Question 15.8: Flow through a Converging-Diverging Nozzle In Example 15-6, ...

Do an isentropic analysis before and after the shock wave with the normal shock table relating properties on the two sides of the shock.

Assumption
One-dimensional flow of a perfect gas. Isentropic flow everywhere except across the shock wave.

Analysis
Represent the various states along the nozzle by State-1 through State-5, with State-2 being the throat (critical state), as shown in Figure 15.37(a). The flow is isentropic until State-3; therefore, M_{3}  can be obtained from the isentropic table (or the table panel of the gas dynamics TESTcalc). From A_{3} / A_{*}=A_{3} / A_{2}=1.5, \text { we obtain } M_{3}=1.854 . Therefore:

p_{3}=\left(\frac{p}{p_{t}}\right)_{@ M_{3}} p_{t 3}=(0.1602) p_{t 1}=(0.1602)(404.4)=64.78 kPa

Using the normal shock table (or the gas dynamics TESTcalc), we can obtain State-4:

p_{t 4}=\frac{p_{t 4}}{p_{t 3}} p_{t 3}=\left(\frac{p_{t e}}{p_{t i}}\right)_{@ M_{3}} p_{t 3}=(0.7884)(404.64)=318.83 kPa ;

and,

A_{*_{4}}=\frac{A_{* 4}}{A_{* 3}} A_{*_{3}}=\left(\frac{A_{*_{e}}}{A_{*_{i}}}\right)_{@ M_{3}} A_{2}=(1.268) A_{2}

The flow being isentropic after the shock, the critical area and total pressure remain constant downstream of State-4. Therefore:

\left(\frac{A}{A_{*}}\right)_{@ M_{5}}=\frac{A_{5}}{A_{* 5}}=\frac{A_{5}}{A_{* 4}}=\frac{2 A_{2}}{1.268 A_{2}}=1.577, \text { producing } M_{5}=0.404

and ,  \frac{p_{5}}{p_{t 5}}=\left(\frac{p}{p_{t}}\right)_{M_{5}}=0.894 ; \quad \Rightarrow \quad p_{5}=(0.894)(318.83)=285.03 kPa

When the shock is at the exit, the states are renumbered, as shown in Figure 15.37(b), where p_{4} is the desired unknown. Realizing that the flow is isentropic up to the shock location, properties at State-3 can be obtained from the isentropic table:

\begin{aligned}\left(\frac{A}{A_{*}}\right)_{@ M_{3}} &=\frac{A_{3}}{A_{* 3}}=\frac{A_{3}}{A_{2}}=2 ; \quad \Rightarrow \quad M_{3}=2.20 ; \quad p_{3}=\left(\frac{p}{p_{t}}\right)_{@ M_{3}} p_{t 3} \\&=(0.0939)(404.4)=37.97 kPa\end{aligned}

From the shock table, we can obtain the static pressure ratio for M_{i}=2.20 to find:

p_{4}=\frac{p_{4}}{p_{3}} p_{3}=\left(\frac{p_{e}}{p_{i}}\right)_{@ M_{3}} P_{3}=(5.466)(37.97)=207.5 kPa

TEST Analysis
Launch the gas dynamics TESTcalc and select air as the working fluid. Evaluate the five states, as described in the TEST-code (see TEST > TEST-codes), to verify the manual results.

Discussion
The mass flow rate can be calculated if the throat area is known. The presence of a shock wave cannot change the mass flow rate which is a function of chamber pressure, chamber temperature, and throat area. The flow is choked at the throat; therefore, what happens in the diverging section cannot affect any property upstream of the throat.