Chapter 17.7, Problem 105P

Air is heated as it flows subsonically through a duct. When the amount of heat transfer reaches 67 kJ/kg, the flow is observed to be choked, and the velocity and the static pressure are measured to be 680 m/s and 270 kPa. Disregarding frictional losses, determine the velocity, static temperature, and static pressure at the duct inlet.

To determine

The static temperature in the duct.

The static pressure in the duct.

The velocity in the duct.

Answer to Problem 105P

The static temperature in the duct is $\underset{\_}{1172\text{K}}$.

The static pressure in the duct is $\underset{\_}{351.3\text{kPa}}$.

The velocity in the duct is $\underset{\_}{\text{533}\text{.9}\text{m}/\text{s}}$.

Explanation of Solution

Determine the speed of sound at the exit.

$\begin{array}{l}{\text{Ma}}_2=\frac{V_2}{c_2}\\ c_2=\frac{V_2}{{\text{Ma}}_2}\end{array}$ (I)

The exit velocity of the air flow in the device is $V_2$ and the exit Mach number is ${\text{Ma}}_2$.

Determine the relation of ideal gas speed of sound at the exit.

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

Here, the specific heat ratio of air is $k$, the gas constant of the air is $R$, and the exit temperature of the air is $T_2$.

Determine the exit stagnation temperature of air.

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

Here, the exit static temperature of ideal gas is $T_2$, the specific heat of pressure for ideal gas is $c_p$, and the exit velocity of the ideal gas flow is $V_2$.

Determine the inlet stagnation temperature from energy equation.

$\begin{array}{l}q=c_p\left(T_{02}-T_{01}\right)\\ T_{01}=T_{02}-\frac{q}{c_p}\end{array}$ (IV)

Here, the heat transfer to the duct is $q$.

Determine the stagnation temperature ratio at the inlet.

${\text{Ratio}}_{\text{stagnation}}=\frac{T_{01}}{T_0^{\ast }}$ (V)

Here, the maximum value of stagnation temperature is $T_0^{\ast }$.

Determine the static temperature in the duct.

$\frac{T_2}{T_1}=\frac{T_2/T^{\ast }}{T_1/T^{\ast }}$ (VI)

Here, the ratio of Rayleigh flow for inlet temperature is $T_1/T^{\ast }$ and the ratio of Rayleigh flow for outlet temperature is $T_2/T^{\ast }$.

Determine the static pressure in the duct.

$\frac{P_2}{P_1}=\frac{P_2/P^{\ast }}{P_1/P^{\ast }}$ (VII)

Here, the ratio of Rayleigh flow for inlet pressure is $P_1/P^{\ast }$ and the ratio of Rayleigh flow for outlet pressure is $P_2/P^{\ast }$.

Determine the velocity in the duct.

$\frac{V_2}{V_1}=\frac{V_2/V^{\ast }}{V_1/V^{\ast }}$ (VIII)

Here, the ratio of Rayleigh flow for inlet velocity is $V_1/V^{\ast }$ and the ratio of Rayleigh flow for outlet velocity is $V_2/V^{\ast }$.

Conclusion:

From the Table A-2, “Ideal-gas specific heats of various common gases” to obtain value of universal gas constant, specific heat of pressure, and the specific heat ratio of air at $300\text{K}$ temperature as $0.287\text{kJ}/\text{kg}\cdot \text{K}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ and $1.400$.

Substitute 1 for ${\text{Ma}}_{\text{2}}$ and $680\text{m}/\text{s}$ for $V_2$ in Equation (I).

$\begin{array}{c}{c}_2=\frac{\left(680\text{m}/\text{s}\right)}{1}\\ =680\text{m}/\text{s}\end{array}$

Substitute 1.400 for k, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, and $680\text{m}/\text{s}$ for $c_2$ in Equation (II).

$\begin{array}{l}\left(680\text{m}/\text{s}\right)=\sqrt{\left(1.400\right)\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\times T_2}\\ \left(680\text{m}/\text{s}\right)=\sqrt{\left(1.400\right)\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(\frac{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{kJ}/\text{kg}}\right)\times T_2}\\ T_2=1151\text{K}\end{array}$

Substitute $1151\text{K}$ for $T_2$, $680\text{m}/\text{s}$ for $V$, and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in Equation (III).

$\begin{array}{c}{T}_0=\left(1151\text{K}\right)+\frac{{\left(680\text{m}/\text{s}\right)}^2}{2\times \left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =\left(1151\text{K}\right)+\frac{{\left(680\text{m}/\text{s}\right)}^2}{\left(2.01\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =\left(1151\text{K}\right)+\frac{\left(462400{\text{m}}^2/{\text{s}}^2\right)\times \left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)}{\left(2.01\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =1381\text{K}\end{array}$

Substitute $1381\text{K}$ for $T_{02}$, $67\text{kJ}/\text{kg}$ for $q$ and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in Equation (V).

$\begin{array}{c}{T}_{01}=1381\text{K}-\frac{67\text{kJ}/\text{kg}}{1.005\text{kJ}/\text{kg}\cdot \text{K}}\\ =1381\text{K}-1314\text{K}\\ =67\text{K}\end{array}$

Substitute $1314\text{K}$ for $T_{01}$ and $1381\text{K}$ for $T_0^{\ast }$ in Equation (VI).

$\begin{array}{c}{\text{Ratio}}_{\text{stagnation}}=\frac{1314\text{K}}{1381\text{K}}\\ =0.95148\\ \cong 0.9515\end{array}$

Refer to Table A-34, “Rayleigh flow function for an ideal gas with k=1.4”, to obtain the value inlet Mach number at $0.9515$ ratio of stagnation temperature using interpolation method of two variables.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (IX)

Here, the variables denote by x and y is ratio of stagnation temperature and Mach number.

Show the ratio of stagnation temperature at $0.7$ and $0.8$ as in Table (1).

S. No	ratio of stagnation temperature $\left(x\right)$	Mach number $\left(y\right)$
1	$0.9085$	$0.7$
2	$0.9516$	$y_2=?$
3	$0.9639$	$0.8$

Calculate inlet Mach number at $0.9515$ ratio of stagnation temperature using interpolation method.

Substitute $0.9085$ for $x_1$, $0.9516$ for $x_2$, $0.9639$ for $x_3$, $0.7$ for $y_1$, and $0.8$ for $y_3$ in Equation (IX).

$\begin{array}{c}{y}_2=\frac{\left(0.9516-0.9085\right)\left(0.8-0.7\right)}{\left(0.9639-0.9085\right)}+0.7\\ =0.777798\\ \cong 0.7778\end{array}$

From above calculation the inlet Mach number at $0.9515$ ratio of stagnation temperature is $0.7778$.

Repeat the Equation (IX), to obtain the value of inlet ratio of temperature, pressure, and velocity at 0.7778 inlet Mach number as:

$\begin{array}{l}{T}_1/T^{\ast }=1.018\\ P_1/P^{\ast }=1.301\\ V_1/V^{\ast }=0.7852\end{array}$

From the Table A-34, “Rayleigh flow function for an ideal gas with k=1.4”, to obtain the value of the outlet ratio of temperature, pressure, and velocity at 1 outlet Mach number as:

$\begin{array}{l}{T}_2/T^{\ast }=1\\ P_2/P^{\ast }=1\\ V_2/V^{\ast }=1\end{array}$

Substitute 1151 K for $T_2$, 1 for $T_2/T^{\ast }$, and $1.018$ for $T_1/T^{\ast }$ in Equation (VI).

$\begin{array}{l}\frac{\left(1151\text{K}\right)}{T_1}=\frac{1}{1.018}\\ T_1=\left(1.018\right)\times \left(1151\text{K}\right)\\ T_1=1171.7\text{K}\\ T_1\cong 1172\text{K}\end{array}$

Thus, the static temperature in the duct is $\underset{\_}{1172\text{K}}$.

Substitute 270 kPa for $P_2$, 1 for $P_2/P^{\ast }$, and $1.301$ for $P_1/P^{\ast }$ in Equation (VII).

$\begin{array}{l}\frac{\left(270\text{kPa}\right)}{P_1}=\frac{1}{1.301}\\ P_1=\left(1.301\right)\times \left(270\text{kPa}\right)\\ P_1=351.27\text{kPa}\\ P_1\cong 351.3\text{kPa}\end{array}$

Thus, the static pressure in the duct is $\underset{\_}{351.3\text{kPa}}$.

Substitute $680\text{m}/\text{s}$ for $V_2$, 1 for $V_2/V^{\ast }$, and $0.7852$ for $P_1/P^{\ast }$ in Equation (VIII).

$\begin{array}{l}\frac{\left(680\text{m}/\text{s}\right)}{V_1}=\frac{1}{0.7852}\\ V_1=\left(0.7852\right)\times \left(680\text{m}/\text{s}\right)\\ V_1=533.9\text{m}/\text{s}\end{array}$

Thus, the velocity in the duct is $\underset{\_}{\text{533}\text{.9}\text{m}/\text{s}}$.