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Question 17.CSGP.84: At what Mach number will the normal shock occur in the nozzl......

At what Mach number will the normal shock occur in the nozzle of Problem 17.53 if the back pressure is 1.4 MPa? (trial and error on M_x)

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Relate the inlet and exit conditions to the shock conditions with reversible flow before and after the shock. It becomes trial and error.

Assume  M _{ x }=1.8 \Rightarrow M _{ y }=0.6165 ; \quad P _{\text {oy }} / P _{\text {ox }}=0.8127

\begin{aligned}& A _{ E } / A _{ x }^*= A _2 / A ^*=0.002435 / 0.001516=1.6062 \\& A _{ x } / A _{ x }^*=1.439 ; A _{ x } / A _{ y }^*=1.1694 \\& A _{ E } / A _{ y }^*=\left( A _{ E } / A _{ x }^*\right)\left( A _{ x } / A _{ y }^*\right) /\left( A _{ x } / A _{ x }^*\right)=\frac{1.6062(1.1694)}{1.439}=1.3053 \\& \Rightarrow M _{ E }=0.5189 ; P _{ E } / P _{\text {oy }}=0.8323 \\& P _{ E }=\left( P _{ E } / P _{\text {oy }}\right)\left( P _{\text {oy }} / P _{\text {ox }}\right) P _{\text {ox }}=0.8323 \times 0.8127 \times 2000=1353\, kPa <1.4 \,MPa\end{aligned}

So select the mach number a little less

\begin{aligned}& M _{ x }=1.7 \Rightarrow M _{ y }=0.64055 ; P _{\text {oy }} / P _{\text {ox }}=0.85573 \\& A _{ x } / A _{ x }^*=1.3376 ; A _{ x } / A _{ y }^*=1.1446 \\& A _{ E } / A _{ y }^*=\left( A _{ E } / A _{ x }^*\right)\left( A _{ x } / A _{ y }^*\right) /\left( A _{ x } / A _{ x }^*\right)=\frac{1.6062(1.1446)}{1.3376}=1.3744 \\& \Rightarrow M _{ E }=0.482 ; P _{ E } / P _{\text {oy }}=0.853 \\& P _{ E }=\left( P _{ E } / P _{\text {oy }}\right)\left( P _{\text {oy }} / P _{\text {ox }}\right) P _{\text {ox }}=0.853 \times 0.85573 \times 2000=1459.9\,kPa\end{aligned}

Now interpolate between the two

M _{ x }= 1 . 7 5 6 \text { and we check } \Rightarrow M _{ y }=0.6266 ; P _{\text {oy }} / P _{ ox }=0.832

\begin{aligned}& A _{ x } / A _{ x }^*=1.3926 ; A _{ x } / A _{ y }^*=1.1586 \\& A _{ E } / A _{ y }^*=1.6062 \times 1.1586 / 1.3926=1.3363 \\& \Rightarrow M _{ E }=0.5 ; P _{ E } / P _{\text {oy }}=0.843 \\& P _{ E }=0.843 \times 0.832 \times 2000=1402.7 \,kPa \quad \text { OK }\end{aligned}

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