Chapter 21, Problem 136RQ

(a)

To determine

The rate of heat transfer by natural convection.

Explanation of Solution

Given:

The diameter of first sphere $\left(D_1\right)$ is $15\text{cm}$.

The diameter of second sphere $\left(D_2\right)$ is $25\text{cm}$.

The surface temperature of first sphere $\left(T_1\right)$ is $350\text{K}$.

The surface temperature of second sphere $\left(T_2\right)$ is $275\text{K}$.

Calculation:

Calculate the film temperature.

    $\begin{array}{c}{T}_f=\frac{T_s+T_{\infty }}{2}\\ =\frac{350\text{K}+275\text{K}}{2}\\ \text{=}\left(312.5\text{K}-\text{273}\right)°\text{C}\\ =\text{39}\text{.5}°\text{C}\end{array}$

Refer Table A-22, “Properties of air at $1\text{atm}$ pressure”.

Obtain the following properties of air corresponding to $\text{39}\text{.5}°\text{C}$ as follows.

$\begin{array}{l}k=0.02658\text{W}/\text{m}\cdot \text{K}\\ v=1.697\times {10}^{-5}{\text{m}}^2/\text{s}\\ \beta =0.0032{\text{K}}^{-1}\\ \mathrm{Pr}=0.7256\end{array}$

Calculate the characteristic length.

    $\begin{array}{c}{L}_c=\frac{D_0-D_i}{2}\\ =\frac{\left(25\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)-\left(15\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{2}\\ =0.05\text{m}\end{array}$

Calculate the Rayleigh number.

    $\begin{array}{c}Ra=\frac{g\beta \left(T_1-T_2\right)L_c^3}{v^2}\mathrm{Pr}\\ =\frac{\left(9.81\text{m}/{\text{s}}^2\right)\left(0.0032{\text{K}}^{-1}\right)\left(350\text{K}-275\text{K}\right){\left(0.05\text{m}\right)}^3}{{\left(1.697\times {10}^{-5}{\text{m}}^2/\text{s}\right)}^2}0.7256\\ =7.415\times {10}^5\end{array}$

Calculate the view function.

    $\begin{array}{c}{F}_{\text{sph}}=\frac{L_c}{{\left(D_i{D}_0\right)}^4{\left(D_i^{-7/5}+D_0^{-7/5}\right)}^5}\\ =\frac{0.05\text{m}}{{\left(15\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\times 25\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^4{\left(\begin{array}{l}{\left(15\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^{-7/5}\\ +{\left(25\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^{-7/5}\end{array}\right)}^5}\\ =0.005900\end{array}$

Calculate the thermal conductivity.

    $\begin{array}{l}{k}_{eff}=0.74k{\left(\frac{\mathrm{Pr}}{0.861+\mathrm{Pr}}\right)}^{1/4}{\left(F_{\text{sph}}Ra\right)}^{1/4}\\ =0.74\left(0.02658\text{W}/\text{m}\cdot \text{K}\right){\left(\frac{0.7256}{0.861+0.7256}\right)}^{1/4}{\left(0.005900\times 7.415\times {10}^5\right)}^{1/4}\\ =0.1315\text{W}/\text{m}\cdot \text{K}\end{array}$

Calculate rate of heat transfer by natural convection.

    $\begin{array}{c}\dot{Q}=k_{eff}\pi \left(\frac{D_i{D}_0}{L_c}\right)\left(T_i-T_0\right)\\ =\left(0.1315\text{W}/\text{m}\cdot \text{K}\right)\pi \left(\frac{\left(15\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\times \left(25\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{0.05\text{m}}\right)\left(350\text{K}-275\text{K}\right)\\ =23.2\text{W}\end{array}$

Thus, rate of heat transfer by natural convection is $23.2\text{W}$.

(b)

To determine

The rate of heat transfer by radiation.

Explanation of Solution

Calculation:

Calculate the rate of heat transfer by radiation.

    $\begin{array}{c}{\dot{Q}}_{12}=\frac{\left(\pi D_1^2\right)\sigma \left(T_2^4-T_1^4\right)}{\frac{1}{{\epsilon }_1}+\frac{1-{\epsilon }_2}{{\epsilon }_1}{\left(\frac{D_1}{D_2}\right)}^2}\\ =\frac{\left(\pi {\left(15\text{cm}\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right)\left[{\left(350\text{K}\right)}^4-{\left(275\text{K}\right)}^4\right]}{\frac{1}{0.75}+\frac{1-0.75}{0.75}{\left(\frac{15\text{cm}}{25\text{cm}}\right)}^2}\\ =25.6\text{W}\end{array}$

Thus, the rate of heat transfer by radiation is $25.6\text{W}$.