In a solar water heater, energy from the Sun is gathered by water that circulate
ID: 1896591 • 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 29.0% (that is, 71% of the incident solar energy is lost from the system). What collector area (in square meters) is necessary to raise the temperature of 400 L of water in the tank from 18Explanation / Answer
total heat required to raise the temperature of water is :
(Q=M*C*Delta T)
where M is the mass of water, (M= ho*V)
volume is 400 L = 400*103 cm3
Thus, (M = 1*400*10^3=400*10^3 g = 400 kg)
thus, (Q=400*4186*(41-18)=3.85*10^{7}J)
Now, time is 2.8h=3600*2.8=10080 s
Also, let A be the area required, then
Power = 780*A
Thus, heat from sun = Q' = Power*10080=7.86*106A Joule
since efficiency is 29%, thus, Q = 0.29Q' = 2.28*106A Joule
from above calculations, Q = 3.85*107 Joule
thus, (2.28*10^{6}A=3.85*10^{7})
thus, (A=16.8859 m^2)