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DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
CHAPTER 7
ENTROPY
OBJECTIVES
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
At the end of t his chapter , the s tudent should be able to:
Define a new property called entropy
Establish the increase of entropy p rinciple
Determine the entropy changes during processes for
pure substances and ideal gas
Develop the property relations fo r isentropic processes
Derive the reversible steady flow work relations
Determine the isentropic efficiencies for various steady
flow devices
Determine the entropy generation for closed and open
system
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CLAUSIUS INEQUALITY
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
0T
Q First stated by German physicist Rudolf J.E. Clausius (1822-1888) in 1850 and is
expressed as
S
S
P
P
T
Q
T
Q = 0T
Q
T
Q
T
Q
S
S
P
P =
CHAPTER 7 : ENTROPY
This inequality is valid for all cycles, reversible or
irreversible
The cyclic integral of Q/T is always less than zero(irreversible cycles) and equal to zero (reversiblecycles)
Consider a reversible heat engine and a irreversible
heat engine that operating between two similar heatreservoirs
For the reversible heat engine, we can write,
High temp reservoir
TH
Low temp reservoir
TL
HErev
Wrev
QH,rev
QL, rev
Wirrev
QH,irrev
QL,irrev
HEirrev
Thermodynamicstemp scale
CLAUSIUS INEQUALITY
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
From 1st law of thermodynamics, W = QH - QL then,
So, we can conclude that QL,irrev > QL,rev , thus,
S
rev,L
S
irrev,L
T
Q
T
Q> 0
L
irrev,L
H
H
T
Q
T
Q
T
Q
0T
Q0
T
Q
Both heat engines receive heat from the same heat reservoir, then
QH,rev = QH,irrev and Wrev > Wirrev
Thus, QL,rev QL,irrev
0
L
L
H
H
T
Q
T
Q
T
Q
QH QL,rev > QH QL,irrev
CHAPTER 7 : ENTROPY
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DEFINITION OF ENTROPY CHANGE
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
1
2
B
C
A
T
Q
T
Q
B
2
1
rev
A
2
1
rev Let assumes that the Clausius inequality of A is not equal to B
0=T
Q
T
Q
T
Q
C
1
2
rev
A
2
1
rev
AC
rev = 121
0=T
Q+
T
Q
T
Q
C
1
2
rev
B
2
1
rev
BC
rev = 121
0T
Q
T
Q
BC
rev
AC
rev == 121121
T
Q
T
Q
B
2
1
rev
A
2
1
rev =
For cycle 1-A-2-C-1
For cycle 1-B-2-C-1
kJ/KT
QdS
rev
= TQSS2
1 rev
= 12
Property I
P
r
o
p
e
rt
y
II
CHAPTER 7 : ENTROPY
DEFINITION OF ENTROPY CHANGE
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
kJ/KT
QdS
revint
=
CHAPTER 7 : ENTROPY
In 1865, Clausius introduced a new thermodynamics property called
ENTROPY and designated S and defined as
Entropy is an extensive property and the unit is kJ/K
The entropy change of a system during a process can be determined by
integrating above equation,
kJ/KT
QSSS
revint
=2
1
12
The entropy change S between two states isthe same for reversible and irreversibleprocess but,
The integral of Q/T gives a value of entropychange only if the integration is carried out
along an internally reversible process.
Thus the Q/T during an irreversible process isnot a property (entropy)
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INTERNALLY REVERSIBLE ISOTHERMAL
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CHAPTER 7 : ENTROPY
The heat transfer processes are internally reversible
Thus, the heat transfer can be determined by
kJ/KT
TT
Q
T
QS
orevint
orevintrevint
=== 21 2121 1 To is the constant temperature of the system during the heat transfer process
EXAMPLE 7-1
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CHAPTER 7 : ENTROPY
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THE INCREASE OF ENTROPY PRINCIPLE
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CHAPTER 7 : ENTROPY
From the Clausius inequality0T
Q
For the cycle in the diagram
0T
Q
T
Q
T
Q
revint,
2121 Rearrange, then
T
QdSor
T
QSS
21
12
processleirreversib-T
Q
dS
>
processreversible-T
QdS
=
ENTROPY GENERATION, Sgen
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
The entropy change of a closed system during an irreversible process is always
greater than the entropy transfer.
So, entropy is generated or created during an irreversible process called entropy
generation, Sgen
gensytem ST
QSSS 2
1
12
For an isolated (adiabatic) system, Q = 0, then Sisolated 0 Thus, the entropy of adiabatic system during a process always increases and never
decreases
The entropy change of the adiabatic system is the sum of the entropy changes of the
system and its surroundings which equal to the entropy generation
Sgen = Stotal = Ssys + Ssurr 0 The increase of entropy principle can be summarized as
Sgen
> 0 Irreversible process
= 0 Reversible process< 0 Impossible process
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EXAMPLE 7-2
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
ENTROPY CHANGE OF PURE SUBSTANCE
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
The value entropy at a specified state is
determined just like any other property.
In the superheated vapor regions :
can be obtained directly from the
tables
In saturated vapor :
s = sf+ xsfg
In compressed liquid :
s = sfat specified temperature
The entropy change,
S = m(s2 s1) kJ/K
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EXAMPLE 7-3
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
EXAMPLE 7-4
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
A piston cylinder device initially contains 1.5 kg of liquid water at 150 kPa and 20oC. Thewater is now heated at constant pressure by the addition of 4000 kJ of heat. Determine the
entropy change of the water during the process.
Since T1 < Ts at 150 kPa, the water exists as compressed liquid
P1 = 150 kPaT
1= 20oC
s1 = sf@ 20oC = 0.2965 kJ/kg
h1
= hf@ 20oC = 83.915 kJ/kgK
From energy balance for closed system, Q inWb = UQ
in= H = m(h2 h1)
4000 kJ = (1.5 kg)(h2 83.915 kJ/kg)
h2
= 2750.6 kJ/kg
Since h2 = 2750.6 kJ/kg > hg at 150 kPa = 2693.1 kJ/kg, state 2 is s.h. vapor
P1 = 150 kPa
h2 = 2750.6 kJ/kgs2 = 7.3674 kJ/kgK (interpolation)
Thus, S = m(s2 s1) = 1.5(7.3674 0.2965) = 10.61 kJ/K
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ISENTROPIC PROCESSES
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
An internally reversible and adiabatic process, which the entropy
remains constant is called an isentropic process : s = 0 or s2 = s1.Many devices such as pumps, turbines, nozzles and diffusers are
essentially adiabatic in their operation
Using isentropic processes enable us to define the efficiencies of above
devices and to compare the actual performance of these devices to the
performance under idealized conditions
EXAMPLE 7-5
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
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T-S DIAGRAM
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CHAPTER 7 : ENTROPY
One of the diagram commonly used in the second law analysis
Rearrange the entropy change definition : Qintrev = Tds Thus, the area under the process curve on a T-S diagram represents heat transfer
during an internally reversible process, which is21
TdsQ revint
For internally reversible isothermal process,
Qintrev = ToS (kJ) or q intrev = To s (kJ/kgK) During an isentropic processes, Q = O, thus the area under process curve is zero
Carnot Cycle on T-s Diagram
Tds RELATIONS
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
From energy balance for a closed system, an internally reversible process can beexpressed as
Qint rev - Wint rev = dUSubstituting Qint rev = T ds and Wint rev = P dV, then
T dS = dU + P dV or Tds = du + P d
Writing the enthalpy definition (h = u + p) in differential form,dh = du + P dv + dP or du = dh P dv + dP
Substituting into the 1st
Tds equation, then the 2nd
Tds equation
Tds = (dh P dv + dP) + P dv= dh - dP
Solving for ds, then
T
dP
T
dhdsand
T
Pd
T
duds
This equation is known as the first Tds or Gibbs equation
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ENTROPY CHANGE OF IDEAL GASES
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
Substituting du = cvdT and P = RT/ into the 1st Tds equation, for an ideal gas,
Rd
T
dTc
T
RTd
T
dTcds vv
=1
2
1
2
2
1
2
1
12
lnR
T
Tlnc
dR
T
dTcss vv
Substituting dh = cpdT and = RT/p into the 2nd Tds equation, then
p
Rdp
T
dTc
pT
RTdp
T
dTcds
pp
=1
2
1
2
2
1
2
1
12p
PlnR
T
Tlnc
P
dPR
T
dTcss pp
Constant volume process
Constant pressure process
Constant temperature process
=1
2
12T
Tlncss v
= 1212 TTlncss p
=1
2
1
2
12P
PlnRlnRss
EXAMPLE 7-9
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
Air is compressed from an initial state of 100 kPa and 17oC to a final state of 600 kPa and
57oC. Determine the entropy change of air during this compression process.
The cp and R of air can be taken as 1.006 kJ/kgK and 0.287 kJ/kgK respectively
8/13/2019 Chap7 Entropy Handout
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1kR
c
1k
Rc vv
ISENTROPIC PROCESSES OF IDEAL GASES
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
During an isentropic process, ds = 0,
0lnRT
Tlnc
1
2
1
2v =
Rearrange
vcR
2
1
1
2
v1
2 lnlnc
R
T
Tln
=
1
2
1
= kln
1k
2
1
1
2
T
T
=
pcR
1
2
1
2
p1
2
P
Pln
P
Pln
c
R
T
Tln
=
=
0p
PlnR
T
Tlnc
1
2
1
2p =
1k
k
R
c
1k
kRc
pp k
1k
1
2
1
2
P
P
T
T
=
k
2
1
1
2
1k
2
11k
k
1
2
P
P
P
P
=
=
Tk-1 = constant
k
1k
1
2
P
Pln
= TP(1-k/k) = constant
Pk = constant
Sme 9/3
EXAMPLE 7-10
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
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EXAMPLE 7-11
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
Helium gas is compressed by an adiabatic compressor from an initial state of 100 kPa
and 10oC to a final temperature of 160oC in a reversible manner. Determine the exitpressure of helium. [For helium, k = 1.667]
kPa289283
433100
T
TPP
1667.1
667.1
1k
k
1
212 ==
=
REVERSIBLE STEADY FLOW
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
The energy balance for a steady a steady flow device undergoing an
internally reversible process can be expressed in differential form as
qrev - wrev = dh +dke + dpeqrev = TdsTds = dh -dprev = dh - dp dh -dP - wrev = dh + dke + dpe
- wrev = dP + dke + dpeIntegrating, 21rev pe-ke-dPw Neglecting the kinetic and potential energy, then 21rev dPw For constant volume process (incompressible fluid),
pekePPdPw 1221
rev For any device involves no work interactions such as a nozzle and pipe,
0zzg2
VVPP0pekePP 12
21
22
1212 =This equation is known as the Bernoulli equation in Fluid mechanics
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Determine the compression work required to compress steam isentropically from 100 kPa
to 1 MPa, assuming that the steam exists as (a) saturated liquid and (b) saturated vapor atthe inlet state.
EXAMPLE 7-12
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
We take first the pump and then the compressor as the system
(a) Steam is a saturated liquid initially, = f@ 100 kPa= 0.001043 m3/kg (Table A-5)
The volume remains constant during the process (incompressible),
thus
kJ/kg0.94
10010000.001043
PPdPW 121
2
1rev
=
EXAMPLE 7-12 cont.
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
(b) The steam is a saturated vapor initially and remains a
vapor during the entire compression process. The
specific volume of the vapor changes considerably
during the compression process. From the 2nd Tds
relations,
T ds = dh - dP = 0 (isentropic process, ds = 0) dP = dh
1
2
12
2
1rev hhdhdPW
P1 = 100 kPaSat. vapor
h1
= 2675.0 kJ/kg
S1 = 7.3589 kJ/kgK
h2 = 3194.5 kJ/kg (Table A-6)P2 = 1 MPas2 = s1
(Table A-5)
kJ/kg5.5190.26755.3194w rev =
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PROOF THAT STEADY FLOW REVERSIBLE DEVICES
DELIVER THE MOST AND CONSUME THE LEAST WORK
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
Consider two steady flow devices, one reversible and the other irreversible, operating
between the same inlet and exit states.
Actual (irreversible) qact - wact = dh + dke + dpeReversible qrev - wrev = dh + dke + dpe
Comparing both equations, we can write
qact - wact = qrev - wrev or wrev - wact = qrev - qactFor reversible process, qrev = T ds, and dividing by T, then
T
qds
T
ww actactrev Since
T
qds act> Then we can write,
actrev
actrevactrev
ww
ww0
T
ww
>>
ISENTROPIC EFFICIENCY OF TURBINES
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
The isentropic efficiency of turbine is defined as
s
aT
w
w
workturbineIsentropic
workturbineAct ual =Neglecting the changes in kinetic and potentialenergies, then
s21
a21T
hh
hh =A well-designed, large turbines have isentropicefficiencies above 90% and small turbines have can
be below 70%
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EXAMPLE 7-14
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
Steam enters an adiabatic turbine steadily at 3 MPa and 400oC and leaves at 50 kPa and100oC. If the power output of the turbine is 2 MW, determine (a) the isentropic efficiency of
the turbine and (b) the mass flow rate of the steam flowing through the turbine.
Ass ump tions : Steady flow and the changes in kinetic and potential energies are negligible.
P1 = 3 MPa
T1 = 300oC
h1 = 3231.7 kJ/kg
s1 = 6.9235 kJ/kgK
h2a = 2682.4 kJ/kg (Table A-6)P2a = 50 kPa
T2a = 100oC
(Table A-6)
Anal ysi s :
State 1
State 2a
sf= 1.0912 kJ/kgK (Table A-5)
sg = 7.5931 kJ/kgK (sat mixture)P2a = 50 kPa
s2s = s1State 2s
kg/kJ9.2407)7.2304(897.054.340hxhh
897.05019.6
0912.19235.6
s
ssx
fgs2fs2
fg
fs2s2
====
66.7%or667.09.24077.3231
4.26827.3231
hh
hh
s21
a21T ===
EXAMPLE 7-14 Cont
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
(b) From the energy balance for steady-flow systems,
kg/s64.3kJ/kg)4.26827.3231(
kW10x2m
kJ/kg)4.26827.3231(mkW10x2
)hh(mW
hmWhm
3
3
a21a
a2a1
=
&
&
&&
&&&
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ISENTROPIC EFFICIENCIES OF COMPRESSORS
AND PUMPS
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
The isentropic efficiency of a compressor is defined as
1a2
1s2
a
sC
hh
hh
w
w
workcompressorAct ual
workcompressorIsentropic
==
The value of c greatly depends on the design of thecompressor.
Well designed compressors have isentropic efficienciesthat range from 80 to 90 percent
Similarly, the isentropic efficiency of a pump is defined
as
1a2
12
1a2
1s2
a
sP
hh
)PP(V
hh
hh
w
w
workpumpAct ual
workpumpIsentropic
==
EXAMPLE 7-15
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
Air is compressed by an adiabatic compressor from 100 kPa and 12oC to a pressure of
800 kPa at a steady rate of 0.2 kg/s. If the isentropic efficiency of the compressor is 80percent, determine (a) the exit temperature of air and (b) the required power input to the
compressor.
Assumpti ons : Steady flow, the changes in kinetic and potentialenergies are negligible and air is an ideal gas.
Anal ysi s :
K3.516285X100
800xT
P
PT
4.1
4.0
1
k
1k
1
2s2 ==
=
8.0)TT(
)TT(
)TT(c
)TT(c
hh
hh
1a2
1s2
1a2p
1s2p
1a2
1s2C ==
==
K1.5742858.0
2853.516
T8.0
)TT(
T1s2
a2 ===(b) From energy balance for steady flow devices,
kW58.11
2851.574005.10.2
)TT(cm
)hh(mWhmWhm
1a2p
1a2aa2a1
=
&
&&&&&
(a)
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ISENTROPIC EFFICIENCIES OF NOZZLE
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CHAPTER 7 : ENTROPY
The isentropic efficiency of a nozzle is defined as
s21
a212s2
2a2
Nhh
hh
V
V
exitatKEIsentropic
exitatKEAct ual = Isentropic efficiency of a nozzles are typically above 90 percent
EXAMPLE 7-16
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
Air at 200 kPa and 950 K enters an adiabatic nozzle at low velocity and is discharged at a
pressure of 80 kPa. If the isentropic efficiency of the nozzle is 92 percent, determine (a)the maximum possible exit velocity, (b) the exit temperature and (c) the actual exit
velocity of the air.
Assumpti ons : Steady flow, the inlet kinetic energy isnegligible and air is an ideal gas.
Anal ysi s : The cp and k of the air can be taken as 1.099kJ/kg and 1.354 respectively
(a) The exit velocity will be maximum when the nozzle
operates as reversible device. Then for isentropic
process
K748950X200
80
xTP
P
T
354.1
354.0
1
k
1k
1
2s2 ==
=
From energy balance,
2
V0)TT(c
2
VVhh
2s2
1s2p
2s2
21
1s2
==
m/s666
748950099.110x2)TT(c2V 3s21ps2 =
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EXAMPLE 7-16 Cont
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
(b) The exit temp is determined from
92.0
TTc
TTc
hh
hh
s21p
a21p
s21
a21N =
== K76474895092.0950TT92.0TT s211a2 =
(c) The actual exit velocity is determined from,
2s2
2a2
NV
V= m/s63966692.0VV 22s2Na2 =
ENTROPY BALANCE
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
The second law of thermodynamics states that entropy can be created but it cannot
be destroyed.
The increase of entropy principle (entropy balance) for any system is expressed as
=
systemtheof
entropytotal
theinChange
systemthe
withingenerated
entropyTotal
system
theleaving
entropyTotal
system
theentering
entropyTotal
sysgenoutin SSSS Entropy change of system = Entropy at final Entropy at initial state
Ssys = Sfinal Sinitial = S2 S1Entropy can be transferred to or from a system by 2
mechanisms, heat transfer and mass flow (entropy transfer
for an adiabatic closed system is zero)Entropy transfer by heat transfer :
constant)(TT
QS =
Entropy transfer by mass flow
Entropy entering the control volume at state 1 = S1 = m1s1Entropy leaving the control volume at state 2 = S2 = m2s2
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ENTROPY GENERATION, Sgen
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
Irreversibilities such friction cause the entropy of a system to increase
Entropy generation is a measure of the entropy created by this effectsFor reversible process, the entropy generation is zero, thus the entropy change of a
system is equal to the entropy transfer by heat transfer and mass.
Entropy balance for any system undergoing any process can be expressed as,
(kJ/K)SSSS systemgenoutin
Net entropy transfer
by heat and massEntropy
generation
Change in
entropy
(kW/K)dt
dSSSS
systemgenoutin =&&&
Rate of net
entropy transfer by
heat and massRate of Entropy
generation
Rate of change
in entropy
The entropy balance for an adiabatic closed system : Sgen = Ssystem
ENTROPY GENERATION FOR
CLOSED SYSTEM
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
The entropy change of a closed system (fixed mass) can be expressed as
systemgenk
k SST
Q gssurroundinsystemgen
k
k12gen SSSSor
T
QSSS
For an adiabatic process (Q = 0),
12systemgen ssmSS
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ENTROPY GENERATION FOR
OPEN SYSTEM
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CHAPTER 7 : ENTROPY
Outlet
Inlet
1
2
misi
Q
W
SystemS = (S2 S1)CVmese
The entropy change of a open system can be expressed as
kJ/KSSSsmsmT
QCV12geneeii
k
k Or in the rate form
kW/Kdt
dSSsmsm
T
Q CVgeneeii
k
k = &&&&For steady flow process, dScv/dT = 0,
kW/KT
Q-sm-smS
k
kiieegen &&&&
For steady flow, single stream, kW/KT
Q-ssmS
k
kiegen &&&
For steady flow, single stream and adiabatic, kW/KssmS iegen &&
Sme 15/3
EXAMPLE 7-18
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
kJ/kgK0.36936.6353-7.0046ssssmS
12ie
gen =&&
Steam at 7 MPa and 450oC is throttled in a valve to a pressure of 3 MPa during a steady-
flow process. Determine the entropy generated during this process and check if the
increase of entropy principle is satisfied.
Ass ump tions : Steady flow, the kinetic and potential energy changes and the heat transfer
is negligible.
Anal ysi s : For throttling device, h2 = h1
State 1 P1 = 7 MPa
T1 = 450oC
h1 = 3288.3 kJ/kgs1 = 6.6353 kJ/kgK
State 2 P2 = 3 MPa
h2 = h1s2 = 7.0046 kJ/kgK
The increase of entropy principle is satisfied since the s > 0
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EXAMPLE 7-20
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
Water at 200 kPa and 10oC enters a mixing chamber at a rate of 150 kg/min where it is
mixed steadily with steam entering at 200 kPa and 150oC. The mixture leaves the chamberat 200 kPa and 70oC and heat is lost to the surrounding air at 20 oC at a rate of 190 kJ/min.
Neglecting the changes in kinetic and potential energies, determine the rate of entropy
generation during this process.
Assumpti ons : Steady flow, the kinetic and potential energy changes are negligible.
Analys is : Two inlets and one exit
Mass balance : 321 mmm &&& =Energy balance (W = 0)
)hmhm(h)mm(
)hmhm(hmQ
1122321
112233
&&&&
&&&&
State 1 P1 = 200 kPa
T1 = 10oC
h1 = hf@ 10oC = 42.022 kJ/kg
s1
= sf@ 10oC = 0.1511 kJ/kgK
State 2 P2
= 200 kPa
T2 = 150o
C
h2 = 2769.1 kJ/kg
s2 = 7.0046 kJ/kgKState 3 P3 = 200 kPa
T3 = 70oC
h3
= hf@ 70oC = 293.07 kJ/kg
s3 = sf@ 70oC = 0.9551 kJ/kgK
T1 = 10oC
150 kg/min
T2 = 150oC
T3 = 70oC
EXAMPLE 7-20
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
The rate of entropy generation,
kJ/min.K53.24
293
190--1015.29x7.28-150x0.1511-551165.29x0.9
T
Q)smsm(sm
kW/KT
Q-sm-smS
221133
k
kiieegen
==
=
&&&
&
&&&
]kg/min29.15m
022.42x1501.2769xm07.293xm150190
)hmhm(h)mm(Q
2
22
1122321
= &&&
&&&&&
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EXAMPLE 7-21
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004
CHAPTER 7 : ENTROPY
Assumpti ons : No irreversibilities internal reversible.
Anal ysi s : The entropy of the system decreases during this process because of heat loss.
A frictionless piston-cylinder device contains a saturated liquid-vapor mixture of water at
100o
C. During a constant pressure process, 600 kJ of heat is transferred to thesurrounding air at 25oC. As a result, part of the water vapor contained in the cylindercondenses. Determine (a) the entropy change of the water and (b) the total entropygeneration during this heat transfer process.
kJ/K40.027325
60061.1
T
QSS
k
ksystemgen =
(a) For internally reversible isothermal process,
kJ/K61.1273100600
T
QS
systemsystem =
(b) The entropy generation for closed system,
Note : If we reverse the heat transfer direction, then,
kJ/K40.027325
60061.1
T
QSS
k
ksystemgen
kJ/K61.1
273100
600
T
QS
systemsystem =
Impossible, the
process cannotbe reversed