Answer:
The specific heat addition is 773.1 kJ/kg
Explanation:
from table A.5 we get the properties of air:
k=specific heat ratio=1.4
cp=specific heat at constant pressure=1.004 kJ/kg*K
We calculate the pressure range of the Brayton cycle, as follows
n=1-(1/(P2/P1)^(k-1)/k))
where n=thermal efficiency=0.5. Clearing P2/P1 and replacing values:
P2/P1=(1/0.5)^(1.4/0.4)=11.31
the temperature of the air at state 2 is equal to:
P2/P1=(T2/T1)^(k/k-1)
where T1 is the temperature of the air enters the compressor. Clearing T2
11.31=(T2/290)^(1.4/(1.4-1))
T2=580K
The temperature of the air at state 3 is equal to:
P2/P1=(T3/T4)^(k/(k-1))
11.31=(T3/675)^(1.4/(1.4-1))
T3=1350K
The specific heat addition is equal to:
q=Cp*(T3-T2)=1.004*(1350-580)=773.1 kJ/kg
