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Question 17.CSGP.83: The nozzle in Problem 17.46 will have a throat area of 0.001......

The nozzle in Problem 17.46 will have a throat area of 0.001272 m² and an exit area 2.896 times as large. Suppose the back pressure is raised to 1.4 MPa and that the flow remains isentropic except for a normal shock wave. Verify that the shock mach number \left( M _{ x }\right) is close to 2 and find the exit mach number, the temperature and the mass flow rate through the nozzle.

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(a) From Table A.12:    M_E=2.6

\begin{aligned}& P _{ E }=2.0 \times 0.05012=0.1002 \,MPa \\& T ^*=423.15 \times 0.8333= 3 5 2 . 7 ~ K \\& P ^*=2.0 \times 0.5283= 1 . 0 5 7 \,M P a\end{aligned}

\begin{aligned}& c ^*=\sqrt{1.4 \times 1000 \times 0.287 \times 352.7}=376.5 \,m / s \\& v ^*=0.287 \times 352.7 / 1057=0.0958 \,m ^3 / kg \\& A ^*=5 \times 0.0958 / 376.5= 1 . 2 7 2 \times 1 0^{-3}\, m ^2 \\& A _{ E }=1.272 \times 10^{-3} \times 2.896= 3 . 6 8 \times 1 0 ^{-3} \,m ^2 \\& T _{ E }=423.15 \times 0.42517=179.9 \,K\end{aligned}

Assume M_x=2 then

\begin{aligned}& M _{ y }=0.57735, P _{\text {oy }} / P _{\text {ox }}=0.72088, A _{ E } / A _{ x }^*=2.896 \\& A _{ x } / A _{ x }^*=1.6875, A _{ x } / A _{ y }^*=1.2225, \\& A _{ E } / A _{ y }^*=2.896 \times 1.2225 / 1.6875=2.098 \\& \Rightarrow M _{ E }= 0 . 2 9 3 , P _{ E } / P _{\text {oy }}=0.94171\end{aligned}

P _{ E }=0.94171 \times 0.72088 \times 2.0=1.357 \,MPa  OK close to the 1.4 MPa

T _{ E }=0.98298 \times 423.15= 4 1 6 \,K , \quad \dot{ m }= 5 \,k g / s

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