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MJ m-2 d-1 conversion to Wm-2

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  • MJ m-2 d-1 conversion to Wm-2

    I need to convert MJ m-2 d-1 c to Wm-2 for comparing incoming radiation, does anyone have a formula for that?

    Thanks

  • #2
    Re: MJ m-2 d-1 conversion to Wm-2

    Hello Poster,

    I am assuming the letter 'c' in your question was an extraneous typo.

    Code:
     1 MJ m[SUP]-2[/SUP] d[SUP]-1[/SUP] = 1000000 J m[SUP]-2[/SUP] d[SUP]-1[/SUP] / 86400 s d[SUP]-1[/SUP]
                 = 11.574 J m[SUP]-2[/SUP] s[SUP]-1[/SUP]
                 = 11.574 W m[SUP]-2[/SUP]
    The last step makes use of the definition 1 W = 1 J s-1.

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    • #3
      Re: MJ m-2 d-1 conversion to Wm-2

      I need to do a similar conversion and am feeling over my head. First of all the conversion factor I found for the question was 0.864MJ m-2 d-1 while your conversion factor yielded 0.0864. Which is correct? Obviously I am trying to convert from Wm-2 to MJ m-2 d-1. What if I need to calculate it per hour? Excuse me if I am being overcautious. Need a good grade. Do I just multiply by 24? This coarse has me a bit overwhelmed and I am trying my best. Please help.

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      • #4
        Re: MJ m-2 d-1 conversion to Wm-2

        Originally posted by Unregistered View Post
        I need to do a similar conversion and am feeling over my head. First of all the conversion factor I found for the question was 0.864MJ m-2 d-1 while your conversion factor yielded 0.0864. Which is correct? Obviously I am trying to convert from Wm-2 to MJ m-2 d-1. What if I need to calculate it per hour? Excuse me if I am being overcautious. Need a good grade. Do I just multiply by 24? This coarse has me a bit overwhelmed and I am trying my best. Please help.
        Assuming the watts are constant, there are clearly 86400 seconds in a day (24*60*60), and "mega" means 10^6

        So watts/meter² x 86400 s/day x 1 MJ/10^6 J = MJ/(meter²*day)

        However, this is only valid if the watts are constant or have been properly averaged for the 24-hour period. In the case of solar power, peak power at high noon times 86400 s/day is nowhere near the correct solar energy for the day. energy is the integral of P(t)dt, where P(t) is the instantaneous power varying throughout the day.

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        • #5
          Re: MJ m-2 d-1 conversion to Wm-2

          [QUOTE= I have used .66 W/m2 for 20 hrs how to convert it value in to KJ/m2 and Mj/m2.

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          • #6
            Re: MJ m-2 d-1 conversion to Wm-2

            [QUOTE=Unregistered;96329][QUOTE= I have used .66 W/m2 for 20 hrs how to convert it value in to KJ/m2 and Mj/m2.[/QUOTE]

            0.66 W/m² x 20 h x 3600 s/h = 47.52 kJ/m² or 0.04752 MJ/m²

            1 J = 1 W x 1 s

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            • #7
              Re: Convert metric tonne to litre

              Quantity :02.200 ( metric tonne ) change to litre .

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              • #8
                Hi I'm currently trying to convert W m^-2 to MJ m^-2 d^-1 for radiation. I've read the reply above but still not quite sure I understand how to convert.
                If my original was 25.303 W m^-2 what would I do to the value to convert it to MJ m-2 d-1?
                Thanks

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                • #9
                  Just following on from that last post am I right in thinking to convert from W m-2 to MJ m-2 d-1 you multiply the number by 24*60*60 and then divide by 10^6?

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                  • #10
                    Originally posted by turnka View Post
                    Just following on from that last post am I right in thinking to convert from W m-2 to MJ m-2 d-1 you multiply the number by 24*60*60 and then divide by 10^6?
                    Yes, IF the power is constant for 24 h. If you are looking at solar radiation, please read the "However" note in post 4. The sun sets at night, and because the angle varies during the day the total energy is considerable less than assuming the "high noon" value for even the daylight hours. Using the 24 h day assumes the power is constant for 24 h. The site data in MJ/day is based on carrying out a proper integral of the instantaneous values (including zero watts all night). Similar problems exist with wind power because the wind is never constant for 24 h/day.

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                    • #11
                      Hi, thank you for your reply!
                      I am looking at solar radiation, so therefore the power is not constant.
                      I've seen you've wrote about using P(t)dt, what exactly does this stand for/how do I use it?
                      Thanks

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                      • #12
                        Originally posted by turnka View Post
                        Hi, thank you for your reply!
                        I am looking at solar radiation, so therefore the power is not constant.
                        I've seen you've wrote about using P(t)dt, what exactly does this stand for/how do I use it?
                        Thanks
                        It is the integrand of an integral. If you have not had calculus, you could approximation it with a summation instead. You need a plot of the power P(t) as a function of time during the day. Chop it up in many small samples which are fairly constant over their duration. Multiply the approximately constant power by the duration and sum.

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