Question

Two rectangular surfaces both having 1 m² area are placed perpendicular to each other with a common edge. One surface is hot, having a temperature of 1000 K and emissivity of 0.4, while the other is insulated and in radiant balance with a large surrounding room at 300 K. If the fraction of radiation leaving the hot surface which reaches the cold surface is 0.2, then the equivalent overall resistance for the radiation heat loss from the hot surface is _______ m^-2 (rounded off to 2 decimal places).

Hint:

Model this as a three-surface radiation network where the perpendicular surface is a reradiating node.

Answer 1

Correct Answer: 2.54

Step 1: Identify the three-surface radiation network.
Surface 1 is the hot surface (\(A_1=1\) m\(^2\), \(\varepsilon_1=0.4\)), surface 2 is the perpendicular reradiating surface that is insulated so its net heat flux is zero, and surface 3 is the large surrounding room, which acts like a black body at fixed radiosity.

Step 2: Get the view factors.
Given \(F_{12}=0.2\). Since \(A_1=A_2=1\) m\(^2\), reciprocity gives \(F_{21}=F_{12}=0.2\). All remaining radiation from each surface must reach the room, so \(F_{13}=1-F_{12}=0.8\) and \(F_{23}=1-F_{21}=0.8\).

Step 3: Write the surface and space resistances.
Surface resistance of 1: \(R_1=\dfrac{1-\varepsilon_1}{\varepsilon_1A_1}=\dfrac{0.6}{0.4}=1.5\) m\(^{-2}\).
Space resistances: \(R_{12}=\dfrac{1}{A_1F_{12}}=\dfrac{1}{0.2}=5\) m\(^{-2}\), \(R_{13}=\dfrac{1}{A_1F_{13}}=\dfrac{1}{0.8}=1.25\) m\(^{-2}\), \(R_{23}=\dfrac{1}{A_2F_{23}}=\dfrac{1}{0.8}=1.25\) m\(^{-2}\). Surface 3's own resistance is essentially zero since it is a very large surface.

Step 4: Combine the network.
Because surface 2 carries no net heat flow, the path through it (\(R_{12}\) in series with \(R_{23}\)) acts in parallel with the direct path \(R_{13}\): \(R_{12}+R_{23}=5+1.25=6.25\) m\(^{-2}\), and \(R_{parallel}=\dfrac{(6.25)(1.25)}{6.25+1.25}=\dfrac{7.8125}{7.5}=1.04\) m\(^{-2}\).

Step 5: Add the surface resistance of the hot surface.
\(R_{total}=R_1+R_{parallel}=1.5+1.04=2.54\) m\(^{-2}\).

Final Answer:
This total resistance, from the hot surface all the way to the surrounding room, sets the radiative heat loss once the temperature difference is known. \[ \boxed{R_{total} = 2.54 \ \text{m}^{-2}} \]