How to include heat input from the sun and radiative heat exchange between walls

PDF slides

The total heat $\Phi_{SW}$ brought by the sun to a surface $S$ with an absorption coefficient $\alpha$ is the sum of direct irradiance $E_{dir}$, diffuse irradiance $E_{dif}$ and reflected irradiance $E_{ref}$. Each of these variables can be calculated with a little trigonometry as functions of the angle of the wall and the position of the sun.

The unit of $\Phi_{SW}$ is [W], and not [W/m$^2$], since irradiances have been multiplied by and area $S$.

PDF slides

### Methodology

View factors $F_{ij}$ are required to solve the radiative heat exchange between walls. Here are two formulas for view factors that cover two widespread specific cases:

Two parallel planes of equal sizes: $$F_{12} = \frac{1}{\pi x y} \left[ \ln \frac{x_1^2 y_1^2}{x_1^2+y_1^2-1} + 2x\left(y_1 \arctan \frac{x}{y_1}-\arctan x \right) + 2y\left(x_1 \arctan \frac{y}{x_1}-\arctan y \right) \right]$$ $$\mathrm{with} \: x_1=\sqrt{1+x^2} \: ; \: y_1=\sqrt{1+y^2} \: ; \: x=W/H \: ; \: y=L/H$$
Two adjacent perpendicular rectangles: $$F_{12} = \frac{1}{\pi w} \left[ h \arctan \left(\frac{1}{h} \right) + w \arctan \left(\frac{1}{w} \right) - \sqrt{h^2+w^2} \arctan \left(\frac{1}{\sqrt{h^2+w^2}} \right) + \frac{1}{4} \ln \left( a \, b^{w^2} \, c^{h^2}\right) \right]$$ $$\mathrm{with} \: a = \frac{(1+h^2)(1+w^2)}{1+h^2+w^2} \: ; \: b = \frac{w^2(1+h^2+w^2)}{(1+w^2)(h^2+w^2)} \: ; \: c = \frac{h^2(1+h^2+w^2)}{(1+w^2)(h^2+w^2)}$$ $$h=H/L \: ; \: w=W/L$$

The problem can be solved either by using an equivalent electrical circuit as shown on the video, or by solving a linear system of equations with the method shown below.

Either the net heat flux or the temperature of each wall $i$ should be known. For each wall, write one of the following equations:

• If the temperature $T_i$ is known :
• If the net heat flux $\Phi_i$ is known :

We can write a system of equations of the type $Ax=b$ where $A$ is a $n \times n$ matrix ($n$=number of surfaces) and $x$ is the vector of radiosities. Once the system is solved and the radiosities are known, the following formula can be used to calculate the temperatures and heat fluxes that were initially unknown:

An example of this methodology is proposed in the following exercise.

## Exercise

Consider this room:

• The wall $S_5$ (on the left) is a window with a temperature of $T_5 = 8°C$.
• The surface $S_0$ is a radiator covering half the height of the wall on the right. Its temperature is $T_0 = 60°C$.
• The surface $S_2$ (the ground) is adiabatic.
• All other surfaces have a temperature of $T = 20°C$.

Calculate the radiative heat loss through the window, the net heat flux given by the radiator and the temperature of the ground. The emission coefficient of all surfaces is $\varepsilon=0.85$

All the walls that have a temperature of $T = 20°C$ can be considered as a single surface. The question is therefore equivalent to heat exchange balance between 4 surfaces: we need to write 4 equations.

First, the view factors are required. The formulas shown above only need to be used three times, to find the following values:

All other view factors can be calculated from the simple formulas: $S_iF_{ij}=S_jF_{ji}$ et $\sum_j F_{ij}=1$.

We then need to write one equation for each surface. Surface 0 (the radiator) has a known temperature, which means:

Surface 2 (the floor) has a known heat flux of zero since it is adiabatic:

Surfaces 3 and 5 have the same kind of boundary condition as surface 2 (prescribed temperature): their equations are therefore similar. We end up with the following linear system of equations for the radiosities:

Solving this system results in the following values in W/m$^2$:

The last step is to use this formula:

in order to find the target values of the exercise: floor temperature $T_2$, radiator heat flux $\Phi_0$ and net heat flux on the window $\Phi_5$: