chapter 7 for students.pdf
TRANSCRIPT
1
Chapter 7
Entropy
by Asst.Prof. Dr.Woranee Paengjuntuek
and Asst. Prof. Dr.Worarattana Pattaraprakorn
Reference: Cengel, Yunus A. and Michael A. Boles,
Thermodynamics: An Engineering Approach,
5th ed., New York, McGraw-Hill: 2006.
2
Entropy and the Clausius Inequality
The second law of thermodynamics leads to the definition of a new property called
entropy, a quantitative measure of microscopic disorder for a system. Entropy is a
measure of energy that is no longer available to perform useful work within the
current environment. To obtain the working definition of entropy and, thus, the
second law, let's derive the Clausius inequality.
Consider a heat reservoir giving up heat to a reversible heat engine, which in turn
gives up heat to a piston-cylinder device as shown below.
3
E E E
Q W W dE
in out c
R rev sys c
( )
W W W
Q W dE
c rev sys
R c c
We apply the first law on an incremental basis to the combined system composed of
the heat engine and the system.
where Ec is the energy of the combined system.
Let Wc be the work done by the combined system. Then the first law becomes
If we assume that the engine is totally reversible, then
Q
T
Q
T
Q TQ
T
R
R
R R
The total net work done by the combined system becomes
W TQ
TdEc R c
4
Now the total work done is found by taking the cyclic integral of the incremental work.
If the system, as well as the heat engine, is required to undergo a cycle, then
and the total net work becomes
If Wc is positive, we have a cyclic device exchanging energy with a single heat
reservoir and producing an equivalent amount of work; thus, the Kelvin-Planck
statement of the second law is violated. But Wc can be zero (no work done) or
negative (work is done on the combined system) and not violate the Kelvin-Planck
statement of the second law. Therefore, since TR > 0 (absolute temperature), we
conclude
5
or
Here Q is the net heat added to the system, Qnet.
This equation is called the Clausius inequality. The equality holds for the reversible
process and the inequality holds for the irreversible process.
6
Example 7-1
For a particular power plant, the heat added and rejected both occur at constant
temperature and no other processes experience any heat transfer. The heat is
added in the amount of 3150 kJ at 440oC and is rejected in the amount of 1950 kJ
at 20oC. Is the Clausius inequality satisfied and is the cycle reversible or
irreversible?
7
Calculate the net work, cycle efficiency, and Carnot efficiency based on TH and TL for
this cycle.
The Clausius inequality is satisfied. Since the inequality is less than zero,
the cycle has at least one irreversible process and the cycle is irreversible.
8
Example 7-2
For a particular power plant, the heat added and rejected both occur at constant
temperature; no other processes experience any heat transfer. The heat is added in
the amount of 3150 kJ at 440oC and is rejected in the amount of 1294.46 kJ at 20oC.
Is the Clausius inequality satisfied and is the cycle reversible or irreversible?
The Clausius inequality is satisfied. Since the cyclic integral is equal to zero, the
cycle is made of reversible processes. What cycle can this be?
9
Calculate the net work and cycle efficiency for this cycle.
10
Definition of Entropy
Let’s take another look at the quantity
If no irreversibilities occur within the system as well as the reversible cyclic device,
then the cycle undergone by the combined system will be internally reversible. As
such, it can be reversed. In the reversed cycle case, all the quantities will have the
same magnitude but the opposite sign. Therefore, the work WC, which could not be a
positive quantity in the regular case, cannot be a negative quantity in the reversed
case. Then it follows that
WC,int rev = 0 since it cannot be a positive or negative quantity, and therefore
Thus we conclude that the equality in the Clausius inequality holds for totally or just
internally reversible cycles and the inequality for the irreversible ones.
For internally reversible cycles.
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To develop a relation for the definition of entropy, let us examine this last equation
more closely. Here we have a quantity whose cyclic integral is zero. Let us think for a
moment what kind of quantities can have this characteristic. We know that the cyclic
integral of work is not zero. (It is a good thing that it is not. Otherwise, heat engines
that work on a cycle such as steam power plants would produce zero net work.)
Neither is the cyclic integral of heat.
Now consider the volume occupied by a gas in a piston-cylinder device undergoing a
cycle, as shown below.
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When the piston returns to its initial position at the end of a cycle, the volume of the
gas also returns to its initial value. Thus the net change in volume during a cycle is
zero. This is also expressed as
We see that the cyclic integral of a property is zero. A quantity whose cyclic integral
is zero depends only on the state and not on the process path; thus it is a property.
Therefore the quantity (Qnet/T)int rev must be a property.
13
Consider the cycle shown below composed of two reversible processes A and B.
Apply the Clausius inequality for this cycle. What do you conclude about these two
integrals?
B
A
2
V
P
1
A cycle composed of two reversible processes.
Apply the Clausius inequality for the cycle made of two internally reversible
processes:
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You should find:
Since the quantity (Qnet/T)int rev is independent of the path and must be a property,
we call this property the entropy S.
The entropy change occurring during a process is related to the heat transfer and the
temperature of the system. The entropy is given the symbol S (kJ/K), and the specific
entropy is s (kJ/kgK).
The entropy change during a reversible process, sometimes called an internally
reversible process, is defined as
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Consider the cycle 1-A-2-B-1, shown below, where process A is arbitrary that is, it
can be either reversible or irreversible, and process B is internally reversible.
B
A
2
V
P
1
A cycle composed of reversible and irreversible processes.
The integral along the internally reversible path, process B, is the entropy change
S1 –S2. Therefore,
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or
In general the entropy change during a process is defined as
dSQ
T
net
where = holds for the internally reversible process
> holds for the irreversible process
Consider the effect of heat transfer on entropy for the internally reversible case.
dSQ
T
net
Which temperature T is this one? If
Q then dS
Q then dS
Q then dS
net
net
net
0 0
0 0
0 0
,
,
,
17
This last result shows why we have kept the subscript net on the heat transfer Q. It is
important for you to recognize that Q has a sign depending on the direction of heat
transfer. The net subscript is to remind us that Q is positive when added to a system
and negative when leaving a system. Thus, the entropy change of the system will
have the same sign as the heat transfer in a reversible process.
From the above, we see that for a reversible, adiabatic process
dS
S S
0
2 1
The reversible, adiabatic process is called an isentropic process.
Entropy change is caused by heat transfer and irreversibilities. Heat transfer to a
system increases the entropy; heat transfer from a system decreases it. The effect of
irreversibilities is always to increase the entropy. In fact, a process in which the heat
transfer is out of the system may be so irreversible that the actual entropy change is
positive. Friction is one source of irreversibilities in a system.
The entropy change during a process is obtained by integrating the dS equation over
the process:
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Here, the inequality is to remind us that the entropy change of a system during an
irreversible process is always greater than , called the entropy transfer. That is,
some entropy is generated or created during an irreversible process, and this
generation is due entirely to the presence of irreversibilities. The entropy generated
during a process is called entropy generation and is denoted as Sgen.
We can remove the inequality by noting the following
Sgen is always a positive quantity or zero. Its value depends upon the process and
thus it is not a property. Sgen is zero for an internally reversible process.
The integral is performed by applying the first law to the process to obtain the
heat transfer as a function of the temperature. The integration is not easy to perform,
in general.
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Definition of Second Law of Thermodynamics
Now consider an isolated system composed of several subsystems exchanging
energy among themselves. Since the isolated system has no energy transfer across
its system boundary, the heat transfer across the system boundary is zero.
Applying the definition of entropy to the isolated system
The total entropy change for the isolated system is
0isolatedS
20
This equation is the working definition of the second law of thermodynamics. The
second law, known as the principle of increase of entropy, is stated as
The total entropy change of an isolated system during a process
always increases or, in the limiting case of a reversible process,
remains constant.
Now consider a general system exchanging mass as well as energy with its
surroundings.
S S S Sgen total sys surr 0
21
where = holds for the totally reversible process
> holds for the irreversible process
Thus, the entropy generated or the total entropy change (sometimes called the
entropy change of the universe or net entropy change) due to the process of this
isolated system is positive (for actual processes) or zero (for reversible processes).
The total entropy change for a process is the amount of entropy generated during
that process (Sgen), and it is equal to the sum of the entropy changes of the system
and the surroundings. The entropy changes of the important system (closed system
or control volume) and its surroundings do not both have to be positive. The entropy
for a given system (important or surroundings) may decrease during a process, but
the sum of the entropy changes of the system and its surroundings for an isolated
system can never decrease.
Entropy change is caused by heat transfer 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 increase in entropy principle can be summarized as follows:
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Some Remarks about Entropy
1. Processes can occur in a certain direction only, not in just any direction, such that
Sgen≥0.
2. Entropy is a nonconserved property, and there is no such thing as the
conservation of entropy principle. The entropy of the universe is continuously
increasing.
3. The performance of engineering systems is degraded by the presence of
irreversibilities, and entropy generation is a measure of the magnitudes of the
irreversibilities present during that process.
Heat Transfer as the Area under a T-S Curve
For the reversible process, the equation for dS implies that
dSQ
T
Q TdS
net
net
or the incremental heat transfer in a process is the product of the temperature and
the differential of the entropy, the differential area under the process curve plotted on
the T-S diagram.
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In the above figure, the heat transfer in an internally reversible process is shown as
the area under the process curve plotted on the T-S diagram.
Isothermal, Reversible Process
For an isothermal, reversible process, the temperature is constant and the integral to
find the entropy change is readily performed. If the system has a constant
temperature, T0, the entropy change becomes
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For a process occurring over a varying temperature, the entropy change must be
found by integration over the process.
Adiabatic, Reversible (Isentropic) Process
For an adiabatic process, one in which there is no heat transfer, the entropy change
is
S S S
S S
2 1
2 1
0
sS
m
s s
2 1
If the process is adiabatic and reversible, the equality holds and the entropy change
is
or on a per unit mass basis
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The adiabatic, reversible process is a constant entropy process and is called
isentropic. As will be shown later for an ideal gas, the adiabatic, reversible process is
the same as the polytropic process where the polytropic exponent n = k = Cp/Cv.
The principle of increase of entropy for a closed system exchanging heat with its
surroundings at a constant temperature Tsurr is found by using the equation for the
entropy generated for an isolated system.
Qout, sys
A general closed system (a
cup of coffee) exchanging
heat with its surroundings
Surroundings
Tsurr
System
Boundary
S S S S
S S S
SQ
T
gen total sys surr
sys sys
surr
net surr
surr
0
2 1( )
,
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S S m s sQ
Tgen total sys
net surr
surr
( ),
2 1 0
where
Q Q Q Qnet surr net sys out sys out sys, , , ,( ) 0
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Effect of Heat Transfer on Entropy
Let's apply the second law to the following situation. Consider the transfer of heat
from a heat reservoir at temperature T to a heat reservoir at temperature T - T > 0
where T > 0, as shown below.
Q
HR
at
T
HR
at
T-T
Two heat reservoirs
exchanging heat
over a
finite temperature
difference
The second law for the isolated system composed of the two heat reservoirs is
S S S S
S S S S
gen total sys surr
gen total HR T HR T T
0
@ @
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In general, if the heat reservoirs are internally reversible
S SQ
T
Q
T Tgen Total
Now as T 0, Sgen 0 and the process becomes totally reversible.
Therefore, for reversible heat transfer T must be small.
As T gets large, Sgen increases and the process becomes irreversible.
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Example 7-3
Find the total entropy change, or entropy generation, for the transfer of 1000 kJ of
heat energy from a heat reservoir at 1000 K to a heat reservoir at 500 K.
Q=1000 kJ
HR
at
T=1000 K
HR
at
T-T = 500K0 1 2 S, kJ/K
1000 K
500 K
T Areas
= 1000 kJ
The second law for the isolated system is
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What happens when the low-temperature reservoir is at 750 K?
…………
The effect of decreasing the T for heat transfer is to reduce the entropy
generation or total entropy change of the universe due to the isolated
system and the irreversibilities associated with the heat transfer process.
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Third Law of Thermodynamics
The third law of thermodynamics states that the entropy of a pure crystalline
substance at absolute zero temperature is zero. This law provides an absolute
reference point for the determination of entropy. The entropy determined relative to
this point is called absolute entropy.
Entropy as a Property
Entropy is a property, and it can be expressed in terms of more familiar properties
(P,v,T) through the Tds relations. These relations come from the analysis of a
reversible closed system that does boundary work and has heat added. Writing the
first law for the closed system in differential form on a per unit mass basis
Wint rev, out
Qint rev
System used to find expressions for ds
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Q W dU
Q T dS
W PdV
TdS PdV dU
int rev int rev, out
int rev
int rev, out
=
Tds du Pdv
T ds dh vdP
On a unit mass basis we obtain the first Tds equation, or Gibbs equation, as
Recall that the enthalpy is related to the internal energy by h = u + Pv. Using this
relation in the above equation, the second Tds equation is
These last two relations have many uses in thermodynamics and serve as the
starting point in developing entropy-change relations for processes. The successful
use of Tds relations depends on the availability of property relations. Such relations
do not exist in an easily used form for a general pure substance but are available for
incompressible substances (liquids, solids) and ideal gases. So, for the general pure
substance, such as water and the refrigerants, we must resort to property tables to
find values of entropy and entropy changes.
33
The temperature-entropy and enthalpy-entropy diagrams for water are shown below.
34
Shown above are the temperature-entropy and enthalpy-entropy diagrams for water.
The h-s diagram, called the Mollier diagram, is a useful aid in solving steam power
plant problems.
35
Example 7-4
Find the entropy and/or temperature of steam at the following states:
P T Region s kJ/(kg K)
5 MPa 120oC
1 MPa 50oC
1.8 MPa 400oC
40 kPa Quality, x = 0.9
40 kPa 7.1794
36
Answer to Example 7-4
Find the entropy and/or temperature of steam at the following states:
37
Example 7-5
Determine the entropy change of water contained in a closed system as it changes
phase from saturated liquid to saturated vapor when the pressure is 0.1 MPa and
constant. Why is the entropy change positive for this process?
System: The water contained in the system (a piston-cylinder device)
Steam
s
T
Property Relation: Steam tables
Process and Process Diagram: Constant pressure (sketch the process relative to
the saturation lines)
Conservation Principles: Using the definition of entropy change, the entropy
change of the water per mass is
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The entropy change is positive because: (Heat is added to the water.)
Example 7-6
Steam at 1 MPa, 600oC, 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.
System: The control volume formed by the turbine
Control surface
1
2
Wout
s
T
39
Property Relation: Steam tables
Process and Process Diagram: Isentropic (sketch the process relative to the
saturation lines on the T-s diagram)
Conservation Principles:
Assume: steady-state, steady-flow, one entrance, one exit, neglect KE and PE
Conservation of mass:
m m m1 2
First Law or conservation of energy:
The process is isentropic and thus adiabatic and reversible; therefore Q = 0. The
conservation of energy becomes
40
Since the mass flow rates in and out are equal, solve for the work done per unit mass
Now, let’s go to the steam tables to find the h’s.
The process is isentropic, therefore; s2 = s1 = 8.0311 kJ/(kg K )
At P2 = 0.01 MPa, sf = 0.6492 kJ/kgK, and sg = 8.1488 kJ/(kg K);
thus, sf < s2 < sg.
State 2 is in the saturation region, and the quality is needed to specify the state.
41
Since state 2 is in the two-phase region, T2 = Tsat at P2 = 45.81oC.
42
s s s 2 1
s s2 1
dsdu
T
P
Tdv
du CdT
dv
0
Entropy Change and Isentropic Processes
The entropy-change and isentropic relations for a process can be summarized as
follows:
1.Pure substances:
Any process: (kJ/kgK)
Isentropic process:
2.Incompressible substances (Liquids and Solids):
The change in internal energy and volume for an incompressible substance is
The entropy change now becomes
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s s CT
Tav2 1
2
1
ln
T T2 1
If the specific heat for the incompressible substance is constant, then the entropy
change is
Any process: (kJ/kgK)
Isentropic process:
3. Ideal gases:
a. Constant specific heats (approximate treatment):
Any process: s s CT
TR
v
vv av2 1
2
1
2
1
, ln ln (kJ/kgK)
and 2 22 1 ,
1 1
ln lnp av
T Ps s C R
T P (kJ/kgK)
Or, on a unit-mole basis,
s s CT
TR
v
vv av u2 1
2
1
2
1
, ln ln(kJ/kmolK)
and2 2
2 1 ,
1 1
ln lnp av u
T Ps s C R
T P (kJ/kmolK)
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2 1
1 2.
k
s const
P v
P v
For an isentropic process this last result looks like Pvk = constant which is the
polytropic process equation Pvn = constant with n = k = Cp/Cv.
b. Variable specific heats (exact treatment):
From Tds = dh - vdP, we obtain2
2
11
( )ln
pC T Ps dT R
T P
The first term can be integrated relative to a reference state at temperature Tref.
Isentropic process:
45
The integrals on the right-hand side of the above equation are called the standard
state entropies, so, at state 1, T1, and state 2, T2; so is a function of temperature only.
s s s s RP
P
o o
2 1 2 12
1
ln
s s s s RP
P
o o
u2 1 2 12
1
ln
Therefore, for any process:
(kJ/kgK)
(kJ/kmolK)
or
46
The standard state entropies are found in Tables A-17 for air on a mass basis and
Tables A-18 through A-25 for other gases on a mole basis. When using this variable
specific heat approach to finding the entropy change for an ideal gas, remember to
include the pressure term along with the standard state entropy terms--the tables
don’t warn you to do this.
Isentropic process: s = 0
s s RP
P
o o
2 12
1
ln (kJ/kgK)
If we are given T1, P1, and P2, we find so1 at T1, calculate so
2, and then determine
from the tables T2, u2, and h2.
When air undergoes an isentropic process when variable specific heat data are
required, there is another approach to finding the properties at the end of the
isentropic process. Consider the entropy change written as
22
11
( )ln
pC T Ps dT R
T P
47
Letting T1 = Tref, P1 = Pref = 1atm, T2 = T, P2 = P, and setting the entropy change
equal to zero yield
We define the relative pressure Pr as the above pressure ratio. Pr is the pressure
ratio necessary to have an isentropic process between the reference temperature
and the actual temperature and is a function of the actual temperature. This
parameter is a function of temperature only and is found in the air tables, Table A-17.
The relative pressure is not available for other gases in this text.
The ratio of pressures in an isentropic process is related to the ratio of relative
pressures.
22 2
1 1 1. .
/
/
ref r
ref rs const s const
P PP P
P P P P
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There is a second approach to finding data at the end of an ideal gas isentropic
process when variable specific heat data are required. Consider the following entropy
change equation set equal to zero.
From Tds = du + Pdv, we obtain for ideal gases
Letting T1 = Tref, v1 = vref, T2 = T, v2 = v, and setting the entropy change equal to zero
yield
We define the relative volume vr as the above volume ratio. vr is the volume ratio
necessary to have an isentropic process between the reference temperature and the
actual temperature and is a function of the actual temperature. This parameter is a
function of temperature only and is found in the air tables, Table A-17. The relative
volume is not available for other gases in this text.
49
Example 7-7
Aluminum at 100oC is placed in a large, insulated tank having 10 kg of water at a
temperature of 30oC. If the mass of the aluminum is 0.5 kg, find the final equilibrium
temperature of the aluminum and water, the entropy change of the aluminum and the
water, and the total entropy change of the universe because of this process. Before
we work the problem, what do you think the answers ought to be? Are entropy
changes going to be positive or negative? What about the entropy generated as the
process takes place?
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System: Closed system including the aluminum and water.
Water
AL
Tank insulated
boundary
Property Relation: ?
Process: Constant volume, adiabatic, no work energy exchange between the
aluminum and water.
Conservation Principles:
Apply the first law, closed system to the aluminum-water system.
Using the solid and incompressible liquid relations, we have
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But at equilibrium, T2,AL = T2,water = T2
The second law gives the entropy production, or total entropy change of the universe,
as
Using the entropy change equation for solids and liquids,
Why is Sgen or STotal positive?
52
Why is SAL negative? Why is Swater positive?
53
Example 7-9
Air, initially at 17oC, is compressed in an isentropic process through a pressure ratio
of 8:1. Find the final temperature assuming constant specific heats and variable
specific heats, and using EES.
a. Constant specific heats, isentropic process
For air, k = 1.4, and a pressure ratio of 8:1 means that P2/P1 = 8
b. Variable specific heat method
54
Using the air data from Table A-17 for T1 = (17+273) K = 290 K, Pr1 = 1.2311.
Interpolating in the air table at this value of Pr2, gives T2 = 522.4 K = 249.4oC
c.A second variable specific heat method.
Using the air table, Table A-17, for T1 = (17+273) K = 290 K, soT1 = 1.66802 kJ/kgK.
For the isentropic process
At this value of soT2, the air table gives T2 = 522.4 K= 249.4oC.
This technique is based on the same information as the method shown in part b.
55
Example 7-10
Air initially at 0.1 MPa, 27oC, is compressed reversibly to a final state.
(a) Find the entropy change of the air when the final state is 0.5 MPa, 227oC.
(b) Find the entropy change when the final state is 0.5 MPa, 180oC.
(c) Find the temperature at 0.5 MPa that makes the entropy change zero.
Assume air is an ideal gas with constant specific heats.
56
Show the two processes on a T-s diagram.
a.
b.
57
c.
c
b
a
1
s
T
P1
P22
The T-s plot is
Give an explanation for the difference in the signs for the entropy changes.
58
Example 7-11
Nitrogen expands isentropically in a piston cylinder device from a temperature of 500
K while its volume doubles. What is the final temperature of the nitrogen, and how
much work did the nitrogen do against the piston, in kJ/kg?
System: The closed piston-cylinder device
59
Property Relation: Ideal gas equations, constant properties
Process and Process Diagram: Isentropic expansion
Conservation Principles:
Second law:
Since we know T1 and the volume ratio, the isentropic process, s = 0, allows us to
find the final temperature. Assuming constant properties, the temperatures are
related by
Why did the temperature decrease?
60
First law, closed system:
Note, for the isentropic process (reversible, adiabatic); the heat transfer is zero. The
conservation of energy for this closed system becomes
Using the ideal gas relations, the work per unit mass is
Why is the work positive?
61
Example 7-12
A Carnot engine has 1 kg of air as the working fluid. Heat is supplied to the air at 800
K and rejected by the air at 300 K. At the beginning of the heat addition process, the
pressure is 0.8 MPa and during heat addition the volume triples.
(a) Calculate the net cycle work assuming air is an ideal gas with constant specific
heats.
(b) Calculate the amount of work done in the isentropic expansion process.
(c) Calculate the entropy change during the heat rejection process.
System: The Carnot engine piston-cylinder device.
62
Property Relation: Ideal gas equations, constant properties.
Process and Process Diagram: Constant temperature heat addition.
63
Conservation Principles:
a.
Apply the first law, closed system, to the constant temperature heat addition process,
1-2.
So for the ideal gas isothermal process,
64
But
The cycle thermal efficiency is
For the Carnot cycle, the thermal efficiency is also given by
The net work done by the cycle is
65
b.
Apply the first law, closed system, to the isentropic expansion process, 2-3.
But the isentropic process is adiabatic, reversible; so, Q23 = 0.
Using the ideal gas relations, the work per unit mass is
This is the work leaving the cycle in process 2-3.
66
c.
Using equation (6-34)
But T4 = T3 = TL = 300 K, and we need to find P4 and P3.
Consider process 1-2 where T1 = T2 = TH = 800 K, and, for ideal gases
67
Consider process 2-3 where s3 = s2.
Now, consider process 4-1 where s4 = s1.
68
Now,
69
Reversible Steady-Flow Work
Isentropic, Steady Flow through Turbines, Pumps, and Compressors
Consider a turbine, pump, compressor, or other steady-flow control volume, work-
producing device.
The general first law for the steady-flow control volume is
( ) ( )
E E
Q m hV
gz W m hV
gz
in out
net i
inlets
ii
i net e
exits
ee
e
2 2
2 2
For a one-entrance, one-exit device undergoing an internally reversible process, this
general equation of the conservation of energy reduces to, on a unit of mass basis
rev rev
rev
rev
w q dh dke dpe
But q T ds
w T ds dh dke dp
70
Using the Gibb’s second equation, this becomes
rev
dh T ds vdP
w vdP dke dpe
Integrating over the process, this becomes
Neglecting changes in kinetic and potential energies, reversible work becomes
Based on the classical sign convention, this is the work done by the control volume.
When work is done on the control volume such as compressors or pumps, the
reversible work going into the control volume is
71
Turbine
Since the fluid pressure drops as the fluid flows through the turbine, dP < 0, and the
specific volume is always greater than zero, wrev, turbine > 0. To perform the integral,
the pressure-volume relation must be known for the process.
Compressor and Pump
Since the fluid pressure rises as the fluid flows through the compressor or pump, dP
> 0, and the specific volume is always greater than zero, wrev, in > 0, or work is
supplied to the compressor or pump. To perform the integral, the pressure-volume
relation must be known for the process. The term compressor is usually applied to
the compression of a gas. The term pump is usually applied when increasing the
pressure of a liquid.
Pumping an incompressible liquid
For an incompressible liquid, the specific volume is approximately constant. Taking v
approximately equal to v1, the specific volume of the liquid entering the pump, the
work can be expressed as
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For the steady-flow of an incompressible fluid through a device that involves no work
interactions (such as nozzles or a pipe section), the work term is zero, and the
equation above can be expressed as the well-know Bernoulli equation in fluid
mechanics.
v P P ke pe( )2 1 0
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Example 7-13
Saturated liquid water at 10 kPa leaves the condenser of a steam power plant and is
pumped to the boiler pressure of 5 MPa. Calculate the work for an isentropic
pumping process.
a. From the above analysis, the work for the reversible process can be applied to the
isentropic process (it is left for the student to show this is true) as
Here at 10 kPa, v1 = vf = 0.001010 m3/kg.
The work per unit mass flow is
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b. Using the steam table data for the isentropic process, we have
From the saturation pressure table,
Since the process is isentropic, s2 = s1. Interpolation in the compressed liquid tables
gives
The work per unit mass flow is
“The first method for
finding the pump work is
adequate for this case.”
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Turbine, Compressor (Pump), and Nozzle Efficiencies
Most steady-flow devices operate under adiabatic conditions, and the ideal process
for these devices is the isentropic process. The parameter that describes how a
device approximates a corresponding isentropic device is called the isentropic or
adiabatic efficiency. It is defined for turbines, compressors, and nozzles as follows:
Turbine:
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The isentropic work is the maximum possible work output that the adiabatic turbine
can produce; therefore, the actual work is less than the isentropic work. Since
efficiencies are defined to be less than 1, the turbine isentropic efficiency is defined
as
Ta
s
Actual turbine work
Isentropic turbine work
w
w
1 2
1 2
aT
s
h h
h h
Well-designed large turbines may have isentropic efficiencies above 90
percent. Small turbines may have isentropic efficiencies below 70 percent.
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Compressor and Pump:
The isentropic work is the minimum possible work that the adiabatic compressor
requires; therefore, the actual work is greater than the isentropic work. Since
efficiencies are defined to be less than 1, the compressor isentropic efficiency is
defined as
Compressor
or pump
T1
P1
WC
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Cs
a
Isentropic compressor work
Actual compressor work
w
w
Cs
a
h h
h h
2 1
2 1
Well-designed compressors have isentropic efficiencies in the range from 75 to 85
percent.
Nozzle:
The isentropic kinetic energy at the nozzle exit is the maximum possible kinetic
energy at the nozzle exit; therefore, the actual kinetic energy at the nozzle exit is less
than the isentropic value. Since efficiencies are defined to be less than 1, the nozzle
isentropic efficiency is defined as
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Nozzle
T1
P1T2
P2
Na
s
Actual KE at nozzle exit
Isentropic KE at nozzle exit
V
V
22
2
2
2
2
/
/
80
For steady-flow, no work, neglecting potential energies, and neglecting the inlet
kinetic energy, the conservation of energy for the nozzle is
h hV
aa
1 22
2
2
The nozzle efficiency is written as
Na
s
h h
h h
1 2
1 2
Nozzle efficiencies are typically above 90 percent, and nozzle efficiencies above 95
percent are not uncommon.
81
Example 7-14
The isentropic work of the turbine in Example 7-6 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.
Now to find the actual exit state for the steam.
From Example 7-6, steam enters the turbine at 1 MPa, 600oC, and expands to 0.01
MPa.
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From the steam tables at state 1
At the end of the isentropic expansion process, see Example 7-6.
The actual turbine work per unit mass flow is (see Example 7-6)
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For the actual turbine exit state 2a, the computer software gives
A second method for finding the actual state 2 comes directly from the expression for
the turbine isentropic efficiency. Solve for h2a.
Then P2 and h2a give T2a = 86.85oC.
84
Example 7-15
Air enters a compressor and is compressed adiabatically from 0.1 MPa, 27oC, to a
final state of 0.5 MPa. Find the work done on the air for a compressor isentropic
efficiency of 80 percent.
System: The compressor control volume
Property Relation: Ideal gas equations, assume constant properties.
2a
2s
1
s
T P2
P1
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Process and Process Diagram: First, assume isentropic, steady-flow and then
apply the compressor isentropic efficiency to find the actual work.
Conservation Principles:
For the isentropic case, Qnet = 0. Assuming steady-state, steady-flow, and neglecting
changes in kinetic and potential energies for one entrance, one exit, the first law is
E E
m h W m h
in out
Cs s
1 1 2 2
The conservation of mass gives
m m m1 2
The conservation of energy reduces to
( )
( )
W m h h
wW
mh h
Cs s
CsCs
s
2 1
2 1
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Using the ideal gas assumption with constant specific heats, the isentropic work per
unit mass flow is
w C T TCs p s ( )2 1
The isentropic temperature at state 2 is found from the isentropic relation
The conservation of energy becomes
87
The compressor isentropic efficiency is defined as
88
Property Relation: The ideal gas equations, assuming constant specific heats
Process and Process Diagram: First assume an isentropic process and then apply
the nozzle isentropic efficiency to find the actual exit velocity.
Example 7-16
Nitrogen expands in a nozzle from a temperature of 500 K while its pressure
decreases by factor of two. What is the exit velocity of the nitrogen when the nozzle
isentropic efficiency is 95 percent?
System: The nozzle control volume.
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The conservation of mass gives
m m m1 2
The conservation of energy reduces to
V h hs s2 1 22 ( )
Using the ideal gas assumption with constant specific heats, the isentropic exit
velocity is
V C T Ts p s2 1 22 ( )
The isentropic temperature at state 2 is found from the isentropic relation
( )
E E
m h m hV
in out
ss
1 1 2 22
2
2
Conservation Principles:
For the isentropic case, Qnet = 0. Assume steady-state, steady-flow, no work is
done. Neglect the inlet kinetic energy and changes in potential energies. Then
for one entrance, one exit, the first law reduces to
90
The nozzle exit velocity is obtained from the nozzle isentropic efficiency as
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Entropy Balance
The principle of increase of entropy for any system is expressed as an
entropy balance given by
or
S S S Sin out gen system
The entropy balance relation can be stated as: the entropy change of a system during
a process is equal to the net entropy transfer through the system boundary and the
entropy generated within the system as a result of irreversibilities.
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Entropy change of a system
The entropy change of a system is the result of the process occurring within the
system.
Entropy change = Entropy at final state – Entropy at initial state
Mechanisms of Entropy Transfer, Sin and Sout
Entropy can be transferred to or from a system by two mechanisms: heat transfer and
mass flow. Entropy transfer occurs at the system boundary as it crosses the
boundary, and it represents the entropy gained or lost by a system during the
process. The only form of entropy interaction associated with a closed system is heat
transfer, and thus the entropy transfer for an adiabatic closed system is zero.
Heat transfer
The ratio of the heat transfer Q at a location to the absolute temperature T at that
location is called the entropy flow or entropy transfer and is given as
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Entropy transfer by heat transfer SQ
TTheat: ( ) constant
Q/T represents the entropy transfer accompanied by heat transfer, and the direction
of entropy transfer is the same as the direction of heat transfer since the absolute
temperature T is always a positive quantity.
When the temperature is not constant, the entropy transfer for process 1-2 can be
determined by integration (or by summation if appropriate) as
Work
Work is entropy-free, and no entropy is transferred by work. Energy is transferred by
both work and heat, whereas entropy is transferred only by heat and mass.
Entropy transfer by work Swork: 0
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Mass flow
Mass contains entropy as well as energy, and the entropy and energy contents of a
system are proportional to the mass. When a mass in the amount m enters or leaves
a system, entropy in the amount of ms enters or leaves, where s is the specific
entropy of the mass.
Entropy transfer by mass S msmass:
Entropy Generation, Sgen
Irreversibilities such as friction, mixing, chemical reactions, heat transfer through a
finite temperature difference, unrestrained expansion, non-quasi-equilibrium
expansion, or compression always cause the entropy of a system to increase, and
entropy generation is a measure of the entropy created by such effects during a
process.
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For a reversible process, the entropy generation is zero and the entropy change of a
system is equal to the entropy transfer. The entropy transfer by heat is zero for an
adiabatic system and the entropy transfer by mass is zero for a closed system.
The entropy balance for any system undergoing any process can be expressed in
the general form as
The entropy balance for any system undergoing any process can be expressed in
the general rate form, as
Q
m
where the rates of entropy transfer by heat transferred at a rate of and mass
flowing at a rate of are / S Q T S msheat mass and
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The entropy balance can also be expressed on a unit-mass basis as
( ) ( / )s s s s kJ kg Kin out gen system
The term Sgen is the entropy generation within the system boundary only, and not the
entropy generation that may occur outside the system boundary during the process
as a result of external irreversibilities. Sgen = 0 for the internally reversible process,
but not necessarily zero for the totally reversible process. The total entropy
generated during any process is obtained by applying the entropy balance to an
Isolated System that contains the system itself and its immediate surroundings.
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Closed Systems
Taking the positive direction of heat transfer to the system to be positive, the general
entropy balance for the closed system is
Q
TS S S S kJ Kk
k
gen system 2 1 ( / )
For an adiabatic process (Q = 0), this reduces to
Adiabatic closed system S Sgen adiabatic system:
A general closed system and its surroundings (an isolated system) can be treated as
an adiabatic system, and the entropy balance becomes
System surroundings S S S Sgen system surroundings :
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Control Volumes
The entropy balance for control volumes differs from that for closed systems in that the
entropy exchange due to mass flow must be included.
Q
Tm s m s S S S kJ Kk
k
i i e e gen CV ( ) ( / )2 1
In the rate form we have
( / )
Q
Tm s m s S S kW Kk
k
i i e e gen CV
This entropy balance relation is stated as: the rate of entropy change within the
control volume during a process is equal to the sum of the rate of entropy transfer
through the control volume boundary by heat transfer, the net rate of entropy transfer
into the control volume by mass flow, and the rate of entropy generation within the
boundaries of the control volume as a result of irreversibilities.
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For a general steady-flow process, by setting the entropy balance
simplifies to SCV 0
S m s m sQ
Tgen e e i i
k
k
For a single-stream (one inlet and one exit), steady-flow device, the entropy balance
becomes
( )
S m s sQ
Tgen e i
k
k
100
For an adiabatic single-stream device, the entropy balance becomes
( )S m s sgen e i
This states that the specific entropy of the fluid must increase as it flows through an
adiabatic device since . If the flow through the device is reversible and
adiabatic, then the entropy will remain constant regardless of the changes in other
properties.
Sgen 0
Therefore, for steady-flow, single-stream, adiabatic and reversible process:
s se i
101
Example 7-17
An inventor claims to have developed a water mixing device in which 10 kg/s of water
at 25oC and 0.1 MPa and 0.5 kg/s of water at 100oC, 0.1 MPa, are mixed to produce
10.5 kg/s of water as a saturated liquid at 0.1 MPa. If the surroundings to this device
are at 20oC, is this process possible? If not, what temperature must the surroundings
have for the process to be possible?
System: The mixing chamber control volume.
Property Relation: The steam tables
Process and Process Diagram: Assume steady-flow
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Conservation Principles:
First let’s determine if there is a heat transfer from the surroundings to the mixing
chamber. Assume there is no work done during the mixing process, and neglect
kinetic and potential energy changes. Then for two entrances and one exit, the first
law becomes
103
So, 1996.33 kJ/s of heat energy must be transferred from the surroundings to this
mixing process, or , ,Q Qnet surr net CV
For the process to be possible, the second law must be satisfied. Write the second
law for the isolated system,
Q
Tm s m s S Sk
k
i i e e gen CV
104
SgenSince must be ≥ 0 to satisfy the second law, this process is impossible, and the
inventor's claim is false.
To find the minimum value of the surrounding temperature to make this mixing
process possible, set = 0 and solve for Tsurr. Sgen
For steady-flow . Solving for entropy generation, we have SCV 0
105
One way to think about this process is as follows: Heat is transferred from the
surroundings at 315.75 K (42.75oC) in the amount of 1997.7 kJ/s to increase the
water temperature to approximately 42.75oC before the water is mixed with the
superheated steam. Recall that the surroundings must be at a temperature greater
than the water for the heat transfer to take place from the surroundings to the water.
106
Homework
1. A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle
wheel. The paddle wheel does 200 kJ of work on the ideal gas. It is observed
that the temperature of the ideal gas remains constant during this process as a
result of heat transfer between the system and the surroundings at 30°C.
Determine the entropy change of the ideal gas.
2. Air is compressed by a 12 kW compressor from P1 to P2. The air
temperature is maintained constant at 25°C during this process as a result of
heat transfer to the surrounding medium at 10°C. Determine the rate of
entropy change of the air. Assumed that the process is an isothermal and
internally reversible process .
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3. During the isothermal heat addition process of a Carnot cycle, 900 kJ of heat is
added to the working fluid from a source at 400°C. Determine
(a) the entropy change of the working fluid,
(b) the entropy change of the source,
(c) the total entropy change for the process.
4. The radiator of a steam heating system has a volume of 20 L and is filled with
superheated water vapor at 200 kPa and 150°C. At this moment both the inlet
and the exit valves to the radiator are closed. After a while the temperature of
the steam drops to 40°C as a result of heat transfer to the room air. Determine
the entropy change of the steam during this process.
5. A piston–cylinder device contains 1.2 kg of saturated water vapor at 200°C.
Heat is now transferred to steam, and steam expands reversibly and
isothermally to a final pressure of 800 kPa. Determine the heat transferred
and the work done during this process.
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Have a good
…GRADE!!!