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 18.0% (that is, 82% of the incident solar energy is lost from the system). What collector area is necessary to raise the temperature of 450 L of water in the tank from 17°C to 45°C in 1.8 h when the intensity of incident sunlight is 690 W/m2? The specific heat of water is 4186 J/kg·K. The density of water is 1.00 g/cm3.

Explanation / Answer

P = 1/e *C*m*dT/t

e = efficiency = 18%

C = 4.18

m= 450L = 450*10^3 m^3

dT = 45-17 = 28

t = 1.8 hr = 6480 s

P = 1/0.18 *4.18*450*10^3*28/6480 = 4.515*10^4 watts

I = P/A

A = P/I = 4.515*10^4 /690 = 65.44 m^2

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