Internal combustion engine (ja304) chapter 2

INTERNAL COMBUSTION ENGINE (JA304)

CHAPTER 2
PISTON ANALYSIS
BY

MOHD SAHRIL MOHD FOUZI
MECHANICAL ENGINEERING DEPARTMENT
UNGKU OMAR POLYTECHNIC

Otto Cycle Analysis
Pressure (P) 3
P3
pvγ = cons tan t
Draw the P-V diagram for Otto
cycle. Label all important point in
P2 2
the diagram such as Pressure (P), 4
Volume (V), each point of P4
process.

P1
1
Volume ( v )

ClearanceV2
V1
Volume Swept Volume
TDC BDC

State the Otto cycle process for each points:
• Point 1 to 2 is isentropic compression with compression ratio v1 / v2, or rv .
• Point 2 to 3 is reversible constant volume heating, the heat supplied Q1
• Point 3 to 4 is isentropic expansion, v4 /v3 is similar to v1 /v2.
• Point 4 to 1 is reversible constant volume cooling, the heat rejected Q2.

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,

v1 sweptvolume + clearencevolume
Compression ratio, rv = =
v2 clearencevolume

π
swept volume = x B² x S where, B = cylinder bore @ cylinder
4 diameter @ cylinder piston
S = length of stroke
*Note: Please convert all unit to centimentre

The heat supplied, Q1, at constant volume between T2 and T3 is given by the
following equation ,
Q1 = Cv(T3 ─ T2 )

Similarly the heat rejected per unit mass at constant volume between T4 and
T1 is given by the equation

Q2 = Cv(T4 ─ T1 )

Pressure (P) 3
P3
pvγ = cons tan t
The heat
supplied, Q1
P2 2
4

P4
The heat
P1
rejected, Q2
1
Volume ( v )
V2 V1

The processes 1 to 2 and 3 to 4 are isentropic and therefore there is no heat
flow during these processes.

η = Q2 =
Q1
Cv (T3 – T2) – Cv (T4 – T1)
=
Cv (T3 – T2) – Cv (T4 – T1)

Cv (T3 – T2) Cv (T3 – T2) Cv (T3 – T2)

Cv (T4 – T1)
= 1 –
Cv (T3 – T2)

Thermal efficiency for Otto cycle

Since process 1- 2 and 3 to 4 are isentropic, use the following equation

γ γ γ
P1V1 = P2 V2 P2 =
P1
( ) V1
V2

γ −1 γ −1 γ −1
T 1 V1 = T 2 V2
T2 =
T1
( ) V1
V2

γ −1 γ −1 γ −1
V4
T2 =
T1 ( )
V1
V2
= (V )
3
= T3
T4
= rv

γ −1 γ −1
Where rv is the compression ratio, then T3 = T4 rv and T2 = T 1 rv

η = Q2 =
Q1
Cv (T3 – T2) – Cv (T4 – T1)
= 1 – (T4 – T1)
Cv (T3 – T2) (T3 – T2)

= 1 – (T4 – T1)
γ −1
(T4 – T1) rv

1 – 1
= γ −1
rv


Tutorial 1

Calculate the ideal air standard cycle efficiency based on the Otto cycle in a
petrol engine with a cylinder bore of 55mm, a stroke of 80mm, and a
clearance volume of 23.3 cm 3. Use the information form this compression
ratio to calculate :

Pressure and temperature at a beginning compression process is 101 kN/m2
and 870 C. Temperature at the beginning of an expand process is15020 C.
Calculate the temperature and pressure at the important points based on the
Otto cycle.

Answer for Tutorial 1

Cylinder bore, B = 55mm = 5.5 cm
Abstract the data from the
Stroke, S = 80mm = 8.0 cm question. Convert the data into
Clearance Volume = 23.3 cm 3 centimetre unit.

π
swept volume = x B² x S
4 Substitute the above data into
π this formula to get the swept
= x (5.5)² x 8
4 volume
= 190.07 cm 3

v1 sweptvolume + clearencevolume
Compression ratio, rv = =
v2 clearencevolume

= (190.07 + 23.3) cm 3
23.3 cm 3
= 9.16

Thermal efficiency for Otto Cycle

η= 1 – 1
γ −1
= 1 – 1
rv 9.16 (1.4 −1) Formula used to get the
thermal efficiency for Otto
= 1 - 0.412 Cycle.
= 0.588 @ 58.8 %

Otto Cycle Analysis
Pressure (P) 3
P3
pvγ = cons tan t

1st Step: Draw the P-V
P2 2 diagram for Otto Cycle
4
with complete labeling
P4 for each important point
and axis.
P1
1
Volume ( v )
V2 V1

From the question: P1 = 101 kN/m2
T1 = 870 C + 273 = 360 K

T3 = 15020 C + 273 =1775 K

Compression Process (1-2)

Point 1: Beginning of compression process
P1 = 101 kN/m2
T1 = 870 C + 273 = 360 K

Point 2: End of compression process
γ γ
P1V1 = P2 V2
γ Equal to compression ratio, rv
P2 =
P1
( )
V1
V2
γ 1.4
P2 = P 1 ( ) V1
V2
= 101 kN/m2 (9.16)

P2 = 2243.75 kN/m2

Point 2: End of compression process
γ −1 γ −1
T1V1 = T2V2
Equal to compression ratio, rv
γ −1

T1
( )
T 2 = V1
V2
γ −1 1.4 −1
( )
T 2 = T 1 V1
V2
= 360 (9.16)

T2 = 873.09 K

Heat Process(2-3)
Gas is burning at a constant Volume, V2 = V3. Temperature and pressure
are built-up from T2 to T3..

P2V2 P3V3 P2 P3
= =
T2 T3 T2 T3

Point 3: Beginning of expansion process
T3 is given = 15020 C +273 = 1775 K

P3 = T 3
T2
( )P 2 =
( 1775 K
873.09 K )(2243.75 kN/m 2
)

P3 = 4561.56 kN/m2

Point 4: Expansion process (3 -4 isentropic)
Gas is expanded from V3 to V4 , pressure is reduced from P3 to P4
γ γ
P3 V3 = P4 V4
γ
P4 =
P3
( )
V3
V4
γ 1.4
(r )1 1
P4 = P 3
v
= 4561.56 kN/m 2
( 9.16 )
P4 = 205.3 kN/m2

γ −1 γ −1
T 3 V3 = T 4 V4
γ −1
T4 =
T3
( ) V3
V4
γ −1
T4 =T (V )
3
3

V 4
γ −1 1 1.4 −1
( 1775 K ) (
=T ( ) 9.16 )
T4 1
3 =
r v

T4 = 731.88 K

Diesel Cycle Analysis

Draw the P-V diagram for Diesel
cycle. Label all important point in
the diagram such as Pressure (P), The heat
Volume (V), each point of supplied, Q1
process. Pressure, P

2 3 pv γ = cons tan t
P2 = P3

4
P4
The heat
In diesel cycle, heat is supplied at P1 1
rejected, Q2
a constant pressure and rejected
at a constant volume. The V Volume
V2 V3 V1= V4
process of cycle is:
Process 1 to 2 is isentropic compression
Process 2 to 3 is reversible constant pressure heating.
Process 3 to 4 is isentropic expansion.
Process 4 to 1 is reversible constant volume cooling.

At constant pressure from the equation, per kg of air
Q23 = Cp (T3 ─ T2)

At constant volume per kg of air, the heat being rejected is
Q41 = Cv(T4 ─ T1)

Thermal Efficiency for Diesel Cycle As we know the value of
temperature T4 is higher than T1.
Q23 + Q41
η= So, we change the position
Q23 between T4 & T1 and automatically
the sign of the equation also
Q
= 1 + 41 change.
Q23

C (T − T ) (
Cv T4 − T1 )
= 1+ v 1 4 1 - C (T − T )
C p ( T3 − T2 ) p 3 2

Process 1-2 ( Isentropic)
γ −1
T2  V1  γ −1 V 
=  = rv Because of ratio V1 and V2,  1  is
T1 V2   V2 
 
r
equal to compression ratio, v

γ −1
T2 = T1 × rv

Process 2 – 3 ( constant pressure)

P2V2 P3V3 V2 V3
= =
T2 T3 T2 T3
T3 V3
= = rc Cut of ratio
T2 V2

T3 = T2 rc

Process 3 – 4 ( isentropic)

γ −1 γ −1
T4 =  V3


 =
 V3 V 
 × 2 =


rc  γ −1
T3  V4
 

 V2 V1 
  
 rv 




rc  1



rv 


η diesel 1  rc γ − 1 
γ −1  
= 1 - Thermal efficiency for
rv  γ ( rc − 1)  Diesel Cycle

*Note: In this equation, efficiency of diesel cycle is based on rc and rv

Example
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.

1st Step: Draw the P-V diagram for Diesel
Solution cycle process with complete labeling for each
Pressure, P important points and axis
2 3 pvγ = cons tan t
P2 = P3
2nd Step: Abstract all the data given
in the question :
T1 = 15°C + 273 K = 288 K
4
P4 P1 = 1 bar
T3 = 1100°C + 273 K = 1373 K
P1
compression ratio ,rv = 12
1

V Volume
V2 V3 V1= V4

Process 1-2 ( Isentropic)
γ −1
T2  V1  γ −1
=  = rv
T1 V2 
γ −1
T2 = T1 × rv
T2 = (288 K) x (12) (1.4 −1)
T2 = 778 K

Process 2 – 3 ( constant pressure)
At constant pressure from 2 to 3, pv = RT is for perfect gas, therefore

P2V2 P3V3 T3 V3
= = = rc
T2 T3 T2 V2
1373K V
= 3 = rc
778K V2

rc = 1.765 Cut of ratio

Therefore;
v4 =
v4 × v2 = 12 ×
1
= 6.798
v3 v2 v3 1.765

rv 


1

rc 
 

γ −1
t3 =  v4


 = 6.7980.4 = 2.153
t4  v3




T4 = T3
2.153
1373
T4 = = 638K
2.153

Also per kilogram of air, the heat supplied is
Then from the equation Q1 = Cp ( T3 − T2 ) = 1.005(1373 − 778) = 598kJ / kg.
Also per kilogram of air, the heat supplied is
Then from the equation Q2 = CV ( T4 − T1 ) = 0.718( 638 − 288) = 252kJ / kg.

Therefore; the thermal efficiency for Diesel cycle:

Q2
η = 1−
Q1
251
η = 1− = 0.58 @ 58%
598

The Dual Combustion Cycle
In this cycle, heat is supplied in two parts; the first part at constant volume
and the second in constant pressure. Hence the name ‘dual combustion’.

P Heat supplied at
constant pressure,
Q1
3
P3=p4 4
Heat supplied at
constant p.v γ = conts
volume, Q1 2

5
Heat rejected,
Q2
1
V
V2=V3 V1=V5
P-V Diagram for Dual @ Combined Cycle

The dual combustion cycle process.
Process 1 to 2 is isentropic compression.
Process 2 to 3 is reversible constant volume heating.
Process 3 to 4 is reversible constant pressure heating.
Process 4 to 5 is isentropic expansion.
Process 5 to 1 is reversible constant volume cooling.

rv = compression ratio, V
V 1
2

rc = cut of ratio, V
V
4
3

rp = pressure ratio, P3
P
2

Thermal efficiency for dual cycle
Q2
η = 1−
Q1

The heat supplied, Q1, is calculated using the equation,
Q1 =c v (T3 −T2 ) +c p (T4 −T3 )

The heat rejected, Q2 , is calculated by
Q2 = cv ( T5 − T1 )

Example
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.

γ −1
T2  V1 
=  = 180.4 = 3.18
T1  V2 
 
T2 = 3.18 x T1 (Where T1 = 20 + 273 = 293 K)

T2 = 3.18 x 293 = 931 K.

γ
P2  V1 
=   = 181.4 = 57.2
P  V2 
1  
p2 = 57.2 x 1.01 = 57.8 bar.

from 2-3, the process is at constant volume, hence

P3 T3 P3V3 P2V2
= (since = and V3 =V2 )
P2 T2 T3 T2

P3 69 × 931
T3 = × T2 =
P2 P2
69 × 931
T3 = = 1112 K
57.8

Now the heat added at constant volume is equal to the heat added at
constant pressure in this example, therefore,
cv(T3 – T2 ) = cp (T4 – T3 )

0.718(1112 – 931) = 1.005 (T4 – 1112)

0.718 × 181
T4 = + 1112 = 1241.4 K
1.005

Find T5, it is necessary to know the value of the volume ratio, V5/V4. At
P constant pressure from 3 to 4,

P3V3 P4V4
=
P3=p4 3 T3 T4
4
V4 T4 1241.4
p.v γ = conts
= = = 1.116
2 V3 T3 1112
V5 V1 V1 V3 1
5 = = = 18 × = 16.14
V4 V4 V2 V4 1.116
1
V
V2=V3 V1=V5

γ −1
T4  V5 
=  = 16.140.4 = 3.04
T5  V4 
 
1241.4
T5 = = 408 K
3.04
Now the heat supplied, Q1 = cv (T3 – T2) + cp (T4 – T3 )

Q1 = 0.718 (1112 – 931) + 1.005(1241.4 – 1112)
Q1 = 260 kJ/kg

The heat rejected, Q2, is given by
Q2 = cv (T5 – T1 )
= 0.718(408 – 293) = 82.6 kJ/kg

Q2 82.6kJ / kg Thermal efficiency for
η = 1− = 1− = 0.682 @ 68.2% Dual @ Combined
Q1 260kJ / kg Cycle

TUTORIAL

Question 1
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.

Question 2
One engine operates in diesel cycle, the beginning temperature is 18 °C,
and pressure is 1 bar. The compression ratio is 14:1 , the temperature at
beginning of isentropic expansion is 1320°C. given cp= 1.005 KJ/kg.K,
R=287 KJ/kg.K.
Calculate the temperature and rc and mechanical efficiency.

TUTORIAL

Question 3
In a dual combustion cycle the maximum pressure is 54 bar. Calculate the
thermal efficiency when the pressure and temperature is at the start of
compression 1.01 bar and 17 0 C respectively. The ratio of compression is
16/1.

Internal combustion engine (ja304) chapter 2

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Internal combustion engine (ja304) chapter 2