Question

Two infinitely long parallel plates of width \(w\) are located at \(w\) distance apart, as shown in Fig. P13-51. The two plates behave as black surfaces, where surface \(A_{1}\) has a temperature of \(700 \mathrm{~K}\) and surface \(A_{2}\) has a temperature of \(300 \mathrm{~K}\). Determine the radiation heat flux between the two surfaces.

Short answer

Question: Calculate the radiation heat flux between two infinitely long parallel plates with different temperatures, given the width of the plates (w) and the temperatures: \(T_{1} = 700 \ \text{K}\) and \(T_{2} = 300 \ \text{K}\). Answer: The radiation heat flux between the two surfaces, per unit length, is given by: \(q = 2.29 \times 10^{-1} \ \text{W} \cdot \text{m}^{-1} \cdot w\) This equation represents the radiation heat flux between the two surfaces of the parallel plates, which will vary based on the width (w) of the plates.

Step-by-step solution

Identify the given data

We are given the following information: - Width of plates (w) - Distance between plates (also w) - Temperature of plates: \(T_{1} = 700 \ \text{K}\) and \(T_{2} = 300 \ \text{K}\)

Apply the Stefan-Boltzmann Law

The Stefan-Boltzmann law states that the radiation heat flux \(q\) between two surfaces is given by: \(q = F_{1 \to 2} \cdot A \cdot \sigma \cdot (T_{1}^4 - T_{2}^4)\) Where: - \(F_{1 \to 2}\) is the view factor (the portion of the radiation leaving surface 1 that arrives at surface 2) - \(A\) is the area of the emitting surface (in this case, \(A_{1}\) or \(A_{2}\), since both are infinitely long parallel plates) - \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \ \text{W} \cdot \text{m}^{-2} \cdot \text{K}^{-4}\)) - \(T_{1}\) and \(T_{2}\) are the temperatures of the two surfaces in Kelvin.

Calculate the view factor

As the plates are infinitely long and parallel, the view factor \(F_{1 \to 2}\) simplifies to \(1\). This is because all the radiation leaving surface 1 will reach surface 2.

Determine the area of the emitting surface

Since the plates are infinitely long, we need to determine the area of the emitting surface per unit length (\(A_l\)) of the plates, because the problem involves the width (w) of the plates. \(A_l = w \cdot 1\)

Calculate the radiation heat flux

Now, we can plug in all the values we have into the equation from Step 2: \(q = F_{1 \to 2} \cdot A_l \cdot \sigma \cdot (T_{1}^4 - T_{2}^4)\) \(q = 1 \cdot w \cdot 5.67 \times 10^{-8} \ \text{W} \cdot \text{m}^{-2} \cdot \text{K}^{-4} \cdot (700^4 - 300^4)\) \(q = 5.67 \times 10^{-8} \ \text{W} \cdot \text{m}^{-1} \cdot \text{K}^{-4} \cdot w \cdot ( 40353607)\) Now, the radiation heat flux between the two surfaces (per unit length) is given by: \(q = 2.29 \times 10^{-1} \ \text{W} \cdot \text{m}^{-1} \cdot w\) This equation represents the radiation heat flux between the two surfaces of the parallel plates per unit length, which will vary based on the width (w) of the plates.