UCL ENERGY INSTITUTE BLOG By Andrew ZP Smith, on 21 May 2015
How much power do solar photovoltaic systems produce per unit of land area? And does it matter: is it a constraint in the real world?
At Elon Musk’s glitzy launch of the Tesla PowerWall and PowerPack batteries, the Tesla CEO showed a map of the US, with a small square in the North-West corner of Texas marked in blue, and said that solar panels over that surface area would be enough to enable US electricity to go carbon-free.
Here’s a quick fact-check on that claim.
How the calculation works
We are comparing current US electricity consumption, to see how it compares to the amount of electricity that would be generated from solar PV panels covering the area shown.
I’ll calculate the PV generation by multiplying the surface area by the efficiency of PV in capturing sunlight and by the amount of PV that could be installed per unit area.
We’re talking specifically about current US electricity consumption. Not all energy, just electricity.
The graphic isn’t about building a solar farm there, to supply the whole US: that would be preposterous. Instead, the graphic is designed to describe the physical area of panels required. i.e. it’s a data visualisation; not a project proposal.
Inputs to the calculation
US electricity consumption is about 425 GW on average
The EIA give a figure of 3,725,101 thousand Megawatt Hours of total electricity sales in 2013, i.e. 3725 TWh in a year. That’s equivalent to 425 GW.
The area shown is 10,000 km2 in NW Texas
Looking at the map presented by Elon Musk, and comparing it with a scale map of the US, leads me to an estimate that the square, in North-West Texas, is about 100km along the side, and thus has a total surface area of 100km x 100km, i.e. 10,000 km2
Average PV yield in NW Texas is about 21%
The USA National Renewable Energy Lab (NREL) provides an online calculator for PV yields, called PVWatts. I looked at the data for location TMY2 Amarillo, TX, which seems close to the area in question. A 1 kW system would generate 1,838 kWh per year (on a 10o slope facing south), which is equivalent to 210W; that gives a ratio of generated power to capacity of 21%.
The highest efficiency we currently get from solar modules is about 24%
Table 2 of Green et al’s Solar cell efficiency tables (Version 45) gives the best PV module as being 24% efficient. And that translates to installed capacity of 0.24 GW/km2, given the standard measure of 1 full sun being equal to 1 kW/m2.
10,000 km2 x 0.24 GW/km2 x 21% = 500 GW
Which is more than current US electricity consumption of 425 GW.
Yes, the area shown is reasonable, as a visualisation of the surface area of panels required to generate electricity equal to total US electricity consumption, on a multi-year average: that area of panels would generate about 500 GW, which is above the current US annual average electricity consumption of 425 GW, with enough spare to account for resistance losses. And do bear in mind that the claim wasn’t about whether demand could meet demand second-by-second, but whether the total amount over time could be met. The whole point of the presentation that the claim occurs in was to sell storage, which is there to bridge gaps between generation and demand.
Photo credit: Activ Solar some rights reserved under creative commons license
Comment to the post by Tom Tamarkin
For a detailed analysis of the surface area required to generate all the electricity currently comprising American baseload power generation by solar please see my article “The Green Mirage.” It is on-line at: http://fuelrfuture.com/review-of-forbes-on-line-magazine-article-solar-energy-revolution-a-massive-opportunity/