Effect of Back Pressure on Mass Flow Rate Air at 1 MPa and 600°C enters a converging nozzle, shown in Fig. 12–24, with a velocity of 150 m/s. Determine the mass flow rate through the nozzle for a nozzle throat area of 50 cm² when the back pressure is (a) 0.7 MPa and (b) 0.4 MPa.

Question

Effect of Back Pressure on Mass Flow Rate

Air at 1 MPa and 600°C enters a converging nozzle, shown in Fig. 12–24, with a velocity of 150 m/s. Determine the mass flow rate through the nozzle for a nozzle throat area of 50 cm² when the back pressure is (a) 0.7 MPa and (b) 0.4 MPa.

Accepted Answer

Air enters a converging nozzle. The mass flow rate of air through the nozzle is to be determined for different back pressures.
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 constant pressure specific heat and the specific heat ratio of air are [latex] c_{p}=1.005 kJ / kg \cdot K ext { and } k =1.4 [/latex].
Analysis We use the subscripts i and t to represent the properties at the nozzle inlet and the throat, respectively. The stagnation temperature and pressure at the nozzle inlet are determined from Eqs. 12–4 and 12–5:

[latex]\begin{gathered}T_{0 i}=T_{i}+\frac{V_{i}^{2}}{2 c_{p}}=873 K +\frac{\left(150 m / s ^{2} ight)^{2}}{2(1.005 kJ / kg \cdot K )}\left(\frac{1 kJ / kg }{1000 m ^{2} / s ^{2}} ight)=884 K \P_{0 i}=P_{i}\left(\frac{T_{0 i}}{T_{i}} ight)^{k /(k-1)}=(1 MPa )\left(\frac{884 K }{873 K } ight)^{1.4 /(1.4-1)}=1.045 MPa\end{gathered}[/latex]

[latex] T_{0}=T+\frac{V^{2}}{2 c_{p}} [/latex] (12–4)

[latex] \frac{P_{0}}{P}=\left(\frac{T_{0}}{T} ight)^{k /(k-1)} [/latex] (12–5)

These stagnation temperature and pressure values remain constant throughout the nozzle since the flow is assumed to be isentropic. That is,

[latex] T_{0}=T_{0 i}=884 K \quad ext { and } \quad P_{0}=P_{0 i}=1.045 MPa [/latex]

The critical-pressure ratio is determined from Table 12–2 (or Eq. 12–22) to be [latex] P^{*} / P_{0}=0.5283 [/latex].
(a) The back pressure ratio for this case is

[latex] \frac{P_{b}}{P_{0}}=\frac{0.7 MPa }{1.045 MPa }=0.670 [/latex]

[latex] \frac{P^{*}}{P_{0}}=\left(\frac{2}{k+1} ight)^{k /(k-1)} [/latex] (12–22)

which is greater than the critical-pressure ratio, 0.5283. Thus the exit plane pressure (or throat pressure [latex] P_{t} [/latex]) is equal to the back pressure in this case.
That is, [latex] P_{t}=P_{b}=0.7 MPa , ext { and } P_{t} / P_{0}=0.670 [/latex]. Therefore, the flow is not choked. From Table A–13 at [latex] P_{t} / P_{0}=0.670 [/latex], we read [latex] Ma _{t}=0.778 ext { and } T_{t} / T_{0} = 0.892 [/latex].
The mass flow rate through the nozzle can be calculated from Eq. 12–24.
But it can also be determined in a step-by-step manner as follows:

[latex]\begin{aligned}T_{t} &=0.892 T_{0}=0.892(884 K )=788.5 K \ ho_{t} &=\frac{P_{t}}{R T_{t}}=\frac{700 kPa }{\left(0.287 kPa \cdot m ^{3} / kg \cdot K ight)(788.5 K )}=3.093 kg / m ^{3} \V_{t} &= Ma _{t} c_{t}= Ma _{t} \sqrt{k R T_{t}} \&=(0.778) \sqrt{(1.4)(0.287 kJ / kg \cdot K )(788.5 K )\left(\frac{1000 m ^{2} / s ^{2}}{1 kJ / kg } ight)} \&=437.9 m / s\end{aligned}[/latex]

[latex] \dot{m}=\frac{A Ma P_{0} \sqrt{k /\left(R T_{0} ight)}}{\left[1+(k-1) Ma ^{2} / 2 ight]^{(k+1)[2(k-1)]}} [/latex] (12–24)

Thus,

[latex] \dot{m}= ho_{t} A_{t} V_{t}=\left(3.093 kg / m ^{3} ight)\left(50 imes 10^{-4} m ^{2} ight)(437.9 m / s )=6.77 kg / s [/latex]

(b) The back pressure ratio for this case is

[latex] \frac{P_{b}}{P_{0}}=\frac{0.4 MPa }{1.045 MPa }=0.383 [/latex]

which is less than the critical-pressure ratio, 0.5283. Therefore, sonic conditions exist at the exit plane (throat) of the nozzle, and Ma = 1. The flow is choked in this case, and the mass flow rate through the nozzle can be calculated from Eq. 12–25:

[latex]\begin{aligned}\dot{m} &=A^{*} P_{0} \sqrt{\frac{k}{R T_{0}}}\left(\frac{2}{k+1} ight)^{(k+1)[2(k-1)]} \&=\left(50 imes 10^{-4} m ^{2} ight)(1045 kPa ) \sqrt{\frac{1.4}{(0.287 kJ / kg \cdot K )(884 K )}}\left(\frac{2}{1.4+1} ight)^{2.4 / 0.8} \&=7.10 kg / s\end{aligned}[/latex]

[latex] \dot{m}_{\max }=A^{*} P_{0} \sqrt{\frac{k}{R T_{0}}}\left(\frac{2}{k+1} ight)^{(k+1) /[2(k-1)]} [/latex] (12–25)

since [latex] kPa \cdot m ^{2} / \sqrt{ kJ / kg }=\sqrt{1000} kg / s [/latex].
Discussion This is the maximum mass flow rate through the nozzle for the specified inlet conditions and nozzle throat area.

TABLE 12–2

The critical-pressure, critical-temperature, and critical-density ratios for isentropic flow of some ideal gases

[latex]\begin{array}{ccccc}& \begin{array}{c} ext { Superheated } \ ext { steam, } \k=1.3\end{array} & \begin{array}{c} ext { Hot products } \ ext { of combustion, } \k=1.33\end{array} & \begin{array}{c} ext { Air, } \k=1.4 \end{array} & \begin{array}{c} ext { Monatomic } \ ext { gases, } \k=1.667 \end{array} \\hline \frac{P^{*}}{P_{0}} & 0.5457 & 0.5404 & 0.5283 & 0.4871 \\frac{T^{*}}{T_{0}} & 0.8696 & 0.8584 & 0.8333 & 0.7499 \\frac{ ho^{*}}{ ho_{0}} & 0.6276 & 0.6295 & 0.6340 & 0.6495\end{array}[/latex]

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