OTTO CYCLE (OR) CONSTANT VOLUME CYCLE
The present day petrol engine operates on the otto cycle. This cycle was introduced in
practical form by scientist Otto in 1876. An I.C. Engine does not undergo a cycle change but it is
assumed here that the working medium is pure air which does not undergo chemical change. The
air is simply heated and cooled to undergo a cycle. The working fluid is also assumed to be a
perfect gas (having constant specific heats). Further it is assumed that the ideal indicator diagram is
strictly followed.
This cycle consists of two reversible adiabatic(isentropic) processes and two constant volume
processes.
The P-V and T-S diagrams are shown in fig (1.2)
The otto cycle consists of
Process 1-2: Isentropic compression Q  0
Air is compressed isentropically.
Process 2-3: Constant volume heat addition Qs  mCv T3  T2 
Heat is added (spark ignites) while volume remains giant.
Process 3-4: Isentropic Expansion Q  0
Expansion of air takes place isentropically
Process 4-1: Constant volume heat rejection Qr  mCv T4  T1 
Heat rejected to the surroundings at constant volume.
Net work done  Wnet  Qs  Qr  Heat supplied – Heat rejected
 air s tan dard   thermal 
 1
Wnet Qs  Qr

Qs
Qs
mCv (T4  T1 )
Qr
 1
Qs
mCv (T3  T2 )
 1
r = compression ratio 

(T4  T1 )
(T3  T2 )
…(1)
V1
V2
Volume at the beginning of compressio n V4

Volume at the end of compressio n
V3
1-2 Isentropic process :
T2  V1 
 
T1  V2 
 1
 r 
 1
3-4 Isentropic process :
T3  V4

T4  V3



 1
V
  1
 V2



 1
 r 
 1
T2 T3 T3 T4
 , 
T1 T4 T2 T1
T3
T
1  4 1
T2
T1
T3  T2 T4  T1

T2
T1
T4  T1 T1  V2


T3  T2 T2  V1
T  T V 
Substitute 4 1   2 
T3  T2  V1 
V
  1   2
 V1



 1
 1
1
r
 1



 1
 1
in equation 1, we get
V1  V4
and V2  V3 
 otto  1 

V1 v1 
 
 Note : r 
V2 v2 

1
r
 1
This expression in which =1.4 is called air standard efficiency of the Otto cycle.
So the efficiency is the function of compression ratio only. If the compression ratio is
more, then efficiency will be more. However, we cannot increase the compression ratio beyond
the limit since detonation (noisy and destructive combustion problem) occur in petrol engine at
higher compression ratio.
Work dome in otto cycle
We know that the work done per kg in otto cycle is given by
 P V  P4V4 P2V2  P1V1 

W   3 3

 1
  1 

Also
P3 P2

 r
P4 P1
Also
P3 P4

  p (Pressure ratio)
P4 P1
…(3)
 V V

Alsowe know V1  rV2 and V4  rV3  1  4  r 
 V2 V3

Substituting these in (2) we have
W 

 P3V3

 PV

1 
 1  P1V1  2 2  1
 P4V4 
 1 
 P4V4

 P1V1

 P3

P

1 
 1  P1V1  2  1 
 P4V4 
 1 
 P4 r 
 P1 r  
 V1 V4

 r
 
 V2 V3


…(2)

r

V1   r 
 P4   1  P1   1 
 1   r

r



P3 P2
 r 
 from (3) 
P2 P1


Work done
W  




V1
P4 r  1  1  P1 r  1  1
 1

V1

r  1  1P4  P1 
 1




P1V1  1
r  1rp  1
 1
…(4)
 P4

  rp 
 P1

Mean effective pressure (M.E.P)
Mean effective pressure is defined as constant pressure to which the cylinder is subjected
to so that same work is developed.
Expression For Mean Effective Pressure of Otto cycle
Mean effective pressure is the ratio of net work done to the swept volume. Therefore
Mean effective pressure (M.E.P)
 Pm 
Net work done
W

Swept volume V1  V2
From (2) we have
 P3V3  P4V4 P2V2  P1V1


 1
 1

Pm 

V1  V2








Now from (4) we have



P1V1  1
r  1 rp  1
 1
Pm 
V1  V2
Now V1  V2  V1 
V1
r
…(5)
 V1

  r 
 V2

 r  1
V1  V2  V1 
 r 
…(6)
Substituting (6) in (5) we get
Pm 



P1V1  1
r  1 rp  1 
 1
1
 r 1
V1 

 r 
Therefore we get
Mean effective pressure
Pm 


  P r
P1 r r  1  1 rp  1
  1  r  1
1
p

 1 r   r
r  1    1

DIESEL CYCLE OR CONSTANT PRESSURE CYCLE
Dr.Rudholph Diesel introduced diesel cycle in 1897. In case of otto cycle heat is supplied
to working fluid at constant volume while in case of Diesel cycle heat is supplied at constant
pressure. Diesel cycle consists of the following four operations.
Process 1-2 = Adiabatic compression
Process 2-3 = Constant pressure heat
Process 3-4 = Adiabatic expansion
Process 4-1 = Heat rejection at constant
The limitation on compression ratio in the S.I. engine can be overcome by compressing
air alone, instead of the fuel-air mixture, and then injecting fuel into the cylinder in spray form
when combustion is desired.' The temperature of air after compression must be high enough so
that the fuel sprayed into the hot air burns spontaneously. The rate of burning can, to some
extent, be controlled by the rate of injection of fuel. An engine operating in this way is called a
compression ignition (C.I.) engine. The sequence of processes in the elementary operation of a
C.I. engine is, shown in Fig. 1.4
Process 1-2, Intake. The air valve is open. The piston moves out admitting air into the cylinder
at constant pressure.
Process 2-3, Compression. The air is then compressed by the piston to the minimum volume
with all valves closed.
Process 3-4, Fuel injection and combustion. The fuel valve is open, fuel is sprayed into the hot
air, and combustion take place at constant pressure.
Process 4-5, Expansion. The combustion products expand, doing work on the piston which
moves out to the maximum volume.
Process 5-6, Blow-down. The exhaust valve opens, and the pressure drops to the initial pressure.
Pressure 6-1, Exhaust. With the exhaust valve open, the piston moves towards the cylinder
cover driving away the combustion products from the cylinder at constant pressure.
The above processes constitute an engine cycle, which is completed in four strokes of the
piston or two revolutions of the crank shaft.
Figure 1.5 shows the air standard cycle, called the Diesel cycle, corresponding to the C.I.
engine, as described above. The cycle is composed of:
Two reversible adiabatic, one reversible isobar, and one reversible isochore.
Air is compressed reversibly and adiabatically in process 1-2. Heat is then added to it
from an external source reversibly at constant pressure in process 2-3. Air then expands
reversibly and adiabatically in process 3-4. Heat is rejected reversibly at constant volume in
process 4-1, and the cycle repeats itself.
For m. kg of air in the cylinder, the efficiency analysis of the cycle can be made as given
below.
Heat supplied Q1  Q23  mc p T3  T2 
Heat rejected Q2  Q41  mcv T4  T1 
Efficiency   1 

  1
mcv T4  T1 
Q2
 1
Q1
mc p T3  T2 
T4  T1 
 T3  T2 
Let compression ratio, r 
 (cut-off ratio) 
V3 v3

V2 v 2
Process 1-2 (Compression)
…(1)
V1 v1

V2 v2
T2  V1

T1  V2



 1
 r 
 1
or T2
 T1 r 
 1
Process 2-3 (Constant pressure)
T3  V3 
 
T2  V2 
 1
  or T3   .T2   .T1 r 
 1
Process 3-4 (Expansion)
T3  V4 
 
T4  V3 
Now T4 
 1
r
  

T3
r /   1
 1
 V4 V1 V1 V2 r 


  

 V3 V3 V2 V3  
T1 r  1

 T1 . r
 1
r /  
Substituting values of T2 , T3 , T4 in (1) we get

T1 .   T1
 1
 1 
 T1 .r
  .T1 .r


 diesel  1  


Air Standard Efficiency
 diesel
 1     1 
   1 


 1    1
  1 
 1 


1



r


1



r





 
…(2)
1    1
 is also greater than unity.
As   1, 
    1 
Therefore, the efficiency of the diesel engine is less than that of the Otto cycle for the same
compression ratio.
Work done in Diesel cycle
The net work done in diesel cycle in terms of P, V is given as follow:
The work done
W   P2 V3  V2   P3V3  P4V4
 1
 P2 V2  V2  

P2V2  P1V1
 1
P3 V2  P4 rV2 P2V2  P1 rV2

 1
 1
 V3

V
  ;V3  V2 ; 1  r;V1  rV2 ;V1  V4 V4  rV2 

V2
 V2

W  P2V2   1 

V2
P2   1  1  P3   P4 r  P2  P1r 
 1
W 

P3 V2  P4 rV2 P2V2  P1 rV2

 1
 1

P2V2 
P4
P1 
  1  1    r  1  r  P3  P2 
 1 
P3
P2 



P2V2
  1r  1       r 1  1  r 1
 1

 P  V    

 4   3       r  r  
r
 P3  V4 



P1V1r  1
  1  1       r 1  1  r 1
W
 1

 P  V 

V
 2   1  orP2  P1  r  and 1  r or V2  V1r 1 
V2
 P1  V2 

Work Done W  


P1V1r  1
   1  r 1    1
 1

…(3)
Expression for Mean Effective Pressure (Pm) for Diesel cycle
We know that Mean Effective Pressure (Pm) is given by
Net Work done
W

Swept volume V1  V2
…(4)
 1
 r 1
V1  V2  V1 1    V1 

 r
 r 
…(5)
Pm 
Substituting (3), (5) in (4) we get
Pm 


P1r 
   1  r 1    1
  1r  1
