Question 9.CSGP.25: The exit nozzle in a jet engine receives air at 1200 K, 150 ......
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with neglible kinetic energy. The exit pressure is 80 kPa and the process is reversible and adiabatic. Use constant heat capacity at 300 K to find the exit velocity.
C.V. Nozzle, Steady single inlet and exit flow, no work or heat transfer.
Energy Eq.6.13: h _{ i }= h _{ e }+ \mathbf{V} _{ e }^2 / 2 \quad\left( Z _{ i }= Z _{ e }\right)
Entropy Eq.9.8: s _{ e }= s _{ i }+\int dq / T + s _{ gen }= s _{ i }+0+0
Use constant specific heat from Table A.5, C _{ Po }=1.004 \frac{ kJ }{ kg K }, \quad k =1.4
The isentropic process \left( s _{ e }= s _{ i }\right) gives Eq.8.23
\Rightarrow \quad T_e=T_i\left(P_e / P_i\right)^{\frac{k-1}{k}}=1200(80 / 150)^{0.2857}=1002.7 \,K
The energy equation becomes
