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Chapter 12: Problem 60

A small circular plate with a surface \(A_{1}\) can be approximated as a blackbody. A radiometer with a surface \(A_{2}\) is placed normal to the direction of viewing from the plate at a distance \(L\). Would the irradiation on the radiometer from the plate decrease if the distance \(L\) is doubled, and if so, by how much?

Short Answer

Expert verified
Answer: Yes, the irradiation on the radiometer decreases when the distance L is doubled. It decreases by a factor of 1/4.

Step by step solution

01

Determine the irradiance at the original distance L

Find the irradiance E₁ received by the radiometer from the plate at the original distance L using the equation E₁ = σ * T^4 * (A₁ * A₂) / (4 * π * L^2).
02

Determine the irradiance at the doubled distance 2L

Find the irradiance E₂ received by the radiometer from the plate at the doubled distance 2L using the equation, E₂ = σ * T^4 * (A₁ * A₂) / (4 * π * (2L)^2).
03

Compare the irradiance at the original and doubled distances

Determine if the irradiance decreases by finding the ratio of E₂ to E₁. If the ratio is less than 1, then the irradiance has decreased. Calculate the ratio as follows: E₂ / E₁ = {σ * T^4 * (A₁ * A₂) / (4 * π * (2L)^2)} / {σ * T^4 * (A₁ * A₂) / (4 * π * L^2)}.
04

Simplify the ratio

Simplify the ratio E₂ / E₁ by canceling out the common terms: E₂ / E₁ = (L^2) / (4L^2).
05

Calculate the factor by which the irradiance decreases

Calculate the factor by which the irradiance decreases when the distance L is doubled: Factor = (L^2) / (4L^2) = 1/4. So, the irradiation on the radiometer from the plate decreases by a factor of 1/4 when the distance L is doubled.

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Most popular questions from this chapter

Consider a surface at \(500 \mathrm{~K}\). The spectral blackbody emissive power at a wavelength of \(50 \mu \mathrm{m}\) is (a) \(1.54 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\) (b) \(26.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\) (c) \(108.4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}(d) 2750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}(e) 8392 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\)

A small circular plate has a diameter of \(2 \mathrm{~cm}\) and can be approximated as a blackbody. To determine the radiation from the plate, a radiometer is placed normal to the direction of viewing from the plate at a distance of \(50 \mathrm{~cm}\). If the radiometer measured an irradiation of \(85 \mathrm{~W} / \mathrm{m}^{2}\) from the plate, determine the temperature of the plate.

A surface is exposed to solar radiation. The direct and diffuse components of solar radiation are 350 and \(250 \mathrm{~W} / \mathrm{m}^{2}\), and the direct radiation makes a \(35^{\circ}\) angle with the normal of the surface. The solar absorptivity and the emissivity of the surface are \(0.24\) and \(0.41\), respectively. If the surface is observed to be at \(315 \mathrm{~K}\) and the effective sky temperature is \(256 \mathrm{~K}\), the net rate of radiation heat transfer to the surface is (a) \(-129 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(-44 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(0 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(129 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(537 \mathrm{~W} / \mathrm{m}^{2}\)

When the variation of spectral radiation quantity with wavelength is known, how is the corresponding total quantity determined?

The electromagnetic spectrum that lies between \(0.40\) and \(0.76 \mu \mathrm{m}\) is what we call visible light. Within this spectrum, the color violet has the shortest wavelength while the color red has the longest wavelength. Determine which of these colors, violet \((\lambda=0.40 \mu \mathrm{m})\) or red \((\lambda=0.76 \mu \mathrm{m})\), propagates more photon energy.
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