Question

Chapter 18, Problem 029 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 14.0% (that is, 86% of the incident solar energy is lost from the system). What collector area is necessary to raise the temperature of 340 L of water in the tank from 24°C to 49°C in 1.6 h when the intensity of incident sunlight is 550 W/m2? The specific heat of water is 4186 J/kg.K. The density of water is 1.00 g/cm3. Number Units the tolerance is +/-2%

Answer 1

Specific heat of water is 4187 JKG^-1 k^-1 So, the energy required to change temperature is, Q = E_m(T_2 - T_1) Density of water is S = 1 kg/l So, mass of water, m = 340 times 1 m = 340 kg therefore energy required to change temperature in Q_r = 1487 times 340 times (49 - 24) 12.6395 times 10^6 J This is the 14% of total energy . Hence the total energy is Q_tot = 12.6395 times 10^6/14 times 100J Q_tot = 90.282 times 10^6 J so, the rate at which the solar energy should fall is dQ_tot/dt = 90.282 times 10^6/5760 w [1.6 hr = 5760 sec] 15.673 times 10^3 W so, the required surface area is A = 15.673 times 10^3/550 A = 28.496 m^2