Chapter 19, Problem 102P

(a)

To determine

The exit temperature of the hot air leaving the basement.

Explanation of Solution

Given:

The length $\left(L\right)$ of the duct is $12\text{m}$.

The cross section area $\left(A_c\right)$ of the duct is $20\text{cm}\times 20\text{cm}$.

The average velocity $\left(V\right)$ of the hot air in the duct is $4\text{m}/\text{s}$

The temperature $\left(T_1\right)$ of the hot air leaving the furnace is $60°\text{C}$.

The emissivity $\left(\epsilon \right)$ of the outer surface of duct is $0.3$.

The temperature $\left(T_0\right)$ of the cold air is $10°\text{C}$.

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 bulk mean temperature of $50°\text{C}$ as follows:

$c_p=1007\text{J}/\text{kg}\cdot \text{K}$

$k=0.02735\text{W}/\text{m}\cdot \text{K}$

$\nu =1.798\times {10}^{-5}{\text{m}}^2/\text{s}$

$\rho =1.092\text{kg}/{\text{m}}^3$

$\mathrm{Pr}=0.7228$

Calculate the cross sectional area of the duct by using the relation.

    $\begin{array}{c}{A}_s=4aL\\ =4\left(0.2\text{m}\right)\left(12\text{m}\right)\\ =9.6{\text{m}}^{\text{2}}\end{array}$

Calculate the hydrodynamic diameter by using the relation.

    $\begin{array}{c}{D}_h=\frac{4A_c}{P}\\ =\frac{4\left(0.2\text{m}\times 0.2\text{m}\right)}{4\left(0.2\text{m}\right)}\\ =0.2\text{m}\end{array}$

Calculate the Reynolds number by using the relation.

    $\begin{array}{c}\mathrm{Re}=\frac{V{D}_h}{\nu }\\ =\frac{\left(4\text{m}/\text{s}\right)\left(20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{1.798\times {10}^{-5}{\text{m}}^2/\text{s}}\\ =44494\end{array}$

The Reynolds number is greater than $10000$ so the flow in the tube is turbulent.

Calculate the length of hydrodynamic layer by using the relation.

    $\begin{array}{c}{L}_h=10D_h\\ =10\left(0.2\text{m}\right)\\ =2\text{m}\end{array}$

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

    $\begin{array}{c}{L}_t=10D_h\\ =10\left(0.2\text{m}\right)\\ =2\text{m}\end{array}$

The length of hydrodynamic layer and thermal boundary layer is less than the length of duct.

Calculate the heat transfer coefficient by using the relation.

    $\begin{array}{c}\text{Nu}=0.023{\mathrm{Re}}^{0.8}{\mathrm{Pr}}^{0.4}\\ \frac{hD}{k}=0.023{\left(44494\right)}^{0.8}{\left(0.7228\right)}^{0.4}\\ h=\left(105.68\right)\left(\frac{0.02735\text{W}/\text{m}\cdot \text{K}}{0.2\text{m}}\right)\\ =14.45\text{W}/{\text{m}}^{\text{2}}\text{K}\end{array}$

Calculate the mass flow rate by using the relation.

    $\begin{array}{c}\dot{m}=\rho A_{c}V\\ =\left(1.092\text{kg}/{\text{m}}^3\right)\left(\left(20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\times \left(20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\right)\left(4\text{m}/\text{s}\right)\\ =0.1747\text{kg}/\text{s}\end{array}$

Calculate the heat transfer due to convection by using the relation.

    $\begin{array}{c}{Q}_{con}=h_1{A}_s\left(\Delta T_{lm}\right)\\ =h_1{A}_s\left[\frac{\left(T_2-T_1\right)}{\ln \left(\frac{T_s-T_2}{T_s-T_1}\right)}\right]\\ =\left(14.45\text{W}/{\text{m}}^{\text{2}}\text{K}\right)\left(9.6{\text{m}}^{\text{2}}\right)\left[\frac{\left(T_2-60°\text{C}\right)}{\ln \left(\frac{T_s-T_2}{T_s-60°\text{C}}\right)}\right]\\ =138.72\text{W}/\text{K}\left[\frac{\left(T_2-60°\text{C}\right)}{\ln \left(\frac{T_s-T_2}{T_s-60°\text{C}}\right)}\right]\end{array}$

Calculate the heat transfer due to convection and radiation by using the relation.

    $\begin{array}{c}{Q}_{con+rad}=h_2{A}_s\left(T_s-T_o\right)+\epsilon A_s\sigma \left(T_s^4-T_o^4\right)\\ =\left[\begin{array}{l}\left(10\text{W}/{\text{m}}^{\text{2}}\text{K}\right)\left(9.6{\text{m}}^{\text{2}}\right)\left(T_s-10°\text{C}\right)+\left(0.3\right)\left(9.6{\text{m}}^{\text{2}}\right)\\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^4\right)\left[{\left(T_s\right)}^4-{\left(10°\text{C}+273\right)}^4{\text{K}}^4\right]\end{array}\right]\end{array}$

Calculate the total heat transfer by using the relation.

    $\begin{array}{c}{Q}_t=\dot{m}{c}_p\left(T_2-T_1\right)\\ =\left(0.1747\text{kg}/\text{s}\right)\left(1007\text{J}/\text{kg}\cdot \text{K}\right)\left(60°\text{C}-T_1\right)\end{array}$

In the steady operation heat transfer from the hot air to duct must be equal to heat transfer due to convection and radiation, which also equal to energy loss of the hot air in the duct.

    $\begin{array}{c}{Q}_{con}=Q_{con+rad}\\ =Q_t\end{array}$

From above condition

$Q=2622\text{W}$

$T_2=45.1°\text{C}$

$T_s=33.3°\text{C}$

Thus, the exit temperature of the hot air leaving the basement is $45.1°\text{C}$.

(b)

To determine

The rate of heat loss from the hot air in the duct.

Explanation of Solution

Calculation:

Calculate the total heat transfer by using the relation.

    $\begin{array}{c}{Q}_t=\dot{m}{c}_p\left(T_2-T_1\right)\\ =\left(0.1747\text{kg}/\text{s}\right)\left(1007\text{J}/\text{kg}\cdot \text{K}\right)\left(\begin{array}{l}\left(60°\text{C}+273\right)\text{K}-\\ \left(45.1°\text{C}+273\right)\text{K}\end{array}\right)\\ =2621.25\text{W}\end{array}$

Thus, the rate of heat loss from the hot air in the duct is $2621.25\text{W}$.