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.

a)r= 7.27

b)

c)P₂=1449.21 KPa

d)W=540 KJ/kg

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

r=Compression ratio

By putting the values

r= 7.27

By putting the values

P₂=1449.21 KPa

This is maximum pressure

The efficiency η

We also know that

W=540 KJ/kg

This is the net work output.

YE

Explanation:

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