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Water heater and area of sunray gatherer

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A water heater gathers energy with the help of solar collectors on the roof, where water circulates in pipes and are provided heat from the sunrays that get through a transparent cover. The water is then pumped down in a tank inside the house.

    What area must the gatherer have to warm the water in a 200 L (litres) tank from 20 deg. Celsius to 40 deg. Celsius in one hour, if we assume that the system has a coefficient of performance of 20%? (80% of the energy in sunlight is lost, or: ends up other places than the water heater.)

    [Answer: 33 m^2]


    2. Relevant equations

    Given: the intensity of the incoming sunlight is 700 W/m^2 and the specific heat capacity of water is C_water = 4186 J kg^(-1) K^(-1).


    3. The attempt at a solution

    Q_water = C_water * m_water * ΔT = 4186 J kg^(-1) * 200 kg * 20 K = 16.7 MJ


    I know coefficient of a heat pump given by: η = abs(ΔQ)/ΔW

    Right now I'm a little stuck here. Thanks for any help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 29, 2009 #2

    LowlyPion

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    Homework Helper

    You figured the total energy that needs collecting.

    You have 1 hour or 3600 seconds.

    And you can collect at 20% of 700 w/m2 every second right?
     
  4. Mar 29, 2009 #3
    Hi there!

    ur on the right track by finding the energy required to raise the temperature of the water by 20 degrees celsius. next remember that the power required to raise the temp of the water is equal to the integral of the intensity of light hitting the solar collectors with respect to the area of the solar collectors. Assuming the intensity of the light over the solar collectors to be constant we can say:

    A=P/lIl (the lines on either side of the I represent the modulus)

    Power is the energy transferred over the time taken, thus the equation becomes:

    A=E/(I*t)

    You have already worked out E (the energy required to raise the temp of the water) and the intensity of the light will be 700*0.2 (because only 20% of the light is reaching the solar collectors. Substituting in values:

    A=(16.7*10^6)/((700*0.2)*(60*60))
    A=33m^2

    hope that helps
     
  5. Mar 30, 2009 #4
    Thanks a lot for the help, both of you.
     
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