large solar farm in california

How Many Solar Panels Would be Needed to Power the U.S.?

Aniket

Written by qualified solar engineer Aniket. Last updated:

About 7.86 billion solar panels would be needed to power the U.S. on solar energy. This is derived from the fact that every year the U. S. consumes around 4000 billion kWh of electricity. This means an astounding consumption of 12,000 kWh per year per capita.

If each one of the 7.5 billion humans on the planet start consuming this amount of energy, all of the oil in the world would last only nine years. In addition to this, climate change data repeatedly points to a probable 2°C rise in average temperatures in this century, bringing catastrophes like never before. These are reasons enough to wonder how the entire U.S. can become solar powered. Let’s find out how.

The Total Installed Capacity Required

If we were to think of the solar panels needed to power the entire world, a significantly large portion of that would be needed for the U.S. alone. Currently, the existing infrastructure in the nation churns out over 4000 billion kWh every year. That means a daily average of almost 11 billion kWh.

Dividing the above number by the number of hours of energy consumption would give us hourly energy data, or simply the power consumption (kilowatt-hours ÷ hours = kilowatts). Now, unlike conventional power generation sources, solar cannot operate 24 hours a day. This is why the hourly consumption and hourly generation numbers for solar power cannot be the same.

For example, if one needs 24 kWh per day, his hourly average consumption would be 1 kW, but a 1 kW solar plant would not generate energy for 24 hours. This is due to the day-night cycles. Owing to this, the number of hours of peak sunshine at any location matters.

In our case, the full sunshine hours in the U.S. vary from 3.5 to 5.5. For our calculation, we will assume a national average or 4. This can, of course, be different based on the latitudes where the majority of the solar plants are located. Going back to our calculation, 4000 billion kWh with a 4-hour full generation every day gives us 11 billion kWh ÷ 4 hours = 2,750 million kilowatts, or better said 2.5 TW.

Annual PV Solar Radiation in the United States

Annual PV Solar Radiation in the United States (Source – NREL)

The current standard size of panels used in a large sized solar plant is easily over 350W. Assuming this power rating, we would need to divide 2.75 TW by 350W, which gives us the gigantic number of 7.85 billion (7,857,142,857, to be precise) panels required. This number does look intimidating, but with a large number of solar companies in U.S., it can be turned into reality.

Area and Costs Required

It is nearly impossible to install all these panels at a single location for more than one reason. We can, however, hope for a cumulative installed capacity crossing the required number (with plants spread throughout various states). A megawatt of solar power plant requires about 5 acres of land. 2.75 TW, or 2,750,000 MW would require 13,750,000 acres.

An important point to note here is that not all the panels have to be installed on vast areas of land. Rooftop solar plants have rapidly gained traction. The number of solar panels required to power the average house, in almost all cases, would find sufficient roof space on homes in the U.S.

For many decades, high cost remained a serious obstacle in the widespread adoption of solar. In the past few years, costs of solar panels have experienced a nosedive. The installation costs don’t burn a hole in your wallet, thanks to a well-established industry with considerable competition. Focusing on large scale deployment of solar, costs can be controlled even more by purchasing panels made in the U.S.

In the end, we cannot afford to ignore the climate story. We need to make sincere efforts towards remarkably large deployment of solar power, starting today.

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2 Comments

  1. Avatar

    That’s not exactly the right calculation to consider.

    The right calculation is to start with the amount of solar radiation per square meter hitting the source of the Earth under ideal conditions, and when the Sun is at its zenith in relation to the solar panel (for the easy of calculation, approximate 1000w/m². However, cutting-edge solar technology is directly achieving approximately 50% conversion efficiency.

    Two other related variables must be considered. The amount of solar radiation per square meter drops with the angle on incidence to the panel, unless the panel is tracking the Sun like a sunflower. The power usage uncovered in tracking, introduces a new inefficiency. In addition, solar panel efficiency drops 5% in the first year and 2% annually, during the ~20 year life is the panel.

    Now repeat your calculation, and you’ll find the answer is closer to approximately 240,000 square miles or approximately an area the size of Arizona and New Mexico combined!

    Then there’s the issue of dependence on China for anything lanthanide related, or a massive increase in open pit mining in the U.S., but you can tackle that one in the next article.

    1. Carlos

      Hello dear Ico Uce,

      Thank you for your comment,

      So as you say there are many factors into account to consider when sizing a PV array whether it is small or large scale. To summarize the most important ones we have far and near shading, thermal losses, irradiance losses (DNI and DHI), conversion efficiencies(best available ones around 20%), incident angle modifier (IAM) losses which you mentioned, ohmic losses, mismatch, degradation, LID losses, clipping losses of the inverter, and in large scale size arrays as this one we must also consider the PID losses of PV systems, among others.

      Of course there is a high background of technical studies needed to approach a more precise number of modules, for instance, for a large scale project like this 4h of sunlight may no be just enough, developers might look more for 5h or 6h of sunlight which I’m sure would cover many states across the US. Connection and transmission of this high amount of solar power would need to be distributed throughout several states simply to avoid high power losses during distribution and transmission of electricity.

      So we are aware of this high technical background needed to develop an enterprise of such magnitude, but that’s why we’ve taking a 4h average to approach this technical losses in a simple way for demonstration purposes and to provide an approximate estimate of the number of panels that would be needed to power the US.

      Thank you so much for your insight and valuable comments,

      Best
      Carlos

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