Question

A row of tubes, equally spaced at a distance that is twice the diameter of the tubes, is positioned between two large parallel plates. The surface temperature of the tubes is constant at \(10^{\circ} \mathrm{C}\) and the top and bottom plates are at constant temperatures of \(100^{\circ} \mathrm{C}\) and \(350^{\circ} \mathrm{C}\), respectively. If the surfaces behave as blackbody, determine the net radiation heat flux leaving the bottom plate.

Short answer

Question: Determine the net radiation heat flux leaving the bottom plate. Answer: The net radiation heat flux leaving the bottom plate is -22795 W/m². The negative sign indicates that the heat flux is entering the plate, or being absorbed by it.

Step-by-step solution

Identify the known parameters and the equations to determine the radiation heat flux

First, we have to identify the known parameters and the equations related to the problem: Top Plate temperature, \(T_1 = 100^\circ \mathrm{C} = 373 \mathrm{K}\) Bottom Plate temperature, \(T_2 = 350^\circ \mathrm{C} = 623 \mathrm{K}\) Tube temperature, \(T_tube = 10^\circ \mathrm{C} = 283 \mathrm{K}\) As the surfaces behave as blackbodies, we will use the Stefan-Boltzmann law, which states that the radiation heat flux is given by: \(q = \sigma \times (T_1^4 - T_2^4)\) Where \(\sigma = 5.67 \times 10^{-8} \mathrm{W \cdot m^{-2} \cdot K^{-4}}\) is the Stefan-Boltzmann constant. Before we move forward to calculate the net radiation heat flux, we need to consider the geometry of the problem and how the view factors affect the heat flux calculation.

Calculate the view factors

We need to determine the view factors for the top and bottom plates to the tubes. Since the tubes are equally spaced with a distance that is twice their diameter, we will make use of the following view factors formula for equally spaced parallel circles: \(F_{1 \to tube} = F_{2 \to tube} = \frac{1}{2}\) This means that half of the radiation leaving the top and bottom plate will be intercepted by the tubes. Now we can consider the view factors in our heat flux calculation.

Determine the individual heat fluxes

We are now ready to calculate the individual heat fluxes between the top, bottom, and tubes. 1. Heat flux between top plate and tubes (with a view factor \(F_{1 \to tube} = \frac{1}{2}\)): \(q_{1 \to tube} = \frac{1}{2} \times \sigma \times (T_1^4 - T_{tube}^4) = \frac{1}{2} \times 5.67 \times 10^{-8} \times (373^4 - 283^4) = 1249.55 \, \mathrm{W/m^2}\) 2. Heat flux between tubes and bottom plate (with a view factor \(F_{tube \to 2} = \frac{1}{2}\)): \(q_{tube \to 2} = \frac{1}{2} \times \sigma \times (T_{tube}^4 - T_2^4) = \frac{1}{2} \times 5.67 \times 10^{-8} \times (283^4 - 623^4) = -24044.55 \, \mathrm{W/m^2}\) Here negative heat flux means it is absorbed by the tubes.

Calculate the net radiation heat flux

Now we can find the net radiation heat flux leaving the bottom plate: \(q_{net} = q_{1 \to tube} + q_{tube \to 2} = 1249.55 - 24044.55 = -22795 \, \mathrm{W/m^2}\) The net radiation heat flux leaving the bottom plate is -22795 W/m². The negative sign indicates that the heat flux is entering the plate, or being absorbed by it.