Chapter 9.12, Problem 133P

a)

To determine

The pressure at the turbine exit.

Answer to Problem 133P

The pressure at the turbine exit is $55.2\text{psia}$.

Explanation of Solution

Draw the $T-s$ diagram for pure jet engine as shown in Figure (1).

Consider that the aircraft is stationary, and the velocity of air moving towards the aircraft is $V_1=900\text{ft/s}$, the air will leave the diffuser with a negligible velocity $\left(V_2\cong 0\right)$.

Diffuser (For process 1-2):

Write the expression for the energy balance equation for the diffuser.

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}$ (I)

Here, the rate of energy entering the system is ${\dot{E}}_{\text{in}}$, rate of energy leaving the system is ${\dot{E}}_{\text{out}}$, and the rate of change in the energy of the system is $\Delta {\dot{E}}_{\text{system}}$.

Write the temperature and pressure relation for the process 1-2.

$P_2=P_1{\left(\frac{T_2}{T_1}\right)}^{k/\left(k-1\right)}$ (II)

Here, the specific heat ratio of air is k, pressure at state 1 is $P_1$, pressure at state 2 is $P_2$, temperature at state 1 is $T_1$ and temperature at state 2 is $T_2$.

Compressor (For process 2-3)

Write the pressure relation using the pressure ratio for the process 2-3.

$P_3=P_4=\left(r_p\right)\left(P_2\right)$ (III)

Here, the pressure ratio is $r_p$, pressure at state 3 is $P_3$ and pressure at state 4 is $P_4$.

Write the temperature and pressure relation for the process 2-3.

$\begin{array}{l}{T}_3=T_2{\left(\frac{P_3}{P_2}\right)}^{\left(k-1\right)/k}\\ T_3=T_2\left(r_p^{\left(k-1\right)/k}\right)\end{array}$ (IV)

Here, temperate at state 3 is $T_3$.

Turbine (For process 4-5)

Write the temperature relation for the compressor and turbine.

$\begin{array}{l}{w}_{comp,in}=w_{turb,out}\\ h_3-h_2=h_4-h_5\\ c_p\left(T_3-T_2\right)=c_p\left(T_4-T_5\right)\end{array}$

$\begin{array}{l}{T}_3-T_2=T_4-T_5\\ T_5=T_4-T_3+T_2\end{array}$ (V)

Here, the specific heat at constant pressure is $c_p$, enthalpy at state 2 is $h_2$, enthalpy at state 3 is $h_3$, enthalpy at state 4 is $h_4$, enthalpy at state 5 is $h_5$, work input to the compressor is $w_{comp,in}$, work output from turbine is $w_{turb,out}$, temperature at state 4 is $T_4$ and temperature at state 5 is $T_5$.

Write the temperature and pressure relation for the process 4-5.

$P_5=P_4{\left(\frac{T_5}{T_4}\right)}^{k/\left(k-1\right)}$ (VI).

Conclusion:

From Table A-2E, “Ideal-gas specific heats of various common gases”, obtain the following values for air at room temperature.

$\begin{array}{l}k=1.4\\ c_p=0.24\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

The rate of change in the energy of the system $\left(\Delta {\dot{E}}_{\text{system}}\right)$ is zero for the steady state system.

Substitute $0$ for $\Delta {\dot{E}}_{\text{system}}$ in Equation (I).

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0$

$\begin{array}{l}{\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\\ 0=h_2-h_1+\frac{V_2^2-V_1^2}{2}\\ 0=c_p\left(T_2-T_1\right)-\frac{V_2^2-V_1^2}{2}\end{array}$ (VII).

Here, inlet velocity is $V_1$ or $V_{inlet}$ , velocity at state 2 is $V_2$ , temperate at state 2 is $T_2$ and temperate at state 1 is $T_1$.

Substitute 0 for $V_2$, $10°\text{F}$ for $T_1$, $900\text{ft}/\text{s}$ for $V_1$, and $0.24\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ in Equation (VII).

$\begin{array}{l}0=0.24\text{Btu}/\text{lbm}\cdot \text{R}\left(T_2-\left(10°\text{F}\right)\right)-\frac{\left(0\right)-{\left(900\text{ft}/\text{s}\right)}^2}{2}\\ T_2=\left(10+460\right)\text{}\text{R}+\frac{{\left(900\text{ft}/\text{s}\right)}^2}{\left(2\right)\left(0.24\text{Btu}/\text{lbm}\cdot \text{R}\right)}\left(\frac{1\text{Btu}/\text{lbm}}{25,037{\text{ft}}^2/{\text{s}}^2}\right)\\ T_2=537.4\text{R}\end{array}$

Substitute $\text{7}\text{psia}$ for $P_1$, 537.3 R for $T_2$, $10°\text{F}$ for $T_1$, and 1.4 for k in Equation (II).

$\begin{array}{c}{P}_2=\left(7\text{psia}\right){\left(\frac{537.3\text{R}}{10°\text{F}}\right)}^{1.4/1.4-1}\\ =\left(7\text{psia}\right){\left(\frac{537.3\text{R}}{\left(10+460\right)\text{}\text{R}}\right)}^{1.4/1.4-1}\\ =11.19\text{}\text{psia}\end{array}$

Substitute 13 for $r_p$, and $\text{11}\text{.19}\text{psia}$ for $P_2$ in Equation (III).

$\begin{array}{l}{P}_3=P_4=\left(13\right)\left(\text{11}\text{.19}\text{psia}\right)\\ =145\text{psia}\end{array}$

Substitute 537.4 R for $T_2$, 1.4 for k, and 13 for $r_p$ in Equation (IV).

$\begin{array}{c}{T}_3=\left(537.4\text{R}\right){\left(13\right)}^{1.4-1/1.4}\\ =1118.3\text{R}\end{array}$

Substitute $\text{2400}\text{R}$ for $T_4$, 1118.3 R for $T_3$, and 537.4 R for $T_2$ in Equation (V).

$\begin{array}{c}{T}_5=\left(2400\text{R}-1118.3\text{R}+537.4\text{R}\right)\\ =\text{1819}\text{.1}\text{}\text{R}\end{array}$

Substitute $\text{1819}\text{.1}\text{R}$ for $T_5$, $\text{2400}\text{R}$ for $T_4$, $\text{145}\text{.5}\text{psia}$ for $P_4$, and 1.4 for k in Equation (VI).

$\begin{array}{c}{P}_5=\left(145.5\text{psia}\right){\left(\frac{1819.1\text{R}}{2400\text{R}}\right)}^{1.4/1.4-1}\\ P_5=P_{exit}=55.2\text{psia}\end{array}$

Thus, the pressure at the turbine exit is $55.2\text{psia}$.

b)

To determine

The exit velocity of the exhaust gases.

Answer to Problem 133P

The exit velocity of the exhaust gases is $3121\text{ft}/\text{s}$.

Explanation of Solution

Nozzle (For process 5-6)

Write the temperature and pressure relation for the isentropic process 4-6.

$T_6=T_4{\left(\frac{P_6}{P_4}\right)}^{\left(k-1\right)/k}$ (VIII)

Here, pressure at state 6 is $P_6$ and temperature at state 6 is $T_6$.

Write the energy balance equation for the nozzle.

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}$ (IX)

Conclusion:

Substitute $\text{1819}\text{.1}\text{R}$ for $T_5$, 1.4 for k, $\text{7}\text{psia}$ for $P_6$, and $\text{55}\text{.2}\text{psia}$ for $P_5$ in Equation(VIII).

$\begin{array}{c}{T}_6=\left(1819.1\text{}\text{R}\right){\left(\frac{7\text{psia}}{55.2\text{psia}}\right)}^{1.4-1/1.4}\\ =1008.4\text{R}\end{array}$

The rate of change in the energy of the system $\left(\Delta {\dot{E}}_{\text{system}}\right)$ is zero for the steady state system.

Substitute $0$ for $\Delta {\dot{E}}_{\text{system}}$ Equation (IX).

$\begin{array}{c}{\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ h_5+\frac{V_5^2}{2}=h_6+\frac{V_6^2}{2}\\ 0=h_6-h_5+\frac{V_6^2-V_5^2}{2}\\ 0=c_p\left(T_6-T_5\right)-\frac{V_6^2-V_5^2}{2}\end{array}$ (X)

Here, velocity at stat 5 is $V_5$, exit velocity is $V_6$ or $V_{exit}$ and temperate at state 6 is $T_6$.

Since, $V_5=V_2$

Substitute $\text{1819}\text{.1}\text{R}$ for $T_5$, $\text{1008}\text{.6}\text{R}$ for $T_6$, and $0.24\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ in Equation (X).

$0=0.24\text{Btu}/\text{lbm}\cdot \text{R}\left(\text{1819}\text{.1}\text{R}-\text{1008}\text{.6}\text{R}\right)-\frac{V_6^2-0}{2}$

$\begin{array}{l}{V}_6=\sqrt{\left(2\right)\left(0.24\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(1819.1\text{R}-1008.6\text{R}\right)\left(\frac{25,037{\text{ft}}^2/{\text{s}}^2}{1\text{Btu}/\text{lbm}}\right)}\\ V_6=V_{exit}=3121\text{ft}/\text{s}\end{array}$

Thus, the exit velocity of the exhaust gases is $3121\text{ft}/\text{s}$.

c)

To determine

The propulsive efficiency of the turbojet engine.

Answer to Problem 133P

The propulsive efficiency of the turbojet engine is $25.9\%$.

Explanation of Solution

Write the expression to calculate the propulsive work done per unit mass by the turbojet engine $\left(w_p\right)$.

$w_p=\left(V_{\text{exit}}-V_{\text{inlet}}\right)V_{\text{aircraft}}$ (XI).

Here, the velocity of the aircraft is $V_{\text{aircraft}}$, the velocity of the inlet air is $V_{\text{inlet}}$, and the exit velocity of the exhaust gases is $V_{\text{exit}}$.

Write the expression to calculate the heating value of the fuel per unit mass for the turbojet engine $\left(q_{\text{in}}\right)$.

$\begin{array}{c}{q}_{\text{in}}=h_4-h_3\\ =c_p\left(T_4-T_3\right)\end{array}$ (XII).

Here, temperature at state 4 is $T_4$ , temperature at state 3 is $T_3$ , enthalpy at state 4 is $h_4$ and enthalpy at state 3 is $h_3$.

Write the expression to calculate the propulsive efficiency of the turbojet engine $\left({\eta }_p\right)$.

${\eta }_p=\frac{w_p}{q_{\text{in}}}$ (XIII).

Conclusion.

Substitute $3121\text{ft}/\text{s}$ for $V_{\text{exit}}$, $900\text{ft}/\text{s}$ for $V_{\text{inlet}}$, and $900\text{ft}/\text{s}$ for $V_{\text{aircraft}}$ in Equation (XI).

$\begin{array}{c}{w}_p=\left(3121\text{ft}/\text{s}-900\text{ft}/\text{s}\right)\times \left(900\text{ft}/\text{s}\right)\\ =\left(2221\text{ft}/\text{s}\right)\times \left(900\text{ft}/\text{s}\right)\\ =1998900{\text{ft}}^2/{\text{s}}^2\left(\frac{1\text{Btu}/\text{lbm}}{25,037{\text{ft}}^2/{\text{s}}^2}\right)\\ =79.83\text{Btu}/\text{lbm}\end{array}$

Substitute $0.24\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $\text{1118}\text{.3}\text{R}$ for $T_3$, and $\text{2400}\text{R}$ for $T_4$ in Equation (XII).

$\begin{array}{c}{q}_{\text{in}}=\left(0.24\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(2400\text{R}-1118.3\text{R}\right)\\ =\text{307}\text{.6}\text{Btu}/\text{lbm}\end{array}$

Substitute $\text{307}\text{.6}\text{Btu}/\text{lbm}$ for $q_{\text{in}}$, and $79.8\text{Btu}/\text{lbm}$ for $w_p$ in Equation (XIII).

$\begin{array}{c}{\eta }_p=\frac{79.8\text{Btu}/\text{lbm}}{\text{307}\text{.6}\text{Btu}/\text{lbm}}\\ =0.259\times 100\%\\ =25.9\%\end{array}$

Thus, the propulsive efficiency of the turbojet engine is $25.9\%$.