Question 15.10: Non-isentropic Nozzle Analysis Carbondioxide, at 250 kPa, 60...

Determine if the nozzle is choked and, accordingly, obtain the isentropic solution for the exit state, then use the nozzle efficiency to find the actual exit conditions.

Assumptions
One-dimensional, adiabatic flow of a perfect gas.
Analysis
Different principal states of the nozzle are shown in Figure 15.42. State-1 is the inlet flow state with its total properties represented by a separate (total) state, State-t1. State-2 is the actual exit state. And State-3 is the isentropic exit state at the pressure of the exit \left(p_{2}\right) . From the PG table (Table C-1), or the gas dynamics TESTcalc, obtain the gas properties for CO _{2} .

M_{1}=\frac{V_{1}}{\sqrt{(1000 N / kN ) k R T_{1}}}=\frac{50}{\sqrt{(1000)(1.289)(0.189)(600)}}=0.131

Using the isentropic table or (gas dynamics TESTcalc),

T_{t 1}=601.5 K ; \quad p_{t 1}=252.8 kPa

State-2, -3 (given A_{2}, p_{2} ): First check if the back pressure is low enough to choke the nozzle. To do so, assume M_{2}=1 , determine the expected exit pressure, and compare it with the actual back pressure, 100 kPa.

Using the energy equation, T_{t 1}=T_{t 2} , which is also valid for a non-isentropic nozzle, T_{2}  can be obtained with the help of the isentropic table:

\frac{T_{2}}{T_{t 2}}=\left(\frac{T}{T_{t}}\right)_{M_{2}=1}=0.874 ; \Rightarrow T_{2}=0.874\left(T_{t 1}\right)=(0.874)(601.5)=525.7 K

However, to evaluate p_{2} , we must evaluate the isentropic exit state, State-3, first and use the fact that p_{2}=p_{3} . \text { Toward that goal, } T_{3} is first determined from the known isentropic efficiency:

\eta_{\text {nozzle }}=\frac{h_{t 1}-h_{2}}{h_{t 1}-h_{3}}=\frac{T_{t 1}-T_{2}}{T_{t 1}-T_{3}} ; \quad \Rightarrow \quad T_{3}=T_{t 1}-\left(T_{t 1}-T_{2}\right) / 0.9=517.5 K

State-3 is isentropic to State-1; therefore, p_{t 3}=p_{t 1}=252.8 kPa . The energy equation also produces T_{t 3}=T_{t 1}=601.5 K . The isentropic exit state now can be determined as follows:

\begin{gathered}\frac{T_{3}}{T_{t 3}}=\left(\frac{T}{T_{t}}\right)_{M_{3}}=0.86 ; \quad \Rightarrow \quad M_{3}=1.063 \\p_{3}=\frac{p_{3}}{p_{t 3}} p_{t 3}=\left(\frac{p}{p_{t}}\right)_{M_{3}} p_{t 3}=(0.51)(252.8)=128.9 kPa \\V_{3}=M_{3} \sqrt{(1000 N / kN ) k R T_{3}}=377.0 m / s\end{gathered}

The actual and isentropic exit states share the same pressure; therefore, p_{2}=128.9 kPa for an exit Mach number of unity.
Because the back pressure, 100 kPa, is less than the calculated exit pressure p_{2} , the nozzle is choked, and State-2 is the actual exit state. From the known exit temperature, the exit velocity now can be calculated as:

T_{2}=525.7 K ; \quad V_{2}=M_{2} \sqrt{(1000 N / kN ) k R T_{2}}=357.7 m / s

The velocity coefficient can be obtained from Eq. (15.42) or directly from the values of V_{2} and V_{3} as:

C_{V} \equiv \frac{V_{e}}{V_{e s}}=\sqrt{\frac{V_{e}^{2}}{V_{e s}^{2}}}=\sqrt{\frac{ ke _{e}}{ ke _{e s}}}=\sqrt{\eta_{\text {nozzle }}}                        (15.42)

C_{V} \equiv \frac{V_{2}}{V_{3}}=\sqrt{\eta_{\text {nozzle }}}=0.949

To obtain the discharge coefficient, the mass flow rates at State-2 and -3 are calculated first:

\begin{aligned}&\dot{m}_{2}=\rho_{2} A_{2} V_{2}=\frac{p_{2}}{R T_{2}} A_{2} V_{2}=\frac{(128.9)(0.001)(357.7)}{(0.189)(525.7)}=0.464 \frac{ kg }{ s } \\&\dot{m}_{3}=\rho_{3} A_{3} V_{3}=\frac{p_{3}}{R T_{3}} A_{2} V_{3}=\frac{(128.9)(0.001)(377.0)}{(0.189)(517.5)}=0.497 \frac{ kg }{ s }\end{aligned}

Therefore,

C_{D}=\frac{\dot{m}_{2}}{\dot{m}_{3}}=\frac{0.464}{0.497}=0.933

TEST Solution
Launch the gas dynamics TESTcalcs and select Air as the working fluid. Calculate States 1, 2, and 3 as described in the TEST-code (see TEST > TEST-codes). State-2 is fully evaluated only after State-3 is found. The velocity coefficient and discharge coefficients are calculated in the I/O panel to confirm the manual results.

Discussion
If the nozzle is not choked, p_{3} will be found to be less than the back pressure, 100 kPa. In that case, we have to reject M_{3}=1 , and repeat the isentropic calculations with p_{3}=100 kPa . Note that the definition of isentropic efficiency requires the pressure of the isentropic exit state to be the same as the actual exit pressure. Although the actual nozzle is convergent, the hypothetical isentropic nozzle may be convergent-divergent type as in this case with M_{3}>1 .