chap7 entropy handout

8/13/2019 Chap7 Entropy Handout

1/22

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2004

CHAPTER 7 : ENTROPY

CHAPTER 7

ENTROPY

OBJECTIVES


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


2/22

CLAUSIUS INEQUALITY


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


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


3/22

DEFINITION OF ENTROPY CHANGE


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


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)


4/22

INTERNALLY REVERSIBLE ISOTHERMAL


CHAPTER 7 : ENTROPY

The heat transfer processes are internally reversible

Thus, the heat transfer can be determined by

kJ/KT

QQ

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


CHAPTER 7 : ENTROPY


5/22

THE INCREASE OF ENTROPY PRINCIPLE


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


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


6/22

EXAMPLE 7-2


CHAPTER 7 : ENTROPY

ENTROPY CHANGE OF PURE SUBSTANCE


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


7/22

EXAMPLE 7-3


CHAPTER 7 : ENTROPY

EXAMPLE 7-4


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


8/22

ISENTROPIC PROCESSES


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


CHAPTER 7 : ENTROPY


9/22

T-S DIAGRAM


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


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


10/22

ENTROPY CHANGE OF IDEAL GASES


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


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


11/22

1kR

c

1k

Rc vv

ISENTROPIC PROCESSES OF IDEAL GASES


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


CHAPTER 7 : ENTROPY


12/22

EXAMPLE 7-11


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


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


13/22

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


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.


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 =


14/22

PROOF THAT STEADY FLOW REVERSIBLE DEVICES

DELIVER THE MOST AND CONSUME THE LEAST WORK


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


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%


15/22

EXAMPLE 7-14


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


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

=

&

&

&&

&&&


16/22

ISENTROPIC EFFICIENCIES OF COMPRESSORS

AND PUMPS


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


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)


17/22

ISENTROPIC EFFICIENCIES OF NOZZLE


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


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 =


18/22

EXAMPLE 7-16 Cont


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


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


19/22

ENTROPY GENERATION, Sgen


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


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


20/22

ENTROPY GENERATION FOR

OPEN SYSTEM


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


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


21/22

EXAMPLE 7-20


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


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

= &&&

&&&&&


22/22

EXAMPLE 7-21


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

chap7 entropy handout

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