Question

An Otto Cycle begins the isentropic compression processes with air at 290 K and 90 kPa. During heat addition 1000 kJ/kg of heat is added, after which the temperature is 2050 K. Using the air-standard with specific heats determined at 300 K determine:

a) the compression ratio.
b) the thermal efficiency.
c) the highest pressure in the cycle.
d) the net work out per kg of working fluid.

Answer 1

a)r= 7.27

b)
`\eta =0.54`

c)P₂=1449.21 KPa

d)W=540 KJ/kg

Step-by-step explanation:

Given that

T₁= 290 K

P₁=90 KPa

Qa = 1000 KJ/kg

T₃=2050 K

For air

specific heat at constant volume Cv

Cv=0.71 KJ/kg.k

γ = 1.4

Qa= Cv ( T₃ - T₂)

Bu putting the values

1000 = 0.71 x ( 2050 - T₂)

T₂ = 641.54 K

`(T_2)/(T_1)=r^(\gamma -1)`

r=Compression ratio

By putting the values

`(T_2)/(T_1)=r^(\gamma -1)`

`(641.54)/(290)=r^(1.4 -1)`

r= 7.27

`(P_2)/(P_1)=r^(\gamma)`

By putting the values

`(P_2)/(P_1)=r^(\gamma)`

`(P_2)/(90)=7.27^(\gamma)`

P₂=1449.21 KPa

This is maximum pressure

The efficiency η

`\eta =1-(1)/(r^(\gamma -1))`

`\eta =1-(1)/(7.27^(1.4-1))`

`\eta =0.54`

We also know that

`\eta=(W)/(Q_a)`

`0.54=(W)/(1000)`

W=540 KJ/kg

This is the net work output.