Shock Wave in a Converging–Diverging Nozzle If the air flowing through the converging–diverging nozzle of Example 12–6 experiences a normal shock wave at the nozzle exit plane (Fig. 12–30), determine the following after the shock: (a) the stagnation pressure, static pressure, static temperature

Question

Shock Wave in a Converging–Diverging Nozzle

If the air flowing through the converging–diverging nozzle of Example 12–6 experiences a normal shock wave at the nozzle exit plane (Fig. 12–30), determine the following after the shock: (a) the stagnation pressure, static pressure, static temperature, and static density; (b) the entropy change across the shock; (c) the exit velocity; and (d) the mass flow rate through the nozzle. Approximate the flow as steady, one-dimensional, and isentropic with k = 1.4 from the nozzle inlet to the shock location.

Accepted Answer

SOLUTION 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 [latex]c_{p}[/latex] = 1.005 kJ/kg·K 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 12–6 at the nozzle exit to be

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

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 [latex]Ma _{1}=2.0[/latex], we read

[latex]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{ ho_{2}}{ ho_{1}}=2.6667[/latex]

Then the stagnation pressure [latex]P_{02}[/latex], static pressure [latex]P_{2}[/latex], static temperature [latex]T_{2}[/latex], and static density [latex] ho_{2}[/latex] after the shock are

[latex]\begin{aligned}&P_{02}=0.7209 P_{01}=(0.7209)(1.0 MPa )=0.721 MPa \&P_{2}=4.5000 P_{1}=(4.5000)(0.1278 MPa )=0.575 MPa \&T_{2}=1.6875 T_{1}=(1.6875)(444.5 K )=750 K \& ho_{2}=2.6667 ho_{1}=(2.6667)\left(1.002 kg / m ^{3} ight)=2.67 kg / m ^{3}\end{aligned}[/latex]

(b) The entropy change across the shock is

[latex]\begin{aligned}s_{2}-s_{1} &=c_{ ho} \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 kJ / kg \cdot K\end{aligned}[/latex]

Thus, the entropy of the air increases as it passes through a normal shock, which is highly irreversible.

(c) The air velocity after the shock is determined from [latex]V_{2}= Ma _{2} c_{2}[/latex], where [latex]c_{2}[/latex] is the speed of sound at the exit conditions after the shock:

[latex]\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 } ight)} \&=317 m / s\end{aligned}[/latex]

(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 12–6:

[latex]\dot{m}=2.86 kg / s[/latex]

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.

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