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I am working to build an economic model that includes hourly solar output for solar photovoltaic installations built at several different geographic locations. I have data on irradiation (Direct Normal Irradiation kj/m2, Diffuse Horizontal Irradiation kj/m2) for each site and the location of the sites (latitude, longitude). I've followed the instructions laid out in Chapter 7 of Gilbert Masters' book Renewable and Efficient Electric Power Systems to calculate the amount of irradiation striking a given panel at a given hour, expressed in kj/m2. I would now like to use this irradiation data to calculate the hourly power output (in kWh) of a 1 kilowatt solar installation. As background, I'll use this information in an optimization procedure which will calculate the optimal investment in new solar capacity for each location to provide low-cost, reliable electricity in an integrated electricity system.

Here is a sample of the hourly irradiation numbers. Can anyone tell me the calculation to turn these irradiation numbers into kWh/kw?

Irradiation 1557 kj/m2 2591 kj/m2 1494 kj/m2 987 kj/m2

I understand that it is necessary to assume an efficiency for the solar photovoltaic installation. For the purposes of this example, could you use 16%? Also, I understand that the output I am looking for is DC output. To convert to AC I recognize there will be losses associated with the inverter.

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    \$\begingroup\$ Uh, just winging a unit analysis.. \$kJ/m^2 \times panel\,area\,m^2 \times {1\,hour \over 3600s} \times 0.16\$. This does not take into account any real factors though.... \$\endgroup\$
    – Daniel
    Commented Aug 31, 2016 at 5:18
  • \$\begingroup\$ Thanks Daniel, is it necessary to assume an area for the panel? I would prefer to work in terms of kilowatts installed solar-PV. My understanding is that a kilowatt rating means that the panel produces energy at a rate of 1 kw under standard test conditions of 1000 W/m2. Do I assume the panel area for a 1 kw panel is 1 meter squared? \$\endgroup\$
    – SolarSon
    Commented Aug 31, 2016 at 5:21
  • \$\begingroup\$ Wouldn't it be 1 meter squared / 0.16? Obviously there are scaling factors involved here that I know nothing about. Good luck! \$\endgroup\$
    – Daniel
    Commented Aug 31, 2016 at 5:23
  • \$\begingroup\$ kJ/m^2 does not make any sense. kJ/(m^2*s) = kW/m^2 does or potentially kWh/(m^2*h) = kW as a convoluted way to express average power irradiation. \$\endgroup\$
    – winny
    Commented Aug 31, 2016 at 5:40
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    \$\begingroup\$ @Daniel an "hourly irradiation of kJ/m^2" is again a convoluted way of saying kJ/(m^2*hr) or throwing your readers off. \$\endgroup\$
    – winny
    Commented Aug 31, 2016 at 8:28

1 Answer 1

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There's nothing complicated here, it's no more than book-keeping, and multiplication sums.

Output = Input * efficiency, when everything is expressed consistently.

If your hourly irradiance is expressed in kJ/m2, then you need to multiply by m2 of panel. If your panels are rated in kw each at a reference irradiance, then you need to scale the panel output by the ratio of actual irraadiance to the reference.

Amongst the efficiency terms are light to electricity, oblique incidence effects, panel fouling reducing light, they may be a temperature dependence, and DC to AC conversion losses. You will get out of your simulation an accuracy that relates to the care with which you find all the efficiency terms, and account for them.

If you calculate for a single panel, at reference irradiance, you will get a nominal figure. That's rather like taking your hourly pay rate times your daily hours and arriving at a nominal income for the day.

But you have to pay income taxes, travel to and from work, buy work tools, all of which reduces your disposable income. But these are just book-keeping, and simple sums. In the same way you lose power through non-standard irradiance, oblique illumination, dirty panels, DC-AC losses. The more detailed you are keeping track of them, the more realistic power output estimate you'll have.

If your sole problem is turning kJ into kWh, then 1kWh = 3600kJ = 3.6MJ, as one hour = 3600 seconds, and 1J = 1 watt.second.

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  • \$\begingroup\$ You're assuming the panel output is linear with incident energy. Is it? \$\endgroup\$
    – Daniel
    Commented Aug 31, 2016 at 7:10
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    \$\begingroup\$ That's yet another detail to be taken into account. I rolled it into efficiency, without comment. If the OP has to implement a flux dependent efficiency, then it does get slightly more complicated than just sums. \$\endgroup\$
    – Neil_UK
    Commented Aug 31, 2016 at 8:35
  • \$\begingroup\$ This response was particularly useful, "If your panels are rated in kw each at a reference irradiance, then you need to scale the panel output by the ratio of actual irradiance to the reference". Am I right to think that I can first convert kj/m2 to kWh/m2 by dividing by 3600. For the numbers I provided above this would result in .433 kwh/m2 .72 kwh/m2 .415 kwh/m2 .274 kwh/m2. A 1 kw panel rated at 1000 W/m2 for full output would produce 43% of its capacity in the first hour, then 72%, then 41.5% then 27.4%. These capacity factors offer a way to generalize power output from any size PV unit. \$\endgroup\$
    – SolarSon
    Commented Aug 31, 2016 at 21:00

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