Chapter 9.12, Problem 134P

A turbojet is flying with a velocity of 900 ft/s at an altitude of 20,000 ft, where the ambient conditions are 7 psia and 10°F. The pressure ratio across the compressor is 13, and the temperature at the turbine inlet is 2400 R. Assuming ideal operation for all components and constant specific heats for air at room temperature, determine (a) the pressure at the turbine exit, (b) the velocity of the exhaust gases, and (c) the propulsive efficiency.

9–134E Repeat Prob. 9–133E accounting for the variation of specific heats with temperature.

To determine

The pressure at the turbine exit.

Answer to Problem 134P

The pressure at the turbine exit is $56.6\text{psia}$.

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 pressure and relative pressure relation for the process 1-2.

$P_2=P_1\left(\frac{P_{r_2}}{P_{r_1}}\right)$ (II)

Here, the specific heat ratio of air is k, pressure at state 1 is $P_1$, pressure at state 2 is $P_2$, relative pressure at state 1 is $P_{r_1}$ and relative pressure at state 2 is $P_{r_2}$.

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 pressure and relative pressure relation for the process 2-3.

$P_{r_3}=P_{r_2}\left(\frac{P_3}{P_2}\right)$ (IV)

Here, pressure at state 3 is $P_3$ and relative pressure at state 3 is $P_{r_3}$.

Turbine (For process 4-5)

Write the temperature relation for the compressor and turbine.

$\begin{array}{l}{w}_{comp,in}=w_{turb,out}\\ h_3-h_2=h_4-h_5\end{array}$ (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 $w_{comp,in}$ and work output from turbine is $w_{turb,out}$.

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

$P_5=P_4\left(\frac{P_{r_5}}{P_{r_4}}\right)$ (VI)

Here, pressure at state 5 is $P_5$, pressure at state 4 is $P_4$, relative pressure at state 5 is $P_{r_5}$ and relative pressure at state 4 is $P_{r_4}$.

Conclusion:

From Table A-17E, “Ideal-gas properties of air”, obtain the following properties at the temperature of $470\text{R}$.

$\begin{array}{l}{h}_1=112.20\text{Btu}/\text{lbm}\\ P_{r_1}=0.8548\end{array}$

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).

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

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

Here, inlet velocity is $V_1$ or $V_{inlet}$ , velocity at state 2 is $V_2$ , enthalpy at state 2 is $h_2$ and enthalpy at state 1 is $h_1$.

From Table A-17E, “Ideal-gas properties of air”, obtain the following properties at the temperature of $2400\text{R}$.

$\begin{array}{l}{h}_4=617.22\text{Btu}/\text{lbm}\\ P_{r_4}=367.6\end{array}$

Substitute 0 for $V_2$, $112.20\text{Btu}/\text{lbm}$ for $h_1$ and $900\text{ft}/\text{s}$ for $V_1$, in Equation (VII).

$\begin{array}{c}0=h_2-112.20\text{Btu}/\text{lbm}+\frac{\left(0\right)-{\left(900\text{ft}/\text{s}\right)}^2}{2}\\ h_2=\left(112.20\text{Btu}/\text{lbm}\right)+\frac{{\left(900\text{ft}/\text{s}\right)}^2}{2}\left(\frac{1\text{Btu}/\text{lbm}}{25,037{\text{ft}}^2/{\text{s}}^2}\right)\\ h_2=128.37\text{Btu}/\text{lbm}\end{array}$

Substitute $\text{7}\text{psia}$ for $P_1$, 0.8548 for $P_{r_1}$, and 1.3698 for $P_{r_2}$ in Equation (II).

$\begin{array}{c}{P}_2=\left(7\text{psia}\right)\left(\frac{1.3698}{0.8548}\right)\\ =11.22\text{}\text{psia}\end{array}$

Substitute 13 for $r_p$, and $11.22\text{}\text{psia}$ for $P_2$ in Equation (III).

$\begin{array}{l}{P}_3=P_4=\left(13\right)\left(11.22\text{}\text{psia}\right)\\ =145.8\text{psia}\end{array}$

Substitute $\text{145}\text{.8}\text{psia}$ for $P_3$, $\text{11}\text{.22}\text{psia}$ for $P_2$, and 1.3698 for $P_{r_2}$ in Equation (IV).

$\begin{array}{c}{P}_{r_3}=\left(1.3698\right)\left(\frac{145.8\text{psia}}{11.22\text{psia}}\right)\\ =17.8\end{array}$

From the Table A-17, “Ideal-gas properties of air” obtain the values of enthalpy on $\left(h_3\right)$ at the relative pressure $\left(P_{r_3}\right)$ of $17.8$ as $267.56\text{Btu}/\text{lbm}$.

Substitute $617.22\text{Btu}/\text{lbm}$ for $h_4$, $267.56\text{Btu}/\text{lbm}$ for $h_3$, and $128.48\text{Btu}/\text{lbm}$ for $h_2$.

in Equation (V).

$\begin{array}{c}{h}_5=617.22\text{Btu}/\text{lbm}-267.56\text{Btu}/\text{lbm}+128.48\text{Btu}/\text{lbm}\\ \text{=478}\text{.14}\text{Btu}/\text{lbm}\end{array}$

From the Table A-17, “Ideal-gas properties of air” obtain the values of relative pressure $\left(P_{r_4}\right)$ enthalpy $\left(h_4\right)$ of $478.14\text{Btu}/\text{lbm}$ as $142.7$.

Substitute $\text{145}\text{.8}\text{psia}$ for $P_4$, 142.7 for $P_{r_5}$, and 367.6 for $P_{r_4}$ in Equation (VI).

$\begin{array}{c}{P}_5=\left(145.8\text{psia}\right)\left(\frac{142.7}{367.6}\right)\\ =56.6\text{}\text{psia}\end{array}$

Thus, the pressure at the turbine exit is $56.6\text{psia}$.

To determine

The exit velocity of the exhaust gases.

Answer to Problem 134P

The exit velocity of the exhaust gases is $3252\text{ft}/\text{s}$.

Explanation of Solution

Nozzle (For process 5-6)

Write the pressure and relative pressure relation for the process 5-6.

$P_{r_6}=P_{r_5}\left(\frac{P_6}{P_5}\right)$ (VIII)

Here, pressure at state 6 is $P_6$ and relative pressure at state 6 is $P_{r_6}$.

Write the energy balance equation for the nozzle.

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

Conclusion:

Substitute $\text{7}\text{psia}$ for $P_6$, $\text{56}\text{.6}\text{psia}$ for $P_5$, and $\text{142}\text{.7}$ for $P_{r_5}$ in Equation(VIII).

$\begin{array}{c}{P}_{r_6}=\left(142.7\right)\left(\frac{7\text{psia}}{56.6\text{psia}}\right)\\ =17.65\end{array}$

From the Table A-17, “Ideal-gas properties of air” obtain the values of enthalpy on $\left(h_6\right)$ at the relative pressure $\left(P_{r_6}\right)$ of $17.66$ as $266.93\text{Btu}/\text{lbm}$.

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).

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

Here, velocity at stat 5 is $V_5$, exit velocity is $V_6$ or $V_{exit}$ and enthalpy at state 6 is $h_6$.

Since, $V_5=V_2$

Substitute $0$ for $V_5$, $478.14\text{Btu}/\text{lbm}$ for $h_5$ and $266.93\text{Btu}/\text{lbm}$ for $h_6$, in Equation (X).

$0=\left(478.14\text{Btu}/\text{lbm}-266.93\text{Btu}/\text{lbm}\right)-\frac{V_6^2-0}{2}$

$\begin{array}{l}{V}_6=\sqrt{\left(2\right)\left(478.14\text{Btu}/\text{lbm}-266.93\text{Btu}/\text{lbm}\right)\left(\frac{25,037{\text{ft}}^2/{\text{s}}^2}{1\text{Btu}/\text{lbm}}\right)}\\ V_6=V_{exit}=3252\text{ft}/\text{s}\end{array}$

Thus, the exit velocity of the exhaust gases is $3252\text{ft}/\text{s}$.

To determine

The propulsive efficiency of the turbojet engine.

Answer to Problem 134P

The propulsive efficiency of the turbojet engine is $24.2\%$.

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 $\left(q_{\text{in}}\right)$.

$q_{\text{in}}=h_4-h_3$ (XII).

Here, 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 $3252\text{ft}/\text{s}$ 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).

$\begin{array}{c}{w}_p=\left(3252\text{ft}/\text{s}-900\text{ft}/\text{s}\right)\times \left(900\text{ft}/\text{s}\right)\\ =\left(2352\text{ft}/\text{s}\right)\times \left(900\text{ft}/\text{s}\right)\\ =2116800{\text{ft}}^2/{\text{s}}^2\left(\frac{1\text{Btu}/\text{lbm}}{25,037{\text{ft}}^2/{\text{s}}^2}\right)\\ =84.546\text{Btu}/\text{lbm}\end{array}$

Substitute $\text{267}\text{.56}\text{Btu}/\text{lbm}$ for $h_3$, and $\text{617}\text{.22}\text{Btu}/\text{lbm}$ for $h_4$ in Equation (XII).

$\begin{array}{c}{q}_{\text{in}}=\text{617}\text{.22}\text{Btu}/\text{lbm}-\text{267}\text{.56}\text{Btu}/\text{lbm}\\ =349.66\text{Btu}/\text{lbm}\end{array}$

Substitute $\text{349}\text{.66}\text{Btu}/\text{lbm}$ for $q_{\text{in}}$, and $84.546\text{Btu}/\text{lbm}$ for $w_p$ in Equation (XII).

$\begin{array}{c}{\eta }_p=\frac{84.546\text{Btu}/\text{lbm}}{\text{349}\text{.66}\text{Btu}/\text{lbm}}\\ =0.2417\times 100\%\\ =24.17\%\\ =24.2\%\end{array}$

Thus, the propulsive efficiency of the turbojet engine is $24.2\%$.