Question

Consider two rectangular surfaces perpendicular to each other with a common edge which is \(1.6 \mathrm{~m}\) long. The horizontal surface is $0.8 \mathrm{~m}\( wide, and the vertical surface is \)1.2 \mathrm{~m}$ high. The horizontal surface has an emissivity of \(0.75\) and is maintained at $450 \mathrm{~K}\(. The vertical surface is black and is maintained at \)700 \mathrm{~K}$. The back sides of the surfaces are insulated. The surrounding surfaces are at \(290 \mathrm{~K}\) and can be considered to have an emissivity of \(0.85\). Determine the net rate of radiation heat transfer between the two surfaces and between the horizontal surface and the surroundings.

Short answer

#Answer# To determine the net rate of radiation heat transfer between the horizontal surface and the vertical surface and between the horizontal surface and the surroundings, we first calculated the radiation heat transfers emitted by the surfaces using the Stefan-Boltzmann Law and the given emissivities. We then calculated the absorbed radiation heat transfer by each surface, followed by the calculation of the net radiation heat transfer between the two surfaces. Finally, we determined the net radiation heat transfer between the horizontal surface and its surroundings.

Step-by-step solution

Calculate the radiation heat transfers emitted by the surfaces

We need to first calculate the radiation heat transfers emitted by the horizontal and vertical surfaces. We can use the Stefan-Boltzmann Law and the emissivities given to do this. For the horizontal surface: \(q_{H,emitted} = \epsilon_H * \sigma * T_H^4\) For the vertical surface: \(q_{V,emitted} = \epsilon_V * \sigma * T_V^4\) where \(\epsilon_H = 0.75\), \(T_H = 450 K\), \(\sigma = 5.67\times10^{-8} W/m^2 K^4\) (Stefan-Boltzmann constant), \(\epsilon_V = 1\), and \(T_V = 700 K\).

Calculate the absorbed radiation heat transfer

Next, we need to calculate the absorbed radiation heat transfer by each surface. The horizontal surface absorbs the radiation emitted by the vertical surface and vice versa. For the horizontal surface: \(q_{H,absorbed} = q_{V,emitted}\) For the vertical surface: \(q_{V,absorbed} = q_{H,emitted}\)

Calculate the net radiation heat transfer between the two surfaces

Now, we can calculate the net radiation heat transfer between the two surfaces. \(q_{net,HV} = q_{H,emitted} - q_{H,absorbed}\)

Calculate the net radiation heat transfer between the horizontal surface and the surroundings

Finally, we will determine the net radiation heat transfer between the horizontal surface and its surroundings. For the horizontal surface: \(q_{H,surroundings} = \epsilon_s * \sigma * T_s^4\) where \(\epsilon_s = 0.85\) and \(T_s = 290 K\). \(q_{net,HS} = q_{H,emitted} - q_{H,surroundings}\) Now, we have the net radiation heat transfer between the two surfaces and between the horizontal surface and the surroundings.