In a solar water heater, energy from the Sun is gathered by water that circulate

Question

In a solar water heater, energy from the Sun is gathered by water that circulates through tubes in a rooftop collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes; this water is pumped into a holding tank. Assume that the efficiency of the overall system is 20% (that is, 80% of the incident solar energy is lost from the system). What collector area is necessary to raise the temperature of 200 L of water in the tank from 20°C to 50°C in 1.0 h when the intensity of incident sunlight is 600 W/m2? m2

Explanation / Answer

If P0 = power absorbed, this is

P0 = W0 / t where work absorbed during t is

W0 = CmT (C = specific heat, m = mass, T = temp change)

Also,
P = solar radiation incident
P0 = EP where E = efficiency

So, P = 1/E * CmT/t

= 1/0.20 * (4.18)(200*10^3)(30) / (3600)

= 3.483 * 10^4 Watts

Since Intensity is I = P/A

So, A = P/I
= 3.483 * 10^4 / 600
= 58.05 m^2

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