Applied Thermodynamics Pamphlet 2nd edition, 2018 by musadoto

BPE 211 APPLIED THERMODYNAMICS
Musadoto ISBN-IWR-D-2016-0011
SOKOINE UNIVERSITY OF AGRICULTURE
Department of Engineering sciences and Technology
BPE211/AE213 :APPLIED THERMODYNAMICS
A COMPREHESIVE SIMPLIFIED TO THE APPLIED THERMODYNAMICS
2nd
EDITION
2017
musadoto

About the author:
musadoto is a student at Sokoine University Of Agriculture undertaking bachelor of
science in Irrigation and Water Resources engineering second year with full registration number
iwr/d/2016/0011.Samuye and Njombe secondary were his ordinary and advanced education
level respectively.
i

Forewords
This pamphlet is a 2nd
edition of the series in A COMPREHESIVE SIMPLIFIED TO THE
APPLIED THERMODYNAMICS.It consists of three topics with solved supplementary problems
which based on tutorials, lecture notes ,summaries and the university past papers. Please don‘t
forget to deeply study by yourself to improve your understanding for winning your open book
exams.
ii

Acknowledgement
Much thanks to my God who laid all the following;
Bsc .of irrigation and water resources engineering students for their infinity support both
materially and time. Never the less my friend Kitowe Mariam for her inspiration to complete this
pamphlet. BSc.Age/Bpe for their intentions. I appreciate too Mr Deus,Heri,Donath,Ole David
and Mashauri (Age) for their materials as far as many supporters participated in the preparation
of this pamphlet.
God bless you.
iii

Dedication
This pamphlet is dedicated to all Engineering students at Sokoine University of Agriculture
taking BPE211/AE213(Applied thermodynamic)-2017 and all upcoming youngs at this
institution.
iv

Table of content
About the author…………………………………………………………………….i
Foreword………………………………………………………………………..….ii
Acknowledgement………………………………………………………………....iv
Dedication…………………………………………………………………………v
4. Second law of thermodynamics and Entropy……………………………....1
5. Carnot cycle and Gas power system…………………………………………
6. Steam/Vapour power…………………………………………………………
7. Refrigeration cycles………………………………………………………….
8. Properties of mixtures……………………………………………………….
9. Psychrometry…………………………………………………………………
V.

TOPIC 4
SECOND LAW OF THERMODYNAMICS AND ENTROPY
Introduction
The first law of thermodynamics relates the several variables involved in a physical process but
does not give any information as to the direction of the process. It is the second law of
thermodynamics which helps us to establish the direction of a particular process.
Here down, the first law states that the work done by the falling weight is converted to internal
energy of the air contained in the fixed volume, provided the volume is insulated so that Q = 0. It
would not be a violation of the first law if we postulated that an internal energy decrease of the
air is used to turn the paddle and raise the weight. This, however, would be a violation of the
second law of thermodynamics and would thus be impossibility.
These above processes cannot occur even though they are not in violation of the first law.
Remember to note: A process must satisfy the first law in order to occur BUT Satisfying
the first law alone does not ensure that the process will take place.
Transferring heat to a
paddle wheel will not
cause it to rotate.
Transferring heat to a wire will not
generate electricity.
A cup of hot coffee does not get hotter
in a cooler room.

The following are the usefulness of second law of Thermodynamics:
 Provide means for predicting the direction of processes,
 Establishing conditions for equilibrium,
 Determining the best theoretical performance of cycles, engines and other devices.
STATEMENTS OF THE SECOND LAW OF THERMODYNAMICS
The second law of thermodynamics can be stated in a variety of ways. Here we present
two: the Clausius statement and the Kelvin-Planck statement. Neither is presented in
mathematical terms. We will, however, provide a property of the system, entropy, which
can be used to determine whether the second law is being violated for any particular
situation.
The first statement of the second law is:
Clausius Statement It is impossible to construct a device which operates in a cycle and whose
sole effect is the transfer of heat from a cooler body to a hotter body. This statement relates to a
refrigerator (or a heat pump). It states that it is impossible to construct a refrigerator that
transfers energy from a cooler body to a hotter body without the input of work.
The second statement of the second law takes the following form: Kelvin-Planck Statement. It
is impossible to construct a device which operates in a cycle and produces no other effect than
the production of work and the transfer of heat from a single body. In other words, it is
impossible to construct a heat engine that extracts energy from a reservoir, does work, and does
not transfer heat to a low-temperature reservoir. This rules out any heat engine that is 100
percent efficient, like the one shown

HEAT ENGINES, HEAT PUMPS, AND REFRIGERATORS
A device operating on a cycle is referred as a heat engine, a heat pump, or a refrigerator
depending on the objective of the particular device. defined below:
 If the objective of the device is to perform work it is a heat engine
 if its objective is to supply energy to a body it is a heat pump
 if its objective is toextract energy from a body it is a refrigerator
HEAT ENGINES
As it is known that Work can easily be converted to other forms of energy, but Heat engine differ
considerably from one another, but all can be characterized as follows
 they receive heat from a high-temperature source,
 they convert part of this heat to work,
 they reject the remaining waste heat to a low-temperature sink atmosphere,
 they operate on a cycle.
A schematic diagram of a simple heat engine

The work-producing device that best fit into the definition of a heat engine is the steam
power plant, which is an external combustion engine.
The net work produced by the engine in one cycle would be equal to the net heat transfer, a
consequence of the first law.
W = QH-QL (Heat in –Heat out)
One additional note concerning heat engines is appropriate. There are devices that we will refer
to as heat engines which do not strictly meet our definition; they do not operate on
thermodynamic cycle but instead exhaust the working fluid and then intake new fluid. The
internal combustion engine is an example. Thermal efficiency, is defined below, remains a
quantity of interest for such devices.

THERMAL EFFICIENCY FOR HEAT ENGINE
This is a measure of the performance of engine. It represents the magnitude of the energy
wasted in order to complete the cycle. For a heat engine the desired result is the net work
done and the input is the heat supplied to make the cycle operate.
Heat engine can be expressed in terms of the desired output and the required input. i.e.
= W/QH
The thermal efficiency is always less than 1 or less than 100 percent.
,
 Applying the first law to the cyclic heat engine
 The cycle thermal efficiency may be written as
th 
Desired Result
Required Input
th
net out
in
W
Q

, W W W
Q Q
net out out in
in net
,  

Q W U
W Q
W Q Q
net in net out
net out net in
net out in out
, ,
, ,
,
 

 

th
net out
in
in out
in
out
in
W
Q
Q Q
Q
Q
Q



 
,
1

The second law of thermodynamics will place limits on the above measure of performance.
The law would allow a maximum of unity for the thermal efficiency and an infinite coefficient of
performance. The second law, however, establishes limits that are surprisingly low, limits that
cannot be exceeded regardless of the cleverness of proposed designs.
REVERSIBILITY
In the study of the first law we made use of the concept of equilibrium and we defined
equilibrium, or quasi-equilibrium, with reference to the system only. We must now introduce the
concept of reversibility so that we can discuss the most efficient engine that can possibly be
constructed, an engine that operates with reversible processes only. Such an engine is called a
reversible engine.
A reversible process is defined as a process which, have taken place, can be reversed and in so
doing, it leaves no change in either the system or the surroundings. Observe that our definition
of a reversible process refers to both the system and the surroundings. The process obviously has
to be a quasi-equilibrium process; additional requirements are:
1. No friction is involved in the process.
2. Heat transfer occurs due to an infinitesimal temperature difference only.
3. Unrestrained expansion does not occur
Therefore, The mixing of different substances and combustion also lead to irreversibility. To
explain that friction makes a process irreversible consider the system of block plus inclined plane
shown in fegures below. Weights are added until the block is raised to the position shown in
part( b ) .Now, to return the system to its original state some weight must be removed so that the
block will slide back down the plane, as shown in part (c). Note that the surroundings have
experienced a significant change; the weights must be raised, which requires a work input. Also,
the block and plane are at a higher temperature due to the friction, and heat must be transferred
to the surroundings to return the system to its original state. This will also change the
surroundings. Because there has been a change in the surroundings as a result of the process and
the reversed process, we conclude that the process was irreversible.
For an example of unrestrained expansion, consider the high-pressure gas contained in the
cylinder of figure below. Pull the pin and let the piston suddenly move to the stops shown. Note
that the only work done by the gas on the surroundings is to move the piston against atmospheric
pressure. Now, to reverse this process it is necessary to exert a force on the piston. If the force is
sufficiently large, we can move the piston to its original position, shown in part ( d ) .

This will demand a considerable amount of work, to be supplied by the surroundings. In
addition, the temperature will increase substantially, and this heat must be transferred to the
surroundings to return the temperature to its original value. The net result is a significant change
in the surroundings, a consequence of irreversibility.
HEAT PUMPS AND REFRIGERATORS
 A device that transfers heat from a low temperature medium to a high temperature one is
the heat pump.
 Refrigerator operates exactly like heat pump except that the desired output is the amount
of heat removed out of the system
 The index of performance of a heat pumps or refrigerators are expressed in terms of the
coefficient of performance

THE CARNOT ENGINE
The heat engine that operates the most efficiently between a high-temperature reservoir and a
low-temperature reservoir is the Camot engine. It is an ideal engine that uses reversible
processes to form its cycle of operation; thus it is also called a reversible engine. We will
determine the efficiency of the Carnot engine and also evaluate its reverse operation. The Carnot
engine is very useful, since its efficiency establishes the maximum possible efficiency of any real
engine. If the efficiency of a real engine is significantly lower than the efficiency of a Carnot
engine operating between the same limits, then additional improvements may be possible.
COP
Q
W
Q
Q Q
HP
H
net in
H
H L
 
,
COP
Q
W
R
L
net in

,

ANALYSIS OF CARNOT HEAT ENGINES
Here, we will present what amounts to a version of the discussion of the 1850s inspired by the
original work of Carnot11 depicted along with his writings in Figures below, for heat engines
Sadi Nicolas Léonard Carnot (1796-1832) French engineer
The cycle associated with the Carnot engine is shown in figure below using an ideal gas as the
working substance. It is composed of the following four reversible processes
Process Action Description
1→2 An isothermal expansion. Heat is transferred reversibly from the high
temperature reservoir at the constant
temperature TH. The piston in the cylinder is
withdrawn and the volume increases.
2→3 An adiabatic reversible
expansion.
The cylinder is completely insulated so that no
heat transfer occurs during this reversible
process. The piston continues to be withdrawn,
with the volume increasing.
3→4 An isothermal compression. Heat is transferred reversibly to the low-
temperature reservoir at the constant
temperature TL. The piston compresses the
working substance, with the volume decreasing.
4→1 An adiabatic reversible
compression.
The completely insulated cylinder allows no heat
transfer during this reversible process. The
piston continues to compress the working
substance until the original volume, temperature,
and pressure are reached, thereby completing the
cycle.

Execution of Carnot cycle in a piston cylinder device
The following three postulates observed from a circle.
Postulate 1. It is impossible to construct an engine, operating between two given temperature
reservoirs, that is more efficient than the Carnot engine.
Postulate 2 The efficiency of a Carnot engine is not dependent on the working substance used or
any particular design feature of the engine.
Postulate 3. All reversible engines, operating between two given temperature reservoirs, have
the same efficiency as a Carnot engine operating between the same two temperature
reservoirs.

For example, let‘s show that the efficiency of a Carnot engine is the maximum possible
efficiency. Assume that an engine exists; operating between two reservoirs, that has an efficiency
greater than that of a engine, also, assume that a Carnot engine operates as a refrigerator between
the same two reservoirs, as sketched in Figure(a). Let the heat transferred from the high-
temperature reservoir to the engine be equal to the heat rejected by the refrigerator; then the work
produced by the engine will be greater than the work required by the refrigerator (that is, Q’L <
QL)since the efficiency of the engine is greater than that of a Carnot engine. Now, our system can
be organized as shown in figure(b) below. The engine drives the refrigerator using the rejected
heat from the refrigerator. But, there is some net work (W’ - W )that leaves the system. The net
result is the conversion of energy from a single reservoir into work, a violation of the second
law. Thus, the Carnot engine is the most efficient engine operating between two particular
reservoirs.
Another example to prove the postulate above, lets Show that the efficiency of a Carnot engine
operating between two reservoirs is independent on the working substance used by the engine.
Suppose that a Carnot engine drives a Carnot refrigerator as shown in Figure (a) below. Let the
heat rejected by the engine be equal to the heat required by the refrigerator. Suppose the working
fluid in the engine results in QH being greater than Q’H then W would be greater than W' (a
consequence of the first law) and we would havethe equivalent system shown in Figure (b). The
net result is a transfer of heat ( Q H - QL) from a single reservoir and the production of work, a
clear violation of the second law. Thus, the efficiency of a Carnot engine is not dependent on the
working substance.

CARNOT ENGINE EFFICIENCY
Since the efficiency of a Carnot engine is dependent only on the two reservoir temperatures, the
Objective of this concept will be to determine that relationship. We will assume the working
substance to be an ideal gas and simply perform the required calculations for the four processes
The heat transfer for each of the four processes is as follows:
We should note that we want QL to be a positive quantity, as in the thermal efficiency
relationship; hence, the negative sign. The thermal efficiency is then: Equation1 be
During the reversible adiabatic processes 2 →3 and 4 → 1, we know that:
Thus, we see that
Substituting the above equation into equation 1 we obtain the result recognizing that
(In V2/V1 = -In V1/V2)
We have simply replaced QL/QH with TL/TH. We can do this for all reversible engines or
refrigerators. We see that the thermal efficiency of a Carnot engine is dependent only on the high
and low absolute temperature of the reservoirs. The fact that we used an ideal gas to perform the
calculations is not important since we have shown that Carnot efficiency is independent of the
working substance. Consequently, the relationship in equation above is applicable for all
working substances, or for all reversible engines, regardless of the particular design
characteristics. The Carnot engine, when operated in reverse, becomes a heat pump or a
refrigerator, depending on the desired heat transfer.

The coefficient of performance for a heat pump becomes
The coefficient of performance for a refrigerator takes the form
 The thermal efficiencies of actual and reversible heat engines operating between the same
temperature limits compare as follows
 The coefficients of performance of actual and reversible refrigerators operating between
the same temperature limits compare as follows
ENTROPY
Introduction
The second law states that process occur in a certain direction, not in any direction.It often leads
to the definition of a new property called entropy, which is a quantitative measure of disorder for
a system.
In a property of the system, entropy can be used to determine whether the second law is being
violated for any particular situation. Entropy can also be explained as a measure of the
unavailability of heat to perform work in a cycle .This relates to the 2nd
law since the 2nd
law
predicts that not all heat provided to a cycle can be transformed into an equal amount of work,
some heat rejection must take place
ENTROPY CHANGE
The entropy change during a reversible process is defined as

For a reversible, adiabatic process
Note, The reversible, adiabatic process is called an isentropic process
ENTROPY CHANGE AND ISENTROPIC PROCESSES
The entropy-change and isentropic relations for a process can be summarized as follows:
dS
S S


0
2 1

for all process
for isentropic process
i. Pure substances:
Any process: Δs = s
2
– s
1
(kJ/kgK)
Isentropic process: s
2
= s
1
s s C
T
T
R
v
v
v av2 1
2
1
2
1
  , ln ln
2 2
2 1 ,
1 1
ln lnp av
T P
s s C R
T P
  
2 1
1 2.
k
s const
P v
P v
   
   
   

THE INEQUALITY OF CLAUSIUS
The Carnot cycle is a reversible cycle and produces work which we will refer to as Wrev.
Consider an irreversible cycle operating between the same two reservoirs, shown in Figure below
obviously, since the Carnot cycle possesses the maximum possible efficiency, the efficiency of
the irreversible cycle must be less than that of the Carnot cycle. In other words, for the same
amount of heat addition QH we must have Wirr < Wrev
From the first law applied to a cycle (W = QH- QL) we see that, assuming that (QH)irr and (QH)rev
are the same,(Q L ) rev < (QL)irr
since the above integral for a reversible cycle is zero. If we were considering an irreversible refrigerator
rather than an engine, we would require more work for the same amount of refrigeration QL.By applying the
first law to refrigerators, we would arrive at the same inequality.Hence, for all cycles, reversible or
irreversible, we can write
This is known as the inequality of Clausius. It is a consequence of the second law of
thermodynamics.
ENTROPY CHANGE FOR AN IRREVERSIBLE PROCESS
Take a Consideration of a cycle to be composed of two reversible processes, shown in figure
below .Suppose that we can also return from state 2 to state 1 along the irreversible process
marked by path C. For the reversible cycle we have:

THE SECOND LAW APPLIED TO A CONTROL VOLUME
The second law has been applied thus far in this chapter to a system, a particular collection of
mass particles. We now wish to apply the second law to a control volume, following the same
strategy used in our study of the first law. In Figure below a control volume is enclosed by the
control surface shown with the dashed lines surrounding some device or volume of interest. The
second law can then be expressed over a time increment ∆ t as
Mathematically
,
If we divide the above equation by ∆ t and use dots to denote rates, we arrive at the rate equation

The equality is associated with a reversible process. The inequality is associated with
irreversibility such as viscous effects, which are always present in a material flow; separations of
the flow from boundaries where abrupt changes in geometry occur; and shock waves in high-
speed compressible flow.
For a steady-flow process the entropy of the control volume remains constant with time. We can
then write, recognizing that
ISENTROPIC EFFICIENCY FOR TURBINE
ISENTROPIC EFFICIENCY FOR COMPRESSOR

QUESTIONS FOR CHAPTER 4
QN 1/4
A steam power plant produces 50 MW of net work while burning fuel to produce 150 MW of
heat energy at the high temperature. Determine the cycle thermal efficiency and the heat rejected
by the cycle to the surroundings.
SOLUTION
QN 2/4
A Carnot heat engine receives 500 kJ of heat per cycle from a high-temperature heat reservoir at
652ºC 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
QN 3/4
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
th
net out
H
W
Q
MW
MW

 
,
.
50
150
0 333 or 33.3%
W Q Q
Q Q W
MW MW
MW
net out H L
L H net out
,
,
 
 
 

150 50
100
ii.
Inc
iii.
QL
WO
QH=652o
c
QL = 30O
C
th rev
L
H
T
T
K
K
or
,
( )
( )
. .
 
 



1
1
30 273
652 273
0 672 67 2%
Q
Q
T
T
K
K
Q kJ
kJ
L
H
L
H
L







( )
( )
.
( . )
30 273
652 273
0 328
500 0 328
164

QN 4/4
A 600 MW steam power plant, which is cooled by a river, has a thermal efficiency of 40
percent. Determine the rate of heat transfer to the river water. Will the actual heat transfer
rate be higher or lower than this value? Why? [900 MW]
TL
=
o
HE
QL
Win
COP
Q
Q Q
T
T T
K
K
R
L
H L
L
H L








( )
( )
.
2 273
25 2
1196
QH

QN 5/4
A steam power plant receives heat from a furnace at a rate of 280 GJ/h. Heat losses to the
surrounding air from the steam as it passes through the pipes and other components are
estimated to be about 8 GJ/h. If the waste heat is transferred to the cooling water at a rate of
145 GJ/h, determine (a) net power output and (b) the thermal efficiency of this power plant.
SOLUTION
QN6/4
An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing
electric power at a rate of 6 kW. Determine (a) the COP of this air conditioner and (b) the
rate of heat transfer to the outside air.
QN 7/4
A heat pump is used to heat a house and maintain it at 24o
C. On a winter day when the
outdoor air temperature is -5o
C, the house is estimated to lose heat at a rate of 80,000 kJ/h.
Determine the minimum power required to operate this heat pump.

QN 8/4
An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat
from a house at a rate of 750 kJ/min to maintain its temperature at 24o
C. If the outdoor air
temperature is 35o
C, determine the power required to operate this air-conditioning system.
QN 9/4
An inventor claims to have developed a heat engine that receives 700 kJ of heat from a
source at 500 K and produces 300 kJ of net work while rejecting the waste heat to a sink at
290 K. Is this reasonable claim?

QN 10/4
Determine the COP of a heat pump that supplies energy to a house at a rate of 8000 kJ/h for
each kW of electric power it draws. Also, determine the rate of energy absorption from the
outdoor air.
QN 11/4
Steam at 1 MPa, 600o
C, expands in a turbine to 0.01 MPa. If the process is isentropic, find
the final temperature, the final enthalpy of the steam, and the turbine work.
SOLUTION
 Since that the process is isentropic, s
2
=s
1
1
1
1
1 .
1
sup
1
3698.6
600
8.0311
kJ
kgo
kJ
kg K
State
erheated
P MPa
h
T C
s
 

 

 
1 2
1 1 2 2
1 2
:
in out
out
out
massbalance m m m
energybalance
E E
m h m h W
W m h h
 

 
 
 
2
2
2 . 2
2
2 @
2
0.01 .
8.0311 0.984
191.8 0.984 2392.1
2545.6
45.81
kJ
kg K
kJ
kg
o
sat P
State
P MPa sat mixture
s x
h
T T C
 

 
 

 
1 2
3698.6 2545.6
1153
out
kJ
kg
W h h 
 


QN 12/4
Steam at 1 MPa, 600°C, expands in a turbine to 0.01 MPa. The isentropic work of the turbine
is 1152.2 kJ/kg. If the isentropic efficiency of the turbine is 90 percent, calculate the actual
work. Find the actual turbine exit temperature or quality of the steam.
SOLUTION
QN 13/4
 
1 2
,
1 2
,
0.9 1153
1037.7
a a
isen T
s s
a isen T s
kJ
kg
w h h
w h h
w w



 

 



QN 14/4
The radiator of a steam heating system has a volume of 20 L and is filled with the
superheated water vapor at 200 kPa and 150o
C. At this moment both inlet and exit valves
to the radiator are closed. After a while the temperature of the steam drops to 40o
C as a
result of heat transfer to the room air. Determine the entropy change of the steam during
this process.
QN 15/4
A heavily insulated piston-cylinder device contains 0.05 m3
of steam at 300 kPa and
150o
C. Steam is now compressed in a reversible manner to a pressure of 1 MPa.
Determine the work done on the steam during this process. [ 16 kJ ]

QN 16/4
A piston –cylinder device contains 1.2 kg of nitrogen gas at 120 kPa and 27o
C. The gas is
now compressed slowly in a polytropic process during which PV1.3
=constant. The
process ends when the volume is reduced by one-half. Determine the entropy change of
nitrogen during this process [ -0.0617 kJ/kg.K ]

QN 16/4
Steam enters an adiabatic turbine at 8 MPa and 500o
C with a mass flow rate of 3 kg/s and
leaves at 30 kPa. The isentropic efficiency of the turbine is 0.90. Neglecting the kinetic
energy of the steam, determine (a) the temperature at the turbine exit and (b) the power
output of the turbine [ 69.09o
C,3054 Kw]

QN 17/4
Refrigerant-R134a enters an adiabatic compressor as saturated vapor at 120 kPa at a rate
of 0.3 m3
/min and exits at 1 MPa pressure. If the isentropic efficiency of the compressor
is 80 percent, determine (a) the temperature of the refrigerant at the exit of the
compressor and (b) the power input, in kW. Also, show the process on a T-s diagram
with respect to the saturation lines. [ 58.9o
C,1.70 kW ]

QN 18/4
A heat pump receives energy by heat transfer from the outside air at 0 °C and discharges
energy by heat transfer to a dwelling at 20 °C. Is this in violation of the second law of
thermodynamics? Explain.

QN 19/4
Two kilograms of water execute a Carnot power cycle. During the isothermal expansion,
the water is heated until it is a saturated vapour from an initial state where the pressure is
40 bar and the quality is 15%. The vapour then expands adiabatically to a pressure of 1.5 bar
while doing 491.5 kJ/kg of work.
(a.)Sketch the cycle on p–v coordinates.
(b.)Evaluate the heat and work for each process, in kJ.
(c.)Evaluate the thermal efficiency.

QN 20/4
One kilogram of air as an ideal gas executes a Carnot power cycle having a thermal
efficiency of 60%. The heat transfer to the air during the isothermal expansion is 40 kJ.
At the end of the isothermal expansion, the pressure is 5.6 bar and the volume is 0.3 m3
.
Determine:
(a.) The maximum and minimum temperatures for the cycle, in K.
(b.) The pressure and volume at the beginning of the isothermal expansion in bar and
m3
, respectively.
(c.) The work and heat transfer for each of the four processes, in kJ.
(d.) Sketch the cycle on p–v coordinates.

QN 21/4
Steam is the working fluid in an ideal Rankine cycle. Saturated vapour 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 mass flow rate of the steam, in kg/h, (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 15°C and
exits at 35°C.
Assumptions:
 Each component of the cycle is analyzed as a control volume at steady state. The control
volumes are shown on the
 accompanying sketch by dashed lines.
 All processes of the working fluid are internally reversible.
 The turbine and pump operate adiabatically.
 Kinetic and potential energy effects are negligible.
 Saturated vapor enters the turbine. Condensate exits the condenser as saturated liquid.

QN 22/4
Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa,
and saturated vapour enters the turbine at (a) 18 MPa and (b) 4 MPa. The net power
output of the cycle is 100 MW. Determine for each case the mass flow rate of steam, in
kg/h, the heat transfer rates for the working fluid passing through the boiler and
condenser, each in kW, and the thermal efficiency.

QN 23/4
Water is the working fluid in an ideal Rankine cycle. Superheated vapour enters the turbine
at 8 MPa, 480°C. The condenser pressure is 8 kPa. The net power output of the cycle is 100
MW. Determine for the cycle
(a.)The rate of heat transfer to the working fluid passing through the steam generator, in kW.
(b.)The thermal efficiency.
(c.)The mass flow rate of condenser cooling water, in kg/h, if
(d.) The cooling water enters the condenser at 15°C and exits at 35°C with negligible pressure
change.

QN 24/4
Water is the working fluid in a Carnot vapour power cycle. Saturated liquid enters the boiler at a
pressure of 8 MPa, and saturated vapour enters the turbine. The condenser pressure is 8 kPa.
Determine.
(a.)The thermal efficiency.
(b.)The back work ratio.
(c.)The heat transfer to the working fluid per unit mass passing through the boiler, in kJ/kg.
(d.) The heat transfer from the working fluid per unit mass passing through the condenser, in
kJ/kg.
a.

QN 25/4
A Carnot engine operates between two temperature reservoirs maintained at 200 o
C and
20 o
C, respectively. If the desired output of the engine is 15 kW, as shown in Figure
below, determine the heat transfer from the high-temperature reservoir and the heat
transfer to the low-temperature reservoir.
SOLUTION
QN 26/4
A refrigeration unit is cooling a space to -5°C by rejecting energy to the atmosphere at 20°C.It
is desired to reduce the temperature in the refrigerated space to -25°C. Calculate the minimum
percentage increase in work required, by assuming a Carnot refrigerator, for the same amount
of energy removed.
SOLUTION

QN 27/4
A Carnot engine operates with air, using the cycle shown in Figure below. Determine the
thermal efficiency and the work output for each cycle of operation.
SOLUTION
The thermal efficiency is found to be
QN 28/4
A refrigerator is rated at a COP of 4. The refrigerated space that it cools requires a peak cooling
rate of 30000 kJ/h. What size electrical motor (rated in horsepower) is required for the
refrigerator?
SOLUTION

QN 29/4
An inventor proposes an engine that operates between the 27°C warm surface layer of the ocean
and a 10°C layer a few meters down. The inventor claims that the engine produces 100 kW by
pumping 20 kg/s of seawater. Is this possible?
SOLUTION
sw
QN 30/4
A power utility company desires to use the hot groundwater from a hot spring to power a heat
engine. If the groundwater is at 95O
C, estimate the maximum power output if a mass flux of 0.2
kg/s is possible. The atmosphere is at 20°C.
SOLUTION
The maximum possible efficiency is
QN 31/4
An ideal Rankine cycle operates between pressures of 30 kPa and 6 MPa. The temperature of the
steam at the inlet of the turbine is 550°C. Find the net work for the cycle and the thermal
efficiency

solution

QN 32/4
A Carnot engine operating on air accepts 50 kJ/kg of heat and rejects 20 kJ/kg. Calculate the
high and low reservoir temperatures if the maximum specific volume is 10 m3
/kg and the
pressure after the isothermal expansion is 200 kPa.
SOLUTION
QN 32/4
A heat engine operates on a Carnot cycle with an efficiency of 75 percent. What COP would a
refrigerator operating on the same cycle have? The low temperature is 0°C.
SOLUTION

QN33/4
Two Carnot refrigerators operate in series between two reservoirs maintained at 20°C and
200O
C, respectively. The energy output by the first refrigerator is used as the heat energy input to
the second refrigerator. If the COPs of the two refrigerators are the same, what should the
intermediate temperature be?
SOLUTION
QN 34/4
Air is contained in an insulated, rigid volume at 20 O
C and 200 kPa. A paddle wheel, inserted in
the volume, does 720 kJ of work on the air. If the volume is 2 m3
, calculate the entropy increase
assuming constant specific heats.
SOLUTION
QN 35/4
After combustion process in a cylinder the pressure is 1200 kPa and the temperature is 350 O
C.
The gases are expanded to 140 kPa with a reversible adiabatic process. Calculate the work done
by the gases, assuming they can be approximated by air with constant specific
SOLUTION
QN 36/4
Using QN 34/4 above, assuming variable specific heats.

SOLUTION
QN 37/4
After combustion process in a cylinder the pressure is 1200 kPa and the temperature is 350°C.
The gases are expanded to 140 kPa in a reversible, adiabatic process. Calculate the work done by
the gases, assuming they can be approximated by air with variable specific heats.
SOLUTION
QN 38/4
It is proposed to operate a simple steam power plant as shown in Figure below The water is
completely vaporized in the boiler so that the heat transfer QB takes place at constant
temperature. Does this proposal comply with the inequality of Clausius? Assume no heat transfer
occurs from the pump or the turbine.

SOLUTION
QN 39/4
Two kg of superheated steam at 400°C and 600 kPa is cooled at constant pressure by transferring
heat from a cylinder until the steam is completely condensed. The surroundings are at 25°C.
Determine the net entropy change of the universe due to this process.
SOLUTION
QN 40/4
A preheater is used to preheat water in a power plant cycle, as shown in Figure below. The
superheated steam is at a temperature of 250 o
C and the entering water is subcooled at 45 o
C. All
pressures are 600 kPa. Calculate the rate of entropy production.
SOLUTION

Sf, at Tl = 45o
C,or s1 = 0.639 kJ/kg - K. Finally, to account for two inlets, we have
QN 41/4
A Carnot engine delivers 100 kW of power by operating between temperature reservoirs at100°C
and 1000°C. Calculate the entropy change of each reservoir and the net entropy change of the
two reservoirs after 20 min of operation.
SOLUTION
The efficiency of the engine is
This is zero, except for round-off error,
QN 42/4
Two kg of air is heated at constant pressure of 200 kPa to 500°C. Calculate the entropy change
if the initial volume is 0.8 m3
.

SOLUTION
The initial temperature is found to be
The entropy change is then found,
QN 43/4
A piston allows air to expand from 6 MPa to 200 kPa. The initial volume and temperature are
500 cm3
and 800°C. If the temperature is held constant, calculate the heat transfer and the
entropy change.
SOLUTION
QN 44/4
A paddle wheel provides 200 kJ of work to the air contained in a 0.2-m3
rigid volume, initially
at 400 kPa and 40°C. Determine the entropy change if the volume is insulated.
SOLUTION
QN 45/4
Air expands from 200 to 1000 cm3
in a cylinder while the pressure is held constant at 600 kPa.
If the initial temperature is 20O
C, calculate the heat transfer assuming
( a ) constant specific heat and
( b ) variable specific heat.

SOLUTION
QN 46/4
Water is maintained at a constant pressure of 400 kPa while the temperature changes from 20°C
to 400°C. Calculate the heat transfer and the entropy change.
SOLUTION
QN 47/4.
The steam in a Carnot engine is compressed adiabatically from 10 kPa to 6 MPa with saturated
liquid occurring at the end of the process. If the work output is 500 kJ/kg, calculate the quality at
the end of the isothermal expansion.
SOLUTION
QN 48/4
Two kg of steam is contained in a 6-liter tank at 60°C. If 1 MJ of heat is added, calculate the
final entropy.
SOLUTION

QN 49/4
The Freon 12 in a Carnot refrigerator operates between saturated liquid and vapor during the
heat rejection process. If the cycle has a high temperature of 50°C and a low temperature of-
2OO
C, calculate the heat transfer from the refrigerated space and the quality at the beginning of
the heat addition process.
SOLUTION
QN 50/4
Show that the inequality of Clausius is satisfied by a Carnot engine operating with steam
between pressures of 40 kPa and 4 MPa. The work output is 350 kJ/kg, and saturated vapor
enters the adiabatic expansion process.
SOLUTION

QN 51/4
Two kg of saturated steam is contained in 0.2-m3
rigid volume. Heat is transferred to the
surroundings at 30°C until the quality reaches 20 percent. Calculate the entropy change of the
universe.
SOLUTION
QN 52/4
A steam turbine accepts 2 kg/s of steam at 6 MPa and 600 O
C and exhausts saturated steam at
20 kPa while producing 2000 kW of work. If the surroundings are at 30°C and the flow is
steady, calculate the rate of entropy production.
SOLUTION

QN 53/4
A rigid tank is sealed when the temperature is 0 O
C.On a hot day the temperature in the tank
reaches 50°C. If a small hole is drilled in the tank, estimate the velocity of the escaping air.
SOLUTION
QN 54/4
Steam expands isentropically through a turbine from 6 MPa and 600°C to 10 kPa. Calculate
the power output if the mass flux is 2 kg/s.
SOLUTION
QN 55/4
Calculate the efficiency of the rankine cycle shown in fegure below if the maximum temperature
Is 700°c. The pressure is constant in the boiler and condenser.
SOLUTION

QN 56/7
An inventor claims to have developed a power cycle capable of delivering a net work output of
410 kJ for an energy input by heat transfer of 1000 kJ. The system undergoing the cycle receives
the heat transfer from hot gases at a temperature of 500 K and discharges energy by heat transfer
to the atmosphere at 300 K. Evaluate this claim..

QN 57/4
By steadily circulating a refrigerant at low temperature through passages in the walls of the
freezer compartment, a refrigerator maintains the freezer compartment at -5o
C when the air
surrounding the refrigerator is at 22o
C. The rate of heat transfer from the freezer compartment to
the refrigerant is 8000 kJ/h and the power input required to operate the refrigerator is 3200 kJ/h.
Determine the coefficient of performance of the refrigerator and compare with the coefficient of
performance of a reversible refrigeration cycle operating between reservoirs at the same two
temperatures.
Note : The difference between the actual and maximum coefficients of performance suggests that
there may be some potential for improving the thermodynamic performance. This objective
should be approached judiciously, however, for improved performance may require increases in
size, complexity, and cost.

QN 58/4
A dwelling requires 5 X105 kJ per day to maintain its temperature at 22o
C when the outside
temperature is 10o
C. (a) If an electric heat pump is used to supply this energy, determine the
minimum theoretical work input for one day of operation, in kJ.
QN 59/4
Air as an ideal gas expands isothermally at 20O
C from a volume of 1m3
to 2m3
. During this
process there is heat transfer to the air from the surrounding atmosphere, modeled as a thermal
reservoir, and the air does work. Evaluate the work and heat transfer for the process, in kJ/kg. Is
this process in violation of the second law of thermodynamics? Explain.
(see end of pamphet )
QN 60/4
An inventor claims to have developed a device that undergoes a thermodynamic cycle while
communicating thermally with two reservoirs. The system receives energy QC from the cold
reservoir and discharges energy QH to the hot reservoir while delivering a net amount of work to
its surroundings. There are no other energy transfers between the device and its surroundings.
Using the second law of thermodynamics, evaluate the inventor‘s claim.

QN 61/4
Methane gas within a piston–cylinder assembly is compressed in a quasiequilibrium process. Is
this process internally reversible? Is this process reversible?

QN 62/4
Water within a piston–cylinder assembly cools isothermally at 120O
C from saturated vapor to
saturated liquid while interacting thermally with its surroundings at 20O
C. Is the process
internally reversible? Is it reversible? Discuss.

QN 63/4
A power cycle I and a reversible power cycle R operate between the same two reservoirs, as
shown in Figure below. Cycle I has a thermal efficiency equal to two-thirds of that for cycle
R.Using the Kelvin-Planck statement of the second law, prove that cycle I must be irreversible.

QN 64/4
A reversible power cycle R and an irreversible power cycle I operate between the same two
reservoirs.
(a) If each cycle receives the same amount of energy QH from the hot reservoir, show that cycle I
necessarily discharges more energy QC to the cold reservoir than cycle R. Discuss the
implications of this for actual power cycles.
(b) If each cycle develops the same net work, show that cycle I necessarily receives more energy
QH from the hot reservoir than cycle R. Discuss the implications of this for actual power cycles.
63.

QN 65/4
Two reversible power cycles are arranged in series. The first cycle receives energy by heat
transfer from a reservoir at temperature TH and rejects energy to a reservoir at an intermediate
temperature T. The second cycle receives the energy rejected by the first cycle from the reservoir
at temperature T and rejects energy to a reservoir at temperature TC lower than T. Derive an
expression for the intermediate temperature T in terms of TH and TC when
(a) the net work of the two power cycles is equal.
(b) the thermal efficiencies of the two power cycles are equal.

Qn 66/4
A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25°C by absorbing heat
from geothermal water that enters the evaporator at 50°C at a rate of 0.065 kg/s and leaves at 40°C. The
refrigerant enters the evaporator at 20°C with a quality of 23 percent and leaves at the inlet pressure as
saturated vapor. The refrigerant loses 300 W of heat to the surroundings as it flows through the
compressor and the refrigerant leaves the compressor at 1.4 MPa at the same entropy as the inlet.
Determine (a) the degrees of sub-cooling of the refrigerant in the condenser, (b) the mass flow rate of the
refrigerant, (c) the heating load and the COP of the heat pump, and (d) the theoretical minimum power
input to the compressor for the same heating load

QN 67/4
A heat pump with refrigerant R-134a as the working fluid is used to keep a space at 25°C by absorbing
heat from geothermal water that enters the evaporator at 50°C at a rate of 0.065 kg/s and leaves at 40°C.
Refrigerant enters the evaporator at 20°C with a quality of 15 percent and leaves at the same pressure as a
saturated vapor. If the compressor consumes 1.2 kW of power, determine the mass flow rate of the
refrigerant, the rate of heat supplied to the 25°C space, the coefficient of performance of the heat pump,
and the minimum power input to the compressor for the same rate of heat supplied to the 25°C space

QN 68/4
A reversible power cycle receives QH from a hot reservoir at temperature TH and rejects energy
by heat transfer to the surroundings at temperature T0. The work developed by the power cycle is
used to drive a refrigeration cycle that removes QC from a cold reservoir at temperature TC and
discharges energy by heat transfer to the same surroundings at T0.
(a) Develop an expression for the ratio QC/QH in terms of the ratios TH/T0 and TC/T0.
(b) Plot QC/QH versus TH/T0 for TC/T0 =0.85, 0.9, and 0.95, and versus TC/T0 for TH/T0 = 2, 3,
and 4.

QN 69/4
A reversible power cycle receives energy QH from a reservoir at temperature TH and rejects QC to
a reservoir at temperature TC. The work developed by the power cycle is used to drive a
reversible heat pump that removes energy from a reservoir at temperature and rejects energy to a
reservoir at temperature .

(a) Develop an expression for the ratio Q‘H/QH in terms of the temperatures of the four
reservoirs.
(b) What must be the relationship of the temperatures TH, TC, T’ C, and T’H for Q‘H/QH
to exceed a value of unity?
(c) T‘H = TC =T0, plot versus TH/T0 for 0.85, 0.9, and 0.95, and versus T0 for TH/T0 = 2, 3, and 4.

QN 70/4
Water, initially a saturated liquid at 100o
C, is contained in a piston–cylinder assembly. The water
undergoes a process to the corresponding saturated vapor state, during which the piston moves
freely in the cylinder. If the change of state is brought about by heating the water as it undergoes
an internally reversible process at constant pressure and temperature, determine the work and
heat transfer per unit of mass, each in kJ/kg.
solution

Assumptions:
1. The water in the piston–cylinder assembly is a closed system.
2. The process is internally reversible.
3. Temperature and pressure are constant during the process.
4. There is no change in kinetic or potential energy between the two end states.
As shown in the accompanying figure, the work and heat transfer can be represented as areas on
p–v and T–s diagrams, respectively.

QN 71/5
Water initially a saturated liquid at 100o
C is contained within a piston–cylinder assembly. The
water undergoes a process to the corresponding saturated vapor state, during which the piston
moves freely in the cylinder. There is no heat transfer with the surroundings. If the change of
state is brought about by the action of a paddle wheel, determine the net work per unit mass, in
kJ/kg, and the amount of entropy produced per unit mass, in kJ/kg.k.
solution
Assumptions:
1. The water in the piston–cylinder assembly is a closed system.
2. There is no heat transfer with the surroundings.
3. The system is at an equilibrium state initially and finally. There is no change in kinetic or
potential energy between these
two states.

QN 72/5
Refrigerant 134a is compressed adiabatically in a piston–cylinder assembly from saturated vapor
at 0o
C to a final pressure of 0.7 MPa. Determine the minimum theoretical work input required
per unit mass of refrigerant, in kJ/kg.
Assumptions:
1. The Refrigerant 134a is a closed system.
2. There is no heat transfer with the surroundings.
3. The initial and final states are equilibrium states. There is no change in kinetic or potential
energy between these states.
Analysis: An expression for the work can be obtained from an energy balance. By applying
assumptions 2 and 3

QN73/4

QN 74/4
Steam enters a turbine with a pressure of 30 bar, a temperature of 400o
C, and a velocity of 160
m/s. Saturated vapor at 100o
C exits with a velocity of 100 m /s. At steady state, the turbine
develops work equal to 540 kJ per kg of steam flowing through the turbine. Heat transfer
between the turbine and its surroundings occurs at an average outer surface temperature of 350
K.Determine the rate at which entropy is produced within the turbine per kg of steam flowing, in
Neglect the change in potential energy between inlet and exit.

Assumptions:
1. The control volume shown on the accompanying sketch is at steady state.
2. Heat transfer from the turbine to the surroundings occurs at a specified average outer surface
temperature.
3. The change in potential energy between inlet and exit can be neglected.
Analysis: To determine the entropy production per unit mass flowing through the turbine, begin
with mass and entropy rate balances for the one-inlet, one-exit control volume at steady state:

QN 75/4
An inventor claims to have developed a device requiring no energy transfer by work or heat
transfer, yet able to produce hot and cold streams of air from a single stream of air at an
intermediate temperature. The inventor provides steady-state test data indicating that when air
enters at a temperature of 39O
C and a pressure of 5.0 bars, separate streams of air exit at
temperatures of -18O
C and 79O
C, respectively, and each at a pressure of 1 bar. Sixty percent of
the mass entering the device exits at the lower temperature. Evaluate the inventor‘s claim,
employing the ideal gas model for air and ignoring changes in the kinetic and potential energies
of the streams from inlet to exit.
Assumptions:
2. For the control volume ,
3. Changes in the kinetic and potential energies from inlet to exit can be ignored.
4. The air is modeled as an ideal gas with constant cp= 1.0 kJ/kg .K.
Analysis: For the device to operate as claimed, the conservation of mass and energy principles
must be satisfied. The second law of thermodynamics also must be satisfied; and in particular the
rate of entropy production cannot be negative. Accordingly, the mass, energy and entropy rate
balances are considered in turn. With assumptions 1–3, the mass and energy rate balances
reduce, respectively, to

QN 76/4
Components of a heat pump for supplying heated air to a dwelling are shown in the schematic
below. At steady state, Refrigerant 22 enters the compressor at -5O
C, 3.5 bar and is compressed
adiabatically to 75O
C, 14 bar. From the compressor, the refrigerant passes through the condenser,
where it condenses to liquid at 28O
C, 14 bar. The refrigerant then expands through a throttling
valve to 3.5 bar. The states of the refrigerant are shown on the accompanying T–s diagram.
Return air from the dwelling enters the condenser at 20O
C, 1 bar with a volumetric flow rate of
0.42 m3
/s and exits at 50O
C with a negligible change in pressure. Using the ideal gas model for
the air and neglecting kinetic and potential energy effects, (a) determine the rates of entropy
production, in kW/K, for control volumes enclosing the condenser,compressor, and expansion
valve, respectively. (b) Discuss the sources of irreversibility in the components considered in
part (a).

Assumptions:
1. Each component is analyzed as a control volume at steady state.
2. The compressor operates adiabatically, and the expansion across the valve is a throttling process.
3. For the control volume enclosing the condenser
4. Kinetic and potential energy effects can be neglected.
5. The air is modeled as an ideal gas with constant cp = 1.005 kJ/kg . K.

QN 77/4
A rigid, well-insulated tank is filled initially with 5 kg of air at a pressure of 5 bar and a
temperature of 500 K. A leak develops, and air slowly escapes until the pressure of the air
remaining in the tank is 1 bar. Employing the ideal gas model, determine the amount of mass
remaining in the tank and its temperature.

QN 77/4
A steam turbine operates at steady state with inlet conditions of p1 =5 bar, T1 = 320O
C. Steam
leaves the turbine at a pressure of 1 bar. There is no significant heat transfer between the turbine
and its surroundings, and kinetic and potential energy changes between inlet and exit are
negligible. If the isentropic turbine efficiency is 75%, determine the work developed per unit
mass of steam flowing through the turbine, in kJ/kg.

QN 78/4
A turbine operating at steady state receives air at a pressure of p1 =3.0 bar and a temperature of
T1 =390 K. Air exits the turbine at a pressure of p2 =1.0 bar. The work developed is measured as
74 kJ per kg of air flowing through the turbine. The turbine operates adiabatically, and changes
in kinetic and potential energy between inlet and exit can be neglected. Using the ideal gas
model for air, determine the turbine efficiency.

Assumptions:
2. The expansion is adiabatic and changes in kinetic and potential energy between inlet and exit can be
neglected.
3. The air is modeled as an ideal gas.
Analysis: The numerator of the isentropic turbine efficiency is known. The denominator is evaluated as
follows.
The work developed in an isentropic expansion from the given inlet state to the specified exit pressure is
QN 79/4
Steam enters a nozzle operating at steady state at p1=1.0 MPa and T1=320O
C with a velocity of
30 m/s. The pressure and temperature at the exit are p2 = 0.3 MPa and T2= 180O
C. There is no
significant heat transfer between the nozzle and its surroundings, and changes in potential energy
between inlet and exit can be neglected. Determine the nozzle efficiency.

QN 80/4
An air compressor operates at steady state with air entering at p1=1 bar, T1=20o
C, and exiting at
p2=5 bar. Determine the work and heat transfer per unit of mass passing through the device, in
kJ/kg, if the air undergoes a polytropic process with n= 1.3. Neglect changes in kinetic and
potential energy between the inlet and the exit. Use the ideal gas model for air.

QN 81/4
One kilogram of Refrigerant 22 undergoes a process from 0.2 MPa, 20O
C to a state where the
pressure is 0.06 MPa. During the process there is a change in specific entropy, s2 -s1 = -0.55
kJ/kg K. Determine the temperature at the final state, in O
C, and the final specific enthalpy, in
kJ/kg.

QN 90/4
A quantity of air amounting to 2.42 x10-2
kg undergoes a thermodynamic cycle consisting of
three internally reversible processes in series.
Process 1–2: constant-volume heating at V =0.02 m3
from p1= 0.1 MPa to p2=0.42 MPa
Process 2–3: constant-pressure cooling
Process 3–1: isothermal heating to the initial state Employing the ideal gas model with cp =1
kJ/kg K, evaluate the change in entropy, in kJ/K, for each process. Sketch the cycle on p–v and
T–s coordinates.

QN 91/4
One kilogram of water initially at 160O
C, 1.5 bar undergoes an isothermal, internally reversible
compression process to the saturated liquid state. Determine the work and heat transfer, each in
kJ. Sketch the process on p–v and T–s coordinates. Associate the work and heat transfer with
areas on these diagrams.
QN 92/4
A gas initially at 14 bar and 60O
C expands to a final pressure of 2.8 bar in an isothermal,
internally reversible process. Determine the heat transfer and the work, each in kJ per kg of gas,
if the gas is (a) Refrigerant 134a, (b) air as an ideal gas. Sketch the processes on p–v and T–s
coordinates.

QN 93/4
Nitrogen (N2) initially occupying 0.5 m3
at 1.0 bar, 20O
C undergoes an internally reversible
compression during which pV1.30
= constant to a final state where the temperature is 200O
C.
Determine assuming the ideal gas model
(a) the pressure at the final state, in bar.
(b) the work and heat transfer, each in kJ.
(c) the entropy change, in kJ/K.

QN 94/4
Air initially occupying a volume of 1 m3
at 1 bar, 20O
C undergoes two internally reversible
processes in series
Process 1–2: compression to 5 bar, 110O
C during which pV n
=constant
Process 2–3: adiabatic expansion to 1 bar
(a) Sketch the two processes on p–v and T–s coordinates.
(b) Determine n.
(c) Determine the temperature at state 3, in O
C.
(d) Determine the net work, in kJ.

QN 95/4
One-tenth kilogram of water executes a Carnot power cycle. At the beginning of the isothermal
expansion, the water is a saturated liquid at 160O
C. The isothermal expansion continues until the
quality is 98%. The temperature at the conclusion of the adiabatic expansion is 200
C.
(a) Sketch the cycle on T–s and p–v coordinates.
(b) Determine the heat added and net work, each in kJ.
(c) Evaluate the thermal efficiency.

QN 96/4
A quantity of air undergoes a thermodynamic cycle consisting of three internally reversible
processes in series.
Process 1–2: isothermal expansion from 6.25 to 1.0 bar
Process 2–3: adiabatic compression to 550 K, 6.25 bar
Process 3–1: constant-pressure compression
Employing the ideal gas model,
(a) sketch the cycle on p–v and T–s coordinates.
(b) determine T1, in K
(c) If the cycle is a power cycle, determine its thermal efficiency. If the cycle is a refrigeration
cycle, determine its coefficient of performance.

QN 97/4
A piston–cylinder assembly initially contains 0.04 m3 of water at 1.0 MPa, 320o
C. The water
expands adiabatically to a final pressure of 0.1 MPa. Develop a plot of the work done
by the water, in kJ, versus the amount of entropy produced, in kJ/K.

QN 98/4
One kilogram of air is initially at 1 bar and 450 K. Can a final state at 2 bar and 350 K be
attained in an adiabatic process?
QN 99/4
Air as an ideal gas is compressed from a state where the pressure is 0.1 MPa and the temperature
is 27o
C to a state where the pressure is 0.5 MPa and the temperature is 207o
C. Can this process
occur adiabatically? If yes, determine the work per unit mass of air, in kJ/kg, for an adiabatic
process between these states. If no, determine the direction of the heat transfer.

QN 100/4
Two kilograms of Refrigerant 134a initially at 1.4 bar, 60o
C are compressed to saturated vapor at
60o
C. During this process, the temperature of the refrigerant departs by no more than 0.01o
C
from 60o
C . Determine the minimum theoretical heat transfer from the refrigerant during the
process,
in kJ.
QN 101/4
One-half kilogram of propane initially at 4 bar, 30o
C undergoes a process to 14 bar, 100o
C
while being rapidly compressed in a piston–cylinder assembly. Heat transfer with the
surroundings at 20o
C occurs through a thin wall. The net work is measured as -72.5 kJ. Kinetic
and potential energy effects can be ignored. Determine whether it is possible for the work
measurement to be correct.

QN 102/4
A system consists of 2 m3
of hydrogen gas (H2), initially at 35o
C , 215 kPa, contained in a
closed rigid tank. Energy is transferred to the system from a reservoir at 300oC until the
temperature of the hydrogen is 160oC . The temperature at the system boundary where heat
transfer occurs is 300o
C . Modeling the hydrogen as an ideal gas, determine the heat transfer,
in kJ, the change in entropy, in kJ/K, and the amount of entropy produced, in kJ/K. For the
reservoir, determine the change in entropy, in kJ/K. Why do these two entropy changes differ?

QN 103/4
An isolated system consists of a closed aluminum vessel of mass 0.1 kg containing 1 kg of used engine
oil, each initially at 55o
C , immersed in a 10-kg bath of liquid water, initially at 20o
C .The system is
allowed to come to equilibrium. Determine
(a) the final temperature, in degrees centigrade.
(b) the entropy changes, each in kJ/K, for the aluminum vessel, the oil, and the water.
(c) the amount of entropy produced, in kJ/K.

QN 104/4
An insulated cylinder is initially divided into halves by a frictionless, thermally conducting
piston. On one side of the piston is 1 m3
of a gas at 300 K, 2 bar. On the other side is 1 m3
of the
same gas at 300 K, 1 bar. The piston is released and equilibrium is attained, with the piston
experiencing no change of state. Employing the ideal gas model for the gas, determine
(a) the final temperature, in K.
(b) the final pressure, in bar.
(c) the amount of entropy produced, in kJ/kg.

QN 105/4
A system consisting of air initially at 300 K and 1 bar experiencesthe two different types of
interactions described below. In each case, the system is brought from the initial state to a state
where the temperature is 500 K, while volume remains constant.
(a) The temperature rise is brought about adiabatically by stirring the air with a paddle wheel.
Determine the amount of entropy produced, in kJ/kg K.
(b) The temperature rise is brought about by heat transfer from a reservoir at temperature T. The
temperature at the system boundary where heat transfer occurs is also T. Plot the amount of
entropy produced, in kJ/kg K, versus T for T ≥ 500 K. Compare with the result of (a) and discuss.

QN 106/4
Steam at 10 bar, 600o
C , 50 m/s enters an insulated turbine operating at steady state and exits at
0.35 bar, 100 m/s. The work developed per kg of steam flowing is claimed to be (a) 1000
kJ/kg, (b) 500 kJ/kg. Can either claim be correct? Explain.

QN 107/4
Air enters an insulated turbine operating at steady state at 6.5 bar, 687o
C and exits at 1 bar,
327o
C . Neglecting kinetic and potential energy changes and assuming the ideal gas model,
determine
(a) the work developed, in kJ per kg of air flowing through the turbine.
(b) whether the expansion is internally reversible, irreversible, or impossible.

QN 108/4
Methane gas (CH4) at 280 K, 1 bar enters a compressor operating at steady state and exits at 380
K, 3.5 bar. Ignoring heat transfer with the surroundings and employing the ideal gas model with
Ćp (T) from Table A-21, determine the rate of entropy production within the compressor, in kJ/kg
. K

QN 109/4
Ammonia enters a valve as a saturated liquid at 7 bar with a mass flow rate of 0.06 kg/min and is
steadily throttled to a pressure of 1 bar. Determine the rate of entropy production inkW/K. If the
valve were replaced by a power-recovery turbine operating at steady state, determine the
maximum theoretical power that could be developed, in kW. In each case, ignoreheat transfer
with the surroundings and changes in kinetic and potential energy. Would you recommend using
such a turbine?

QN 110/4
Air enters an insulated diffuser operating at steady state at 1 bar, -3O
C, and 260 m/s and exits
with a velocity of 130 m/s. Employing the ideal gas model and ignoring potential energy,
determine
(a) the temperature of the air at the exit, in o
C .
(b) The maximum attainable exit pressure, in bar.

QN 111/4
Steam enters a horizontal 15-cm-diameter pipe as a saturated vapor at 5 bar with a velocity of 10
m/s and exits at 4.5 bar with a quality of 95%. Heat transfer from the pipe to the surroundings at
300 K takes place at an average outer surface temperature of 400 K. For operation at steady state,
determine
(a) the velocity at the exit, in m/s.
(b) the rate of heat transfer from the pipe, in kW.
(c) the rate of entropy production, in kW/K, for a control volume comprising only the pipe and
its contents.
(d) the rate of entropy production, in kW/K, for an enlarged control volume that includes the
pipe and enough of its immediate surroundings so that heat transfer from the control volume
occurs at 300 K. Why do the answers of parts (c) and (d) differ?

QN 112/4
Steam enters a turbine operating at steady state at a pressure of 3 MPa, a temperature of 400o
C ,
and a velocity of 160 m/s. Saturated vapor exits at 100o
C , with a velocity of100 m/s. Heat
transfer from the turbine to its surroundings takes place at the rate of 30 kJ per kg of steam at a
location where the average surface temperature is 350 K.
(a) For a control volume including only the turbine and its contents, determine the work
developed, in kJ, and the rate at which entropy is produced, in kJ/K, each per kg of steam
flowing.
(b) The steam turbine of part (a) is located in a factory where the ambient temperature is 27o
C .
Determine the rate of entropy production, in kJ/K per kg of steam flowing, for an enlarged
control volume that includes the turbine and enough of its immediate surroundings so that heat
transfer takes place from the control volume at the ambient temperature. Explain why the
entropy production value of part (b) differs from that calculated in part (a).

QN 113/4
Carbon dioxide (CO2) enters a nozzle operating at steady state at 28 bar, 267o
C , and 50 m/s. At
the nozzle exit, the conditions are 1.2 bar, 67o
C , 580 m/s, respectively.
(a) For a control volume enclosing the nozzle only, determine the heat transfer, in kJ, and the
change in specific entropy, in kJ/K, each per kg of carbon dioxide flowing through the nozzle.
What additional information would be required to evaluate the rate of entropy production?
(b) Evaluate the rate of entropy production, in kJ/K per kg of carbon dioxide flowing, for an
enlarged control volume enclosing the nozzle and a portion of its immediate surroundings so that
the heat transfer occurs at the ambient temperature, 25o
C .

QN 114/4
Air enters a compressor operating at steady state at 1 bar, 22o
C with a volumetric flow rate of 1
m3
/min and is compressed to 4 bar, 177o
C . The power input is 3.5 kW. Employing the ideal gas
model and ignoring kinetic and potential energy effects, obtain the following results:
(a) For a control volume enclosing the compressor only, determine the heat transfer rate, in kW,
and the change in specific entropy from inlet to exit, in What additional information would be
required to evaluate the rate of entropy production?
(b) Calculate the rate of entropy production, in kW/K, for an enlarged control volume enclosing
the compressor and a portion of its immediate surroundings so that heat transfer occurs at the
ambient temperature, 22o
C .

QN 115/4
Air is compressed in an axial-flow compressor operating at steady state from 27o
C , 1 bar to a
pressure of 2.1 bar. The work input required is 94.6 kJ per kg of air flowing through the
compressor. Heat transfer from the compressor occurs at the rate of 14 kJ per kg at a location on
the compressor‘s surface where the temperature is 40o
C . Kinetic and potential energy changes
can be ignored. Determine
(a) the temperature of the air at the exit, in o
C .
(b) the rate at which entropy is produced within the compressor, in kJ/K per kg of air flowing.

QN 116/4
Ammonia enters a counter flow heat exchanger at -20o
C , with a quality of 35%, and leaves as
saturated vapor at -20o
C . Air at 300 K, 1 atm enters the heat exchanger in a separate stream with
a flow rate of 4 kg/s and exits at 285 K, 0.98 atm. The heat exchanger is at steady state, and there
is no appreciable heat transfer from its outer surface. Neglecting kinetic
and potential energy effects, determine the mass flow rate of the ammonia, in kg/s, and the rate
of entropy production within the heat exchanger, in kW/K.

QN 117/4
A counterflow heat exchanger operates at steady state with negligible kinetic and potential
energy effects. In one stream, liquid water enters at 15o
C and exits at 23o
C with a negligible
change in pressure. In the other stream, Refrigerant 22 enters at 12 bar, 90o
C with a mass flow
rate of 150 kg/h and exits at 12 bar, 28o
C . Heat transfer from the outer surface of the heat
exchanger can be ignored. Determine
(a) the mass flow rate of the liquid water stream, in kg/ h.
(b) the rate of entropy production within the heat exchanger, in kW/K.

QN 118/4
Steam at 0.7 MPa, 355o
C enters an open feed water heater operating at steady state. A separate
stream of liquid water enters at 0.7 MPa, 35o
C . A single mixed stream exits as saturated liquid at
pressure p. Heat transfer with the surroundings and kinetic and potential energy effects can be
ignored.
(a) If p = 0.7 MPa, determine the ratio of the mass flow rates of the incoming streams and the
rate at which entropy is produced within the feedwater heater, in kJ/K per kg of liquid exiting.
(b) Plot the quantities of part (a), each versus pressure p ranging from 0.6 to 0.7 MPa.

QN 119/4
Air as an ideal gas flows through the compressor and heat exchanger shown in Figure below A
separate liquid water stream also flows through the heat exchanger. The data given are for
operation at steady state. Stray heat transfer to the surroundings can be neglected, as can all
kinetic and potential energy changes. Determine
(a) the compressor power, in kW, and the mass flow rate of the cooling water, in kg/s.
(b) the rates of entropy production, each in kW/K, for the compressor and heat exchanger.

QN 120/4
A rigid well-insulated tank having a volume of 0.2 m3
is filled initially with Refrigerant 134a
vapor at a pressure of 10 bar and a temperature of 40o
C . A leak develops and refrigerant
slowly escapes until the pressure within the tank becomes 1 bar. Determine
(a) the final temperature of the refrigerant within the tank, in o
C .
(b) the amount of mass that exits the tank, in kg.
QN 121/4
Air enters a 3600-kW turbine operating at steady state with a mass flow rate of 18 kg/s at 800o
C
, 3 bar and a velocity of 100 m/s. The air expands adiabatically through the turbine
and exits at a velocity of 150 m/s. The air then enters a diffuser where it is decelerated
isentropically to a velocity of 10 m/s and a pressure of 1 bar. Employing the ideal gas model,
determine
(a) the pressure and temperature of the air at the turbine exit, in bar and o
C , respectively.
(b) the rate of entropy production in the turbine, in kW/K.
(c) Show the processes on a T–s diagram.

QN 122/4
Air enters an insulated compressor operating at steady state at 0.95 bar, 27o
C with a mass flow
rate of 4000 kg/h and exits at 8.7 bar. Kinetic and potential energy effects are negligible.
(a) Determine the minimum theoretical power input required, in kW, and the corresponding exit
temperature, in o
C .

(b) If the exit temperature is 347o
C , determine the power input, in kW, and the isentropic
compressor efficiency.
QN 123/4
Refrigerant 134a enters a compressor operating at steady state as saturated vapor at -4o
C and
exits at a pressure of 14 bar. The isentropic compressor efficiency is 75%. Heat
transfer between the compressor and its surroundings can be ignored. Kinetic and potential
energy effects are also negligible. Determine
(a) the exit temperature, in o
C .
(b) the work input, in kJ per kg of refrigerant flowing.

QN 124/4
A compressor operating at steady state takes in atmospheric air at 20o
C , 1 bar at a rate of 1 kg/s
and discharges air at 5 bar. Plot the power required, in kW, and the exit temperature,
in o
C , versus the isentropic compressor efficiency ranging from 70 to 100%. Assume the ideal
gas model for the air and neglect heat transfer with the surroundings and changes in kinetic and
potential energy.

QN 125/4
In a gas turbine operating at steady state, air enters the compressor with a mass flow rate of 5
kg/s at 0.95 bar and 22o
C and exits at 5.7 bar. The air then passes through a heat exchanger
before entering the turbine at 1100 K, 5.7 bar. Air exits the turbine at 0.95 bar. The compressor
and turbine operate adiabatically and kinetic and potential energy effects can be ignored.
Determine the net power developed by the plant, in kW, if
(a) the compressor and turbine operate without internal irreversibilities.
(b) the compressor and turbine isentropic efficiencies are 82 and 85%, respectively.

QN 126/4
Air enters a compressor operating at steady state at 17o
C , 1 bar and exits at a pressure of 5 bar.
Kinetic and potential energy changes can be ignored. If there are no internal irreversibilities,
evaluate the work and heat transfer, each in kJ per kg of air flowing, for the following cases:
(a) isothermal compression.
(b) polytropic compression with n =1.3.
(c) adiabatic compression.
Sketch the processes on p–v and T–s coordinates and associate areas on the diagrams with the
work and heat transfer in each case. Referring to your sketches, compare for these cases
the magnitudes of the work, heat transfer, and final temperatures, respectively.

QN 127/4
1 m3
of air is heated reversibly at a constant pressure from 20 to 400 oC,and is then cooled
reversibly at a constant volume back to the initial temperature.The initial pressure is 1.03 bars.
(a) calculate (i) net heat flow (ii)The over all change of entropy
(b)sketch yhe process in (a) above on T-S diagram given that Cp=1.005 KJ/Kg.K ,
Cv=0.718KJ/Kg and KR=0.287KJ/Kg.

TOPIC 5
POWER AND REFRIGERATION CYCLES
Hello, before I take you to the main Topic 5 let‘s see the following useful concepts
REVERSIBLE WORK, IRREVERSIBILITY.AND AVAILABILITY
Basic concepts:
Reversible work for a process is defined as the work associated by taking a reversible-process
path state A to state B. As stated previously, a reversible process is a process that, having taken
place, can be reversed and, having been reversed, leaves no change in either the system or the
surroundings.
A reversible process must be a quasiequilibrium process and is subject to the following
restrictions:
 No friction exists.
 Heat transfer is due only to an infinitesimal temperature difference.
 Unrestrained expansion does not occur.
 There is no mixing.
 There is no turbulence.
 There is no combustion or chemical reaction.
It can be easily shown that the reversible work or the work output from a reversible process
going from state A to state B is the maximum work that can be achieved for the state change
from A to B . It is of interest to compare the actual work for a process to the reversible work for a
process. This comparison is done in two ways. First, a second-law eficiency for a process or a
device can be defined as
Where Wa is the actual work and Wrev is the reversible work for the fictitious reversible process.
Second-law efficiency is different from the adiabatic efficiency of a device introduced in Chapter
4 above. It is generally higher and provides a better comparison to the ideal.
Second, irreversibility is defined as the difference between the reversible work and the actual
work for a process, or
on a unit per mass basis

Both irreversibility and second-law efficiency will allow us to consider how close an actual
process or device is to the ideal. Once the irreversibilities for devices in an actual engineering
system, such as a steam power cycle, have been calculated, attempts to improve the performance
of the system can be guided by attacking the largest irreversibilities. Similarly, since the
maximum possible work will be reversible work, irreversibility can be used to evaluate the
feasibility of a device. If the irreversibility of a proposed device is less than zero, the device is
not feasible.
Availability is defined as the maximum amount of reversible work that can be extracted from a
system:
On a per unit mass
REVERSIBLE WORK AND IRREVERSIBILITY
To obtain expressions for reversible work and irreversibility, we will consider a transient process
with specified work output and heat input and a uniform through-flow. We begin by allowing
this to be an irreversible process. Consider the control volume shown in Figure below. The first
law for this control volume can be written as
the above equations and obtain the following equation

When the above equation is set equals to zero we obtain
AVAILABILITY AND EXERGY
availability
For steady flow process

In carrying out a second-law analysis, it is often useful to define a new thermodynamic function
(analogous to enthalpy), called exergy :
Certain engineering devices have useful outputs or inputs that are not in the form of work; a
nozzle is an example. Consequently, we generalize the notion of second-law efficiency to that of
second-law effectiveness
SECOND-LAW ANALYSIS OF A CYCLE
In applying second-law concepts to a cycle two approaches may be employed. The first is simply
to evaluate the irreversibilities associated with each device or process in the cycle; this will
identify sources of large irreversibilities which will adversely affect the efficiency of the cycle.
The second is to evaluate Ɛπ for the whole cycle.
Let‘s continue with our main Topic………………….
(THERMODYNAMIC CYCLES)
POWER AND REFRIGERATION CYCLES
Strong Introduction
Two important areas of application for thermodynamics are power generation and refrigeration.
Both power generation and refrigeration are usually accomplished by systems that operate on a
thermodynamic cycle. Thermodynamic cycles can be divided into two general categories:
power cycles and refrigeration cycles.

The devices or systems used to produce a net power output are often called engines, and the
thermodynamic cycles they operate on are called power cycles. The devices or systems used to
produce refrigeration are called refrigerators, air conditioners, or heat pumps, and the cycles
they operate on are called refrigeration cycles.
Thermodynamic cycles can be categorized as gas cycles or vapor cycles, depending on the
phase of the working fluid—the substance that circulates through the cyclic device. In gas
cycles, the working fluid remains in the gaseous phase throughout the entire cycle, whereas in
vapor cycles the working fluid exists in the vapor phase during one part of the cycle and in the
liquid phase during another part.
Thermodynamic cycles can be categorized yet another way: closed and open cycles. In closed
cycles, the working fluid is returned to the initial state at the end of the cycle and is recalculated.
In open cycles, the working fluid is renewed at the end of each cycle instead of being
recalculated. In automobile engines, for example, the combustion gases are exhausted and
replaced by fresh air–fuel mixture at the end of each cycle. The engine operates on a mechanical
cycle, but the working fluid in this type of device does not go through a complete
thermodynamic cycle.
Heat engines are categorized as internal combustion or external combustion engines, depending
on how the heat is supplied to the working fluid. In external combustion engines (such as steam
power plants), energy is supplied to the working fluid from an external source such as a furnace,
a geothermal well, a nuclear reactor, or even the sun. In internal combustion engines (such as
automobile engines), this is done by burning the fuel within the system boundary. In this chapter,
various gas power cycles are analyzed under some simplifying assumptions.
Steam is the most common working fluid used in vapor power cycles because of its many
desirable characteristics, such as low cost, availability, and high enthalpy of vaporization. Other
working fluids used include sodium, potassium, and mercury for high-temperature applications
and some organic fluids such as benzene and the freons for low-temperature applications.
Steam power plants are commonly referred to as coal plants, nuclear plants, or natural gas
plants, depending on the type of fuel used to supply heat to the steam. But the steam goes
through the same basic cycle in all of them. Therefore, all can be analyzed in the same manner.
Wet Steam- Steam below it's saturation temperature which contains droplets of water is called
wet steam
Dry Steam- Steam above or at it's saturation temperature is called dry steam.
The most frequently used refrigeration cycle is the vapor compression refrigeration cycle in
which the refrigerant is vaporized and condensed alternately and is compressed in the vapor
phase.

POWER AND REFRIGERATION VAPOR CYCLES
INTRODUCTION
CARNOT CYCLE
The ideal Carnot cycle is used as a model to compare all real and all other ideal cycles against.
The efficiency of a Carnot power cycle is the maximum possible for any power cycle; it is given
by;
 The performance of power plant s using wet vapour can be described using
thermodynamics property diagram.
 The Carnot cycle is the most efficient cycle operating between two specified temperature
limits.
 For wet it will appear as shown in Figure below
Shortcoming of Carnot circle
The Carnot circle 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.
Note that the efficiency is increased by raising the temperature TH at which heat is added or by
lowering the temperature TL at which heat is rejected. We will observe that this carries over to
real cycles: the cycle efficiency can be maximized by using the highest maximum temperature
and the lowest minimum temperature.
THE RANKINE CYCLE
The first class of power cycles that we consider are those utilized by the electric power
generating industry, namely, power cycles that operate in such a way that the working fluid
changes phase from a liquid to a vapor. The simplest vapor power cycle is called the Rankine
cycle, shown schematically in Figure (a) below. A major feature of such a cycle is that the pump
requires very little work to deliver high-pressure water to the boiler.
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

A possible disadvantage is that the expansion process in the turbine usually enters the quality
region, resulting in the formation of liquid droplets that may damage the turbine blades.
The Rankine cycle is an idealized cycle in which losses in each of the four components are
neglected. The losses usually are quite small and will be neglected completely in our initial
analysis. The Rankine cycle is composed of the four ideal processes shown on the T-s diagram in
Figure (b) below.
1 → 2: Isentropic compression in a pump
2 →3: Constant-pressure heat addition in a boiler
3 → 4: Isentropic expansion in a turbine
4 →1: Constant-pressure heat extraction in a condenser
The thermal efficiency ῃ of the Rankine cycle is

 Thermal efficiency is the ratio of the net work output to the net heat supplied
   
 
   
 
3 4 2 1 3 4 2 1
3 2 3 2
R
h h h h h h v P P
h h h h

     
 
 

 Sometimes work input by the feed pump is neglected because is very small to the
work output by the turbine
IT IS TRUE THAT carnot cycle is a ideal cycle. It uses some assumptions like cylinder piston
is frictionless only isentropic, adiabatic processes take place.Cylinder head act as heat souce as
well as sink but inside temperature remain constant even during heat rejection and
absorption. But rankine cycle use stream as working fluid. Further modifed rankine cycle
represents real steam power plant process. It has polytropic process. It has entropy, heat
losses.Thats why you get lesser efficiency for rankine cycle.Simply carnot is ideal cycle and
rankine is real cycle
Note that the efficiency of the Rankine cycle is less than that of a Carnot cycle operating
between the high temperature T3 and the low temperature Tl since most of the heat transfer from
a high-temperature reservoir occurs across large temperature differences
REMEMBER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
It is possible for the efficiency of a Rankine cycle to be equal to that of a Carnot cycle if the
cycle is designed to operate as shown in Figure (a). However, the pump would be required to
pump a mixture of liquid and vapor, a rather difficult and work-consuming task compared to
pumping all liquid. In addition, the condensation of liquid droplets in the turbine would result in
severe damage, To avoid the damage from droplets, one could propose superheating the steam at
constant temperature, as shown in Figure (b). This, however, requires that the pressure for the
constant-temperature superheated portion of the process decrease from the saturated vapor point
to state 3. To achieve such a decrease, the flow in the boiler pipes would have to be accelerated,
a task that would require pipes of decreasing diameter. This would be expensive, should it even
be attempted. Thus it is proposed that P2 and T3 be quite large (T3 being limited by the
temperature-resistance characteristics of the pipe metal, typically about 600°C).(figure (c)). It is
also proposed that the condenser outlet pressure be very low (it can be quite close to absolute
zero). This would, however, result in state 4 being in the quality region (a quality of 90 percent is
too low) causing water droplets to form. To avoid this problem it is necessary to reheat the
steam.
 
 
3 4
3 2
R
h h
h h





Relation used for the thermal efficiency of the Rankine cycle is;
Where
RANKINE CYCLE EFFICIENCY INCREASE
The efficiency of the Rankine cycle can be improved by increasing the boiler pressure while
maintaining the maximum temperature and the minimum pressure. The net increase in work
output is the crosshatched area minus the dotted area of Figure (a) relatively small change;
the added heat, however, decreases by the dotted area minus the crosshatched area of Figure (b).
This is obviously a significant decrease, and it leads to a significant increase in efficiency.
Increasing the maximum temperature also results in an improvement in thermal efficiency of the
Rankine cycle. In Figure (a) below the net work is increased by the crosshatched area and the
heat input is increased by the sum of the crosshatched area and the dotted area, a smaller
percentage increase than the work increase.

A decrease in condenser pressure, illustrated in Figure (b) above, will also result in increased
Rankine cycle efficiency. The net work will increase a significant amount, represented by the
crosshatched area, and the heat input will increase a slight amount because state 1' will move to a
slightly lower entropy than that of state 1;this will result in an increase in the Rankine cycle
efficiency
THE REHEAT CYCLE
It is apparent from the previous section that when operating a Rankine cycle with a high boiler
pressure or a low condenser pressure it is difficult to prevent liquid droplets from forming in the
low-pressure portion of the turbine. Since most metals cannot withstand temperatures above
about 600o
C, the reheat cycle is often used to prevent liquid droplet formation: the steam passing
through the turbine is reheated at some intermediate pressure, thereby raising the temperature to
state 5 in the T-s diagram of Figure below. The steam then passes through the low-pressure
section of the turbine and enters the condenser at state 6. This controls or completely eliminates
the moisture problem in the turbine. Often the turbine is separated into a high-pressure turbine
and a low-pressure turbine. The reheat cycle does not significantly influence the thermal
efficiency of the cycle, but it does result in a significant additional work output, represented in
the figure by area 4-5-6-4'-4. The reheat cycle demands a significant investment in additional
equipment, and the use of such equipment must be economically justified by the increased work
output.
THE REGENERATIVE CYCLE
In the conventional Rankine cycle, as well as in the reheat cycle, a considerable percentage of the
total energy input is used to heat the high-pressure water from T2 to its saturation temperature.
The crosshatched area in Figure (a) below represents this necessary energy. To reduce this
energy, the water could be preheated before it enters the boiler by intercepting some of the steam
as it expands in the turbine

and mixing it with the water as it exits the first of the pumps, thereby preheating the water from
T2 to T6. This would avoid the necessity of condensing all the steam, thereby reducing the
amount of energy lost from the condenser. (Note that the use of cooling towers would allow
smaller towers for a given energy output.) A cycle which utilizes this type of heating is a
regenerative cycle, and the process is referred to as regeneration. A schematic representation of
the major elements of such a cycle is shown in Figure below. The water entering the boiler is
often referred to as feedwater, and the device used to mix the extracted steam and the condenser
water is called a feedwater heater. When the condensate is mixed directly with the steam, it is
done so in an open feedwater heater, as sketched in Figure down.
In analyzing a regenerative cycle we must consider a control volume surrounding the feedwater
heater, see Figure below. A mass balance would result in
An energy balance, assuming an insulated heater, neglecting kinetic and potential energy
changes, gives
Combining the above two equations gives

Applied Thermodynamics Pamphlet 2nd edition, 2018 by musadoto

Applied Thermodynamics Pamphlet 2nd edition, 2018 by musadoto

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Applied Thermodynamics Pamphlet 2nd edition, 2018 by musadoto