Chapter 17.7, Problem 128RP

An aircraft flies with a Mach number Ma₁ = 0.9 at an altitude of 7000 m where the pressure is 41.1 kPa and the temperature is 242.7 K. The diffuser at the engine inlet has an exit Mach number of Ma₂ = 0.3. For a mass flow rate of 38 kg/s, determine the static pressure rise across the diffuser and the exit area.

To determine

The static pressure rise across the diffuser and the exit area.

Answer to Problem 128RP

The pressure rise across the diffuser in the aircraft is $24.2\text{kPa}$.

The exit area of the diffuser is $0.463{\text{m}}^2$.

Explanation of Solution

Write the formula for velocity of sound at the inlet conditions.

$c_1=\sqrt{kR{T}_1}$ (I)

Here, speed of sound at the inlet condition is $c_1$, gas constant of air is R, actual temperature of air at the inlet is $T_1$ and ratio of specific heats is k.

Write formula for the velocity of air at inlet.

$V_1={\text{Ma}}_1{c}_1$ (II)

Here, velocity of air after the normal shock is $V_2$.

Write the formula stagnation temperature of air at the inlet.

$T_{01}=T_1+\frac{V_1^2}{2c_p}$ (III)

Here, actual (static) temperature of air at the inlet is $T_1$, specific heat at constant pressure is $c_p$ and stagnation temperature of air at inlet is $T_{01}$.

Consider the flow is isentropic.

Write the formula for stagnation pressure of air at inlet (isentropic condition).

$P_{01}=P_1{\left(\frac{T_{01}}{T_1}\right)}^{k/\left(k-1\right)}$ (IV)

Here, the static pressure of air at inlet is $P_1$, and stagnation pressure of air at inlet is $P_{01}$.

Consider the diffuser is adiabatic and the inlet and exit enthalpies are equal.

$\begin{array}{l}{h}_{01}=h_{02}\\ c_p{T}_{01}=c_p{T}_{02}\\ T_{01}=T_{02}\end{array}$

When the stagnation temperatures are equal, the stagnation pressure are also equal.

$P_{01}=P_{02}$

Here, the static pressure of air at inlet is $P_1$, and stagnation pressure of air at exit is $P_{02}$.

Write the formula for velocity of sound at the exit conditions.

$c_2=\sqrt{kR{T}_2}$ (V)

Here, speed of sound at the exit condition is $c_2$, gas constant of air is R.

Write formula for the velocity of air at exit.

$\begin{array}{c}{V}_2={\text{Ma}}_2{c}_2\\ ={\text{Ma}}_2\sqrt{kR{T}_2}\end{array}$ (VI)

Here, the exit Mach number is ${\text{Ma}}_2$.

Write the formula stagnation temperature of air at the exit.

$T_{02}=T_2+\frac{V_2^2}{2c_p}$ (VII)

Here, actual (static) temperature of air at the inlet is $T_1$, specific heat at constant pressure is $c_p$ and stagnation temperature of air at inlet is $T_{01}$.

Write the formula for static pressure of air at exit (isentropic condition).

$P_2=P_{02}{\left(\frac{T_2}{T_{02}}\right)}^{k/\left(k-1\right)}$ (VIII)

Here, the static pressure of air at inlet is $P_1$, and stagnation pressure of air at exit is $P_{02}$.

Write the formula for static pressure difference of air.

$\Delta P=P_2-P_1$ (IX)

Write the formula for mass flow rate of air at exit condition.

$\begin{array}{c}\dot{m}=\frac{A_2{V}_2}{v_2}\\ =\frac{A_2{V}_2{P}_2}{R{T}_2}\end{array}$ (X)

Here, the exit cross sectional area is $A_2$ and exit specific volume is $v_2$.

Rearrange the Equation (X) to obtain $A_2$.

$A_2=\frac{\dot{m}R{T}_2}{V_2{P}_2}$ (XI)

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant $\left(R\right)$ of air is $0.287\text{kJ/kg}\cdot \text{K}$.

Refer Table A-2, “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure $\left(c_p\right)$ of air is $1.005\text{kJ/kg}\cdot \text{K}$.

The specific heat ratio $\left(k\right)$ of air is $1.4$.

Conclusion:

Substitute $0.287\text{kJ/kg}\cdot \text{K}$ for R, 1.4 for k, and $242.7\text{K}$ for $T_1$ in Equation (I).

$\begin{array}{c}{c}_1=\sqrt{1.4\times 0.287\text{}\text{kJ}/\text{kg}\cdot \text{K}\times 242.7\text{K}}\\ =\sqrt{97.51686\text{kJ/kg}\times \frac{1000{\text{m}}^2{\text{/s}}^2}{1\text{kJ/kg}}}\\ =312.2769\text{m}/\text{s}\end{array}$

Substitute $312.2769\text{m}/\text{s}$ for $c_1$, and 0.9 for ${\text{Ma}}_1$ in Equation (II).

$\begin{array}{c}{V}_1=0.9\times 312.2769\text{m}/\text{s}\\ =281.0492\text{m}/\text{s}\\ \simeq 281\text{m}/\text{s}\end{array}$

Substitute $242.7\text{K}$ for $T_1$, $281\text{m}/\text{s}$ for $V_1$ , and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in Equation (III).

$\begin{array}{c}{T}_{01}=242.7\text{K}+\frac{{\left(281\text{m}/\text{s}\right)}^2}{2\times 1.005\text{kJ}/\text{kg}\cdot \text{K}}\\ =242.7\text{K}+\frac{78961{\text{m}}^2{\text{/s}}^2\times \frac{1\text{kJ/kg}}{1000{\text{m}}^2{\text{/s}}^2}}{2.01\text{kJ}/\text{kg}\cdot \text{K}}\\ =242.7\text{K+39}\text{.28}\text{K}\\ =282\text{K}\end{array}$

Substitute $41.1\text{kPa}$ for $P_1$, $282\text{K}$ for $T_{01}$, 1.4 for k and $242.7\text{K}$ for $T_1$ in Equation (IV).

$\begin{array}{c}{P}_{01}=41.1\text{kPa}{\left(\frac{282\text{K}}{242.7\text{K}}\right)}^{1.4/\left(1.4-1\right)}\\ =41.1\text{kPa}{\left(1.1619\right)}^{3.5}\\ =41.1\text{kPa}\left(1.6908\right)\\ =69.5\text{kPa}\end{array}$

Here,

$\begin{array}{l}{T}_{01}=T_{02}=282\text{K}\\ P_{01}=P_{02}=69.5\text{kPa}\end{array}$

Substitute 0.3 for ${\text{Ma}}_2$, $0.287\text{}\text{kJ}/\text{kg}$ for R, and 1.4 for $k$ in Equation (VI).

$\begin{array}{c}{V}_2=0.3\times \sqrt{1.4\times 0.287\text{}\text{kJ}/\text{kg}\cdot \text{K}\times T_2}\\ =0.3\times \sqrt{0.4018\text{kJ/kg}\cdot \text{K}\times \frac{1000{\text{m}}^2{\text{/s}}^2}{1\text{kJ/kg}}\times T_2}\\ =0.3\times 20.04\sqrt{T_2}\\ =6.01\sqrt{T_2}\text{m}/\text{s}\end{array}$ (XII)

Substitute 282 K for $T_{02}$, $6.01\sqrt{T_2}\text{m}/\text{s}$ for $V_2$, and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in

Equation (VII).

$\begin{array}{l}282\text{K}=T_2+\frac{{\left(6.01\sqrt{T_2}\text{m}/\text{s}\right)}^2}{2\times 1.005\text{kJ}/\text{kg}\cdot \text{K}}\\ 282\text{K}=T_2+\frac{36.12T_2{{\text{m}}^2/\text{s}}^2\times \frac{1\text{kJ/kg}}{1000{\text{m}}^2{\text{/s}}^{\text{2}}}}{2.01\text{kJ}/\text{kg}\cdot \text{K}}\\ 282\text{K}=T_2+0.01797T_2\\ 282=1.01797T_2\end{array}$

$\begin{array}{c}{T}_2=\frac{282}{1.01797}\\ =277.0219\text{K}\\ \simeq \text{277}\text{K}\end{array}$

Substitute 1.4 for k, $282\text{K}$ for $T_{02}$, $\text{277}\text{K}$ for $T_2$ and $69.5\text{kPa}$ for $P_{02}$ in

Equation (VIII).

$\begin{array}{c}{P}_2=69.5\text{kPa}{\left(\frac{277\text{K}}{282\text{K}}\right)}^{1.4/\left(1.4-1\right)}\\ =69.5\text{kPa}{\left(0.9823\right)}^{3.5}\\ =69.5\text{kPa}\left(0.9393\right)\\ =65.28\text{kPa}\end{array}$

Substitute $65.28\text{kPa}$ for $P_2$, and $41.1\text{kPa}$ for $P_1$ in Equation (IX).

$\begin{array}{c}\Delta P=65.28\text{kPa}-41.1\text{kPa}\\ =24.2\text{kPa}\end{array}$

Thus, the pressure rise across the diffuser in the aircraft is $24.2\text{kPa}$.

Substitute $\text{277}\text{K}$ for $T_2$ in Equation (XII).

$\begin{array}{c}{V}_2=6.01\sqrt{\text{277}\text{K}}\text{m}/\text{s}\\ =100.0263\text{m}/\text{s}\\ \simeq 100\text{m}/\text{s}\end{array}$

Substitute $38\text{kg}/\text{s}$ for $\dot{m}$, $65.28\text{kPa}$ for $P_2$, $100\text{m}/\text{s}$ for $V_2$, $0.287\text{}\text{kJ}/\text{kg}\cdot \text{K}$ for R, and $\text{277}\text{K}$ for $T_2$ in Equation (XI).

$\begin{array}{c}{A}_2=\frac{\left(38\text{kg}/\text{s}\right)\left(0.287\text{}\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\text{277}\text{K}\right)}{\left(100\text{m}/\text{s}\right)\left(65.28\text{kPa}\right)}\\ =\frac{\left(38\text{kg}/\text{s}\right)\left(0.287\text{}\text{kJ}/\text{kg}\cdot \text{K}\times \frac{1\text{kPa}\cdot {\text{m}}^3}{1\text{kJ}}\right)\left(\text{277}\text{K}\right)}{6528\text{kPa}\cdot \text{m}/\text{s}}\\ =\frac{3020.962\text{kPa}\cdot {\text{m}}^3\text{/s}}{6528\text{kPa}\cdot \text{m}/\text{s}}\\ =0.4628{\text{m}}^2\end{array}$

$\simeq 0.463{\text{m}}^2$

Thus, the exit area of the diffuser is $0.463{\text{m}}^2$.