Question 9.PS.26: Do the previous problem using the air tables in A.7 The exit......
Do the previous problem using the air tables in A.7
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
Process: q = 0, s _{\text {gen }}=0 \quad \text { as used above leads to } s _{ e }= s _{ i }
Inlet state: h _{ i }=1277.8 \,kJ / kg , \quad P _{ ri }=191.17
The constant s is done using the P_r function from A.7.2
P _{ r\,e }= P _{ r\,i }\left( P _{ e } / P _{ i }\right)=191.17(80 / 150)=101.957
Interpolate in A.7 ⇒
From the energy equation we have \mathbf{V} _{ e }^{2} / 2= h _{ i }- h _{ e } , so then
\mathbf{V} _{ e }=\sqrt{2\left( h _{ i }- h _{ e }\right)}=\sqrt{2(1277.8-1076.2) \,kJ / kg \times 1000 \,J / kJ }= 6 3 5 \,m / s
