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 13.0% (that is, 87% of the incident solar energy is lost from the system). What collector area (in square meters) is necessary to raise of the temperature 240 L of water in the tank from 22°C to 35°C in 2.8 h when the intensity of incident sunlight is 600 W/m2? The specific heat of water is 4186 J/kg·K. The density of water is 1.00 g/cm3.
Explanation / Answer
Let P_o be the power absorbed is P_0 = W_0 / t Work absorbed during time t is W_0 = c m t Let P is the incident solar radiation P = e P where e is the efficiency P = 1/ e * c m t / t = 1/ 0.13 * 4.186 J / g. K ( 240 * 10^3 cm^3 )( 35C - 22C ) / 2.8 *60*60 = 9.96*10^3 W Intensity is I = P / A collector Area A = P / I = 9.96*10^-3 W / 600 W / m^2 = 16.6 m^2