Chapter 9, Problem 9.8CP

To determine

Find the exit conditions and thrust produced of the given system by using given data.

Answer to Problem 9.8CP

The values of exit conditions and thrust produced is determined below.

Explanation of Solution

Given information:

From the properties of air altitude of $10000m$ from the table

Inlet pressure $p_1=26416Pa$

Inlet temperature $T_1=233.16K$

Density at inlet ${\rho }_1=0.4125kg/m^3$

Velocity at this temperature is

$\begin{array}{l}{a}_1=\sqrt{kRT}\\ a_1=\sqrt{1.4\times 287\times 233.16}\\ a_1=299.5m/s\end{array}$

Find the inlet velocity

$\begin{array}{l}{V}_1=M{a}_1{a}_1\\ =6\times 299.5\\ =1797m/s\end{array}$

Find the mass flow rate

$\begin{array}{l}\dot{m}={\rho }_1{A}_1{V}_1\\ =0.4125\times 1.0\times 1797\\ =741kg/s\end{array}$

Find the stagnation temperature at 2.

$\begin{array}{l}{T}_{o1}=T_{o2}\\ =T_1\left[1+0.2{\left(6\right)}^2\right]\\ =1830K\end{array}$

Find the stagnation pressure at 2.

$\begin{array}{l}{p}_{o1}=p_{o2}\\ =p_1{\left[1+0.2M{a}_1^2\right]}^{\frac{k}{k-1}}\\ =26416{\left[1+0.2{\left(6\right)}^2\right]}^{\frac{1.4}{1.4-1}}\\ =4.17\times {10}^{7}Pa\end{array}$

And then find the Mach number by using below relation:

$\begin{array}{l}\frac{A_2}{A^{*}}=\left(\frac{A_1}{A^{*}}\right)\left(\frac{0.2m^2}{1.0m^2}\right)\\ =\left(53.18\right)\left(\frac{0.2m^2}{1.0m^2}\right)\\ =10.64\end{array}$

Find the temperature at section 2.

$\begin{array}{l}{T}_2=\frac{T_{o2}}{\left[1+0.2M{a}_2^2\right]}\\ =\frac{1830}{\left[1+0.2{\left(3.991\right)}^2\right]}\\ =437K\end{array}$

Find the pressure at section 2.

$\begin{array}{l}{p}_2=\frac{p_{o2}}{{\left[1+0.2M{a}_2^2\right]}^{\frac{k}{k-1}}}\\ =\frac{4.17\times {10}^7}{{\left[1+0.2{\left(3.991\right)}^2\right]}^{\frac{1.4}{1.4-1}}}\\ =278000Pa\end{array}$

Apply the frictionless heat addition theory

$\begin{array}{l}\frac{T_{o2}}{T_o^{*}}=0.5895\\ \frac{1830}{T_o^{*}}=0.5895\\ T_o^{*}=3104K\end{array}$

Temperature:

$\begin{array}{l}\frac{T_2}{T^{*}}=0.169\\ T^{*}=\frac{437}{0.169}\\ T^{*}=2587K\end{array}$

Pressure:

$\begin{array}{l}\frac{p_2}{p^{*}}=0.103\\ p^{*}=\frac{278000}{0.03}\\ p^{*}=2.698\times {10}^{6}Pa\end{array}$

Find the stagnation temperature at section 3

$\begin{array}{l}{T}_{o3}=T_{o2}+\frac{Q}{c_p}\\ =1830+\frac{500000}{1005}\\ =2327K\end{array}$

Find the temperature ratio

$\begin{array}{l}\frac{T_{o3}}{T_o^{*}}=\frac{2327}{3104}\\ =0.7497\\ M{a}_3=2.238\\ \frac{p_3}{p^{*}}=0.2996\\ p_3=0.2996\times 2.698\times {10}^6\\ p_3=808500Pa\end{array}$

Find the stagnation pressure at states 3 and 4:

$\begin{array}{l}{p}_{o3}=p_{o4}\\ =p_3{\left[1+0.2M{a}_3^2\right]}^{\frac{k}{k-1}}\\ =808500{\left[1+0.2{\left(2.238\right)}^2\right]}^{\frac{1.4}{1.4-1}}\\ =9.17\times {10}^{6}Pa\end{array}$

$\frac{A_3}{A_3{}^{*}}=2.073$

Corresponding ratio at section 4 is

$\begin{array}{l}\frac{A_4}{A_3{}^{*}}=\left(\frac{A_3}{A_3{}^{*}}\right)\left(\frac{1.0m^2}{0.2m^2}\right)\\ =\left(2.037\right)\left(\frac{1}{0.2}\right)\\ =10.37\end{array}$

Then the value of Mach number is $M{a}_4=3.963$

Find the temperature at section 4 is

$\begin{array}{l}{T}_4=\frac{T_{o4}}{\left[1+0.2M{a}_4^2\right]}\\ =\frac{2327}{\left[1+0.2{\left(3.963\right)}^2\right]}\\ =562K\end{array}$

Find the pressure at section 4

$\begin{array}{l}{p}_4=\frac{p_{o4}}{{\left[1+0.2M{a}_4^2\right]}^{\frac{k}{k-1}}}\\ =\frac{9.17\times {10}^6}{{\left[1+0.2{\left(3.963\right)}^2\right]}^{\frac{1.4}{1.4-1}}}\\ =63480Pa\end{array}$

Find the density of air at section 4 is

$\begin{array}{l}{\rho }_4=\frac{p_4}{R{T}_4}\\ =\frac{63480}{287\times 562}\\ =0.934kg/m^3\end{array}$

Find the velocity at section 4 is

$\begin{array}{l}{V}_4=M{a}_4\sqrt{kR{T}_4}\\ =3.963\sqrt{1.4\times 287\times 562}\\ =1883m/s\end{array}$

Find the thrust at given section.

$\begin{array}{l}F=\dot{m}\left(V_4-V_1\right)+A_1\left(p_4-p_1\right)\\ =741\left(1883-1797\right)+1.00\left(63480-26416\right)\\ =64150+37060\\ =101200N\end{array}$.