Chapter 9.12, Problem 170RP

To determine

The specific impulse of the jet engine.

Answer to Problem 170RP

The specific impulse of the jet engine is $49\text{ft}/\text{s}$.

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 $V_1=1200\text{ft}/\text{s}$, 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 impulse of the jet engine.

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

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-1E, “Molar mass, gas constant, and critical-point properties”, obtain the

value of gas constant $\left(R\right)$ for air substance as $0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$.

From Table A-2Ea, “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 $\Delta {\dot{E}}_{\text{system}}$ Equation (I).

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

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

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

Substitute 0 for $V_2$, $30°\text{F}$ for $T_1$, $1200\text{ft}/\text{s}$ for $V_1$, and $0.24\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ in Equation (IX).

$\begin{array}{l}0=0.24\text{Btu}/\text{lbm}\cdot \text{R}\left(T_2-\left(30°\text{F}\right)\right)-\frac{V_2^2-{\left(1200\text{ft}/\text{s}\right)}^2}{2}\\ T_2=\left(30+460\right)\text{}\text{R}+\frac{{\left(1200\text{ft}/\text{s}\right)}^2}{\left(2\right)\left(0.24\text{Btu}/\text{lbm}\cdot \text{R}\right)}\left(\frac{1\text{Btu}/\text{lbm}}{25037{\text{ft}}^2/{\text{s}}^2}\right)\\ T_2=609.8\text{R}\end{array}$

Substitute 10psia for $P_1$, 609.8 R for $T_2$, $30°\text{F}$ for $T_1$, and 1.4 for k in Equation (II).

$\begin{array}{c}{P}_2=\left(10\text{psia}\right){\left(\frac{609.8\text{R}}{30°\text{F}}\right)}^{1.4/1.4-1}\\ =\left(10\text{psia}\right){\left(\frac{609.8\text{R}}{\left(30+460\right)\text{}\text{R}}\right)}^{1.4/1.4-1}\\ =21.5\text{}\text{psia}\end{array}$

Substitute 9 for $r_p$, and 21.5psia for $P_2$ in Equation (III).

$\begin{array}{l}{P}_3=P_4=\left(9\right)\left(21.5\right)\\ =193.5\text{psia}\end{array}$

Substitute 609.8 R for $T_2$, 1.4 for k, and 9 for $r_p$ in Equation (IV).

$\begin{array}{c}{T}_3=\left(609.8\right){\left(9\right)}^{1.4-1/1.4}\\ =1142.4\text{R}\end{array}$

Substitute $700°\text{F}$ for $T_4$, $\text{1142}\text{.4 R}$ for $T_3$, and $\text{609}\text{.8 R}$ for $T_2$ in Equation (V).

$\begin{array}{c}{T}_5=\left(\left(700°\text{F}\right)-1142.4\text{R}+609.8\text{R}\right)\\ =\left(700+460\right)\text{R}-1142.4\text{R}+609.8\text{R}\\ \text{=}\text{627}\text{.4}\text{}\text{R}\end{array}$

Substitute $700°\text{F}$ for $T_4$, 1.4 for k, $\text{10}\text{psia}$ for $P_6$, and $\text{193}\text{.5psia}$ for $P_4$ in Equation (VI).

$\begin{array}{c}{T}_6=\left(700°\text{F}\right){\left(\frac{10\text{psia}}{193.5\text{psia}}\right)}^{1.4-1/1.4}\\ =\left(700+460\right)\text{R}{\left(\frac{10\text{psia}}{193.5\text{psia}}\right)}^{1.4-1/1.4}\\ =497.5\text{R}\end{array}$

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 (VII).

$\begin{array}{c}{\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ h_5+\frac{V_5^2}{2}=h_6+\frac{V_6^2}{2}\\ 0=h_6-h_5+\frac{V_6^2-V_5^2}{2}\\ 0=c_p\left(T_6-T_5\right)-\frac{V_6^2-V_5^2}{2}\end{array}$ (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 $0$ for $V_5$, $\text{627}\text{.4}\text{}\text{R}$ for $T_5$, $497.5\text{R}$ for $T_6$, and $0.24\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ to find $V_6$ in Equation (X).

$\begin{array}{l}0=0.24\text{Btu}/\text{lbm}\cdot \text{R}\left(497.5\text{R}-\text{627}\text{.4}\text{}\text{R}\right)-\frac{V_6^2-0}{2}\\ V_6=\sqrt{2\left(0.24\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(\text{627}\text{.4}\text{}\text{R}-497.5\text{R}\right)\left(\frac{25037{\text{ft}}^2/{\text{s}}^2}{1\text{Btu}/\text{lbm}}\right)}\\ V_6=1249\text{ft}/\text{s}\\ V_6=V_{exit}=1249\text{ft}/\text{s}\end{array}$

Substitute $1200\text{ft}/\text{s}$ for $V_{inlet}$ and $1249\text{ft}/\text{s}$ for $V_{exit}$ in Equation (VIII).

$\begin{array}{c}\frac{F}{\dot{m}}=\left(1249\text{ft}/\text{s}-1200\text{ft}/\text{s}\right)\\ =49\text{ft}/\text{s}\end{array}$

Thus, the specific impulse of the jet engine is $49\text{ft}/\text{s}$.