Question 15.5: Flow Through a Converging Nozzle Air is discharged from a re...
Flow Through a Converging Nozzle
Air is discharged from a reservoir, at 500 kPa and 500 K, through an isentropic converging nozzle with an exit area of 10 cm². Determine the mass flow rate for a back pressure of (a) 0 kPa, (b) 100 kPa, (c) 200 kPa, and (d) 300 kPa.
Determine if the flow is choked for a given back pressure and use Eq. (15.28) to determine the mass flow rate.
\dot{m}=A_{*} p_{t} \sqrt{\frac{(1000 J / kJ ) k}{R T_{t}}}\left(\frac{2}{k+1}\right)^{(k+1) /[2(k-1)]} ;\left[\frac{ kg }{ s }\right] (15.28)
Assumption
Isentropic, one-dimensional flow of a perfect gas.
Analysis
From the isentropic table of air, Table H-1, or the table panel of the gas dynamics TESTcalc with air selected as the working fluid, the critical pressure ratio p_{*} / p_{t} \text { for } M=1 can be obtained as 0.5283 so that:
p_{*}=\frac{p_{*}}{p_{t}} p_{t}=\left(\frac{p}{p_{t}}\right)_{@ M=1} p_{r}=(0.5283)(500)=264.15 kPa
For back pressures less than the critical pressure 264.15 kPa—0 kPa, 100 kPa, and 200 kPa—the flow is choked.
For choked flow: T_{ th }=T_{*}=\left(\frac{T_{*}}{T_{t}}\right) T_{t}=\left(\frac{T}{T_{t}}\right)_{@ M=1} T_{t}=(0.8333)(500)=416.7 K
The mass flow rate for the choked flow can be directly calculated:
\begin{aligned}\dot{m} &=\rho_{*} A_{*} V_{*}=\rho_{*} A_{*} c_{*}=\frac{p_{*}}{R T_{*}} A_{ th } \sqrt{(1000 N / kN ) k R T_{*}} \\&=\frac{(264.15)(0.001) \sqrt{1000(1.4)}}{\sqrt{(0.287)(416.7)}} \\&=0.904 \frac{ kg }{ s }\end{aligned}
The same answer can be obtained by application of Eq. (15.28).
For a back pressure of 300 kPa, the flow is subsonic and the back pressure must be equal to the exit pressure. For p_{ th } / p_{t}=300 / 500=0.6 , the isentropic table produces M_{\text {th }}=0.886 and T_{\text {th }}=(0.864)(500)=432 K . Therefore:
\begin{aligned}\dot{m} &=\rho_{ th } A_{ th } V_{ th }=\frac{p_{ th }}{R T_{ th }} A_{ th } M_{ th } \sqrt{(1000 N / kN ) k R T_{\text {th }}} \\&=\frac{(300)(0.001)(0.886) \sqrt{1000(1.4)}}{\sqrt{(0.287)(432.1)}} \\&=0.893 \frac{ kg }{ s }\end{aligned}
TEST Analysis
Launch the gas dynamics TESTcalc. Select Air as the working fluid. Evaluate State-1 with p1 = 500 kPa, T1 = 500 K, and Vel1 = 0 as the reservoir (total) state. For the choked throat state, evaluate State-2, with M2 = 1, T_t2 = T1, p_t2 = p1, and A2 = 10 cm2. For the nonchoked state, State-3, use p3, T_t3, p_t3, and A3 = A2. Mass flow rates are calculated as part of the complete flow states. For more details, see TEST-code (posted in TEST > TEST-codes module).
Discussion
By gradually increasing p3, and recalculating State-3, mdot3 can be plotted as a function of the back pressure, as shown in Figure 15.23. A similar analysis can be used to determine the exit area required to create a desired mass flow rate for a given total pressure and temperature. We could use Eq. (15.28) to obtain the mass flow rates, since p_{t} \text { and } T_{t} are known from the reservoir conditions and A_{*} can be obtained from Table H-1 for a given Mach number.
