Chapter 19: Q. 68 (page 546)

You would like to put a solar hot water system on your roof, but you're not sure it's feasible. A reference book on solar energy shows that the ground-level solar intensity in your city is $800 W / m^{2}$ for at least 5 hours a day throughout most of the year. Assuming that a completely black collector plate loses energy only by radiation, and that the air temperature is $20 ° C$ , what is the equilibrium temperature of a collector plate directly facing the sun? Note that while a plate has two sides, only the side facing the sun will radiate because the opposite side will be well insulated.

Short Answer

Expert verified

The equilibrium temperature of a collector plate directly facing the sun is $383 K .$

Step by step solution

Given Information

Solar Intensity in the city $= 800 W / m^{2}$

Number of hours $= 5 h r s$

Temperature in the air $= 20^{\circ} C$

Explanation

The power radiated from a body at temperature $T$ when the surroundings is at temperature $T_{a}$ will be given by

$P = A ϵ σ (T^{4} - T_{a}^{4})$

We can therefore express the temperature of the radiating object as

$A ϵ σ T^{4} = P + A ϵ σ T_{a}^{4}$

Assuming an emissivity of 1 and simplifying by the area and Stefan-Boltzmann constant, we have

$T = \sqrt[4]{\frac{P}{A σ} + T_{a}^{4}}$

Given our numerical case when $T_{a} = 293 K$ , and remembering that the value of the Stefan-Boltzmann constant is $σ = 5.67 \cdot 10^{- 8} \frac{W}{m^{2} K^{4}}$ , we have

$T = \sqrt[4]{\frac{800}{1 \cdot 5.67 \cdot 10^{- 8}} + 293^{4}} = 383 K$

Final Answer

Therefore, the equilibrium temperature of a collector plate directly facing the sun is $383 K .$

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