Airflow through a Converging–Diverging Nozzle Air enters a converging–diverging nozzle, shown in Fig. 12–28, at 1.0 MPa and 800 K with a negligible velocity. The flow is steady, one-dimensional, and isentropic with k = 1.4. For an exit Mach number of Ma = 2 and a throat area of 20 cm², determine

Question

Airflow through a Converging–Diverging Nozzle

Air enters a converging–diverging nozzle, shown in Fig. 12–28, at 1.0 MPa and 800 K with a negligible velocity. The flow is steady, one-dimensional, and isentropic with k = 1.4. For an exit Mach number of Ma = 2 and a throat area of 20 cm², determine (a) the throat conditions, (b) the exit plane conditions, including the exit area, and (c) the mass flow rate through the nozzle.

Accepted Answer

Air flows through a converging–diverging nozzle. The throat and the exit conditions and the mass flow rate are 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.
Properties The specific heat ratio of air is given to be k = 1.4. The gas constant of air is 0.287 kJ/kg ⋅ K.
Analysis The exit Mach number is given to be 2. Therefore, the flow must be sonic at the throat and supersonic in the diverging section of the nozzle. Since the inlet velocity is negligible, the stagnation pressure and stagnation temperature are the same as the inlet temperature and pressure, [latex] P_{0}=1.0 [/latex] MPa and [latex] T_{0}=800 [/latex] K. Assuming ideal-gas behavior, the stagnation density is

[latex] ho_{0}=\frac{P_{0}}{R T_{0}}=\frac{1000 kPa }{\left(0.287 kPa \cdot m ^{3} / kg \cdot K ight)(800 K )}=4.355 kg / m ^{3} [/latex]

(a) At the throat of the nozzle Ma = 1, and from Table A–13 we read

[latex] \frac{P^{*}}{P_{0}}=0.5283 \quad \frac{T^{*}}{T_{0}}=0.8333 \quad \frac{ ho^{*}}{ ho_{0}}=0.6339 [/latex]

Thus,

[latex]\begin{aligned}P^{*} &=0.5283 P_{0}=(0.5283)(1.0 MPa )=0.5283 MPa \T^{*} &=0.8333 T_{0}=(0.8333)(800 K )=666.6 K \ ho^{*} &=0.6339 ho_{0}=(0.6339)\left(4.355 kg / m ^{3} ight)=2.761 kg / m ^{3}\end{aligned}[/latex]

Also,

[latex]\begin{aligned}V^{*} &=c^{*}=\sqrt{k R T^{*}}=\sqrt{(1.4)(0.287 kJ / kg \cdot K )(666.6 K )\left(\frac{1000 m ^{2} / s ^{2}}{1 kJ / kg } ight)} \&=517.5 m / s\end{aligned}[/latex]

(b) Since the flow is isentropic, the properties at the exit plane can also be calculated by using data from Table A–13. For Ma = 2 we read

[latex] \frac{P_{e}}{P_{0}}=0.1278 \quad \frac{T_{e}}{T_{0}}=0.5556 \frac{ ho_{e}}{ ho_{0}}=0.2300 \quad Ma _{e}^{*}=1.6330 \quad \frac{A_{e}}{A^{*}}=1.6875 [/latex]

Thus,

[latex]\begin{aligned}P_{e} &=0.1278 P_{0}=(0.1278)(1.0 MPa )=0.1278 MPa \T_{e} &=0.5556 T_{0}=(0.5556)(800 K )=444.5 K \ ho_{e} &=0.2300 ho_{0}=(0.2300)\left(4.355 kg / m ^{3} ight)=1.002 kg / m ^{3} \A_{e} &=1.6875 A^{*}=(1.6875)\left(20 cm ^{2} ight)=33.75 cm ^{2}\end{aligned}[/latex]

and

[latex] V_{e}= Ma _{e}^{*} c^{*}=(1.6330)(517.5 m / s )=845.1 m / s [/latex]

The nozzle exit velocity could also be determined from [latex] V_{e}= Ma _{e} c_{e} ext {, where } c_{e} [/latex] is the speed of sound at the exit conditions:

[latex]\begin{aligned}V_{e} &= Ma _{e} c_{e}= Ma _{e} \sqrt{k R T_{e}}=2 \sqrt{(1.4)(0.287 kJ / kg \cdot K )(444.5 K )\left(\frac{1000 m ^{2} / s ^{2}}{1 kJ / kg } ight)} \&=845.2 m / s\end{aligned}[/latex]

(c) Since the flow is steady, the mass flow rate of the fluid is the same at all sections of the nozzle. Thus it may be calculated by using properties at any cross section of the nozzle. Using the properties at the throat, we find that the mass flow rate is

[latex] \dot{m}= ho^{*} A^{*} V^{*}=\left(2.761 kg / m ^{3} ight)\left(20 imes 10^{-4} m ^{2} ight)(517.5 m / s )= 2 . 8 6 \mathrm { kg } / \mathrm { s } [/latex]

Discussion Note that this is the highest possible mass flow rate that can flow through this nozzle for the specified inlet conditions.

Question 12.7: Airflow through a Converging–Diverging Nozzle Air enters a c...

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