1

Applied Thermodynamics

2

1. Air Standard Power Cycles Introduction

Two important applications of thermodynamics are power generation and refrigeration. Both are usually accomplished by systems that operate on thermodynamic cycles.Hence the thermodynamic cycles are usually divided into two general categories, viz., “power cycles” and “ refrigeration cycles”; Power or refrigeration cycles are further classified as “ gas cycles” and “ vapour cycles” ;

3

In case of gas cycles, the working substance will be in gaseous phase throughout the cycle, where as in vapour cycles, the working substance will be in liquid phase in one part of the cyclic process and will be in vapour phase in some other part of the cycle;

Thermodynamic cycles are also classified as “ closed cycles” and “ open cycles”.

In closed cycles, the working fluid is returned to its original state at the end of each cycle of operation and is recirculated.

4

In an open cycle, the working substance is renewed at the end of each cycle instead of being re-circulated.

In automobile engines, the combustion gases are exhausted and replaced by fresh air-fuel mixture at the end of each cycle.

Though the engine operates in a mechanical cycle, the working substance does not go through a complete thermodynamic cycle.

5

Basic Considerations in the Analysis of Power Cycles The cycles encountered in actual devices are difficult to

analyze because of the presence of friction, and the absence of sufficient time for establishment of equilibrium conditions during the cycle.

In order to make an analytical study of a cycle feasible, we have to make some idealizations by neglecting internal Irreversibilities and complexities.

Such cycles resemble the actual cycles closely but are made up of internal reversible processes.

These cycles are called ideal cycles.

6

Air Standard Cycles In gas power cycles, the working fluid will be in gaseous

phase throughout the cycle. Petrol engines (gasoline engines), diesel engines and gas

turbines are familiar examples of devices that operate on gas cycles.

All these devices are called “ Internal combustion engines” as the fuel is burnt within the boundaries of the system.

Because of the combustion of the fuel, the composition of the working fluid changes from a mixture of air and fuel to products of combustion during the course of the cycle.

7

However, considering that air is predominantly nitrogen which hardly undergoes any chemical reaction during combustion, the working fluid closely resembles air at all times.

The actual gas power cycles are complex.

In order that the analysis is made as simple as possible, certain assumptions have to be made.

These assumptions result in an analysis that is far from correct for most actual combustion engine processes, but the analysis is of considerable value for indicating the upper limit of performance.

8

Air standard assumptions 1. The working medium is a perfect gas with constant specific

heats and molecular weight corresponding to values at room temperature.

2. No chemical reactions occur during the cycle. The heat addition and heat rejection processes are merely heat transfer processes.

3. The processes are reversible.4. Losses by heat transfer from the apparatus to the atmosphere

are assumed to be zero in this analysis.5. The working medium at the end of the process (cycle) is

unchanged and is at the same condition as at the beginning of the process (cycle).

i.e Changes in kinetic and potential energies of the working substance are very small and hence negligible.

Air standard Carnot CycleThe Carnot cycle is represented on P-v and T-s diagrams as in Fig.

The Carnot cycle is composed of four totally reversible processes: isothermal heat addition, isentropic expansion, isothermal heat rejection, and isentropic compression.

The Carnot cycle is the most efficient cycle that can be executed between a heat source at temperature and a sink at temperature , and its thermal efficiency is expressed as

9

10

Process 1 – 2: Reversible Adiabatic CompressionProcess 1-2: In this, air is compressed isentropically from volume

During this process heat rejected is zero. i.e., P: Increases from p1 to p2

V: Decreases from V1 to V2

T: Increases from T1 to T2

S: Remains same.

1W2 = =

1Q2 = 0 or

12211

VPVP

1

)( 21

TTmR

11

Process 2 -3: Isothermal Heat AdditionIn this air is heated isothermally so that volume increases and Temperature remains constant.Amount of heat supplied is equalto the work done by the gas.P: Decreases from p2 to p3

V: Increases from V2 to V3

T: Remains same.S: Increases from S2 to S3

2W3= p2V2 ln = mRT2 ln

2Q3 = p2V2 ln

12

Process 3 – 4: Reversible Adiabatic Expansion

This is isentropic(Adiabatic) expansion

process.

Heat supplied during the process is zero. i.e., P:

Decreases from p3 to p4

V: Increases from V3 to V4

T: Decreases from T3 to T4

S: Remains same.

3W4 = =

3Q4 = 0

14433

VPVP

1

)( 43

TTmR

13

Process 4 – 1:

Isothermal Heat Rejection

P: Increases from p4 to p1

V: Decreases from V4 to V1

T: Remains same.

S: Decreases from S4 to S1

4W1= p4V4 ln = mRT4 ln

4Q1 = p4V4 ln

14

15

16

max

minmax

max

minmax

ln

ln

T

TT

TrR

TTrRth

And also,

17

Mean Effective Pressure

Mean effective pressure may

be defined as the theoretical

pressure which, if it is maintained

constant throughout the volume

change of the cycle, would give the

same work output as that obtained from the cycle.

Or it is the constant pressure which produces the same work output while causing the piston to move through the same swept volume as in the actual cycle.

18

Mean effective Pressure:When the piston moves from TDC to BDC, the air inside expands resulting in work output. If Pm1 is the average pressure on the piston during

this stroke, the average force on the piston is

Where d = diameter of piston or cylinder bore Work output = average force on piston X stroke length During the return stroke, as the piston moves from BDC to TDC, air is compressed requiring work input of the average pressure on the piston during this stroke is Pm2, the work input is given by;

Where Pm is known as mean effective pressure and is the swept volume.

Usually the net work output is in kJ, volume in m3 and mean effective pressure in bar.

19

Stirling cycleWhen a confined body of gas (air, helium, whatever) isheated, its pressure rises. This increased pressure can push on a piston and do work. The body of gas is then cooled, pressure drops, and thepiston can return. The same cycle repeats over and over, using the samebody of gas. That is all there is to it. No ignition, no carburetion, no valvetrain, no explosions. Many people have a hard time understanding the Stirlingbecause it is so much simpler than conventional internalcombustion engines.

20

Stirling Cycle:The Stirling cycle is represented on P-v and T-s diagrams as in Fig.

It consists of two isothermal processes and two isochors.

Process 1-2: In this air is heated isothermally so that volume increases from Temperature remains constant.

Amount of heat supplied is equal to the work done by the gas.

21

Stirling Cycle:Process 2-3: This is constant volume heat rejection process. Temperature decreases from pressure decreases from the heat rejected during the process is given by,

Process 3-4: In this air is compressed isothermally from volume

During this process heat rejected is equal to the work done by the gas.

22

Stirling Cycle: Process 4-1: This is constant volume heat

addition process. Temperature increase from

The heat added during the process is given by,

23

24

The Efficiency of the cycle:Due to heat transfers at constant volume processes, the efficiency of the Stirling cycle is less than that of the Carnot cycle.

However if a regenerative arrangement is used such that, i.e., the area under 2-3 is equal to the area

under 4 -1 on T-s diagram, then the efficiency,

25

Otto cycle OR Constantvolume cycle: The Otto cycle is the ideal

cycle for spark-ignition reciprocating engines.

It is named after Nikolaus A. Otto, who built a successful four-stroke engine in 1876.

This cycle is also known as constant volume cycle as the heat is received and rejected at constant volume.

The cycle consists of two adiabatic processes and two constant volume processes as shown in P-v and T-s diagrams.

26

Otto cycle OR Constant volume cycle: Process 1-2: In this air is compressed

isentropically from V1 to V2 Temperature increases from T1 to T2.

Since this is an adiabatic process heat rejected is zero. i.e.

Process 2-3: In this air is heated at constant volume

and temperature increases from T2 to T3.

Heat supplied during this process is given by,

27

Otto cycle OR Constant volume cycle: Process 3-4: In this air is expanded isentropically from

V3 to V4 and temperature decreases from T3 to T4. Since this is an adiabatic process, the heat supplied is zero. i.e., Process 4-1:

In this air is cooled at constant volume and temperature decreases from T4 to T1. Heat rejected during this process is equal to change in internal energy and is given by,

28

The Efficiency of the cycle: Efficiency of the cycle is given by,

Considering isentropic expansion process 3-4,

Or

Considering isentropic compression process 1-2,

Or

Substituting for in eqn (1)

Or

Where, r = compression OR expansion ratio and

.

129

Mean effective pressure: We know that for Otto cycle,

the pressure ratio

30

31

Diesel cycle OR Constant pressure cycle:

The Diesel cycle is the ideal cycle for Compression Ignition reciprocating engines.

The CI engine was first proposed by Rudolph Diesel.

The Diesel cycle consists of one constant pressure heating process, one constant volume cooling process and two adiabatic processes as shown in P-v and T-s diagrams.

This cycle is also known as constant pressure cycle because heat is added at constant pressure.

32

Diesel cycle OR Constant pressure cycle:

Process 1-2: During this process air is compressed

adiabatically and volume decreases from V1 to V2 Heat rejected during this process is zero. i.e.,

Process 2-3: During this process air is heated at

constant pressure and temperature rises from T2 to T3 Heat supplied during this process is given by,

33

Diesel cycle OR Constant pressure cycle:

Process 3-4: During this process air is expanded

adiabatically and volume increases from V3 to V4

.

Heat supplied during the process is zero. i.e.,

Process 4-1: In this air is cooled at constant volume

and temperature decreases from T4 to T1 .

Heat rejected during this process is given by,

34

35

The Efficiency of the cycle:

The efficiency of the cycle is given by,

Let, compression ratio, Cut-off ratio,

Expansion ratio,

Considering process 1-2,

36

Considering process 3-4,

Substituting for in eqn (1), we get

Considering process 2-3,

37

Mean effective pressure:

we know that work done per kg in Diesel cycle is given by,

And the mean effective pressure is given by:

38

Expression for cut-off ratio:Let ‘k’ be the cut-off in percentage of stroke (from

We know that,

Dual combustion or Limited pressure or Mixed cycle:

This cycle is a combination of Otto and Diesel cycles.

It is also called semi-diesel cycle because semi-diesel engines work on this cycle.

In this cycle heat is absorbed partly at constant volume and partly at constant pressure.

It consists of two reversible adiabatic or isentropic, two constant volume and a constant pressure processes as shown in P-v and T-s diagrams.

4

5

3

39

Dual combustion or Limited pressure or Mixedcycle:

Process 1-2: The air is compressed

reversibly and adiabatically from temperature T1 to T2 .

No heat is rejected or absorbed by the air.

Process 2-3: The air is heated at constant

volume from T2 to T3.

Heat absorbed by the air is given by,

3 4

5

40

Dual combustion or Limited pressure or Mixed cycle:

Process 3-4: The air heated at constant pressure from

temperature T3 to T4.

The heat supplied by the fuel or heat absorbed by the air is given by,

Process 4-5: The air is expanded reversibly and

adiabatically from temperature T4 to T5 .

No heat is absorbed or rejected during the process.

Process 5-1: The air is now cooled at constant volume from temperature T5 to T1 . Heat rejected by the air is given by,

3 4

5

41

42

The Efficiency of the cycle:

The efficiency of the cycle is given by,

Let, compression ratio,

Cut-off ratio,

Pressure ratio,

Expansion ratio,

5

43

Considering process 1-2,

Considering process 2-3,

Considering process 3-4,

Considering process 4-5,

Substituting for in (1)

44

Mean effective pressure:

We know that work done per kg in dual cycle is given by,

And the mean effective pressure is given by:

Note:1) For Otto cycle

2) For Diesel cycle

45

Comparison between Otto, Diesel and Dual combustion cyclesThe important variables which are used as the basis for comparison of the cycles are compression ratio, peak pressure, heat supplied, heat rejected and the net work output.

In order to compare the performance of the Otto, Diesel and Dual combustion cycles some of these variables have to be fixed.

46

Comparison with same compression ratio and heat supply:

The comparison of these cycles for the same compression ratio and same heat supply are shown in on both p – V and T – S diagrams.

In these diagrams, cycle 1-2-3-4-1 represents Otto Cycle, cycle 1-2-3’-4’-1 represents diesel cycle and cycle 1-2”-3”-4”-1 represents the dual combustion cycle for the same compression ratio and heat supply.

47

From the T-S diagram, it can be seen that area 5236 = area 522”3”6” = area 523’6’ as this area represents the heat supply which is same for all the cycles.

All the cycles start from the same initial point 1 and the air is compressed from state 1 to state 2 as the compression ratio is same.

48

It is seen from the T-s diagram, that for the same heat supply, the heat rejection in Otto cycle (area 5146) is minimum and heat rejection in Diesel cycle (area 514’6’) is maximum. Consequently Otto cycle has the highest work output and efficiency. Diesel cycle has the least efficiency and dual cycle has the efficiency between the two.

49

50

Therefore for the same compression ratio and same heat rejection, Otto cycle is the most efficient while the Diesel cycle is the least efficient.

It can also be seen from the same diagram that q3>q2>q1We know that thermal efficiency is given by 1 – heat rejected/heat suppliedThermal efficiency of these engines under given circumstances is of the following orderDiesel>Dual>OttoHence in this case it is the diesel cycle which shows greater thermal efficiency.

51

Problem 1 In an Otto cycle, the upper and lower limits for

the absolute temperature respectively are T1 and T2.

Show that for the maximum work, the ratio of compression should have the value

25.1

1

3

TTrc

52

Solution: Process 1-2 is reversible adiabatic

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

12

2

1

1

2

c

c

rTT

rVV

TT

Process 3-4 is reversible adiabatic

(2)..........

33

4

2

1

3

4

cc

c

rTrTT

rVV

TT

53

Work done = Heat added - Heat rejected

In the above equation T3, T1 and Cv are constants. Therefore for maximum work

11

313

1423

- C - - C

- C - - C

TrTrTT

TTTT

cc

0 drdW

54

25.1

1

3

14.1

1

31

1

1

3

TTr

TT

TTr

c

c

1

2

1

2

32

1

32

1

131

11

313

0 1C - C-

0 1C - C-

0 - C - - C

c

ccc

c

cc

cc

cc

cc

r

rrrr

TT

rTrT

rTrT

rTrT

TrTrTTdrd

55

Problem 2 An engine working on Otto cycle in which

salient points are 1,2,3 and 4 has upper and lower temperature limits T3 and T1.

If the maximum work per kg of air is to be done, show that the intermediate temperatures are given by

3142 TTTT

56

Solution: For maximum work/kg in an Otto cycle•

31

21

1

31

11

1

3 112

11

1

3

1) problemin proved (as

TTTTT

TTTrTT

TTr

c

c

57

11

1

3

334

TT

TrTTc

3142

313

13

1

TTTT

TTTTT

Again

58

Problem 3 An engine working on the otto cycle has a suction

pressure of 1 bar and a pressure of 14 bar at the end of compression.

Find Compression ratio, Clearance volume as a percentage of cylinder volume

The ideal efficiency and MEP if the pressure at the end of combustion is 21 bar.

Solution: Given: P1 = 1 bar, P2 = 14 bar, P3 = 21 bars

59

11

1

2

2

1

2

1

1

2

14

P

Prv

v

vv

PP

c

53% 58.6

11

11 efficiency Ideal

15.18%

1006.58

1 100 x

4.

1

2

1

2

1

c

cc

r

xvv

vv

vvr

60

bar 1.65 )158.6)(14.1(

)11)(6.58-1x6.58(1.5

1

1..

1.5 1421

ratio

1-1.4

1

2

3

c

c

r

rpPEM

pppressureExplosion

61

Problem 4 In a constant volume cycle the pressure at the end of compression is 15

times that at the start, the temperature of air at the beginning of compression is 37° C and the maximum temperature attained in the cycle is 1950°C. Find,

(i) the compression ratio(ii) thermal efficiency of the cycle(iii) heat supplied per kg of air(iv) the work done per kg of airSolution:Given:P2/P1 = 15 , T1 = 37ºC = 310 KT3 = 1950ºC = 2223 K

62

6.91

15

r

11

1

2

c2

1

1

2

c

c

r

P

Pr

vv

PP

54%

0.54 91.611

11

4.

cr

K 671.66 (31096.91) 1-1.412

crTT

63

Heat supplied = Cv(T3-T2) = 0.72(2223 - 671.66)

=1116.96 KJ/kg of air

Work done = 0.54 x 1116.96 = 603.16 KJ/kg of air

suppliedHeat doneWork

64

Problem 5 An air standard Diesel cycle has a compression ratio

of 18 and the heat transferred to the working fluid per cycle is 2000 kJ/kg.

At the beginning of the compression stroke, the pressure is 1 bar and the temperature is 300 K.

Calculate the thermal efficiency. Given: rc = 18

P1 = 1 bar

T1 = 300 K

K 953.3 300(18) 1-1.4

12

crTT

65

Heat transferred = Cp(T3 – T2)

2000 = 1.005(T3 -953.3)]T3 = 2943.34 K

3.08 953.3

2943.34 ratio offcut 2

3 TT

58.6% 0.586 08

108.181

11

4.

cr

66

Problem 6

An engine with 200 mm cylinder diameter and 300 mm stroke length, works on the theoretical Diesel cycle. The initial pressure and temperature of air are 1 bar and 27° C. The cut off is at 8% of the stroke and compression ratio is 15. Determine

(i) Pressure and temperatures at all salient points of the cycle.

(ii) theoretical air standard efficiency.(iii) mean effective pressure.(iv) power developed if there are 400 working strokes

per minute.

67

Solution: Given:rc = 15,

P1 = 1 bar,

T1 = 27º C

d = 200 mm, L = 300 mm

3m .009424

.3 2.0 sV

V

2

s

x

dLvolumeSwept

68

3

23

23

2sc2

cc

s

c

s

c

sc

2

1c

c2

m 0.001427

009424.008.00006731.008.0 100

8 )(

stroke of 8%at place takesoff

m 0.0006731 14

0.009424

14

V V

14 1)-(15 )1(r V

V

V

V1

V

VV

V

V r

volumeclearance V

xVVV

VVV

Cut

V

V

s

s

69

kPaPEM

x

cr

r

rpPEM

c

c

cr

cut

2

4.114.1

4.11

2

3

10 x 7.41bar 7.14 ..

112.1512.4.115.

151

11

1..

59.8% 0.598

12.

112.15

1 1

1

2.12 0.00067310.001427

VV

ratio off

70

bar 44.3 3

2

3.441.4

1x15 1

2

25.8861-1.4300x15 1

2

kW 46.53 Power 60

400 x 6.98 Power

sec / cycles ofNumber x cycle / done Power Work

kJ/cycle 6.98 0.00942 10 x 7.41 cycle / done

meSwept volu

cycledonework ..

2

PP

barcrPP

KcrTT

xWork

PEM

71barP

rV

V

V

V

V

V

V

V

V

V

P

P

x

VceV

V

V

V

V

V

V

V

V

V

c

86.215

12.23.44

x x

K 858.99 15

12.285.1878

r

1

V sin x x

K 1878.85 2.12 x 886.25

4.1

14.1

1

14

111

72

Problem 7 In a dual combustion cycle the compression ratio is

14, maximum pressure is limited to 55 bar. The cut-off ratio is 1.07. Air is admitted at a pressure

of 1 bar. Find the thermal efficiency and M.E.P of the cycle.

Solution: (Given): rc = 14

P1 = 1 bar

P3 = 55 bar

Cut off ratio = =1.07

73

62.18%

)107.1(1.4 367.1)1367.1(

11.07 x 367.1

14

11

)1()1(

111

367.123.40

55

P

P ratio pressureExplosion

bar 40.23

)14(1

4.1

14.1

1

2

3

4.112

x

r

rPP

c

c

74

112

2

323

3

33

2

22

112

2

32

3

43p

2334p

TP

PP

process lumecontant vo is 3-2 Pr

T T

1 1C

)()(C added

c

c

rTP

TT

T

V

T

V

ocess

r

T

TTC

T

TT

TTCTTHeat

75

kgm

c

cc

rP

RTSwept

xWork

T

rTCrT

/21

21

1

1

1

2121

1

1

1

14.11

14.11

11

11p

T 0.003091

14

11

10 x 1

T 0.287

11

V

V1VVV Volume

T 0.6439

6218.0T 1.0356 x addedHeat done

1.0356T

1367.11472.0107.114T 1.367 x 1.005

11C addedHeat

76

bar 083.2 ..

kPa 208.3

0.003091T

0.6439T

volume/kg

/kgdoneWork ..

1

1

PEM

SweptPEM

77

Problem 8 From the PV diagram of an engine working on the Otto

cycle, it is found that the pressure in the cylinder after 1/8th of the compression stroke is executed is 1.4 bar. After 5/8th of the compression stroke, the pressure is 3.5bar. Compute the compression ratio and the air standard efficiency. Also if the maximum cycle temperature is limited to 1000.C, find the net work out put

78

ration compressio where-(1)--- 8

1

8

788

78

18

1

300)(27

12732731000

5.3 ,4.1

2

21

211

211

1

3

cca

a

a

a

ba

rrV

V

VVV

VVVV

VVVV

Solution

KassumedT

KT

barPbarP

Given

79

85

83

81

87

have we2 and 1equation From

-(1)--- 8

5

8

38

58

5

2

211

211

c

c

b

a

cb

b

b

r

r

V

V

rV

V

VVVV

VVVVAgain

80

7

924.1

85

83

81

87

4 and 3 from

(4)----- 1.924

4.1

5.3

PP

4.1

11

ba

c

c

c

b

a

b

a

b

a

ba

r

r

r

V

V

P

P

V

V

VVBut

81

kJ/kg 240.6

445 x 0.5408

suppliedheat x output Network

kJ/kg 445

653.4)-0.718(1273

)(C addedHeat

K 653.4

7300

%08.547

11

11

23

14.1

1

2

112

14.11

TT

V

VTT

rc

82

Problem 9An air standard diesel cycle has a compression ratio of 16. The

temperature before compression is 27°C and the temperature after expansion is 627°C. Determine:

i) The net work output per unit mass of airii) Thermal efficiencyiii) Specific air consumption in kg/kWh.

83

(1)--- T x T

PP and PP

have we3-2 processFor

K 909.43

x16300T

TT have we2-1 process

300)(27

900273627

16

22

33

323

33

2

22

0.4

1

2

112

122

111

1

4

2

1

V

V

T

V

T

V

V

VTOr

VVFor

Solution

KassumedT

KT

rV

V

Given

c

84

1

2

11243

1

3

2

1

2

143

3

2

1

3

2

2

143

1

3

14

1

3

443

144

133

T

T

have we(1)Eqn from for

T

T

TT have we4-3 process

V

VTT

T

T

V

VT

V

VngSubstituti

V

V

V

VT

V

VT

V

VTOr

VVFor

85

kwhkgW

Specific

W

W

TTCHeat

k

V

VTT

p

/57.55.658

36003600 n consumptioair

%45.606045.03.1089

5.658

q efficiency Thermal

658.5kJ/kg

8.4303.1089 doneWork

g1089.3kJ/k

909.43]-[1993.3 x 1.005

)(q massunit per supplied

3.199361 x 09.439 x 900

T

3-2

233-2

4.11

4.04.0

11

2

11243

86

Problem 10 The compression ratio of a compression ignition

engine working on the ideal Diesel cycle is 16. The temperature of air at the beginning of compression is 300K and the temperature of air at the end of expansion is 900K. Determine

i) cut off ratioii) expansion ratio and iii) the cycle efficiency

87

1

3

2c

4

3

1

3

2

2

1

4

3

1

3

2

2

4

1

3

4

4

3

14.1112

1

4

x T

T

x T

T

x T

T

42.90961 x300T

300)(27

900273627

16

V

Vr

V

V

V

V

V

V

V

V

V

V

KrT

Solution

KassumedT

KT

r

Given

c

881993.28K

)909.42 x 61 x 009(

)T(

T

T x T

x T

T

T

T

PP and PP

4.1114.114.1

112

143

12

143

12

14

133

13

121

1

3

2c

4

3

3

2

3

2

323

33

2

22

TrT

TrT

TrT

T

Tr

T

Tr

V

V

T

V

T

V

c

c

c

c

89

29.7900

28.1993

r ratioExpansion

60.46%

119.2

119.2

4.1

61-1

1

1r-1

19.2

42.909

28.1993

T

T ratio offcut

14.1

1

1

1

4

3

3

4E

4.11.4-1

1c

2

3

T

T

V

V

90

Problem 11 An air standard limited pressure cycle has a compression

ratio of 15 and compression begins at 0.1 MPa, 40°C. The maximum pressure is limited to 6 MPa and heat added is 1.675 MJ/kg. Compute

(i) the heat supplied at constant volume per kg of air(ii) the heat supplied at constant pressure per kg of air(iii) the work done per kg of air(iv) the cycle efficiency(v) cut off ratio and(vi) the m.e.p of the cycle

91

354.12.4431

6000

P

26.443151 x100P

1675kJ/kgMJ/kg 1.675 added

60006

40

1001.0

15

2

3

14.112

43

1

1

P

KParP

Solution

Heat

KPaMPaPP

CT

KPaMPaP

r

Given

c

c

92

/kg1439.286kJ235.71-1675

olumeconstant vat addedheat -

addedheat Total pressureconstant at addedHeat

g235.71kJ/k

)65.924251.9910.72(

)0.72( olumeconstant vat addedHeat

K 1251/99

1.354 x 6.924

924.65k

)15(313

23

23

14.1112

TT

JTT

rTT C

93

60.56%

11438.21.4 x 354.11354.1

12.1438 x 354.1

15

1-1

11

1

r

1-1 effeciency standard

11.2684T

)99.12511.005(T1439.286

C pressureconstant at added

4.1

14.1

1-c

4

4

34p

Air

K

TTHeat

KPa 2000.13

1.14382 x 354.11511438.2.3541 x 4.11354.111514.1

15 x 100

11111

..

4.14.114.1

11

c

c

rr

rPPEM