Chapter 9.12, Problem 136P

A turbojet aircraft is flying with a velocity of 280 m/s at an altitude of 9150 m, where the ambient conditions are 32 kPa and −32°C. The pressure ratio across the compressor is 12, and the temperature at the turbine inlet is 1100 K. Air enters the compressor at a rate of 50 kg/s, and the jet fuel has a heating value of 42,700 kJ/kg. Assuming ideal operation for all components and constant specific heats for air at room temperature, determine (a) the velocity of the exhaust gases, (b) the propulsive power developed, and (c) the rate of fuel consumption.

To determine

The velocity of the exhaust gases.

Answer to Problem 136P

The velocity of the exhaust gases is $832.7\text{m}/\text{s}$.

Explanation of Solution

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

Consider, the pressure is $P_i$ , the specific volume is $v_i$, the temperature is $T_i$, the entropy is $s_i$, the enthalpy is $h_i$ corresponding to $i^{th}$ state.

Consider that the aircraft is stationary, and the velocity of air moving towards the aircraft is $V_1=280\text{m/s}$, 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 energy entering the system is ${\dot{E}}_{\text{in}}$, 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.

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 $r_p$.

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)

Turbine (For process 4-5)

Write the temperature relation for the compressor and turbine to calculate the temperature at state 5 $\left(T_5\right)$.

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

Nozzle (For process 5-6)

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

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

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

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

Conclusion:

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

$\begin{array}{l}{c}_p=1.005\text{kJ}/\text{kg}\cdot \text{K}\\ c_v=0.718\text{kJ}/\text{kg}\cdot \text{K}\\ R=0.287\text{kJ}/\text{kg}\cdot \text{K}\\ k=1.4\end{array}$

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).

$\begin{array}{l}{\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\end{array}$

$\begin{array}{l}0=h_2-h_1+\frac{V_2^2-V_1^2}{2}\\ 0=c_p\left(T_2-T_1\right)-\frac{V_2^2-V_1^2}{2}\\ T_2=T_1-\frac{V_2^2-V_1^2}{2c_p}\end{array}$ (VIII).

Here, the specific heat at constant pressure of air is $c_p$.

Substitute 0 for $V_2$, $-32°\text{C}$ for $T_1$, $280\text{m}/\text{s}$ for $V_1$, and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ to find $T_2$. Also convert $\text{kJ}/\text{kg}$ to ${\text{m}}^2/{\text{s}}^2$ using suitable conversion factor in Equation (VIII).

$\begin{array}{c}{T}_2=\left(-32°\text{C}\right)+\frac{{\left(280\text{m}/\text{s}\right)}^2}{\left(2\right)\left(\text{1}\text{.005kJ}/\text{kg}\cdot \text{K}\right)}\\ =\left(-32+273\right)\text{K}+\frac{{\left(280\text{m}/\text{s}\right)}^2}{\left(2\right)\left(\text{1}\text{.005kJ}/\text{kg}\cdot \text{K}\right)}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)\\ =280.0\text{K}\end{array}$

Substitute 32 kPa for $P_1$, 280.0 K for $T_2$, $-32°\text{C}$ for $T_1$, and 1.4 for k to find $P_2$ in Equation (II).

$\begin{array}{c}{P}_2=\left(32\text{kPa}\right){\left(\frac{280.0\text{K}}{-32°\text{C}}\right)}^{1.4/1.4-1}\\ =\left(32\text{kPa}\right){\left(\frac{280.0\text{K}}{\left(-32+273\right)\text{K}}\right)}^{1.4/1.4-1}\\ =54.10\text{kPa}\end{array}$

Substitute 12 for $r_p$, and 54.10 kPa for $P_2$ in Equation (III).

$\begin{array}{c}{P}_3=P_4\\ =\left(12\right)\left(54.10\text{kPa}\right)\\ =649.2\text{kPa}\end{array}$

Substitute 280.0 K for $T_2$, 1.4 for k, and 12 for $r_p$ in Equation (IV).

$\begin{array}{c}{T}_3=\left(280.0\text{K}\right){\left(12\right)}^{0.4/1.4}\\ =569.5\text{K}\end{array}$

Substitute 1,100 K for $T_4$, 569.5 K for $T_3$, and 280.0 K for $T_2$ in Equation (V).

$\begin{array}{c}{T}_5=\left(1,100-569.5+280.0\right)\text{K}\\ \text{=810}\text{.5}\text{}\text{K}\end{array}$

For process 5-6 (Nozzle)

Substitute 1100 K for $T_4$, 1.4 for k, 32 kPa for $P_6$, and 649.2 kPa for $P_4$ in Equation (VI).

$\begin{array}{c}{T}_6=\left(1100\text{}\text{K}\right){\left(\frac{32\text{kPa}}{649.2\text{kPa}}\right)}^{1.4-1/1.4}\\ =465.5\text{K}\end{array}$

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 (VII).

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

$\begin{array}{l}0=c_p\left(T_6-T_5\right)-\frac{V_6^2-V_5^2}{2}\\ V_6^2=2c_p\left(T_5-T_6\right)\\ V_6=\sqrt{2c_p\left(T_5-T_6\right)}\end{array}$ (IX)

Substitute 810.5 K for $T_5$, 465.5 K for $T_6$, and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ to find $V_6$ in Equation (IX).

$\begin{array}{c}{V}_6=\sqrt{\left(2\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(810.5-465.5\right)\text{K}}\\ =\sqrt{\left(2\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(810.5-465.5\right)\text{K}\left(\frac{1000{\text{m}}^2/{\text{s}}^2}{1\text{kg}/\text{K}}\right)}\\ =832.7\text{m}/\text{s}\end{array}$

Thus, the velocity of the exhaust gases is $832.7\text{m}/\text{s}$.

To determine

The propulsive power produced by the turbojet engine.

Answer to Problem 136P

The propulsive power produced by the turbojet engine is $7738\text{kW}$.

Explanation of Solution

Nozzle (For process 5-6)

Write the expression to calculate the propulsive power produced by the turbojet engine $\left({\dot{W}}_p\right)$.

${\dot{W}}_p=\dot{m}\left(V_{\text{exit}}-V_{\text{inlet}}\right)V_{\text{aircraft}}$ (X)

Here, the velocity of the aircraft is $V_{\text{aircraft}}$, the velocity of the inlet air is $V_{\text{inlet}}$, the exit velocity of the exhaust gases is $V_{\text{exit}}$ and the mass flow rate of air through the

 engine is $\dot{m}$.

Conclusion:

Substitute $50\text{kg}/\text{s}$ for $\dot{m}$, $832.7\text{m}/\text{s}$ for $V_{\text{exit}}$, $280\text{m}/\text{s}$ for $V_{\text{inlet}}$, and $280\text{m}/\text{s}$ for $V_{\text{aircraft}}$ to find ${\dot{W}}_p$ in Equation (X).

$\begin{array}{c}{\dot{W}}_p=\left(50\text{kg}/\text{s}\right)\left[\left(832.7-280\right)\text{m}/\text{s}\right]\left(280\text{m}/\text{s}\right)\\ =\left(50\text{kg}/\text{s}\right)\left[\left(832.7-280\right)\text{m}/\text{s}\right]\left(280\text{m}/\text{s}\right)\left(\frac{1\text{kJ}/\text{kg}}{1,000{\text{m}}^2/{\text{s}}^2}\right)\\ =7738\text{kW}\end{array}$

Thus, the propulsive power produced by the turbojet engine is $7738\text{kW}$.

To determine

The rate of fuel consumption.

Answer to Problem 136P

The rate of fuel consumption is $0.6243\text{kg}/\text{s}$.

Explanation of Solution

Write the expression to calculate the heating value of the fuel for the turbojet engine $\left({\dot{Q}}_{\text{in}}\right)$.

$\begin{array}{c}{\dot{Q}}_{\text{in}}=\dot{m}\left(h_4-h_3\right)\\ =\dot{m}{c}_p\left(T_4-T_3\right)\end{array}$ (XI)

Write the expression to calculate the mass flow rate of fuel for the turbojet engine $\left({\dot{m}}_{\text{fuel}}\right)$.

${\dot{m}}_{\text{fuel}}=\frac{{\dot{Q}}_{\text{in}}}{\text{HV}}$ (XII)

Here, the calorific value of the fuel is HV.

Conclusion.

Substitute $50\text{kg}/\text{s}$ for $\dot{m}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 569.5 K for $T_3$, and 1100 K for $T_4$ in Equation (XI).

$\begin{array}{c}{\dot{Q}}_{\text{in}}=\left(50\text{kg}/\text{s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(1100-569.5\right)\text{K}\\ \text{=26657}\text{kJ}/\text{s}\end{array}$

Substitute  $42700\text{kJ}/\text{kg}$ for $\text{HV}$, and $26657\text{kJ}/\text{s}$ for ${\dot{Q}}_{\text{in}}$ in Equation (XII).

$\begin{array}{c}{\dot{m}}_{\text{fuel}}=\frac{26657\text{kJ}/\text{s}}{42700\text{kJ}/\text{kg}}\\ =0.6243\text{kg}/\text{s}\end{array}$

Thus, the rate of fuel consumption is $0.6243\text{kg}/\text{s}$.