Question 12.9: Shock Wave in a Converging–Diverging Nozzle If the air flowi...

Air flowing through a converging–diverging nozzle experiences a normal shock at the exit. The effect of the shock wave on various properties is to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane.
Properties The constant-pressure specific heat and the specific heat ratio of air are c_{p}=1.005 kJ / kg \cdot K \text { and } k=1.4 . The gas constant of air is 0.287 kJ/kg ⋅ K.
Analysis (a) The fluid properties at the exit of the nozzle just before the shock (denoted by subscript 1) are those evaluated in Example 12–7 at the nozzle exit to be

P_{01}=1.0 MPa \quad P_{1}=0.1278 MPa \quad T_{1}=444.5 K \quad \rho_{1}=1.002 kg / m ^{3}

The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A–14. For Ma _{1} = 2.0, we read

Ma _{2}=0.5774 \quad \frac{P_{02}}{P_{01}}=0.7209 \quad \frac{P_{2}}{P_{1}}=4.5000 \quad \frac{T_{2}}{T_{1}}=1.6875 \quad \frac{\rho_{2}}{\rho_{1}}=2.6667

Then the stagnation pressure P_{02} , static pressure P_{2} , static temperature T_{2} , and static density \rho_{2} after the shock are

\begin{aligned}P_{02} &=0.7209 P_{01}=(0.7209)(1.0 MPa )= 0 . 7 2 1 \mathrm { MPa } \\P_{2} &=4.5000 P_{1}=(4.5000)(0.1278 MPa )= 0 . 5 7 5 \mathrm { MPa } \\T_{2} &=1.6875 T_{1}=(1.6875)(444.5 K )=750 K \\\rho_{2} &=2.6667 \rho_{1}=(2.6667)\left(1.002 kg / m ^{3}\right)=2.67 kg / m ^{3}\end{aligned}

(b) The entropy change across the shock is

\begin{aligned}s_{2}-s_{1} &=c_{\rho} \ln \frac{T_{2}}{T_{1}}-R \ln \frac{P_{2}}{P_{1}} \\&=(1.005 kJ / kg \cdot K ) \ln (1.6875)-(0.287 kJ / kg \cdot K ) \ln (4.5000) \\&= 0 . 0 9 4 2 ~ k J / kg \cdot K\end{aligned}

Thus, the entropy of the air increases as it experiences a normal shock, which is highly irreversible.
(c) The air velocity after the shock can be determined from V_{2}=M a_{2} c_{2} , where c_{2} is the speed of sound at the exit conditions after the shock:

\begin{aligned}V_{2} &= Ma _{2} c_{2}= Ma _{2} \sqrt{k R T_{2}} \\&=(0.5774) \sqrt{(1.4)(0.287 kJ / kg \cdot K )(750.1 K )\left(\frac{1000 m ^{2} / s ^{2}}{1 kJ / kg }\right)} \\&=317 m / s\end{aligned}

(d) The mass flow rate through a converging–diverging nozzle with sonic conditions at the throat is not affected by the presence of shock waves in the nozzle. Therefore, the mass flow rate in this case is the same as that determined in Example 12–7 :

\dot{m}=2.86 kg / s

Discussion This result can easily be verified by using property values at the nozzle exit after the shock at all Mach numbers significantly greater than unity.