Flow through a Converging-Diverging Nozzle In Example 15-6, determine the back pressure for which a normal shock stands (a) in the diverging section at an area 50% larger than the throat area and (b) at the exit.

Question

Flow through a Converging-Diverging Nozzle

In Example 15-6, determine the back pressure for which a normal shock stands (a) in the diverging section at an area 50% larger than the throat area and (b) at the exit.

Accepted Answer

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, [latex] M_{3} [/latex] can be obtained from the isentropic table (or the table panel of the gas dynamics TESTcalc). From [latex] A_{3} / A_{*}=A_{3} / A_{2}=1.5, ext { we obtain } M_{3}=1.854 [/latex]. Therefore:

[latex] p_{3}=\left(\frac{p}{p_{t}} ight)_{@ M_{3}} p_{t 3}=(0.1602) p_{t 1}=(0.1602)(404.4)=64.78 kPa [/latex]

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

[latex] p_{t 4}=\frac{p_{t 4}}{p_{t 3}} p_{t 3}=\left(\frac{p_{t e}}{p_{t i}} ight)_{@ M_{3}} p_{t 3}=(0.7884)(404.64)=318.83 kPa ;[/latex]

and,

[latex] A_{*_{4}}=\frac{A_{* 4}}{A_{* 3}} A_{*_{3}}=\left(\frac{A_{*_{e}}}{A_{*_{i}}} ight)_{@ M_{3}} A_{2}=(1.268) A_{2} [/latex]

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

[latex] \left(\frac{A}{A_{*}} ight)_{@ M_{5}}=\frac{A_{5}}{A_{* 5}}=\frac{A_{5}}{A_{* 4}}=\frac{2 A_{2}}{1.268 A_{2}}=1.577, ext { producing } M_{5}=0.404 [/latex]

and , [latex] \frac{p_{5}}{p_{t 5}}=\left(\frac{p}{p_{t}} ight)_{M_{5}}=0.894 ; \quad \Rightarrow \quad p_{5}=(0.894)(318.83)=285.03 kPa [/latex]

When the shock is at the exit, the states are renumbered, as shown in Figure 15.37(b), where [latex] p_{4} [/latex] 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:

[latex]\begin{aligned}\left(\frac{A}{A_{*}} ight)_{@ 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}} ight)_{@ M_{3}} p_{t 3} \&=(0.0939)(404.4)=37.97 kPa\end{aligned}[/latex]

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

[latex] p_{4}=\frac{p_{4}}{p_{3}} p_{3}=\left(\frac{p_{e}}{p_{i}} ight)_{@ M_{3}} P_{3}=(5.466)(37.97)=207.5 kPa [/latex]

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.

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

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