Physics Help Forum Finding area of solar collector

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 Mar 1st 2018, 05:05 PM #1 Junior Member   Join Date: Feb 2018 Posts: 8 Finding area of solar collector Hi, I was wondering if someone could help me with this question: The total electric generating capacity of Canada is 81GW. What is the area (in km^2) of horizontal photovoltaic solar collector that could satisfy this need? Note the average insolation on a horizontal surface is 4.8kWh/m^2 for a 24 hour period. Assume the solar cells have an efficiency of 19%
 Mar 1st 2018, 05:33 PM #2 Senior Member   Join Date: Apr 2017 Posts: 498 I think the rules are here you're suppose to show some sort of attempt at the question ... But I'll give you a bit of a push start ... That 'isolation" figure .... Note it's in KWhrs .... that's a measure of Energy .... KW is a measure of Power ... I hope you understand the difference. We are told "average insolation on a horizontal surface is 4.8kWh/m^2 for a 24 hour period" ... that means the Energy received for each square meter is 4.8 KWhrs over a 24 hr period ... If the panel was 100% efficient you could run a 2Kw kettle for 2.4Hrs .... once a day, if the panel was 1 sq m. and laid horizontally. Last edited by oz93666; Mar 1st 2018 at 05:36 PM.
 Sep 20th 2018, 06:12 AM #3 Junior Member   Join Date: Sep 2016 Location: London Posts: 21 I don't have an answer to this post. Sorry about that.
Sep 20th 2018, 06:59 AM   #4
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Join Date: Apr 2017
Posts: 498
 Originally Posted by oz93666 I think the rules are here you're suppose to show some sort of attempt at the question ... But I'll give you a bit of a push start ... That 'isolation" figure .... Note it's in KWhrs .... that's a measure of Energy .... KW is a measure of Power ... I hope you understand the difference. We are told "average insolation on a horizontal surface is 4.8kWh/m^2 for a 24 hour period" ... that means the Energy received for each square meter is 4.8 KWhrs over a 24 hr period ... If the panel was 100% efficient you could run a 2Kw kettle for 2.4Hrs .... once a day, if the panel was 1 sq m. and laid horizontally.
Continuing .... @19% efficiency one square meter generates 912 Watt Hrs in one 24 hr period

We are told the generating capacity of Canada is 81GW , that means if everything is producing the output is 81GW ... at night perhaps output is only 20GW ...

lets gestimate the average output is 50GW that's 24x50 =1200GWHrs in a 24 Hr period

1200,000,000,000 / 912 = 13,000,000,000 square meters . 13K sq km ..

So a square piece of land 114Km x 114Km will supply all canada's electricity needs ( assuming storage is available for nights and cloudy days)

Land area of canada = 9.9 M sq km ...

You would need just over 0.1% of the land area of canada

 Sep 20th 2018, 07:55 AM #5 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 424 Oz's calculation is fine. For a simple back-of-the-envelope calculation, you just have $\displaystyle P = \eta A G_T$ where $\displaystyle \eta$ is the efficiency of the PV panel material, $\displaystyle A$ is the surface area (m2) and $\displaystyle G_T$ is the total radiation captured by the area over the course of the year (J). There are slightly more detailed models where you can include the effect of ambient temperature and cell temperature. The most popular one is called the Nominal Operating Cell Temperature (NOCT) model. I can help with that if you need it. However, that model tends to overestimate the power generated by about 5-10% or so. There are also more detailed models that try to capture the semiconductor physics, circuitry or irradiance mechanisms more closely, but I don't know much about those and they are probably much, much too detailed for the kind of calculation you need to perform.

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