Chapter 9, Problem 9.44P

To determine

To calculate:

The temperature, pressure, and velocity at exit and to compute the thrust.

Answer to Problem 9.44P

$p_e=2266.7lbf/f{t}^2$

$T_e=1587.17°R$

$V_e=5683.17ft/s$

$F_{thrust}=4004.32lbf$

Thrust force is not dependent on stagnation temperature

Because,

$\begin{array}{l}{\rho }_e\infty \frac{1}{T_0}\\ V_e\infty \sqrt{T_0}\end{array}$

Therefore, the stagnation temperature cancels out.

Explanation of Solution

Given information:

Throat diameter is $3in$

$k=1.4$

Molecular weight is equal to $M=26$

$p_a=14.7lbf/i{n}^2$

Speed of sound is defined as,

$a=\sqrt{kRT}$

Where,

$R$ - Gas constant

$k$ - Specific heat capacity

The Mach number is defined as,

$Ma=\frac{V}{a}$

Where,

$V$ - Air velocity

The density at section 1 is defined as,

${\rho }_1=\frac{p_1}{R{T}_1}$

The pressure ratio is defined as,

$\frac{p_0}{p}={\left[1+\frac{1}{2}\left(k-1\right)M{a}^2\right]}^{k/k-1}$

Area change is defined as,

$\frac{A}{A^{\ast }}=\frac{1}{Ma}\frac{{\left(1+0.2M{a}^2\right)}^3}{1.728}$

The temperature ratio is defined as,

$\frac{T_0}{T_e}=\left[1+\frac{1}{2}\left(k-1\right)M{a}_e{}^2\right]$

Calculation:

First of all, find the relevant gas constant,

$R=\frac{49720}{26}=1912.3f{t}^2/s^2°R$

Calculate the Mach number at exit,

The area change can be defined as,

$\frac{A_e}{A^{\ast }}=\frac{1}{M{a}_e}\frac{{\left(1+0.2M{a}_e{}^2\right)}^3}{1.728}$

Substitute,

$\frac{d_e{}^2}{d^{{\ast }^2}}=\frac{1}{M{a}_e}\frac{{\left(1+0.2M{a}_e{}^2\right)}^3}{1.728}$

Substitute for known values,

$\begin{array}{l}{\left(\frac{5.5}{3}\right)}^2=\frac{1}{M{a}_e}\frac{{\left(1+0.2M{a}_e{}^2\right)}^3}{1.728}\\ M{a}_e=2.757\end{array}$

Calculate the pressure at exit,

Convert,

$p_0=400lbf/i{n}^2=57600lbf/f{t}^2$

$\begin{array}{l}\frac{p_0}{p_e}={\left[1+\frac{1}{2}\left(k-1\right)M{a}_e{}^2\right]}^{k/k-1}\\ \frac{57600lbf/f{t}^2}{p_e}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(2.757\right)}^2\right]}^{1.4/1.4-1}\\ p_e=2266.7lbf/f{t}^2\end{array}$

Calculate the exit temperature,

$\begin{array}{l}\frac{T_0}{T_e}=\left[1+\frac{1}{2}\left(k-1\right)M{a}_e{}^2\right]\\ \frac{4000°R}{T_e}=\left[1+\frac{1}{2}\left(1.4-1\right){\left(2.757\right)}^2\right]\\ T_e=1587.17°R\end{array}$

Calculate the exit velocity,

$V_e=M{a}_e\sqrt{kR{T}_e}=\left(2.757\right)\sqrt{1.4\left(1912.3f{t}^2/s^2°R\right)\left(1587.17°R\right)}$

Solve to find exit velocity,

$V_e=5683.17ft/s$

Calculate the exit density,

${\rho }_e=\frac{p_e}{R{T}_e}=\frac{2266.7lbf/f{t}^2}{\left(1912.3f{t}^2/s^2°R\right)\left(1587.17°R\right)}=7.468\times {10}^{-4}lbf/f{t}^3$

For this problem, the thrust force is defined as,

$F_{thrust}=A_e\left({\rho }_e{V}_e{}^2+\left(p_e-p_a\right)\right)$

Substitute for known values,

$F_{thrust}=\left(\pi {\left(\frac{2.75}{12}ft\right)}^2\right)\left[\left(7.468\times {10}^{-4}lbf/f{t}^3\right){\left(5683.17ft/s\right)}^2+\left(2266.7lbf/f{t}^2-2116.8lbf/f{t}^2\right)\right]$

Solve to find thrust force,

$F_{thrust}=4004.32lbf$

Thrust force is not dependent on stagnation temperature

Because,

$\begin{array}{l}{\rho }_e\infty \frac{1}{T_0}\\ V_e\infty \sqrt{T_0}\end{array}$

Therefore, the stagnation temperature cancels out.

Conclusion:

Exit pressure is $p_e=2266.7lbf/f{t}^2$

Exit temperature is $T_e=1587.17°R$

The exit velocity is $V_e=5683.17ft/s$

The thrust force is equal to $F_{thrust}=4004.32lbf$

Thrust force is not dependent on stagnation temperature

Because,

$\begin{array}{l}{\rho }_e\infty \frac{1}{T_0}\\ V_e\infty \sqrt{T_0}\end{array}$

Therefore, the stagnation temperature cancels out.