Chapter 21, Problem 114P

(a)

To determine

The net rate of radiation heat transfer between the base and the side surfaces.

Explanation of Solution

Given:

The side of the cubic furnace $\left(S\right)$ is $10\text{ft}$.

The emissivity of the base surface $\left(\epsilon \right)$ is $0.7$.

The temperature of the base $\left(T_1\right)$ is $800\text{R}$.

The temperature of the top $\left(T_2\right)$ is $1600\text{R}$.

The temperature of the side $\left(T_3\right)$ is $2400\text{R}$.

Calculation:

The below figure represent the required diagram.

Figure-(1)

Calculate the area of the base surface $\left(A_1\right)$ using the relation.

  $\begin{array}{c}{A}_1=S\times S\\ =\left(10\text{ft}\right)\times \left(10\text{ft}\right)\\ =100{\text{ft}}^{\text{2}}\end{array}$

Calculate the area of the top surface $\left(A_2\right)$ using the relation.

  $\begin{array}{c}{A}_2=S\times S\\ =\left(10\text{ft}\right)\times \left(10\text{ft}\right)\\ =100{\text{ft}}^{\text{2}}\end{array}$

Calculate the area of the side surface $\left(A_3\right)$ using the relation.

  $\begin{array}{c}{A}_3=S\times S\\ =\left(10\text{ft}\right)\times \left(10\text{ft}\right)\\ =100{\text{ft}}^{\text{2}}\end{array}$

Calculate the emissive power of the base surface $\left(E_{b1}\right)$ using the relation.

  $\begin{array}{c}{E}_{b1}=\sigma T_1{}^4\\ =\left(0.1714\times {10}^{-8}\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\cdot {\text{R}}^{\text{4}}\right){\left(800\text{R}\right)}^4\\ =702.05\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\end{array}$

Calculate the emissive power of the top surface $\left(E_{b2}\right)$ using the relation.

  $\begin{array}{c}{E}_{b2}=\sigma T_2{}^4\\ =\left(0.1714\times {10}^{-8}\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\cdot {\text{R}}^{\text{4}}\right){\left(1600\text{R}\right)}^4\\ =11232.87\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\end{array}$

Calculate the emissive power of the side surface $\left(E_{b3}\right)$ using the relation.

  $\begin{array}{c}{E}_{b3}=\sigma T_2{}^4\\ =\left(0.1714\times {10}^{-8}\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\cdot {\text{R}}^{\text{4}}\right){\left(2400\text{R}\right)}^4\\ =56866.4\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\end{array}$

Consider, the view factor of the cube from base to the top surface $\left(F_{12}\right)$ is $0.2$ and the view factor of the cube from base surface $\left(F_{11}\right)$ is $0$.

Calculate the view factor from the base or top to the side surfaces $\left(F_{13}\right)$ using the relation.

  $\begin{array}{c}{F}_{13}=1-F_{12}-F_{11}\\ =1-0.2-0\\ =0.8\end{array}$

Calculate the radiation resistance of the base surface $\left(R_1\right)$ using the relation.

  $\begin{array}{c}{R}_1=\frac{1-\epsilon }{A_1\epsilon }\\ =\frac{1-0.7}{\left(100{\text{ft}}^{\text{2}}\right)\left(0.7\right)}\\ =4.285\times {10}^{-3}{\text{ft}}^{-2}\end{array}$

Calculate the radiation resistance between base and top surface $\left(R_{12}\right)$ using the relation.

  $\begin{array}{c}{R}_{12}=\frac{1}{A_1{F}_{12}}\\ =\frac{1}{\left(100{\text{ft}}^{\text{2}}\right)\left(0.2\right)}\\ =0.05{\text{ft}}^{-2}\end{array}$

Calculate the radiation resistance between top and side surface $\left(R_{13}\right)$ using the relation.

  $\begin{array}{c}{R}_{13}=\frac{1}{A_1{F}_{13}}\\ =\frac{1}{\left(100{\text{ft}}^{\text{2}}\right)\left(0.8\right)}\\ =0.0125{\text{ft}}^{-2}\end{array}$

Calculate the radiosity $\left(J_1\right)$ using the relation.

  $\begin{array}{c}\frac{E_{b1}-J_1}{R_1}+\frac{E_{b2}-J_1}{R_{12}}+\frac{E_{b3}-J_1}{R_{13}}=0\\ \left[\begin{array}{l}\frac{\left(702.05\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\right)-J_1}{\left(4.285\times {10}^{-3}{\text{ft}}^{-2}\right)}+\\ \frac{\left(11232.87\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\right)-J_1}{\left(0.05{\text{ft}}^{-2}\right)}+\\ \frac{\left(56866.4\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\right)-J_1}{\left(0.0125{\text{ft}}^{-2}\right)}\end{array}\right]=0\\ J_1=14811.7\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\end{array}$

Calculate the net rate of radiation heat transfer between the base and the side surfaces $\left({\dot{Q}}_{31}\right)$ using the realtion.

  $\begin{array}{c}{\dot{Q}}_{31}=\frac{E_{b3}-J_1}{R_{13}}\\ =\frac{\left(56866.4\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\right)-\left(14811.7\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\right)}{\left(0.0125{\text{ft}}^{-2}\right)}\\ =3364376\text{Btu}/\text{h}\end{array}$

Thus, the net rate of radiation heat transfer between the base and the side surfaces is $3364376\text{Btu}/\text{h}$.

(b)

To determine

The net rate of radiation heat transfer between the base and the top surfaces.

The net rate of radiation heat transfer to the base surface.

Explanation of Solution

Calculation:

Calculate the net rate of radiation heat transfer between the base and the top surfaces $\left({\dot{Q}}_{12}\right)$ using the relation.

  $\begin{array}{c}{\dot{Q}}_{12}=\frac{J_1-E_{b2}}{R_{12}}\\ =\frac{\left(14811.7\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\right)-\left(11232.87\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\right)}{\left(0.05{\text{ft}}^{-2}\right)}\\ =71576.6\text{Btu}/\text{h}\end{array}$

Thus, the net rate of radiation heat transfer between the base and the top is $71576.6\text{Btu}/\text{h}$.

Calculate the net rate of radiation heat transfer to the base surface $\left({\dot{Q}}_1\right)$ using the relation.

  $\begin{array}{c}{\dot{Q}}_1=\frac{J_1-E_{b1}}{R_1}\\ =\frac{\left(14811.7\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\right)-\left(702.05\text{Btu}/\text{h}\cdot {\text{ft}}^{\text{2}}\right)}{\left(4.285\times {10}^{-3}{\text{ft}}^{-2}\right)}\\ =3292800.5\text{Btu}/\text{h}\end{array}$

Thus, the net rate of radiation heat transfer to the base surface is $3292800.5\text{Btu}/\text{h}$.