12.9: Combination of the Oblique Shock and Isentropic Expansion

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Example 12.18

Consider two–dimensional flat thin plate at an angle of attack of $4^\circ$ and a Mach number of 3.3. Assume that the specific heat ratio at stage is $k=1.3$, calculate the drag coefficient and lift coefficient.

Solution 12.18

For $M=3.3$, the following table can be obtained:

Prandtl — Meyer		Input: $M$			k = 1.4
$M$	$\nu$	$\dfrac{P}{P_0}$	$\dfrac{T}{T_0}$	$\dfrac{\rho}{\rho_0}$	$\mu$
3.300	62.3113	0.01506	0.37972	0.03965	73.1416

With the angle of attack the region 3 will be at $\nu \sim 62.31+4$ for which the following table can be obtained (Potto-GDC)

Prandtl — Meyer		Input: $M$			k = 1.4
$M$	$\nu$	$\dfrac{P}{P_0}$	$\dfrac{T}{T_0}$	$\dfrac{\rho}{\rho_0}$	$\mu$
3.4996	66.3100	0.01090	0.35248	0.03093	74.0528

On the other side, the oblique shock (assuming weak shock) results in

Oblique Shock		Input: $M_x$				k = 1.4
$M_x$	${{M_y}_s}$	${{M_y}_w}$	$\theta_{s}$	$\theta_{w}$	$\delta$	$\dfrac{{P_0}_y}{{P_0}_x}$
3.300	0.43534	3.1115	88.9313	20.3467	4.0000	0.99676

and the additional information, by clicking on the minimal button, provides

Oblique Shock		Input: $M_x$				k = 1.4
$M_x$	${{M_y}_w}$	$\theta_{w}$	$\delta$	$\dfrac{{P}_y}{{P}_x}$	$\dfrac{{T}_y}{{T}_x}$	$\dfrac{{P_0}_y}{{P_0}_x}$
3.300	3.1115	20.3467	4.0000	1.1157	1.1066	0.99676

The pressure ratio at point 3 is
\begin{align*}
\dfrac{P_3 }{ P_1} = \dfrac{P_3 }{ P_{03} } \, \dfrac{P_{03} }{ P_{01}} \, \dfrac{P_{01} }{ P_{1}}
= 0.0109 \times 1 \times \dfrac{1 }{ 0.01506} \sim 0.7238
\end{align*}
The pressure ratio at point 4 is
\begin{align*}
{P_3 \over P_1} = 1.1157
\end{align*}
\begin{align*}
d_L = {2 \over k P_1 {M_1}^2 }(P_4 - P_3) \cos \alpha =
{2 \over k {M_1}^2 } \left( {P_4 \over P_1} - {P_3 \over P_1} \right) \,\cos \alpha
\end{align*}
\begin{align*}
d_L = { 2 \over 1.3 3.3^2 } \left( 1.1157 - 0.7238 \right) \cos 4^\circ \sim .054
\end{align*}
\begin{multline*}
d_d = \dfrac{2 }{ k\, {M_1}^2 } \left( \dfrac{P_4 }{ P_1} - \dfrac{P_3 }{ P_1} \right) \sin \alpha =
\dfrac{ 2 }{ 1.3\, 3.3^2 } \left( 1.1157 - 0.7238 \right) \sin 4^\circ
\sim .0039
\end{multline*}
This shows that on the expense of a small drag, a large lift can be obtained. Discussion on the optimum design is left for the next versions.

Fig. 12.31 Schematic of the nozzle and Prandtl–Meyer expansion.

Example 12.19

To understand the flow after a nozzle consider a flow in a nozzle shown in Figure 12.31. The flow is choked and additionally the flow pressure reaches the nozzle exit above the surrounding pressure. Assume that there is an isentropic expansion (Prandtl–Meyer expansion) after the nozzle with slip lines in which there is a theoretical angle of expansion to match the surroundings pressure with the exit. The ratio of exit area to throat area ratio is 1.4. The stagnation pressure is 1000 [kPa]. The surroundings pressure is 100[kPa]. Assume that the specific heat, $k=1.3$. Estimate the Mach number after the expansion.

Solution 12.19

The Mach number a the nozzle exit can be calculated using Potto-GDC which provides

Isentropic Flow		Input: $\dfrac{A}{A^{\star}}$				k = 1.3
$M$	$\dfrac{T}{T_0}$	$\dfrac{\rho}{\rho_0}$	$\dfrac{A}{A^{\star}}$	$\dfrac{P}{P_0}$	$\dfrac{{A\,P}_y}{{A^{\star}\,P_0}_x}$	$\dfrac{F}{F_0}$
1.7285	0.69052	0.29102	1.4000	0.20096	0.28134	0.59745

Thus, the exit Mach number is 1.7285 and the pressure at the exit is
\begin{align*}
P_{exit} = P_0 \dfrac{P_{\text{exit}}}{P_0} = 1000 \times 0.20096 = 200.96 [kPa]
\end{align*}
This pressure is higher than the surroundings pressure and an expansion must occur. This pressure ratio is associated with a expansion angle that Potto-GDC provide as

Oblique Shock		Input: $M_x$				k = 1.3
$M_x$	${{M_y}_w}$	$\theta_{w}$	$\delta$	$\dfrac{{P}_y}{{P}_x}$	$\dfrac{{T}_y}{{T}_x}$	$\dfrac{{P_0}_y}{{P_0}_x}$
3.300	3.1115	20.3467	4.0000	1.1157	1.1066	0.99676

The final pressure ratio ultimately has to be
\begin{align*}
\dfrac{ P_{\text{surroundings}}} {P_{0}} =
\dfrac{100}{1000} = .1
\end{align*}
Hence the information for this pressure ratio can be provided by Potto-GDC as

Oblique Shock		Input: $M_x$				k = 1.3
$M_x$	${{M_y}_w}$	$\theta_{w}$	$\delta$	$\dfrac{{P}_y}{{P}_x}$	$\dfrac{{T}_y}{{T}_x}$	$\dfrac{{P_0}_y}{{P_0}_x}$
3.300	3.1115	20.3467	4.0000	1.1157	1.1066	0.99676

The change of the angle is
\begin{align*}
\Delta \text{angle} = 30.6147 - 20.0641 = 10.5506
\end{align*}
Thus the angle, $\beta$ is
\begin{align*}
\beta = 90 - 10.5506 \sim 79.45
\end{align*}
The pressure at this point is as the surroundings. However, the stagnation pressure is the same as originally was enter the nozzle! This stagnation pressure has to go through serious of oblique shocks and Prandtl-Meyer expansion to match the surroundings stagnation pressure.

Contributors and Attributions

Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.

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Prandtl — Meyer		Input: \(M\)			k = 1.4
\(M\)	\(\nu\)	\(\dfrac{P}{P_0}\)	\(\dfrac{T}{T_0}\)	\(\dfrac{\rho}{\rho_0}\)	\(\mu\)
3.300	62.3113	0.01506	0.37972	0.03965	73.1416

Oblique Shock		Input: \(M_x\)				k = 1.4
\(M_x\)	\({{M_y}_s}\)	\({{M_y}_w}\)	\(\theta_{s}\)	\(\theta_{w}\)	\(\delta\)	\(\dfrac{{P_0}_y}{{P_0}_x}\)
3.300	0.43534	3.1115	88.9313	20.3467	4.0000	0.99676

Isentropic Flow		Input: \(\dfrac{A}{A^{\star}}\)				k = 1.3
\(M\)	\(\dfrac{T}{T_0}\)	\(\dfrac{\rho}{\rho_0}\)	\(\dfrac{A}{A^{\star}}\)	\(\dfrac{P}{P_0}\)	\(\dfrac{{A\,P}_y}{{A^{\star}\,P_0}_x}\)	\(\dfrac{F}{F_0}\)
1.7285	0.69052	0.29102	1.4000	0.20096	0.28134	0.59745