Chapter four

Reversible and Irreversible Processes

(A) Reversible non-flow processes:-

1-Constant volume process:-

For Steam:

Q=(u2-u1)+W

2

1

0

pdW

Q=u2-u1

For mass; m of the working substance

Q=U2-U1

For Ideal gas:

W=0

Q=u2-u1=cυ(T2-T1)

Note:- All the heat supplied in a constant volume process

goes to increasing the internal energy.

2-Constant pressure process:-

For Steam:

)()(

)()(

)(

)(

1122

1212

12

12

2

1

2

1

pupuQ

puuQ

WuuQ

pdppdW

For mass; m of a fluid ; Q=H2-H1

Q=h2-h1

For Ideal gas:

)(

)(

1212

12

TTchhQ

pW

p

Example(4.1):

A mass of 0.05 kg of a certain fluid is healed at a constant pressure of 2 bar until the volumeoccupied is 0.0658 m3. Calculate the heat supplied and the work done:(i) when the fluid is steam, initially dry saturated;(ii) when the fluid is air, initially at 130°C.

Solution:

m =0.05kg, p1=p2=2bar, V2=0.0658 m3, Q=?, W=?

(i):

kgJW

kgmm

V

kgm

pW

g

/86080)8856.0316.1(10*2

/316.105.0

0658.0

/8856.0

)(

5

32

31

12

Work done by the total mass=86080*0.05=4304 J

Q=h2-h1

h1 =hg=2707kJ/kg at p=2bar

h2 =3072kJ/kg at p=2bar and υ=1.316 m3/kg

Q=3072-2707=365 kJ/kg

Heat supplied =365*0.05=18.25 kJ

(ii): air; T1=130°C

35

1

11

111

12

0289.010*2

403*287*05.0

)(

mp

mRTV

mRTVp

pW

υ1=0.578 m3/kg

υ2=1.316 m3/kg

W=2*105(1.316-0.578)=147600 J/kg

Total work=147600*0.05=7380 J=7.38 kJ

Or

kJTTmcQ

kJTTmRW

KTmRTVp

TTRpW

p 83.25)403917(*005.1*05.0)(

38.7)403917(287.0*05.0)(

917287*05.0

0658.0*10*2

)()(

12

12

5

2222

1212

3-Constant temperature process (Isothermal process):-

A process at constant temperature is called an isothermal process.

For Steam:

W=shaded area

Q=(u2-u1)+W

For Ideal gas:

1

211

332211

ln

........

tan

2

1

2

1

2

1

pd

cdc

pdW

ppp

cptconsp

RTp

1

222 ln

pW

1

211 ln

pW

12121212

12

2

1

11

2

121

2

1

1

2

)(

)(

ln

ln

uuTbutTTTcuu

WuuQ

p

pRTW

RTp

p

ppW

p

p

Example(4.2):

Steam at 7 bar and dryness fraction 0.9 expands in a cylinder behind a piston isothermallyand reversibly to a pressure of 1.5 bar. Calculate the change of internal energy and thechange of enthalpy per kg of steam, The heat supplied during the process is found to be 400kJ/kg, calculate the work done per kilogram of steam.Solution:At p=7 bar, Ts=165°CAt p=1.5 bar and T=165°C is superheatedBecause at p=1.5 bar, Ts=111.4°C

kgkJu

xuuxu gf

/3.23852573*9.0696*)9.01(

)1(

1

1

Interpolation from superheated tables at 1.5 bar and 165°C

kgkJu

kgkJu

u

/8.2602

/8.26022580)25802656(50

152580

150165

25802656

150200

2

Gain in internal energy=u2-u1=217.5 kJ/kg

kgkJxhhh fgf /3.25572067*9.06971

Interpolation from superheated tables

WQ

kgkJhh

kgkJh

kgkJh

/7.245

/2803

/28032773)27732873(50

15

12

2

kgkJW

WuuQ

/5.1825.217400

)( 12

Example(4.3):

1 kg of nitrogen (molar mass 28 kg / kmol) is compressed reversibly and isothermally from1.01 bar, 20°C to 4.2 bar. Calculate the work done and the heat flow during the process.Assume nitrogen to be a perfect gas.

Solution:m =1kg, N2, M=28, T=C, p1=1.01 bar, T1=T2=20°Cp2=4.2 bar, W=?, Q=?

kgkJW

KkgJM

RR

p

pRTW

o

/1242.4

01.1ln293*297

./29728

8314

ln2

1

The work input=124kJ/kgQ=W=-124kJ/kg

The heat is rejected

4-Adiabatic process (Q=0):-An adiabatic process is one in which no heat is transferred to or from the fluid during theprocess. Such a process can be reversible or irreversible.For reversible adiabatic non-flow process:-For Steam:

21

12

0

)(

uuW

Q

WuuQ

Note:- For an adiabatic process to take place, perfect thermal insulation for the system mustbe available.

For Ideal gas:

dWdudQ

For a reversible process, pddW 0 pddudQ

For a perfect gas ; dTcduRTp

0

)()(

pddu

RdTdppd

RTdpd

Adiabatic process for perfect gas

2211 pp

tconsR

C

p

TC

p

RTp

p

RTRTp

tan)(1

tconsep

Cp

CpCp

d

p

dp

ppddp

pdpddppd

pddppd

pddppd

c

RRc

pddppdR

c

pddTc

C tan

)ln(

lnlnlnln

0

0

0

0)1(

01

1

11

0

0

Cp

)(1

2

2

1 p

p

tconsR

CTC

RT

RTpRTp

tan1

CT 1

1

1

2

2

1 )(

T

T

The work done in an adiabatic process per kg of gas

1

)(

)(

2121

12

Rc

TTcuuW

WuuQ

222111 , RTpRTpRTp

**: A reversible adiabatic process for a perfect gas is shown on a p-υ diagram.

1

11

2211

2211

12

11

1 2

1

2

1

2

1

2

1

ppW

ppc

ccd

cdc

pdW

Example(4.4):

1 kg of steam at 100 bar and 375°C expands reversibly in a perfectly thermally insulatedcylinder behind a piston until the pressure is 38 bar and the steam is then dry saturated.Calculate the work done by the steam.Solution:m =1kg, p1=100 bar, T1=375°C, Q=0, p2=38 bar, W=?at p=100 bar Ts=311°C < T1(375°C)

the steam is superheated at point (1)W=u1-u2

From superheated table we find u1

At 100 bar and 375°Cυ =υ1=0.02453 m3/kg, h=h1=3017 kJ/kg

)1

(

2

1

2

1 )(

p

p

T

T

1

)( 21

TTRW

12211

ppW

Example(4.5):

Air at 1.02 bar, 22°C, initially occupying a cylinder volume of 0.015 m3, is compressedreversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate the finaltemperature, the final volume, and the work done on the mass of air in the cylinder.Solution:air, p1=1.02 bar, T1=22°C, V1=0.015 m3

p2=6.8 bar, T2=?, V2=?, W=?for adiabatic process

KT

p

pTT

p

p

T

T

3.507)02.1

8.6(295

)(

)4.1

4.0(

2

)1

(

1

212

)1

(

2

1

2

1

The final temperature =507.3-273=234.3°C

kgRT

Vpm

mRTpV

kgkJTTcuuW

mV

p

pVV

V

V

p

p

0181.0295*287

015.0*10*02.1

/4.152)3.23422(718.0)(

00387.0)8.6

02.1(015.0

)()(

5

1

11

2121

3)4.11(

2

)1(

2

112

1

2

2

1

The total work=-152.4*0.0181=2.76kJ

kgkJW

kgkJu

baratpuu

kgkJu

phu

g

/7.16926027.2771

/2602

38

/7.27711000

02453.0*10*1003017

2

2

5

1

5- Polytropic processes:-

It is found that many processes in practice approximate to a reversible law of the form

Cp n ; where n is a constant. Both vapors and perfect gases obey this type of law closely

in many non-flow processes. Such processes are internally reversible.

For Steam:

For any reversible process, pdW

cp n

Where; n and c : constant

1

11

1222

1111

2211

12

11

1 2

1

2

1

n

ppW

ppc

nc

ncd

cW

nnnn

nn

nnn

n

For Ideal gas:

pυ=RT

for polytropic process; Cp n

122

111

1)(

nn

nn

TT

cR

cTc

RT

RTp

………………………….(1)

12211

n

ppW

n

p

p)(

1

2

2

1

1

1

2

2

1 )( n

T

T

n

n

n

n

n

nn

n

nn

nn

p

T

p

T

c

p

Tc

p

T

tconsR

c

p

Tc

p

RTp

p

RT

1

2

21

1

1

)1

(1

1tan)(

…………………………(2)

Note: It can be seen that these equations are exactly similar to the equations for a reversibleadiabatic process for a perfect gas. This is mean that the reversible adiabatic process for aperfect gas is a particular case of a polytropic process with the index; n equal to γ.

For polytropic process,1

2211

n

ppW

222111 RTpRTp

Or for mass; m

)1)(1()(

)1)(1(

)1(1)(

1

1

1

1)(

)(1

)(1

1

)()()(

1

)(

21

21

21

2121

211212

21

n

nTTRQ

n

nTTRQ

nTTRQ

TTR

TTn

RQ

n

TTRTTcWuuQ

n

TTmRW

1

)()

1( 21

n

TTRnQ

)1

(

2

1

2

1 )( n

n

p

p

T

T

1

)( 21

n

TTRW

Example(4.6):

In a steam engine the steam at the beginning of the expansion process is at 7 bar, drynessfraction 0.95, and the expansion follows the law tconsp tan1.1 , down to a pressure of

0.34bar. Calculate the work done per kg of steam during the expansion, and the heat flow perkg of steam to or from the cylinder walls during the expansion.

Solution:

p1=7 bar, x=0.95, cp 1.1 , p2=0.34bar, W=?, Q=?

kgmbarp

kgmp

p

p

ppp

kgm

baratkgm

xn

ppW

g

n

nnn

g

g

/649.434.0

/05.4)34.0

7(259.0)(

)(

/259.0)2728.0(95.0

)7(/2728.0

1

32

31.111

2

112

1

2

1

1

22211

31

3

1

2211

υg> υ2

the steam is wet at point 2

kgkJQ

kgkJuuxuu

xx

kgkJuuxuu

WuuQ

kgkJW

fgf

gg

fgf

/9.148436)15.24791.2192(

/1.2192)3022472(871.0302)(

871.0649.4

05.4

/15.2479)6962573(95.0696)(

)(

/43611.1

05.4*10*34.0259.0*10*7

22222

2222

11111

12

55

Heat is supplied

Example(4.7):

1 kg of a perfect gas is compressed from 1.1 bar, 27 °C according to law, tconsp tan3.1 until the pressure is 6.6 bar. Calculate the heat flow to or from the cylinder walls.

Wn

Q )1

(

(i) When the gas is ethane (molar mass 30 kg/kmol), which has cp=1.75 kJ/kg K.(ii) When the gal is argon (molar mass 40 kg/kmol), which has cp = 0.515 kJ/kl K.Solution:(i)/ M=30, cp=1.75kJ/kg.K

kgkJQ

kgkJW

KT

p

pTT

p

p

T

Tc

p

T

n

TTRW

KkgkJRcc

KkgkJM

RR

ccRc

c

Wn

Q

n

n

n

n

n

n

p

o

pp

/5.84)8.141)(1188.1

3.1188.1(

/8.14113.1

)6.453300(277.0

6.453)1.1

6.6)(27327(

)()(

1

)(

188.1473.1

75.1

./473.1277.075.1

./277.030

314.8

)1

(

3.1

3.0

2

1

1

212

1

1

2

1

21

21

The heat is supplied

(ii)/ use the same method for argon

Heat is supplied

Note:- The various processes considered in previous sections are cases of the polytropicprocess for a perfect gas.

kgkJQ

kgkJW

KkgkJc

KkgkJR

/4.59)5.106)(1678.1

3.1678.1(

/5.10613.1

)6.453300(208.0

678.1307.0

515.0

./307.0208.0515.0

./208.040

314.8

Where;

cpn

tconsTcpn

tconscpcpn

tconspcpn

,

tan,1

tan,

tan,011

0

State 1→A (constant pressure cooling; n=0)

State 1→B (Isothermal compression; n=1)

State 1→C (Reversible adiabatic compression; n=γ)

State 1→D (Constant volume heating; n=∞)

State 1→A' (constant pressure heating; n=0)

State 1→B' (Isothermal expansion; n=1)

State 1→C' (Reversible adiabatic expansion; n=γ)

State 1→D' (Constant volume cooling; n=∞)

Note:- (γ) is always greater than unity, then process 1→C must lie between processes 1→Band 1→D similarly, process 1→C' must lie between processes 1→B' and 1→D'.

Note:- A vapor may undergo a process according to a law pυ=c. In this case the process isnot isothermal because the characteristic equation of state, pυ=RT , does not apply to vapor.Therefore tables must be used to find the properties at the end states, making use of the factthat 2211 pp .

Example(4.8):

In the cylinder of a steam engine the steam expands from 5.5bar to 0.75bar according to ahyperbolic law. If the steam is initially dry saturated, calculate the work done per kg ofsteam, and the heat flow to or from the cylinder walls.

Solution:

p1=5.5 bar, p2=0.75bar, cp , , W=?, Q=?

kgmkgmbarp

kgmp

p

barpkgkJuu

WuuQ

kgkJp

ppW

p

ppp

barpkgm

pcdc

pdW

g

g

g

/217.2/513.2,75.0

/513.2)75.0

5.5(3427.0)(

5.5/2565

)(

/5.37575.0

5.5ln3427.0*10*5.5ln

5.5/3427.0

ln]ln

3322

3

2

112

11

12

5

2

111

2

1

1

22211

13

1

1

211

2

1

2

1

2

1

υ2 > υg the steam is superheated at point 2

kgkJQ

kgkJu

u

/8.3775.375)25653.2567(

/3.2567271.2513.2

2510

271.2588.2

25102585

2

(B) Irreversible processes:-

1-Unresisted, or free expansion:

**: In a free expansion process, the internal energy initially equals the internal energyfinally.

(For ideal gas

21 TcTc

Tcu

21 uu

0

0

0

)(

12

12

uu

W

Q

WuuQ

21 TT

2-Throttling:

A flow fluid is said to be throttling when there is some restriction to the flow, such as, partlyopen valve, an orifice or any sudden reduction in the cross section of the flow.

22

0,022

22

2

21

1

22

2

21

1

Ch

Ch

WQ

WC

hQC

h

When the velocities C1 and C2 are small, or when C1 is very nearly equal to C2, then thekinetic energy can be neglected.

Therefore for throttling process, the enthalpy initially is equal to the enthalpy finally.

(For ideal gas

21 TcTc

Tch

pp

p

3- Adiabatic mixing:

The mixing of two streams of fluid is quite common in engineering practices, and canusually be assumed to occur adiabatically.

0

0

W

Q

Neglecting change in kinetic energy.

332211

321

hmhmhm

HHH

21 hh

21 TT

3212211 )( hmmhmhm

(For ideal gas

Tch p

3212211 )( TcmmTcmTcm ppp

Example(4.9):

Air at 20 bar is initially contained in vessel A as shown in figure below, the volume of thisvessel is 1m3. The valve is opened and the air expands to fill vessels A and B. Assuming thatthe vessels are of equal volume ,calculate the final pressure of the air.Solution:Air is a perfect gasThe process is free expansion

barV

Vpp

V

V

p

p

mVV

VpVp

TT

uu

10)2

1(20)(

21*22

2

112

2

1

1

2

312

2211

21

21

Note:- The point 1 and 2 lie on isothermal line, but the

process between 1 and 2 cannot be called isothermal, since

the intermediate temperatures are not the same throughout the

process.

Example(4.10):

Steam at 19 bar is throttled to 1 bar and the temperature after throttling is found to be 150°C.Calculate the initial dryness fraction of the steam.

Solution:

The process is throttling, 21 hh

Point 2 at superheated, at p=1bar Ts= 99.6°C

Ts < T

3212211 )( TmmTmTm

From superheated table at p=1bar and T=150°C.

h=2777kJ/kg

989.0)1901(8972777

/2777

1

21

xx

xhhh

kgkJhh

fgf

Note:- States 1 and 2 are fixed but the intermediate states are indeterminate, therefore for avapor, throttling can be used as a means of finding the dryness fraction of wet steam, as inexample.

Reversible Flow Processes:-

)2

()(

022

22

21

21

22

2

21

1

CChhW

QWC

hQC

h

If the kinetic energy terms are negligible small;

)( 21 hhW

For ideal gas: )( 21 TTcW p

Note:- The work done in a reversible adiabatic flow process between two states )( 21 hhW

is not equal to the work done in a reversible adiabatic non-flow process between the samestates )( 21 uuW

Example(4.11):

A gas turbine receives gases from the combustion chamber at7 bar and 650°C, with velocityof 9 m/s. The gases leave the turbine at 1 bar with a velocity of 45 m/s. Assuming that theexpansion is adiabatic and reversible in the ideal case, calculate the work done per kg of gas, For the gases take γ= 1.333 and cp= 1.11 kJ/kg K.Solution:

)2

()(

022

22

21

21

22

2

21

1

CChhW

QWC

hQC

h

For gas; Tch p

kgkJW

Kp

pTT

p

P

T

T

CCTTcW p

/4.393)459(2

1)7.567923(10*11.1

7.567)7

1)(273650()()(

)2

()(

223

333.1

333.01

1

212

1

2

1

2

1

22

21

21