SlideShare a Scribd company logo
1 of 51
Download to read offline
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.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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)
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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)
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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).
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION 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
=
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION 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
Point 2 (1 to 2 is isentropic compression )

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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 3 (2 to 3 is reversible constant volume heating )

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
Point 4 (3 to 4 is isentropic expansion )
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
Solution for Exercise 2.5
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.5
Qin, Q1 = Cv (T3 ─ T2 ) = 0.718 kJ/kg.K x ( 1793 K – 857.33 K)
Qin, Q1 = 671.81 kJ/kg
Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 744.53 K – 356 K)
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 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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:
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,
4 to 1 is reversible constant volume cooling,
the heat rejected Q2.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Formula to find the thermal efficiency for Diesel Cycle
Point 2 (1 to 2 is isentropic compression )

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
=
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 4 (3 to 4 is isentropic expansion )

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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
=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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 2 (1 to 2 is isentropic compression )

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
Point 3 (2 to 3 is reversible constant pressure heating )
Constant
pressure
P2V2 P3V3
T2 T3
=
T3
T2
= rc
V3
V2
= 1373 K
778.15 K
= = 1.764
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 4 (3 to 4 is isentropic expansion )

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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Qin, Q1 = Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1373 K – 778.15 K)
Qin, Q1 = 597.82 kJ/kg
Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 637.67 K – 288 K)
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.7
= 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
Point 2 (1 to 2 is isentropic compression )
T1V1
1
= T2V2
1
T2 = x T1
V1
V2
[ ]
1
T2 =
40.
[14] x 291 K
T2 = 836.27 K
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 3 (2 to 3 is reversible constant pressure heating )
Constant
pressure
P2V2 P3V3
T2 T3
=
T3
T2
= rc
V3
V2
= 1593 K
836.27 K
= = 1.905
Solution for Exercise 2.7

P3V3

P4V4=
P4 = V3
V4
[ ] x P3

=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
1
2=V3
V4
[ ] xrc[ ] 1
rv
[ ]
Point 4 (3 to 4 is isentropic expansion )
Put 12 into
P4 = rc
rv
[ ] x P3

P4 = 1.905
14[ ] x 40.23 bar
41.
P4 = 2.47 bar
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
T3V3 T4V4=
1 1
T4 = V3
V4
[ ] x T3
1
3
=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
=V3
V4
[ ] xrc[ ] 1
rv
[ ] 4
Solution for Exercise 2.7
Put 34 into
T4 = rc
rv
[ ] x T3
1
T4 = 1.905
14[ ] x 1593 K
40.
T4 = 717.34 K
Thermal Efficiecy for Diesel Cycle
Qin, Q1 = Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1593 K – 836.27 K)
Qin, Q1 = 760.51 kJ/kg
Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 717.34 K – 291 K)
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%
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.8
= 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
Point 2 (1 to 2 is isentropic compression )
T1V1
1
= T2V2
1
T2 = x T1
V1
V2
[ ]
1
T2 =
40.
[13.5] x 291 K
T2 = 821.36 K
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 3 (2 to 3 is reversible constant pressure heating )
Constant
pressure
P2V2 P3V3
T2 T3
=
T3
T2
= rc
V3
V2
= 1563 K
821.36 K
= = 1.903
Solution for Exercise 2.8

P3V3

P4V4=
P4 = V3
V4
[ ] x P3

=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
1
2=V3
V4
[ ] xrc[ ] 1
rv
[ ]
Point 4 (3 to 4 is isentropic expansion )
Put 12 into
P4 = rc
rv
[ ] x P3

P4 = 2.71 bar
P4 = 1.903
13.5[ ] x 40.23 bar
41.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
T3V3 T4V4=
1 1
T4 = V3
V4
[ ] x T3
1
3
=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
=V3
V4
[ ] xrc[ ] 1
rv
[ ] 4
Solution for Exercise 2.8
Put 34 into
T4 = rc
rv
[ ] x T3
1
T4 = 1.903
13.5[ ] x 1563 K
40.
T4 = 713.84 K
Thermal Efficiecy for Diesel Cycle
Qin, Q1 = Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1563 K – 821.36 K)
Qin, Q1 = 745.35 kJ/kg
Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 713.84 K – 290 K)
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%
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.9

P1V1

P2V2=
P2 =

V1
V2
[ ] x P1
P2 =
41.
[12.5] x 1.05 bar
P2 = 36.05 bar
Point 2 (1 to 2 is isentropic compression )
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 3 (2 to 3 is reversible constant pressure heating )
Solution for Exercise 2.9

P3V3

P4V4=
P4 = V3
V4
[ ] x P3

=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
1
2=V3
V4
[ ] xrc[ ] 1
rv
[ ]
Point 4 (3 to 4 is isentropic expansion )
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
T3V3 T4V4=
1 1
T4 = V3
V4
[ ] x T3
1
3
=V3
V4
[ ] xV3
V2
[ ] V2
V1
[ ]
=V3
V4
[ ] xrc[ ] 1
rv
[ ] 4
Solution for Exercise 2.9
Put 34 into
T4 = rc
rv
[ ] x T3
1
T4 = 1.75
12.5[ ] x 1392.84 K
40.
T4 = 634.38 K
Thermal Efficiecy for Diesel Cycle
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 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Dual/Combined Cycle Analysis
QinP
V
1
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,
5 to 1 is reversible constant volume cooling,
the heat rejected Q2.
Figure 2.11: P-V Diagram for Dual/Combined Cycle
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Formula to find the thermal efficiency for Dual/Combined Cycle
Point 2 (1 to 2 is isentropic compression )

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 volume heating )
Constant
volume
P2V2 P3V3
T2 T3
= P3 = T3
T2
x P2
T3 = P3
P2
x T2
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 4 (3 to 4 is reversible constant pressure heating )
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 5 (4 to 5 is isentropic expansion )

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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 2 (1 to 2 is isentropic compression )

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
Point 3 (2 to 3 is reversible constant volume heating )
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
Point 4 (3 to 4 is reversible constant pressure heating )
Constant
pressure
P3V3 P4V4
T3 T4
=
T4
T3
= rc
V4
V3
=
rc =
1241.30 K
1112.05 K
rc = 1.116
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 5 (4 to 5 is isentropic expansion )

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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION 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.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
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 %
Answer for Exercise 2.12
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 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Answer for Exercise 2.13
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 %
Answer for Exercise 2.14
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 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE

More Related Content

What's hot

Internal Combustion Engines - Construction and Working (All you need to know,...
Internal Combustion Engines - Construction and Working (All you need to know,...Internal Combustion Engines - Construction and Working (All you need to know,...
Internal Combustion Engines - Construction and Working (All you need to know,...Mihir Pai
 
A.thermo Mc conkey ch12 solution-pb
A.thermo Mc conkey ch12 solution-pbA.thermo Mc conkey ch12 solution-pb
A.thermo Mc conkey ch12 solution-pbM SAQIB
 
Engine type and classification.
Engine type  and classification. Engine type  and classification.
Engine type and classification. Manmeet Singh
 
Load test on a perkins diesel engine
Load test on a perkins diesel engineLoad test on a perkins diesel engine
Load test on a perkins diesel engineLahiru Dilshan
 
Internal combustion engine (ja304) chapter 4
Internal combustion engine (ja304) chapter 4Internal combustion engine (ja304) chapter 4
Internal combustion engine (ja304) chapter 4mechanical86
 
Testing and performance of IC engine
Testing and performance of IC engineTesting and performance of IC engine
Testing and performance of IC engineMustafa Bzu
 
torque , power, volumetric efficiency and their dependence on unit air charge...
torque , power, volumetric efficiency and their dependence on unit air charge...torque , power, volumetric efficiency and their dependence on unit air charge...
torque , power, volumetric efficiency and their dependence on unit air charge...mp poonia
 
Acutal Cycles and Their Analysis - Unit-I
Acutal Cycles and Their Analysis - Unit-IAcutal Cycles and Their Analysis - Unit-I
Acutal Cycles and Their Analysis - Unit-IS.Vijaya Bhaskar
 
Valve timing diagram for - four stroke & two stroke - diesel & petrol engine ...
Valve timing diagram for - four stroke & two stroke - diesel & petrol engine ...Valve timing diagram for - four stroke & two stroke - diesel & petrol engine ...
Valve timing diagram for - four stroke & two stroke - diesel & petrol engine ...Satish Patel
 
Performance testing of IC engine
Performance testing of IC enginePerformance testing of IC engine
Performance testing of IC engineINDRAKUMAR PADWANI
 
Disassemble and reassemble a single cylinder or multi cylinder
Disassemble and reassemble a single cylinder or multi cylinderDisassemble and reassemble a single cylinder or multi cylinder
Disassemble and reassemble a single cylinder or multi cylinderSunil Kumar
 
Air compressor
Air compressorAir compressor
Air compressorsureshkcet
 

What's hot (20)

Mc conkey 13-pb
Mc conkey 13-pbMc conkey 13-pb
Mc conkey 13-pb
 
Internal Combustion Engines - Construction and Working (All you need to know,...
Internal Combustion Engines - Construction and Working (All you need to know,...Internal Combustion Engines - Construction and Working (All you need to know,...
Internal Combustion Engines - Construction and Working (All you need to know,...
 
DJA3032 CHAPTER 6
DJA3032   CHAPTER 6DJA3032   CHAPTER 6
DJA3032 CHAPTER 6
 
A.thermo Mc conkey ch12 solution-pb
A.thermo Mc conkey ch12 solution-pbA.thermo Mc conkey ch12 solution-pb
A.thermo Mc conkey ch12 solution-pb
 
Engine type and classification.
Engine type  and classification. Engine type  and classification.
Engine type and classification.
 
Load test on a perkins diesel engine
Load test on a perkins diesel engineLoad test on a perkins diesel engine
Load test on a perkins diesel engine
 
Performance Parameters of IC engines
Performance Parameters of  IC enginesPerformance Parameters of  IC engines
Performance Parameters of IC engines
 
Internal combustion engine (ja304) chapter 4
Internal combustion engine (ja304) chapter 4Internal combustion engine (ja304) chapter 4
Internal combustion engine (ja304) chapter 4
 
Testing and performance of IC engine
Testing and performance of IC engineTesting and performance of IC engine
Testing and performance of IC engine
 
torque , power, volumetric efficiency and their dependence on unit air charge...
torque , power, volumetric efficiency and their dependence on unit air charge...torque , power, volumetric efficiency and their dependence on unit air charge...
torque , power, volumetric efficiency and their dependence on unit air charge...
 
Acutal Cycles and Their Analysis - Unit-I
Acutal Cycles and Their Analysis - Unit-IAcutal Cycles and Their Analysis - Unit-I
Acutal Cycles and Their Analysis - Unit-I
 
Valve timing diagram for - four stroke & two stroke - diesel & petrol engine ...
Valve timing diagram for - four stroke & two stroke - diesel & petrol engine ...Valve timing diagram for - four stroke & two stroke - diesel & petrol engine ...
Valve timing diagram for - four stroke & two stroke - diesel & petrol engine ...
 
Performance testing of IC engine
Performance testing of IC enginePerformance testing of IC engine
Performance testing of IC engine
 
IC engine Mathematical problem & solution
 IC engine Mathematical problem & solution IC engine Mathematical problem & solution
IC engine Mathematical problem & solution
 
Performance parameters
Performance parametersPerformance parameters
Performance parameters
 
Disassemble and reassemble a single cylinder or multi cylinder
Disassemble and reassemble a single cylinder or multi cylinderDisassemble and reassemble a single cylinder or multi cylinder
Disassemble and reassemble a single cylinder or multi cylinder
 
Air compressor
Air compressorAir compressor
Air compressor
 
DJA3032 CHAPTER 5
DJA3032   CHAPTER 5DJA3032   CHAPTER 5
DJA3032 CHAPTER 5
 
Diesel engine
Diesel engineDiesel engine
Diesel engine
 
Compressor
CompressorCompressor
Compressor
 

Viewers also liked

Internal combustion engine (ja304) chapter 3
Internal combustion engine (ja304) chapter 3Internal combustion engine (ja304) chapter 3
Internal combustion engine (ja304) chapter 3mechanical86
 
INTERNAL COMBUSTION ENGINES PPT
INTERNAL COMBUSTION ENGINES PPT INTERNAL COMBUSTION ENGINES PPT
INTERNAL COMBUSTION ENGINES PPT AKASH1001
 
Thermodynamic Chapter 5 Air Standard Cycle
Thermodynamic Chapter 5 Air Standard CycleThermodynamic Chapter 5 Air Standard Cycle
Thermodynamic Chapter 5 Air Standard CycleMuhammad Surahman
 
solution manual to basic and engineering thermodynamics by P K NAG 4th edition
solution manual to basic and engineering thermodynamics by P K NAG 4th editionsolution manual to basic and engineering thermodynamics by P K NAG 4th edition
solution manual to basic and engineering thermodynamics by P K NAG 4th editionChandu Kolli
 
Dynamic otto-cycle-analysis
Dynamic otto-cycle-analysisDynamic otto-cycle-analysis
Dynamic otto-cycle-analysisEbra21
 
Chapter 1 internal combustion engine
Chapter 1 internal combustion engineChapter 1 internal combustion engine
Chapter 1 internal combustion engineMuhd Taufik Muhammad
 
gas reheat and intercooling
gas reheat and intercoolinggas reheat and intercooling
gas reheat and intercoolingCik Minn
 
Engineering fundamentals of_the_internal_combustion_engine Erdi Karaçal Mecha...
Engineering fundamentals of_the_internal_combustion_engine Erdi Karaçal Mecha...Engineering fundamentals of_the_internal_combustion_engine Erdi Karaçal Mecha...
Engineering fundamentals of_the_internal_combustion_engine Erdi Karaçal Mecha...Erdi Karaçal
 
heat engine information detail
heat engine information detailheat engine information detail
heat engine information detailpratik darji
 
Ic engine and its types,applications
Ic engine and its types,applicationsIc engine and its types,applications
Ic engine and its types,applicationsYuvaraja MM
 

Viewers also liked (13)

JA304 chapter 1
JA304 chapter 1JA304 chapter 1
JA304 chapter 1
 
Internal combustion engine (ja304) chapter 3
Internal combustion engine (ja304) chapter 3Internal combustion engine (ja304) chapter 3
Internal combustion engine (ja304) chapter 3
 
INTERNAL COMBUSTION ENGINES PPT
INTERNAL COMBUSTION ENGINES PPT INTERNAL COMBUSTION ENGINES PPT
INTERNAL COMBUSTION ENGINES PPT
 
Thermodynamic Chapter 5 Air Standard Cycle
Thermodynamic Chapter 5 Air Standard CycleThermodynamic Chapter 5 Air Standard Cycle
Thermodynamic Chapter 5 Air Standard Cycle
 
solution manual to basic and engineering thermodynamics by P K NAG 4th edition
solution manual to basic and engineering thermodynamics by P K NAG 4th editionsolution manual to basic and engineering thermodynamics by P K NAG 4th edition
solution manual to basic and engineering thermodynamics by P K NAG 4th edition
 
Dynamic otto-cycle-analysis
Dynamic otto-cycle-analysisDynamic otto-cycle-analysis
Dynamic otto-cycle-analysis
 
Chapter 1 internal combustion engine
Chapter 1 internal combustion engineChapter 1 internal combustion engine
Chapter 1 internal combustion engine
 
3
33
3
 
gas reheat and intercooling
gas reheat and intercoolinggas reheat and intercooling
gas reheat and intercooling
 
Engineering fundamentals of_the_internal_combustion_engine Erdi Karaçal Mecha...
Engineering fundamentals of_the_internal_combustion_engine Erdi Karaçal Mecha...Engineering fundamentals of_the_internal_combustion_engine Erdi Karaçal Mecha...
Engineering fundamentals of_the_internal_combustion_engine Erdi Karaçal Mecha...
 
heat engine information detail
heat engine information detailheat engine information detail
heat engine information detail
 
Ic engine and its types,applications
Ic engine and its types,applicationsIc engine and its types,applications
Ic engine and its types,applications
 
Ic engine
Ic engineIc engine
Ic engine
 

Similar to DJA3032 CHAPTER 2

Itenas termodinamika ii bab 9a
Itenas termodinamika ii bab 9aItenas termodinamika ii bab 9a
Itenas termodinamika ii bab 9aNoviyantiNugraha
 
gas power plant problem.pdf
gas power plant problem.pdfgas power plant problem.pdf
gas power plant problem.pdfMahamad Jawhar
 
Gas Power Cycles.ppt
Gas Power Cycles.pptGas Power Cycles.ppt
Gas Power Cycles.pptWasifRazzaq2
 
2 gas turbinepp
2 gas turbinepp2 gas turbinepp
2 gas turbineppskdass23
 
Air standard cycles
Air standard cyclesAir standard cycles
Air standard cyclesSoumith V
 
Lecture 7.pptx
Lecture 7.pptxLecture 7.pptx
Lecture 7.pptxNelyJay
 
Thermodynamics Examples and Class test
Thermodynamics Examples and Class testThermodynamics Examples and Class test
Thermodynamics Examples and Class testVJTI Production
 
chap5airstandardcycle2010-130703012738-02.pdf
chap5airstandardcycle2010-130703012738-02.pdfchap5airstandardcycle2010-130703012738-02.pdf
chap5airstandardcycle2010-130703012738-02.pdf21M220KARTHIKEYANC
 
Mcconkey Chapter 9 solution
Mcconkey Chapter 9 solutionMcconkey Chapter 9 solution
Mcconkey Chapter 9 solutionAzeem Waqar
 
Aircraft propulsion non ideal cycle analysis
Aircraft propulsion   non ideal cycle analysisAircraft propulsion   non ideal cycle analysis
Aircraft propulsion non ideal cycle analysisAnurak Atthasit
 
Thermodynamic cycles4.ppt
Thermodynamic cycles4.pptThermodynamic cycles4.ppt
Thermodynamic cycles4.pptMehtab Rai
 
GAS POWER CYCLES PRESENTATION FOR STUDENT UNIVERSITY
GAS POWER CYCLES PRESENTATION FOR STUDENT UNIVERSITYGAS POWER CYCLES PRESENTATION FOR STUDENT UNIVERSITY
GAS POWER CYCLES PRESENTATION FOR STUDENT UNIVERSITYssuser5a6db81
 
Chap 8: Internal Combustion Engine Power Plant
Chap 8: Internal Combustion Engine Power PlantChap 8: Internal Combustion Engine Power Plant
Chap 8: Internal Combustion Engine Power PlantMulugeta Wotango
 
Gaspowercycle by- anshuman pptt
Gaspowercycle by- anshuman ppttGaspowercycle by- anshuman pptt
Gaspowercycle by- anshuman ppttanahuman singh
 
Aircraft propulsion non ideal turbofan cycle analysis
Aircraft propulsion   non ideal turbofan cycle analysisAircraft propulsion   non ideal turbofan cycle analysis
Aircraft propulsion non ideal turbofan cycle analysisAnurak Atthasit
 

Similar to DJA3032 CHAPTER 2 (20)

Itenas termodinamika ii bab 9a
Itenas termodinamika ii bab 9aItenas termodinamika ii bab 9a
Itenas termodinamika ii bab 9a
 
Gas power-09
Gas power-09Gas power-09
Gas power-09
 
13. The Otto Cycle.pdf
13. The Otto Cycle.pdf13. The Otto Cycle.pdf
13. The Otto Cycle.pdf
 
gas power plant problem.pdf
gas power plant problem.pdfgas power plant problem.pdf
gas power plant problem.pdf
 
Ch19 ssm
Ch19 ssmCh19 ssm
Ch19 ssm
 
Gas Power Cycles.ppt
Gas Power Cycles.pptGas Power Cycles.ppt
Gas Power Cycles.ppt
 
2 gas turbinepp
2 gas turbinepp2 gas turbinepp
2 gas turbinepp
 
Air standard cycles
Air standard cyclesAir standard cycles
Air standard cycles
 
Lecture 7.pptx
Lecture 7.pptxLecture 7.pptx
Lecture 7.pptx
 
Thermodynamics Examples and Class test
Thermodynamics Examples and Class testThermodynamics Examples and Class test
Thermodynamics Examples and Class test
 
chap5airstandardcycle2010-130703012738-02.pdf
chap5airstandardcycle2010-130703012738-02.pdfchap5airstandardcycle2010-130703012738-02.pdf
chap5airstandardcycle2010-130703012738-02.pdf
 
Dual cycle
Dual cycleDual cycle
Dual cycle
 
Ambrish
AmbrishAmbrish
Ambrish
 
Mcconkey Chapter 9 solution
Mcconkey Chapter 9 solutionMcconkey Chapter 9 solution
Mcconkey Chapter 9 solution
 
Aircraft propulsion non ideal cycle analysis
Aircraft propulsion   non ideal cycle analysisAircraft propulsion   non ideal cycle analysis
Aircraft propulsion non ideal cycle analysis
 
Thermodynamic cycles4.ppt
Thermodynamic cycles4.pptThermodynamic cycles4.ppt
Thermodynamic cycles4.ppt
 
GAS POWER CYCLES PRESENTATION FOR STUDENT UNIVERSITY
GAS POWER CYCLES PRESENTATION FOR STUDENT UNIVERSITYGAS POWER CYCLES PRESENTATION FOR STUDENT UNIVERSITY
GAS POWER CYCLES PRESENTATION FOR STUDENT UNIVERSITY
 
Chap 8: Internal Combustion Engine Power Plant
Chap 8: Internal Combustion Engine Power PlantChap 8: Internal Combustion Engine Power Plant
Chap 8: Internal Combustion Engine Power Plant
 
Gaspowercycle by- anshuman pptt
Gaspowercycle by- anshuman ppttGaspowercycle by- anshuman pptt
Gaspowercycle by- anshuman pptt
 
Aircraft propulsion non ideal turbofan cycle analysis
Aircraft propulsion   non ideal turbofan cycle analysisAircraft propulsion   non ideal turbofan cycle analysis
Aircraft propulsion non ideal turbofan cycle analysis
 

Recently uploaded

Nosfdsfsdfasdfasdfasdfsadf asdfasdfasdfasdf
Nosfdsfsdfasdfasdfasdfsadf asdfasdfasdfasdfNosfdsfsdfasdfasdfasdfsadf asdfasdfasdfasdf
Nosfdsfsdfasdfasdfasdfsadf asdfasdfasdfasdfJulia Kaye
 
Study on Financing of zero-emission trucks and their infrastructure
Study on Financing of zero-emission trucks and their infrastructureStudy on Financing of zero-emission trucks and their infrastructure
Study on Financing of zero-emission trucks and their infrastructureEuropeanCleanTruckin
 
Commercial Extractor fan repair services
Commercial Extractor fan repair servicesCommercial Extractor fan repair services
Commercial Extractor fan repair servicesmb1294198
 
Work Experience - A Love Supreme.ghgfhdgfhfgpptx
Work Experience - A Love Supreme.ghgfhdgfhfgpptxWork Experience - A Love Supreme.ghgfhdgfhfgpptx
Work Experience - A Love Supreme.ghgfhdgfhfgpptxLewisJB
 
Modern Trams, Light rail transit systems.pdf
Modern Trams, Light rail transit  systems.pdfModern Trams, Light rail transit  systems.pdf
Modern Trams, Light rail transit systems.pdfmaputi
 
Lakshitha maduranga CV - for data entry clerck
Lakshitha maduranga CV - for data entry clerckLakshitha maduranga CV - for data entry clerck
Lakshitha maduranga CV - for data entry clerckLakshanMadhushanka3
 

Recently uploaded (8)

EVAT - Future Mobility Transformation in Thailand
EVAT - Future Mobility Transformation in ThailandEVAT - Future Mobility Transformation in Thailand
EVAT - Future Mobility Transformation in Thailand
 
Nosfdsfsdfasdfasdfasdfsadf asdfasdfasdfasdf
Nosfdsfsdfasdfasdfasdfsadf asdfasdfasdfasdfNosfdsfsdfasdfasdfasdfsadf asdfasdfasdfasdf
Nosfdsfsdfasdfasdfasdfsadf asdfasdfasdfasdf
 
Reinventing the Car - as I reported it in 1985!
Reinventing the Car - as I reported it in 1985!Reinventing the Car - as I reported it in 1985!
Reinventing the Car - as I reported it in 1985!
 
Study on Financing of zero-emission trucks and their infrastructure
Study on Financing of zero-emission trucks and their infrastructureStudy on Financing of zero-emission trucks and their infrastructure
Study on Financing of zero-emission trucks and their infrastructure
 
Commercial Extractor fan repair services
Commercial Extractor fan repair servicesCommercial Extractor fan repair services
Commercial Extractor fan repair services
 
Work Experience - A Love Supreme.ghgfhdgfhfgpptx
Work Experience - A Love Supreme.ghgfhdgfhfgpptxWork Experience - A Love Supreme.ghgfhdgfhfgpptx
Work Experience - A Love Supreme.ghgfhdgfhfgpptx
 
Modern Trams, Light rail transit systems.pdf
Modern Trams, Light rail transit  systems.pdfModern Trams, Light rail transit  systems.pdf
Modern Trams, Light rail transit systems.pdf
 
Lakshitha maduranga CV - for data entry clerck
Lakshitha maduranga CV - for data entry clerckLakshitha maduranga CV - for data entry clerck
Lakshitha maduranga CV - for data entry clerck
 

DJA3032 CHAPTER 2

  • 1. 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
  • 2. 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. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 3. 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) © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 4. 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) © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 5. 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. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 6. 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). © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 7. 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 8. 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 9. 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 = © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 10. 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 11. 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 12. 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 13. 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 14. 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 15. 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. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 16. 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 Point 2 (1 to 2 is isentropic compression )  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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 17. Point 3 (2 to 3 is reversible constant volume heating )  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 Point 4 (3 to 4 is isentropic expansion ) 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 Solution for Exercise 2.5 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 18. Solution for Exercise 2.5 Qin, Q1 = Cv (T3 ─ T2 ) = 0.718 kJ/kg.K x ( 1793 K – 857.33 K) Qin, Q1 = 671.81 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 744.53 K – 356 K) 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 % © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 19. 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: 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, 4 to 1 is reversible constant volume cooling, the heat rejected Q2. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 20. Formula to find the thermal efficiency for Diesel Cycle Point 2 (1 to 2 is isentropic compression )  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 = © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 21. Point 4 (3 to 4 is isentropic expansion )  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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 22. =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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 23. 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 24. Point 2 (1 to 2 is isentropic compression )  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 Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure P2V2 P3V3 T2 T3 = T3 T2 = rc V3 V2 = 1373 K 778.15 K = = 1.764 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 25. Point 4 (3 to 4 is isentropic expansion )  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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 26. Qin, Q1 = Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1373 K – 778.15 K) Qin, Q1 = 597.82 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 637.67 K – 288 K) 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 27. 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. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 28. Solution for Exercise 2.7 = 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 Point 2 (1 to 2 is isentropic compression ) T1V1 1 = T2V2 1 T2 = x T1 V1 V2 [ ] 1 T2 = 40. [14] x 291 K T2 = 836.27 K © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 29. Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure P2V2 P3V3 T2 T3 = T3 T2 = rc V3 V2 = 1593 K 836.27 K = = 1.905 Solution for Exercise 2.7  P3V3  P4V4= P4 = V3 V4 [ ] x P3  =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] 1 2=V3 V4 [ ] xrc[ ] 1 rv [ ] Point 4 (3 to 4 is isentropic expansion ) Put 12 into P4 = rc rv [ ] x P3  P4 = 1.905 14[ ] x 40.23 bar 41. P4 = 2.47 bar © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 30. T3V3 T4V4= 1 1 T4 = V3 V4 [ ] x T3 1 3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] =V3 V4 [ ] xrc[ ] 1 rv [ ] 4 Solution for Exercise 2.7 Put 34 into T4 = rc rv [ ] x T3 1 T4 = 1.905 14[ ] x 1593 K 40. T4 = 717.34 K Thermal Efficiecy for Diesel Cycle Qin, Q1 = Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1593 K – 836.27 K) Qin, Q1 = 760.51 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 717.34 K – 291 K) 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% © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 31. Solution for Exercise 2.8 = 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 Point 2 (1 to 2 is isentropic compression ) T1V1 1 = T2V2 1 T2 = x T1 V1 V2 [ ] 1 T2 = 40. [13.5] x 291 K T2 = 821.36 K © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 32. Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure P2V2 P3V3 T2 T3 = T3 T2 = rc V3 V2 = 1563 K 821.36 K = = 1.903 Solution for Exercise 2.8  P3V3  P4V4= P4 = V3 V4 [ ] x P3  =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] 1 2=V3 V4 [ ] xrc[ ] 1 rv [ ] Point 4 (3 to 4 is isentropic expansion ) Put 12 into P4 = rc rv [ ] x P3  P4 = 2.71 bar P4 = 1.903 13.5[ ] x 40.23 bar 41. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 33. T3V3 T4V4= 1 1 T4 = V3 V4 [ ] x T3 1 3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] =V3 V4 [ ] xrc[ ] 1 rv [ ] 4 Solution for Exercise 2.8 Put 34 into T4 = rc rv [ ] x T3 1 T4 = 1.903 13.5[ ] x 1563 K 40. T4 = 713.84 K Thermal Efficiecy for Diesel Cycle Qin, Q1 = Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1563 K – 821.36 K) Qin, Q1 = 745.35 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 713.84 K – 290 K) 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% © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 34. Solution for Exercise 2.9  P1V1  P2V2= P2 =  V1 V2 [ ] x P1 P2 = 41. [12.5] x 1.05 bar P2 = 36.05 bar Point 2 (1 to 2 is isentropic compression ) 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 35. Point 3 (2 to 3 is reversible constant pressure heating ) Solution for Exercise 2.9  P3V3  P4V4= P4 = V3 V4 [ ] x P3  =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] 1 2=V3 V4 [ ] xrc[ ] 1 rv [ ] Point 4 (3 to 4 is isentropic expansion ) 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 36. T3V3 T4V4= 1 1 T4 = V3 V4 [ ] x T3 1 3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] =V3 V4 [ ] xrc[ ] 1 rv [ ] 4 Solution for Exercise 2.9 Put 34 into T4 = rc rv [ ] x T3 1 T4 = 1.75 12.5[ ] x 1392.84 K 40. T4 = 634.38 K Thermal Efficiecy for Diesel Cycle 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 % © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 37. Dual/Combined Cycle Analysis QinP V 1 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, 5 to 1 is reversible constant volume cooling, the heat rejected Q2. Figure 2.11: P-V Diagram for Dual/Combined Cycle © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 38. Formula to find the thermal efficiency for Dual/Combined Cycle Point 2 (1 to 2 is isentropic compression )  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 volume heating ) Constant volume P2V2 P3V3 T2 T3 = P3 = T3 T2 x P2 T3 = P3 P2 x T2 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 39. Point 4 (3 to 4 is reversible constant pressure heating ) 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 40. Point 5 (4 to 5 is isentropic expansion )  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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 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 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 42. 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 43. Point 2 (1 to 2 is isentropic compression )  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 Point 3 (2 to 3 is reversible constant volume heating ) 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 44. 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 Point 4 (3 to 4 is reversible constant pressure heating ) Constant pressure P3V3 P4V4 T3 T4 = T4 T3 = rc V4 V3 = rc = 1241.30 K 1112.05 K rc = 1.116 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 45. Point 5 (4 to 5 is isentropic expansion )  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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 46. 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 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 47. 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. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 48. 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. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 49. 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 % Answer for Exercise 2.12 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 % © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 50. Answer for Exercise 2.13 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 % Answer for Exercise 2.14 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 % © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
  • 51. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE