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Chapter 12: Problem 82

An opaque horizontal plate is well insulated on the edges and the lower surface. The irradiation on the plate is \(3000 \mathrm{~W} / \mathrm{m}^{2}\), of which \(500 \mathrm{~W} / \mathrm{m}^{2}\) is reflected. The plate has a uniform temperature of \(700 \mathrm{~K}\) and has an emissive power of \(5000 \mathrm{~W} / \mathrm{m}^{2}\). Determine the total emissivity and absorptivity of the plate.

Short Answer

Expert verified
Answer: The total emissivity of the plate is 0.375 or 37.5%, and the absorptivity of the plate is 0.833 or 83.3%.

Step by step solution

01

Find the amount of absorbed radiation.

The total energy incident on the plate per unit area is the irradiation, which is given as 3000 W/m^2. Also, we are given that 500 W/m^2 is reflected. So, the amount of radiation absorbed by the plate can be found by subtracting the incident radiation from the reflected radiation. Amount of absorbed radiation = Incident radiation - Reflected radiation Amount of absorbed radiation = \(3000 \ \text{W/m}^2 - 500 \ \text{W/m}^2 = 2500 \ \text{W/m}^2\)
02

Determine the absorptivity of the plate.

Using the absorbed radiation, we can find the absorptivity (α) of the plate. Absorptivity is the ratio between the amount of absorbed radiation and the amount of incident radiation. Absorptivity (α) = (Amount of absorbed radiation) / (Incident radiation) α = 2500 W/m^2 / 3000 W/m^2 α = 0.833 The absorptivity of the plate is 0.833 or 83.3%.
03

Determine the total emissivity of the plate.

We are given that the emissive power of the plate is 5000 W/m^2, and the plate has a uniform temperature of 700 K. To find the emissivity, we will use the Stefan-Boltzmann law that relates the emitted radiation to temperature: E = εσT^4 Here, ε is the emissivity, σ is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \ \text{W/m}^2 \text{K}^4\)), and T is the temperature. We are given E = 5000 W/m^2 and T = 700 K. We can now solve for ε: ε = E / (σT^4) ε = \((5000 \ \text{W/m}^2) / (5.67 \times 10^{-8} \ \text{W/m}^2 \text{K}^4 \times (700 \ \text{K})^4)\) ε = 0.375 The total emissivity of the plate is 0.375 or 37.5%.

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