DJA3032 CHAPTER 2

CHAPTER 2: PISTON ENGINE PROCESS ANALYSIS

byMOHD SAHRIL BIN MOHD FOUZI, Grad. IEM (G 27763)

DEPARTMENT OF MECHANICAL ENGINEERING

© MSF @ POLITEKNIK UNGKU OMAR

(DJA3032) INTERNAL COMBUSTION ENGINE

General objective:To understand the structure and various types of cycle engines. Specific objectives:At the end of this unit you should be able to: define the air standard cycle. define constant pressure (cp) and constant volume (cv). draw p-v diagram of Otto cycle, Diesel cycle and Combined cycle. explain Otto cycle, Diesel cycle and Combine/dual cycle.



Introduction :

In this unit we are to discuss the meaning of standard cycle, heat supplied at constant volume and heat supplied at constant pressure. An internal combustion engine can be classified into three different cycles and they are Otto cycle, Diesel cycle and Dual-combustion cycle.

Pressure (P) 3

24

1

P3

P2

P4

P1

Volume ( v )

tconspv tan

V1V2

Qin

Qout

V

4

1

2 3P2 = P3

P1

V1= V4

V3V2

P4

PPressure,

Volume

tconspv tan

Qin

Qout

Figure 2.1: P-V Diagram (Otto Cycle) Figure 2.2: P-V Diagram (Diesel Cycle)



QinP

V

1

2

34

5

P3=p4

V2=V3V1=V5

contsvp .Qin

Qout

Figure 2.3: P-V Diagram (Dual/Combined Cycle)



Air Standard Cycles

The air standard cycle is a cycle followed by a heat engine which uses air as the working medium. Since the air standard analysis is the simplest and most idealistic, such cycles are also called ideal cycles and the engine running on such cycles are called ideal engines.

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.

The analysis is also a simple means for indicating the relative effects of principal variables of the cycle and the relative size of the apparatus.



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).



Figure 2.4: T-S Diagram for Otto Cycle

Figure 2.5: T-S Diagram for Diesel Cycle

Figure 2.6: T-S Diagram for Combined/Dual Cycle

T-S Diagram



Compression ratio

To give direct comparison with an actual engine the ratio of specific volume, v1 / v2, is taken to be the same as the compression ratio of the actual engine,

Compression ratio, rv = =2

1

vv

olumeclearencevolumeclearencevesweptvolum

Pressure (P) 3

24

1

P3

P2

P4

P1

Volume ( v )

tconspv tan

V1V2

Clearance volume

Swept volume

Minimum volume

Maximum volume

TDC

BDC

Swept volume

Clearance volume

Figure 2.7: P-V Diagram for Otto Cycle Figure 2.8: Animated 4 Stroke Engine © MSF @ POLITEKNIK UNGKU

OMAR


Example 2.1

An Otto cycle in a petrol engine with a cylinder bore of 55mm, a stroke of 80mm, and a clearance volume of 23.3 cm3 is given. Find the compression ratio of this engine.

Solution:

Cylinder bore, B = 55mm = 5.5 cmStroke, S = 80mm = 8.0 cmClearance Volume = 23.3 cm 3

Abstract the data from the question. Convert the data into centimetre unit.

swept volume = x B² x S4

= x (5.5)² x 8

= 190.07 cm 3 4

Compression ratio, rv = 2

1

vv

olumeclearencevolumeclearencevesweptvolum

23.3 cm 3

= (190.07 + 23.3) cm 3

= 9.16

=



Exercise 2.1A petrol engine with a cylinder bore of 73 mm, a stroke of 95 mm, and a clearance volume of 26.3 cm 3 is given. Find the compression ratio of this engine.

[Ans: 16.12]Exercise 2.2An air engine is operated with cylinder bore 65 mm with the stroke of 73 mm. The clearance volume for this engine is 1/10 of swept volume. Calculate the compression ratio for this air engine.

[Ans: 11.00]Exercise 2.3Given compression ratio for a petrol engine is 10.5.The cylinder bore and stroke length for this engine are 69 mm and 83 mm. Calculate the clearance volume for this engine.

[Ans: 32.66 cm3 ]Exercise 2.4An otto cycle engine with compression ratio 9.5 operating with clearrance volume 23.45 cm 3. The stroke length for this engine is 83 mm, find the cylinder bore for this engine.

[Ans: 5.53 cm]

Self-Exercise



Otto Cycle AnalysisPressure (P) 3

24

1

P3

P2

P4

P1

Volume ( v )

tconspv tan

V1V2

Qin

Qout

Figure 2.9: P-V Diagram for Otto Cycle

Otto Cycle Process:

1 to 2 is isentropic compression with compression ratio v1 / v2, or rv .

2 to 3 is reversible constant volume heating, the heat supplied Q1

3 to 4 is isentropic expansion, v4 /v3 is similar to v1 /v2.

4 to 1 is reversible constant volume cooling, the heat rejected Q2.

Formula to find the thermal efficiency for Otto Cycle

P1V1

P2V2=

T1V11 = T2V2

1

Qin, Q1 = Cv(T3 ─ T2 )

Qout, Q2 = Cv(T4 ─ T1 ) Qout, Q2 = 1 -

Qin, Q1



Example 2.2 One petrol engine is working at a constant volume, the compression ratio is 8.5:1. Pressure and temperature at a beginning compression process is 101 kN/m2 and 840 C. Temperature at the beginning of an expand process is14960 C. Calculate the temperature and pressure at the important points based on the Otto cycle.

Solution:Pressure (P) 3

24

1

P3

P2

P4

P1

Volume ( v )

tconspv tan

V1V2

Qin

Qout

P-V Diagram for Otto Cycle

3

4

2

1v V

VVVr = 8.5

*T1 = 84°C + 273K = 357 K

P1 = 101 kN/m²

*T3 = 1496°C + 273K = 1769 K

* Temperature must be convert into Kelvin



Point 2 (1 to 2 is isentropic compression )

P1V1

P2V2=

P2 =V1

V2[ ] x P1

P2 =41.

[8.5] x 101 kN / m²

P2 = 2020.73 kN / m²

T1V11

= T2V21

T2 = x T1V1V2

[ ]1

T2 =40.

[8.5] x 357 K

T2 = 840.30 K

Point 3 (2 to 3 is reversible constant volume heating )

Constant volume

P2V2 P3V3T2 T3

=

P3 = T3T2

x P2

P3 = 1769840.30

x 2020.73 kN / m²

P3 = 4254.04 kN / m²



Point 4 (3 to 4 is isentropic expansion )

P3V3

P4V4=

P4 = V3V4[ ] x P3

P4 = 212.63 kN / m²

P4 =41.

x 4254.04 kN / m²[ 18.5]

T3V31

= T4V41

T4 = x T3V3V4

[ ]1

T4 =40.

x 1769 K

T4 = 751.55 K

[ ]18.5

Qin, Q1 = Cv (T3 ─ T2 ) = 0.718 kJ/kg.K x ( 1769 K – 840.30 K)

Qin, Q1 = 666.81 kJ/kg

Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 751.55 K – 357 K)

Qout, Q2 = 283.29 kJ/kg



Exercise 2.5

A four cylinder engine operates in Otto cycle, the volume is constant and the compression ratio is 9:1, beginning pressure is 105 KN/m2 , temperature is 83 ° C and final temperature is 1520 °C. Draw a p-v diagram and find the temperature and pressure for each point. Lastly calculate the efficiency of Otto cycle.

Exercise 2.6

One petrol engine is working at a constant volume, the compression ratio is 8.5:1. Pressure and temperature at a beginning compression process is 101 kN/m2 and 84 0C. Temperature at the beginning of an expand process is14960 C. Calculate the temperature and pressure at the important points based on the Otto cycle. Hence, calculate thermal efficiency for this engine.



Solution for Exercise 2.5

3

4

2

1v V

VVVr = 9

T1 = 83°C + 273K = 356 K

P1 = 105 kN/m²

T3 = 1520°C + 273K = 1793 K

Pressure (P) 3

24

1

P3

P2

P4

P1

Volume ( v )

tconspv tan

V1V2

Qin

Qout

P-V Diagram for Otto Cycle


P1V1

P2V2=

P2 =V1

V2[ ] x P1

P2 =41.

[9] x 105 kN / m²

P2 = 2275.77 kN / m²

T1V11

= T2V21

T2 = x T1V1V2

[ ]1

T2 =40.

[9] x 356 K

T2 = 857.33 K




P3V3

P4V4=

P4 = V3V4[ ] x P3

P4 =41.

x 4759.49 kN / m²[ 19 ]

Constant volume

P2V2 P3V3T2 T3

=

P3 = T3T2

x P2

P3 = 1793857.33

x 2275.77 kN / m²

P3 = 4759.49 kN / m²


P4 = 219.59 kN / m²

T3V31

= T4V41

T4 = x T3V3V4

[ ]1

T4 =40.

x 1793 K

T4 = 744.53 K

[ ]19





Qin, Q1 = Cv (T3 ─ T2 ) = 0.718 kJ/kg.K x ( 1793 K – 857.33 K)

Qin, Q1 = 671.81 kJ/kg


Qout, Q2 = 278.96 kJ/kg

Thermal Efficiecy for Otto Cycle

Qout, Q2 th, Otto = 1 - Qin, Q1

= 1 - 278.96 kJ.K/kg 671.81 kJ.K/kg

= 0.585 @ 58.5 %



Diesel Cycle Analysis

V

4

1

2 3P2 = P3

P1

V1= V4V3V2

P4

PPressure,

Volume

tconspv tan

Qin

Qout

Figure 2.10: P-V Diagram for Diesel Cycle

Diesel Cycle Process:

1 to 2 is isentropic compression with compression ratio v1 / v2, or rv .

2 to 3 is reversible constant pressure heating, the heat supplied Q1, v3 / v2, cut off

ratio, rc

3 to 4 is isentropic expansion,




Formula to find the thermal efficiency for Diesel Cycle


P1V1

P2V2=

Point 1 Get the data from the question.

T1V11 = T2V2

1To get pressure

To get temperature (Kelvin)

Point 3 (2 to 3 is reversible constant pressure heating )

Constant pressure

P2V2 P3V3T2 T3

=

T3T2

= rcV3V2

=




P3V3

P4V4=

P4 = V3V4[ ] x P3

=V3V4[ ] xV3

V2[ ] V2V1[ ]

V

4

1

2 3P2 = P3

P1

V1= V4V3V2

P4

PPressure,

Volume

tconspv tan

Qin

Qout

1

2

Put 12 into

=V3V4[ ] xrc[ ] 1

rv[ ]

P4 = rcrv[ ] x P3

T3V3 T4V4=1 1

T4 = V3V4[ ] x T3

13



=V3V4[ ] xV3

V2[ ] V2V1[ ]

=V3V4[ ] xrc[ ] 1

rv[ ] 4

Put 34 into

T4 = rcrv[ ] x T3

1

Qin, Q1 = Cp(T3 ─ T2 )

Qout, Q2 = Cv(T4 ─ T1 )Thermal Efficiency

Qout, Q2 = 1 - Qin, Q1

th, Diesel

V

4

1

2 3P2 = P3

P1

V1= V4V3V2

P4

PPressure,

Volume

tconspv tan

Qin

Qout



Example 2.3

Diesel engine has an inlet temperature and a pressure at 15°C and 1 bar respectively. The compression ratio is 12/ 1 and the maximum cycle temperature is 1100°C. Calculate the air standard thermal efficiency based on the diesel cycle.

= 12

T1 = 15°C + 273K = 288 K

P1 = 1 bar

T3 = 1100°C + 273K = 1373 K

rv

Solution




P1V1

P2V2=

P2 =V1

V2[ ] x P1

P2 =41.

[12] x 1 bar

P2 = 32.42 bar

T1V11

= T2V21

T2 = x T1V1V2

[ ]1

T2 =40.

[12] x 288 K

T2 = 778.15 K


Constant pressure

P2V2 P3V3T2 T3

=

T3T2

= rcV3V2

=

1373 K778.15 K

= = 1.764




P3V3

P4V4=

P4 = V3V4[ ] x P3

=V3V4[ ] xV3

V2[ ] V2V1[ ]

1

2

Put 12 into

=V3V4[ ] xrc[ ] 1

rv[ ]

P4 = rcrv[ ] x P3

P4 = 1.76412[ ] x 32.42 bar

41.

T3V3 T4V4=1 1

T4 = V3V4[ ] x T3

13

=V3V4[ ] xV3

V2[ ] V2V1[ ]

=V3V4[ ] xrc[ ] 1

rv[ ] 4

Put 34 into

T4 = rcrv[ ] x T3

1

T4 = 1.76412[ ] x 1373 K

40.

P4 = 2.21 bar

T4 = 637.67 K © MSF @ POLITEKNIK UNGKU

OMAR


Qin, Q1 = Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1373 K – 778.15 K)

Qin, Q1 = 597.82 kJ/kg


Qout, Q2 = 251.06 kJ/kg

Qout, Q2 th, Diesel = 1 - Qin, Q1

= 1 - 251.06 kJ/kg 597.82 kJ/kg

= 0.580 @ 58 %

Thermal Efficiecy for Diesel Cycle



Exercise 2.7

One engine operates in diesel cycle, the beginning temperature is 18 °C, and pressure is 1 bar. The compression ratio is 14:1 , maximun temperature is 1320°C. Given cp = 1.005 KJ/kg.K, R=287 KJ/kg.K. Calculate the temperature and rc and mechanical efficiency.

Exercise 2.8

A diesel cycle’s engine running with the beginning of compression process temperature 17 °C and pressure 1.1 bar. The compression ratio for this engine is 13.5 and the temperature for beginning expansion process is 1290 °C. Calculate thermal efficiency for this diesel cycle engine.

Exercise 2.9

An engine was operated with diesel cycle process with the beginning temperature and pressure at 16.8 °C and 1.05 bar. The compression ratio for this engine is 12.5 and cut of ratio for this engine is 1.75. Calculate thermal efficiency for this engine.




= 14

T1 = 18°C + 273K = 291 K

P1 = 1 bar

T3 = 1320 °C + 273K = 1593 K

rv

P1V1

P2V2=

P2 =V1

V2[ ] x P1

P2 =41.

[14] x 1 bar

P2 = 40.23 bar


T1V11

= T2V21

T2 = x T1V1V2

[ ]1

T2 =40.

[14] x 291 K

T2 = 836.27 K




Constant pressure

P2V2 P3V3T2 T3

=

T3T2

= rcV3V2

=

1593 K836.27 K

= = 1.905


P3V3

P4V4=

P4 = V3V4[ ] x P3

=V3V4[ ] xV3

V2[ ] V2V1[ ]

1

2=V3V4[ ] xrc[ ] 1

rv[ ]


Put 12 into

P4 = rcrv[ ] x P3

P4 = 1.90514[ ] x 40.23 bar

41.

P4 = 2.47 bar



T3V3 T4V4=1 1

T4 = V3V4[ ] x T3

13

=V3V4[ ] xV3

V2[ ] V2V1[ ]

=V3V4[ ] xrc[ ] 1

rv[ ] 4


Put 34 into

T4 = rcrv[ ] x T3

1

T4 = 1.90514[ ] x 1593 K

40.

T4 = 717.34 K



Qin, Q1 = 760.51 kJ/kg


Qout, Q2 = 306.11 kJ/kg


= 1 - 306.11 kJ/kg 760.51 kJ/kg

= 0.597 @ 59.7%




= 13.5

T1 = 17°C + 273K = 290 K

P1 = 1.1 bar

T3 = 1290 °C + 273K = 1563 K

rv

P1V1

P2V2=

P2 =V1

V2[ ] x P1

P2 =41.

[13.5] x 1.1 bar

P2 = 42.06 bar


T1V11

= T2V21

T2 = x T1V1V2

[ ]1

T2 =40.

[13.5] x 291 K

T2 = 821.36 K




Constant pressure

P2V2 P3V3T2 T3

=

T3T2

= rcV3V2

=

1563 K821.36 K

= = 1.903


P3V3

P4V4=

P4 = V3V4[ ] x P3

=V3V4[ ] xV3

V2[ ] V2V1[ ]

1

2=V3V4[ ] xrc[ ] 1

rv[ ]


Put 12 into

P4 = rcrv[ ] x P3

P4 = 2.71 bar

P4 = 1.90313.5[ ] x 40.23 bar

41.



T3V3 T4V4=1 1

T4 = V3V4[ ] x T3

13

=V3V4[ ] xV3

V2[ ] V2V1[ ]

=V3V4[ ] xrc[ ] 1

rv[ ] 4


Put 34 into

T4 = rcrv[ ] x T3

1

T4 = 1.90313.5[ ] x 1563 K

40.

T4 = 713.84 K



Qin, Q1 = 745.35 kJ/kg


Qout, Q2 = 304.32 kJ/kg


= 1 - 304.32 kJ/kg 745.35 kJ/kg

= 0.592 @ 59.2%




P1V1

P2V2=

P2 =V1

V2[ ] x P1

P2 =41.

[12.5] x 1.05 bar

P2 = 36.05 bar


T1V11

= T2V21

T2 = x T1V1V2

[ ]1

T2 =40.

[12.5] x 289.8 K

T2 = 795.91 K

= 12.5

T1 = 16.8°C + 273K = 289.8 K

P1 = 1.05 bar

rc

rv

= 1.75





P3V3

P4V4=

P4 = V3V4[ ] x P3

=V3V4[ ] xV3

V2[ ] V2V1[ ]

1

2=V3V4[ ] xrc[ ] 1

rv[ ]


Put 12 into

P4 = rcrv[ ] x P3

P4 = 2.30 bar

P4 = 1.7512.5[ ] x 36.05 bar

41.

Constant pressure

P2V2 P3V3T2 T3

=

T3T2

= rcV3V2

= 795.91 K

= = 1.75T3

T3 = 1392.84 K



T3V3 T4V4=1 1

T4 = V3V4[ ] x T3

13

=V3V4[ ] xV3

V2[ ] V2V1[ ]

=V3V4[ ] xrc[ ] 1

rv[ ] 4


Put 34 into

T4 = rcrv[ ] x T3

1

T4 = 1.7512.5[ ] x 1392.84 K

40.

T4 = 634.38 K


Qin, Q1 = Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1392.84 K – 795.91 K)

Qin, Q1 = 599.91 kJ/kg

Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 634.38 K – 289.8 K)

Qout, Q2 = 247.41 kJ/kg


= 1 - 247.41 kJ/kg 599.91 kJ/kg

= 0.588 @ 58.8 %



Dual/Combined Cycle Analysis

QinP

V1

2

3 4

5

P3=P4

V2=V3 V1=V5

contsvp .Qin

Qout

Dual/Combined Cycle Process:

1 to 2 is isentropic compression with compression ratio v1 / v2, or rv ,

2 to 3 is reversible constant volume heating, the heat supplied Q1

3 to 4 is reversible constant pressure heating, the heat supplied Q1, v4/ v3, cut off

ratio, rc

4 to 5 is isentropic expansion,


Figure 2.11: P-V Diagram for Dual/Combined Cycle



Formula to find the thermal efficiency for Dual/Combined Cycle


P1V1

P2V2=

Point 1 Get the data from the question.

T1V11 = T2V2

1To get pressure

To get temperature (Kelvin)


Constant volume

P2V2 P3V3T2 T3

=

P3 = T3T2

x P2

T3 = P3P2

x T2




Constant pressure

P3V3 P4V4T3 T4

=

T4T3

= rcV4V3

=

Assume the heat added at constant volume is equal to the heat added at constant pressure;

cv(T3 – T2 ) = cp (T4 – T3 )

T4 = cv(T3 – T2 ) + cp. (T3) cp




P4V4

P5V5=

P5 = V4V5[ ] x P4

=V4V5[ ] xV4

V3[ ] V2V1[ ]

1

2

Put 12 into

=V4V5[ ] xrc[ ] 1

rv[ ]

P5 = rcrv[ ] x P4

QinP

V1

2

3 4

5

P3=P4

V2=V3 V1=V5

contsvp .Qin

Qout



T4V4 T5V5=1 1

T5 = V4V5[ ] x T4

13

=V4V5[ ] xV4

V3[ ] V2V1[ ]

=V4V5[ ] xrc[ ] 1

rv[ ] 4

Put 34 into

T5 = rcrv[ ] x T4

1

QinP

V1

2

3 4

5

P3=P4

V2=V3 V1=V5

contsvp .Qin

Qout

Qin, Q1 = Cv(T3 ─ T2 ) + Cp(T4 ─ T3 )

Qout, Q2 = Cv(T5 ─ T1 )

Thermal Efficiency

Qout, Q2 = 1 - Qin, Q1

th, Dual



Example 2.4

An oil engine takes in air at 1.01 bar, 200C and the maximum cycle pressure is 69 bar. The compression ratio is 18/1. Draw the p-v diagram and calculate the air standard thermal efficiency based on the dual combustion cycle. Assume that the heat added at constant volume is equal to the heat added at constant pressure.

Solution

= 18

T1 = 20°C + 273K = 293 K

P1 = 1.01 bar

P3 = 69 bar

rv




P1V1

P2V2=

P2 =V1

V2[ ] x P1

P2 =41.

[18] x 1.01 bar

P2 = 57.77 bar

T1V11

= T2V21

T2 = x T1V1V2

[ ]1

T2 =40.

[18] x 293 K

T2 = 931.06 K


Constant volume

P2V2 P3V3T2 T3

=

P3 = T3T2

x P2

T3 = P3P2

x T2

T3 = 69 bar57.77 bar x 931.06 K

T3 = 1112.05 K



Assume the heat added at constant volume is equal to the heat added at constant pressure;

cv(T3 – T2 ) = cp (T4 – T3 )

T4 = cv(T3 – T2 ) + cp. (T3) cp

T4 = 0.718 kJ/kg.K (1112.05 K – 931.06 K) + 1.005 kJ/kg.K(1112.05 K)

1.005 kJ/kg.K

T4 = 1241.30 K


Constant pressure

P3V3 P4V4T3 T4

=

T4T3

= rcV4V3

=

rc =

1241.30 K1112.05 K

rc =

1.116




P4V4

P5V5=

P5 = V4V5[ ] x P4

=V4V5[ ] xV4

V3[ ] V2V1[ ]

1

2=V4V5[ ] xrc[ ] 1

rv[ ]

Put 12 into

P5 = rcrv[ ] x P4

P5 = 1.41 bar

P5 = 1.11618[ ] x 69 bar

41.

T4V4 T5V5=1 1

T5 = V4V5[ ] x T4

13

=V4V5[ ] xV4

V3[ ] V2V1[ ]

=V4V5[ ] xrc[ ] 1

rv[ ] 4

Put 34 into

T5 = rcrv[ ] x T4

1

T5 = 1.11618[ ] x 1241.30 K

40.

T5 = 408.16 K © MSF @ POLITEKNIK UNGKU

OMAR


Qin, Q1 = 0.718 kJ/kg.K(1112.05 K ─ 931.06 K ) + 1.005 kJ/kg.K (1241.30 K ─1112.05 K)

Qin, Q1 = Cv(T3 ─ T2 ) + Cp(T4 ─ T3 )

Qin, Q1 = 259.85 kJ/kg

Qout, Q2 = Cv (T5 ─ T1 )

Qout, Q2 = 82.68 kJ/kgQout, Q2 = 0.718 kJ/kg.K x (408.16 K– 293 K)

Qout, Q2 th, Dual = 1 - Qin, Q1

= 1 - 82.68 kJ/kg 259.85 kJ/kg

= 0.682 @ 68.2 %

Thermal Efficiecy for Dual/Combined Cycle



Self –Assessment

Exercise 2.11

In a dual combustion cycle, the maximum pressure is 64 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.01 bar and 18 0 C respectively. The compression ratio is 17/1.

Exercise 2.10

The pressure and temperature of air standard dual combustion cycle are given below, i. T1 = 290 Kii. P1 = 1.01 bariii. T2 = 871.1Kiv. T3 = 1087.5 Kv. T4 = 1236.3 Kvi. T5 = 429.3 K

and ratio of compression is 16:1. Calculate the thermal efficiency for this dual cycle engine.



Exercise 2.12

In a dual combustion cycle, the maximum pressure is 62 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.01 bar and 16.7 °C respectively. The compression ratio is 17.7.

Exercise 2.13

In a combine cycle, the maximum pressure is 67 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.03 bar and 17 °C respectively. The compression ratio is 16.9.

Exercise 2.14

An engine is operating in dual combustion cycle, the maximum pressure is 66 bar. Calculate the thermal efficiency for this engine when the pressure and temperature at the beginning of the compression process are 1.03 bar and 20.3 °C respectively. The compression ratio for this engine is 17.9.



Answer for Exercise 2.11

T1 = 291 K P1 = 1.01 bar

T2 = 903.80 K P2 = 53.33 bar

T3 = 1084.68 K P3 = 64 bar

T4 = 1213.91 K P4 = 64 barrc = 1.119

T5 = 408.85 K P5 = 1.42 bar

Qin = 259.75 kJ/kg Qout = 84.61 kJ/kg ɳth, dual = 0.674 @ 67.4 %


T1 = 289.7 K P1 = 1.01 bar

T2 = 914.40 K P2 = 56.43 bar

T3 = 1004.72 K P3 = 62 bar

T4 = 1069.25 K P4 = 62 barrc = 1.064

T5 = 347.30 K P5 = 1.21 bar





T1 = 290 K P1 = 1.03 bar

T2 = 898.57 K P2 = 53.94 bar

T3 = 1116.22 K P3 = 67 bar

T4 = 1271.71 K P4 = 67 barrc = 1.139

T5 = 432.40 K P5 = 1.54 bar



T1 = 293.3 K P1 = 1.03 bar

T2 = 929.94 K P2 = 58.46 bar

T3 = 1049.94 K P3 = 66 bar

T4 = 1135.68 K P4 = 66 barrc = 1.082

T5 = 369.62 K P5 = 1.30 bar






DJA3032 CHAPTER 2

Automotive