GANADIPATHY TULSIS JAIN ENGINEERING COLLEGEChittoor Cuddalore Road, Kaniyambadi (Post), Vellore 632102.

Unit: 1 GAS POWER CYCLES

Part-B (16 Marks)1) Derive an expression for the air standard efficiency of diesel diesel cycle and then deduce

it for mean effective pressure.(Nov 2010)This cycle was introduced by Dr. R. Diesel in 1897. It differs from Otto cycle in that heat

is supplied at constant pressure instead of at constant volume. This cycle comprises of thefollowing operations :(i) 1-2......Adiabatic compression.(ii) 2-3......Addition of heat at constant pressure.(iii) 3-4......Adiabatic expansion.(iv) 4-1......Rejection of heat at constant volume.

Point 1 represents that the cylinder is full of air. Let p1, V1 and T1 be the correspondingpressure, volume and absolute temperature.Working Process:

During this addition of heat let volume increases from V2 to V3 and temperature T2 to T3,corresponding to point 3. This point (3)is called the point of cut-off. The air then expandsadiabatically to the conditions p4, V4 and T4 respectively corresponding to point 4. Finally, the airrejects the heat to the cold body at constant volume till the point 1 where it returns to its originalstate.

P-V and T-S diagram for Diesel Cycle

Department of Mechanical Engineering

ME 6404 THERMAL ENGINEERING

Consider 1 kg of air.Heat supplied at constant pressure = Cp (T3 T2)Heat rejected at constant volume = Cv(T4 T1)Work done = Heat supplied heat rejected

= Cp(T3 T2) Cv(T4 T1) Efficiency = Work done / Heat supplied

= Cp(T3 T2) Cv (T4 T1)/ Cp(T3 T2)= 1- { (T4 T1)/ (T3 T2)}

Where Cp/Cv=

It may be observed that for efficiency of diesel cycle is different from that of the Ottocycle only in bracketed factor. This factor is always greater than unity, because > 1. Hence fora given compression ratio, the Otto cycle is more efficient.The net work for diesel cycle can be expressed in terms of pv as follows :

2) A six cylinder four stroke petrol engine has a swept volume of 300cubic cm per cylinder,a compression ratio of 10 and operates at a speed of 35000rpm. If the engine is required todevelop an output of 73.5kw at this speed, calculated the cycle efficiency, the necessaryRate of heat addition,the mean effective pressure , maximum temperature of the cycle andefficiency ratio. The pressure and temperature before isentropic compression are 1.0barand 15 C respectively, take = 0.72 and = 1.4 (nov 2010)

3 6 3

10

1

:

10300 300 10

3500073.51.015 288

0.721.4

:

s

v

Given datar

v cm m

n rpmw Kwp barT C Kc

solution

6

6

6

1

1.4 1

,

300 1010

300 10 1

33.33 101

, 1

1110

0.602 60.2%

,

73.5 122.090.602

s c

c

c

c

c

c

s

s

v vcompression ratio r

v

v

v

v

v

air standard efficiencyr

work outputheat supplied Q

Q Kw

3 6

2

12 1

1 2

11.4 11

2 12

,

603500073.5 10 300 10 6

6070000 / 0.7

288 10 723.4

m s

m

m

Nwork output w p v Z

p

p KN m bar

T VT V

VT T K

V

3) A four stroke petrol , four cylinder petrol engine of 250 mm bore and 375 mm strokeworks on Otto cycle. The clearance volume is 0.01052 .The initial pressure andtemperature are 1bar and 47 C. if the maximum pressure is limited to 25 bar, find the

i. the air standard efficiency of the cycleii. mean effective pressure. Take = 1.005 kJ/kgk and = 1.4 (Nov- 2011)

Given data:

3

2

0

32

1

3

1

22

, 0.250, 0.375

,

100 /

2500 /:

, 0.250 0.3

0.01052

47 3

754 40.01839

20

c

s

s

cylinder diameter d mstrokelength l mclearancevolume vp KN

KN msolution

dstrokevolu

m

m

T C

me

K

p

m

v l

v

compre

250 mm375 mm

1bar

25 bar

1 1.4 1

1

2

0.01052 0.01839, 2.74

0.010521 1

, 1 16.888

53.78%. .

2.74

c s

c

v vssion ratio r

v

air standard efficiencyr

w k tv

v

1.4

2

3

2

100 (2.74)410.0631 /

2500, 6.096

410.063

KN mp

pressure ratio kp

1.4 16.096 1 2.74100 2.741.4 1 2.74 1

993.43mp Kpa

4) In an air standard dual cycle, the pressure and temperature at the beginning ofcompression are 1 bar and 57 C respectively. The heat supplied in the cycle is 1250 Kj.kg,two third of this being added at constant volume and rest at constant pressure. If thecompression ratio is 16, determine the maximum pressure, temperature in the cycle,thermal efficiency and mean effective pressure. (Nov 2011)

12 1

2

1 2( );consider process adiabatic compressionv

p pv

10

1

1

2

311 5

13

2

:

147 320161250 /2 / 3 833.33 /1 / 3 416.67 /

:

287 320, 0.9184 /

1 100.0574 /

s

s s

s

Given datap barT C kr

Q KJ kgQ Q KJ KgQs Q KJ Kgsolution

rTspecific volume v m kg

p

v m kg

2 11.4

11

1.4 1

3 2

3

33 2

2

1 2( )16 1 48.5

( )16 320 970.06

2 3( )

0.718( 970.06)2130.69

. .

2130.69970.06

v

isentropic compression processesp r p

barT r T

Kconstant volume heat addition process

c T TT

Kw k t

Tp p

T

2 4 3

4

4

48.5

106.533 4

( )416.67 1.005( 2130.69)

2545.29

s

barconstant pressure heat addition process

Q C T TT

T K

44 3

33

4

1

14

1.4 1

5 1

2545.29 0.05742130.690.0686 /

0.0686, 0.0747

2130.694 5

(0.0747) 2545.29901.71

,

( )0.718

e

e

r v

Tv v

T

m Kgv

expansion ratio rv

isentropic expansion processT r T

Kheat rejected fromthecycle

Q C T T

2

(901.71 320)417.67 /

1250 417.67 832.33 /832.33

, 66.59%1250

,

832.33 9.670.9184 0.0574

s r

s

m

KJ Kgworkdone W Q Q

KJ Kgw

cycleefficiency Qmeaneffective pressure

wp barv v

5)In the engine working on dual cycle , the temperature and pressure at the beginning ofthe cycle are 90 C and 1 bar respectively. The compression ratio is 9 . The maximumpressure is limited to 68 bar and heat supplied per Kg of air is 1750KJ. determine :

i. Pressure and temperature at all salient pointsii. Air standard efficiency

iii. Mean effective pressure. (may 2012)

313 2

44 3

3

3

4

3

3

2

1.04181 0.11576 /93149 0.115762743

0.132894 /0.13289

, 1.1480.11576

68, 3.138

21.67

vv v m Kg

r

Tv v

T

m Kgv

cut off ratiov

ppressure ratio K

pefficiency of thecycle

10

1

3 4

.1.42 1

1 0.42 1

33

2

190

689

1750 /:

1 2 : .9 1

21.679 363

8742 3

68 87421.67

27433

s

p barT Cp p barr

Q KJ Kgsolution

isentropic comp processp r p

barT r T

Kconstant volume heat addition process

pT T

p

K

3 2 4 3

4

4

31

1

4 :( ) ( )

1750 0.718(2743 874) 1.005( 2743)3149

287 363 1.04181 /1 105

s v p

constant pressureheat addition processQ c T T c T T

TT K

RTv m Kg

p

11.4

1.4 1

1 2

1 11 ( 1) ( 1)1 3.138 1.148 11 (3.138 1) 3.138 1.4(1.148 1)9

58.19%,

0.5819 17501018.33 /

,

1018.331.0418

net s

netm

kkr

net work of thecyclew Q

KJ Kgmean effective pressure

wp

v v

10.981 0.11576 bar

6) a. Consider an air standard cycle in which the air enters the compression at 1 bar and 20C. The pressure of air leaving the compressor is 3.5 bar and the temperature at turbineinlet is 600 C .determine per Kg of air. (May 2012)

i. Efficiency of the cycleii. Heat supplied to air

iii. Work available at the shaftiv. Heat rejected in the cooler andv. Temperature of air leaving the turbine

Given data:

1

1

3

2

1206003.5

p barT CTP

12 2

1 1

1

22 1

1

1.4 11.4

1

4 4

3 3

1

44 3

3

:

1 2

3.5 2931

4193 4 exp

13.5

solutionconsider the process adiabatic compression

T pT p

pT T

p

Kconsider the process adiabatic ansion

T pT p

pT T

p

1.4 11.4

1

1.4 11.4

3 2

4 1

873 674.3

1, 1

11 0.303

30%( ) 1.005(873 419) 456.2 /( ) 1.005(610.3 293) 318.8 /

,

p

s p

r p

c

K

air standard efficiencyr

heat supplied Q c T T KJ Kgheat rejected Q c T T KJ Kgcompressor work w

2 1

3 4

( ) 1.005(419 293) 126.63, ( ) 1.005(873 610) 264.31

264.31 126.63 137.68674.3

p

e p

e c

c T T KJsimilarly for exapander W c T T KJwork output w W wtemperature of leaving theturbine K

b. The efficiency of an Otto cycle is 60% and = 1.5. What is the compression ratio

Solution:

Efficiency of Otto cycle, = 60%

Ratio of specific heats, = 1.5

Compression ratio, r = ?

Efficiency of Otto cycle is given

by,7) a. A spark ignition engine working on ideal Otto cycle has the compression ratio 6. Theinitial pressure and temperature of air are 1 bar and 37 C. the maximum pressure in thecycle is 30bar. For unit mass flow, calculated

i. P , v and T at various salient points of the cycle andii. The ratio of heat supplied to the heat rejected. Assume = 1.4 and R = 8.314KJ/k

mol K (Nov 2012)

21

01

3

2 1

1 2

12 1

2

1.4 22

12 1

1 2

12 1

2

1.4 1

:

61 100 /37 37 273 31030

:

1 2( ) :

6 100 1228.6 /

6

Given datar

p bar KN mT C kp barsolutionconsider process adiabatic process

p Vp V

Vp p

V

p KN m

T VT V

VT T

V

2

3 3

2 2

310634.78

2 3( );T Kconsider process constant volume processp Tp T

33 2

2

3

34

3 41.4

234 3

41

34

3 4

1 0.43

4 34

3000 634.78100

19043.43 4( ) :

1 3000 244.18 /6

1 19043.4 93006

:

pT T

pT Kconsider process adiabaticprocess

vpp v

vp p KN m

v

vTT v

vT T K

v

heat supplied Q

3 2

4 1

( ) 1 0.718 (19043.3634.78)13217.39 /( ) 1 0.718 (9300 310) 6454.82 /

13217.39 2.0486454.82

s v

s

v

s

r

mC T TQ KJ Kg

heat rejected mC T T KJ KgQQ

8) An air standard dual cycle has a compression ratio of 18, and compression begins at 1bar 40 C. The maximum pressure is 85 bar. The heat transferred to air at constantpressure is equal to that at constant volume. Estimate:

i. The pressure and temperatures at the cardinal points of the cycle.ii. The cycle efficiency and

iii. Mean effective pressure of the cycle (Nov 2012)

1 2

10

1

3

311 5

13

2

1.42 1

:

18140 31385

1.005 /0.718 /

:

287 313, 0.92701 /

1 100.049906 /

1 218 1 5

s s

p

v

Given datar

p barT C kp barQ QC KJ KgKC KJ KgKsolution

RTspceific volume V m Kg

P

V m Kgisentropiccompression processp r p

1

1 1.4 12 1

33 2

2

3 2

7.1918 313 994.61

2 3 tan85 994 1478.26

48.5( )s v

barT r T Kcons volume heat addition process

pT T K

pQ C T T

1 2 4 3

4

4

344

3

4

1

0.718(1478 994.61)347.26 /

3 4( )

347.26 1.005( 1478.26)1832.79

1832.79 0.04996 0.061919 /1478.26

0.061919, 0

0.89831

s s p

e

KJ Kgconstant pressure heat addition

Q Q C T TT

T KT

v v m KgT

vexpansion ratio r

v

5 41.4

1 1.4 15 4

4

3

3

2

.06892

4 5 exp

0.06892 85 2.0090.06892 1832.79 628.7

0.06892, 1.380

0.049906

85, 1.486

57.19

e

e

isentropic ansion processp r p

barT r T K

vcut off ratio

v

ppressure ratio K

pthe cycl

e efficiency

1 2

.1.4

11 1.486 1.6380 11 (1.486 1) 1.4 1.486 (1.380 1)

67.83%

311.612 311.612 692.521 /,

692.52 0.6783 469.73 /

s s s

s

r

net heat supplied tocycleQ Q Q KJ Kg

network donetocycleW Q KJ Kg

the meanefficitive pr

1 2

413.45 5.570.8931 0.049906m

essure

Wp barv v

9) In an oil engine working on dual cycle the heat supplied at constant pressure is twicethat of heat supplied at constant volume. The compression and expansion ratios are 8 and5.3. The pressure and temperature at the beginning of cycle are 0.93 bar and 27 C. findthe efficiency of the cycle and mean effective pressure. Take = 1.005 KJ/kgK and =0.718 KJ/kgK. (May 2013)

Given data:

2 1

10

1

5 1

4 4

311 5

312

33 2

2

0.9327

82

5.3

1.005 /0.718 /

:

,

287 300 0.926 /0.93 10

0.926 0.11572 /8

0.11572 /1 2

s s

p

v

P barT Cr

Q Qv vkv v

C KJ KgKC KJ KgKsolution

specificvolumesRT

v m Kgp

vv m Kg

r

v v m Kgcompression process

p

1.41

1 1.4 12 1

8 0.93 17.0938 300 689

r p barT r T K

10) Air standard cycle consists of the following process.(a).Isentropic compression from 15oc and 1 bar to 5 bar.

(b).2500KJ/Kg of heat is added at constant volume.

(c).Isentropic expansion to initial volume.(d).Heat rejection at constant volume.Calculate the ideal efficiency, mean effective pressure and peak pressure. (Nov 2013)