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solar radiation on a cylinder

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  • solar radiation on a cylinder

    I am making a batch water heater out of an old immersion heater.

    Does anyone know the mathematical formula for calculating the solar radiation in watts.

    I would assume that the cylinder is in the perfect perpendicular alignment to the sun.

    We know the radius and the length. The tank is circular.

  • #2
    Wouldn't it be the cross-sectional area of the times times 1000 watts per square meter times the effeciency rating of the tank?

    Dia x length x 1000 x ??% = energy input

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    • #3
      You would actually want to work your solution in BTUs but you would need to know quite a few thing to work it out and have to use additional formulas to figure those out. I.E. if you desire to know the time it would take to heat it up, you'd have to know the beginning and ending temps, heat transfer rate and any flow rates but I'm not sure exactly what you're trying to figure out.
      1150W, Midnite Classic 200, Cotek PSW, 8 T-605s

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      • #4
        Originally posted by thastinger View Post
        You would actually want to work your solution in BTUs but you would need to know quite a few thing to work it out and have to use additional formulas to figure those out. I.E. if you desire to know the time it would take to heat it up, you'd have to know the beginning and ending temps, heat transfer rate and any flow rates but I'm not sure exactly what you're trying to figure out.
        If a flat panel points directly at the sun it gets around 1000 watts per square meter.

        But a cylinder is more complicated.
        The bit that faces the sun would get more solar radiation than the area 90 degrees around the cylinder.

        I thought that someone might know a formula.

        If I knew the wattage then I could start to calculate the amount of time required to increase the temp.
        The sort of calculation that is used here.
        http://crazycalculations.com/temp_increase/

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        • #5
          Originally posted by cortijo View Post
          If a flat panel points directly at the sun it gets around 1000 watts per square meter.

          But a cylinder is more complicated.
          The bit that faces the sun would get more solar radiation than the area 90 degrees around the cylinder.

          I thought that someone might know a formula.

          If I knew the wattage then I could start to calculate the amount of time required to increase the temp.
          The sort of calculation that is used here.
          http://crazycalculations.com/temp_increase/
          There are a lot of formulas, etc. Unfortunately, they are more complicated than space or this format allows. On the other hand and fortunately, folks have been building breadbox heaters successfully for a long time using common sense and watching the movements of the sun. It ain't rocket science.

          For solar radiation purposes, the projected area is what is used. So, for a .5 m dia., by 2 m long cylinder, the projected area is .5 X 2.0 = 1.0 m^2.

          If the cylinder is painted flat black, the absorption of solar radiation will be "mostly" constant until the local angle of incidence on the cylinder is about 70+ degrees or so.

          That 1,000 W/m^2 is a probable max. Over an average day, expect solar flux to be something like, very roughly, 600 to 800 Watts/m^2 per hour in the plane of the cylinder. and expect that, again, very roughly, about 40 to 70% of the daylight hours.

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