Chapter 13, Problem 13.1P

Note: The purpose of the following problem is to provide an exercise in carrying out a unit process for the method of characteristics. A more extensive application to a complete flow field is left to your specific desires. Also, an extensive practical problem utilizing the finite-difference method requires a large number of arithmetic operations and is practical only on a digital computer. You are encouraged to set up such a problem at your leisure. The main purpose of the present chapter is to present the essence of several numerical methods, not to burden the reader with a lot of calculations or the requirement to write an extensive computer program.

Consider two points in a supersonic flow. These points are located in a cartesian coordinate system at $\left(x_1,y_1\right)=\left(0,0.0684\right)$ and $\left(x_2,y_2\right)=\left(0.0121,0\right)$, where the units are meters. At point $\left(x_1,y_1\right):u_1=\text{639 m}/\text{s},v_1=\text{232}.\text{6 m}/\text{s},p_1=\text{1 atm},T_1=\text{288 K}$. At point $\left(x_2,y_2\right):u_2=\text{680 m}/\text{s},v_2=0,p_2=\text{1 atm},T_2=\text{288 K}$. Consider point 3 downstream of points 1 and 2 located by the intersection of the $C_{+}$ characteristic through point 2 and the $C_{-}$ characteristic through point 1. At point 3, calculate: $u_3,v_3,p_3$, and $T_3$. Also, calculate the location of point 3, assuming the characteristics between these points are straight lines.

To determine

The numerical value of $u_3$ .

The numerical value of $v_3$ .

The numerical value of $p_3$ .

The numerical value of $T_3$ .

The location of point 3.

Answer to Problem 13.1P

The value of $u_3$ is $735.3\text{m}/\text{s}$ .

The value of $v_3$ is $129.6\text{m}/\text{s}$ .

The value of $p_3$is $0.535\text{atm}$ .

The value of $T_3$ is $240.9\text{K}$ .

The location of point 3 is $\left(0.0894,0.0489\right)$ .

Explanation of Solution

Given:

The Cartesian coordinate system at $\left(x_1,y_1\right)=\left(0,0.0684\right)$ .

The Cartesian coordinate system at $\left(x_2,y_2\right)=\left(0.0121,0\right)$ .

The numerical value of $u_1=639\text{m}/\text{s}$ .

The numerical value of $u_2=680\text{m}/\text{s}$ .

The numerical value of $v_1=232.6\text{m}/\text{s}$ .

The numerical value of $v_2=0\text{m}/\text{s}$ .

The numerical value of $p_1=1\text{atm}$ .

The numerical value of $p_2=1\text{atm}$ .

The numerical value of $T_1=288\text{K}$ .

The numerical value of $T_2=288\text{K}$ .

Formula used:

The expression for the Mach number is given as,

  $M=\frac{V}{a}$

Here, $V$ is the object speed and $a$ is the sound speed in that medium.

The expression for speed of object is given as,

  $V=\sqrt{u^2+v^2}$

The expression for speed of sound is given as,

  $a=\sqrt{\gamma RT}$

The expression for angle of object is given as,

  $\theta ={\tan }^{-1}\left(\frac{v}{u}\right)$

Calculation:

The speed of sound at point 1 can be calculated,

  $\begin{array}{c}{a}_1=\sqrt{\gamma R{T}_1}\\ =\sqrt{\left(1.4\right)\left(287\text{kJ}/\text{kg}\text{.K}\right)\left(288\text{K}\right)}\\ =340\text{m}/\text{s}\end{array}$

The speed of object at point 1 can be calculated as,

  $\begin{array}{c}{V}_1=\sqrt{u_1^2+v_1^2}\\ =\sqrt{{\left(639\text{m}/{\text{s}}^{\text{2}}\right)}^2+{\left(232.6\text{m}/{\text{s}}^{\text{2}}\right)}^2}\\ =680\text{m}/\text{s}\end{array}$

The Mach number at point 1 can be calculated as,

  $\begin{array}{c}{M}_1=\frac{V_1}{a_1}\\ =\frac{680\text{m}/{\text{s}}^{\text{2}}}{340\text{m}/{\text{s}}^{\text{2}}}\\ =2\end{array}$

The angle of object at point 1 can be calculated as,

  $\begin{array}{c}{\theta }_1={\tan }^{-1}\left(\frac{v_1}{u_1}\right)\\ ={\tan }^{-1}\left(\frac{232.6\text{m}/{\text{s}}^{\text{2}}}{639\text{m}/{\text{s}}^{\text{2}}}\right)\\ =20°\end{array}$

The constant at point 1 can be calculated as,

  $\begin{array}{c}{\upsilon }_1=M_1\\ =26.38°\end{array}$

The flow constant at point 1 can be calculated as,

  $\begin{array}{c}{K}_{-}=\theta +\upsilon \\ =20+26.38\\ =46.38°\end{array}$

The speed of sound at point 2 can be calculated as,

  $\begin{array}{c}{a}_2=\sqrt{\gamma R{T}_1}\\ =\sqrt{\left(1.4\right)\left(287\text{kJ}/\text{kg}\text{.K}\right)\left(288\text{K}\right)}\\ =340\text{m}/\text{s}\end{array}$

The speed of object at point 2 can be calculated as,

  $V_2=680\text{m}/\text{s}$

The Mach number at point 2 can be calculated as,

  $\begin{array}{c}{M}_2=\frac{V_2}{a_2}\\ =\frac{680\text{m}/\text{s}}{340\text{m}/\text{s}}\\ =2\end{array}$

The angle of object at point 2 can be calculated as,

  ${\theta }_2=0°$

The constant at point 2 can be calculated as,

  ${\upsilon }_2=26.38°$

The flow constant at point 2 can be calculated as,

  $\begin{array}{c}{K}_{+}=\theta -\upsilon \\ =-26.38°\end{array}$

The angle of object at point 3 can be calculated as given bellow,

  $\begin{array}{c}{\theta }_3=\frac{1}{2}\left[{\left(K_{-}\right)}_1+{\left(K_{+}\right)}_2\right]\\ =\frac{1}{2}\left(46.38°-26.38°\right)\\ =10°\end{array}$

The constant at point 3 can be calculated as,

  $\begin{array}{c}{\upsilon }_3=\frac{1}{2}\left[{\left(K_{-}\right)}_1+{\left(K_{+}\right)}_2\right]\\ =36.38°\end{array}$

The Mach number at point 3 can be calculated as,

  $M_3=2.4$

To obtain the other flow variables at point 3, expression is given as,

  $\frac{P_{o_1}}{P_1}=7.824$ and $\frac{P_{o_3}}{P_3}=14.62$

The pressure at point 2 can be calculated as,

  $\begin{array}{c}{P}_3=\frac{P_3}{P_{o_3}}\frac{P_{o_3}}{P_{o_1}}\frac{P_{o_1}}{P_1}{P}_1\\ =\left(\frac{1\text{atm}}{14.62\text{atm}}\right)\left(1\right)\left(7.824\right)\left(1\text{atm}\right)\\ =0.535\text{atm}\end{array}$

The temperature at point 3 is given as below,

  $\frac{T_{o_1}}{T_1}=1.8$ and $\frac{T_{o_3}}{T_3}=2.152$

Expression for the temperature,

  $\begin{array}{c}{T}_3=\frac{T_3}{T_{o_3}}\frac{T_{o_3}}{T_{o_1}}\frac{T_{o_1}}{T_1}{T}_1\\ =\left(\frac{\text{1}\text{K}}{\text{2}\text{.152}\text{K}}\right)\left(\text{1}\right)\left(\text{1}\text{.8}\right)\left(\text{288}\text{K}\right)\\ =240.9\text{K}\end{array}$

The speed of sound at point 3 can be calculated as,

  $\begin{array}{c}{a}_3=\sqrt{\gamma R{T}_3}\\ =\sqrt{1.4\times \left(287\text{kJ}/\text{kg}\text{.K}\right)\times \left(240.9\text{K}\right)}\\ =211.1\text{m}/\text{s}\end{array}$

The speed of object at point 3 can be calculated as,

  $\begin{array}{c}{V}_3=M_3{a}_3\\ =2.4\times \left(311.1\text{m}/\text{s}\right)\\ =746.6\text{m}/\text{s}\end{array}$

The initial velocity of object at point 3 can be calculated as,

  $\begin{array}{c}{u}_3=V_3\cos {\theta }_3\\ =\left(746.6\text{m}/\text{s}\right)\cos 10°\\ =735.3\text{m}/\text{s}\end{array}$

The final velocity of object at point 3 can be calculated as,

  $\begin{array}{c}{v}_3=V_3\sin {\theta }_3\\ =\left(746.6\text{m}/\text{s}\right)\sin 10°\\ =129.6\text{m}/\text{s}\end{array}$

To locate point 3 expression is,

The average angle at point 2 and 3 along the $C_{+}$ Characteristics is calculated as,

  $\begin{array}{c}{\theta }_{avg}=\frac{1}{2}\left({\theta }_2+{\theta }_3\right)\\ =\frac{1}{2}\left(0°+10°\right)\\ =5°\end{array}$

The average angle at point 2 and 3 along the $C_{-}$ Characteristics is calculated as,

  $\begin{array}{c}{\mu }_{avg}=\frac{1}{2}\left({\mu }_2+{\mu }_3\right)\\ =\frac{1}{2}\left(30°+24.62°\right)\\ =27.31°\end{array}$

The final angle at point 2 and 3 is given as,

  $\begin{array}{c}\frac{dy}{dx}=\tan \left({\theta }_{avg}+{\mu }_{avg}\right)\\ =\tan \left(5°+27.31°\right)\\ =0.6324°\end{array}$

The equation for the point on they-axis is given as,

  $y=0.6324x-0.00765$   ....... (1)

The average angle at point 1 and 3 along the $C_{+}$ Characteristics is calculated as,

  $\begin{array}{c}{\theta }_{avg}=\frac{1}{2}\left({\theta }_1+{\theta }_3\right)\\ =\frac{1}{2}\left(20°+10°\right)\\ =15°\end{array}$

The average angle at point 1 and 3 along the $C_{-}$ Characteristics is calculated as,

  $\begin{array}{c}{\mu }_{avg}=\frac{1}{2}\left({\mu }_1+{\mu }_3\right)\\ =\frac{1}{2}\left(30°+24.62°\right)\\ =27.31°\end{array}$

The final angle at point 1 and 3 is given as below,

  $\begin{array}{c}\frac{dy}{dx}=\tan \left({\theta }_{avg}-{\mu }_{avg}\right)\\ =\tan \left(15°+27.31°\right)\\ =-0.2182°\end{array}$

The equation for the point on the x-axis is given as below,

  $y=-0.2182x-0.0684$   ....... (2)

On solving equation (1) and (2), we will get the point 3 as,

  $\begin{array}{l}x=0.0894\\ y=0.0489\end{array}$

Thus, $\left(x_3,y_3\right)=\left(0.0894,0.0489\right)$

Conclusion:

Therefore, the value of $u_3$ is $735.3\text{m}/\text{s}$ .

Therefore, the value of $v_3$ is $129.6\text{m}/\text{s}$ .

Therefore, the value of $p_3$is $0.535\text{atm}$.

Therefore, the value of $T_3$ is $240.9\text{K}$.

Therefore, the location of point 3 is $\left(0.0894,0.0489\right)$.