So I am a little confused on how to deal with the Kilowatt hours unit of power, I have only ever used Kilowatts and I have to design a residential fuel cell used as a backup generator for one day.

The average power consumption of a US household is 8,900 kW-hr per year and 25 kW-hr per day and approximate 1 kW-hr per hour. Does this mean that the power output of my fuel cell is 1 kW and if I wanted to use it for the entire day would it have to be designed to be 25 kW?

  • $\begingroup$ A backup generator has to handle your peak load, not your average load. Unless you have something like a very large accumulator. 1KW is very roughly 8A at 120 V (though you'd ideally take into account power factors etc). Just sum the wattage of the items you want to power simultaneously and allow a margin for error. Some devices occasionally consume much more power than their average consumption (e.g. laser printers). Think about kettles etc. $\endgroup$ Nov 21, 2012 at 22:27
  • $\begingroup$ My question was not aiming towards information about a generator. It was more towards a better understanding of a kilowatt hour. $\endgroup$ Nov 21, 2012 at 22:36
  • $\begingroup$ this is not a physics question $\endgroup$
    – yca
    Nov 22, 2012 at 0:19
  • $\begingroup$ This is an important question, exactly because it is so basic, yet so many (especially the press) get it wrong. $\endgroup$ Nov 22, 2012 at 6:03
  • $\begingroup$ It is also (maybe deliberately) underspecified. A house that uses 24 Kw-hr in 24 hrs probably does not have the lights on all the time. You have to guess! If they have the power on only half the time, that would be 2Kw peak you need. If they run an electric dryer for 1 hour, and then have the lights off the rest of the time, that would be a 24Kw peak. @yca, think of it as an experimental physics question. $\endgroup$ Nov 22, 2012 at 6:42

6 Answers 6


A kilowatt is a unit of power, which has the dimensions of energy over time.

A kilowatt-hour, then, has dimensions of energy.

As a simple example, if you wanted to charge up a battery so as to operate a 1,000-watt (DC) heater for one hour, you'd need one kilowatt-hour of energy (assuming the mythical world of perfectly efficient batteries, lossless wires, etc.)

In terms of SI units, this is

1000 J/s $\times$ 3600 s = 3.6 MJ.

  • $\begingroup$ So is my above solution correct? If the average consumtion of a house per day is 24 kW-hr then should my fuel cell be designed as a 1 kW fuel cell or a 24 kW fuel cell? $\endgroup$ Nov 21, 2012 at 22:43
  • $\begingroup$ 24 kW (in simple terms) means that you can handle a 24 kW load. Like a water heater, electric stove, five microwaves, and 600 feet of incandescent Christmas lights. If you wanted to power all of that stuff for a full day, you'd need about 576 kWh of energy to do that. Now, it's a bit more complicated than that, because we're not considering the efficiency, or how many phases your fuel cell has, whether you're converting to AC or not, etc. $\endgroup$
    – John
    Nov 21, 2012 at 22:52
  • $\begingroup$ I guess i understand what you're saying, I am just having a hard time trying to apply it to the design of a fuel cell. I am trying to size a fuel cell so that it can run consistently for a day, essentially proving 24 kW-hr so I thought i could just design a 1 kW fuel cell and provide it with enough fuel and air so that it could run for 24 hours $\endgroup$ Nov 21, 2012 at 23:01
  • $\begingroup$ It's a matter of capacity. A battery (or fuel cell) only has so much capacity for pumping electrons into something to power it. If you expect to power something for longer, all other things being equal, you'll need a bigger fuel cell. $\endgroup$
    – John
    Nov 21, 2012 at 23:12
  • $\begingroup$ Thanks anyway, I will try to deal with this in another way to find a solution $\endgroup$ Nov 21, 2012 at 23:14

What RedGrittyBrick said in his comment is correct, but as you need to understand $kWh$ I'll try to explain a bit more.

There are two main things you need to take into account: the total energy in your fuel cell (in $kWh$) and the maximum power it can deliver (in $kW$). Power is the rate at which energy is used. In electricity, $1W$ is equal to $1V times 1A$, so 1kW is equivalent to $220V$ at about $4.5A$ and $1kWh$ is that amount of power used for 1 hour.

A $1kWh$ battery can deliver $1kW$ for 1 hour, or $0.5kW$ for 2 hours, etc. It does get problematic though, if we try to deliver more power over a shorter time. Batteries can only deliver so much current. For example, a $1kWh$ battery is unlikely to be able to supply $60kW$ for 1 minute. It may not be able to supply the full current required, or at those high currents it may not be able to supply it for the expected time.

If your fuel cell has to last a day, then it obviously needs to store enough energy to last that time. So, if your average use is $1kW$, then you need a $24kWh$ fuel cell to go the distance. But your usage will not be constant during the day. At some time most appliances may be switched off, but at other times you may turn on heaters or airconditioners, etc, and use lots of power. The power used by appliances can also vary a lot. A laser printer can use $1kW$ while printing, but only $10W$ in standby. Your fuel cell has to be able to deliver the maximum current required under the worst circumstances.

On top of all that, the above is only valid if you are using DC appliances - and I have not noticed any of those recently... If your appliances run AC you also need to allow for things like the "power factor".

So you need to know a lot more than just the power or energy usage.

  • $\begingroup$ Thank you. Since the course I am taking deals with fuel cells and their use and not on electronics then I was told to just simplify the problem which is why I don't include other factors. I was just reading something and it said "A 3 kW fuel cell ... ". So I guess that means it supplies 72 kW-hr of energy. So if I am supplying 24 kW-hr, I was wondering if I could just call it a 1 kW fuel cell $\endgroup$ Nov 21, 2012 at 23:30
  • 1
    $\begingroup$ A 3KW fuel cell may not be able to supply 72 kWh, as the total amount of energy stored in it may not be large enough. You need to remember the difference between power and energy. A fuel cell has limits on both of them, and the relationship between them is not necessarily "24 hours" $\endgroup$
    – hdhondt
    Nov 21, 2012 at 23:36
  • $\begingroup$ @GregHarrington To put this another way, an $n\ kWH$ power power source can supply $1 kW$ for $n$ hours before being exhausted. If you're supplying $24\ kWH$, your battery needs to contain $24\ kWH$, regardless of the rate at which it dispenses this energy. $\endgroup$ Dec 11, 2013 at 1:24

A kilowatt-hour is a unit of energy.

It's the energy of one kilowatt for one hour. Which is equivalent to the energy of 2 kilowatts for half an hour, 4 kilowatts for quarter of an hour, or 2 kilowatts for quarter of an hour plus 1 kilowatt for half an hour.

A house might use 24kWh in a day. But that doesn't mean 1kW continuous output would meets needs, hour by hour. It just means that 1kW continuous output would produce equivalent energy to the household consumption. It would only meet the house's needs, if the household demand were perfectly constant at 1kW power.

To meet power needs and energy needs, you need three things:

  1. enough fuel to provide 24kWh of energy
  2. the fuel cell has to have enough peak power to meet the household's peak demand. This is likely to be of the order of 3-5kW, unless it has an electric shower, in which case peak demand could be 10-13kW.
  3. the fuel cell has to be able to match the slew rate - the rate of change of power - that demand has. For an electric shower, that might mean going from 0 to 10kW in a few seconds, and back down again in the same time.

Your question doesn't have enough information on points 2 and 3 to specify the fuel cell.

A 1kW fuel cell, if it could run continuously for 24 hours, would produce equivalent energy to a household that needs 24 kWh in a day - because the household's mean power requirement is 1kW, the same output as the fuel cell. So the house would not be energy-independent; but it would be some sort of self-sufficient, as long as it could export its surplus at times of surplus, and import extra power at times of deficit. In such a case, it would on average have zero net imports (but it would have N kWh of absolute imports each day, and N kWh of absolute exports each day, where N is some number between 0 and 24).


The average power output is about 1kW (which is the same as 1KW-hour per hour). The energy capacity required for 1 day's operation at this power level is 25 kW-hours.

  • $\begingroup$ If I need to supply 25 kW-hr per day then I am just going to simplify it and design a 1 kW fuel cell $\endgroup$ Nov 21, 2012 at 23:51
  • $\begingroup$ I don't think it's that simple. You need enough fuel to last through a day (24 times as much as for 1 hour). Other answers have also made this point: the power rating tells you how fast you can extract energy without over-stressing the cell (think rated car horsepower, for example), while the energy capacity tells you the total amount of work the cell can perform (the size of the gas tank in the car analogy). Of course, if you have hoses supplying fuel from external reservoirs, your capacity is unlimited. $\endgroup$
    – Art Brown
    Nov 21, 2012 at 23:56
  • $\begingroup$ I am not getting anywhere on here. I am trying to make a parallel between my homework problems so that I can do the rest of the project. All of my homework problems just say "A ___ kW fuel cell ....". If I could determine how much power my fuel cell provides then I could do the rest of the homework but this is not helping me $\endgroup$ Nov 22, 2012 at 0:00
  • $\begingroup$ Sorry... Looking at wikipedia, fuel cell technologies are indeed rated by power (e.g. kW), as your problems state. I guess you just strap on an appropriately sized tank to get the total energy you need. Good luck. $\endgroup$
    – Art Brown
    Nov 22, 2012 at 0:14
  • $\begingroup$ Thanks for the help. I did not mean to be rude before. Just getting stressed about starting a project last minute. Thanks again! $\endgroup$ Nov 22, 2012 at 0:50

OK, I think you have it right, in the context and scope of the question, the fuel cell should be able to deliver 1Kw, and do it all day.

From a more practical interpretation, the problem is under-specified. You need to know the peak load. What is the highest power the house needs? It might sometimes draw 5Kw, maybe 20Kw. Your fuel cell should be able to handle that.

But the homework did not mention that. So set the fuel cell to deliver 1Kw all the time. It will produce 24Kw-Hrs in a day.


So I am a little confused on how to deal with the Kilowatt hours unit of power

Kilowatt-hours is not a unit of power, it is a unit of energy.

I think the easiest way to clarify this may be to think in terms of energy and power and try to avoid terms like kilowatt-hours per hour.

The average power consumption of a US household is 8,900 kW-hr per year and 25 kW-hr per day and approximate 1 kW-hr per hour. Does this mean that the power output of my fuel cell is 1 kW and if I wanted to use it for the entire day would it have to be designed to be 25 kW?

Answer 1

A fuel-cell with a power output of 1042 Watt operated for a day will deliver the 25 kWh of energy you mention. (Assuming you have a hydrogen-cylinder delivery man who visits regularly with fresh supplies)

It will not allow you to use a kettle.

It may not allow you to turn all the lights on and watch TV

Answer 2

It may help if you translate everything into SI units

1 kW is 1,000 Watts of power
1 kWh is 1,000 Watts for 3,600 seconds = 3,600,000 Joules of energy

So your target house uses 90,000,000 Joules in a day. If the rate of energy usage is constant and never varies you can supply that amount of energy at a constant rate of 1,042 Joules per second (otherwise known as 1,042 Watts or just over 1 kW)

A typical kettle for boiling drinking water needs over 2 kW of power.

Answer 3

I eat 2,500 kCalories in 24 hours. (10,460,000 Joules)

That is an average of 121 Joules every second.

However I don't eat a tiny speck of food every second 24 hours a day. I spend some of that time sleeping and on activities where I do not necessarily want to concurrently eat blueberry pie at that rate.


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