**Summary:** 375 MW of electricity, an amount currently provided by fossil fuel burning power plants for a population of about 320,000, requires a 2 GW solar farm composed of 5 million 200 W/m^{2 }solar panels taking up 10 km^{2} (4 mi^{2}) or 2500 acres of land (just under 1/10 the area of a city of 150,000 such as Fort Collins, Colorado). This will cost nearly $2.5 billion to construct and, if amortized over 20 years, will add $0.03 per kWh to the operational cost of electricity. This assumes some storage technology is available and takes into account intermittency and losses from storage.

*Back of the Envelope* is a blog series exploring issues related to transitioning from fossil fuel based electricity to renewable electricity production. There are two big issues: first, the scale, i.e. how many renewable energy sources are needed to replace the fossil fuel sources currently in use; second, because of intermittency issues, i.e. that the sun doesn’t shine at night or that the wind doesn’t blow all the time, extra energy needs to be produced during production times and storage needs to be developed.

I am from Fort Collins, Colorado. Fort Collins gets its electricity from the Platte River Power Authority (PRPA), a public power utility that services Loveland, Longmont, and Estes Park in addition to Fort Collins. The total population served is nearly 320,000. This is a nice number because it represents approximately 1/1000 of the US population. Roughly speaking, we need to multiply whatever we find here by 1000 to cover the whole United States.

PRPA’s sources include coal (about 75%), hydropower (about 20%), wind (about 5%), and some other sources including natural gas for summer peak demand. The daily load profile is typical with a peak load around 7pm and a low around 3am. The average daily load for January 2016 is around 400 MW. Let’s call it 500 MW to account for summer load and in the spirit of a back-of-the-envelope calculation.

Some of PRPA’s energy (25%) already comes from renewable sources. Nationwide the number is between 10% and 15%, but is nearly 35% if you include nuclear. For this calculation I will assume that we need 75% of 500 MW, 375 MW, to replace the fossil fuel sources. I will also do the calculation using solar PV only. Obviously, a real solution will be a combination of solar PV, solar thermal (CSP), and wind. But the land use/resource requirements are similar among the three, so we can get our back-of-the-envelope result by just using solar PV.

Here’s where we’re heading. We need to meet the daytime load. We will assume that daytime is 6 hours per day (a 25% capacity factor). (Perhaps with a combination of wind and solar we could get up to 8 hours per day (closer to a 35% capacity factor).) We will need to produce enough energy while the sun is shining brightly to cover the 18 hours when it’s not. This will be produced during the day and stored somehow. Since storage will not be 100% efficient, we will need to produce additional energy to account for a round trip efficiency, which we assume to be 75%.

**Daytime Load**

Let’s be generous with our solar panel efficiency and say we can have 200 W/m^{2} solar panels. That would be a 400 W 1 m x 2 m panel (2 m^{2} panels).

375 MW x 1,000,000 W/MW x 1 solar panel/400 W = 937,500 panels. Let’s just call it 1,000,000. A million solar panels takes up 2 million m^{2} of space. A km^{2} is 1,000,000 m^{2} so we’re talking 2 km^{2} (0.77 mi^{2}). That’s a square of land 1.4 km (0.9 mi) on a side. If you think in acres, that’s about 500 acres.

**Nighttime Load**

We have to multiply everything times 3 to cover when the sun is not shining. So that’s 3 million more solar panels taking up 6 km^{2} (2.3 mi^{2}) or 1500 acres.

**Storage Efficiency Issues**

Since the nighttime load has to be stored we will have to increase our numbers to accommodate inefficiencies in round trip storage. Since our efficiency is assumed to be 75% we can divide our nighttime load numbers by 0.75 to get the amount that needs to be stored: 4 million solar panels taking up 8 km^{2} (3 mi^{2}) or 2000 acres.

**Totals**

Adding in the daytime load now gives a total of 5 million solar panels and 10 km^{2} (4 mi^{2}) or 2500 acres. That’s a 2 GW solar farm.

The area of the city of Fort Collins is 56 mi^{2} so we’re talking about covering an area equivalent to 7% of the entire city with solar panels. These solar panels cost about $200 per panel for a home installation. Since we’re buying 5 million of them, let’s say we can get half off for a total cost of $500,000,000. If we estimate $2/W for installation that’s an additional $2 billion for a total of $2.5 billion.

Assuming our 2 GW solar farm produces electricity for 6 hours a day, 365 days a year for 20 years, we will produce 2 GW x 1,000,000 kW/GW x 6 hours/day x 365 days/year x 20 years = 87.6 billion kWh. That would be $0.03/kWh to cover the initial building of the solar farm. While that’s a lot of money up front, over the long run it’s a decent construction cost per kWh of electricity produced.

The largest solar PV plants in the US today are around 500 MW (Solar Star, Topaz, Desert Sunlight, Copper Mountain) each costing in the $2-$3 billion range. Prices of solar installations continue to come down, so it’s conceivable that we could get a 4x larger plant for nearly the same price.

The general result is that if we switch from fossil fuel sources to intermittent sources with storage that we need to a solar or wind facility that produces 5x the amount of electricity that we get from the fossil fuel plant. In order to make this work we need to store that electricity somehow, not a trivial problem today. We’ll discuss that in future *Back of the Envelope* posts. And, remember, we need 1000 of these nationwide.

Check out *Energy: What the World Needs Now* by Terry M. Gray and Anthony K. Rappé.