synergyinstitute.ac.in · ENGINEERING THERMODYNAMICS 1 SIT, BBSR, DR. R. C. Nayak, ETD, 4th Sem, Mech. THEORY PARAMETERS : P - Pressure (bar) 1 bar = 1 105 N /m2 1 N /m2 = 1 Pascal

Engineering Thermodynamics

Fourth Semester

Dr. Ramesh Ch. Nayak

(Associate Professor, Mechanical Engineering Department)

Synergy Institute of Technology, Bhubaneswar

At: Bhimpur, Near Phulnakhara, PO: Pahala,

Bhubaneswar, Odisha-752101, INDIA.

ENGINEERING THERMODYNAMICS 1

SIT, BBSR, DR. R. C. Nayak, ETD, 4th Sem, Mech.

THEORYPARAMETERS :

P - Pressure (bar)

1 bar = 25 /101 mN

2/1 mN = 1 Pascal

610MicroM

310millim

310kilok

610Mega

910GigaG

)( 3mvolumeV

)/( smvelocityV

)/(int kgJenergyernalU

)//( kgkgorkJJEnthalpyh

)/( kgKkJEntropyS

STATE POINT RELATION :

mRcTPV

tconsgassticCharacteriRc tan

kgKJC pair /10005.1 3

kgKJC vair /10718.0 3

kgKJR cair /10287.0 3

4.1air

J

RCC c

vp & v

p

C

C

Where, J = mechanical equivalent of heat [ J= 1(SI)]

Process Relation :

T

PV= Constant

2

22

1

11

` T

VP

T

VP

nCompressio Pressure increases &

volume decreases

Expainsion Pressure decreases & vol-ume in creases.

For P = Constant :

CPPP 21

1

2

1

2

V

V

T

T

For Volume constant :

21 VV

1

2

1

2

P

P

T

T

For t = Constant (Isothermal) :

21 TT

2211 VPVP

For t = Constant (Isothermal) :

CVP n

1

2

1

1

1

2

1

2

nn

n

V

V

P

P

T

T

2


ENGINEERING THERMODYNAMICSAdiabatic :

1

2

1

1

1

2

1

2

V

V

P

P

T

T

2211 VPVP

GAS CYCLE :

cycleOtto constat heat addition volumecycle

cycleDiesel constant pressure cycle

cycleDual Both Pressure & volume re-mains constant

Engine Parameter & arrangement :

Fig-1

Spark Plug = Petrol

Nozzle = Diesel

TDC = Top Dead center

BDC = Bottom Dead Centre

sV = Swept volume

cV = Clearance volume

VolumeSwept The volume under whichpiston can compress the mixture

VolumeClearance The Volume where pis-ton can’t go thruogh it 3

vol = Volumetric efficiency

mech = Mechanical efficiency

the = Thermal efficiency

Fig-2

enginePetrol Spark Ignition (SI) engine.

engineDiesel Compression Ignition (C.I)engine

doneWorkWd

payweWhat

gotweWhat

PI

PO

/

/

enginestroke4 Cycle begins when pistonis at TDC

enginestroke2 Cycle begins when pistonis at BDC

FOR 4- STROKE ENGINE :

(1) Suction : Piston moves from TDC toBDC IV open and EV Closed.

(2) Compression : Piston moves from BDCto TDC EV and IV both closed.

(3) Expansion : Piston moves from TDCto BDC, IV & EV both closed.

(4) Exhaust : Piston moves from BDC toTDC, EV open.



IC. Through nozzle fine spray of fuelis injected & mixed with compressed air whichis at high temp & pressure.

IS. By using of spark plug air fuel mix-ture is ignited.

PARAMETER :

D = d = Bore dia / cylinder dia / piston dia :

L = Stroke length

= Length of connecting rod.

cV = Clearance volume

sV = Swept volume

r = Compression ratio

volsmall

volbig

= Cut off ratio.

OTTO CYCLE :

• Constant Volume cycle

• In this cycle heat addition & rejectiontakes place const volume

Ex : Petrol engine :

Fig-3

2

1

V

Vr

21 Adiabatic compression

32 Heat addition at constant volume

43 Adiabatic expansion

14 Heat rejection at constant volume

AH

RH

AH

RHAH

AH

Wd

p

pO

.

.1

.

..

./1

/

RHAHWd ..

)(. 23 TTCAH v

)(. 14 TTCRH v

Now, )(

)(1

.

.1

23

14

TTC

TTC

AH

RH

v

v

)(

)(1

23

14

TT

TT

........(1)

volsmaller

volbigger

V

Vrationcompressior

2

1

Process 1-2 :

1

2

1

1

2

V

V

T

T

4


ENGINEERING THERMODYNAMICS

)........(....................)( 1

1

2 irT

T

Process 3-4 :-

2314

1

2

1

3

4

4

3 &

1

VVVVV

V

V

V

T

T

).........(....................)( 1

4

3 iirT

T

from (i) & (ii)

1

4

3

1

2 )( rT

T

T

T

1

41

32 )(

rrTT

TT

1

14

23 )(

rrTT

TT

23

141TT

TT

1)(

11

rr

DIESEL CYCLE :

It is known as constant pressure cycle i.e.H.A takes place at const. Pr.

Fig-4

Compression ratio,2

1

V

Vr

Cut- off ratio,2

3

V

V

Expansion ratio, r

V

V

V

V

V

V

V

Vre

3

2

2

1

3

1

3

4

HA

HR

HA

HRHA

HA

Wddisel

1

)( 23 TTCHA p

)( 14 TTCHR v

)(

)(11

)(

)(1

23

14

23

14

TT

TT

TTC

TTCdisel

p

v

1-2 :-

1

1

2

1

1

2 )(

rV

V

T

T

13121 )(

1

)(

1

rrT

rTT

2-3 :-

1

33

232

2

3

2

3 TV

VTT

V

V

T

T



3-4 :-

11

3

4

4

3

r

V

V

T

T

1

34

r

TT

Substituting values of neqinTTT 421 &,,

33

13

1

3 )(1

11

TT

r

T

rT

11

1)(1

11

3

113

T

rT

1

1)(1

11

1

1rr

r

1

1

)(

111

1

rr

1

1

)(

111

1

rdiesel

DUEL CYCLE :

HA in both P = Const & V = Const

Fig-5

)(. 23 TTCmAH vcv

)(. 34 TTCmAH pcp

)(. 15 TTCmRH vcv

Compression ratio,2

1

V

Vr

Cut off ratio,2

4

3

4

V

V

V

V

Explosion ratio,2

3

Expansion ratio,4

1

4

5

V

V

V

V

=4

2

2

1

V

V

V

V

6



= r

r 1

AH

RH

AH

HRHA

AH

dW

.

.1

..

.

= )()(

)(1

3423

15

TTmCTTCm

TTCm

pv

v

)()(

)(1

3423

15

TTTT

TT

(Divide Cv on numerator & denomi-nator)

Process 4-5 :-

1

5

4

4

5

V

V

T

T 1

45

r

TT

Process 3-4 :-

4

3

4

3

V

V

T

T

1

43 TT

Process 2-3 :-

3

2

3

2

P

P

T

T

1

32 TT

14

2 T

T

Process 1-2 :-

1

1

2

2

1

V

V

T

T

121 )(

1 r

TT

14

1 )(

11 r

TT

44

44

14

1

4 )(1

1T

TTT

r

T

rT

11

11

1)(1

1

44

114

TT

rT

111

1)(1

1

1

1r

1

11

1

1

r

1

11

1

1

r

)1(1

11

1

1

rdual

Procedure for sloving problem :

1. Read the question & confirm (Medium :



air, Cycle : otto, diesel, dualword is there)

2. Then check above cycle.

3. Draw P-V & T-S diagram

4. Write the data given

5. If medium is air then assume pC air, vC

air, cR & Vair

.

6. If cut off ratio word is there.

[Ex. cut off takes place at 5% of stroke]

Fig-6

sc VVV100

53

21 VVVs

Work done/cycle =sm VP

mP = Mean effective pressure

sV = Stroke volume = Ld 2

4

.sec/./. cycleofNocycledWPower

Q.A disel cycle having bore dia 25 cm & strok

length 40 cm, the vol. after compression is1.2 lit, the pr at the end of suction is 1 bar, cutoff takes place 5% of stroke. Calculate the & MEP.

Ans.

Fig-7

d = 25 cm = 0.25 cm

l = 40 cm = 0.4 m

332 102.1 mV

251 /101 mNP

323 05.0 VVV

242 100

5VVV

4

4.025.0

4

22

LdVs

331063.19 mvs

333 1063.1905.02.105.0 sVVV

331018.2 m

1

1)(

)(

111

1 P

r

rr

r

rdielel

8



sc vvvv 41

332 10)63.192.1( vv

331083.20 m

19.17102.1

1083.203

3

2

1

V

Vr

sVVV 05.023

22

2

2

3 05.0

v

v

v

v

v

v s

2

05.01

v

vs 3

3

102.1

1063.1905.01

82.1

1

1

)(

111

1

r

rdiesel rr

182.1

1)82.1(

)19.17(

1

4.1

11

4.1

4.0

581395.3

131259.21

%34.63diesel

sm vpcyclewd /

14433221/ wdwdwdwdcyclewd

1-2 2211 vPvP

1

2

2

1

V

V

P

P

2P

12211

21

vPvP

Wd

Process 2-3 23 PP

)( 23232 VVPW

Process 4-1 rr VPVP 4433

r

P

P

P

P

3

4

4

3

________4 P

14 VV

14433

43

r

VPVPWd

014 Wd

14433221/ WdWdWdWdcycleWd

3_______m

JPVPWd msm

Q. Dual cycle having comp. ratio = 9, expensionratio =5, the inlet temp & pr is 30oC & 1 barheat liberated at const pressure is twice then ofat const Vol. If bore dia is 250 mm & stroklength 400mm having working cycle per secas 8 cal pr. & temp at all state points, , MEP..

FIG-8

92

1 v

vr

54

5 v

vre

25

1 101m

NP



34 mv

323 mVV

_______3

4 v

v

3-43

4

3

4

V

V

T

T

_________________ 34 TT

3243 2 HAHA

)(2)( 2334 TTmCvTTmCp

___________3 T

___________4 T

2-32

3

2

3

P

P

T

T

3P

2

3

P

P

5-1 15 VV

1

5

1

5

P

P

T

T

_____5 P

4-5

1

4

5

5

4

r

V

V

T

T

5T

10



_____dual

vs

cyclewd

Pm

Power = w.d/cycle No of cycle/sec.

= _______watt.

3432211 0

1/ vvp

vpvpccwd

BRAYTON CYCLE :

It is a simple cycle used for gas turbinepower plant it consist of compressor, turbine,combustion chamber.

Initially medium (air) is allowed to be com-pressed in compressor, now this compressedair is ignited in combustion chamber by add-ing of fuel, where heat addition to the mediumtakes place, Now this heated air in expanded isturbine, after expanding heat is rejected fromthe medium by provision of heat exchanger.

Fig-9

Fig-10

HA

HR

HA

HRAH

AH

dw

PI

PO

1.

.

.

)(. 2332 TTmCAH p

)(. 1414 TTmCRH p

No loss so no effect of for & comp, consid-ered.

)(

)(1

)(

)(1

23

14

14

23

TT

TT

TTCpm

TTCpm

1-2

1

1

2

1

2

P

P

T

T

3-4

1

1

2

1

4

3

4

3

P

P

P

P

T

T

1

4

3

4

3

1

2

P

P

T

T

T

T

r

r

p

r

r

rP

P

TT

TT 1

1

1

2

14

23



where1

2PrP

Pratioessurer p

r

r

p

brayton

r1

11

due to loss of Tur & comp comes inpicture.

Effect of irrenversibility in Turbine & compres-sor in brayton cycle If of turbine and com-

pressor is given i.e. t ( of turbine), c (

of compressor)

Fig-11

Initially 11 4321

)&(4321 CTefficiencytodue

12

112

12

11

2

hh

hh

TT

TTc

143

431

43

43

Th

hh

TT

TTc

h = Enthalpy

TCm p

bigger

smaller

CTnet WWW

43 hhWT

12 hhWC

23... hhAHQ

14... hhRHQ

AH

Wnet

.

(whenever in question comp & expensiongiven)

Q. A gas turbine power plant being supplied at1 b a r , 2 7

oC to be compressed upto 5 bar withisentropic efficiency of 25%, compressed airis heated upto 1000K in combustion chamber,where pressure drop of 0.2 bar. determine isen-

tropic efficiency of the turbine, if cycle is 20%.

Fig-12

Solution :

251

441 1011m

NbarPPP

251

`22 1055m

NbarPP

12



kT 300273271

,25.0c kT 10003

(as pressure drop of 0.2 bar)

?T

2.0%20 Brayton

for 1-21 process

1

1

12

1

12

P

P

T

T

kT 14.47512

12

11

2

TT

TTc

KT 7.11752

for 3-a1 process

1

3

14

3

14

P

P

T

T

kT 79.63814

Now 143

43

TT

TTT

AH

dw netBrayton .

).(

32.

AH

WW CT

)(

)()(

23

1243

hh

hhhh

)(

)()(

23

1243

TT

TTTTtonBary

4T =950.23 k

136.01

43

43

TT

TTT

Effect of inter cooling on Brayton cycle :

In this cycle there are provision of & com-pressor & 1 turbine as shown in figure below.

Fig-13

Fig-14



45. hhAH

],[],[ 3213 PPTT

16 hhHR

56 hhWT

14 hhWC

HA

HR1

Generally Intercooling process takes placein between 1st & 2nd compressor & it is ofconst pressure process, Intercooling is re-quired to avoid high pressure & high tempera-ture at the end of single compression.

Effect of reheat and inter cooling :

Fig-15

8765 ...... AHAHAH

)()( 7865 TTCmTTCm pp

)(.. 110 TTCmRH p

975 TTT

1042 TTT

8732 PPPP

96418732 PPPPPPPP

HA

HR1

T

net

W

WdratioWork

Regeneration on Brayton cycle :

Fig-17

Fig-18

14



1616.. TThhRH

3434.. TThhAH

62 TT

53 TT

34

1611TT

TT

HA

HR

54

121TT

TT

53

62

TT

TT

4

54

1

21

1

1

1

T

TT

T

TT

4

5

2

1

1

2

4

1

1

1

1

T

T

T

T

T

T

T

T

For 21 process

1

11

1

1

2

1

2

prP

P

T

T

pr = pressure ratio

1

2

P

P

For 54 process

1

11

1

1

21

1

5

4

5

4

r

p

rr

rP

P

P

P

T

T

4

5

2

1

4

2

1

1

1

T

T

T

T

T

T

1

1

1

1

4

2

11

11

1

p

p

r

r

T

T

4

21T

T

4

24

T

TT

In regeneration cycle of the Braytoncycle is only dependent on the temp after com-pression & the temp of the medium enteringinto the turbine.



Procedure for solving problem :

1) Read the question & confirm (cycle usedin power plant, H.A & H.R. at cont pr.)

2) Then check about of compressor & Tur--bine is given or not is given.

AH

Wnet

.

CTnet WWWd

3) Then check about the conditions (Re-heat, inter coooling)

4) Draw T-S diagram

5) Write the formula & then solve theproblem.

Q. A Brayton cycle operates with air enteringthe compressor at 1 bar & 25oC The pressureratio across the compressor is 3 to 1 and themax. temp in the cycle is 650oC. Determine thecompressure work & turbine work, thermal &work ratio of the cycle.

Solution:

Cycle :- Brayton

FIG-19

kgkJC airP /10005.1 3

kgkJC airV /10718.0 3

kgkJR airC /10287.0 3

4.1air

Data Given :

25

1 101m

NP

kT 298273251

31

3

1

2 P

P

.9232736503 kT

?cW

?TW

?th

work ratio = ?

1

11

p

brayton

r

3pr

pC CTThhW 1212

kgkj/1.110005.1)29888.407(

pT CTThhW 4343

kgkj /90.249

005.1)29834.674923(

21

r

r

P

P

T

T1

1

2

1

2

4.1

4.0

2 )3(298T

KT 88.4072

16



43

r

r

P

P

T

T1

4

3

4

3

.34.6744 KT

3

&

4

3

1

2

1423

P

P

P

P

PPPP

kgKJWWW CTnet /)43.11090.249(

= 139.47 KJ/kg

269.069.517

47.139

. 23

TTCp

W

AH

W netnet

= 26.9%

work ratio 558.09.249

47.139

T

net

W

Wd

Solution :

Fig -A

1.4

0.287

0.718

1.005

air

air

air

air

γkj/kgkRC

kj/kgkCV

kj/kgkCP

251 N/m101bar1P

k300T1

5P

P

P

P

1

12

1

2

?Wd

0.85η0.80η

k1075T

net

tur

com

3

cTnet WdWdWd

43T hhWd

338.58738.18)(10751.005

)T(TC 43p

kj/kg220.02300)(518.931.005

hhWd 12c

)T(TCP 12

121

1.4

0.412

1

1

12

1

12 (5)300T

P

P

T

T

k475.14T12

k518.93TTT

TTη 212

112

c

143

1.4

0.414

1

14

314

3

5

1075T

P

P

T

T



k678.74T14

143

43T TT

TTη

k738.18T4

kj/kg118.48WdWdWd

kj/kg220.02Wd

kj/kg338.5Wd

cTnet

c

T

%21.20.212558.85

118.48

)T(TCp

Wd

HA

Wdη23

netnet

0.35338.6

118.48RatioWork

Tur

net

W

Wd

Q. In an air standard Brayton cycle the mintemp T

1 is governed by the ambient atmosphere

& max temp T3 is dictated by the material of

construction of turbine blade. For fixed val-ues of T

1 & T

3 determine the Pr ratio (r

p) for

obtaining max net work per unit mass of air.

Fig-B

Pressure ratio,4

3

1

2p P

P

P

Pr

Wd = HA - HR

= Cp (T3 - T

2) - Cp (T

4 - T

1)

for maximum work,

0dr

dW

p

for 21 process

1

p12

1

1

2

1

3 )(rTTP

P

T

T

for 43 process

1

p

34

1

p43

1

4

3

4

3

)(r

1TT

)(rTT

P

P

T

T

11

p

3

1

p13 T

)(r

1TCp)(rTTCpWd

1

1

1

p3

1

p13 T)(rTCp)(rTTCp

Given T3 & T

1 as fixed values)

So, T3 = T

1

(we convert T2 & T

4 in terms of rp but not T

3

& T1 as in quetion its given to keep T

3 & T

1

fixed)

1rTCpr1TCpWd1

p3

1

p3

18


ENGINEERING THERMODYNAMICS (T

3 = T

1 as given fixed value)

Now, 0dr

dW

p

11

p33 )(r1

TCp0TCp

00TCp)(r1

TCp 3

11

p3

21

p

1

p

21

p

1

p

r

1

p

1

p

)(r)(r

)(r1

)(r1

0)(r1

)(r1

0

Reciprocating Air Compressor :

In reciprocating air compressor the work-ing medium is air i.e. atmospheric air entersinto the cylinder piston arrangement & this airis compressed by change of position due tocrank rotation.

Reciprocating air compressor may be singlestage or double stage depending on the pres-sure requirement.

Fig-C

The figure above shows the arrangement ofa single acting reciprocating air compressor.

It consist of crank, connecting rod, piston,cylinder, exhaust valve, inlet valve.

In reciprocating air compressor maximumamount of pressure of air is the requirement.

In double stage reciprocating air compres-sor maximum pressure of air is possible butthat amount of temp is not required. sointercooling is used in such arrangement.

For working of a reciprocating air compres-sor work is done on the compressor i.e. workdone on system i.e. indicating -ve symbol.

Equation for work done per kg of air com-pressed in compressor (single stage) withoutclearance :

Here,

23

41

PP

PP

Fig-D

322114 WdWd(Wd

14321

diagramVPunderAreaWd



)V(VP1

VPVP)V(VP 232

2211411 n

)VP(1

VPVPVP 22

221111 n

1

VPVP)VPVP( 2211

2211 n

1

VPVPVPVP 2211

2211 n

1

VPVPVPVP 1122

1122 n

1

11VPVP 1122 n

1

11VPVP 1122 n

n

12

1122

TmRcTmRc1n

n

1n

nVPVP

TRcmPV

Rc1n

nWd

12 TT

kg1m

etemperaturncompressio&'2''1'oftermsIn

Again,

1122 VPVP1n

nWd

1

VP

VPVP

1n

n

11

2211

1T

TTRcm

1n

n

1TRcm

TRcmTRcm

1n

n

1

21

1

21

1P

PTRc

1n

nWd n

1n

1

21

unnit of Wd = J/kg.

kgm 1

Equation for work per kg of air conpressedin conpressor (single state) with clearance :

Fig-E

Here, 2341 PP&PP

56435underAreaWd

16521underAreaWd

WdWdWd

e

c

expansionncompressio

20



1P

PVP

1n

n1

P

PVP

1n

nWd

n

1n

4

344

n

1n

1

211

11

1

1

2411

n

n

P

PVVP

n

n

as P1 = P

4 & P

2 = P

3

Let V1 - V

4 = V

a

where, Va = Actual Vol. of air delivered.

11

11

1

1

21

1

1

21

n

n

n

n

a

P

PTRcm

n

n

P

PVP

n

nWd

kg1m1P

PTRc

1n

nWd

n

1n

1

21clearance

Hence it is concluded that for work per kgof air compressed in air compressor (singlestage) without clearance and with clearance hasnot efect i.e. same value of Wd.

Volumetric Efficiency :

It is ratio of actual free air delivered to thedisplacement of the compressor. displacementof the compressor = (Vs)

s

a

31

41v V

V

VV

VV

In terms of pressure :

1

P

P

V

V1

n

1

1

2

s

cvolumetric

Fig-F

Isothermal Efficincy of Reciprocating

Air compressor :-

In such arrangement the compression pro-cess 12 is isothermal (const temp) i.e. PV =Const.

Fig -G

41

3122

PP

PPP

12

12

1

12

e1111

322114

V0PV

VlogVP0VP

WdWdWdWd 11



(-ve sign as we give work on the system)

1112

121

2

1e11 VPVP

V

VlogVP

But, 12

1211 VPVP

12

1e11 V

VlogVPWd

12

1e1isothermal V

VlogTRcmWd

kg1mfor

12

1e1isothermal V

VlogTRcWd

Isothermal Efficiency :

It is the ratio between isothermal work tothe actual work.

11

log

1

1

21

12

11

iso

n

n

c

e

P

PTR

n

n

V

VTRc

11

log

1

1

2

1

12

n

n

e

P

P

n

n

P

P

12

1

1

12

V

V

P

P


1. Read the question & confirm stage.

2. Check about single or double stage.

3. Then check about compression process (ifit given a law (PVn = const., n= any val-ues) then con sider it as actual work done

(or Wdactual

or Wdpolytropic

)).

4. Write the data given :

Pistom Speed = 2 LN

Where, L = Stroke length

N = Speed in rpm

LD4

πV 2

s

Input of the compressor :

Input of the compressor

1000

compressorondonework

Sometimes the indicated power of a com-

pressor (IP) is calculated in terms of mean Ef-

fective Pressure(Pn).

IP kwNALPm 100

where, N = revolution per sec of compressor

Pm = MEP in 2m

N

L = Length of the stroke in m.

A = Area of the piston in m2.

Related Formula :

Single acting single stage :

Fig-H

22



11

1

1

1

21

n

VP

P

P

n

nWd a

41a VVV

11

1

1

211

n

n

P

PVP

n

n

31s VVV

11

1

1

21

n

n

s P

PVP

n

nwithout clear--

ance

( I.P. = Indicated Power)

Fig-I

kw1000

RPS1

P

PnVP

1n

n

kw1000

RPS1

P

P

1n

nIP

1

1n

1

2vs1

n

1n

1

21

aVP

kw1000

11

P

PVP

1n

n n

1n

1

211

Volumetric volη

sV

4V

1V

sV

aV

sV

cV

k1P

Pk1η

n

1

1

2v

Mean Effective pressure -

11

1

1

21

n

n

m P

PP

n

nP

* F r e e a i r d e l i v e r e d ( Va)per min = sv VN

( N in RPM )

* deliveredAirFreeaV = sv V

* Piston speed (rpm) = 2 LN

* LdVs 2

4

SINGLE ACTING WITH TWO STAGE :

Q. Determine the size of the cylinder of adouble acting air compressor of 32 kw inputpower in which air is drawn at 1 bar & com-pressed to 16 bar according to law PV1.25 = c,

RPM = 300, piston speed 180 m/min, v = 0.8

(volumetric m)

Solution :

Indicated Power, IP= 32 kw = watt1032 3



60

300RPSrpm300N

1.25)(nConstantPV

N/m1016PP

N/m101PP

1.25

2523

2514

0.8

m/min180speedPiston

v

21000

RPS1

P

PVP

1n

nIP

n

1n

1

2a1

(as given double action)

31032

1000

2

6

3001

1

16101

25.0

25.1 25.1

25.0

5

aV

3s

s

41

s

av

3a

m10.8V

64.8

V

VV

V

Vm

m8.64V

sV

m6.77d

Ld4

πV

n0.3L

LN2speedPiston

2s

Q.A single stage single acting compressordellvers 15 m3 of the free air per minute from1 bar to 8 bar, that compression & expansionfollow the law PV1.3= C and clearance is 1/

16th of teh swept vol. Find the dia & stroke ofthe compressor.

Take 1.5D

L .

Solution : 1.5D

L

sc

1.3

252

251

3a

V16

1V

1.3)(nCPV

rpm300N

N/m108P

N/m101P

min/m15V

Find D = ?

L = ?

11

1

1

2n

S

CV P

P

V

V

753.01

8

16

11

3.1

1.1

%3.75

Free air delivered / min (Va) =

SSV VVN 753.0300

066.0SV

5.1

42

D

LLDVS

24



,383.0 mD

mL 574.0383.05.1

60

1

1000

11

1.

1

1

21

n

n

a P

PVP

n

nPI

= 66.4 kw.

Q. A double 2 stage compressor delivors airat 25 bar, the pressure and temp of air at thebeginning of compression in L.P cylinder is1 bar & C020 . The temp of air coming out

from intercooler nbet the 2 stage is C040 andthe P is 7 bar, the diameter & stoke of LP cyl-inder are 60cm & 80cm respectively, rpm ofcompressor is 100. The volumetric of bothstage is 80% neglecting the pr loss in the sys-tem, find the brake power, indicated power ifmechanical is 85% . Take the compression

& expansion follow tPV cos35.1 .

Solution Double stage

V is given so both comp & ex-

pansion

( If no V is not given then no ex-

pansion)

Fig-J

ercoolerint32

13 TTquestionthisIn

even if intercooler present

KT 313273403

25

1 101m

NP

KT 273201

25

23 107m

NPP

25

4 1025m

NP

85.0%85 mech

100,80.0%80 Nvol

Find, IP & BP =?

As 1T and 3T are not same

Work done

11

21

1

211

n

n

P

PVP

n

n

11

1

3

433

n

n

P

PVP

n

n

As it is given double acting

n

n

P

PTmRc

n

n1

1

211



1

1

3

43

n

n

P

PT

11 TMRCVP a

NVV Sa

VLd

2

4

= 0.181m3

11 TmRcVP a

kgm 215.0293287

181.0101 5

1000.

speedWdPI

kw244

PI

PBmech .

.

kwPI

PBmech

287..

..

Q. A two stage compressor dlivers 32mfree air per min, the temp & pr of air atthe suction are C027 & 1 bar the pr at thedelivery is 50 bar. The clearance is 5% ofthe stroke in LP cylinders as well as inHP cylinder, assuming perfect inter cool-ing nbent the 2 stage, find the min powerrequired to run the compressor at 200rpm.also find the dia & stroke (assuming stoke= both bore dia) if compression & expan-sion both follows tPV cos35.1

Fig-K

stageDouble

presentExpComp .&

It is case of with clearance

min23mVa

25

18 101m

NPP

KT 300273271

rP at delivery = 4P

25

543 1050m

NPPP

barPPP 07.7312

SC VV10

5

cPVrpmN 35.1100 n=1.35

Given its perfect intercooling

So 13 TT

111

2

1

2

3

1

1

21

n

n

n

n

P

P

P

PmRT

n

nWd

Double stage, N=2

26



1

1

RT

VPm a

= 2.32 kg/min = 0.038 kg/sec

kwWd

PI 965.161000

.

Free air delivered = NV VS 11

3011.01

mVs

11

1

1

21

n

S

CV P

P

V

V

= 83.75

LdVS 21 4

dd 2

4

cmd 8.24

cmLd 8.24

Refrigeration Cycle

(Vapur Compression cycle)

Fig-L

Parameters : -

* P Pressure (bar)

* T Temp (K)

* t Temp ( C0 )

* v Specific volume

kgm3

* h Enthalpy (kJ/kg)

* SEntropy (kJ/kg K)

* U Internal energy (kJ/kg)

* x Dryness fraction

* Dry Steam

* Wet Water

Steam

wet steam

saturated steam

superheated steam

WET STEAM Denoted by ‘X’

Where X : Dryness fraction

If given condition of steam of less thansatoration condition (Found from steamtable) then that steam is wet steam.

wetwetwetwet VUSh ,,,

gwet

fgfwet

fgfwet

XVV

XSSS

Xhhh

Satorated Steam :

If given condition of steam is equal tosturation nCond then given steam becomessatrated steam.



gfgfgf VVSShh ,,,,, (directly found

from steam table)

Superheated Steam :

If given condition of steam is greater thansaturation condition.

supsupsupsup ,,, UVSh

* satPsteamg TTChh supsup

*

satPsteamg T

TeCSS sup

sup log

*

satV

sat

g

T

V

T

V

sup

sup

foundbecanV sup

kgKkJCPsteam 1.2

kgKkJCPwater 2.4

wpwwater TCh

Fig-M

Fig-N

Fig-O

Q. Steam of pressure 10 bar saturated is cooledat constant volume till pressure drops to 2 barwith a neat stetch find nCond of steam of pres-sure 2 bar.

Steam :

Fig-P

at P=10 bar

kgmVVg

3

1 1944.0

and kgmVVV wcts

3

21 1944.0

28



Now, gswct VXV 2

at P = 2bar

kgmVg

38857.0

8857.01944.0 X

22.02 XX

22222 fgfwct hXhhh

9.220122.07.504

kgkJ91.986

22222 fgfwct SXSSS

5970.52.25301.1

kgKkJ7558.2

Vapour Compression Cycle :

Fig-Q

For engine or power cycle

PI

PO

AH

dW

.

.

For vapoor compression cycle

dW

ExtractedHeatCOP

.

dW

EH

.

.

Working principle :

Initially refrigerant is placed in the compres-sor. Due to compression process its pressureand temp incrases. Now this higher pr of temprefrigerant is passed through the condenserwhere it rejects heat to the cooling medium(like atmosphere). Now, the reprigerant be-comes cool. This cooled refrigerant is againpassed through the expansion valve where italso rejects heat. Now, this cooled refrigerantpasses to the evaporator where it extracts heatfrom the substance to be cooled & it becomesheated. This heated refrigerant passes to thecompressor & the cycle is completd

(1) Saturated Vapour after Compression :

Fig-R

12

u1

hh

hh

Wd

HEC.O.P.

(2) Superheated Vapour after Compres-sion :



Fig-S

(3) Superheated Vapour before and afterCompression :

Fig-T

(4) Wet steam before and after Compression:

Fig-U

Q. A Vapour compression cycle working be-tween 2 pressure limit 5 bar and 10 bar. Thesteam after compression is dry saturated findcop, work required to run the compressor.

Condition After compression dry saturated

Fig-V

25

41 1055m

NbarPP

25

32 101010m

NbarPP

21

14

12

41

Wd

HE

hh

hhCOP

At 10 bar

kgkjhh g 1.277822

kgkjhhh f 81.762334

kgkkjSS g 5865.62

& 11111 fgfwct hXhhh

But, 112 wctSSS

1111 fgfwct SXSS

At 5 bar

kgkjhkgk

kjS ff 23.640,8607.1 11

kgkjhkgk

kjS gfg 5.2108,9606.4 11

1111 fgfwct SXSS

9606.48607.15865.6 1X

95.01 X

kgkjhXhhh fgfwct 3.264311111

95.133.26431.2778

81.7623.2643

12

41

hh

hhCOP

kgkjhhcompressorWd 8.13412

30


ENGINEERING THERMODYNAMICSQ. A vapour compression cycle working be-tween 2 pr. limit 11 bar and 15 bar, the tempafter compression is find the COP &Heat exchanged.

Sol - After compression superheated As at 15

bar, but given =

Fig-W

so of steam is superheated

after compression.

We take of steam as saturated be-fore compression as no condition is given& before superheated the of steamis saturated

At P= II bar

At P= 15 bar

At P=15 bar :

or P=15 bar

from superheated steam table

Q. A vapour compression cycle working be-tween 2pr limit 5 bar & 10 bar, the steam aftercompression is 60% dry draw the arrangment& find the cop, cal the power required to runthe compressor if 5 kg/sec of steam flows incompressor.

Here its given steam is 60% dry so drynesspraction is 0.6.

Fig-X



barPP 541

barPP 1032

6.0%602 X

12

41

hh

hhcop

At P= 10bar

2334 81.762 ff hkgkjhhh

kgkkjSkg

kjh ff 1387.2,81.762 22

kgkkjSkg

kjh fgfg 4478.4,3.2015 22

kgkjhXhhh fgfwct 99.197122222

kgkkjSXSSS fgfwct 80738.422222

kgkkjSSS wct 80738.4112

1111 fgfwct SXSS

P=5bar

kgkjhkgk

kjS ff 23.640,8607.1 11

kgkjhkgk

kjS fgfg 5.2108,9606.4 11

59.01111 fgfwc SXSS

25.188411111 fgfwct hXhhh

25.188499.1971

81.76225.1884

12

41

hh

hhCop

78.1274.87

44.1121

Power of compressor mhh 12

kwS 87.4374.87

Q. A vapour compression cycle working bytaking refrigerant a R-12. The properties aregiven below it table.

Fig-Y

Refrigerant before and after compression

is superheated, having temp CC 00 370&250

12

41

hh

hhCop

Fig-Z

P= 5 bar

22sup12222 satPRgsyp TTChhh

3003705.2360

kgk

kjhhkgkjh gg 360535 22

CTkgk

kjC satPR0

212 300,5.2

32



11sup1211 satPRg TTChh

P=10 bar

CT 01sup 250

CTsat0200

kgkjhg 2601

200205.22601 h

kgkj385

P= 15 bar

kgkjhhh f 110334

833.1150

275

385535

110385

12

41

hh

hhCop

Steam Power Cycle :

Carnot Cycle : (Reversed Carnot Cycle)

The cornot cycle using vapour as workingfluid (steam).

It consist of 2 reversible isothermal & 2reversible adibatic process.

The Figure below about P-v & T-s diagrm.

Fig-1 Fig-2

12 Isentropic compression

23 Isothermal compression

34 Isentropic compression

41 Isothermal compression

W.d = H.A - H.R.

)S(STH.A. 232

)S(STH.R. 141

)S(ST

)S(ST)S(ST

H.A

W.dη232

141232

)(TSS

)T(TSS

223

1223

]SS&SS[ 1243

2

12

T

TT th

etemperaturminimumT

etemperaturmaxiumT

1

2

TdSdQor

dTCdQ p

T

dQdSas,

Drawback of Cornot Cycle :



(1) During Isothermal condensation process,the condensation is stopped at point 1 i.e. beforesaturated liquid condition is reached.

(2) Heat addition & Heat rejection take placeat const. temperature.

(3) of the carnot vapour cycle is 100% whichis practically impossible.

(4) As non-homogeneous mixture(water+steam) enter into the compressor socompressor size must be increased & workinput have to be large.

Rankine Cycle :

Fig- 3

Fig 4

3 4 Pump work

4 4 boiler (H.A)

1 2 turbine (Expansion)

2 3 Pump work

The figure above shows the arrangementof a rankine cycle. Initially saturated water fromthe condenser passes into the pump where itsPr & temp. slightly increase. Now, thisincreased Pr & temp. water passes into theboiler where heat is supplied from externalsource which is shown in process 23 in P-V& T-S diagram. Now this high Pr & temp.steam enter into the turbine where it expandswhich is shown in process 3-4. Now this wetsteam inter into the condenser which is showin process 4-1. Now the saturated water fromcondenser inter into the pump & the cycle iscompleted.

From this concept it is found that pumpwork acts as one of the loss.

Fig -5

=H.A

W.d1HA

21Wd

4

p31

p21

Wdhh

Wd)h(h

workPumpWd p

)P(PVm 34 f

p31

p21

Wd)hf(h

Wd)h(h

34


ENGINEERING THERMODYNAMICSV

f = Saturated water Sp volume.

Steam Consumption =TurbineWd

3600

Work ratio =T

pT

T

net

Wd

WdWd

Wd

Wd

Efficiency ratio =Rankine

ith

Where ith is the indicated thermal

efficiency.

)hf(hm

3600

31sith

Where ms is the mass of steam.

Hint : If in quest it is not asked about workratio then we can neglect pump work.

Q. A rankine cycle works between 2 pr. limit12 bar & 0.01 bar. The steam that exits fromboiler is dry saturated. Find the rankine η .

Solution : Rankine cycle : since work ratio isnot given so we can neglect pump work.

Fig - 6

Now, P4 = P

1 = 12 bar

P2 = P

3 = 0.01 bar

η rankine = ?

η rankine =31

21

hh

hh

P = 12 bar.

h1 = hg

1 2784.8 kj/kg

S1 = S

g1 = 6.5233 kj/kgk

P = 0.01 bar

hf3 = h

3= 29.3 kj/kg

S1=S

2 =Swet

2 =6.5233 kj/kgk

Swet2 = Sf

2 + X

2 Sfg

2

P = 0.01 bar

Sf2 = 0.1059 kj/kgk,

hf2 = 29.3 kj/kg

Sfg2 = 8.8697 kj/kgk,

kgkjhgf /9.2484

2

kj/kg1818.43

2484.90.7229.3

hfgxhfh

0.72x

8.8697x0.10596.5233

2222

2

2

rankineη31

21

hfh

hh

3.298.2784

43.18188.2784

5.2755

37.966 .3507.0



Q : A rankine cycle working between 2 pr. limit10 bar & 0.2 bar, find the η neglecting pumpwork if the steam after boiler has 90% drynesspraction.

Fig-7

Solution :

P4 = P1 = 10 bar.

P2 = P3 = 0.2 bar

rankineη = ?

31

21

frankine hh

hh

9.01 x

hf1 = 762.81 kj/kg Sf

1 = 2.1387 kj/kgk

hfg1 = 2015.3 kj/kg Sfg

1=4.4478 kj/kgk

h1 = 52576.58 kj/kg.

Now, S1 = Swet

1 = Sf

1 + x

1 Sfg

1

= 6.1417 kj/kgk

At 0.2 bar,

h3 = hf

3 = 251.40 kj/kg

hf2 = 251.40 kj/kg

Sf2=0.8320 kj/kgk

hfg2 = 2358.3 kj/kg

Sfg2 = 7.0766 kj/kgk

S2 = S

1 = Sf

2 + x

2 Sfg

2

6.1417 = 0.8320 + x2 7.0766

x2 = 0.75

h2 = hf

2 + x

2 hfg

2

= 2020.12 kj/kg.

η =31

21

hfh

hh

251.40-2576.58

2020.12-2576.58

2393.018.2325

46.556

Q. A rankine cyle working between 2 pr limit10 bar and 0.1 bar Temp after the boiler is 4000Cfind the work ratio steam consumption and alsofind rankine η and draw the total arrangementin T-S dia.

Fig-8

Solution :

Pump work is considered.

P4 = P

1 = 10 bar

P2=P

3 = 0.1 bar

tsat

= 179.910C

tgiven

= 4000C

tgiven

> tsat

, condition of

36


ENGINEERING THERMODYNAMICSsteam is superheated so

point 1 is in superheated region.

P = 10 bar,

tsup

= 4000C

in superheated steam table

h1 = h

sup1 = 3263.9 kj/kg

S1 = S

sup1 = 7.4651 kj/kgk

P = 0.1 bar

hf3 = h

3 = 29.30 kj/kg

vf = vf3 = 0.001000 m3/kg

S2= S

1 = 7.4651 kj/kgk

S2 = S

wet2= Sf

2 + x

2 Sfg

2

P = 0.1 bar

Sf2 = 0.1059 kj/kgk

hf2 = 29.3 kj/kg

Sfg2= 8.8697

x2 = 0.83

h2 = hf

2 + x

2 hfg

2

kj/kg2089.28

2484.9)(0.8329.3

p31

p21

Wd)hp(h

Wd)h(hη

0.9929.3)(3263.9

0.992089.28)(3263.9

= 0.36

kj0.99

100.1)(100.0011

)P(PVmWd2

34fpump

0.991174.62

0.991174.62

Wd

WdWdRatioWork

T

pT

kj/kg1174.62

hhWd 21T

Stem consumption = kg/kj3.064Wd

3600

T

Problem on Carnot Cycle :

Q.A ideal carnot cycle working between perlimit 10 bar & 1.5 bar, the steam after adiabatic

expansion is 90% dry cal. thermalη power

develkoped if 0.5 kg/cycle mass flow takesplace, there are 200 cycle/min.

Solution :

condition : Carnot Cycle,

G i v e n , P

2 = P

3 = 10 bar

P1 = P

4 = 1.5 bar

x4 = 0.9 = 90%

?η thermal

Power developed = ?



Fig - 9

2

12

T

TTη

at 10 bar 452.91K273179.91tt 2sat

at 1.5 bar K37.84327337.111tt 1sat

0.151452.91

68.74

452.91

384.37452.91

t

ttη2

12

kj/kg307.142470.962778.1

hh4WdWd 433T

P = 10 bar ;

kj/kg2778.1hh g3

4444 hfgxhfh

P = 1.5 bar

kj/kg2226.5hgf

kj/kg467.11hf

4

1

kj/kg2470.96

2226.50.9467.11h4

Power,

cycle/min200kg/cycle0.5kj/kg(307.14)

m)h(hP 43

= (30714) kj/min

= kw511.960

30714

REHEAT CYCLE :

The efficiency of the Rankine cycle canbe increased by use of reheat and Regenerativecycle.

Fig - 10

Fig - 11

Here,

ReheatTT

PP

PPP

31

54

176

Initially water is supplied to the boiler,so it is converted to saturated steam. Now, thissteam is allowed to flow through the super

38


ENGINEERING THERMODYNAMICSheater where it is converted to superheatedsteam. This super heated steam is allowed to flowthrough the 1st turbine where it expands. Nowthis steam is again allowed to flow through the2nd tirbine where it expands again. Now thissteam flows through the condenser where it iscooled to saturated liquid. Now this saturatedliquid passes to the boiler through the pump.

*HA

Wdη

p23f51

p4321

Wd)h(h)h(hHA

Wd)h(h)h(hWd

REGENERATION CYCLE :In case of regeneration cycle the η of the

Rankine cycle is increased. By using this cycleheat lost from the 2nd turbine is utilized to heatthe water before boiler.Problem :

Q. Steam at 90 bar and C4800 is supplied to a

steam turbine, the steam is reheated to itsoriginal temp, passing the steam throughreheater at 12 bar. The expansion afterreheating takes place to condenser pressure0.07 bar. Find the η of the reheat cycle, workavailable if the flow of steam is 5 kg/sec.

Neglecting the pr. loss in the system andassume the expansion through the turbine isisentropic, don’t neglect pmp work.Solution :

Fig -12

P6 = P

1 = 90 bar

P3 = P

2 = 12 bar

P5 = P

4 = 0.07 bar

kg/sec.5m

HA

Wdη

p2351

p4321

Wd)h(h)hf(hHA

Wd)h(h)h(hWd

P= 0.07 bar

kj/kg163.40hf5

/kgm0.001007Vf 3

kj/kg2360h

kj/kg3440h

kj/kg2820h

kj/kg3350h

4

3

2

1

kj/kg9.055

100.07)(900.0010041

)P(PVf1Wd2

56p

Fig- 13



kw8050

m(1610)

kj/kg)]h(h)h[(hP 4321

kj/kg1600.945

9.0551610

Wd)h(h)h(h

WdWdWd

P4321

pTnet

STEADY FLOW ENERGY EQUATION :

A flow is said to be steady if sum of energyat input is equal to sum of energy at output andalso, mass at i/p side is equal to mass at o/pside.

Fig -14

System Related :

(i) Nozzle

(ii) Diffuser

(iii) Turbine

(iv) Compressor

Mixing System :

Fig-15

321 mmm

332211 hmhmhm

Terms Related :

P Pressure (KN/m2)

A Cross sectinal area,

v Velocity (m/sec)

V Sp. Volume (m3/kg)

h Enthalpy (kj/kg) ( h = U+J

PV)

m mass (kg or kg/sec)

z datum height (m)

g Accelation due to gravity (g = 9.81m)

Energy balance equation :

dt

dQ

1000

g

2000

1hm 1

2

11

z

dt

dw

1000

g

2000

2hm 2

2

22

z

kg/sec.ism

mass heat work

kg dQ/dm dW/dm

40



kg/sec dQ/dt. dW/dt.

Fig-17

dt

dQ

1000

g

2000hm

1000

g

2000hm 2

22

221

21

11

zVzV

dt

dW

1000

g

2000hm 3

23

33

zV

If nothing given like term (dQ, dW, Z,V)neglect them

332211 hmhmhm

Mass Balance Equation :

Fig-18

21 mm

2

22

1

113

2

/

sec/

V

VA

V

VA

kgm

mm

(if mass in kg/sec)

= kg/sec

21 mm


(1) Read the question and confirm

(2) Then check about the system.

(Tuebine, nozzle)

(3) Then check about the medium

watermedium

waterwwater TCPh

kj/kgk4.2CPwater

airmedium

airairair TCPh

kj/kgk1.005CPair

steammedium

“h” can be found from steam table by consideing

which type of srteam.

(4) Then write energy balance equation :

(If some terms are not given, not calculated, not asked neglect them)



THERMODYNAMIC RELATION..

Reciprocal and Cyclic Relations:

Thermodynamic relations pertain to the re-lations between different thermodynamic prop-erties of a system. Such relations help to deter-mine changes in certain properties which arerather difficult to measure experimentally. Fur-ther, through these relations, the limited experi-mental data can be used to derive equtions ofstate for different substances which do not un-dergo chemical reactions or a charge in phase.

Before establishing some general thermo-dynamic relations, it is desirable to review afew principles of partical ifferential calculus.

Consider to relationship f (x, y, z) = 0among the three variables x, y and z. Then

x = f (y, z)

oryz

x xdx dy dz

y z

The subscript denotes the variable heldconstant.

Also y = f (x, z)

orz x

y ydy dx dz

x z

When dy from equation is substituted inequation one obtains.

z x yz

x y y xdx dx dz dz

y x z z

or,

zz

x yz

x ydx dx

y x

x y xdz

y z z

x and z are independent variables in theabove expression and accordingly.

1zz

x y

y x

and 0x yz

x y x

y z z

Equation can be rewritten as

1

/z z

x

y y x

and this is called the reciprocal relation.

Further, the equation can be recast as

1

x yz

y

x y xzy z zx

or 1x yz

x y z

y z x

Equation is called cyclic relation.

Example:-

Prove that the ideal gas equation p = RTTsatisfies the cyclic relation.

Ans.From the given ideal gas equation

p =RT

, i.e., the pressure changes due to

changes in volume and temperature (the gasonstant R being constant for a particular gas).

The cyclic relation then slipulates that

1r p

p T

T p

42 ENGINEERING THERMODYNAMICS


The three partial derivates can be deter-mined separately and we get

2

1

r

p RTRT

v v

P

R RT

T p T p

Tp

p R p R

Substituting these values in left hand sideof the cyclic relation, we obtain

21

RT R RT

p R p

p RT

and it equals the right hand side. Thatproves that the given gas equation satisfies thecyclic relation.

Property relations:

For a unit mass of the system, the firstand second law of thermodynamics are pre-scribed by the relations.

q w du p du du

and q T ds

T ds = p du + du

or du = T ds - p du

The enthalpy is defined as the sum of in-ternal energy and the flow work (i.e., pressurevolume product). Thus for a unit mass,

h = u + pu

or dh = du + p du + u dp

= (T ds - p du) + p du + u dp

= T ds + u dp

The Helmholtz free energy is defined asthe difference between the internal energy andthe product to temperature and entropy. Thusfor a unit mass,

f = u - Ts

or df = du - t ds - s dT

= (T ds - p dv) - T ds - s dT

= - p dv - s dT

The Gibbs free energy is defined as thedifference between enthalpy and the productof temperature and entropy. Thus for a unitmass

g = h - Ts

or dg = dh - T ds - s dT

= (T ds + v dp) - T ds - s dT

= v dp - s dT

Since u, h, f and g are properties of thesystem, the relations are often referred to asthe property relations (also called the Gibbsrelations). These relations are applicable toall processes whether reversible orirreversible, and whether in a closed systemor in an open system.

Check for a property:

Consider z to be a function of twoindependent variables x and y. Mathematically

z = f(x, y)

ory x

z zdz dx dy

x y

where z is an exact differential, i.e., itrepresents a thermodynamic property. Letthe partial derivative in the above identity



be written as

y x

z zM and N

x y

Equation may then be written as

dz = M dx + N dy

Further:2

y

M z z

y y x x y

and2

x

N z z

x x y x y

Thusyx

M N

y x

This leads us to conclude that consideringthe expression: dz = M dx + N dy

If z represents thermodynamic propertyof the system, then

yx

M N

y x

Maxwell Relations:

Consider the expression: du = T ds - p dv

Here internal energy u is athermodynamic property and the expressionindicates that internal energy varies withentropy s and volume v. That is

u = f(s, v)

orv s

u udu ds dv

s v

Equating the coefficients of ds and dv, wemay write

v s

u uT and p

s v

When differentiated, we obtain

2 2

s v

T u p uand

v v s s v s

s v

T p

v s

Alternatively: Comparing the givenidentity with dz = M dx + N dy

M = T N = -p

x = s y = v

Since z corresponds to u which is athermodynamic property, the followingidentity must hold good:

, . .,y s vx

M N T pi e

y x v s

Consider the expression:

dh = T ds + v dp

Here enthalpy h is a thermodynamicproperty and the expression indicates thatenthalpy is a function of entropy and pressure.That is

h = f(s, p)

orp s

h hdh ds dp

s p

Equating the coefficients of ds and dv, wemay write

p s

h hT and v

s p

When differentiated, we obtain



2 2

ps

T h v hand

p s p s s p

ps

T v

p s

Alternatively: Comparing the givenidentity with dz = M dx + N dy

M = T N = v

x = s y = p

Since z corresponds to h which is athermodynamic property, the followingidentity must hold good,

, . .,y px s

M N T vi e

y x p s

T-ds Equations( , )u f T v

orv T

du udu dT dv

T v

vT

uc dT dv

v

... (a)

Invoking property relation :

du T ds p dv ... (b)

Comparing identities (a) and (b), andequating the coefficients

vv

sc T

T

Introducing total differential of ds interms of ( , )s f T v

v T

s sds dT dv

T v

vT

dT sc dv

T v

From Maxwell relation :

v T

p s

T v

vv

dT pds c dv

T T

The term v

dTc

T gives change in entropy

at constant volume and the term

v

pdv

T

represents change in entropy

at constant temperature.

Equation is known as first T ds equationand is often recast as

....vv

pT ds c dT T dv

T

(ii) Consider variation of enthalpy h withtemperature T and pressure p. Then

( , )h f T p

orp T

h hdh dT dp

T p

... ( )pT

hc dT dv c

v

Invoking property relation :

... ( )dh T ds v dp d



Comparing identities (c) and (d), andequating the coefficients,

pp

sc T

T

Introducing total differential of ds interms of ( , )s f T p

p T

s sds dT dp

T p

p

T

dT sc dp

T p


pT

s v

p T

pp

dT vds c dp

T T

The term p

dTc

T gives change in entropy

at constant pressure and the termp

vdp

T

represents change in entropy at constanttemperature.

Equation is known as second T dsequation and is often recast is

pp

vT ds c dT T dp

T

HEAT CAPACITY RELATIONS

Equating the first and second T dsequations, we get

vv

pc dT T dv

T

pp

vc dT T dp

T

( )p vor c c dT

v p

p vT dv T dp

T T

pv

p v p v

vp T dpT dvTT

or dTc c c c

From the fundamental relation ( , )T f p v

pv

T TdT dp dv

p v

Comparing identities (i) and (ii), and equatingthe coefficients

v

pp v

pT

TT

c c v

p vv p

p vc c T ... (iii)

T T

Following cyclic relationship holdsgood among the thermodynamic variables p,T and v

v p T

p T v1 .... (iv)

T v p

From identities (iii) and (iv), we have

p vp

p T

1 vc c T

TT vv p



2

p T

v pT

T v

The following important conclusions canbe drawn from the above thermodynamicrelation :

(i) For any substanceT

p

v

is negative,

and the term

2

p

v

T

is always positive.

Obviously p v(c c ) is positive and

accordingly cp is always greater than c

v.

(ii) p vc c as T 0. Apparently cp

equals cv at zero absolute temperature.

(iii) For solids and liquids, the volumechange with temperature is negligible underconstant pressure conditions. That is

p

v0.

T

That gives p vc c 0, i.e.,

p vc c . For water, such a situation exists at

04 C where specific volume is minimum or

density is maximum.(iv) For an ideal gas: pv = RT

p

v R v

T p T

2T p

p RT RT

v v v v

2

p v 2

v RTc c T R

T v

(v) Quite often, it is considered worthwhile to express equation in terms of volume

expansivity and isothermal compressibility

.• Volume expansivity (coefficient of

volume expansion) is defined as the change

in volume with change in temperature per unitvolume, keeping the pressure constant.

p

1 v

v T

where the subscript p indicates that thepartial differential is to be taken at constantpressure.

PAGE – 284 FIGURE

With reference to volume-temperaturediagram at constant pressure as shown in Fig.the slope at any point A is given by

p

v

T

This slope divided by volume at that point

gives the volume expansivity .For a perfect gas : pv = RT

orRT

vp

and therefore

p

dv R

T P



1 R 1 v 1That gives :

v P v T T

Apparently, the volume expansivity of aperfect gas is a function of temperature only.It varies inversely with absolute temperatureand is independent of both pressure andvolume.

* Isothermal compressibility

T and adiabittic compressibility S

In thermodynamics, compressibility isa measure of the relative volume change of afluid in response to a pressure change. That is

1 v

v p

This statement is, however, incompletebecause for any fluid system, the magnitudeof compressibility depends strongly onwhether the process is isothermal or adiabatic.Accordingly we define the isothermalcompressibility as

T

T

1 v

v p

where the subscript T indicates that the partialdifferential is to be taken at constanttemperature. The negative sign implies that apositive pressure increment results in anegative volume increment, i.e., an increasein pressure causes a decrease in volume.Isothermal compressibility thus representsthe change in volume with the change inpressure per unit volume, keeping thetemperature constant. With reference topressure-volume diagram at constanttemperature as shown in Fig. the slope at any

point is given byT

v

p

. This slope divided

by volume at that point gives the isothermal

compressibilityT .

For a perfect gas : pv = RT

orRT

vp

and therefore

2T

v 1RT

p p

That gives :

T 2

1 1RT

v p

2

1 1 1pv

v p p

FIGURE – 285

Apparently the isothermalcompressibility of a perfect gas variesinversely with pressure and is independent ofboth temperature and volume.

The adiabatic compressibility is defined as



S

S

1 v

v p

where the subscript s indicates that the partialdifferential is to be taken at constant entropy.The adiabatic compressibility thus representsthe change in volume with change in pressureper unit volume when entropy is kept constant.

Considering changes in volume due tochanges in pressure and temperature,

v f (p, T)

orP T

v vdv dT dp

T p

p T

v 1 v 1 vdT dp

v v T v p

dT dp

The above correlation expresses volumechange in terms of volume expansivity andisothermal compressibility.

Equation can be rewritten as

2

2p

p v

T

1 vv T

c c Tv Tv1 vv p

The above identity is called the Mayerrelation.

Ration of specific heats : The ratio of

p vc and c is given as

p p p

v

v v

s sT

c T T

s sc TT T

From the cyclic correlations

p Ts

s T p1

T p s

v s T

s T vand 1,

T v s

p s T

s p s

T T v

s sT

s T T

p s pT p vv s pT v v

Since is greater than 1,

s T

p p

v v

The indicates that on p – v plot, the slopeof an isentropic is greater than that of anisotherm (Fig....)

FIGURE – 286

Volume change in terms of volumeexpansivi ty and isothermalcompressibility :

Considering changes in volume due tochanges in pressure and temperature,

v f (p, T)



p T

v vor dv dT dp

T p

p T

dv 1 v 1 vdT dp

v v T v p

Since v

T T

1 v vor

v p p

... (i)

T T

T T

1 v vor v

v p p

...(ii)

T

dv 1 1v dT ( v) dp

v v v

TdT dp Frome cyclic property of differentials,

v p T

p T v1

T v p

Substituting the values ofp

vand

T

T

v

p

from identities (i) and (ii), we obtain

Tv

p 1( v) 1

T v

v T T

p vor

T v

That is, the ratio of coefficient of volume

expansion and isothermal compressibility

T represents the change in pressure withtemperature when volume is kept constant.

Further, from expression

s T

ST

p / v v / p

p / v ( v / p)

T T

S S

1/ v ( v / p)

( 1 / v) ( v / dp)

Apparently the adiabatic exponent (the

ratio of specific heats p vc / c ) is equal to the

ratio of isothermal compressibility and

adiabatic compressibility. Since is always

greater than unity, the above expression

indicates that T is always greater than SRELATIONS FOR INTERNAL ENERGYAND ENTHALPY

(a) When a system undergoes aninfinitesimal reversible process, the changein internal energy between two equilibriumstates is given by

du T ds p dv The first T ds equation gives

vv

pT ds c dT T dv

T

vv

pdu c dT T dv p dv

T

vv

pc dT T p dv .... (i)

T

Considering the functional relationship

u f (T, v) which expresses internal energy

u as a function of temperature T and volume v.

u f (T, v)

v T

u uor du dT dv .... (ii)

T v

The equivalence between (i) and (ii)gives

vv

uc

T

T V

u pand T p

v T



The above identity is called the energyequation.

(b) When a system undergoes aninfinitesimal reversible process, the changein enthalpy between two equilibrium states isgiven by

dh T ds v dp The second T ds equation gives

pp

vT ds c dT T dp

T

pp

vdh c dT T dp vdp

T

pp

vc dT v T dp ... (iii)

T

Considering the functional relationship

h = f(T, p) which expresses enthalpy as afunction of temperature T and pressure p.

h f (T, p)

p T

h hor dh dT dp ... (iv)

T p

The equivalence between (iii) and (iv)gives

pp

hc

T

pT

h vand v T

p T

EXAMPLESet up the following expression for the

specific heats at constant volume and atconstant pressure :

2v

2T v

c pT

v T

2p

2

T p

c vand T

p T

Further, proceed to show that for aperfect gas obeying pv = RT, the specific heats

p vc and c are function of temperature alone.

Solution : The specific heat at constantvolume is

vv

sc T

T

2v

T

c sor T

v T v


T v

s p

v T

2 2

2

v

s por

v T T

2v

2T v

c pT

v T

which is the required result.

For the perfect gas :RT

pv

2

2

p R por and 0

T v T

v

T

cHence 0

v

vc is independent of volume and is a

function of temperature alone.(b) The specific heat at constant

pressure is

pp

sc T

T



2p

T

c sor T

p T p


pT

s v

p T

2 2

2

p

s vor

p T T

2p

2

T p

c v

p T


RT For a perfect gas,RT

vp

2

2

v R vor and 0

T P T

p

T

cHence 0

p

pc is independent of pressure and is

a function of temperature alone.EXAMPLE

Determine the difference in

p vc and c for water at 300 K for which

coefficient of volume exp ansion 4 12 10 K

isothermal compressibility a4 14.85 10 MPa and

specific volume 3v 0.001003 m / kg(b) What would be the value of

p v(c c ) for an ideal gas.

Solution : The difference in specificheats for any substance is given by

2

p vc c T v

Therefore, for liquid water

4 2

p v 4 6

(2 10 )c c

4.85 10 10

300 0.001003

24.816 J / kg K

0.0248 kJ / kg K(b) For an ideal gas : pv = RT

p

1 v p RT

v T RT T p

p R 1

RT p T

T

1 v

v p

T

p RT

RT p p

2

p RT 1

RT p p

2

p v 2

p pvc c T v Tv R

T R

EXAMPLE

Show that for a perfect gas, thedifference between the specific heats

p v(c c ) can be expressed as

p vT p

u vc c p

v T

T

upv v

v



where is the coefficient of volume

expansion.Solution : The property relations for

internal energy u and enthalpy h areT ds = du + p dv

and T ds = dh – v dp du + p dv = dh – v dp ...(i)From the functional relations u = (T, v)

and h = f (T, p), we have

v T

u udu dT

T v

vT

udv c dT dv

v

p T

h hand dh dT

T p

p

T

hdp c dT dp

p

Substituting the values of du and dh inexpression (i), we get

vT

uc dT dv p dv

v

p

T

hc dT dp v dp

p

vT

uor c dT p dv

dv

p

T

hc dT v dp

p

The above equation is valid for anyprocess. Writing it for the case when dp = 0,we have

p v p pT

u(c c ) (dT) p (dv)

v

p vT p

u vor c c p

v T

which is the required result.(b) The coefficient of volume expansion

p is defined as

p

1 v

v T

and accordingly, we can write

p vT

uc c p v

v

T

upv v

v

EXAMPLE 11.5

Set up a T ds relation in the followingform :

v

TT ds c dT dv


expansion, a is the isothermal compressibility,and the other symbols have their usualmeanings.

Solution : The given relation expressesentropy s as a function of temperature T andvolume v. That is

s f (T, v)

v T

s sor ds dT dv

T v

It is known that:

T v

s p

v T

.... Maxwell relation

vv

sand T c

T



Making these substitutions, we get

vv

pT ds c dT T dv

T

The parameters and are defined as

p

1 v

v T

T

1 vand

v p

p T

1 v v

v T ( v / p)

p T

v p.... (ii)

T v

The cyclic relation between p, v and Tgives

T p v

p v T1

v T p

p T v

v p 1or

T v ( T / p)

v

p... (iii)

T

From relations (ii) and (iii),

v

p

T

When the value ofv

p

T

as obtained

above is substituted in expression (i), weget

v

TT ds c dT dv

which is the required relation.EXAMPLE

Set up a T ds relation in the followingform

pT ds c dT v T dp


expansion. The other symbols have their usualmeanings.

Solution : The given relation expressesentropy s as a function of temperature T andpressure p. That is

s f (T, p)

p T

s sor ds dT dp

T p

It is known that :

pT

s v

p T

.... Maxwell relation

pp

sand T c

T

Making these substitutions, we get

pp

vT ds c dT T dp ... (i)

T

The coefficient of volume expansion isdefined as

p p

1 v vor v

v T T

When this value ofp

v

T

is substituted

in expression (i), we get

p pT ds c dT v T d

which is the required relation.



EXAMPLESet up the following relation for the

change in internal energy of a substance fromthe measured values of temperature andvolume

v

Tdu c dT p dv


expansion, a is the isothermal compressibility,and the other symbols have their usualmeanings.

Proceed therefrom to show that internalenergy of an ideal gas is independent ofpressure and volume and is a function oftemperature only.

Solution : The given relation expressesinternal energy u as a function of temperatureT and volume v. That is

u f (T, v)

v T

u uor du dT dv

T v

Since vv

udefines c ,

T

we can write

vT

udu c dT dv ... (i)

v

Consider the property relation :

du T ds p dv Keeping T constant, let this relation be

divided by dv. That gives

T T

u sT p

v v

Substituting the Maxwell relation

T v

s p, we get

v T

T v

u pT p ... (ii)

v T

The cyclic relation between p, v and T gives

T p v

p v T1

v T p

v T p

p p vor ... (iii)

T v T

The parameters P and a are defined as

p

1 v

v T

T

1 vand

v p

p T

1 v v

v T ( v / p)

p T

v p... (iv)

T v

Relations (iii) and (iv) are then combinedto give

v

p

T

When the value ofv

p

T

as obtained

above is substituted in expression (ii), we get

T

u Tp

v

The above correlation is then substituted

in expression (i) and we obtain

v

Tdu c dT p dv



which is the required relation.(b) For an ideal gas pv = RT

p

1 v

v T

p

p RT

RT T p

p R 1

RT p T

T

1 v

v p

T

p RT

RT p p

2

p RT 1

RT p p

v v

Tpdu c dT p dv c T

T

Thus the internal energy of an ideal gasis independent of pressure and volume and isa function of temperature only.EXAMPLE

Set up the following relation for thechange in enthalpy of a substance from themeasured values of temperature and pressure

pdh c dT v (1 T) dp

where is is the coefficient of volumeexpansion. The other symbols have their usualmeanings.

Proceed therefrom to show thatenthalpy of an ideal gas is independent ofpressure and volume and depends only ontemperature.Solution : The given relation expressesenthalpy h as a function of temperature T and

pressure p. That is

h f (T, p)

p T

h hor dh dT dp

T p

pp

hSince defines c ,

T

we can write

p

T

hdh c dT dp ... (i)

p

From the property relation :dh = T ds + v dp

Keeping T constant, let this relation bedivided by dp. That gives

T T

h sT v

p p

Substituting the Maxwell relation , we get

pT

s v

p T

, we get

pT

h vT v

p T

The coefficient of volume expansion isdefined as

p

1 v

v T

T

hv t v

p

v (1 T) ...(ii)

When the value ofT

h

p

as obtained

above is substituted in expression (i), we get

pdh c dT v (1 T) dp



which is the required expression.(b) For an ideal gas pv = RT

p

1 v

v T

p

p RT

RT T p

p R 1

RT p T

p p

1dh c dT v 1 T c dT

T

Thus the enthalpy of an ideal gas isindependent of pressure and volume and is afunction of temperature only.EXAMPLE

Over a certain range of pressures andtemperatures, the equation of state of a certaingas is prescribed by the relation

3

RT Cv

p T where C is a constant

Set up expressions for a change inenthalpy and entropy of the gas. Consider thechange to occur under isothermal conditions.Solution : From the relations

dh = T ds + v dp

pp

vand T ds c dT T

T

dp; we have

pp

vdh c dT T dp v dp

T

For an isothermal process (T = constant)

Tp

v(dh) v T dp ... (i)

T

From the given equation of state :

3

RT Cv

p T

4p

v R 3C

T p T

p

vand v T dp

T

3 3

RT C RT 3Cdp

p T p T

3

4Cdp

T

The identity (i) may then be written as

T 3

4C(dh) dp

T

Upon integration between the state point1 and state point 2, we have

2 1 2 1 T3

4Ch h (p p )

T

1 2 T3

4C(p p )

T

(b) From the T ds relation :

pp

vT ds c dT T dp,

T

we have

pp

dT vds c dp

T T

For an isothermal (constant temperatureprocess),

Tp

v(ds) dp

T

From the given equation of state:

3

RT Cv

p T

4p

v R 3C

T p T



T 4

R 3C(ds) dp

p T

Upon integration between the state point

1 and state point 2,

22 1 T e 2 1 T4

1

p 3C(s s ) R log (p p )

p T

1e 1 2 T4

2

p 3CR log (p p )

p T

EXAMPLESet up the T ds equation in the form

v ppv

T TT ds c dp c dv

p v

Further, proceed to show that this rela-tion can be recast as

pv

v

cT ds c dp dv

where is the coefficient of volume ex-

pansion and a is the isothermal compressibil-ity factor.

Solution : Consider variation of entropys with pressure p and specific volume v.

Then s f (p, v)

pv

s sor ds dp dv

p v

v v

s Tdp

T p

p p

s Tdv

T v

Substituting the relations:

pv

v p

ccs sand ,

T T T T

we can write

pv

pv

cc T Tds dp dv

T p T v

v ppv

T Tor T ds c dp c dv ... (i)

p v

which is the required expression.This identity is often known as Third T

ds equation.(b) The parameters (3 and a are defined

as

p

1 v

v T

T

1 vand

v p

p p

v T 1Then : v or

T v v

From the cyclic relationship

T pv

T p v1

p v T

v

T p

T 1or

p vpv T

T

p

vp

vT

Making the substitution forv

T

p

and

p

T

v

in expression (/), we get



v p

1T ds c dp c dv

pv

cc dp dv

which is the required result.EXAMPLE

Set up ffe following relations for a gaswhich conforms to the van der Waal’s equa-tion of state.

(a) p v 2 3

Rc c

1 2a (v b) RT v

(b) v

RcT(v b) constant for an isen-

tropic process where the symbols have theirusual meanings.

Solution : From generalised relation

for p vc and c , we have

p vv p

p vc c T

T T

The van der Waals equation of state is

2

ap (v b) RT

v

2

RT aor p

v b v

Differentiating p with respect to T atconstant volume, we get

v

p R

T v b

p vp

RT vc c

v b T

Again when volume v in the van derWaal’s equation is differentiated with respectto temperature T treating pressure p as con-stant, we get

2 3p p

a v 2a vp (v b) R

v T v T

3p p

RT v 2a vor (v b) R

v b T v T

p2 3

v Ror

T RT 2a[v b]

(v b) v

3

3 2

R(v b) v

RT v 2a (v b)

3

p v 3 2

RT R(v b) vc c

v b RTv 2a (v b)

3

3 2 3

R vRT

RT v [1 2a (v b) / RT v ]]

2 3

R

1 2a (v b) / RT v

(b) Consider the first T ds equation

vv

pT ds c dT T dv

T

The van der Waal’s equation of state is

2

ap (v b) RT

v

2

RT aor p

v b v

Differentiating p with respect to T treat-ing v as constant, we get

v

p R

T v b



v

RTThen T ds c dT dv

v b

v

dT Ror ds c dv

T v b

At constant entropy ds = 0 and, there-

fore v

dT Rc dv 0

T v b

v

dT R dvor 0

T c v b

By integration:

e e ev

Rlog T log (v b) log

c constant

R / Ce eor log T (v b) log constant

vR / cor T(v b) cons tan t EXAMPLE

Set up the expressions for changes in in-ternal energy, enthalpy and entropy for a gaswhich obeys the van der Waal’s equation

2

ap (v b) RT

v Solution : When a system undergoes an

infinitesimal reversible process, the changein internal energy between two equilibriumstates is given by

vv

pdu c dT T p dv

T

From the given van der Waal’s equation

2

RT ap

v b v

v

p R Rand 0

T v b v b

2

2 1

1

u u du

2 2

v

1 1

RTc dT p dv

v b

2

v

1

c dT 2

21

RT RT adv

v b v b v

2 2 2

v 21 1

ac dT dv

v

v 2 11 2

1 1c (T T ) a

v v

which is the desired expression for

change in internal energy.Under constant temperature conditions,

v 2 1c (T T ) 0 and accordingly

2 1 T1 2

1 1(u u ) a

v v

(b) When a system undergoes an

infinitesimal reversible process, the changein enthalpy between two equilibrium states is

pp

vdh c dT v T dp

T

Under constant temperature conditions,

dT = 0 and then

T Tp

dv(dh) v T (dp) ... (a)

dt

Considering the functional relation

p f (v, T)

T v

p pdp dv dT

v T



TT

por (dp) dv 0 as dT 0

v

Substituting -the above value of (dp)

T in

identity (a), we get

Tp T

v p(dh) v T dv

T v

T p T

p v pv dv ... (b)

v T T

Using the cyclic relation for p, v and T,we have

p Tv

v T p1

T p v

p T v

v p 1or

T v T / p

v

p

T

Substituting this value in expression (b),

we get

T

T v

p pdh v T dv

v T

From van der Waal’s equation

2T T

p RT a

v v v b v

2 3

RT 2a

(v b) v

2v

p RT a R

T T v b v (v b)

T(dh)

2 3

RT 2a Rv T dv

(v b) v v b

Upon integration between state points1 and 2,

2

2 1 T 21

v(h h ) RT dv

(v b)

2 2

21 1

dv dv2a RT

v v b

2e

1 2 1

v b 1 1RT log b

v b v b v b

2e

2 1 1

1 1 v b2a RT log

v v v b

2 1 2 1

1 1 1 1bRT 2a

v b v b v v

which is the desired expression forchange in enthalpy.

(c) Consider the first T ds equation,

vv

pT ds c dT T dv

T

The van der Waal’s equation of state is

2

ap (v b) RT

v

2

RT aor p

v b v

2v v

p RT a RThen

T T v b v v b

v

RTT ds c dT dv

v b

v

dT Ror ds c dv

T v b

Upon integration between state points

1 and 2



2 2

2 1 v

1 1

dT dvs s c R

T v b

2 2v e e

1 1

T v bc log R log

T v b

which is the desired expression forchange in entropy.EXAMPLE

Show that for any fluid

(a)T T v

h p pv T

v v T

(b)pT

h vv T

p T

Further, proceed to prove that for anypure substance obeying van der Waal’sequation, the enthalpy is given by

RTb 2ah f (T)

v b v

where f(T) is arbitrary and R, a, b are theconstants appearing in van der Waal’sequation.

Solution : For unit mass, the enthalpychange is given by

dh T ds v dp and from first T ds equation

vv

pT ds c dT T dv

T

vv

pdh c dT T dv v dp

T

Under constant temperature conditionsdT = 0 and therefore

Tv

p(dh) T dv v dp

T

T v T

h p por T v

v T v


(b)TT T

h h v

p v p

v T T

p p vT v

T v p

T T

h v

v p

v T T

p p vT v

T v p

v T

p vT v

T p

From cyclic relation :

v p T

p T v1

T v p

v pT

p v vor

T p T

Making this substitution, we get

pT

h vT v

p T

p

vv T

T

which is the required result.(c) From van der Waal’s equation

2

RT ap

v b v

The differentiations with respect to v atconstant temperature T, and with respect to 7at constant v give

2 3T

p RT 2a

v (v b) v

v

p Rand

T v b



2 3T

h RT 2av

v (v b) v

RT

v b

v2 2

RT 2a RT

(v b) v v b

2 2

RTv RT(v b) 2a

(v b) v

2 2

RTb 2a

(v b) v

That gives :

RTbh 2a f (T)

v b

Obviously h depends on T and v.EXAMPLE

The following expressions for theequation of state and specific heat c areobeyed by a certain gas:

2RTv T

p

and c

p = A + BT + Cp

where , a, B and C are constants.

Obtain an expression for the joule-Thomsoncoefficient.

Solution : Joule-Thomson coefficientrepresents the numerical value of the slopeof an isenthalpy on T-p diagram at any point

and is denoted by ju . Mathematically,,

j

h

Tu

p

The property relation for dh is :

dh T ds v dp From second T ds equation, we have

pp

vT ds c dT T dp

T

pp

vdh c dT T dp v dp

T

pp

vc dT T v dp

T

For a constant enthalpy process dh - 0and therefore,

p h0 (c dT)

ph

vT v dp

T

jpph

T 1 vor u T v ... (a)

p c T

For the given gas:

2RTv T

p

p

v R2 T

T p

Substuting the given values of c , v and

pp

vc , v and

T

as calculated above in

identity (a), we get

j

1u

A BT Cp

2R RTT 2 T T

p p

1

A BT Cp

2RT RT2 T T

p p

221 T

( T )A BT CP A BT Cp

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synergyinstitute.ac.in · ENGINEERING THERMODYNAMICS 1 SIT, BBSR, DR. R. C. Nayak, ETD, 4th Sem, Mech. THEORY PARAMETERS : P - Pressure (bar) 1 bar = 1 105 N /m2 1 N /m2 = 1 Pascal