Chapter 9.12, Problem 132P

A pure jet engine propels an aircraft at 240 m/s through air at 45 kPa and −13°C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557°C. Determine the velocity at the exit of this engine’s nozzle and the thrust produced. Assume ideal operation for all components and constant specific heats at room temperature.

To determine

The velocity at the exit of this engine’s nozzle and the thrust produced.

Answer to Problem 132P

The velocity at the exit of this engine’s nozzle is $564.8\text{m/s}$.

The thrust produced by the engine is $94520\text{N}$.

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 $V_1=240\text{m}/\text{s}$, the air will leave the diffuser with a negligible velocity $\left(V_2\cong 0\right)$.

Diffuser:

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:

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:

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 $T_5$.

Nozzle:

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)

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

Write the energy balance equation for the nozzle.

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

Write the expression to calculate the specific volume at state 1 $v_1$.

$v_1=\frac{R{T}_1}{P_1}$ (VIII)

Here, gas constant is $R$ , temperature at state 1 is $T_1$ and pressure at state 1 is $P_1$.

Write the expression to calculate the mass flow rate of propeller ${\dot{m}}_p$.

$\begin{array}{c}{\dot{m}}_p=\frac{A_1{V}_{inlet}}{v_1}\\ =\frac{\pi D^2}{4}\frac{V_{inlet}}{v_1}\end{array}$ (IX)

Here, propeller diameter is $D$, area is $A_1$ and inlet velocity is $V_{inlet}$.

Write the expression to calculate the thrust force generated by the propeller $\left(F\right)$.

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

Here, the thrust force produced by engine is $F$, mass flow rate of air is $\dot{m}$, inlet velocity of air is $V_{\text{inlet}}$, and outlet velocity of air is $V_{\text{exit}}$.

Conclusion.

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

$\begin{array}{l}k=1.4\\ c_p=1.005\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

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 $\Delta {\dot{E}}_{\text{system}}$ Equation (I).

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

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

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

Substitute 0 for $V_2$, $-13°\text{C}$ for $T_1$, $240\text{m}/\text{s}$ for $V_1$, and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in Equation (XI).

$\begin{array}{l}0=1.005\text{kJ}/\text{kg}\cdot \text{K}\left(T_2-\left(-13°\text{C}\right)\right)-\frac{V_2^2-{\left(240\text{m}/\text{s}\right)}^2}{2}\\ T_2=\left(-13+273\right)\text{}\text{K}+\frac{{\left(240\text{m}/\text{s}\right)}^2}{\left(2\right)\left(\text{1}\text{.005kJ}/\text{kg}\cdot \text{K}\right)}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)\\ T_2=288.7\text{K}\end{array}$

Substitute $45\text{kPa}$ for $P_1$, $288.7\text{K}$ for $T_2$, $-13°\text{C}$ for $T_1$, and 1.4 for k to find $P_2$ in Equation (II).

$\begin{array}{c}{P}_2=\left(45\text{kPa}\right){\left(\frac{288.7\text{K}}{-13°\text{C}}\right)}^{1.4/1.4-1}\\ =\left(45\text{kPa}\right){\left(\frac{288.7\text{K}}{\left(-13+273\right)\text{}\text{K}}\right)}^{1.4/1.4-1}\\ =64.88\text{kPa}\end{array}$

Substitute 13 for $r_p$, and $64.88\text{kPa}$ for $P_2$ in Equation (III).

$\begin{array}{l}{P}_3=P_4=\left(13\right)\left(64.88\text{kPa}\right)\\ =843.5\text{kPa}\end{array}$

Substitute $288.7\text{K}$ for $T_2$, 1.4 for k, and 13 for $r_p$ in Equation (IV).

$\begin{array}{c}{T}_3=\left(288.7\text{K}\right){\left(13\right)}^{1.4-1/1.4}\\ =600.7\text{K}\end{array}$

Substitute $557°\text{C}$ for $T_4$, $600.7\text{K}$ for $T_3$, and $288.7\text{K}$ for $T_2$ in Equation (V).

$\begin{array}{c}{T}_5=\left(557°\text{C}-600.7\text{K}+288.7\text{K}\right)\\ =\left(557+273\right)\text{K}-600.7\text{K}+288.7\text{K}\\ =\text{518}\text{K}\end{array}$

Substitute $557°\text{C}$ for $T_4$, 1.4 for k, $45\text{kPa}$ for $P_6$, and $843.5\text{kPa}$ for $P_4$ in Equation (VI).

$\begin{array}{c}{T}_6=\left(557°\text{C}\right){\left(\frac{45\text{kPa}}{843.5\text{kPa}}\right)}^{1.4-1/1.4}\\ =\left(557+273\right)\text{K}{\left(\frac{45\text{kPa}}{843.5\text{kPa}}\right)}^{1.4-1/1.4}\\ =359.3\text{K}\end{array}$

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 $\Delta {\dot{E}}_{\text{system}}$ Equation (VII).

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

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 $0$ for $V_5$, $\text{518}\text{K}$ for $T_5$, $359.3\text{K}$ for $T_6$, and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ to find $V_6$ in Equation (XII).

$\begin{array}{l}0=1.005\text{kJ}/\text{kg}\cdot \text{K}\left(359.3\text{K}-\text{518}\text{K}\right)-\frac{V_6^2-0}{2}\\ V_6=\sqrt{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\text{518}\text{K}-359.3\text{K}\right)\left(\frac{1000{\text{m}}^2/{\text{s}}^2}{1\text{kJ}/\text{kg}}\right)}\\ V_6=564.8\text{m/s}\\ V_6=V_{exit}=564.8\text{m/s}\end{array}$

Thus, the velocity at the exit of this engine’s nozzle is $564.8\text{m/s}$.

Substitute $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$ , $-13°\text{C}$ for $T_1$ and $45\text{kPa}$ for $P_1$ in Equation (VIII).

$\begin{array}{c}{v}_1=\frac{0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\left(-13°\text{C}\right)}{45\text{kPa}}\\ =\frac{0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\left(-13+273\right)\text{K}}{8\text{psia}}\\ =1.658{\text{m}}^3/\text{kg}\end{array}$

Substitute $1.6{\text{m}}^3$ for $D$, $240\text{m}/\text{s}$ for $V_{inlet}$ and $1.658{\text{m}}^3/\text{kg}$ for $v_1$ in Equation (IX).

$\begin{array}{c}{\dot{m}}_p=\frac{\pi {\left(1.6{\text{m}}^3\right)}^2}{4}\frac{240\text{m}/\text{s}}{1.658{\text{m}}^3/\text{kg}}\\ =291\text{kg/s}\end{array}$

Substitute $291\text{kg/s}$ for ${\dot{m}}_p$, $240\text{m}/\text{s}$ for $V_{inlet}$ and $564.8\text{m/s}$ for $V_{exit}$ in Equation (X).

$\begin{array}{c}F=291\text{kg/s}\left(564.8\text{m/s}-240\text{m}/\text{s}\right)\\ =94520\text{kg}\cdot \text{m}/{\text{s}}^2\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\\ =94520\text{N}\end{array}$

Thus, the thrust produced by the engine is $94520\text{N}$.