Chapter 7 Thermo

8/13/2019 Chapter 7 Thermo

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Chapter 7

Air Standard Cycles

The power cycles can be classified into two important fields. The first is the

power generation which the work done output of the system such as Heat Engine. The

second is the refrigeration and air conditioning which the work done input to the system

such as Heat Pump. Both of it are usually operating on a thermodynamic cycle. In the

power generation systems, we will use the air standard cycles which the working fluid is

returned to the initial state at the end of the cycle. Some assumptions were made to

study the air standard cycles such as,

(a)The working fluid is air and always ideal gas.(b)All processes are internally reversible.(c)Heat added to the cycle is from external heat source.(d)Heat rejected from the cycle is to the surrounding.

The simplified model of heat engine is a piston cylinder device which called

reciprocating engine and the basic components are shown in Fig. 7-1.

Fig. 7-1 Basic components of reciprocating engine

The piston reciprocates in the cylinder between two fixed positions called the top dead

center (TDC)and the bottom dead center (BDC). The distance between the TDC and

the BDC is called the stroke, Lst. The diameter of the piston is called the bore, d. The


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thermodynamic cycle of reciprocating engine is four processes in two strokes as shown

in Fig. 7-2. The maximum volume when the piston is at BDC. The minimum volume

when the piston is at TDC which is called the clearance volume. The volume between

TDC and BDC is called the displacement volume or swept volume. The ratio between

maximum to minimum volume is called the compression ratio, r.

Fig. 7.2 Thermodynamic cycle of reciprocating engine

min

max

V

Vr=

The mean effective pressure, MEP, is the exerted pressure on the piston during the

power stroke to produce the same net work done.

minmax VV

W

V

WMEP net

stroke

net

==

The mean effective pressure can be used as a parameter to compare the performance of

reciprocating engines of equal size.

Reciprocating engines are classified as spark-ignition (SI) engines and

compression-ignition (CI)engines. In spark-ignition engines, the combustion of the air

fuel mixture is initiated by a spark plug. But in CI engines, the air fuel mixture is self


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149

ignited as a result of the compression temperature. The Otto and Diesel cycles are the

ideal cycles for SI and CI reciprocating engines.

Otto Cycle

The Otto cycle is the ideal cycle for spark ignition, (SI), reciprocating engines.

The thermodynamic analysis of four-stork cycle can be simplified if the air standard

assumptions are used. The Otto cycle is consists of four internally reversible processes

as shown in Fig. 7-3.

Fig. 7-3 Otto cycle and P-v, T-s diagrams

The Otto cycle is executed in a closed system and the change of kinetic and potential

energies are disregarded. The energy balance for any of the processes is expressed in a

unit mass basis as,

Process 12, isentropic compression and ris the compression ratio,


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1

12

1

1

2

1

1

2

2

1

.,

==

=

=

kk

k

rTTrv

v

T

T

v

vr

Process 23, constant volume heat addition, v2= v3,

)( 23 TTCq vin =

Process 34, isentropic expansion,

1

43

1

1

3

4

4

3

3

4

2

1

.,

==

=

==

kk

k

rTTrv

v

T

T

v

v

v

vr

Process 41, constant volume heat rejection, v4= v1,

)( 14 TTCq vout =

Otto cycle net work done is,

outinnet qqw =

The thermal efficiency of Otto cycle is,

)(

)(11

23

14,

TTC

TTC

q

q

q

qq

q

w

v

v

in

out

in

outin

in

net

Ottoth

==

==

1.

14

14

11

1

1

4

14

23

14.

11

11

..11

=

=

=

=

kOttoth

kkkOttoth

r

TT

TT

rrTrT

TT

TT

TT

From the relation of thermal efficiency for Otto cycle, it is clear that the thermal

efficiency is very dependent on the compression ratio, r, and the specific heat ratio, k =

Cp/Cvas shown in Fig. 7-4. The Otto cycle thermal efficiency increases with increasing

the compression ratio rand the specific heat ratio k.


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Fig. 7-4 Dependent of Otto thermal efficiency on rand k

Diesel Cycle

The diesel cycle is the ideal cycle for compression ignition, (CI), reciprocating

engines. The thermodynamic analysis of four-stork cycle can be simplified if the air

standard assumptions are used. The Diesel cycle is consists of four internally reversible

processes as shown in Fig. 7-5.

Fig. 7-5 Diesel cycle P-v and T-s diagrams

Process 12, isentropic compression and ris the compression ratio,

1

12

1

1

2

1

1

2

2

1

.,

==

=

=

kk

k

rTTrv

v

T

T

v

vr

Process 23, constant pressure heat addition, P2= P3, and the cut-off ratio, rc,


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)(

...,//,

23

1

1232323

2

3

TTCq

rrTrTTrTTvvv

vr

pin

c

k

ccc

=

=====


k

c

k

c

c

k

k

c

k

c

k

c

k

rTr

rrrTT

r

rTT

r

r

vr

v

v

v

T

T

....

,

1

1

1

14

1

34

11

2

1

1

3

4

4

3

=

=

=

=

=

=


)( 14 TTCq vout =

Diesel cycle net work done is,

outinnet qqw =

The thermal efficiency of Diesel cycle is,

)(

)(1

)(

)(11

23

14

23

14

,

TTk

TT

TTC

TTC

q

q

q

qq

q

w

p

v

in

out

in

outin

in

net

Dieselth

=

==

==

[ ]11

1

1

11

23

14.

...

.1

)(1

=

=

k

c

k

k

c

DieselthrTrrTk

TrT

TTk

TT

=

)1(

111

1.

c

k

c

kDieselth rk

r

r

From the relation of thermal efficiency for Diesel cycle, it is clear that the thermal

efficiency is very dependent on the compression ratio, r, cut-off ratio, rc, and the

specific heat ratio, k = Cp/Cvas shown in Fig. 7-6. The Diesel cycle thermal efficiency

increases with increasing the compression ratio rand the specific heat ratio k. But the

Diesel cycle thermal efficiency increases with decreasing the cut-off ratio rcand


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Fig. 7-6 Dependent of Diesel thermal efficiency on r, rc, and k

From Fig. 7-6, when the cut-off ratio equal 1, it becomes the thermal efficiency of Otto

cycle. Therefore,

DieselthOttoth ,, >

When both cycles operate on the same compression ratio r. Also, as the cut-off ratio

decreases the thermal efficiency of diesel cycle increased as shown in Fig. 7-6.

Dual cycle

Fig. 7-7 Dual cycle P-v and T-s diagrams


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The ideal power cycle which the heat addition partially at constant volume and constant

pressure processes is called Dual cycle as shown in Fig. 7-7. Compression ratio, r =

v1/v2, cut-off ratio, rc= v4/v3, pressure ratio, rp= P3/P2.

Process 12, isentropic compression,

1

12

1

2

1

1

2 .,

=

= k

k

rTTV

V

T

T

Process 23, heat added at constant volume, v3= v2, or P/T=C

)(

...,

23,

1

12

2

3

23

2

2

3

3

TTCq

rrTrTP

PTT

T

P

T

P

vcvin

k

pp

=

====

=

Process 34, heat added at constant pressure, P = C, or V/T=C

)(

...,.,

34,

1

143

3

434

3

3

4

4

TTCq

rrrTTrTV

VTT

T

V

T

V

pcpin

cp

k

c

=

====

=


k

cp

k

c

cp

k

k

c

k

c

k

c

k

c

k

rrTr

rrrrTT

r

rTT

r

r

V

Vr

V

Vr

V

V

T

T

.....

,

1

1

1

15

1

45

11

1

2

1

5

3

1

5

4

4

5

=

=

=

=

=

=

=

Process 51, heat rejection at constant volume,

)( 15, TTCq vcvout ==

The thermal efficiency of Dual cycle is,

)()(

1

3423,, TTCTTCqqq

q

q

q

qq

q

w

pvcpincvinin

in

out

in

outin

in

net

th

+=+=

=

==

==

)()(

)(1

)()(

)(1

3423

15

3423

15

,TTkTT

TT

TTCTTC

TTC

pv

v

Dualth+

=

+

=

+

=

+

=

+

=

)1(.)1(

111

)1(..)1(

)1.(1

)()(

)(1

1,

1

1

1

1

1

3423

15,

cpp

k

cp

kDualth

cp

k

p

k

k

cp

Dualth

rrkr

rr

r

rrrkTrrT

rrT

TTkTT

TT

For Otto cycle, the cut-off ratio, rc= 1, the thermal efficiency yields,


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155

111,1

11

111

)1(.)1(

111

=

=

+

=

k

p

p

k

cpp

k

cp

kOttothrr

r

rrrkr

rr

r

For Diesel cycle, the pressure ratio, rp= 1, the thermal efficiency yields,

=

+

=

)1(

111

)1(.)1(

111

11,

c

k

c

k

cpp

k

cp

kdieselth rk

r

rrrkr

rr

r

Brayton Cycle

The Brayton cycle is used for gas turbines only where both the compression and

expansion processes take place in a rotating machinery and can be modeled as a closedcycle as shown in Fig. 7-8.

Fig. 7-8 A closed cycle of gas turbine engine

The air standard assumptions are used for ideal cycle and the compression and

expansion processes are internally reversible adiabatic. The heat addition and heat

rejection processes are at constant pressure and occurred in heat exchangers. The four

processes are shown in Fig. 7-9 as follows,

Process 12, isentropic compression in a compressor and the pressure ratiorp= P2/P1,

kk

p

kk

rP

P

T

T /)1(/)1(

1

2

1

2

=

=


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156

Fig. 7-9 Brayton cycle P-v and T-s diagrams

Process 23, heat addition at constant pressure, P2= P3,

)( 23 TTCq pin = Process 34, isentropic expansion in a turbine, and rp= P3/P4,

kk

p

kk

rP

P

T

T /)1(/)1(

4

3

4

3

=

=

4

3

1

23241 ,,

T

T

T

TPPPP ===Q

Process 41, heat rejection at constant pressure, P4= P1,

)( 14 TTCq pout = The Brayton thermal efficiency is,

kk

p

kk

p

kk

p

th

in

out

in

outin

in

net

th

rrTrT

TT

TT

TT

q

q

q

qq

q

w

/)1(/)1(

1

/)1(

4

14

23

14

11

..1

11

=

=

==

==

Examples of Air Standard Cycles

1. The compression ratio of air-standard Otto cycle is 8. At the beginning of thecompression stroke the pressure is 1 bar and the temperature is 17 C. The heat of

800 kJ/kg is added during the constant volume process. Determine:

a. The pressure and temperature at each corner of the cycle,b. The thermal efficiency, andc. The mean effective pressure.


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157

Solution

Process 12, isentropic compression, r = 8.

barrPPrv

v

P

P

KrTTrv

v

T

T

kgmvv

vr

kgmP

RTv

kk

k

kk

k

38.1881.,

245.6668290.,

/10404.08/8323.0,

/8323.01001

290287.0

4.1

12

2

1

1

2

4.01

12

1

1

2

1

1

2

3

2

2

1

3

1

11

====

=

====

=

===

=

==

Process 23, constant volume heat addition, v2= v3,

barT

TPP

KT

TTTCq vin

118.49245.666

45.178038.18

45.1780

)245.666(718.0800),(

2

323

3

323

===

=

==


barrPPrv

v

P

P

KrTTrv

v

T

T

k

k

k

k

k

k

672.28/118.49/,1

986.7748/45.1780/,1

4.1

34

4

3

3

4

4.01

341

1

4

3

3

4

====

=

====

=


kgkJTTCq vout /22.348)290986.774(718.0)( 14 ===

The work done net is,


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158

kgkJqqw outinnet /78.45122.348800 ===

%47.565647.0800

78.451, ====in

netOttoth

q

w

barKPavv

wMEP net 204.6369.620

10404.08323.0

78.451

21

==

=

=

2- An engine operates on the theoretical diesel cycle with a compression ratio of 15. The

heat of 1300 kJ/kg is added at constant pressure for 10 % of the stroke volume. The

pressure and temperature of the air at the beginning of compression are 100 kPa and

27 C. Determine, (a) The cut-off ratio, (b) The pressure and temperature at the end

of each process, (c) The thermal efficiency of the cycle, (d) The mean effective

pressure, and the output power from the engine if the mass flow rate equal 0.15 kg/s.

Solution


kgmvvvr

kgmP

RTv

/0574.015/8323.0,

/861.0100

300287.0

32

2

1

3

1

11

===

=

==

KrTTrv

v

T

T kkk

253.88615300.,4.01

12

1

1

2

1

1

2 ====

=

KParPPrv

v

P

P kkk

265.443115100., 4.1122

1

1

2 ====

=

Process 2

3, constant pressure heat addition, P2= P3,


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159

==

=

==

11.01),(1.0

785.2179

)253.886(005.11300),(

2

1

2

3

2123

3

323

v

v

v

vvvvv

KT

TTTCq pin

( ) ( )

kgmvrv

rr

c

c

/13776.00574.04.2

4.21151.0111.01

3

23 ===

=+=+=


kPaP

r

r

v

rv

v

v

P

P

KT

r

r

v

rv

v

v

T

T

k

c

k

c

k

k

c

k

c

k

64.34015

4.2265.4431

.

277.104715

4.2785.2179

.

4.1

4

1

2

4

3

3

4

14.1

4

11

1

2

1

4

3

3

4

=

=

=

=

=

=

=

=

=

=


kgkJTTCq vout /346.536)300277.1047(718.0)( 14 ===

The work done net is,

kgkJqqw outinnet /654.763346.5361300 ===

%74.585874.01300

654.763, ====

in

net

Dieselthq

w

kWmwPower

barKPavv

wMEP

airnet

net

548.11415.0654.763

5029.929.9500574.0861.0

654.763

21

===

==

=

=

&

3. An air-standard Dual cycle operates with a compression ratio of 14. The conditionsat the beginning of compression are 100 kPa and 300 K. The maximum temperature

in the cycle is 2200 K and the heat added at constant volume is twice the heat added

at constant pressure. Determined, (a) The pressure, temperature, and specific volume


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160

at each corner of the cycle, (b) The thermal efficiency of the cycle, and (c) The mean

effective pressure.

Solution


kgmP

RTv

kgmP

RTv

kParPPv

v

P

P

KrTTv

v

T

T

k

k

k

k

/0615.0271.4023

129.862287.0

/861.0100

300287.0

271.402314100.,

129.86214300.,

3

2

22

3

1

1

1

4.1

12

2

1

1

2

4.01

12

1

2

1

1

2

=

==

=

==

===

=

===

=

Process 23, heat added at constant volume, v3= v2, or P/T=C

KT

TT

TTCTTCqq pvvpincvin

877.1847

)2200(005.12)129.862(718.0

)(2)(,2

3

33

3423,,

=

=

== ==

( ) kgkJTTCq

kgmvv

kPaT

TPP

T

P

T

P

vcvin /767.707129.862877.1847718.0)(

/0615.0

43.8623129.862

877.1847271.4023,

23,

3

23

2

3

23

2

2

3

3

===

==

====

=

Process 34, heat added at constant pressure, P = C, or V/T=C


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161

( ) kgkJTTCq

v

vr

kgmP

RT

v

KTkPaPP

pcpin

c

/884.353877.18472200005.1)(

19.10615.0

0732.0

/0732.043.8623

2200287.0

2200,43.8623

34,

3

4

3

4

4

4

434

===

===

=

==

===

=


kgmvv

kPaPv

v

v

v

P

P

KTv

v

v

v

T

T

kk

kk

/861.0

139.731861.0

0732.043.8623,

88.819861.0

0732.02200,

3

15

4.1

5

1

4

5

4

4

5

14.1

5

1

1

4

1

5

4

4

5

==

=

=

=

=

=

=

=

=

Process 51, heat rejection at constant volume,

( ) kgkJTTCq vcvout /273.37330088.819718.0)( 15, ====

The thermal efficiency of Dual cycle is,

kgkJqqw

kgkJqqq

outinnet

cpincvinin

/377.688273.37365.1061

/651.1061884.353767.707,,

===

=+=+= ==

barkPavv

wMEP

q

q

q

qq

q

w

net

in

out

in

outin

in

net

Dualth

61.801.8610615.0.861.0

377.688

%84.646484.0651.1061

273.37311

21

,

==

=

=

====

==

4. A gas turbine power plant operates on a simple ideal Brayton cycle that has apressure ratio of 12. The compressor inlet pressure and temperature are 1 bar and 300

K. The inlet temperature to the turbine is 1000 K. Determine the thermal efficiency

and the air mass flow rate required for a net power of 3 MW.

Solution


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162

Process 12, isentropic compression,rp= P2/P1=12,

barPrP

KTrP

P

T

T

p

kk

p

kk

12112

181.61012300,

12

4.1/)14.1(

2

/)1(

/)1(

1

2

1

2

===

===

=

Process 23, heat addition at constant pressure, P2= P3, T3= 1000 K.

kgkJTTCq pin /768.391)181.6101000(005.1)( 23 ===

Process 34, isentropic expansion, rp= P3/P4= 12.

KrTT

rP

P

T

T

kk

p

kk

p

kk

657.49112/1000/ 4.1/)14.1(/)1(

34

/)1(

/)1(

4

3

4

3

===

=

=

Process 41, heat rejection at constant pressure, P4= P1,

kgkJqqw

kgkJTTCq

outinnet

pout

/153.199615.192768.391

/615.192)300657.491(005.1)( 14

===

===

skgmmwPower

r

q

w

airairnet

kk

p

th

in

net

th

/064.15153.199/103,

%83.5012

11

11

%83.505083.0

768.391

153.199

3

4.1/)14.1(/)1(

===

===

====

&&

Problems

1- The compression ratio in air-standard Otto cycle is 8. At the beginning of the

compression stroke the pressure is 1 bar and the temperature is 25 C. The heat


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163

transfer to the air per cycle is 1800 kJ/kg. Determine; the pressure and temperature at

the end of each process of the cycle, the thermal efficiency, and the mean effective

pressure.

2- 1 kg of air undergoes an Otto cycle with a compression ratio of 8.7. Initial state of air

is 1 bar and 15 C. The heat added is 1500 kJ. Determine; the pressure, volume, and

temperature at the beginning of each process, the expansion work, compression

work, and net work, the thermal efficiency of the cycle, and the mean effective

pressure.

3- An air-standard diesel cycle has compression ratio of 15 and the heat transferred to

the working fluid per cycle is 1800 kJ/kg. At the beginning of the compression

process the pressure is 1 bar and the temperature is 25 C. Determine; the pressure

and temperature at each point of the cycle, the thermal efficiency of the cycle, and

the mean effective pressure

4- An engine operates on the theoretical diesel cycle with a compression ratio of 15. The

heat is added for 10% of the stroke. The pressure and temperature of the air at thebeginning of compression are 98 kPa and 17 C. Determine; the cut-off ratio, the

pressure and temperature at the end of each process, the amount of heat added, the

thermal efficiency of the cycle, and the volume flow rate of air, measured at the

beginning of compression, needed to produce 200 kW.

5- The intake conditions for an air-standard Dual cycle operating with a compression

ratio of 15 are 1bar and 300 K. The pressure ratio during constant volume heating is

1.5 and the volume ratio during the constant pressure part of the heating process is

1.8. Calculate; the temperatures and pressures around the cycle, the heat input and

the heat rejection, and the thermal efficiency of the cycle.

6- An air-standard Dual cycle operates with a compression ratio of 18. The conditions at

the beginning of compression are 1bar and 300 K. the maximum temperature in the


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164

cycle is 2500 K. The heat added at constant volume is twice the heat added at

constant pressure. Plot the cycle on P-V and T-S diagrams and find; the pressure,

temperature, specific volume at each corner of the cycle, the thermal efficiency of

the cycle, and the mean effective pressure.

7- Air is used as the working fluid in the simple Brayton cycle that has a pressure ratio

of 12. The compressor inlet pressure and temperature are 0.96 bar and 300 K. The

inlet temperature to the turbine is 1000 K. Determine the thermal efficiency and the

mass flow rate of air required for a net power of 30 MW.

8- Air is used as the working fluid in the simple Brayton cycle that has a pressure ratio

of 14. The compressor inlet temperature is 300 K, and a turbine inlet temperature of

1100 K. Determine the mass flow rate of air for a net power output of 100 MW

assuming that the adiabatic efficiency of the compressor is 90 % and the adiabatic

efficiency of the turbine is 85 %. Calculate also the work ratio, back work ratio.

Documents

Chapter 7 Thermo