DJA3032 CHAPTER 2

CHAPTER 2: PISTON ENGINE PROCESS
ANALYSIS
by
MOHD 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
2
4
1
P3
P2
P4
P1
Volume ( v )
tconspv tan
V1V2
Qin
Qout
V
4
1
2 3
P2 =
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)

Qin
P
V
1
2
3
4
5
P3=p4
V2=V3
V1=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
for Diesel Cycle
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
v
v
olumeclearencev
olumeclearencevesweptvolum 
Pressure (P) 3
2
4
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

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 cm
Stroke, S = 80mm = 8.0 cm
Clearance Volume = 23.3 cm 3
Abstract the data from the question.
Convert the data into centimetre unit.
swept volume = x B² x S
4

= x (5.5)² x 8
= 190.07 cm 3
4

Compression ratio, rv =
2
1
v
v
olumeclearencev
olumeclearencevesweptvolum 
23.3 cm 3
= (190.07 + 23.3) cm 3
= 9.16
=

Exercise 2.1
A 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 2.2
An 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 2.3
Given 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 2.4
An 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 Analysis
Pressure (P) 3
2
4
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=
T1V1
1 = 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
2
4
1
P3
P2
P4
P1
Volume ( v )
tconspv tan
V1V2
Qin
Qout
P-V Diagram for Otto Cycle
3
4
2
1
v
V
V
V
V
r  = 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
T1V1
1
= T2V2
1
T2 = x T1
V1
V2
[ ]
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 P3V3
T2 T3
= P3 = T3
T2
x P2
P3 = 1769
840.30
x 2020.73 kN / m²
P3 = 4254.04 kN / m²

Point 4 (3 to 4 is isentropic expansion )

P3V3

P4V4=
P4 = V3
V4
[ ] x P3
P4 = 212.63 kN / m²
P4 =
41.
x 4254.04 kN / m²[ 1
8.5 ]
T3V3
1
= T4V4
1
T4 = x T3
V3
V4
[ ]
1
T4 =
40.
x 1769 K
T4 = 751.55 K
[ ]1
8.5
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
1
v
V
V
V
V
r  = 9
T1 = 83°C + 273K = 356 K
P1 = 105 kN/m²
T3 = 1520°C + 273K = 1793 K
Pressure (P) 3
2
4
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
T1V1
1
= T2V2
1
T2 = x T1
V1
V2
[ ]
1
T2 =
40.
[9] x 356 K
T2 = 857.33 K


P3V3

P4V4=
P4 = V3
V4
[ ] x P3
P4 =
41.
x 4759.49 kN / m²[ 1
9 ]
Constant
volume
P2V2 P3V3
T2 T3
= P3 = T3
T2
x P2
P3 = 1793
857.33
x 2275.77 kN / m²
P3 = 4759.49 kN / m²
P4 = 219.59 kN / m²
T3V3
1
= T4V4
1
T4 = x T3
V3
V4
[ ]
1
T4 =
40.
x 1793 K
T4 = 744.53 K
[ ]1
9

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 3
P2 = P3
P1
V1= V4V3V2
P4
PPressure,
Volume
tconspv tan
Qin
Qout
Figure 2.10: P-V Diagram for Diesel Cycle
Diesel Cycle Process:
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.
T1V1
1 = T2V2
1
To get pressure
To get temperature (Kelvin)
Point 3 (2 to 3 is reversible constant pressure heating )
Constant
pressure
P2V2 P3V3
T2 T3
=
T3
T2
= rc
V3
V2
=


P3V3

P4V4=
P4 = V3
V4
[ ] x P3

=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
V
4
1
2 3
P2 = P3
P1
V1= V4V3V2
P4
PPressure,
Volume
tconspv tan
Qin
Qout
1
2
Put 12 into
=V3
V4
[ ] xrc[ ] 1
rv
[ ]
P4 = rc
rv
[ ] x P3

T3V3 T4V4=
1 1
T4 = V3
V4
[ ] x T3
1
3

=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
=V3
V4
[ ] xrc[ ] 1
rv
[ ] 4
Put 34 into
T4 = rc
rv
[ ] 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 3
P2 = 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
T1V1
1
= T2V2
1
T2 = x T1
V1
V2
[ ]
1
T2 =
40.
[12] x 288 K
T2 = 778.15 K
Constant
pressure
P2V2 P3V3
T2 T3
=
T3
T2
= rc
V3
V2
= 1373 K
778.15 K
= = 1.764


P3V3

P4V4=
P4 = V3
V4
[ ] x P3

=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
1
2
Put 12 into
=V3
V4
[ ] xrc[ ] 1
rv
[ ]
P4 = rc
rv
[ ] x P3

P4 = 1.764
12[ ] x 32.42 bar
41.
T3V3 T4V4=
1 1
T4 = V3
V4
[ ] x T3
1
3
=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
=V3
V4
[ ] xrc[ ] 1
rv
[ ] 4
Put 34 into
T4 = rc
rv
[ ] x T3
1
T4 = 1.764
12[ ] x 1373 K
40.
P4 = 2.21 bar
T4 = 637.67 K

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
T1V1
1
= T2V2
1
T2 = x T1
V1
V2
[ ]
1
T2 =
40.
[14] x 291 K
T2 = 836.27 K

Constant
pressure
P2V2 P3V3
T2 T3
=
T3
T2
= rc
V3
V2
= 1593 K
836.27 K
= = 1.905

P3V3

P4V4=
P4 = V3
V4
[ ] x P3

=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
1
2=V3
V4
[ ] xrc[ ] 1
rv
[ ]
Put 12 into
P4 = rc
rv
[ ] x P3

P4 = 1.905
14[ ] x 40.23 bar
41.
P4 = 2.47 bar

T3V3 T4V4=
1 1
T4 = V3
V4
[ ] x T3
1
3
=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
=V3
V4
[ ] xrc[ ] 1
rv
[ ] 4
Put 34 into
T4 = rc
rv
[ ] x T3
1
T4 = 1.905
14[ ] x 1593 K
40.
T4 = 717.34 K
Qin, Q1 = 760.51 kJ/kg
Qout, Q2 = 306.11 kJ/kg
 Qout, Q2
th, Diesel = 1 -
Qin, Q1
= 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
T1V1
1
= T2V2
1
T2 = x T1
V1
V2
[ ]
1
T2 =
40.
[13.5] x 291 K
T2 = 821.36 K

Constant
pressure
P2V2 P3V3
T2 T3
=
T3
T2
= rc
V3
V2
= 1563 K
821.36 K
= = 1.903

P3V3

P4V4=
P4 = V3
V4
[ ] x P3

=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
1
2=V3
V4
[ ] xrc[ ] 1
rv
[ ]
Put 12 into
P4 = rc
rv
[ ] x P3

P4 = 2.71 bar
P4 = 1.903
13.5[ ] x 40.23 bar
41.

T3V3 T4V4=
1 1
T4 = V3
V4
[ ] x T3
1
3
=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
=V3
V4
[ ] xrc[ ] 1
rv
[ ] 4
Put 34 into
T4 = rc
rv
[ ] x T3
1
T4 = 1.903
13.5[ ] x 1563 K
40.
T4 = 713.84 K
Qin, Q1 = 745.35 kJ/kg
Qout, Q2 = 304.32 kJ/kg
 Qout, Q2
th, Diesel = 1 -
Qin, Q1
= 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
T1V1
1
= T2V2
1
T2 = x T1
V1
V2
[ ]
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 = V3
V4
[ ] x P3

=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
1
2=V3
V4
[ ] xrc[ ] 1
rv
[ ]
Put 12 into
P4 = rc
rv
[ ] x P3

P4 = 2.30 bar
P4 = 1.75
12.5[ ] x 36.05 bar
41.
Constant
pressure
P2V2 P3V3
T2 T3
=
T3
T2
= rc
V3
V2
=
795.91 K
= = 1.75T3
T3 = 1392.84 K

T3V3 T4V4=
1 1
T4 = V3
V4
[ ] x T3
1
3
=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
=V3
V4
[ ] xrc[ ] 1
rv
[ ] 4
Put 34 into
T4 = rc
rv
[ ] x T3
1
T4 = 1.75
12.5[ ] 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
 Qout, Q2
th, Diesel = 1 -
Qin, Q1
= 1 - 247.41 kJ/kg
599.91 kJ/kg
= 0.588 @ 58.8 %

Dual/Combined Cycle Analysis
QinP
V
1
2
3 4
5
P3=P4
V2=V3 V1=V5
contsvp 
.
Qin
Qout
Dual/Combined Cycle Process:
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.
T1V1
1 = T2V2
1
To get pressure
To get temperature (Kelvin)
Constant
volume
P2V2 P3V3
T2 T3
= P3 = T3
T2
x P2
T3 = P3
P2
x T2

Constant
pressure
P3V3 P4V4
T3 T4
=
T4
T3
= rc
V4
V3
=
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 = V4
V5
[ ] x P4

=V4
V5
[ ] xV4
V3
[ ] V2
V1
[ ]
1
2
Put 12 into
=V4
V5
[ ] xrc[ ] 1
rv
[ ]
P5 = rc
rv
[ ] x P4

QinP
V
1
2
3 4
5
P3=P4
V2=V3 V1=V5
contsvp 
.
Qin
Qout

T4V4 T5V5=
1 1
T5 = V4
V5
[ ] x T4
1
3
=V4
V5
[ ] xV4
V3
[ ] V2
V1
[ ]
=V4
V5
[ ] xrc[ ] 1
rv
[ ] 4
Put 34 into
T5 = rc
rv
[ ] x T4
1
QinP
V
1
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
T1V1
1
= T2V2
1
T2 = x T1
V1
V2
[ ]
1
T2 =
40.
[18] x 293 K
T2 = 931.06 K
Constant
volume
P2V2 P3V3
T2 T3
= P3 = T3
T2
x P2
T3 = P3
P2
x T2
T3 = 69 bar
57.77 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 P4V4
T3 T4
=
T4
T3
= rc
V4
V3
=
rc =
1241.30 K
1112.05 K
rc = 1.116


P4V4

P5V5=
P5 = V4
V5
[ ] x P4

=V4
V5
[ ] xV4
V3
[ ] V2
V1
[ ]
1
2=V4
V5
[ ] xrc[ ] 1
rv
[ ]
Put 12 into
P5 = rc
rv
[ ] x P4

P5 = 1.41 bar
P5 = 1.116
18[ ] x 69 bar
41.
T4V4 T5V5=
1 1
T5 = V4
V5
[ ] x T4
1
3
=V4
V5
[ ] xV4
V3
[ ] V2
V1
[ ]
=V4
V5
[ ] xrc[ ] 1
rv
[ ] 4
Put 34 into
T5 = rc
rv
[ ] x T4
1
T5 = 1.116
18[ ] x 1241.30 K
40.
T5 = 408.16 K

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/kg
Qout, 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 K
ii. P1 = 1.01 bar
iii. T2 = 871.1K
iv. T3 = 1087.5 K
v. T4 = 1236.3 K
vi. 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 bar
rc = 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 bar
rc = 1.064
T5 = 347.30 K
P5 = 1.21 bar
Qin = 129.70 kJ/kg
Qout = 41.35 kJ/kg
ɳth, dual = 0.681 @ 68.1 %

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 bar
rc = 1.139
T5 = 432.40 K
P5 = 1.54 bar
Qin = 312.54 kJ/kg
Qout = 102.25 kJ/kg
ɳth, dual = 0.673 @ 67.3 %
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 bar
rc = 1.082
T5 = 369.62 K
P5 = 1.30 bar
Qin = 172.33 kJ/kg
Qout = 54.79 kJ/kg
ɳth, dual = 0.682 @ 68.2 %

DJA3032 CHAPTER 2

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DJA3032 CHAPTER 2