Chapter 12, Problem 12.1P

Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is $2500W/m^2$ , of which $500W/m^2$ is reflected. The plate is at ${227}^{\circ }C$ and has an emissive power of $1200W/m^2$ . Air at ${127}^{\circ }C$ flows over the plate with a heat transfer convection coefficient of $15W/m^{2}K$ . Determine the emissivity, absorptivity, and radiosity of the plate. What is the net heat transfer rate per unit area?

To determine

The value of emissivity, absorptivity and the radiosity of the plate.

The value of the net heat transfer rate per unit area.

Answer to Problem 12.1P

The value of emissivity is, $\epsilon =0.33$

The value of absorptivity is, $\alpha =0.8$

The value of radiosity is, $J=1700\frac{W}{m^2}$

The value of the net heat transfer rate per unit area is, $q_{net}=700\frac{W}{m^2}$.

Explanation of Solution

Given:

The irradiation on the plate is, $G=2500W/m^2$

The reflected irradiation is, $G_{ref}=\rho .G=500W/m^2$

The emissive power of any surface, $E=1200W/m^2$

The temperature of the plate,

  $\begin{array}{l}{T}_{Plate}={227}^{\circ }C\\ T_{Plate}=227+273\\ T_{Plate}=500K\end{array}$

The temperature of the air over the plate

  $\begin{array}{l}{T}_{air}={127}^{\circ }C\\ T_{air}=127+273\\ T_{air}=400K\end{array}$

Calculation:

The energy balance equation as per the incident radiation on a surface is,

  $\alpha +\rho +\tau =1$    (1)

Where,

  $\alpha =$ absorptivity

  $\rho =$ reflectivity

  $\tau =$ transmissivity

For an opaque surface transmissivity, $\tau =0$

Now the equation (1) become as,

  $\alpha +\rho =1$ (2)

The emissive power of a black surface, $E_b=\sigma T_{plate}^4$.

Where,

The Stefan Boltzmann constant, $\sigma =5.67\times {10}^{-8}W/m^2{K}^4$

Emissivity$\left(\epsilon \right)$-It is the ratio of emissive power of any surface (E) to the emissive power of the black surface.

Mathematically,

  $\begin{array}{l}\epsilon =\frac{E}{E_b}\\ \epsilon =\frac{E}{\sigma T_{plate}^4}\\ \epsilon =\frac{1200}{5.67\times {10}^{-8}\times {\left(500\right)}^4}\\ \epsilon =\frac{1200}{3543.75}\\ \epsilon =0.33\end{array}$

Reflectivity$\left(\rho \right)$-It is the ratio of the reflected radiation $\left(G_{ref}\right)$ to the irradiation $\left(G\right)$ on the surface.

Mathematically,

  $\begin{array}{l}\rho =\frac{G_{ref}}{G}\\ \rho =\frac{500}{2500}\\ \rho =0.2\end{array}$

Absorptivity$\left(\alpha \right)$-The amount of irradiation that is absorbed by the surface is known as the absorptivity of that surface.

From equation (2), absorptivity will be calculated as,

  $\begin{array}{l}\alpha +\rho =1\\ \alpha =1-\rho \\ \alpha =1-0.2\\ \alpha =0.8\end{array}$

Radiosity $\left(J\right)$ -It is a radiant flux, at which radiation leaves from the surface.

  $\begin{array}{l}J=E+\rho G\\ J=1200+0.2\times 2500\\ J=1700\frac{W}{m^2}\end{array}$

The net heat transfer rate per unit area is calculated as:

  $\begin{array}{l}{q}_{net}=q_{in}-q_{out}\\ q_{net}=G-\left(G_{ref}+E+q_{convection}\right)\\ q_{net}=G-\left[G_{ref}+E+\left(h\Delta t\right)\right]\\ q_{net}=G-\left[G_{ref}+E+h\left(T_{plate}-T_{air}\right)\right]\\ q_{net}=2500-\left[500+1200+15\left(500-400\right)\right]\\ q_{net}=-700\frac{W}{m^2}\end{array}$

Therefore, the value of the net heat transfer rate per unit area is, $700\frac{W}{m^2}$.