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12.8: Examples For Prandtl–Meyer Function

Example 12.16

Fig. 12.29 Schematic for Example.

A flow of air with a temperature of \(20^\circ C\) and a speed of \(U = 450 m/sec\) flows (see Figure 12.29). Calculate the pressure reduction ratio, and the Mach number after the bending point. If the air flows in an imaginary two–dimensional tunnel with width of 0.1\([m]\) what will the width of this imaginary tunnel after the bend? Calculate the "fan'' angle. Assume the specific heat ratio is \(k=1.4\).

Solution 12.16

First, the initial Mach number has to be calculated (the initial speed of sound).   
\begin{align*}
    c = \sqrt{k\,R\,T} = \sqrt{1.4*287*293} = 343.1 m/sec
\end{align*}
The Mach number is then
\begin{align*}
    M = \dfrac{450 }{ 343.1} = 1.31  
\end{align*}
this Mach number is associated with

Prandtl — Meyer Input: \(M\) k = 1.4
\(M\) \(\nu\) \(\dfrac{P}{P_0}\) \(\dfrac{T}{T_0}\) \(\dfrac{\rho}{\rho_0}\) \(\mu\)
1.3100 6.4449 0.35603 0.74448 0.47822 52.6434

The "new" angle should be  
\begin{align*}
    \nu_2 = 6.4449 + 20 = 26.4449^\circ  
\end{align*}
and results in

Prandtl — Meyer Input: \(M\) k = 1.4
\(M\) \(\nu\) \(\dfrac{P}{P_0}\) \(\dfrac{T}{T_0}\) \(\dfrac{\rho}{\rho_0}\) \(\mu\)
2.0024 26.4449 0.12734 0.55497 0.22944 63.4620

Note that \({P_0}_1 = {P_0}_2\)  
\begin{align*}
    \dfrac{P_2 }{ P_1} = \dfrac{ {P_0}_1 }{ P_1}  \, \dfrac{ P_2 }{ {P_0}_2 }   
    = \dfrac{0.12734 }{ 0.35603} = 0.35766  
\end{align*}
The "new" width can be calculated from the mass conservation equation.      
\begin{align*}
     \rho_1 x_1 M_1 c_1 =
     \rho_2 x_2 M_2 c_2  \Longrightarrow x_2 = x_1 \,\dfrac{\rho_1 }{  \rho_2 }
    \dfrac{M_1 }{ M_2} \sqrt{\dfrac{T_1 }{ T_2} }  
\end{align*}
\begin{align*}
     x_2 =
    0.1 \times \dfrac{0.47822 }{ 0.22944}
    \times \dfrac{1.31 }{ 2.0024} \sqrt{ \dfrac{0.74448 }{ 0.55497 } }
    = 0.1579 [m]  
\end{align*}
Note that the compression "fan'' stream lines are note and their function can be obtain either by numerical method of going over small angle increments. The other alternative is using the exact solution. The expansion "fan'' angle  changes in the Mach angle between the two sides of the bend
\begin{align*}
    \text{fan angle} = 63.4 + 20.0 - 52.6 = 30.8^\circ   
\end{align*}

Reverse the example, and this time the pressure on both sides are given and the angle has to be obtained.

Example 12.17

Fig. 12.30 Schematic for Example .

Gas with \(k=1.67\) flows over bend (see Figure 12.17). The gas flow with Mach 1.4 and Pressure 1.2[\(Bar\)]. It is given that the pressure after the turning is 1[\(Bar\)]. Compute the Mach number after the bend, and the bend angle.

Solution 12.17

The Mach number is determined by satisfying the condition that the pressure downstream and the Mach are given. The relative pressure downstream can be calculated by the relationship
\begin{align*}
    \dfrac{P_2 }{ {P_0}_2} = \dfrac{P_2 }{ P_1}  \,\dfrac{P_1 }{ {P_0}_1}  
    = \dfrac{ 1 }{ 1.2}\times{ 0.31424 } = 0.2619
\end{align*}

Prandtl — Meyer Input: \(M\) k = 1.67
\(M\) \(\nu\) \(\dfrac{P}{P_0}\) \(\dfrac{T}{T_0}\) \(\dfrac{\rho}{\rho_0}\) \(\mu\)
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.

Prandtl — Meyer Input: \(M\) k = 1.67
\(M\) \(\nu\) \(\dfrac{P}{P_0}\) \(\dfrac{T}{T_0}\) \(\dfrac{\rho}{\rho_0}\) \(\mu\)
1.4576 9.1719 0.26190 0.58419 0.44831 55.5479

For the rest of the calculation the initial condition is used. The Mach number after the bend is \(M= 1.4576\). It should be noted that specific heat isn't \(k=1.4\) but \(k=1.67\).  The bend angle is
\begin{align*}
    \Delta\nu  = 9.1719 - 7.7720 \sim 1.4^\circ  
\end{align*}
\begin{align*}
    \Delta\mu = 55.5479  -  54.4623 = 1.0^\circ  
\end{align*}

Contributors

  • 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.