Question 17.5: Effect of Back Pressure on Mass Flow Rate Air at 1 MPa and 6...

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 c_{p}=1.005 kJ / kg \cdot K and k = 1.4, respectively (Table A–2a).
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. 17–4 and 17–5:

T_{0}=T+\frac{V^{2}}{2 c_{p}}

\frac{\bar{P}_{0}}{P}=\left(\frac{T_{0}}{T}\right)^{k /(k-1)}

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

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

T_{0}=T_{0 i}=884 K \quad \text { and } \quad P_{0}=P_{0 i}=1.045 MPa

The critical-pressure ratio is determined from Table 17–2 (or Eq. 17–22) to be P^{*} / P_{0}=0.5283 .
(a) The back pressure ratio for this case is

\frac{P^{*}}{P_{0}}=\left(\frac{2}{k+1}\right)^{k /(k-1)}

\frac{P_{b}}{P_{0}}=\frac{0.7 MPa }{1.045 MPa }=0.670

which is greater than the critical-pressure ratio, 0.5283. Thus the exit plane pressure (or throat pressure P_{t} ) is equal to the back pressure in this case. That is, P_{t}=P_{b}=0.7 MPa , \text { and } P_{t} / P_{0}=0.670 . Therefore, the flow is not choked.
From Table A–32 at P_{t} / P_{0}=0.670, \text { we read } Ma _{t}=0.778 \text { and } T_{t} / T_{0}=0.892 . The mass flow rate through the nozzle can be calculated from Eq. 17–24. But it can also be determined in a step-by-step manner as follows:

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

\begin{aligned}T_{t} &=0.892 T_{0}=0.892(884 K )=788.5 K \\\rho_{t} &=\frac{P_{t}}{R T_{t}}=\frac{700 kPa }{\left(0.287 kPa \cdot m ^{3} / kg \cdot K \right)(788.5 K )}=3.093 kg / m ^{3} \\V_{t} &= Ma _{t} c_{t}= Ma _{t} \sqrt{ kRT _{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 }\right)} \\&=437.9 m / s\end{aligned}

Thus,

\dot{m}=\rho_{t} A_{t} V_{t}=\left(3.093 kg / m ^{3}\right)\left(50 \times 10^{-4} m ^{2}\right)(437.9 m / s )=6.77 kg / s

(b) The back pressure ratio for this case is

\frac{P_{b}}{P_{0}}=\frac{0.4 MPa }{1.045 MPa }=0.383

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. 17–25:

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

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

since kPa \cdot m ^{2} / \sqrt{ kJ / kg }=\sqrt{1000} kg / s .

Discussion This is the maximum mass flow rate through the nozzle for the specified inlet conditions and nozzle throat area.