Engineering Thermodynamics Combustion products very similar
Engineering Thermodynamics:
Combustion products (very similar properties to hot air) enter a nozzle of a turbojet at 900K and a negligible velocity. To provide thrust, the gasses pass through a nozzle which increases their velocity to 500 m/s. What is the temperature of the jet leaving the nozzle?
Solution
We have by energy balance,
h1 + V1^2 /2 = h2 + V2^2 /2
h1 - h2 = (V2^2 - V1^2) /2
Cp(T1 - T2) = (V2^2 - V1^2) /2
1120.9*(900 - T2) = (500^2 - 0) /2..........where Cp of air at 900 K is taken as 1120.9 J/kg-K
T2 = 788.48 K
