Chapter 20, Problem 94RQ

To determine

The rate of heat loss from all surfaces of the tank by natural convection and radiation.

Explanation of Solution

Given:

The diameter of the cylinder $\left(d\right)$ is $40\text{cm}$.

The height of the cylinder $\left(l\right)$ is $110\text{cm}$.

The temperature of the surrounding $\left(T_o\right)$ is $20°\text{C}$.

The temperature of the tank surface $\left(T_s\right)$ is $44°\text{C}$.

The emissivity of the tank $\left(\epsilon \right)$ is $0.4$.

Calculation:

Calculate the bulk mean temperature $\left(T_m\right)$ using the relation.

    $\begin{array}{c}{T}_m=\frac{T_o+T_s}{2}\\ =\frac{\left(20°\text{C}\right)\text{+}\left(\text{44}°\text{C}\right)}{2}\\ =32°\text{C}\end{array}$

Refer table $\text{A-22}$ “properties of air at $1\text{atm}$ pressure”.

Obtain the following properties of air corresponding to the temperature of $32°\text{C}$.

$\begin{array}{c}k=0.02603\text{W}/\text{m}\cdot \text{K}\\ v=1.627\times {10}^{-5}{\text{m}}^2/\text{s}\\ \mathrm{Pr}=0.7276\end{array}$

Calculate the temperature coefficient $\left(\beta \right)$ using the relation

    $\begin{array}{c}\beta =\frac{1}{T_m}\\ =\frac{1}{\left(32°\text{C}+\text{273}\right)\text{K}}\\ =0.003279{\text{K}}^{-1}\end{array}$

Calculate the Rayleigh number $\left(Ra\right)$ using the relation

    $\begin{array}{c}Ra=\frac{g\beta \left(T_s-T_o\right)D^3}{{\upsilon }^2}\mathrm{Pr}\\ =\frac{\left(9.81\text{m}/{\text{s}}^2\right)\left(0.003279{\text{K}}^{-1}\right)\left(\left(44°\text{C+273}\right)\text{K}-\left(20°\text{C+273}\right)\text{K}\right){\left(1.1\text{m}\right)}^3}{{\left(1.627\times {10}^{-5}{\text{m}}^2/\text{s}\right)}^2}\left(0.7276\right)\\ =2.334\times {10}^9\end{array}$

Calculate the Nusslet number $\left(Nu\right)$ using the relation.

    $\begin{array}{c}Nu={\left\{0.825+\frac{0.387R{a}^{\frac{1}{6}}}{{\left[1+{\left(\frac{0.492}{\text{Pr}}\right)}^{\frac{9}{16}}\right]}^{\frac{8}{27}}}\right\}}^2\\ ={\left\{0.825+\frac{0.387{\left(2.334\times {10}^9\right)}^{\frac{1}{6}}}{{\left[1+{\left(\frac{0.492}{0.7276}\right)}^{\frac{9}{16}}\right]}^{\frac{8}{27}}}\right\}}^2\\ =170.2\end{array}$

The heat transfer coefficient $\left(h\right)$ can be calculated using the relation

    $\begin{array}{c}h=\frac{k}{L}Nu\\ =\frac{0.02603\text{W}/\text{m}\cdot \text{K}}{1.1\text{m}}\left(170.2\right)\\ =4.027\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

The surface area $\left(A_s\right)$ can be calculated using the relation

    $\begin{array}{c}{A}_s=\pi DL\\ =\pi \left(40\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(110\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =1.382{\text{m}}^2\end{array}$

The convective heat transfer from the curved surface is $\left(Q_{side}\right)$ by using the relation

    $\begin{array}{c}{Q}_{side}=h{A}_s\left(T_s-T_o\right)\\ =\left(4.027\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(1.382{\text{m}}^2\right)\left[\left(44°\text{C+273}\right)\text{K}-\left(20°\text{C+273}\right)\text{K}\right]\\ =133.6\text{W}\end{array}$

The characteristic length $\left(L_c\right)$ can be calculated using the relation.

    $\begin{array}{c}{L}_c=\frac{d}{4}\\ =\frac{0.4\text{m}}{4}\\ =0.1\text{m}\end{array}$

The Rayleigh number $\left(\text{Ra}\right)$ can be calculated using the relation

     $\begin{array}{c}Ra=\frac{g\beta \left(T_s-T_o\right)D^3}{{\upsilon }^2}\mathrm{Pr}\\ =\frac{\left(9.81\text{m}/{\text{s}}^2\right)\left(0.003279{\text{K}}^{-1}\right)\left(\left(44°\text{C+273}\right)\text{K}-\left(20°\text{C+273}\right)\text{K}\right){\left(0.1\text{m}\right)}^3}{{\left(1.627\times {10}^{-5}{\text{m}}^2/\text{s}\right)}^2}\left(0.7276\right)\\ =2.123\times {10}^6\end{array}$

The Nusslet number $\left(Nu\right)$ can be calculated using the relation

    $\begin{array}{c}\text{Nu}=0.54{\text{Ra}}^{1/4}\\ =0.54{\left(2.123\times {10}^6\right)}^{1/4}\\ =20.61\end{array}$

The convective heat transfer coefficient $\left(h\right)$ can be calculated using the relation

    $\begin{array}{c}h=\frac{k}{L_c}Nu\\ =\frac{0.02603\text{W}/\text{m}\cdot \text{K}}{0.1\text{m}}\left(20.61\right)\\ =5.635\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

The surface area of the top surface $\left(A_{top}\right)$ can be calculated as using the relation

    $\begin{array}{c}{A}_{top}=\frac{\pi }{4}{d}^2\\ =\frac{\pi }{4}\times {\left(0.4\text{m}\right)}^2\\ =0.1257{\text{m}}^2\end{array}$

The convective heat transfer from the top surface $\left(Q_{top}\right)$ can be calculated as using the relation

    $\begin{array}{c}{Q}_{top}=h{A}_{top}\left(T_s-T_o\right)\\ =\left(5.635\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(0.1257{\text{m}}^2\right)\left[\left(44°\text{C+273}\right)\text{K}-\left(20°\text{C+273}\right)\text{K}\right]\\ =16.2\text{W}\end{array}$

The Nusslet number for the bottom surface $\left(Nu\right)$ can be calculated by using the relation

    $\begin{array}{c}\text{Nu}=0.27R{a}^{1/4}\\ =0.27{\left(2.123\times {10}^6\right)}^{\frac{1}{4}}\\ =10.31\end{array}$

The heat transfer coefficient $\left(h\right)$ can be calculated as using the relation

    $\begin{array}{c}h=\frac{k}{L_c}Nu\\ =\frac{0.02603\text{W}/\text{m}\cdot \text{K}}{0.1\text{m}}\left(10.31\right)\\ =2.683\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

The heat loss from the bottom $\left(Q_{bottom}\right)$ can be calculated as using the relation

    $\begin{array}{c}{Q}_{bottom}=h{A}_{bottom}\left(T_s-T_o\right)\\ =\left(2.683\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(0.1257{\text{m}}^2\right)\left[\left(44°\text{C+273}\right)\text{K}-\left(20°\text{C+273}\right)\text{K}\right]\\ =8.1\text{W}\end{array}$

The total heat loss by convection $\left(Q_{conv}\right)$ can be calculated as using the relation,

    $\begin{array}{c}{Q}_{conv}=Q_{side}+Q_{top}+Q_{bottom}\\ =133.6\text{W}+\text{16}\text{.2}\text{W}+\text{8}\text{.1}\text{W}\\ =\text{157}\text{.9}\text{W}\end{array}$

The heat transfer by radiation $\left(Q_{rad}\right)$ can be calculated as using the relation

    $\begin{array}{c}{Q}_{rad}=\epsilon \left(A_s+A_{top}+A_{bottom}\right)\left(T_s{}^4-T_o{}^4\right)\\ =\left[\begin{array}{l}0.4\times \left(1.382{\text{m}}^{\text{2}}+0.1257{\text{m}}^2+0.1257{\text{m}}^2\right)\\ \left[{\left\{\left(44°\text{C+273}\right)\text{K}\right\}}^4-{\left\{\left(20°\text{C+273}\right)\text{K}\right\}}^4\right]\end{array}\right]\\ =101.1\text{W}\end{array}$

Thus, the rate of heat transfer by convection from all surfaces is $157.9\text{W}$ and by means of radiation is $101.1\text{W}$.