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Examples For Prandtl-Meyer Function

$\begin{examl} % latex2html id marker 28761A wall is included with $20.0^\circ$... ...on{The schematic of Example \ref{pm:ex:simpleExample}} \end{figure}\end{examl}$

Solution

First the initial Mach number has to calculated (the initial speed of sound).

$\displaystyle a= \sqrt{kRT} = \sqrt{1.4*287*293} = 343.1 m/sec$

The Mach number is then

$\displaystyle M = {450 \over 343.1} = 1.31$

This Mach number associated with

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{\nu}$	$\mathbf{P \over P_0}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{\mu }$
1.3100	6.4449	0.35603	0.74448	0.47822	52.6434

The ``new'' angle should be

$\displaystyle \nu_2 = 6.4449 + 20 = 26.4449^\circ$

and results in

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{\nu}$	$\mathbf{P \over P_0}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{\mu }$
2.0024	26.4449	0.12734	0.55497	0.22944	63.4620

Note that ${P_0}_1 = {P_0}_2$

$\displaystyle {P_2 \over P_1} = { {P_0}_1 \over P_1} { P_2 \over {P_0}_2 } = {0.12734 \over 0.35603} = 0.35766$

The ``new'' width can be calculated from the mass conservation equation.

$\displaystyle \rho_1 x_1 M_1 c_1 = \rho_2 x_2 M_2 c_2 \Longrightarrow x_2 = x_1 {\rho_1 \over \rho_2 } {M_1 \over M_2} \sqrt{T_1 \over T_2}$

$\displaystyle x_2 = 0.1 \times {0.47822 \over 0.22944} \times {1.31\over 2.0024} \sqrt{0.74448 \over 0.55497} = 0.1579 [m]$

Note that the compression ``fan'' stream lines are not known and their function can be obtained either by the numerical method of going over small angle increments. The other alternative is using the exact solution^14.1. The expansion ``fan'' angle changes in the Mach angle between the two sides of the bend

$\displaystyle \hbox{fan angle} = 63.4 + 20.0 - 52.6 = 30.8^\circ$

Reverse the example, and this time the pressure on both sides are given and the angle has to be obtained^14.2.
$\begin{examl} % latex2html id marker 28838Gas with $k=1.67$ flows over bend (... ...d{figure}Compute the Mach number after the bend, and the bend angle. \end{examl}$
Solution

The Mach number is determined by satisfying the condition that the pressure downstream are Mach the given one. The relative pressure downstream can be calculated by the relationship

$\displaystyle {P_2 \over {P_0}_2} = {P_2 \over P_1} {P_1 \over {P_0}_1} = { 1\over 1.2}\times{ 0.31424 } = 0.2619$

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{\nu}$	$\mathbf{P \over P_0}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{\mu }$
1.4000	7.7720	0.28418	0.60365	0.47077	54.4623

With this pressure ratio, $\bar{P} = 0.2619$ , require either locking in the table or using the enclosed program is required.

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{\nu}$	$\mathbf{P \over P_0}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{\mu }$
1.4576	9.1719	0.26190	0.58419	0.44831	55.5479

For the rest of the calculation the initial conditions are used. The Mach number after the bend is . It should be noted that specific heat isn't but . The bend angle is

$\displaystyle \Delta\nu = 9.1719 - 7.7720 \sim 1.4^\circ$

$\displaystyle \Delta\mu = 55.5479 - 54.4623 = 1.0^\circ$

Next: Combination of the Oblique Up: Prandtl-Meyer Function Previous: d'Alembert's Paradox Index

Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21

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