In a solar water heater, energy from the Sun is gathered by water that circulate
ID: 1635785 • Letter: I
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 24.0% (that is, 76% of the incident solar energy is lost from the system). What collector area is necessary to raise the temperature of 460 L of water in the tank from 24°C to 47°C in 1.1 h when the intensity of incident sunlight is 650 W/m2? The specific heat of water is 4186 J/kg·K. The density of water is 1.00 g/cm3.
Explanation / Answer
we know that
intensity is I = P/A
collector area is A = P/I
where P is the power absorbed
P = W/t
W is the work done in time t
but W = S*m*dT
S is the specific heat capacity of water
m is mass of water = density * volume = rho*V = 1000*460*10^-3 = 460 Kg
dT = 47-24 = 23 deg C
then
W = S*m*dT = 4186*460*23 = 4.43*10^7 J
then P = W/t = (4.43*10^7)/(1.1*60*60) = 1.12*10^4 W
efficiency is e = power output / input power
input power is P_in = I*A = 600*A
Power output is P_out = 1.12*10^4
e = 0.24
then
0.24 = (1.12*10^4)/(600*A)
A = 78 m^2 is the answer