same: Increase the average temperature at which heat is
transferred to the working fluid in the boiler, or decrease the average
temperature at which heat is rejected from the working fluid in the condenser.

Lowering the Condenser Pressure (Lowers Tlow,avg)

To take advantage of the increased

efficiencies at low pressures, the condensers
of steam power plants usually operate well
below the atmospheric pressure. There is a
lower limit to this pressure depending on the
temperature of the cooling medium
Side effect: Lowering the condenser
pressure increases the moisture content of
the steam at the final stages of the turbine.

The effect of lowering the condenser pressure 
on the ideal Rankine cycle.

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 Superheating the Steam to High Temperatures (Increases Thigh,avg)

Both the net work and heat input increase

as a result of superheating the steam to a
higher temperature. The overall effect is an
increase in thermal efficiency since the
average temperature at which heat is
added increases.
Superheating to higher temperatures
decreases the moisture content of the
steam at the turbine exit, which is
desirable.
The temperature is limited by metallurgical
considerations. Presently the highest
steam temperature allowed at the turbine
The effect of superheating the steam  inlet is about 620°C.
to higher temperatures on the ideal 
Rankine cycle.

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 Increasing the Boiler Pressure (Increases Thigh,avg)
For a fixed turbine inlet temperature, the Today many modern steam power plants
cycle shifts to the left and the moisture operate at supercritical pressures (P >
content of steam at the turbine exit 22.06 MPa) and have thermal efficiencies
increases. This side effect can be of about 40% for fossil-fuel plants and
corrected by reheating the steam. 34% for nuclear plants.

The effect of increasing the boiler pressure  A supercritical Rankine cycle.

on the ideal Rankine cycle.

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 Rankine Cycles (Increases Thigh,avg )

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EXAMPLE 10–3
Consider a steam power plant operating on the ideal Rankine cycle. The steam enters the turbine at
3 MPa and 350°C and is condensed in the condenser at a pressure of 10 kPa. Determine (a) the
thermal efficiency of this power plant, (b) the thermal efficiency if steam is superheated to 600°C
instead of 350°C, and (c) the thermal efficiency if the boiler pressure is raised to 15 MPa while the
turbine inlet temperature is maintained at 600°C.

(a) The thermal efficiency is determined in a similar manner:

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 Therefore, the thermal efficiency increases from 26.0 to 33.4 percent as a result of
lowering the condenser pressure from 75 to 10 kPa. At the same time, however, the
quality of the steam decreases from 88.6 to 81.3 percent (in other words, the moisture
content increases from 11.4 to 18.7 percent).

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 (b) States 1 and 2 remain the same in this case, and the enthalpies at state 3 (3 MPa and
600°C) and state 4 (10 kPa and s4=s3) are determined to be

Therefore, the thermal efficiency increases from 33.4 to 37.3 percent as a result of
superheating the steam from 350 to 600°C. At the same time, the quality of the steam
increases from 81.3 to 91.5 percent (in other words, the moisture content decreases
from 18.7 to 8.5 percent).

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(c) State 1 remains the same in this case, but the other states change. The enthalpies at
state 2 (15 MPa and s2=s1), state 3 (15 MPa and 600°C), and state 4 (10 kPa and s4=s3)
are determined in a similar manner to be

Discussion The thermal efficiency increases from 37.3 to 43.0 percent as a result of
raising the boiler pressure from 3 to 15 MPa while maintaining the turbine inlet
temperature at 600°C. At the same time, however, the quality of the steam decreases from
91.5 to 80.4 % (in other words, the moisture content increases from 8.5 to 19.6 %).

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THE IDEAL REHEAT RANKINE CYCLE
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?
1. Superheat the steam to very high temperatures. It is limited metallurgically.
2. Expand the steam in the turbine in two stages, and reheat it in between (reheat)

The ideal reheat Rankine cycle.

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The single reheat in a modern power plant
improves the cycle efficiency by 4 to 5% by
increasing the average temperature at
which heat is transferred to the steam.
The average temperature during the reheat
process can be increased by increasing the
number of expansion and reheat stages. As
the number of stages is increased, the
expansion and reheat processes approach
an isothermal process at the maximum
temperature. The use of more than two
reheat stages is not practical. The
theoretical improvement in efficiency from
the second reheat is about half of that
which results from a single reheat.
The reheat temperatures are very close or The average temperature at which heat is 
transferred during reheating increases as 
equal to the turbine inlet temperature.
the number of reheat stages is increased.
The optimum reheat pressure is about one-
fourth of the maximum cycle pressure.

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 Rankine Cycles (the ideal reheat rankine cycle)

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EXAMPLE 10–4
Consider a steam power plant operating on the ideal reheat Rankine cycle. Steam enters the high-
pressure turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa.
If the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10.4
percent, determine (a) the pressure at which the steam should be reheated and (b) the thermal
efficiency of the cycle. Assume the steam is reheated to the
inlet temperature of the high-pressure turbine.
Solution A steam power plant operating on the ideal reheat Rankine cycle is considered.
For a specified moisture content at the turbine exit, the reheat pressure and the thermal
efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes
are negligible.

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 (a) The reheat pressure is determined from the requirement that the entropies at states 5
and 6 be the same:

Therefore, steam should be reheated at a pressure of 4 MPa or lower to prevent a

moisture content above 10.4 percent.

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(b) To determine the thermal efficiency, we need to know the enthalpies at all other states:

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 Discussion This problem was solved in Example 10–3c for the same pressure and
temperature limits but without the reheat process. A comparison of the two results
reveals that reheating reduces the moisture content from 19.6 to 10.4 percent while
increasing the thermal efficiency from 43.0 to 45.0 percent.

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THE IDEAL REGENERATIVE RANKINE CYCLE
Heat is transferred to the working fluid during
process 2-2 at a relatively low temperature.
This lowers the average heat-addition
temperature and thus the cycle efficiency.
In steam power plants, steam is extracted
from the turbine at various points. This steam,
which could have produced more work by
expanding further in the turbine, is used to
heat the feedwater instead. The device where
the feedwater is heated by regeneration is
called a regenerator, or a feedwater heater
(FWH).
A feedwater heater is basically a heat
exchanger where heat is transferred from the
The first part of the heat‐addition process  steam to the feedwater either by mixing the
in the boiler takes place at relatively low  two fluid streams (open feedwater heaters) or
temperatures. without mixing them (closed feedwater
heaters).

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Open Feedwater Heaters
An open (or direct-contact)
feedwater heater is basically a mixing
chamber, where the steam extracted
from the turbine mixes with the
feedwater exiting the pump. Ideally, the
mixture leaves the heater as a
saturated liquid at the heater pressure.
fraction of steam extracted

The ideal regenerative Rankine cycle with an open feedwater heater.

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 Rankine Cycles (Open Feedwater Heaters)

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Closed Feedwater Heaters
Another type of feedwater heater frequently used in steam power plants is
the closed feedwater heater, in which heat is transferred from the extracted
steam to the feedwater without any mixing taking place. The two streams now
can be at different pressures, since they do not mix.

The ideal regenerative Rankine cycle with a closed feedwater heater.

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The closed feedwater heaters are more complex because of the internal tubing
network, and thus they are more expensive. Heat transfer in closed feedwater
heaters is less effective since the two streams are not allowed to be in direct contact.
However, closed feedwater heaters do not require a separate pump for each heater
since the extracted steam and the feedwater can be at different pressures.

A steam power plant with one open

and three closed feedwater heaters.
Open feedwater heaters
are simple and inexpensive
and have good heat
transfer characteristics. For
each heater, however, a
pump is required to handle
the feedwater.

Most steam power plants

use a combination of open
and closed feedwater
heaters.

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EXAMPLE 10–5
Consider a steam power plant operating on the ideal regenerative Rankine cycle with one open
feedwater heater. Steam enters the turbine at 15 Mpa and 600°C and is condensed in the
condenser at a pressure of 10 kPa. Some steam leaves the turbine at a pressure of 1.2 MPa and
enters the open feedwater heater. Determine the fraction of steam extracted from the turbine and
the thermal efficiency of the cycle.

Solution A steam power plant operates on the ideal regenerative Rankine cycle with one
open feedwater heater. The fraction of steam extracted from the turbine and the thermal
efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes
are negligible.

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 The energy analysis of open feedwater heaters is identical to the energy analysis of
mixing chambers. The feedwater heaters are generally well insulated (Q=0), and they do
not involve any work interactions (W=0). By neglecting the kinetic and potential energies
of the streams, the energy balance reduces for a feedwater heater to

𝑚6/ 𝑚5). Solving for y and

where y is the fraction of steam extracted from the turbine (=𝑚
substituting the enthalpy values, we find

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Discussion This problem was worked out in Example 10–3c for the same pressure and
temperature limits but without the regeneration process. A comparison of the two results
reveals that the thermal efficiency of the cycle has increased from 43.0 to 46.3 percent as
a result of regeneration. The net work output decreased by 171 kJ/kg, but the heat input
decreased by 607 kJ/kg, which results in a net increase in the thermal efficiency.

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EXAMPLE 10–6
Consider a steam power plant that operates on an ideal reheat–regenerative Rankine cycle with
one open feedwater heater, one closed feedwater heater, and one reheater. Steam enters the
turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa. Some
steam is extracted from the turbine at 4 MPa for the closed feedwater heater, and the remaining
steam is reheated at the same pressure to 600°C. The extracted steam is completely condensed in
the heater and is pumped to 15 Mpa before it mixes with the feedwater at the same pressure.
Steam for the open feedwater heater is extracted from the low-pressure turbine at a pressure of 0.5
MPa. Determine the fractions of steam extracted from the turbine as well as the thermal efficiency
of the cycle.

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Solution A steam power plant operates on an ideal reheat–regenerative Rankine cycle
with one open feedwater heater, one closed feedwater heater, and one reheater. The
fraction of steam extracted from the turbine and the thermal efficiency are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes
are negligible.

The fractions of steam extracted are determined from the mass and energy balances of
the feedwater heaters:

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Closed feedwater heater:

Open feedwater heater:

The enthalpy at state 8 is determined by applying the conservation of mass and energy
equations to the mixing chamber, which is assumed to be insulated:

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 Discussion This problem was worked out in Example 10–4 for the same pressure
and temperature limits with reheat but without the regeneration process.
A comparison of the two results reveals that the thermal efficiency of the cycle
has increased from 45.0 to 49.2 percent as a result of regeneration.

The thermal efficiency of this cycle could also be determined from

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SECOND-LAW ANALYSIS OF VAPOR POWER CYCLES
Exergy destruction for a steady‐flow system

Steady‐flow, one‐inlet, one‐exit

Exergy destruction of a cycle

For a cycle with heat transfer only with a source and a sink

Stream exergy

A second-law analysis of vaporpower cycles reveals where the largest

irreversibilities occur and where to start improvements.
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EXAMPLE 10–7
Determine the exergy destruction associated with the Rankine cycle (all four processes as well as
the cycle) discussed in Example 10–1, assuming that heat is transferred to the steam in a furnace
at 1600 K and heat is rejected to a cooling medium at 290 K and 100 kPa. Also, determine the
exergy of the steam leaving the turbine.

Solution The Rankine cycle analyzed in Example 10–1 is reconsidered. For specified
source and sink temperatures, the exergy destruction associated with the cycle and exergy
of the steam at turbine exit are to be determined.

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Processes 1-2 and 3-4 are isentropic (s1=s2, s3=s4) and therefore do not involve any
internal or external irreversibilities, that is,

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Therefore, the irreversibility of the cycle is

The exergy (work potential) of the steam leaving the turbine is determined from Eq. 10–22.
Disregarding the kinetic and potential energies, it reduces to

Discussion Note that 449.1 kJ/kg of work could be obtained from the steam leaving the
turbine if it is brought to the state of the surroundings in a reversible manner.
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COGENERATION
Many industries require energy input in the form of heat, called process heat.
Process heat in these industries is usually supplied by steam at 5 to 7 atm and
150 to 200°C. Energy is usually transferred to the steam by burning coal, oil,
natural gas, or another fuel in a furnace.

Industries that use large amounts

of process heat also consume a
large amount of electric power.
It makes sense to use the already-
existing work potential to produce
power instead of letting it go to
waste.
The result is a plant that produces
electricity while meeting the
process-heat requirements of
certain industrial processes
A simple process‐heating plant. (cogeneration plant)

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

• The utilization factor of the ideal

steam-turbine cogeneration plant
is 100%.
• Actual cogeneration plants have
utilization factors as high as 80%.
• Some recent cogeneration plants
have even higher utilization
factors.

An ideal cogeneration plant.

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 At times of high demand for process heat,
all the steam is routed to the process-
heating units and none to the condenser
(m7= 0). The waste heat is zero in this
mode.
If this is not sufficient, some steam leaving
the boiler is throttled by an expansion or
pressure-reducing valve to the extraction
pressure P6 and is directed to the
process-heating unit.
Maximum process heating is realized
when all the steam leaving the boiler
passes through the PRV (m5= m4). No
A cogeneration plant with adjustable loads. power is produced in this mode.
When there is no demand for process
heat, all the steam passes through the
turbine and the condenser (m5=m6=0),
and the cogeneration plant operates as an
ordinary steam power plant.

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EXAMPLE 10–8
Consider the cogeneration plant, steam enters the turbine at 7 MPa and 500°C. Some steam is
extracted from the turbine at 500 kPa for process heating. The remaining steam continues to
expand to 5 kPa. Steam is then condensed at constant pressure and pumped to the boiler pressure
of 7 MPa. At times of high demand for process heat, some steam leaving the boiler is throttled to
500 kPa and is routed to the process heater. The extraction fractions are adjusted so that steam
leaves the process heater as a saturated liquid at 500 kPa. It is subsequently pumped to 7 MPa.
The mass flow rate of steam through the boiler is 15 kg/s. Disregarding any pressure drops and
heat losses in the piping and assuming the turbine and the pump to be isentropic, determine (a) the
maximum rate at which process heat can be supplied, (b) the power produced and the utilization
factor when no process heat is supplied, and (c) the rate of process heat supply when 10 % of the
steam is extracted before it enters the turbine and 70 % of the steam is extracted from the turbine at
500 kPa for process heating.

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 Solution A cogeneration plant is considered. The maximum rate of process heat supply,
the power produced and the utilization factor when no process heat is supplied, and the
rate of process heat supply when steam is extracted from the steam line and turbine at
specified ratios are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Pressure drops and heat losses in
piping are negligible. 3 Kinetic and potential energy changes are negligible.

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(a) The maximum rate of process heat is achieved when all the steam leaving the boiler
is throttled and sent to the process heater and none is sent to the turbine (that is,
𝑚4=𝑚7 𝑚1= 15 kg/s and 𝑚3=𝑚5 𝑚6=0)

The utilization factor is 100 % in this case since no heat is rejected in the condenser,
heat losses from the piping and other components are assumed to be negligible, and
combustion losses are not considered.
(b) When no process heat is supplied, all the steam leaving the boiler will pass through
the turbine and will expand to the condenser pressure of 5 kPa (that is, 𝑚3=𝑚 𝑚6 𝑚1=15
kg/s and 𝑚2=𝑚5=0). Maximum power will be produced;

That is, 40.8 percent of the energy is utilized for a useful purpose. Notice that the utilization
factor is equivalent to the thermal efficiency in this case.
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(c) Neglecting any kinetic and potential energy changes, an energy balance on the process
heater yields

Discussion Note that 26,199 kW of the heat transferred will be utilized in the process
heater. We could also show that 10,966 kW of power is produced in this case, and the rate
of heat input in the boiler is 42,970 kW. Thus the utilization factor is 86.5 percent.

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 COMBINED GAS–VAPOR POWER CYCLES
• The continued quest for higher thermal efficiencies has resulted in rather
innovative modifications to conventional power plants.
• A popular modification involves a gas power cycle topping a vapor power
cycle, which is called the combined gas–vapor cycle, or just the combined
cycle.
• The combined cycle of greatest interest is the gas-turbine (Brayton) cycle
topping a steam-turbine (Rankine) cycle, which has a higher thermal
efficiency than either of the cycles executed individually.
• It makes engineering sense to take advantage of the very desirable
characteristics of the gas-turbine cycle at high temperatures and to use the
high-temperature exhaust gases as the energy source for the bottoming cycle
such as a steam power cycle. The result is a combined gas–steam cycle.
• Recent developments in gas-turbine technology have made the combined
gas–steam cycle economically very attractive.
• The combined cycle increases the efficiency without increasing the initial cost
greatly. Consequently, many new power plants operate on combined cycles,
and many more existing steam- or gas-turbine plants are being converted to
combined-cycle power plants.
• Thermal efficiencies over 50% are reported.
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 Combined gas–steam power plant.
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 Combined Power Cycles

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 Thermal Power Cycles

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SUMMARY
 The Carnot vapor cycle
 Rankine cycle: The ideal cycle for vapor power cycles
 Energy analysis of the ideal Rankine cycle
 Deviation of actual vapor power cycles from idealized ones
 How can we increase the efficiency of the Rankine cycle?
 Lowering the condenser pressure (Lowers Tlow,avg)
 Superheating the steam to high temperatures (Increases Thigh,avg)
 Increasing the boiler pressure (Increases Thigh,avg)
 The ideal reheat Rankine cycle
 The ideal regenerative Rankine cycle
 Open feedwater heaters
 Closed feedwater heaters
 Second-law analysis of vapor power cycles
 Cogeneration
 Combined gas–vapor power cycles

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EXAMPLE 1
Steam is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 8.0 MPa
and saturated liquid exits the condenser at a pressure of 0.008 MPa. The net power output of the
cycle is 100 MW. Determine for the cycle
(a) the thermal efficiency,
(b) the back work ratio,
(c) the mass flow rate of the steam, in kg/h,
(d) the rate of heat transfer, 𝑄 𝑖𝑛, into the working fluid as it passes through the boiler, in MW,
(e) the rate of heat transfer 𝑄 𝑜𝑢𝑡, from the condensing steam as it passes through the condenser,
(f) the mass flow rate of the condenser cooling water, in kg/ h, if cooling water enters the
condenser at 15C and exits at 35C.

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h1= 2758.0 kJ/kg and s1= 5.7432 kJ/kg
K.

(a) The net power developed by the cycle

is

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(b) The back work ratio
is

(c) The mass flow rate of the steam

can be obtained from the expression
for the net power given in part (a).

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 (d) With the expression for 𝑄 𝑖𝑛 from part (a) and previously determined specific
enthalpy values

(e) Mass and energy rate balances applied to a control volume enclosing the
steam side of the condenser give

(f) Taking a control volume around the condenser, the mass and energy rate
balances give at steady state

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 𝑚 is the mass flow rate of the cooling water

 Note that the back work ratio is relatively low for the Rankine cycle. In the
present case, the work required to operate the pump is less than 1% of the
turbine output.

 In this example, 62.9% of the energy added to the working fluid by heat
transfer is subsequently discharged to the cooling water. Although
considerable energy is carried away by the cooling water, its exergy is small
because the water exits
at a temperature only a few degrees greater than that of the surroundings.
See Sec. 8.6 for further discussion.

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EXAMPLE 2
Reconsider the vapor power cycle of Example 1, but include in the analysis that the turbine and the
pump each have an isentropic efficiency of 85%. Determine for the modified cycle
(a) the thermal efficiency,
(b) the mass flow rate of steam, in kg/h, for a net power output of 100 MW,
(c) the rate of heat transfer into the working fluid as it passes through the boiler, in MW,
(d) the rate of heat transfer from the condensing steam as it passes through the condenser, in MW,
(e) the mass flow rate of the condenser cooling water, in kg/h, if cooling water enters the condenser
at 15C and exits as 35C. Discuss the effects on the vapor cycle of irreversibilities within the turbine
and pump.

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 (a) the thermal efficiency

(b) the mass flow rate of the steam is

(c) the rate of heat transfer into the working fluid as it passes through the
boiler,

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 (d) The rate of heat transfer from the condensing steam to the cooling water
is

(e) The mass flow rate of the cooling water can be determined from

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EXAMPLE 3 HEAT EXCHANGER EXERGY ANALYSIS
The heat exchanger unit of the boiler of Example 2 has a stream of water entering as a liquid at 8.0
MPa and exiting as a saturated vapor at 8.0 MPa. In a separate stream, gaseous products of
combustion cool at a constant pressure of 1 atm from 1107 to 547C. The gaseous stream can be
modeled as air as an ideal gas. Let T0 22C, p0 1 atm. Determine
(a) the net rate at which exergy is carried into the heat exchanger unit by the gas stream, in MW,
(b) the net rate at which exergy is carried from the heat exchanger by the water stream, in MW,
(c) the rate of exergy destruction, in MW,
(d) the exergetic efficiency given by Eq. 7.45.

1. The control volume shown in the accompanying figure operates at steady state with .
2. Kinetic and potential energy effects can be ignored.
3. The gaseous combustion products are modeled as air as an ideal gas.
4. The air and the water each pass through the steam generator at constant pressure.
5. Only 69% of the exergy entering the plant with the fuel remains after accounting for the
stack loss and combustion exergy destruction.
6. T0= 22C, p0=1 atm.

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(a) The net exergy rate into the heat exchanger;

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 (b) The net exergy
rate of the boiler;

(c) The rate of exergy destruction can be evaluated by reducing the exergy rate
balance to obtain

(d) The exergetic

efficiency
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EXAMPLE 4 TURBINE AND PUMP EXERGY ANALYSIS
Reconsider the turbine and pump of Example 2. Determine for each of these components the rate
at which exergy is destroyed, in MW. Express each result as a percentage of the exergy entering
the plant with the fuel.

The turbine and pump operate adiabatically and each has an efficiency of 85%.

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 the rate of exergy destruction for the
turbine

the net rate at which exergy is supplied by the cooling combustion gases is 231.28 MW.
The turbine rate of exergy destruction expressed as a percentage of this is
(16.72/231.28)(100%)= 7.23 %.

the exergy destruction rate for the pump

(0.11/231.28)(69%) = 0.03%.
The net power output of the vapor power plant of Example 2 is 100 MW.
Expressing this as a percentage of the rate at which exergy is carried into the plant
with the fuel, (100/231.28)(69%) = 30%
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EXAMPLE 5 CONDENSER EXERGY ANALYSIS
The condenser of Example 2 involves two separate water streams. In one stream a two-phase
liquid–vapor mixture enters at 0.008 MPa and exits as a saturated liquid at 0.008 MPa. In the other
stream, cooling water enters at 15C and exits at 35C.
(a) Determine the net rate at which exergy is carried from the condenser by the cooling water, in
MW. Express this result as a percentage of the exergy entering the plant with the fuel.
(b) Determine for the condenser the rate of exergy destruction, in MW.
Express this result as a percentage of the exergy entering the plant with the fuel.

(a) The net rate at which exergy is carried out of the condenser

As a percentage of the exergy entering the plant with the fuel; (2.23/231.28)(69%) = 1%.
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 (b) The rate of exergy destruction for the condenser can be evaluated by
reducing the exergy rate balance. Alternatively, the relationship 𝐸 𝑑 =To 𝜎𝑐𝑣 can
be employed, where 𝜎𝑐𝑣 is the time rate of entropy production for the condenser
determined from
an entropy rate balance. With either approach, the rate of exergy destruction
can be expressed as

Expressing this as a percentage of the exergy entering the plant with the
fuel, (11.56/231.28)(69%) = 3 %

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 Vapor Power Plant Exergy Accounting

Outputs
Net power out 30%
Losses
Condenser cooling water 1%
Stack gases (assumed) 1%
Exergy Destruction
Boiler
Combustion unit (assumed) 30%
Heat exchanger unit 30%
Turbine 5%
Pump —
Condenser 3%

Total 100%

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 Thermodynamics: An Engineering Approach, 5th Edition
Yunus A. Cengel, Michael A. Boles
McGraw-Hill, 2008

Chapter 11
REFRIGERATION CYCLES

Prof. Dr. Ali PINARBAŞI

Yildiz Technical University
Mechanical Engineering Department
Yildiz, ISTANBUL

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1
 REFRIGERATION CYCLES

11–1 Refrigerators and Heat Pumps

11–2 The Reversed Carnot Cycle
11–3 The Ideal Vapor-Compression
Refrigeration Cycle
11–4 Actual Vapor-Compression Refrigeration Cycle
11–5 Selecting the Right Refrigerant
11–6 Heat Pump Systems
11–7 Innovative Vapor-Compression
Refrigeration Systems
Cascade Refrigeration Systems
Multistage Compression Refrigeration Systems
Multipurpose Refrigeration Systems with a Single Compressor
Liquefaction of Gases
11–8 Gas Refrigeration Cycles
11–9 Absorption Refrigeration Systems

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 Objectives

• Introduce the concepts of refrigerators and heat pumps and the

measure of their performance.
• Analyze the ideal vapor-compression refrigeration cycle.
• Analyze the actual vapor-compression refrigeration cycle.
• Review the factors involved in selecting the right refrigerant for an
application.
• Discuss the operation of refrigeration and heat pump systems.
• Evaluate the performance of innovative vapor-compression
refrigeration systems.
• Analyze gas refrigeration systems.
• Introduce the concepts of absorption-refrigeration systems.

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REFRIGERATORS AND HEAT PUMPS
The transfer of heat from a low-temperature
region to a high-temperature one requires
special devices called refrigerators.
Refrigerators and heat pumps are essentially the
same devices; they differ in their objectives only.

for fixed values of QL and QH

The objective of a refrigerator is to remove heat (QL) from the 
cold medium; the objective of a heat pump is to supply heat 
(QH) to a warm medium.
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 Refrigeration Cycle

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THE REVERSED CARNOT CYCLE
The reversed Carnot cycle is the most efficient refrigeration cycle operating
between TL and TH.
However, it is not a suitable model for refrigeration cycles since processes 2-3
and 4-1 are not practical because
Process 2-3 involves the compression of a liquid–vapor mixture, which requires
a compressor that will handle two phases, and process 4-1 involves the
expansion of high-moisture-content refrigerant in a turbine.

Both COPs increase as the

difference between the two
temperatures decreases,
Schematic of a Carnot refrigerator and T‐s diagram of the  that is, as TL rises or TH
reversed Carnot cycle. falls.
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THE IDEAL VAPOR-COMPRESSION REFRIGERATION CYCLE
The vapor-compression refrigeration cycle is the ideal model for refrigeration
systems. Unlike the reversed Carnot cycle, the refrigerant is vaporized
completely before it is compressed and the turbine is replaced with a throttling
device.

This is the most widely used

cycle for refrigerators, A-C
systems, and heat pumps.

Schematic and T‐s diagram for the ideal vapor‐
compression refrigeration cycle.

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The ideal vapor-compression refrigeration cycle involves an irreversible
(throttling) process to make it a more realistic model for the actual systems.
Replacing the expansion valve by a turbine is not practical since the added
benefits cannot justify the added cost and complexity.
Steady-flow energy balance

The P‐h diagram of an ideal vapor‐
An ordinary household refrigerator. compression refrigeration cycle.
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 Vapor‐compression refrigeration cycle

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EXAMPLE 11–1
A refrigerator uses refrigerant-134a as the working fluid and
operates on an ideal vapor-compression refrigeration cycle
between 0.14 and 0.8 MPa. If the mass flow rate of the refrigerant
is 0.05 kg/s, determine
(a) the rate of heat removal from the refrigerated space and the
power input to the compressor,
(b) the rate of heat rejection to the environment, and
(c) the COP of the refrigerator.

Assumptions 1 Steady operating conditions exist.

2 Kinetic and potential energy changes are negligible.

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(a) The rate of heat removal from the refrigerated space and the power input to the
compressor are determined from their definitions:

(b) The rate of heat rejection from the refrigerant to the environment is

(c) The coefficient of performance of the refrigerator is

Discussion It would be interesting to see what happens if the throttling valve were replaced by
an isentropic turbine. The enthalpy at state 4s (the turbine exit with P4s=0.14 MPa, and
s4s=s3=0.35404 kJ/kgꞏK) is 88.94 kJ/kg, and the turbine would produce 0.33 kW of power. This
would decrease the power input to the refrigerator from 1.81 to 1.48 kW and increase the rate
of heat removal from the refrigerated space from 7.18 to 7.51 kW. As a result, the COP of the
refrigerator would increase from 3.97 to 5.07, an increase of 28 %.
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ACTUAL VAPOR-COMPRESSION REFRIGERATION CYCLE
An actual vapor-compression refrigeration cycle differs from the ideal one in
several ways, owing mostly to the irreversibilities that occur in various
components, mainly due to fluid friction (causes pressure drops) and heat transfer
to or from the surroundings. The COP decreases as a result of irreversibilities.

DIFFERENCES
Non-isentropic
compression
Superheated vapor
at evaporator exit
Subcooled liquid at
condenser exit
Pressure drops in
condenser and
evaporator

Schematic and T‐s diagram for the actual vapor‐
compression refrigeration cycle.
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EXAMPLE 11–2
Refrigerant-134a enters the compressor of a refrigerator as
superheated vapor at 0.14 MPa and -10°C at a rate of 0.05 kg/s
and leaves at 0.8 MPa and 50°C. The refrigerant is cooled in the
condenser to 26°C and 0.72 MPa and is throttled to 0.15 MPa.
Disregarding any heat transfer and pressure drops in the
connecting lines between the components, determine
(a) the rate of heat removal from the refrigerated space and the
power input to the compressor,
(b) the isentropic efficiency of the compressor, and
(c) the coefficient of performance of the refrigerator.

Assumptions 1 Steady operating conditions exist.

2 Kinetic and potential energy changes are negligible.

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(a) The rate of heat removal from the refrigerated space and the power input to the
compressor are determined from their definitions:

(b) The isentropic efficiency of the compressor is determined from

(c) The coefficient of performance of the refrigerator is

Discussion This problem is identical to the one worked out in Example 11–1, except that the
refrigerant is slightly superheated at the compressor inlet and subcooled at the condenser
exit. Also, the compressor is not isentropic. As a result, the heat removal rate from the
refrigerated space increases (by 10.4 %), but the power input to the compressor increases
even more (by 11.6 %). Consequently, the COP of the refrigerator decreases from 3.97 to
3.93.

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SELECTING THE RIGHT REFRIGERANT
• Several refrigerants may be used in refrigeration systems such as chlorofluorocarbons
(CFCs), ammonia, hydrocarbons (propane, ethane, ethylene, etc.), carbon dioxide, air
(in the air-conditioning of aircraft), and even water.
• R-11, R-12, R-22, R-134a, and R-502 account for over 90 percent of the market.
• The industrial and heavy-commercial sectors use ammonia (it is toxic).
• R-11 is used in large-capacity water chillers serving A-C systems in buildings.
• R-134a (replaced R-12, which damages ozone layer) is used in domestic refrigerators
and freezers, as well as automotive air conditioners.
• R-22 is used in window air conditioners, heat pumps, air conditioners of commercial
buildings, and large industrial refrigeration systems, and offers strong competition to
ammonia.
• R-502 (a blend of R-115 and R-22) is the dominant refrigerant used in commercial
refrigeration systems such as those in supermarkets.
• CFCs allow more ultraviolet radiation into the earth’s atmosphere by destroying the
protective ozone layer and thus contributing to the greenhouse effect that causes
global warming. Fully halogenated CFCs (such as R-11, R-12, and R-115) do the most
damage to the ozone layer. Refrigerants that are friendly to the ozone layer have been
developed.
• Two important parameters that need to be considered in the selection of a refrigerant
are the temperatures of the two media (the refrigerated space and the environment)
with which the refrigerant exchanges heat.

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 HEAT PUMP SYSTEMS
The most common energy source for heat
pumps is atmospheric air (air-to- air
systems).
Water-source systems usually use well water
and ground-source (geothermal) heat pumps
use earth as the energy source. They
typically have higher COPs but are more
complex and more expensive to install.
Both the capacity and the efficiency of a heat
pump fall significantly at low temperatures.
Therefore, most air-source heat pumps
require a supplementary heating system
such as electric resistance heaters or a gas
furnace.
Heat pumps are most competitive in areas
that have a large cooling load during the
cooling season and a relatively small heating
load during the heating season. In these
areas, the heat pump can meet the entire
A heat pump can be used to heat a house in  cooling and heating needs of residential or
winter and to cool it in summer. commercial buildings.

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INNOVATIVE VAPOR-COMPRESSION REFRIGERATION SYSTEMS

• The simple vapor-compression refrigeration cycle is the most widely used

refrigeration cycle, and it is adequate for most refrigeration applications.
• The ordinary vapor-compression refrigeration systems are simple,
inexpensive, reliable, and practically maintenance-free.
• However, for large industrial applications efficiency, not simplicity, is the
major concern.
• Also, for some applications the simple vapor-compression refrigeration
cycle is inadequate and needs to be modified.
• For moderately and very low temperature applications some innovative
refrigeration systems are used. The following cycles will be discussed:
• Cascade refrigeration systems
• Multistage compression refrigeration systems
• Multipurpose refrigeration systems with a single compressor
• Liquefaction of gases

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Cascade Refrigeration Systems

A two‐stage cascade refrigeration system with the 
same refrigerant in both stages.

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 Some industrial applications require moderately low temperatures, and the
temperature range they involve may be too large for a single vapor-compression
refrigeration cycle to be practical. The solution is cascading.

Cascading improves the COP of a refrigeration system.

Some systems use three or four stages of cascading.

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 Reciproating Compression

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EXAMPLE 11–3
Consider a two-stage cascade refrigeration system operating
between the pressure limits of 0.8 and 0.14 MPa. Each stage
operates on an ideal vapor compression refrigeration cycle with
refrigerant-134a as the working fluid. Heat rejection from the
lower cycle to the upper cycle takes place in an adiabatic
counterflow heat exchanger where both streams enter at about
0.32 MPa. (In practice, the working fluid of the lower cycle will
be at a higher pressure and temperature in the heat exchanger
for effective heat transfer.) If the mass flow rate of the
refrigerant through the upper cycle is 0.05 kg/s, determine
(a) the mass flow rate of the refrigerant through the lower cycle,
(b) the rate of heat removal from the refrigerated space and the
power input to the compressor, and
(c) the coefficient of performance of this cascade refrigerator.

Assumptions 1 Steady operating conditions exist.

2 Kinetic and potential energy changes are negligible.
3 The heat exchanger is adiabatic.

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(a) The mass flow rate of the refrigerant through the lower cycle is determined from the steady-
flow energy balance on the adiabatic heat exchanger,

(b) The rate of heat removal by a cascade cycle is the rate of heat absorption in the evaporator
of the lowest stage. The power input to a cascade cycle is the sum of the power inputs to all of
the compressors:

(c) The COP of a refrigeration system is the ratio of the refrigeration rate to the net power input:

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Multistage Compression Refrigeration Systems
When the fluid used throughout the cascade refrigeration system is the same,
the heat exchanger between the stages can be replaced by a mixing chamber
(called a flash chamber) since it has better heat transfer characteristics.

A two‐stage compression refrigeration system with a flash chamber.
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EXAMPLE 11–4
Consider a two-stage compression refrigeration system operating between the pressure limits of 0.8
and 0.14 MPa. The working fluid is refrigerant-134a. The refrigerant leaves the condenser as a
saturated liquid and is throttled to a flash chamber operating at 0.32 MPa. Part of the refrigerant
evaporates during this flashing process, and this vapor is mixed with the refrigerant leaving the low-
pressure compressor. The mixture is then compressed to the
condenser pressure by the high-pressure compressor. The liquid in the flash chamber is throttled to
the evaporator pressure and cools the refrigerated space as it vaporizes in the evaporator.
Assuming the refrigerant leaves the evaporator as a saturated vapor and both compressors are
isentropic, determine (a) the fraction of the refrigerant that evaporates as it is throttled to the flash
chamber, (b) the amount of heat removed from the refrigerated
space and the compressor work per unit mass of refrigerant flowing through the condenser, and (c)
the coefficient of performance.

Assumptions 1 Steady operating conditions exist.

2 Kinetic and potential energy changes are negligible.
3 The flash chamber is adiabatic.

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(a) The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is
simply the quality at state 6, which is

(b) The amount of heat removed from the refrigerated space and the compressor work
input per unit mass of refrigerant flowing through the condenser are

The enthalpy at state 9 is determined from an

energy balance on the mixing chamber,

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(c) The coefficient of performance is

Discussion This problem was worked out in Example 11–1 for a single-stage refrigeration
system (COP=3.97) and in Example 11–3 for a two-stage cascade refrigeration system
(COP= 4.46). Notice that the COP of the refrigeration system increased considerably
relative to the single-stage compression but did not change much relative to the two-stage
cascade compression.

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Multipurpose Refrigeration Systems with a Single Compressor
Some applications require refrigeration at more than one temperature. A practical
and economical approach is to route all the exit streams from the evaporators to a
single compressor and let it handle the compression process for the entire
system.

Schematic and T‐s diagram for a refrigerator–freezer unit with one compressor.

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 Liquefaction of Gases
Many important scientific and engineering processes at cryogenic temperatures
(below about 100°C) depend on liquefied gases including the separation of
oxygen and nitrogen from air, preparation of liquid propellants for rockets, the
study of material properties at low temperatures, and the study of
superconductivity.

The storage (i.e., hydrogen) and transportation of

some gases (i.e., natural gas) are done after they are
liquefied at very low temperatures. Several innovative
cycles are used for the liquefaction of gases.

Linde‐Hampson system for liquefying gases.
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 GAS REFRIGERATION CYCLES
The reversed Brayton cycle (the gas refrigeration cycle) can be used for
refrigeration.

Simple gas refrigeration cycle.

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The gas refrigeration cycles have lower
COPs relative to the vapor-
compression refrigeration cycles or the
reversed Carnot cycle.
The reversed Carnot cycle consumes a
fraction of the net work (area 1A3B)
but produces a greater amount of
refrigeration (triangular area under B1).

An open‐cycle aircraft cooling system.

Despite their relatively low COPs, the gas

refrigeration cycles involve simple, lighter
components, which make them suitable for
aircraft cooling, and they can incorporate
regeneration

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Without regeneration, the lowest turbine inlet temperature is T0, the temperature
of the surroundings or any other cooling medium.
With regeneration, the high-pressure gas is further cooled to T4 before expanding
in the turbine.
Lowering the turbine inlet temperature automatically lowers the turbine exit
temperature, which is the minimum temperature in the cycle.
Extremely low temperatures can be achieved
by repeating regeneration process.

Gas refrigeration cycle with regeneration.
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ABSORPTION REFRIGERATION SYSTEMS
When there is a source of inexpensive thermal energy at a temperature of 100 to
200°C is absorption refrigeration.
Some examples include geothermal energy, solar energy, and waste heat from
cogeneration or process steam plants, and even natural gas when it is at a
relatively low price.

Ammonia absorption refrigeration cycle.

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• Absorption refrigeration systems (ARS) involve the absorption of a refrigerant by a
transport medium.
• The most widely used system is the ammonia–water system, where ammonia
(NH3) serves as the refrigerant and water (H2O) as the transport medium.
• Other systems include water–lithium bromide and water–lithium chloride systems,
where water serves as the refrigerant. These systems are limited to applications
such as A-C where the minimum temperature is above the freezing point of water.
• Compared with vapor-compression systems, ARS have one major advantage: A
liquid is compressed instead of a vapor and as a result the work input is very small
(on the order of one percent of the heat supplied to the generator) and often
neglected in the cycle analysis.
• ARS are often classified as heat-driven systems.
• ARS are much more expensive than the vapor-compression refrigeration systems.
They are more complex and occupy more space, they are much less efficient thus
requiring much larger cooling towers to reject the waste heat, and they are more
difficult to service since they are less common.
• Therefore, ARS should be considered only when the unit cost of thermal energy is
low and is projected to remain low relative to electricity.
• ARS are primarily used in large commercial and industrial installations.

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 The COP of actual absorption refrigeration
systems is usually less than 1.
Air-conditioning systems based on absorption
refrigeration, called absorption chillers,
perform best when the heat source can supply
heat at a high temperature with little
temperature drop.
Determining the maximum COP of an 
absorption refrigeration system.

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SUMMARY

 Refrigerators and Heat Pumps

 The Reversed Carnot Cycle
 The Ideal Vapor-Compression Refrigeration Cycle
 Actual Vapor-Compression Refrigeration Cycle
 Selecting the Right Refrigerant
 Heat Pump Systems
 Innovative Vapor-Compression Refrigeration Systems

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 Thermodynamics: An Engineering Approach, 5th Edition
Yunus A. Cengel, Michael A. Boles
McGraw-Hill, 2008

Chapter 12
THERMODYNAMIC PROPERTY
RELATIONS

Prof. Dr. Ali PINARBAŞI

Yildiz Technical University
Mechanical Engineering Department
Yildiz, ISTANBUL

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1
 THERMODYNAMIC PROPERTY RELATIONS

12–1 A Little Math—Partial Derivatives and Associated Relations

Partial Differentials
Partial Differential Relations
12–2 The Maxwell Relations
12–3 The Clapeyron Equation
12–4 General Relations for du, dh, ds, cv,and cp
Internal Energy Changes
Enthalpy Changes
Entropy Changes
Specific Heats cv and cp
12–5 The Joule-Thomson Coefficient
12–6 The ∆h, ∆u, and ∆s of Real Gases
Enthalpy Changes of Real Gases
Internal Energy Changes of Real Gases
Entropy Changes of Real Gases

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 Objectives

• Develop fundamental relations between commonly encountered

thermodynamic properties and express the properties that cannot be
measured directly in terms of easily measurable properties.
• Review and use partial derivatives in the development of
thermodynamic property relations.
• Develop the Maxwell relations, which form the basis for many
thermodynamic relations.
• Develop the Clapeyron equation and determine the enthalpy of
vaporization from P, v, and T measurements alone.
• Develop general relations for cv, cp, du, dh, and ds that are valid for all
pure substances under all conditions.
• Discuss the Joule-Thomson coefficient.
• Develop a method of evaluating the ∆h, ∆u, and ∆s of real gases
through the use of generalized enthalpy and entropy departure charts..

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A LITTLE MATH—PARTIAL DERIVATIVES AND ASSOCIATED RELATIONS

the state postulate, which expresses that the state of a simple, compressible substance is

completely specified by any two independent, intensive properties. All other properties at 
that state can be expressed in terms of those two properties.

the derivative of a function f(x) with respect 
to x represents the rate of change of f with x.

The derivative of a function at a specified point 
represents the slope of the function at that 
point.

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Example 12–1
The cp of ideal gases depends on temperature only, and it is expressed as cp(T )=dh(T )/dT.
Determine the cp of air at 300 K, using the enthalpy data from Table A–17, and compare it to the
value listed in Table A–2b.

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Partial Differentials
Now consider a function that depends on two (or more) variables, such as z=z(x, y). This 
time the value of z depends on both x and y. It is sometimes desirable to examine the 
dependence of z on only one of the variables. This is done by allowing one variable to 
change while holding the others constant and observing the change in the function. The 
variation of z(x, y) with x when y is held constant is called the partial derivative of z with 
respect to x, and it is expressed as

The symbol 𝜕 represents differential changes, just like the 

symbol d. They differ in that the symbol d represents the 
total differential change of a function and reflects the 
influence of all variables, whereas 𝜕 represents the 
partial differential change due to the variation of a single 
variable.

Note that the changes indicated by d and 𝜕 are 
Geometric representation of partial identical for independent variables, but not for 
derivative (z/x)y. dependent variables.

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 This equation is the fundamental relation for 
the total differential of a dependent variable 
in terms of its partial derivatives with respect 
to the independent variables. This relation 
can easily be extended to include more
independent variables. Geometric representation of total 
derivative dz for a function z(x, y).

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Example 12–2
Consider air at 300 K and 0.86 m3/kg. The state of air changes to
302 K and 0.87 m3/kg as a result of some disturbance. Using Eq.
12–3, estimate the change in the pressure of air.

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Partial Differential Relations

The order of differentiation is immaterial for properties since they are continuous
point functions and have exact differentials. Therefore, the two relations above are
identical:

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 The first relation is called the reciprocity relation

Demonstration of the reciprocity

relation for the function z+2xy‐3y2z=0.

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Example 12–3
Using the ideal-gas equation of state, verify (a) the cyclic relation and (b) the reciprocity relation at
constant P.

(a) Replacing x, y, and z in Eq. 12–9 by P, v, and T, respectively, we can express the cyclic 
relation for an ideal gas as

(b) The reciprocity rule for an ideal gas at P  constant can be expressed as

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THE MAXWELL RELATIONS
The equations that relate the partial derivatives of properties P, v, T, and s of a simple 
compressible system to each other are called the Maxwell relations.
They are obtained from the four Gibbs equations by exploiting the exactness of the 
differentials of thermodynamic properties.

Helmholtz function
Gibbs function

They are extremely valuable in thermodynamics because they 

provide a means of determining the change in entropy, which 
cannot be measured directly, by simply measuring the changes in 
properties P, v, and T. Note that the Maxwell relations given 
above are limited to simple compressible systems. 

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Example 12–4

Verify the validity of the last Maxwell relation (Eq. 12–19) for steam at 250°C and 300 kPa.

Discussion This example shows that the entropy change of a simple compressible
system during an isothermal process can be determined from a
knowledge of the easily measurable properties P, v, and T alone.

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THE CLAPEYRON EQUATION
The Maxwell relations have far‐reaching implications in thermodynamics and are 
frequently used to derive useful thermodynamic relations. The Clapeyron equation is one 
such relation, and it enables us to determine the enthalpy change associated with a phase 
change (such as the enthalpy of vaporization hfg) from a knowledge of P, v, and T data 
alone.

During a phase‐change process, the pressure is the saturation 
pressure, which depends on the temperature only and is 
independent of the specific

Clapeyron equation

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Example 12–5
Using the Clapeyron equation, estimate the value of the enthalpy of vaporization of
refrigerant-134a at 20°C, and compare it with the tabulated value.

The tabulated value of hfg at 20°C is 182.27 kJ/kg. The small difference between the two 

values is due to the approximation used in determining the slope of the saturation curve at 
20°C.

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 The Clapeyron equation can be simplified for liquid–vapor and solid–vapor phase 
changes by utilizing some approximations. At low pressures vg>>vf, and thus vfg ≅ vg. By 
treating the vapor as an ideal gas, we have vg= RT/P.

Substituting these approximations into Eq. 12–22, we find

Clapeyron–Clausius equation

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Example 12–6
Estimate the saturation pressure of refrigerant-134a at -45°C, using the data available in
the refrigerant tables.

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 GENERAL RELATIONS for du, dh, ds, cv, and cp

The state postulate established that the state of a simple compressible system is 
completely specified by two independent, intensive properties.

Therefore, at least theoretically, we should be able to calculate all the properties of 
a system at any state once two independent, intensive properties are available.

This is certainly good news for properties that cannot be measured directly
such as internal energy, enthalpy, and entropy. 

However, the calculation of these properties from measurable ones depends on the 
availability of simple and accurate relations between the two groups.

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 Internal Energy Changes

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 Enthalpy Changes

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 Entropy Changes

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Specific Heats cv and cp

volume expansivity 𝜷

Mayer relation
𝜶
isothermal compressibility 𝜶,

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 We can draw several conclusions from Mayer
equation:
1. the constant‐pressure specific heat is always greater than or equal to the constant‐
volume specific heat: 𝒄𝒑 𝒄𝒗
2. The difference between cp and cv approaches zero as the absolute temperature
approaches zero. 

3. The two specific heats are identical for truly incompressible substances since v= constant. 

The difference between the two specific heats is very small and is usually disregarded for 
substances that are nearly incompressible, such as liquids and solids.

The volume expansivity (also called the coefficient of volumetric expansion) is a measure of the 

change in volume with temperature at constant pressure.

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Example 12–7
Derive a relation for the internal energy change as a gas that obeys the van der Waals
equation of state. Assume that in the range of interest cv varies according to the relation
cv = c1 + c2T, where c1 and c2 are constants.

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Example 12–8
Show that the internal energy of (a) an ideal gas and (b) an incompressible substance is
a function of temperature only, u = u(T).

(a) For an ideal gas Pv= RT. Then

∝
(b) For an incompressible substance, v = constant and thus dv=0. Also cp= cv= c since ∝
𝛽=0 for incompressible substances.
𝛽

Therefore we conclude that the internal energy of a truly incompressible
substance depends on temperature only.

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 The internal energies and specific heats of ideal gases and
incompressible substances depend on temperature only.

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Example 12–9
Show that cp - cv = R for an ideal gas.

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 THE JOULE-THOMSON COEFFICIENT
When a fluid passes through a restriction such as a porous plug, a capillary tube, or an 
ordinary valve, its pressure decreases.

Thus the Joule‐Thomson coefficient is a measure of 
the change in temperature with pressure during a 
constant‐enthalpy process.

The temperature of a fluid may increase, decrease, 
or remain constant during a throttling process. The development of an h  constant line 
on a P‐T diagram.

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 A throttling process proceeds along a constant‐
enthalpy line in the direction of decreasing pressure, 
that is, from right to left. 

Therefore, the temperature of a fluid will increase
during a throttling process that takes place on the
right‐hand side of the inversion line. 

However, the fluid temperature will decrease during 
a throttling process that takes place on the left‐hand 
side of the inversion line. 

It is clear from this diagram that a cooling effect 
cannot be achieved by throttling unless the fluid is 
below its maximum inversion temperature.

Constant‐enthalpy lines of a substance on 
a T‐P diagram.

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Example 12–10
Show that the Joule-Thomson coefficient of an ideal gas is zero..

Discussion This result is not surprising since 
the enthalpy of an ideal gas is a function of 
temperature only, h = h(T), which requires 
that the temperature remain constant when 
the enthalpy remains constant. Therefore, a 
throttling process cannot be used to lower the 
temperature of an ideal gas. The temperature of an ideal gas
remains constant during a throttling
process since h  constant and T  constant
lines on a T‐P diagram coincide.

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 THE ∆h, ∆u, AND ∆s OF REAL GASES

We have mentioned several times that gases at low pressures behave as ideal
gases and obey the relation Pv= RT. The properties of ideal gases are relatively easy to 
evaluate since the properties u, h, cv, and cp depend on temperature only. 

At high pressures, however, gases deviate considerably from ideal‐gas behavior, and it 
becomes necessary to account for this deviation.

we accounted for the deviation in properties P, v, and T by either using more complex 
equations of state or evaluating the compressibility factor Z from the compressibility 
charts. 

Now we extend the analysis to evaluate the changes in the enthalpy, internal energy, 
and entropy of nonideal (real) gases, using the general relations for du, dh, and ds 
developed earlier.

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 Enthalpy Changes of Real Gases
The enthalpy of a real gas, in general, depends on the pressure as well as on
the temperature. Thus the enthalpy change of a real gas during a process can
be evaluated from the general relation for dh

For an isothermal process dT = 0, and the 

first term vanishes. For a constant‐pressure 
process, dP = 0, and the second term
vanishes.
An alternative process path to evaluate
the enthalpy changes of real gases.

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Using a superscript asterisk (*) to denote an ideal‐gas state, we can express
the enthalpy change of a real gas during process 1‐2 as

The difference between h and h* is called the enthalpy departure, and it represents the 
variation of the enthalpy of a gas with pressure at a fixed temperature. The calculation of 
enthalpy departure requires a knowledge of the P‐v‐T behavior of the gas. In the absence 
of such data, we can use the relation Pv ZRT, where Z is the compressibility factor.

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 enthalpy departure factor

The values of Zh are presented in graphical form as a function of PR and TR in here. This

graph is called the generalized enthalpy departure chart, and it is used to determine the 
deviation of the enthalpy of a gas at a given P and T from the enthalpy of an ideal gas at 
the same T.

the enthalpy change of a gas during a process 1‐2

Internal Energy Changes of Real Gases

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 Entropy Changes of Real Gases
The general relation for ds 
was expressed as

The entropy change during this isothermal 
process can be expressed as

An alternative process path to evaluate
the entropy changes of real gases during
process 1‐2.

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 entropy departure
factor

entropy departure

The values of Zs are presented in graphical form as a function of PR and TR in Fig. A–30. This 

graph is called the generalized entropy departure chart, and it is used to determine the 
deviation of the entropy of a gas at a given P and T from the entropy of an ideal gas at the 
same P and T.

for the entropy change of a gas during a process 1‐2 as

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Example 12–11
Determine the enthalpy change and the entropy change of O2 per unit mole as it
undergoes a change of state from 220 K and 5 MPa to 300 K and 10 MPa (a) by
assuming ideal-gas behavior and (b) by accounting for the deviation from ideal-gas
behavior.
(a) If the O2 is assumed to behave as an ideal gas, its enthalpy will depend on temperature 
only, and the enthalpy values at the initial and the final temperatures can be determined from 
the ideal‐gas table of O2 (A–19) at the specified temperatures:

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(b) The deviation from the ideal‐gas behavior can be accounted for by determining
the enthalpy and entropy departures from the generalized charts at each state:

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 Therefore, in this case, the ideal‐gas assumption would underestimate the
enthalpy change of the oxygen by 2.7 percent and the entropy change by
11.4 percent.

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 Summary

 A Little Math—Partial Derivatives and Associated Relations

 Partial Differentials
 Partial Differential Relations

 The Maxwell Relations

 The Clapeyron Equation

 General Relations for du, dh, ds, cv, and cp
 Internal Energy Changes
 Enthalpy Changes
 Entropy Changes
 Specific Heats cv and cp

 The Joule‐Thomson Coefficient

 The ∆h, ∆u and ∆s of Real Gases
 Enthalpy Changes of Real Gases
 Internal Energy Changes of Real Gases
 Entropy Changes of Real Gases

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 Thermodynamics: An Engineering Approach, 5th Edition
Yunus A. Cengel, Michael A. Boles
McGraw-Hill, 2008

Chapter 13
GAS MIXTURES

Prof. Dr. Ali PINARBAŞI

Yildiz Technical University
Mechanical Engineering Department
Yildiz, ISTANBUL

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1
 GAS MIXTURES

13–1 Composition of a Gas Mixture:

Mass and Mole Fractions
13–2 P-v-T Behavior of Gas Mixtures:
Ideal and Real Gases
Ideal-Gas Mixtures
Real-Gas Mixtures
13–3 Properties of Gas Mixtures:
Ideal and Real Gases
Ideal-Gas Mixtures
Real-Gas Mixtures

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 Objectives

• Develop rules for determining nonreacting gas mixture properties from

knowledge of mixture composition and the properties of the individual
components.
• Define the quantities used to describe the composition of a mixture,
such as mass fraction, mole fraction, and volume fraction.
• Apply the rules for determining mixture properties to idealgas
mixtures and real-gas mixtures.
• Predict the P-v-T behavior of gas mixtures based on Dalton’s law of
additive pressures and Amagat’s law of additive volumes.
• Develop the concept of chemical potential and determine the separation
work of mixtures...

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COMPOSITION OF A GAS MIXTURE: MASS AND MOLE FRACTIONS

To determine the properties of a mixture, we need to know the composition of the mixture 
as well as the properties of the individual components. There are two ways to describe the 
composition of a mixture: either by specifying the number of moles of each component, 
called molar analysis, or by specifying the mass of each component, called gravimetric 
analysis.

the ratio of the mole number of a 
The mass of a mixture is equal to the component to the mole number of the 
sum of the masses of its components. mixture is called the mole fraction y

The ratio of the mass of a component to 
the mass of the mixture is called the
mass fraction mf
The number of moles of a nonreacting
mixture is equal to the sum of the
number of moles of its components.

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 the apparent (or average) molar mass

the sum of the mass fractions or mole 
fractions for a mixture is equal to 1

gas constant

The molar mass of a mixture can also be
expressed as

Mass and mole fractions of a mixture
The sum of the mole fractions of a
mixture is equal to 1.

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Example 13–1
Consider a gas mixture that consists of 3 kg of O2, 5 kg of N2, and 12 kg
of CH4. Determine;
(a) the mass fraction of each component,
(b) the mole fraction of each component, and
(c) the average molar mass and gas constant of the mixture.

(a) the mass fraction of each component,

(b) To find the mole fractions, we need to determine the 
mole numbers of each component first:

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 (c) The average molar mass and gas constant of the mixture are determined from their
definitions,

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P-v-T BEHAVIOR OF GAS MIXTURES: IDEAL AND REAL GASES

The prediction of the P‐v‐T behavior of gas mixtures is usually based on two models:

Dalton’s law of additive pressures: The pressure of a gas mixture is equal to the sum of the 
pressures each gas would exert if it existed alone at the mixture temperature and volume.

Amagat’s law of additive volumes: The volume of a gas mixture is equal to the sum of the 

volumes each gas would occupy if it existed alone at the mixture temperature and pressure.

Dalton’s law of additive pressures for a  Amagat’s law of additive volumes for a 

mixture of two ideal gases. mixture of two ideal gases.

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 Pi component pressure Vi component volume
Pi/Pm pressure fraction Vi /Vm volume fraction

For ideal gases, these two laws are identical 

and give identical results.

The volume a component would occupy if it existed alone at the
mixture T and P is called the component volume (for ideal gases, it is 
equal to the partial volume yiVm).

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 Ideal-Gas Mixtures

This equation is strictly valid for ideal‐gas mixtures since it is derived by

assuming ideal‐gas behavior for the gas mixture and each of its components.

The quantity yiPm is called the partial pressure (identical to the component
pressure for ideal gases), and the quantity yiVm is called the partial volume. 

Note that for an ideal‐gas mixture, the mole fraction, the pressure fraction, and the 
volume fraction of a component are identical. 

The composition of an ideal‐gas mixture (such as the exhaust gases leaving
a combustion chamber) is frequently determined by a volumetric analysis
(Orsat Analysis)

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 Real-Gas Mixtures

Compressibility factor

The compressibility‐factor approach, in 
general, gives more accurate results when 
the Zi’s in below equation are evaluated by 
using Amagat’s law instead of Dalton’s law.

One way of predicting the P‐v‐T behavior of a 
real‐gas mixture is to use compressibility factors.

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 Another approach for predicting the P‐v‐T behavior of a gas mixture is to treat the gas 
mixture as a pseudo pure substance.

One such method, proposed by W. B. Kay in 1936 and called Kay’s rule, involves the use of 

a pseudo critical pressure and pseudo critical temperature for the mixture, defined in 
terms of the critical pressures and temperatures of the mixture components as

Kay’s Rule

The compressibility factor of the mixture Zm is then 
easily determined by using these pseudo critical 
properties. 
Another way of predicting the P‐v‐T
behavior of a real‐gas mixture is to treat
The result obtained by using Kay’s rule is accurate to 
it as a pseudopure substance with critical 
within about 10 % over a wide range of  properties P’cr and T’cr.
temperatures and pressures, which is acceptable for 
most engineering purposes.
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Example 13–2
A rigid tank contains 2 kmol of N2 and 6 kmol of CO2 gases at 300 K
and 15 MPa. Estimate the volume of the tank on the basis of
(a) The ideal-gas equation of state,
(b) Kay’s rule,
(c) Compressibility factors and Amagat’s law, and
(d) Compressibility factors and Dalton’s law.
(a) The ideal‐gas equation of state,.

(b) Kay’s rule

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(c) Compressibility factors and 
Amagat’s law,

Mixture:

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(d) Compressibility factors and Dalton’s law

Discussion Notice that the results obtained in parts (b), (c), and (d ) are very close. But 

they are very different from the values obtained from the ideal‐gas relation. Therefore, 
treating a mixture of gases as an ideal gas may yield unacceptable errors at high 
pressures.

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PROPERTIES OF GAS MIXTURES: IDEAL AND REAL GASES
extensive properties

the changes in internal energy, enthalpy, and entropy

The extensive properties of a 
mixture are determined by 
simply adding the properties
of the components

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 intensive properties of a gas mixture

The intensive 
properties of a mixture
are determined by 
weighted averaging

Notice that properties per unit mass involve mass fractions (mfi) and properties
per unit mole involve mole fractions (yi).

The relations given above are exact for ideal‐gas mixtures, and approximate
for real‐gas mixtures.

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 Ideal-Gas Mixtures
Gibbs–Dalton’s Law: Under the ideal‐gas approximation, 
the properties of a gas are not influenced by the presence 
of other gases, and each gas component in the mixture 
behaves as if it exists alone at the mixture temperature Tm
and mixture volume Vm.

Also, the h, u, cv, and cp of an ideal gas depend on 
temperature only and are independent of the pressure or 
the volume of the ideal‐gas mixture.

Partial pressures (not the 
mixture pressure) are used in 
the evaluation of entropy 
changes of ideal‐gas mixtures.

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Example 13–3
An insulated rigid tank is divided into two compartments by a
partition. One compartment contains 7 kg of oxygen gas at 40°C
and 100 kPa, and the other compartment contains 4 kg of nitrogen
gas at 20°C and 150 kPa. Now the partition is removed, and the two
gases are allowed to mix. Determine (a) the mixture temperature and
(b) the mixture pressure after equilibrium has been established.
(a) the mixture temperature

(b) The final pressure of the mixture is determined from the ideal‐gas relation

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 Discussion We could also determine the mixture pressure by using PmVm= mmRmTm, 
where Rm is the apparent gas constant of the mixture. This would require a knowledge 
of mixture composition in terms of mass or mole fractions.

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Example 13–4
An insulated rigid tank is divided into two compartments by a partition. One compartment
contains 3 kmol of O2, and the other compartment contains 5 kmol of CO2. Both gases
are initially at 25°C and 200 kPa. Now the partition is removed, and the two gases are
allowed to mix. Assuming the surroundings are at 25°C and both gases behave as ideal
gases, determine the entropy change and exergy destruction associated with this
process.

Pm,2=Pi,1=200 kPa

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 The exergy destruction associated with this mixing process is determined from

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 Gerçek Gaz Karışımları

Bir gaz karışımı için T ds bağıntısı

Gerçek gaz karışımlarının davranışlarını

belirlemek zordur, çünkü birbirinden
farklı moleküllerin karşılıklı etkilerini göz
önüne almak gerekir.

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 Denklem entalpi ve entropi için genelleştirilmiş bağıntıları ve diyagramları geliştirmenin
başlangıç noktasıdır. Ayrıca gerçek gazlar için geliştirilen genelleştirilmiş özelik bağıntılarının
ve diyagramlarının, gerçek gaz karışımını oluşturan gazlar için de kullanılabileceğini
belirtmektedir.

Fakat her karışım için indirgenmiş sıcaklık, TR ve indirgenmiş basınç Pm, karışım sıcaklığı Tm

ve karışım basıncı Pm'de hesaplanmalıdır.

Karışım basıncı ve sıcaklığı yerine karışım hacim ve sıcaklığı verilirse izlenecek olan yol, 
Dalton'un toplanan basınçlar yasasını kullanarak yaklaşık bir karışım basıncı belirlemektir.

Gerçek bir gaz karışımının özeliklerini belirlemenin bir başka yolu da, karışımı sanki‐kritik
özelikleri olan, sanki‐saf bir madde gibi ele almaktır. Burada sanki‐kritik özelikler, 
karışanların kritik özeliklerinden Kay kuralını kullanarak belirlenir.

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Örnek 13–5
Hava genellikle N2 ve O2'den oluşan bir karışı›m olarak ele alınır. Bu karışımda N2 ve 02’nin mol
oranları sırasıyla % 79 ve % 21'dir. Sürekli akışlı bir açık sistemde, hava 10 MPa sabit basınçta, 220
K'den 160 K sıcaklığa soğutulmaktadır. Havadan çevreye olan ısı geçişini, (a) mükemmel gaz
yaklaşımıyla, (b) Kay kuralını kullanarak, (c) Amagat yasasını kullanarak hesaplayın..

(a) N2, O2 karışımının mükemmel gaz olduğu kabul edilirse

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 (b) Kay kuralına göre, sanki‐saf bir madde gibi davranan bir gaz karışımının kritik sıcaklığı
ve basıncı;

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(c) N2 ve O2 için, ilk ve son hallerde, entalpi sapma çarpanları

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