FLIP THERMOFLUID

THERMODYNAMIC PROPERTIES

EXERCISE 3.4

1. A piston-cylinder assembly of initial volume 0.6 m3 contains H2O at 500 kPa and
280oC. The system is cooled in a two-step process:
1-2 Constant volume cooling until only saturated vapor remains
2-3 Constant temperature cooling until only saturated liquid remains
Determine:
(a) the process on a T-v diagram
(b) the work done and the heat transferred in process 1-2
(c) the work done in process 2-3.

3.4.6 COMPRESSED LIQUID WATER TABLE

A substance is said to be a compressed liquid when the pressure is
greater than the saturation pressure for the temperature.

It is now noted that state 1 in Figure 3.6 is called a compressed liquid
state because the saturation pressure for the temperature T1 is less than
P1.

Data for water compressed liquid states are found in the compressed
liquid tables. This table is arranged like superheated vapor table. Note
that the data in the table begins at 5 MPa or 50 times atmospheric
pressure.

At pressures below 5 MPa for water, the data are approximately equal to
the saturated liquid data at the given temperature. We approximate
intensive parameter y, which is v, u, h, and s data as

y  yf@T

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THERMODYNAMIC PROPERTIES

Compressed Liquid can be characterized by:

(a) Higher Pressures (P>Psat at a given T)
(b) Lower Temperatures (T<Tsat at a given P)
(c) Lower specific volume (v<vf at a given P or T)
(d) Lower internal energy (u<uf at a given P or T)
(e) Lower enthalpies (h<hf at a given P or T)

Example 3.5:

Determine the internal energy oif compressed liquid water at 80oC and 5
Mpa, using (a) data from the compressed liquid table and (b) saturated
liquid data. What is the error involve in the second case?

Solution:

P  5 MPa
(a) T  80o C

u  333.72 kJ / kg

T  80o C

(b)

u  u f @80o C  334.86 kJ / kg

The error involve is 334.86  333.72 x100%  34.16%
333.72

EXERCISE 3.5

1. Calculate the enthalpy of compressed liquid water at 40oC two ways: using the
approximate relationship for enthalpy of a compressed liquid and using the
compressed liquid tables. Perform the calculation at these pressures:
(a) 10 MPa
(b) 20 MPa
(c) 50 MPa.

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THERMODYNAMIC PROPERTIES

SUMMARY

In this chapter we have studied that:

1. The ideal gas law is not the most accurate way to relate P, T, and v.
2. Thermodynamic tables, which give the most accurate relationship among the

three variables, are available for a variety of substances.
3. When liquid and vapour exist together in equilibrium, the quality of the mixture is

defined as the ratio of vapour mass and total mass (liquid and vapour).
4. The specific volume, internal energy, enthalpy, and entropy of compressed liquid

water may be approximated using values fro the saturated liquid at the same
temperature.

REFERENCES

a. Kaminsky D. A, Jensen M. K., (2005), “Introduction to Thermal and Fluid
Engineering”, John Wiley & Sons, Inc.

b. Eastop, T.D, McConkey, A., (2004), 5th Edition, “Applied Thermodynamics for
Engineering Technologist”, Longman.

c. Yunus, A.C, Michael, A.B., (2002), 4th Edition, “Thermodynamics, an
Engineering Approach”, Mc Graw Hill, New York.

d. D.F. Young, B.R. Munson, T.H. Okiishi, (2004), “Fundamental of Fluid
Mechanics”, 4th Edition, John Wiley & Sons, Inc.

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CHAPTER 4

THERMODYNAMICS CYCLES AND THE
SECOND LAW

THERMODYNAMICS CYCLES AND THE SECOND LAW

ICCHAPTER 4 : THERMODYNAMIC CYCLES AND
THE SECOND LAW

INTRODUCTION

In previous chapter, the conservation of mass and energy has been studied. However,
while these two laws are necessary to solve many problems, for other problems they are
not sufficient to obtain a valid answer. This does not mean conservation of mass or
energy is ever invalid. It means that conservation of energy cannot do certain things. It
cannot determine whether a process is possible or not, whether there is any limitation to
the conversion of heat into work. The conservation of energy also cannot determine in
what direction a process must proceed.

Such limitations to conservation of energy necessitate another fundamental law of
thermodynamics, which is the second law of thermodynamics. The basis of the second
law is experimental evidence and has never been disproved. Fundamentally, the
second law states that heat does not move spontaneously from cold to hot bodies. The
implications of this statement are wide-reaching and include: (a) every process has
losses, (b) we cannot build a perpetual motion machine, and (c) heat cannot be
converted into work with 100% efficiency.

For engineering use, the second law must be expressed on a mathematical basis. In the
course of developing the second law, a new thermodynamic property, which is entropy,
is needed. To derive the second law, an examination of thermodynamic cycles is
initiated first.

LEARNING OBJECTIVES

At the end of this topic, you should be able to:

1. Introduce the concept of Thermodynamic cycles and Second Law of
Thermodynamics (including; Entropy) and explain the related principles.

2. Understand and apply the Second Law of Thermodynamics and the Carnot cycle in
estimating the performance of heat engine and refrigeration system.

3. Describe the turbine, pump, and compressor using the isentropic efficiency.

4.1 THERMODYNAMIC CYCLES

A system has completed a thermodynamics cycle when the system undergoes a
series of processes and then returns to its original state, so that the properties of
the system at the end of the cycle are the same as at its beginning. During
undergoes a series of the processes, there are some energies such as heat is
added to or removed from the system, as well as work is done by or added to the
system.

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THERMODYNAMICS CYCLES AND THE SECOND LAW

Figure 4.1 show a simple power cycle using water as the working fluid, which is
an example of the thermodynamics cycle. The cycle begins at state 1, where
low-pressure liquid water is fed into a pump. Exit from the pump, the pressure
liquid water is raised at state 2. Then enters the boiler, where heat is added, and
the liquid water is vaporized at constant pressure until the water changes into the
superheated vapor at state 3. The high-pressure steam that exits the boiler is
passed through a turbine. The turbine extracts energy from the flowing fluid and
produces work during this process for rotating an electric power generator by
connecting both of turbine and generator shafts. The fluid leaving the turbine
could be either low-pressure superheated vapor or a mixture condition with high
quality. The turbine exhaust enters a condenser at state 4, where heat is
removed, and the steam condenses to the liquid condition, which is an initial
state (state 1).

Generator

Figure 4.1 Power cycle as a thermodynamics cycle

4.1.1 HEAT ENGINE

A heat engine is a thermodynamics system operating in a
thermodynamics cycle to which net heat is transferred and from which net
work is delivered. In the order words, heat engine is a device that can
convert heat energy into work energy.

The system, or working fluid, undergoes a series of processes that
constitute the heat engine cycle.

Figure 4.1 is an example of heat engine, which illustrates a steam power
plant operating in a thermodynamics cycle.

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4.1.2 THERMAL EFFICIENCY

The thermal efficiency is the index of performance of a work-producing
device or a heat engine and is defined by the ratio of the net work output
to the heat input.

For a heat engine, the net work output is produced by the heat supplied to
make the cycle operate. The thermal efficiency is always less than 1 or
less than 100%.

 thermal  the work output  Wnet, out
the heat input Qin

where, Wnet, out  Wout  Win  Qin  Qout

 thermal  Qin  Qout  1  Qout
Qin Qin

Cycle devices such as heat engines, refrigerators, and heat pumps often

operate between a high-temperature thermal reservoir at temperature TH
and a low-temperature thermal reservoir at temperature TL.

TH

QH
Heat Wnet

Engine

QL

TL

The thermal efficiency of the above device becomes

 thermal  1 QL
QH

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THERMODYNAMICS CYCLES AND THE SECOND LAW

Example 4.1:

A steam power plant produces 50 MW of net work while burning fuel to
produce 150 MW of neat energy at the high temperature. Determine the
cycle thermal efficiency and the heat rejected by the cycle to the
surroundings.

Solution:

 thermal  Wnet, out  50 MW  0.333or 33.3%
QH 150 MW

QL  QH  Wnet,out  150 MW  50 MW  100 MW

EXERCISE 4.1

1. A power plant generates 150 MW of electrical power. It uses a supply of 1000 MW
from a geothermal source and rejects energy to the atmosphere. Calculate the
thermal efficiency and the energy rejected to the atmosphere.

4.1.3 REFRIGERATION, AIR CONDITIONING, AND HEAT PUMP

Thermodynamics cycles are also used for refrigeration. A refrigeration
cycle moves heat from a low-temperature environment to a high-
temperature environment by taking advantage of the fact that the
saturation temperature of a fluid or refrigerant depends on its pressure.

Air conditioning is the same function of the refrigeration, only the heat
moves from a low-temperature is less than refrigeration. While, heat
pump operates on a thermodynamics cycle, which rejects the heat to the
high-temperature medium.

The following figure illustrates a refrigeration as a refrigerator and heat
pump operating in a thermodynamics cycle.

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THERMODYNAMICS CYCLES AND THE SECOND LAW

Heat Pump

Refrigerator/Air Cond

Figure 4.2 A Refrigeration Cycle

4.1.4 COEFFICIENT OF PERFORMANCE, COP

The objective of a refrigeration cycle is to remove as much heat as
possible from a cold environment for a given amount of work input. The
performance is defined as

COPR  heat remove from a cold environment Refrigerator /
work input Air Conditioning

COPHP  heat reject to a high temperature Heat Pump
work input

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THERMODYNAMICS CYCLES AND THE SECOND LAW

Cycle devices as refrigeration operate between a high-temperature
thermal reservoir at temperature TH and a low-temperature thermal
reservoir at temperature TL.

TH

QH
Winput

Refrigeration

QL

TL

The COP of the above device becomes

COPR  QL Q
Win 

QH  QL

COPHP  QH  QH
Win QH  QL

Example 4.2:

The food compartment of a refrigerator, is maintained at 4oC by removing
heat from it at a rate of 360 kJ/min. If the required power input to the
refrigerator is 2 kW, determine:
(a) the coefficient of performance (COP) of the refrigerator
(b) the rate of heat rejection to the room that houses the refrigerator.

Solution:



Q L  360 kJ / min  6 kJ / s  6 kW

(a)  6
2
COPR  QL  3
Win

 

(b) Q H  Q L  W net,in  6  2  8 kW  480 kJ / min

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THERMODYNAMICS CYCLES AND THE SECOND LAW

EXERCISE 4.2

1. A refrigerator is capable of delivering 4 kW of cooling while requiring 990 W of
electric power to operate. If the COP of the refrigerator is improved by 20%, how
much electric power would be required to deliver 4 kW of cooling?

4.2 THE CARNOT CYCLE AND SECOND LAW

French engineer Nicolas Sadi Carnot was among the first to study the principles
of the second law of thermodynamics. Carnot was the first to introduce the
concepts of cyclic operation and devised a reversible cycle that is composed of
four reversible processes, two isothermal and two adiabatic.
The four reversible steps in a Carnot cycle are:

Process 1-2 Isothermal heat addition
Process 2-3 Adiabatic expansion
Process 3-4 Isothermal cooling
Process 4-1 Adiabatic compression.

Carnot Heat Engine Cycle Carnot Refrigeration Cycle

Since the thermal efficiency in general is

 thermal  1 QL
QH

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THERMODYNAMICS CYCLES AND THE SECOND LAW

For the Carnot engine, this can be written as

 thermal  1  TL
TH

This maximum possible efficiency of a heat engine operating between two heat
reservoirs at temperature TH and TL, and the temperatures are absolute
temperature.

These statements from the basis for establishing an absolute temperature scale,
also called the Kelvin scale, related to the heat transfers between a reversible
device and the high- and low-temperature heat reservoirs by

QL  TL
QH TH

Then the QH/QL ratio can be replaced by TH/TL for reversible devices, where TH
and TL are the absolute temperature of the high- and low-temperature heat
reservoirs, respectively.

Example 4.3:

A Carnot heat engine receives 500 kJ of heat per cycle from a high-
temperature heat reservoir at 625oC and rejects heat to a low-
temperature heat reservoir at 30oC. Determine:
(a). the thermal efficiency of this Carnot engine
(b). the amount of heat rejected to the low-temperature heat reservoir.

Solution:

(a) th, rev  1  TL 1 (30  273)  0.672  67.2 %
TH (652  273)

(b) QL  TL QL  TL QH  (30  273) 500  164 kJ
QH TH TH (652  273)

EXERCISE 4.3

1. A power cycle operates between temperature limits of 400°C and 15°C. The cycle
requires a heat input of 12.1 kW and rejects 6.5 kW to the low temperature
reservoir. Is the cycle irreversible, reversible, or impossible? Support your answer
with calculations.

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THERMODYNAMICS CYCLES AND THE SECOND LAW

4.2.1 SECOND LAW STATEMENTS

The following two statements of the second law of thermodynamics are
based on the definitions of the heat engine and refrigeration cycle.

4.2.2 KELVIN-PLANKS STATEMENT OF THE SECOND LAW

It is impossible for any device that operates on a cycle to receive heat
from a single reservoir and produce a net amount of work. In the other
word, the maximum possible efficiency is less than 100%.

thermal 100%

4.2.3 CLAUSIUS STATEMENT OF THE SECOND LAW

It is impossible to construct a device that operates in a cycle and
produces no effect other than the transfer of heat from a lower-
temperature body to a higher-temperature body. Thus, the COP of a
refrigerator or heat pump must be less than infinity.

COP < 

4.2.4 PERPETUAL-MOTION MACHINES

Any device that violates the first or second law of thermodynamics is
called a perpetual-motion machine. If the device violates the first law, it is
a perpetual-motion machine of the first kind (PMM1). If the device
violates the second law, it is a perpetual-motion machine of the second
kind (PMM2).

4.3 REVERSIBLE REFRIGERATION CYCLE

If the Carnot device is caused to operate in the reversed cycle, the reversible
refrigeration is created. The COP of reversible refrigerator and heat pump are
given in a similar manner to that of the Carnot heat engine as

COPR  QL Q  1 1
Win  QH 1 TH 1

QH  QL

QL TL

COPHP  QH  QH 1 1
Win QH  QL  

1 QL 1  TL
QH TH

Again, these are the maximum possible COPs for a refrigerator or a heat pump
operating between the temperature limits of TH and TL.

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THERMODYNAMICS CYCLES AND THE SECOND LAW

Example 4.4:

An inventor claims to have developed a refrigerator that maintains the
refrigerated space at 2oC while operating in a room where the
temperature is 25oC and has a COP of 13.5. Is there any truth to his
claim?

Solution:

COPR  1  1  1  11.96
TH 1 (25  273) 0.0836
TL 1
(2  273)

The claim is false since no refrigerator may have a COP larger than the
COP for the reversed Carnot device.
(COP = 13.5 > COPreversed Carnot = 11.96).

EXERCISE 4.4

1. A completely reversible heat pump has a COP of 1.6 and a reservoir high-
temperature of 300 K. Calculate:
(a) the temperature of the reservoir cold-temperature
(b) the rate of heat transfer to the reservoir high-temperature when 1.5 kW of
power is supplied to this heat pump.

4.4 ENTROPY

The second law of thermodynamics leads to the definition of a new property
called entropy, which is a quantitative measure of microscopic disorder for a
system.

The definition of entropy is based on the Clausius inequality, given by

where the equality holds for internally or totally reversible processes and the
inequality for irreversible processes.

Any quantity whose cyclic integral is zero is a property, and entropy is defined as

For the special case of an internally reversible, isothermal process, it gives

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THERMODYNAMICS CYCLES AND THE SECOND LAW

Entropy change is caused by heat transfer, mass flow, and irreversibilities. Heat
transfer to a system increases the entropy, and heat transfer from a system
decreases it. The effect of irreversibilities is always to increase the entropy.

The entropy-change and isentropic relations for a process can be summarized as
follows

Pure substances:

Any process: ∆s = s2 - s1 [kJ/(kg-K)]

Isentropic process: s2 = s1

The entropy-change and isentropic relations for a process can be summarized as
follows:

Ideal gases:
Constant specific heats (approximate treatment):

Any process:

s2 - s1 = Cv,av 1n T2 + R1n v2 [kJ/(kg-K)]
T1 v1

Ts2 - s1 = Cp,av 1n 2 R1n P2 [kJ/(kg-K)]
T1 P1

Isentropic process:

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THERMODYNAMICS CYCLES AND THE SECOND LAW

The entropy-change and isentropic relations for a process can be summarized as
follows:
Ideal gases:

Variable specific heats (exact treatment):

Any process,

S2  S1  o o R ln P2 [kJ/(kg-K)]
P1
S2 S1

Isentropic process

o  o P2 [kJ/(kg-K)]

S2 S 1  R ln P1

0

S

where Pr is the relative pressure and vr is the relative specific volume. The
function so depends on temperature only.

Example 4.5:

A cylinder of volume 300 cm3 contains saturated steam at 0.6 MPa. The
steam is then allowed to expand adiabatically and reversibly to a final
pressure of 0.2 MPa. Determine:
(a) the final quality
(b) the work done.

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THERMODYNAMICS CYCLES AND THE SECOND LAW

Solution:

(a) Because the process is adiabatic and reversible, so the process is an
isentropic, which is given by

P1  0.6 MPa (saturated steam), s1  sg  6.76 kJ / kg.K

P2  0.2 MPa, s2  s1  6.76 kJ / kg.K

s2  s f  x.s fg x  s2  s f  6.76 1.53  0.93
s fg 5.597

(b) The work done is defined as

Q  W  U , sin ce the process is adibatic, Q  0

W  m(u2  u1 )
P1  0.6 MPa (saturated steam),u1  ug  2567.4 kJ / kg

v1  vg  0.3157 m3 / kg

P2  0.2 MPa, u2  u f  xu fg  504.5  0.93x2025  2388 kJ / kg

mV  300 /106  0.00095 kg
v1 0.3157

W  (0.00095)(2388  2567.4)  0.17 kJ

EXERCISE 4.5

1. Steam at 400 kPa, 200oC is contained in a well-insulated piston-cylinder assembly
of initial volume 0.15 m3. How much work is done by the expansion of steam?

4.5 REVERSIBLE AND IRREVERSIBLE PROCESS

4.5.1 REVERSIBLE PROCESS

A system undergoes a reversible process if it can be returned to its initial
state with no net change to the surrounding. A reversible process occurs

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THERMODYNAMICS CYCLES AND THE SECOND LAW

in the limit when all irreversible effects, such as friction, are eliminated
and sets an upper bound on the behavior of real processes.

Some thermodynamics processes can be idealized as reversible. For
examples, a slow, isothermal compression of a gas can be reversed to a
slow, isothermal expansion, and a slow, adiabatic compression of a gas,
which can be reversed to a slow, adiabatic expansion. Other engineering
devices such as ideal turbine, ideal compressor, ideal nozzle, and ideal
diffuser also can be idealized as reversible.

4.5.2 IRREVERSIBLE PROCESS

An irreversible process is a process that is not reversible. All real
processes are irreversible. Irreversible processes occur because of the
following:
1. Friction
2. Unrestrained expansion of a gas
3. Heat transfer through a finite temperature difference
4. Mixing of two different substances
5. Hysteresis effects
6. I2R losses in electrical circuits
7. Any deviation from a quasi-static process, etc.

4.6 SECOND LAW ANALYSIS OF TURBINES, PUMPS, AND
COMPRESSORS

An isentropic process represents the limit of the possible, real processes fall
short of this limit. To evaluate the actual behavior of a real component, such as a
turbine, pump, or compressor, the real process is compared to an idealized,
isentropic process using quantity called the isentropic efficiency.

For a turbine under steady state operation, the inlet state of the working fluid and
the exhaust pressure are fixed. Therefore, the ideal process for an adiabatic
turbine is an isentropic process between the inlet state and the exhaust pressure.

The isentropic efficiency of a turbine is, by definition


m(h1 h2a )
T  Wact  


W ideal m(h1  h2s )

 a

where W act is the actual work



done, and W ideal is the isentropic

work.

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Compressor can also be characterized by an isentropic efficiency. The ideal and
actual compressors have identical inlet states and the same exit pressure. The
isentropic efficiency of a compressor is defined as

a


m(h2s  h1 )
C  Wideal 


W act m(h2a  h1 )

Pumps are analogous to compressors. The work required by a actual pump is
always greater than the work for an ideal pump. Thus the isentropic efficiency of
a pump is define as

 a
m(h2s  h1 )
P  Wideal 


W act m(h2a  h1 )

The enthalpy rise across an ideal pump is define as

h2s  h1  v1 (P2  P1 )

Example 4.6:

Steam at 1 MPa, 600oC, expands in aturbine to 0.01 MPa. If the process
is isentropic, find the final temperature, the final enthalpy of the steam,
and turbine work. If the isentropic of the turbine is 90%, calculate the
actual work.

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Solution:

The final temperature if the process is an isentropic,

P1  1MPa (saturated steam), s1  8.0290 kJ / kg.K

T1  600C h1  3697.9 kJ / kg

P2  0.01MPa, s2  s1  8.0290 kJ / kg.K

s f  0.6491kJ / kg.K

sg  8.1510 kJ / kg.K

s f  s2  sg T2  Tsat @ 0.01MPa  45.8C

The final enthalpy if the process is an isentropic,

s2  s f  x.s fg x  s2  s f  8.0290  0.6491  0.984
s fg 7.5019

h2  h f  x.h fg  191.8  (0.984)(2392.8)  2545.7 kJ / kg

The turbine work if the process is an isentropic,

w  h1  h2  3697.9  2545.7  1152.2 kJ / kg

The turbine work if the isentropic efficiency is 90%,

T  wact wact  T .wideal  (0.9)(1152.2)  1037 kJ / kg
wideal

EXERCISE 4.6

1. Steam enters a turbine at 600 kPa and 300oC and exits at 5 kPa. The turbine
efficiency is 78%. Calculate the work produced per kg of steam flow.

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THERMODYNAMICS CYCLES AND THE SECOND LAW

SUMMARY

In this chapter we have studied that:

1. The principle of a thermodynamic cycle, which is a series of processes that a
working fluid begins at an initial state, moves through one or more intermediate
state, and then returns to its initial state.

2. Heat energy can be converted to work by some devices called heat engines and
the performance can be determined by the thermal efficiency. Refrigerators and
heat pumps are devices that absorb heat from low-temperature media and reject
it to higher-temperature ones. The performance of these devices is expressed in
terms of the coefficient of performance (COP).

3. The Carnot cycle is a reversible cycle that is composed of four reversible
processes, two isothermal and two adiabatic.

4. The second law of thermodynamics leads to the definition of a new property
called entropy, which is a quantitative measure of microscopic disorder fro a
system.

5. Most steady-state devices operate under adiabatic conditions and the ideal
process for these devices is the isentropic process. The parameter that
describes how efficiently a device is called isentropic efficiency.

REFERENCES

1. Deborah A. Kaminsky and Michael K. Jensen, (2005), “Introduction to Thermal and
Fluid Engineering”, John Wiley & Sons, Inc.

2. Eastop, T.D, McConkey, A., (2004), 5th Edition, “Applied Thermodynamics for
Engineering Technologist”, Longman.

3. Yunus, A.C, Michael, A.B., (2002), 4th Edition, “Thermodynamics, An Engineering
Approach”, Mc Graw Hill, New York.

4. D.F. Young, B.R. Munson, T.H. Okiishi, (2004), “Fundamental of Fluid Mechanics”,
4th Edition, John Wiley & Sons, Inc.

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CHAPTER 5

REFRIGERATION, HEAT PUMP AND
POWER CYCLES

常州能源设备厂加热炉

REFRIGERATION, HEAT PUMP AND POWER CYCLES

CHAPTER 5 : REFRIGERATION, HEAT PUMP AND
POWER CYCLES

INTRODUCTION

The Carnot heat power cycle and the Carnot refrigeration cycle discussed in previous
chapter are idealized cycles whose performances cannot be attained in reality.
However, their characteristics can be used to guide the development of practical cycles.
In everyday life, we use refrigeration and heat pump cycles for conditioning the interior of
buildings and vehicles. Whenever we drive a car or fly in an airplane, a heat power
cycle or heat engine is used to provide the motive force. The electricity we use for so
many different tasks is generated by a heat power cycle, excepts for hydroelectric or
wind power engines. In this chapter, we focus only on heat engines, therefore we will
use power cycle to mean heat power cycle.

In each of these applications, the job of the engineer is to devise the cycle that will
accomplish the goals most effectively at lowest cost. In the sections that follow, several
common cycles are examined. We start with the simplest cycle: the vapour-
compression refrigeration cycle. We must get closer and focus on a specific component
or several components of the cycle. After refrigeration cycle, we present power cycles:
first, a vapour power cycle in which the working fluid is alternatively vaporized and
condensed and, second, a gas power cycle in which the working fluid remains a gas
during all processes. Each of these power cycles can be modified in a variety of ways to
improve its performance.

In this chapter, we put all the devices such as turbines, compressors, pumps, heat
exchangers, valves, etc, and processes such as open systems, conversation of mass
and energy equation, together in the analysis of refrigeration and power cycles.

LEARNING OUTCOMES

At the end of this topic, you should be able to:
1. Understand the concept of Refrigeration, heat pump, and power cycles then

explain the related principles.
2. Describe and apply the coefficient of performance of refrigerator and heat pump.
3. Describe the various tapes of gas power cycles and calculate the thermal efficiency

of Brayton, Rankine cycles, Otto cycle and Diesel cycle.

5.1 VAPOR COMPRESSION REFRIGERATION CYCLES

Many of the impracticalities with the reversed Carnot cycle can be eliminated by
vaporizing the refrigerant completely before it is compressed and by replacing
the turbine with a throttling device, such as an expansion valve or capillary tube.

The cycle that results is called the ideal vapor-compression refrigeration
cycle, and it is shown schematically and on T-s diagram in figure 5.1.

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

The vapor-compression refrigeration cycle is the most widely used cycle for
refrigerators, air-conditioning systems, and heat pumps. It consists of four
processes:
a) 1-2 Isentropic compression in a compressor.
b) 2-3 Constant-pressure heat rejection in a condenser.
c) 3-4 Throttling in an expansion device.
d) 4-1 Constant-pressure heat absorption in an evaporator.

Figure 5.1 Schematic and T-s diagram for the ideal vapor-compression
refrigeration cycle

Another diagram frequently used in the analysis of vapor-compression
refrigeration cycle is the P-h diagram, as shown in figure 5.2.

Figure 5.2 The P-h diagram of an ideal vapor-compression refrigeration cycle

As the refrigerant passes through the evaporator, heat is absorbed by the
vaporizing refrigerant, thus cooling the refrigerated space. The heat transfer rate
QL is referred as the refrigerating capacity, expressed in kW.

Another commonly used unit for the refrigerating capacity is the ton of
refrigeration, which is equal to 211 kJ/min. The refrigerant leaving the evaporator

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

is compressed to relatively high pressure and temperature by the compressor.
The compression process results the temperature of refrigerant becomes higher
that the surrounding temperature, as such it is condensed in the condenser by
rejecting heat to the surrounding. At last, the refrigerant is expanded through a
throttling valve to the evaporator pressure.

All four components associated with the vapor-compression refrigeration cycle
are steady-flow devices, and thus all four processes that make up the cycle can
be analyzed as steady-flow processes.

The kinetic and potential energy changes of the refrigerant are usually small
relative to the work and heat transfer terms, and therefore they can be neglected.
Then the steady-flow energy equation on a unit-mass basis reduces to

   qin  qout  win  wout  he  hi

The condenser and the evaporator do not involve any work, and the compressor
can be approximated as adiabatic. Then the COPs of refrigerators operating on
the vapor-compression cycle can be expressed as

COPR  qL  h1  h4
wnet ,in h2  h1

where h1=hg@P1 for the ideal case

Example 5.1:

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
(c) The COP of the refrigerator.

Solution:

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

Assumptions:
1) Steady operating condition exists. 2) Kinetic and potential energy
changes are negligible.

Analysis:
From the refrigerant-134a tables, the enthalpies of the refrigerant at
all four states are determined as follows:

P1=0.14 MPa [email protected]=236.04 kJ/kg
[email protected]=0.9322 kJ/kg.K

P2=0.8 MPa s2=s1 h2=272.05 kJ/kg

P3=0.8 MPa [email protected]=93.42 kJ/kg

h4  h3 (throttling) h4=93.42 kJ/kg

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

  m h1  h4   0.05236.04  93.42  7.13kW

QL

  m h2  h1   0.05272.05  236.04  1.8kW

W in

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

  m h2  h3   0.05272.05  93.42  8.93kW

QH

It could also be determined from

 

Q H  Q L  W in  7.13  1.8  8.93kW

(c) The coefficient of performance of the refrigerator is

 7.13
1.8
COPR  QL   3.96



W in

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EXERCISE 5.1

1. A vapor-compression refrigeration cycle uses R-134a. Liquid at 1200 kPa exits the
condenser at 40 ºC. The evaporator operates at a pressure of 240 kPa. The
compressor isentropic efficiency is 75%. Determine the cycle coefficient of
performance (COP) if the refrigerant leaves the evaporator as superheated vapor
at 0, 5, 10, 15, and 20 ºC above the saturation temperature.

5.2 HEAT PUMPS

A heat pump is a thermodynamics system operating in a thermodynamics cycle
that removes heat from a low-temperature body and delivers heat to a high-
temperature body. To accomplish this energy transfer, the heat pumps receives
external energy in the form of work from the surroundings.

For heat pump, the principle of the cyclic system is similar with refrigerator; the
primary function is to transfer heat from the low-temperature to high-temperature
system.
Therefore, the coefficient of performance for a heat pump is

COPHP  qH  h2  h3
wnet ,in h2  h1

which is referred to the figure 5.2 and h1=hg@P1 and h3=hf@P3 for the ideal case.

Under the same operating condition the COPHP and COPR are related by

COPHP  COPR  1

Example 5.2:

In winter a building requires 94,000 kJ/hr of heat, and an ideal vapor-
compression heat pump is used. R-134a enters the isentropic
compressor of the heat pump at 0.4 MPa, 10 ºC and exits at 1 MPa.
Saturated liquid leaves the condenser. Determine:
(a) the mass flow rate of the refrigerant (in kg/s)
(b) the power input to the compressor (in kW)
(c) the cycle coefficient of performance (COP).

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Solution:

(a) The system is steady and the compressor is isentropic, therefore the
mass can be determine by

P1  0.4 MPa, s1  0.9182 kJ / kg.K

T1  10o C h1  253.35 kJ / kg

P2  1 MPa, s2  s1  0.9182 kJ / kg.K

by int erpolation, h2  272.37 kJ / kg

P3  P2  1 MPa h3  h f  105.29 kJ / kg



Q H  94000 kJ / hr

 (94000 / 3600)
(272.37 105.29)
   QH 
(h2  h3 )
Q H  m(h2  h3 ) m

 0.156 kg / s

(b) The compressor work is defined as



W in  m(h2  h1 )  (0.156)(272.37  253.35)  2.97 kW

(c) The COP of heat pump is defined as

 (94000 / 3600)
2.97
COPHP  QH   8.78



W in

EXERCISE 5.2

1. An ideal vapor-compression heat pump cycle using R-134a is used to heat a
house. The inside temperature is 22ºC; the outside temperature is 0ºC. Saturated
vapor at 2.2 bar enters the compressor, and saturated liquid leaves the condenser
at 8 bar. The mass flow rate is 0.2 kg/s. Determine:
(a) the power input to the compressor (in kW)
(b) the coefficient of performance
(c) the coefficient of performance if the system were used as a refrigeration
cycle

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5.3 THE RANKINE CYCLES

The Rankine cycle is the ideal cycle for vapor power plants. The ideal Rankine
cycle is consisted of the following four processes:

a) 1-2 Isentropic compression in a pump
b) 2-3 Constant pressure heat addition in a boiler
c) 3-4 Isentropic expansion in a turbine
d) 4-1 Constant pressure heat rejection in a condenser

Water enters the pump at state 1 as saturated liquid and is compressed
isentropically to the operating pressure of the boiler. Water enters the boiler as a
compressed liquid at state 2 and leaves as a superheated vapor at state 3.

The boiler is basically a large heat exchanger where the heat originating from
combustion gases, nuclear reactor, or other sources is transferred to the water
essentially at constant pressure. The boiler, together with the section where the
steam is superheated (the superheater), is often called the steam generator.

The superheated vapor at state 3 enter the turbine, where it expands
isentropically and produces work by rotating the shaft connected to an electric
generator. The pressure and the temperature of the steam drop during this
process to the values at state 4, where steam enters the condenser. Steam is
condensed at constant pressure in the condenser, which is basically a large heat
exchanger, by rejecting heat to a cooling medium such as a lake, a river, or the
atmosphere. Steam leaves the condenser as saturated liquid and enters the
pump, completing the cycle.

KKTM Figure 5.3 Rankine Cycle

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

5.3.1 ENERGY ANALYSIS OF THE IDEAL RANKINE CYCLE

All four components associated with the Rankine cycle (the pump, boiler,
turbine, and condenser) are steady-flow devices, and thus all four
processes that make up the Rankine cycle can be analyzed as steady-
flow processes. The kinetic energy and potential energy changes of the
steam are usually small relative to the work and heat transfer terms and
are therefore usually neglected. Then the steady-flow energy equation
per unit mass of steam reduce to

   qin  qout  win  wout  he  hi

The boiler and the condenser do not involve any work, and the pump and
the turbine are assumed to be isentropic. Then the conservation of
energy relation for each device can be expressed as follows:

Pump (q=0): wpump,in  h2  h1

where or
Boiler (w=0):
Turbine (q=0):  wpump,in  v P2  P1

h1  h f @ P1 and v  v1  v f @ P1

qin  h3  h2

wturb,out  h3  h4

Condenser (w=0): qout  h4  h1

The thermal efficiency of the Rankine cycle is determined from

 th  wnet  1  qout
qin qin

Where wnet  qin  qout  wturb,out  wpump,in

Example 5.3:

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.

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Solution:

State 1: P1=75 kPa h1=hf @ 75 kPa = 384.39 kJ/kg
State 2: Saturated liq. v1=vf @ 75 kPa = 0.001037 m3/kg

State 3: P2=3 MPa
State 4: s2=s1
wpump,in= v1(P2-P1) = (0.001037m3/kg)(3000-75kPa)

= 3.03 kJ/kg

h2 = h1 + wpump,in = (384.39 + 3.03) kJ/kg
= 387.42 kJ/kg

P3 = 3 MPa, T3 = 350C

h3 = 3115.3 kJ/kg s3 = 6.7426 kJ/kg.K

P4 = 75 kPa (saturated mixture), s3 = s4

x4  s4  s f  6.7428 1.213  0.8857
s fg 6.2434

h4  h f  x4h fg  384.39  0.8857(2278.6)  2402.6kJ / kg

Thus, qin  h3  h2  3115.3  387.42  2727.9kJ / kg

qout  h4  h1  2402.6  384.39  2018.2kJ / kg

and

 th  1  qout  1 2018.2  0.260 @ 26.0%
qin 2727.9

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

EXERCISE 5.3

1. In an ideal Rankine cycle, saturated water vapor enters the turbine at 20 MPa and
exits at 10 kPa. Saturated liquid exits the condenser. Determine:
(a) the net work per unit mass of steam flow (in kJ/kg)
(b) the heat input per unit mass of steam flow (in kJ/kg)
(c) the cycle thermal efficiency
(d) the heat rejection per unit mass of steam flow (in kJ/kg).

5.4 THE BRAYTON CYCLES

5.4.1 AIR STANDARD ASSUMPTION

The actual gas power cycles are complex. To make it analysis to a
manageable level, following approximation are used, which is called the
Air-Standard Assumption: The assumption are:

a) The working fluid is air, which continuously circulates in a closed
loop and always behaves as an ideal gas.

b) All the processes that make up the cycle are internally reversible.
c) The combustion process is replaced by heat-addition process

from an external source (Figure 5.4).
d) The exhaust process is replaced by a heat rejection process that

restores the working fluid to its initial state.
e) Air has constant specific heats whose values are determined at

room temperature (25C). When this assumption is utilized, the
air-standard assumptions are called cold-air-standard
assumption.

KKTM Figure 5.4 The combustion process is replaced by a heat addition
process in ideal cycles

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

A cycle for which the air-standard assumptions are applicable is
frequently referred to as an air-standard cycle.

5.4.2 BRAYTON CYCLE: THE IDEAL CYCLE FOR GAS-TURBINE
ENGINES

It is used for gas turbines only where both the compression and
expansion processes take place in rotating machinery. Gas turbines
usually operate on an open cycle. The ideal cycle that the working fluid
undergoes in this closed loop is the Brayton Cycle.

Brayton cycle, is made up of four internally reversible processes:
a) 1-2 Isentropic compression (in a compressor)
b) 2-3 Constant-pressure heat addition
c) 3-4 Isentropic expansion (in turbine)
d) 4-1 Constant-pressure heat rejection

KKTM Figure 5.5 A closed gas-turbine engine
Flow process can be expressed, on a unit-mass basis, as

   qin  qout  win  wout  hexit  hinlet

Therefore, heat transfer to and from the working fluid are

 qin  h3  h2  C p T3  T2
 qout  h4  h1  C p T4  T1

The net work net done by the system is defined as

wnet  wout  win  (h3  h4 )  (h2  h1 )
 qin  qout  (h3  h2 )  (h4  h1 )

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

Then the thermal efficiency of the ideal Brayton Cycle under the cold air
standard assumptions becomes

   th,Brayton T1  1
 wnet  1  qout 1 Cp T4  T1  1  T1 T4 T2  1
qin qin Cp T3  T2 T2 T3

Processes 1-2 and 3-4 are isentropic, and P2=P3 and P4=P1. Thus:

T2  P2  k 1 k  P3  k 1 k  T3
T1 P1 P4 T4
 

Example 5.4:

A stationary 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 thermal efficiency.

Solution:

The T-s diagram of the ideal Brayton cycle described is shown below.
We note that the components involved in the Brayton cycle are steady
flow devices.

(a) The air temperatures at the compressor and turbine exits are
determined from isentropic relations.
Process 1-2(isentropic compression of an ideal gas):

T1  300K  h1  300.19 kJ kg Pr1  1.386

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

Pr 2  P2 Pr1  81.386  11.09  T2  540K
P1

h2  544.35 kJ kg

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

T3  1300K  h3  1395.97 kJ kg

Pr3  330.9

Pr 4  P4 Pr 3   1 330.9  41.36  T4  770K
P3  8


h4  789.11kJ kg

wcomp,in  h2  h1  544.35  300.19  244.16 kJ kg
wturb,out  h3  h4  1395.97  789.11  606.86 kJ kg

(b) The thermal efficiency could also be determined from

 th  wnet  606.86  244.16  0.426
qin 1395.97  544.35

EXERCISE 5.4

1. Air at 100 kPa, 27 ºC at a volumetric flow rate of 10 m3/s enters an ideal Brayton
cycle and is compressed to 1750 kPa. The air temperature at the entrance to the
turbine is 1073 ºC. Using a cold-air-standard analysis, determine:
(a) the net power (in kW)
(b) the heat addition (in kW)
(c) the cycle thermal efficiency.

5.5 OTTO AND DIESEL CYCLES

5.5.1 OTTO CYCLE: THE IDEAL CYCLE FOR SPARK IGNITION (SI)
ENGINES

In SI engines, the combustion of the air-fuel mixture is initiated by a spark
plug. The Otto cycle is the ideal cycle for SI reciprocating engines. In term
of stroke, there are 2 categories of engines; 4 strokes engine and 2 stroke
engines:

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

a) 4 stroke engines.
The piston executes 4 complete strokes (2 mechanical cycles) within the
cylinder, and the crankshaft completes 2 revolutions for each
thermodynamic cycle.

Figure 5.6 Actual 4 stroke SI engine

Figure 5.7 Ideal Otto cycle for 4 strokes SI engine

b) 2 stroke engines.
In 2 stroke engines, all 4 functions described above are executed in just 2
strokes:
i) The power stroke
ii) The compression stroke

Figure 5.8 Schematic of a 2- stroke
reciprocating engine

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

The thermodynamics analysis of the actual 4-stroke or 2-stroke cycles
described above is not a simple task. The analysis can be simplified
significantly if the air-standard assumptions are utilized. The resulting
cycle, which closely resembles the actual operating conditions, is the
ideal Otto cycle.

It consists of 4 internally reversible processes:
a) 1-2 Isentropic compression
b) 2-3 Constant-volume heat addition
c) 3-4 Isentropic expansion
d) 4-1 Constant-volume heat rejection
The execution of the Otto cycle in a piston-cylinder device together with a
P-v diagram is illustrated in figure 5.7. The T-s diagram of the Otto cycle
is given in figure 5.9.

Figure 5.9 T-s diagram of the ideal Otto cycle.

The Otto cycle is executed in a closed system, and disregarding the
changes in kinetic and potential energies, the energy balance for any of
the processes is expressed, on a unit-mass basis, as

   qin  qout  win  wout  u (kJ/kg)

No work is involved during the two heat transfer processes since both
take place at constant volume. Therefore, heat transfer to and from the
working fluid can be expressed as

 qin  u3  u2  Cv T3  T2
 qout  u4  u1  Cv T4  T1

Then the thermal efficiency of the ideal Otto cycle under the cold air
standard assumption becomes

 th,Otto T1  1
 wnet  1  qout  1  T4  T1  1  T1 T4 T2  1
qin qin T3  T2 T2 T3

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Processes 1-2 and 3-4 are isentropic, and v2=v3 and v4=v1. Thus,

T1   v2  k 1   v3  k 1  T4
T2 v1 v4 T1

Substituting these equations into the thermal efficiency relation and
simplifying give

 th ,Otto 1 1
r k 1

r  Vmax  V1  v1
Vmin V2 v2

where r is the compression ratio and k is the specific heat ratio ,Cp/Cv.

The thermal efficiency of an ideal Otto cycle depends on the compression
ratio of the engine and the specific heat ratio of the working fluid (if
different from air).

The Mean Effective Pressure (MEP): a fictitious pressure that, if it acted
on the piston during the entire power stroke, would produce the same
amount of net work as that produced during the actual cycles (Figure
5.10).

KKTM Figure 5.10 Mean Effective Pressure (MEP)

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

That is,

Wnet  MEPxPiston _ areaxStroke  MEPxDisplacement _ volume

or

MEP  Wnet  wnet (kPa)
Vmax  Vmin vmax  vmin

Example 5.5:

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
(d) the mean effective pressure for the cycle.

Solution:

(a) T1=290K  u1=206.91 kJ/kg

Vr1=676.1

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

vr2  v2  1  vr2  vr1  676.1 84.51  T2  652.4K / kg
vr1 v1 r r 8 u2  475.11kJ

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P 2 v2  P1v1
T2 T1

P2  P1  T2  v1   100kPa 652.4K 8  1799.7kPa
T1 v2 290K
 

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

qin=u3-u2 T3=795.6K
800 kJ/kg = u3-475.1 kJ/kg vr3=6.108

u3=1275.11 kJ/kg 

P3v3  P2v2  P3  P2  T3  v2 
T3 T2 T2 v3

 1.7997MPa 1575.1K 1  4.345MPa
652.4K
 

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

vr4  v4  r  vr4=rvr3=(8)(6.108)=48.864  T4=795.6K
vr3 v3 u4=588.74 kJ/kg

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

-qout=u1-u4  qout=u4-u1

qout=588.74-206.91=381.83 kJ/kg

Thus,

Wnet = qnet = qin – qout = 800 -381.83= 418.17 kJ/kg
(c) The thermal efficiency of the cycle is determined from its definition

 th,Otto  wnet  418.17kJ / kg  0.523
qin 800kJ / kg

Under the cold-air-standard assumptions (constant specific heat
values at room temperature), the thermal efficiency would be

 th,Otto 1 11.4
1 r k 1  1  r1k 1 8  0.565

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REFRIGERATION, HEAT PUMP AND POWER CYCLES

Which is considerably different from the value obtained above.
Therefore, care should be exercised on utilizing the cold-air
standard assumptions.

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

wnet
 MEP  wnet  wnet 
v1  v2 v1  v1 r 1 1
v1 r

where,

v1  RT1  (0.287kPa.m3kg.K )(290K )  0.832m3 / kg
P1 100kPa

Thus,

418.17kJ / kg  1kPa.m 3 
(0.832m3 / kg)(1 1/ 1kJ
MEP  8)  574.4kPa

EXERCISE 5.5

1. Air at 100 kPa, 300 K enters an ideal Otto cycle. The initial volume is 500 cm3.
The compression ratio is 8.5, and the maximum temperature in the cycle is 2100 K.
Using a cold-air-standard analysis, determine:
(a) the heat addition (in kJ)
(b) the heat rejection (in kJ)
(c) the net work (in kJ)
(d) the cycle thermal efficiency.

5.5.2 DIESEL CYCLE: THE IDEAL CYCLE FOR COMPRESSION-
IGNITION (CI) ENGINES

The diesel cycle is the ideal cycle for CI reciprocating engines. The CI
engine is very similar to the SI engine, differing mainly in the method of
initiating combustion.

In spark ignition engines (gasoline engines), the air-fuel mixture is
compressed to a temperature that is below the auto-ignition temperature
of the fuel, and the combustion process is initiated by firing a spark plug.

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In CI engines (diesel engines), the air is compressed to a temperature
that is above the auto-ignition temperature of the fuel, and the combustion
starts on contact as the fuel is injected into this hot air.
The similarity between the two cycles is also apparent from the T-s and
P-v diagrams of the Diesel cycle, shown in figure 5.11.

P = constant

Figure 5.11 the T-s and P-v diagrams of the Diesel cycle

Noting that the Diesel cycle is executed in a piston-cylinder device, which
forms a closed system, the amount of heat transferred to the working fluid
at constant pressure and rejected from it at constant volume can be
expressed as

qin  wb,out  u3  u2

qin  P2 v3  v2   u3  u2  h3 h2  CP T3  T2 

And

 qout  u1  u4

 qout  u4  u1  CV T4  T1

Then the thermal efficiency of the ideal Diesel cycle under the cold-air
standard assumptions becomes

   th,diesel T1  1
 wnet  1  qout  1  T4  T1  1 T1 T4 T2  1
qin qin k T3  T2 kT2 T3

We now define a new quantity, the cut off ratio, rc, as the ratio of the
cylinder volume after and before the combustion process:

rc  V3  v3
V2 v2

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Utilizing this definition and the isentropic ideal gas relations for process 1-
2 and 3-4, we see that the thermal efficiency relation reduces to

1 1  rck 1 
 th,Diesel r k 1  k rc 1 
 

where r is the compression ratio.

Under the cold-air standard assumptions, the efficiency of a Diesel cycle

differs from the efficiency of an Otto cycle by the quantity in the brackets.

This quantity is always greater than 1. Therefore, th,Otto  th,Diesel when

both cycles operate on the same compression ratio.

Example 5.6:

An ideal Diesel cycle with air as the working fluid has a compression ratio
of 18 and a cut-off ratio of 2. At the beginning of the compression
process, the working fluid is at 101.325 kPa, 27C, and 1916 cm3.
Utilizing the cold air standard assumptions, determine:
(a) the temperature and pressure of the air at the end of each process
(b) the net work output and the thermal efficiency
(c) the mean effective pressure.

Solution:

.

The gas constant of air is R=0.287 kJ/kg.K and its other properties

at room temperature are Cp = 1.005 kJ/kg.K, Cv = 0.718 kJ/kg.K and
k = 1.4.

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(a)

V2  V1  1917  106.5cm3
r 18

V3  rcV2  2106.5  213cm3

V4  V1  1917cm3

Process 1-2 (isentropic compression of an ideal gas, constant
specific heats):

T2  T1  V1  k 1  300K 181.41  953K
V2

P2  P1  V1  k  101.325181.4  5796kPa
V2

Process 2-3 (constant pressure heat addition to an ideal gas):

P3  P2  5796kPa

P2V2  P3V3  T3  T2  V3   953K 2  1906K
T2 T3 V2

Process 3-4 (isentropic expansion of an ideal gas, constant specific
heats):

T4  T3  V3  k 1  1906 213 1.41  792K
V4 1917 


P4  P3  V3  k  5796 213 1.4  267kPa
V4 1917 


(b)  m

 P1V1  101.325kPa1917x106 m3  2.256x103 kg
RT 1 0.287kJ / kg.K 300K 

Process 2-3 is a constant pressure heat addition process, for which
the boundary work and u terms can be combined into h . Thus,

Thus, Qin  mh3  h2   mC p T4  T1 

  2.256x103 0.718792  300  0.797kJ

Wnet  Qin Qout  2.16  0.797  1.36kJ

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Then the thermal efficiency becomes

 th  Wnet  1.36  0.631@ 63.1%
Q net 2.16

(c) The mean effective pressure is determined from its definition

MEP  Wnet  Wnet  1.36  751kPa
Vmax  Vmin V1  V2
1917 106.5

Therefore, a constant pressure of 751kPa during the power stroke
would produce the same net work output as the entire Diesel cycle.

EXERCISE 5.6

1. Air at 100 kPa, 300 K enters a Diesel cycle, which has a compression ratio of 18
and a volume at the beginning of the compression process of 0.05 m3. The
maximum temperature of the cycle is 2200 K. Using an air-standard analysis,
determine:
(a) the net work per cycle (in kJ)
(b) the cycle thermal efficiency
(c) the cutoff ratio.

SUMMARY

1. Refrigeration cycles use input mechanical energy to move low-temperature
energy to a higher temperature, and vapour compression refrigeration cycles
have four processes.

2. The Rankine cycle is a heat power cycle and is used to convert chemical energy
stored in fuel into useful mechanical energy through a combustion process. The
cycle has a minimum of four processes.

3. The Brayton cycle is a heat power cycle that must have a minimum of four
processes, and the working fluid remains a gas through all processes. The cycle
uses a common assumption for simplicity that called the air-standard analysis.

4. The term of internal combustion engine is used to refer to heat power cycles that
use a reciprocating engine. There are two types of these engines, which are the
Otto cycle that defined as a spark-ignition engine and the Diesel cycle that
defined as a compression-ignition engine.

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REFERENCES

a. Kaminsky D. A, Jensen M. K., (2005), “Introduction to Thermal and Fluid
Engineering”, John Wiley & Sons, Inc.

b. Eastop, T.D, McConkey, A., (2004), 5th Edition, “Applied Thermodynamics for
Engineering Technologist”, Longman.

c. Yunus, A.C, Michael, A.B., (2002), 4th Edition, “Thermodynamics, an
Engineering Approach”, Mc Graw Hill, New York.

d. D.F. Young, B.R. Munson, T.H. Okiishi, (2004), “Fundamental of Fluid
Mechanics”, 4th Edition, John Wiley & Sons, Inc.

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CHAPTER 6

FUNDAMENTAL OF FLUIDS MECHANICS

FUNDAMENTAL OF FLUID MECHANICS

CHAPTER 6 FUNDAMENTALS OF FLUID
MECHANICS

INTRODUCTION

The word fluid generally is used to mean a liquid. However, in engineering
terminology, a fluid is a liquid or a gas. The study of fluids mechanics can be divided
into 2 categories ; i) Fluid statics ii) Fluids dynamics

Fluid statics is the study of stationary fluids. Water at rest in an aquarium tank exerts
forces on the sides of the tank. Pressure in the atmosphere varies with height above
sea level, becoming noticeably lower at high altitudes. These different phenomena can
all be understood using the principles of fluid statics. In addition, fluid statics deals with
the buoyancy of floating objects, such as ships, swimmers, and so on.

Fluid dynamics involves moving fluids. The flow of air over a truck, the flow of oil in a
pipeline, and the flow of water issuing from a fire hose are just a few of the many
examples of fluid flow that can be analyzed using fluid dynamics. In this chapter,
equations for conservation of mass of a moving fluid will be introduced which can be
applied to a very wide range of processes and phenomena

LEARNING OBJECTIVE

At the end of this chapter, you should be able to:
1.Understand and apply the concept of fluids static
2.Understand and apply the concept of fluids dynamics by using Bernoulli

Equation and Conservation of mass equations

6.1 FLUID STATICS

THINK :
Imagine that you are standing next to a swimming pool. What is the
difference will you feel about the pressure exerted on you by the
weight of the atmosphere above your head when you are standing
near the swimming pool and and if you are at the bottom of the pool.
How does pressure vary with depth and type of fluid?

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6.1.1 Pressure in a Fluid at Rest

Pressure is a force per unit area:

P = F/A

The pressure force on a surface immersed in a fluid is always normal to the surface. In
a staic fluid of constant density (incompressible), the pressure is a function of depth
,h, according to

P = P atm + ρgh ( Equation 6.1)

Thus, for a constant density fluid (gas or liquid) at rest, the pressure is a function of the
depth and density. Note that pressure is not a function of horizontal location. At any
horizontal location in a stationary fluid, the pressure will be the same.

Example 6.1 :
Question:
A vat in a chemical processing plant contains water at 20°C. The air space at the top of
the closed vat is maintained at 110 kPa. If the depth of the water is 0.8 m, what is the
pressure at the bottom of the tank.
Solution:
At a temperature of 20°C, the density of water (from Table B-6) is
ρ = 1117 kg/m3.
The pressure in an incompressible fluid as a function of depth is
P = P atm + ρgh
P = 110 KPa ( 1000 Pa/ 1 kPa) + (1117 kg/m3 )( 9.81 m/s2 )( 0.8 m)
P = 118,766 Pa = 119 KPa

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The simple relationship between pressure and depth in a static fluid can be used as a
way of measuring pressure differences. The U-tube manometer, shown in Figure 6-1,
is a device that relies on this principle. The gas in the round container is at some
pressure higher than atmospheric pressure. The bent U-tube of the manometer
contains a liquid of known density. The top of the U-tube is open to the atmosphere.

Figure 6.1 A manometer

The pressure difference between point A and point F is being measured. The pressure
at point B is the same as that at point A because there is no vertical distance between
the two and the fluids are at rest; The pressure at point C is related to that at B by

Pc = ρggh1 + PB
where ρg is the density of the gas.
The pressure at point D is related to the pressure at C by

PD = ρLgh2 + Pc = ρLgh2 + ρggh1 + PB
where ρL is the density of the liquid.

The pressure at point D can also be calculated starting the pressure at point F and
working downward. At F, the tube is open to the atmosphere and, therefore, the
pressure there is atmospheric pressure.

The pressure at D can be written
PD = PF + ρLgh3 + ρLgh2 = Patm + ρLgh3 + ρLgh2

Equating the expressions for PD from these last two equations gives
ρLgh2 + ρggh1 + PB = Patm + ρLgh3 + ρLgh2

which, with PA = PB, simplifies to

PA - Patm = ρLgh3 - ρggh1 (Equation 6.2)

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