Chapter 19, Problem 94P

(a)

To determine

The rate of heat transfer of air by Equation $19-79$.

Explanation of Solution

Given:

The diameter $\left(D\right)$ of the circular tube is $12\text{cm}$.

The length $\left(L\right)$ of the circular tube is $5\text{m}$.

The inlet temperature $\left(T_1\right)$ of the air is $10°\text{C}$.

The mass flow rate $\left(\dot{m}\right)$ of air is $0.065\text{kg}/\text{s}$.

The roughness $\left(\epsilon \right)$ of the pipe is $0.22\text{mm}$.

Calculation:

Refer to the table $\text{A}-22$ “Properties of air at $1\text{atm}$ pressure”

Obtain the following properties of air corresponding to the temperature of $10°\text{C}$ as follows:

$\begin{array}{l}{c}_p=1007\text{J}/\text{kg}\cdot \text{K}\\ k=0.02514\text{W}/\text{m}\cdot \text{K}\\ \nu =1.516\times {10}^{-5}{\text{m}}^2/\text{s}\\ {\mu }_s=2.420\times {10}^{-5}\text{kg}/\text{m}\cdot \text{s}\end{array}$

Further,

$\begin{array}{l}\rho =1.204\text{kg}/{\text{m}}^3\\ \mathrm{Pr}=0.7309\end{array}$

Calculate the mean velocity of air by using the relation.

    $\begin{array}{c}V=\frac{\dot{m}}{\rho A_c}\\ =\frac{0.065\text{kg}/\text{s}}{\left(1.204\text{kg}/{\text{m}}^3\right)\left(\frac{\pi }{4}{\left(12\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\right)}\\ =4.773\text{m}/\text{s}\end{array}$

Calculate the Reynolds number by using the relation.

    $\begin{array}{c}\mathrm{Re}=\frac{VD}{\nu }\\ =\frac{\left(5\text{m}/\text{s}\right)\left(12\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{1.516\times {10}^{-5}{\text{m}}^2/\text{s}}\\ =37785\end{array}$

Calculate the length of hydrodynamic boundary layer by using the relation.

    $\begin{array}{c}{L}_h=10D\\ =10\left(12\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =1.2\text{m}\end{array}$

Calculate the length of thermal boundary layer by using the relation.

    $\begin{array}{c}{L}_t=10D\\ =10\left(12\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =1.2\text{m}\end{array}$

The Reynolds number is greater than $100000$ and the length of hydrodynamic boundary layer, thermal boundary layer is less than the length of pipe.

Calculate the friction factor by using Colebrook equation.

    $\begin{array}{c}\frac{1}{\sqrt{f}}=-2\log \left(\frac{\epsilon /D}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=2\log \left(\frac{\left(\frac{0.22\text{mm}\times \frac{1\text{m}}{1000\text{mm}}}{\left(12\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\right)}{3.7}+\frac{2.51}{\left(37785\right)\sqrt{f}}\right)\\ f=0.02695\end{array}$

Calculate the Nusselt number by using the Equation $19-79$.

    $\begin{array}{c}\text{Nu}=0.125f\mathrm{Re}{\mathrm{Pr}}^{1/3}\\ \frac{hD}{k}=0.125\left(0.02695\right)\left(37785\right){\left(0.7309\right)}^{1/3}\\ h=\left(114.7\right)\left(\frac{0.02514\text{W}/\text{m}\cdot \text{K}}{12\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\right)\\ =24.02\text{W}/{\text{m}}^{\text{2}}\text{°C}\end{array}$

Calculate the outlet temperature of the tube by using the relation.

    $\begin{array}{c}{T}_2=T_s-\left(T_s-T_1\right)e^{\left(-\frac{h{A}_s}{\dot{m}{c}_p}\right)}\\ T_2=T_s-\left(T_s-T_1\right)e^{\left(-\frac{h\left(\pi DL\right)}{\dot{m}{c}_p}\right)}\\ =50°\text{C}-\left(50°\text{C}-10°\text{C}\right)e^{\left(\frac{\left(24.02\text{W}/{\text{m}}^{\text{2}}\text{°C}\right)\pi \left(12\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(5\text{m}\right)}{\left(0.065\text{kg}/\text{s}\right)\left(1007\text{J}/\text{kg}\cdot \text{K}\right)}\right)}\\ =30°\text{C}\end{array}$

Calculate the heat transfer by using the relation.

    $\begin{array}{c}\dot{Q}=\dot{m}{c}_p\left(T_2-T_1\right)\\ =\left(0.065\text{kg}/\text{s}\right)\left(1007\text{J}/\text{kg}\cdot \text{K}\right)\left(\left(30°\text{C+273}\right)\text{K}-\left(10°\text{C+273}\right)\text{K}\right)\\ =1307\text{W}\end{array}$

Thus, the rate of heat transfer of air by Equation $19-79$ is $1307\text{W}$.

(b)

To determine

The rate of heat transfer of air by Equation $19-84$.

Explanation of Solution

Calculation:

Calculate the Nusselt number by using the Equation $19-84$.

    $\begin{array}{c}\text{Nu}=\frac{\left(f/8\right)\left(\mathrm{Re}-1000\right)\mathrm{Pr}}{1+12.7{\left(f/8\right)}^{0.5}\left({\mathrm{Pr}}^{2/3}-1\right)}\\ \frac{hD}{k}=\frac{\left(\frac{0.02695}{8}\right)\left(37785-1000\right)\left(0.7309\right)}{1+12.7{\left(\frac{0.02695}{8}\right)}^{0.5}\left({\left(0.7309\right)}^{2/3}-1\right)}\\ h=\left(105.2\right)\left(\frac{0.02514\text{W}/\text{m}\cdot \text{K}}{12\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\right)\\ =22.04\text{W}/{\text{m}}^{\text{2}}\text{°C}\end{array}$

Calculate the outlet temperature of the tube by using the relation.

    $\begin{array}{c}{T}_2=T_s-\left(T_s-T_1\right)e^{\left(-\frac{h{A}_s}{\dot{m}{c}_p}\right)}\\ T_2=T_s-\left(T_s-T_1\right)e^{\left(-\frac{h\left(\pi DL\right)}{\dot{m}{c}_p}\right)}\\ =50°\text{C}-\left(50°\text{C}-10°\text{C}\right)e^{\left(\frac{\left(22.04\text{W}/{\text{m}}^{\text{2}}\text{°C}\right)\pi \left(12\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(5\text{m}\right)}{\left(0.065\text{kg}/\text{s}\right)\left(1007\text{J}/\text{kg}\cdot \text{K}\right)}\right)}\\ =28.8°\text{C}\end{array}$

Calculate the heat transfer by using the relation.

    $\begin{array}{c}\dot{Q}=\dot{m}{c}_p\left(T_2-T_1\right)\\ =\left(0.065\text{kg}/\text{s}\right)\left(1007\text{J}/\text{kg}\cdot \text{K}\right)\left(\left(28.8°\text{C+273}\right)\text{K}-\left(10°\text{C+273}\right)\text{K}\right)\\ =1230\text{W}\end{array}$

Thus, the rate of heat transfer of air by equation $19-84$ is $1230\text{W}$.