How many batteries do you need to store electricity for a night?

Energy storage is a common topic that comes up in discussing energy. A lot of people think it would be easy to just charge some batteries during the day with a solar plant and then use them through the night. Critics of this idea frequently tell proponents to "do the math." This page does the math and tries to help you understand the energy storage issue in concrete, visualizable numbers. We attempted this with pumped storage using aircraft carriers before, but this time we're using batteries.

Robert Stone, the filmmaker who brought us the documentary Pandora's Promise, suggested at a screening in Seattle in October 2013 that he views climate change denial in the same light that he views scale of energy problem deniers. This page is an attempt to better educate the scale-deniers.

The case of interest is the replacement of a single large baseload power plant (1 gigawatt electric) with a large solar array and a bunch of batteries. This is a typical size for a nuclear power plant, and can produce electricity for a city of about 700,000 average US citizens, or 1/440th of the USA. If you want to scale up the battery requirements to the entire USA, just multiply by 440.

While we know that energy demand decreases at night a large baseload remains, and we assume here that the plant continuously operates at 1GWe. This is not uncommon for baseload plants to do.

(This document was created with free and open source ipython notebook)

Doing the math

First we determine how much energy (in Joules) is produced by a 1 GWe power plant over an 8-hour night. This is just energy=power*time

In [1]:
nightEnergy = 1.0e9 * 8.0 * 3600.0  # 1e9 Watts * 8 hours * 3600 seconds/hour
print(nightEnergy)
2.88e+13

Now, to figure out how many batteries that would be, we need to know how much energy is stored in a battery. Different types of batteries can store different amounts of energy. Using information from here, we load up some basic information (including some that will be used later). This table mixes energy units of Joules and Watt-hours, but they are easily related by 3600 Joules = 1 Watt-hour and we'll use this conversion throughout.

In [2]:
batteryTypes=['LeadAcid','Alkaline','NiMH','Li-Ion']
specificEnergy = dict(zip(batteryTypes,[146000,400000,340000,460000])) # J/kg
costPerWh=dict(zip(batteryTypes,[0.17,0.19,0.99,0.47])) # $/Watt*hour
wattHPerLiter=dict(zip(batteryTypes,[100,320,300,230])) # Watt*hours / liter

Now we can figure out the battery masses required to store 8 hours of energy from a 1 GWe plant using each type of battery:

In [3]:
%pylab inline 
massRequired = [nightEnergy/specificEnergy[bType] for bType in batteryTypes]
x = range(len(batteryTypes))
pylab.bar(x,[nightEnergy/specificEnergy[k] for k in batteryTypes],align='center')
pylab.xticks(x,batteryTypes)
dum=pylab.ylabel('Mass of batteries (kg)')
dum=pylab.title('Mass of batteries needed for 1 night')
Populating the interactive namespace from numpy and matplotlib

That seems like a lot, but is hard to visualize. Let's look at the plot in terms of volume. It will still be huge numbers, but we can visualize it better by stacking the batteries up on a full football field, which is 110x49 square meters. How tall will the full football field stacks of batteries be?

In [4]:
areaFootball = 110*49 # area of football field in square meters
# Compute height as height= volume needed/ area available
# Convert nightEnergy from Joules to Watt*hours and liters to m^3. 
heightOnFootballField = [nightEnergy/3600/wattHPerLiter[bType]*0.001/areaFootball for bType in batteryTypes]
pylab.bar(x,heightOnFootballField,align='center')
pylab.xticks(x,batteryTypes)
dum=pylab.ylabel('Height stacked on field (m)')
dum=pylab.title('Size of batteries needed for 1 night')

So, with modern electric-car Lithium-ion batteries, the football field would have to be stacked 6 meters (~18 feet) tall to store enough energy for one night from one power plant. The next question is, how much would this cost to build?

In [5]:
cost = [nightEnergy/3600*(costPerWh[bType]/1e9) for bType in batteryTypes]
pylab.bar(x,cost,align='center')
pylab.xticks(x,batteryTypes)
dum=pylab.ylabel('Cost of batteries (billions $)')
dum=pylab.title('Cost of batteries needed for 1 night')

In all cases, the storage facility adds at least an additional $1 billion to a power plant, just based on battery materials alone. Assuming we don't want to build lots of 14 meter tall lead football fields everywhere (due to heath concerns regarding lead), we're stuck with Li-ion batteries (since Alkalines aren't rechargable), so the estimated battery cost for this facility is $4 billion. This is about the cost of a new 1 GWe nuclear plant.

Other ways to store energy

There are other ways of storing energy besides charging batteries, such as pumped storage. In that case, you have to lift the equivalent of 44 fully loaded aircraft carriers to the top of the Burj Khalifa every night. Another option is to use molten salt, which has high heat capacity.

In the very-cool smartgrid idea, when extra electricity is needed, the grid just temporarily asks for energy back from a fleet of electric vehicles where some percentage are usually plugged into the wall (because people are at home or work, not driving). This might be fine for the daily peaking, but to do this at night, considering that a Tesla Model S comes with a 60kW*hour battery source, it would take:

In [6]:
print(nightEnergy/3600/60000)
133333.333333

133.3k Teslas per 700,000 citizens to get through the night. And then their batteries would all be dead for a few hours after the sun came up. So if 1 in 5 people bought Teslas that they just stored in their garage as batteries for the grid and only drove between 11am and 1pm or so, then the storage problem would be solved.

Conclusion

We determined that to replace a single large baseload power plant with a solar plant that doesn't work at night, we need a very large and expensive storage system. With Lithium-ion batteries, the storage required would be a 6-meter tall stack of batteries on an entire football field and would cost more than the power plant itself. We assumed perfect charging, when in fact the charging efficiency of these batteries is much less than 100%.

Further reading

For additional education, check out our front page at www.whatisnuclear.com.