Chapter 9, Problem 9.124P

To determine

(a)

Value of $M{a}_1$.

Answer to Problem 9.124P

The required value of the $M{a}_1$ is $M{a}_1=1.87$.

Explanation of Solution

Given Information:

$\theta ={20}^o$

Formula used:

$\tan \theta =\frac{2\cot \beta \left(M{a}_1{}^2{\sin }^2\beta -1\right)}{M{a}_1{}^2\left(k+\cos 2\beta \right)+2}$

Calculation:

To solve this problem, for sea level we are taking the values of $p1=101.35kPa$, $T_1=288.16K$ and $\rho 1=1.2255kg/m^3$.

Now we find the Mach number;

We specified that, $\beta ={60}^o$ and $\theta ={20}^o$ ;

We know that, $\tan \theta =\frac{2\cot \beta \left(M{a}_1{}^2{\sin }^2\beta -1\right)}{M{a}_1{}^2\left(k+\cos 2\beta \right)+2}$ ;

Put the values in the above equation for $\theta ={20}^o$ and $\beta ={60}^o$ ;

Iterate, or use EES;

$M{a}_1=1.87$.

To determine

(b)

Value of $p_2$.

Answer to Problem 9.124P

The required value of the $p_2$ is $293kPa$.

Explanation of Solution

Given information:

$\theta ={20}^o$

Formula used:

$\frac{p2}{p1}=\frac{1}{k+1}\left(2kM{a}_1{}^2{\sin }^2\beta -k+1\right)$

Calculation:

We know the equation, $\frac{p2}{p1}=\frac{1}{k+1}\left(2kM{a}_1{}^2{\sin }^2\beta -k+1\right)$ ;

Put the values in the above equation;

$\frac{p2}{p1}=\frac{1}{k+1}\left(2kM{a}_1{}^2{\sin }^2\beta -k+1\right)=2.893$

$\begin{array}{l}p2=2.893\left(101.35kPa\right)\\ \Rightarrow 293kPa\end{array}$.

To determine

(c)

Value of $T_2$.

Answer to Problem 9.124P

The required value of the $T_2$ is $404K$

Explanation of Solution

Given Information:

$\theta ={20}^o$

Formula used:

$\frac{T2}{T1}=\left[2+\left(k-1\right)M{a}_1{}^2{\sin }^2\beta \right]\frac{2kM{a}_1{}^2{\sin }^2\beta -k+1}{{\left(k+1\right)}^{2}M{a}_1{}^2{\sin }^2\beta };$

Calculation:

We know that, $\frac{T2}{T1}=\left[2+\left(k-1\right)M{a}_1{}^2{\sin }^2\beta \right]\frac{2kM{a}_1{}^2{\sin }^2\beta -k+1}{{\left(k+1\right)}^{2}M{a}_1{}^2{\sin }^2\beta };$

Put the values in the above equation;

$\frac{T2}{T1}=\left[2+\left(k-1\right)M{a}_1{}^2{\sin }^2\beta \right]\frac{2kM{a}_1{}^2{\sin }^2\beta -k+1}{{\left(k+1\right)}^{2}M{a}_1{}^2{\sin }^2\beta }=1.401$

$\begin{array}{l}T2=1.401\left(288.16K\right)\\ \Rightarrow 404K\end{array}$.

To determine

(d)

Value of $V_2$.

Answer to Problem 9.124P

The required value of the $V_2$ is $415m/s$

Explanation of Solution

Given information:

$\theta ={20}^o$

Formula used:

$a=\sqrt{kR{T}_1}$

$\frac{V2}{V1}=\frac{\cos \beta }{\cos \left(\beta -\theta \right)}$

$V1=a_{1}M{a}_1$

Calculation:

We know that, $a=\sqrt{kR{T}_1};$

Put the values in the above equation

$\begin{array}{l}a=\sqrt{kR{T}_1}=\sqrt{1.4\left(287\right)\left(288.16\right)}\\ \Rightarrow 340m/s\end{array}$

Now we will find V1;

$\begin{array}{l}V1=a_{1}M{a}_1=\left(340\right)\left(1.87\right)\\ \Rightarrow 636m/s\end{array}$

We know that, $\frac{V2}{V1}=\frac{\cos \beta }{\cos \left(\beta -\theta \right)};$

Put the values in the above equation;

$\begin{array}{l}\frac{V2}{V1}=\frac{\cos \beta }{\cos \left(\beta -\theta \right)}=\frac{\cos {60}^o}{\cos {40}^o}\\ \Rightarrow 0.653\end{array}$

$\begin{array}{l}V2=0.653\left(636m/s\right)\\ \Rightarrow 415m/s\end{array}$.