Problem 1.1:
In an air standard otto cycle the condition of air at beginning of compression is 1 bar,
300K, compressor ratio 6, the maximum temperature is 1000°C. Determine thermal efficiency
th.
Solution
Beginning of compression
: P1 = 1 bar, T1 = 300 K
Compression ratio
:r=6
Tmax  T3  1000C  1273K
 th  1 
1
r 
 1
 1
1
6
0.4 
 0.51164 or 51.16%    1.4
Problem 1.2 :
In an air standard otto cycle the condition of air at beginning of compression is 1 bar,
300K. The temperature at the beginning, end of burning are 400°C, 1000°C. Determine (a)
Compression ratio, (b) thermal efficiency th.
Solution
Beginning of compression
: P1 = 1 bar, T1 = 300 K
Beginning, ending of burning : T2 = 400°C = 693K, T3 =1000°C= 1273K
1 – 2 isentropic
T2  V2 
 
T1  V1 
 673 


 300 
r
 1
1 / 1.41

V1
V2
V1 v1

 7.53762
V2 v2
(a) Compression ratio (r) = 7.53762
(b) Thermal efficiency  th  1 
1
r
 1

1
 1  0.44576
7.53762 1.41
 th  0.55423 or 55.423%
Problem 1.3 :
The condition of air at the beginning of compression of an Otto cycle is 1 bar, 300K.The
max temperature is 1400K. The exhaust temperature is 700K. Determine (a) Compression ratio,
(b) max. pressure, (c) thermal efficiency th.
Solution
Beginning of compression
: P1 = 1 bar, T1 = 300 K
Beginning, ending of burning : Tmax = T3 = 1400K, T4= 700K
T2, P2, r are not given. So we cannot go by the usual method
P4V4  RT4
P1V1  RT1
P1V1  RT1
V1  V4
P4 T4

P1 T1
P4 700

1 300
P4  2.33333bar
P1V1  RT1
V1 
RT1 280  300

 0.861m 3 / kg
P1
1
V1  0.861m 3 / kg
Isentropic Process
 Ideal
gas
T3  V4 
 
T4  V3 
 1400 


 400 
 1
1 / 0.4
V
  4
 V3



V4
0.861
 5.65685 
V3
V3
V3  0.15220m 3 / kg
V2  V3
r
V1
0.861

 0.565685
V2 0.15220
(i) Compression Ratio (r) =5.65685
(ii) Thermal Efficiency
 th  1 
1
r
 1
 1
1
5.656850.4
Pmax  P3 ; P3V3  RT3  P3 
 50%
RT3 287  400

V3
0.15220
P3  26.39947 bar

P2  V1 
 0.861 
    

P1  V2 
 0.15220 
rp 
1.4
 P2  11.3137 bar
P3 26.39947

 2.33333
P2
11.3137
rp  2.33333
(iii) Mean effective pressure (MEP)
MEP 


P1 r   r rp  1
  1r  1


1 5.656851.4  5.65685 2.33333  1
MEP 
1.4  15.65685  1
MEP  4.04912 bar
Problem 1.4 :
The condition of air at the beginning of compression of an diesel cycle is 1 bar, 300K.
The max.temperature is 1400K. The exhaust temperature is 700K. Determine (a) Compression
ratio, (b) max. pressure, (c) thermal efficiency th, (d) MEP
Solution
Beginning of compression
: P1 = 1 bar, T1 = 300 K
Tmax = T3 = 1400K,
T4 = Exhaust temp = 700K
T2, P2, r are not given. So we cannot go by the usual method
P4V4  RT4
…(1)
P1V1  RT1
…(2)
P4 T4

P1 T1
P4 700

P1 300
P4  2.33333bar
P1V1  RT1
V1 
RT1 280  300

 0.861m 3 / kg
P1
1
V1  0.861m 3 / kg
Process (3) – (4) Isentropic
T3  V4 
 
T4  V3 
 T3

 T4



 1
1 /  1

 1400 


 400 
V4
V3
1 / 0.4 1
 0.801 

 
 V3 
V3  0.15220m 3 / kg
P3 
RT3 287  1400

V3
0.15220
P3  26.39865 bar
P3  P2  Pmax  26.39865 bar
Process (1) – (2) Isentropic

P 
P2  V1 
     2 
P1  V2 
 P1 
 26.39865 


1


1 / 1.4

V1
V2
1/ 

V1
V2
(i) Compression Ratio (r)
r
V1
 10.36133
V2
 P2

 P1



 1

0.4
T
T
 26.39865  1.4
 2  2 

T1
300 
1

T2  764.341534K
(ii) Thermal Efficiency
T4  T1 
 T3  T2 
700  300
 th  1 
 0.550522
1.41400  764.341534
 th  1 
 th  55.052%
Qs  C p T3  T2 
Qs  10051400  764.341534
Qs  638518.9291 J / kg
W  Qs  th  638518.9291 0.550522
W  351518.7179 J / kg
(iii) Mean effective pressure (MEP)
MEP 
r
W
W
351518.7179


Vs V1  V2
0.861  V2
V1
 V2  V1 / r  0.083m 3 / kg
V2
MEP 
351518.7179
0.861  0.083
MEP  451880Pa
MEP  4.5188 bar