Chapter 17.7, Problem 147RP

Air is cooled as it flows through a 30-cm-diameter duct. The inlet conditions are Ma₁ = 1.2, T₀₁ = 350 K, and P₀₁ = 240 kPa and the exit Mach number is Ma₂ = 2.0. Disregarding frictional effects, determine the rate of cooling of air.

To determine

The rate of cooling of air.

Answer to Problem 147RP

The required rate of cooling of air is $2370\text{kW}$.

Explanation of Solution

Write the formula of ratio of stagnation temperature to the static temperature at inlet of the duct.

$\begin{array}{l}\frac{T_{01}}{T_1}=1+\frac{k-1}{2}{\text{Ma}}_1^2\\ T_1=T_{01}{\left(1+\frac{k-1}{2}{\text{Ma}}_1^2\right)}^{-1}\end{array}$ (I)

Here, the inlet static temperature is $T_1$, the inlet stagnation temperature is $T_{01}$, the specific heat ratio is $k$, and Mach number of air at the inlet of duct is ${\text{Ma}}_1$.

Write the formula of ratio of stagnation pressure to the static pressure at inlet of the duct.

$\begin{array}{l}\frac{P_{01}}{P_1}={\left(1+\frac{k-1}{2}{\text{Ma}}_1^2\right)}^{k/\left(k-1\right)}\\ P_1=P_{01}{\left(1+\frac{k-1}{2}{\text{Ma}}_1^2\right)}^{-k/\left(k-1\right)}\end{array}$ (II)

Here, the actual (static) pressure at the inlet of duct is $P_1$, and Mach number of air gas at the inlet of duct is ${\text{Ma}}_1$.

Write the formula for inlet density of air.

${\rho }_1=\frac{P_1}{R{T}_1}$ (III)

Here, the pressure of air at the inlet is $P_1$, the temperature of air at the inlet is $T_1$, the density of air at inlet is ${\rho }_1$, and the gas constant of air is R.

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

$c_1=\sqrt{kR{T}_1}$

Here, speed of sound at the inlet condition is $c_1$, gas constant of air is R.

Write formula for the velocity of air at inlet.

$\begin{array}{c}{V}_1={\text{Ma}}_1{c}_1\\ ={\text{Ma}}_1\sqrt{kR{T}_1}\end{array}$ (IV)

Write the formula for mass flow rate of air with inlet conditions of air.

${\dot{m}}_{\text{air}}={\rho }_1{A}_1{V}_1$ (V)

Here, the mass flow rate of air at the inlet is ${\dot{m}}_{\text{air}}$, the density of air at the inlet is ${\rho }_1$, the cross sectional area at the inlet is $A_1$, and the velocity of air at the inlet is $V_1$.

Write the formula for stagnation temperature ratio of exit to inlet.

$\frac{T_0{}_2}{T_{01}}=\frac{T_{02}/T_0{}^{\ast }}{T_{01}/T_0{}^{\ast }}$ (VI)

Write the formula for heat transfer rate $\left(\dot{Q}\right)$.

$\dot{Q}={\dot{m}}_{\text{air}}{c}_p\left(T_{02}-T_{01}\right)$ (VII)

Here, the specific heat at constant pressure is $c_p$, the exit stagnation temperature is $T_{02}$, and inlet stagnation temperature is $T_{01}$.

Refer Table A-34, “Rayleigh flow functions for an ideal gas with $k=1.4$”.

The Rayleigh flow function of inlet stagnation temperature to the critical stagnation temperature corresponding to the inlet Mach number of $1.2$ is $0.9787$.

$\frac{T_{01}}{T_0^{\ast }}=0.9787$

The Rayleigh flow function of exit stagnation temperature to the critical stagnation temperature corresponding to the exit Mach number of $2.0$ is $0.7934$.

$\frac{T_{02}}{T_0^{\ast }}=0.7934$

Refer Table A-, “Molar mass, gas constant, and critical2point 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 ratio $\left(k\right)$ of air is $1.4$ and the specific heat at constant pressure is $1.005\text{kJ/kg}\cdot \text{K}$.

Conclusion:

Substitute $350\text{K}$ for $T_{01}$, 1.4 for $k$, and $1.2$ for ${\text{Ma}}_{\text{1}}$ in Equation (I).

$\begin{array}{c}{T}_1=350\text{K}{\left[1+\frac{\left(1.4-1\right)}{2}{\left(1.2\right)}^2\right]}^{-1}\\ =350\text{K}{\left[1+\left(0.2\right)\left(1.2\right)\right]}^{-1}\\ =350\text{K}{\left(1.288\right)}^{-1}\\ =271.7391\text{K}\end{array}$

$\simeq 271.7\text{K}$

Substitute $240\text{kPa}$ for $P_{01}$, 1.4 for $k$, and $1.2$ for ${\text{Ma}}_{\text{1}}$ in Equation (II).

$\begin{array}{c}{P}_1=240\text{kPa}{\left[1+\frac{\left(1.4-1\right)}{2}{\left(1.2\right)}^2\right]}^{-1.4/\left(1.4-1\right)}\\ =240\text{kPa}{\left[1+\left(0.2\right)\left(1.2\right)\right]}^{-3.5}\\ =240\text{kPa}{\left(1.288\right)}^{-3.5}\\ =98.9705\text{kPa}\end{array}$

$\simeq 98.97\text{kPa}$

Substitute $98.97\text{kPa}$ for $P_1$, $0.287\text{kJ/kg}\cdot \text{K}$ for $R$, and $271.7\text{K}$ for $T_1$ in

Equation (III).

$\begin{array}{c}{\rho }_1=\frac{98.97\text{kPa}}{0.287\text{kJ/kg}\cdot \text{K}\left(271.7\text{K}\right)}\\ =\frac{98.97\text{kPa}}{77.9779\text{kJ/kg}\times \frac{1\text{kPa}\cdot {\text{m}}^3}{1\text{kJ}}}\\ =1.2692{\text{kg/m}}^3\end{array}$

Substitute $1.2$ for ${\text{Ma}}_{\text{1}}$, 1.4 for $k$, $0.287\text{kJ/kg}\cdot \text{K}$ for $R$, and $271.7\text{K}$ for $T_1$ in Equation (IV).

$\begin{array}{c}{V}_1=\left(1.2\right)\sqrt{\left(1.4\right)\left(0.287\text{kJ/kg}\cdot \text{K}\right)271.7\text{K}}\\ =\left(1.2\right)\sqrt{109.1691\text{kJ/kg}\times \frac{1000{\text{m}}^2{\text{/s}}^2}{1\text{kJ/kg}}}\\ =1.2\left(330.4074\text{m/s}\right)\\ =396.4889\text{m/s}\end{array}$

$\simeq \text{396}\text{.5}\text{m/s}$

The cross sectional area $\left(A_1\right)$ of the duct is expressed as follows.

$\begin{array}{c}{A}_1=\frac{\pi D^2}{4}\\ =\frac{\pi {\left(30\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2}{4}\\ =0.07068{\text{m}}^2\end{array}$

Substitute $1.2692{\text{kg/m}}^3$ for ${\rho }_1$, $0.07068{\text{m}}^2$ for $A_1$, and $396.5\text{m/s}$ for $V_1$ in

Equation (V).

$\begin{array}{c}{\dot{m}}_{\text{air}}=\left(1.2692{\text{kg/m}}^3\right)\left(0.07068{\text{m}}^2\right)\left(396.5\text{m/s}\right)\\ =35.5688\text{kg/s}\\ =35.57\text{kg/s}\end{array}$

Thus, the rate of cooling air is $35.57\text{kg/s}$.

Substitute $0.7934$ for $T_{02}/T_0{}^{\ast }$, $0.9787$ for $T_{01}/T_0{}^{\ast }$ and $350\text{K}$ for $T_{01}$ in Equation (VI).

$\begin{array}{l}\frac{T_{02}}{350\text{K}}=\frac{0.7934}{0.9787}\\ T_{02}=\left(350\text{K}\right)\left(0.8107\right)\\ T_{02}=283.7335\text{K}\\ T_{02}\simeq 283.7\text{K}\end{array}$

Substitute $35.57\text{kg/s}$ for ${\dot{m}}_{\text{air}}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $350\text{K}$ for $T_{01}$, and $283.7\text{K}$ for $T_{02}$ in Equation (VII).

$\begin{array}{c}\dot{Q}=\left(35.57\text{kg/s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(283.7\text{K}-350\text{K}\right)\\ =\left(35.57\text{kg/s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(-66.3\text{K}\right)\\ =-2370.0824\text{kJ/s}\times \frac{1\text{kW}}{1\text{KJ/s}}\\ \simeq -2370\text{kW}\end{array}$

Here, the negative sign indicates that the air requires cooling in order to be accelerated.

Thus, the required rate of cooling of air is $2370\text{kW}$.