Chapter 9.12, Problem 134P

a)

To determine

The velocity of the exhaust gases

Answer to Problem 134P

The velocity of the exhaust gases is $677.7\text{m}/\text{s}$.

Explanation of Solution

Draw the $T-s$ diagram for turbojet 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 change in the energy of the system is $\Delta {\dot{E}}_{\text{system}}$.

Write the expression to calculate 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 expression to calculate 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 expression to calculate the temperature and pressure relation for the process 2-3s.

$\begin{array}{l}{T}_{3s}=T_2{\left(\frac{P_3}{P_2}\right)}^{\left(k-1\right)/k}\\ T_{3s}=T_2\left(r_p^{\left(k-1\right)/k}\right)\end{array}$ (IV)

Write the expression for the efficiency of the compressor in the turbojet engine $\left({\eta }_C\right)$.

$\begin{array}{c}{\eta }_C=\frac{h_{3s}-h_2}{h_3-h_2}\\ {\eta }_C=\frac{c_p\left(T_{3s}-T_2\right)}{c_p\left(T_3-T_2\right)}\\ T_3=T_2+\left(T_{3s}-T_2\right)/{\eta }_C\end{array}$ (V)

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

Turbine (For process 4-5)

Write the expression for the temperature relation for the compressor and turbine.

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

Write the expression for the efficiency of the turbine in the turbojet engine $\left({\eta }_T\right)$.

$\begin{array}{c}{\eta }_T=\frac{h_4-h_5}{h_4-h_{5s}}\\ {\eta }_T=\frac{c_p\left(T_4-T_5\right)}{c_p\left(T_4-T_{5s}\right)}\\ T_{5s}=T_4-\left(T_4-T_5\right)/{\eta }_T\end{array}$ (VII)

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

$P_5=P_4{\left(\frac{T_{5s}}{T_4}\right)}^{k/\left(k-1\right)}$ (VIII)

Nozzle (For process 5-6)

Write the expression to calculate the temperature and pressure relation for the isentropic process 4-5.

$T_6=T_5{\left(\frac{P_6}{P_5}\right)}^{\left(k-1\right)/k}$ (IX)

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

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

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).

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

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

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

Substitute 0 for $V_2$, 241 K for $T_1$, $280\text{m}/\text{s}$ for $V_1$, and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in

Equation (XI).

$\begin{array}{c}{T}_2=241\text{}\text{K}+\frac{{\left(280\text{m}/\text{s}\right)}^2}{\left(2\right)\left(\text{1}\text{.005kJ}/\text{kg}\cdot \text{K}\right)}\\ =241\text{}\text{K}+\frac{{\left(280\text{m}/\text{s}\right)}^2}{\left(2\right)\left(\text{1}\text{.005kJ}/\text{kg}\cdot \text{K}\right)}\left(\frac{1\text{kJ}/\text{kg}}{1,000{\text{m}}^2/{\text{s}}^2}\right)\\ =280.0\text{K}\end{array}$

Substitute 32 kPa for $P_1$, 280.0 K for $T_2$, 241 for $T_1$, and 1.4 for k to find $P_2$ in Equation (II).

$\begin{array}{c}{P}_2=\left(32\text{kPa}\right){\left(\frac{280.0\text{K}}{241\text{K}}\right)}^{1.4/1.4-1}\\ =54.10\text{kPa}\end{array}$

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).

$\begin{array}{c}{T}_{3s}=\left(280.0\text{K}\right){\left(12\right)}^{1.4-1/1.4}\\ =569.5\text{K}\end{array}$

Substitute 0.80 for ${\eta }_C$, 280.0 K for $T_2$, and 569.5 K for $T_{3s}$ to find $T_3$ in Equation (V).

$\begin{array}{c}{T}_3=280.0\text{K}+\left(569.5-280.0\right)\text{K}/0.80\\ =641.9\text{K}\end{array}$

Substitute 1100 K for $T_4$, 641.9 K for $T_3$, and 280.0 K for $T_2$ in Equation (VI).

$\begin{array}{c}{T}_5=\left(1100-641.9+280.0\right)\text{K}\\ =\text{738}\text{.1}\text{}\text{K}\end{array}$

Substitute 0.85 for ${\eta }_T$, 1100 K for $T_4$, and 738.1 K for $T_5$ to find $T_{5s}$ in Equation (VII).

$\begin{array}{c}{T}_{5s}=1100\text{K}-\left(1100-738.1\right)\text{K}/0.85\\ =674.2\text{K}\end{array}$

Substitute 674.2 K for $T_{5s}$, 1100 K for $T_4$, 649.2 kPa for $P_4$, and 1.4 for k in Equation (VIII).

$\begin{array}{c}{P}_5=\left(649.2\text{kPa}\right){\left(\frac{674.3\text{}\text{K}}{1100\text{}\text{K}}\right)}^{1.4/1.4-1}\\ =117\text{kPa}\end{array}$

Substitute 738.1 K for $T_5$, 1.4 for k, 32 kPa for $P_6$, and 117.0 kPa for $P_5$ in Equation (IX).

$\begin{array}{c}{T}_6=\left(738.1\text{K}\right){\left(\frac{32\text{kPa}}{117.0\text{kPa}}\right)}^{1.4-1/1.4}\\ =509.6\text{K}\end{array}$

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 (X).

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

$\begin{array}{l}0=h_6-h_5+\frac{V_6^2-V_5^2}{2}\\ 0=c_p\left(T_6-T_5\right)-\frac{V_6^2-V_5^2}{2}\\ V_6^2=2c_p\left(T_5-T_6\right)\\ V_6=\sqrt{2c_p\left(T_5-T_6\right)}\end{array}$ (XII)

Substitute 738.1 K for $T_5$, 509.6 K for $T_6$, and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in Equation (XII).

$\begin{array}{c}{V}_6=\sqrt{\left(2\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(738.1-509.6\right)\text{K}}\\ =\sqrt{\left(2\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(738.1-509.6\right)\text{K}\left(\frac{1,000{\text{m}}^2/{\text{s}}^2}{1\text{kg}/\text{K}}\right)}\\ =677.7\text{m}/\text{s}\end{array}$

Hence, the velocity of the exhaust gases is $677.7\text{m}/\text{s}$.

b)

To determine

The propulsive power produced by the turbojet engine

Answer to Problem 134P

The propulsive power produced by the turbojet engine is $5568\text{kW}$.

Explanation of Solution

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

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

Here, the mass flow rate of air through the engine is $\dot{m}$, 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 $V_{\text{exit}}$.

Conclusion:

Substitute $50\text{kg}/\text{s}$ for $\dot{m}$, $677.7\text{m}/\text{s}$ for $V_{\text{exit}}$, $280\text{m}/\text{s}$ for $V_{\text{inlet}}$, and $280\text{m}/\text{s}$ for $V_{\text{aircraft}}$ in Equation (XIII).

$\begin{array}{c}{\dot{W}}_p=\left(50\text{kg}/\text{s}\right)\left[\left(677.7-280\right)\text{m}/\text{s}\right]\left(280\text{m}/\text{s}\right)\\ =\left(50\text{kg}/\text{s}\right)\left[\left(677.7-280\right)\text{m}/\text{s}\right]\left(280\text{m}/\text{s}\right)\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)\\ =5568\text{kW}\end{array}$

Hence, the propulsive power produced by the turbojet engine is $5568\text{kW}$.

c)

To determine

The rate of fuel consumption.

Answer to Problem 134P

The rate of fuel consumption is $0.5391\text{kg}/\text{s}$.

Explanation of Solution

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

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

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

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

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 $c_p$, 641.9 K for $T_3$, and 1100 K for $T_4$ in Equation (XIV).

$\begin{array}{c}{\dot{Q}}_{\text{in}}=\left(50\text{kg}/\text{s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(1100-641.9\right)\text{K}\\ =\text{23020}\text{kJ}/\text{s}\end{array}$

Substitute $42700\text{kJ}/\text{kg}$ for $\text{HV}$, and $23020\text{kJ}/\text{s}$ for ${\dot{Q}}_{\text{in}}$ in Equation (XV).

$\begin{array}{c}{\dot{m}}_{\text{fuel}}=\frac{23020\text{kJ}/\text{s}}{42700\text{kJ}/\text{kg}}\\ =0.5391\text{kg}/\text{s}\end{array}$

Hence, the rate of fuel consumption is $0.5391\text{kg}/\text{s}$.