Question

In a constant volume ‘Otto cycle’, the pressure at the end of compression is 15 times that at the start, the temperature of air at the beginning of compression is 38°C and maximum temperature attained in the cycle is 1950°C. 

Determine :

(i) Compression ratio. 

(ii) Thermal efficiency of the cycle. 

(iii) Work done. 

Take γ for air = 1.4.

Answer 1

Initial temperature, T₁ = 38 + 273 = 311 K 

Maximum temperature, T₃ = 1950 + 273 = 2223 K. 

(i) Compression ratio, r : 

For adiabatic compression 1-2, 

p₁V1^γ = p₂V₂^γ

Hence compression ratio = 6.9.

(ii) Thermal efficiency :

Thermal efficiency,

(iii) Work done : 

Again, for adiabatic compression 1-2

For adiabatic expansion process 3-4

Heat supplied per kg of air 

= c_v(T₃ – T₂) = 0.717(2223 – 671.7) 

= 1112.3 kJ/kg or air

Heat rejected per kg of air 

= c_v(T₄ – T₁) = 0.717(1029 – 311) 

= 514.8 kJ/kg of air 

∴ Work done per kg of air = Heat supplied – heat rejected 

= 1112.3 – 514.8 

= 597.5 kJ or 597500 N-m.