March 24, 2017

Post a New Question

Posted by on .

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 28.0% (that is, 72% of the incident solar energy is lost from the system). What collector area (in square meters) is necessary to raise the temperature of 450 L of water in the tank from 18°C to 50°C in 2.8 h when the intensity of incident sunlight is 520 W/m2? The specific heat of water is 4186 J/kg·K. The density of water is 1.00 g/cm3.

  • physics - ,

    The intensity of sun light falling on the earth = I'
    Let area of the plate = A
    Power received by the plate, P = I'A
    Efficiency of the over all system = 28 %
    The power given to the tank is given by
    P = 26% * I' A = 0.28 I'A...........(1)
    m= mass of 450 L of water
    S =specific heat of water = 4186 J/kg·K
    4.186 J/g.K
    ΔT = temparature difference = 32 deg C
    The energy needed to raise the temp of water from 18 to 50 C is

    Q = mSΔT = J
    =60278400 J
    power in put needed
    P ' = Q / 1.0 * 60*60 sec
    P =60278400 / 2.8*60*60 sec
    P' = 5980 W -----(2)
    From the equations (1) and (2) we have
    0.28 I'A = 5980
    A =5980/(0.28*520)
    A = 41 m^2

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question