Q. An engine working on the Otto cycle is supplied with air at 0.1 MPa, 35°C. The compression
ratio is 7. The heat supplied is 2500 kJ/kg. Draw the P-V diagram of the Otto cycle. Assuming
Cp=1.005 kJ/kg K, Cv=0.717 kJ/kg K, γ=1.4, and R = 0.287 kJ/kg K, calculate:
(i) the cycle efficiency,
(ii) the maximum pressure and temperature of the cycle,
(iii) the mean effective pressure.
[10 Marks; Test time:15min]
(i) Cycle efficiency
T1= 308 K; P1= 0.1 MPa; r = 7 ; Qs = 2500 kJ/kg
1 mark
1
𝜂𝑜𝑡𝑡𝑜 = 1 − 𝑟 ϒ−1 ;
𝜂𝑜𝑡𝑡𝑜 = 54%
1 mark
(ii) Maximum pressure and temperature of the cycle
𝑣1
=7
𝑣2
𝑇2
𝑣1 𝛾−1
=( )
= (7)0.4 = 2.17
𝑇1
𝑣2
𝑇2 = 2.17 × 308 = 670.8 𝐾
𝑄𝑠 = 𝑐𝑣 (𝑇3 − 𝑇2 ) = 2500
𝑇3 − 670.8 =
1 mark
𝑘𝐽
𝑘𝑔
2500
0.717
𝑇3 = 𝑇𝑚𝑎𝑥 = 3486.7 + 670.8 = 𝟒𝟏𝟓𝟕. 𝟓 𝑲
𝑃2
𝑣1 𝛾
= ( ) = (7)1.4 = 15.24
𝑃1
𝑣2
𝑃2 = 1.524 𝑀𝑃𝑎
𝑃3 𝑃2
=
𝑇3 𝑇2
1 mark
1 mark
𝑃3 = 𝑃𝑚𝑎𝑥 = 1.524 ×
4157.5
= 𝟗. 𝟒𝟒 𝑴𝑷𝒂
670.8
1 mark
(iii) Mean effective pressure
𝑝𝑚 =
𝑊𝑛𝑒𝑡
(𝑣1 − 𝑣2 )
𝑊net = 𝑄𝑠 × 𝜂𝑜𝑡𝑡𝑜 = 2500 × 0.54 = 1350
𝑘𝐽
𝑘𝑔
1 mark
𝑣1 =
𝑅𝑇1 0.287 × 308
3
=
= 0.883 𝑚 ⁄𝑘𝑔
𝑃1
100
𝑣2 =
0.883
3
= 0.126 𝑚 ⁄𝑘𝑔
7
𝑝m =
1350
= 1783 𝑘𝑃𝑎 = 𝟏. 𝟕𝟖𝟑 𝑴𝑷𝒂
(0.883 − 0.126)
1 mark
1 mark
1 mark