Question 15.4: Use of an Isentropic Table In Example 15-2, assume pr = 150 ...

Total temperature remaining constant along an isentropic flow, the ratio of static to total pressure can be found at the exit, and the exit Mach number can be obtained from the isentropic table for air (Table H-1).

Assumptions
Same as in Example 15-2.
Analysis
The exit pressure being equal to the back pressure, the static-to-total pressure ratio at the exit, State-2, is:

\left(\frac{p}{p_{t}}\right)_{2}=\frac{p_{2}}{p_{r}}=\frac{p_{b}}{p_{r}}=\frac{100}{150}=0.667

From Table H-1 (or the table panel of the gas dynamics TESTcalc), the corresponding Mach number and the temperature ratio can be interpolated as:

M_{2}=M_{@ p / p_{t}=0.667}=0.784 ; \quad\left(\frac{T}{T_{t}}\right)_{e}=\left(\frac{T}{T_{t}}\right)_{@ M_{2}}=0.891

Using T_{t 2}=T_{r} :

\begin{aligned}V_{2} &=M_{2} c_{2}=M_{2} \sqrt{(1000 N / kN ) k R T_{2}}=M_{e} \sqrt{(1000) k R \frac{T_{2}}{T_{t 2}} \frac{T_{t 2}}{T_{r}} T_{r}} \\&=(0.784) \sqrt{(1000)(1.4)(0.287)(0.891)(500)}=331.5 m / s\end{aligned}

To calculate the critical pressure, find the pressure ratio corresponding to M = 1 from Table H-1 or the TESTcalc:

\begin{aligned}& \frac{p_{*}}{p_{t}}=\left(\frac{p}{p_{t}}\right)_{@ M=1}=0.5283 \\\Rightarrow \quad p_{*}=& 0.5283 p_{t}=0.5283(150)=79.24 kPa\end{aligned}

Similarly, the critical exit area can be calculated with the help of Table H-1:

\begin{gathered}\frac{A_{2}}{A_{*}}=\left(\frac{A}{A_{*}}\right)_{@ M=0.784}=1.045 \\\Rightarrow \quad A_{*}=A_{2} / 1.045=10 / 1.045=9.57 cm ^{2}\end{gathered}

TEST Analysis
Launch the gas dynamics TESTcalc and select Air. Evaluate States 1 and 2, as in Example 15-2, except p1=150 kPa. To evaluate the critical state, calculate State-3 with Mach3=1,T_t3=T1, p_t3=p1, and A3=Astar2. All critical properties, including pressure, are displayed as part of State-3 and the manual results are verified. See TEST-code (posted in TEST > TEST-codes) for details.

Discussion
If the back pressure is reduced to 79.25 kPa, the flow will be choked at the exit regardless of the exit area. On the other hand, if only the exit area is reduced, the exit Mach number will remain unchanged and only the mass flow rate will change, according to Eq. (15.28). The critical state does not coincide with the nozzle exit, as shown in Figure 15.20. However, if the nozzle is pinched in the middle to create an area of 9.89 cm², the imaginary critical state will then shift to that pinched location, which becomes the new throat of the reshaped converging-diverging duct. How do you think the flow will adjust if the pinched area is reduced below the critical area?

\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)