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I am trying to design a fuel cell that can provide enough energy for 1 day's use a maximum capacity but I am having trouble sizing the fuel cell because I need a better understanding of kilowatt hours.

I found online that the average power consumption of a house is about 8,900 kW-hr per year which amounts to approximately 24 kW-hr per day.

If I wanted to design a fuel cell, would it be considered a 1 kW fuel cell? I am trying to size it.

The reason I ask is that I read an article and it said "A 3 kW fuel cell ..." so i was wondering if I could just consider this a 1 kW fuel cell and then size it based on that assumption

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A side note, its called kWh and not kW-hr. –  Johan Nov 22 '12 at 5:41
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4 Answers 4

  • A joule (J) is a unit of energy. If you lift a one kilogram weight one meter, you have done 9.8 J of work, because the weight of a kilogram is 9.8 Newtons, and a force of one Newton over one meter is a Joule of work.

  • A Watt measures work intensity or power: how much energy is transferred per unit time. One Joule of work done in one second is a watt. If you lift one kilogram by one meter, and you do it in one second, you need 9.8W of power. A Watt is a Joule per second, J/s. If you want to do the same lift in half a second, you need 19.6W: twice the power.

  • A kilowatt-hour is a unit of energy again, like a Joule. We are taking Watts (energy per time) and multiplying them by time, so the time cancels out, leaving energy. A hour is 3600 seconds. So a kilowatt-hour is 1,000 J/s * 3,600 s = 3,600,000 J. If work is being done, or energy being transferred at a power of 1000W, and this is done for one hour, that is a kWh. Or 100W over 10 hours, et cetera. A 100W bulb lit for 10 hours consumes 1 kWh.

  • A food calorie (the big calorie, or the kcal) is 4200J. So a 1 kWh is 3,600,000J / 4,200J/kcal = 857 kcal. This is about the energy in two double cheeseburgers from McDonald's.

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A kilowatt is a measure of power primarily used to measure electric power. Power is the ability to do work. Kilowatt hour is a measure of energy or work. It is measured by multiplying an amount of power by the time period it is being used. If an electric motor draws 1 kilowatt for one hour, then it has consumed a kilowatt hour of energy. You have found that the average home uses 24 kilowatt hours of energy per day. That includes all of the energy of all of the electrical devices in the typical home including lighting, heating, air conditioners,appliances, televisions, radios, computers, etc. However it doesn't tell you how much power the home may be demanding at any one time during the day. For example, in the late evening when everyone is sleeping, the power demanded by the house will be small probably just the refrigerator, heating, clock radios, and some night lights. This may be on the order of 1 kilowatt or less. During the day, if the air conditioner is on and the dryer is on, the house load will be many times 1 kilowatt (an electric dryer probably draws about 6 kilowatts or more). What this means is that any device designed to power a home has to take into account not only the total energy consumed (which determines the amount of fuel used by the device, whether a steam turbine driving an electric generator or a fuel cell) but also the maximum power that the house may demand at any time during the day. Many electric utility companies charge not only for the total energy used (kilowatt hours) but the maximum power demand (peak kilowatts) because that determines how large their generating capacity has to be. Thus you can see that a fuel cell that can provide 1 kilowatt for 24 hours might supply the total energy a house consumes but would never be able to power the house when the electric demand is at a maximum. That maximum value depends on the electric devices in the home and when they are used. It is possible that air conditioners, television, washers and driers, and many lights could be on at the same time. Then the electric demand could easily exceed 10 kilowatts.

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You've got the right idea. Assuming perfect efficiency and some storage (which you don't address--more on that below), a 1kW fuel cell running for 24 hours would produce 24 kWH. (power(kW) * time(h))

The missing piece from your analysis is a power buffer. If you looked at the typical consumption of a home plotted as instantaneous kW as a function of time, it would have a significant diurnal variation. The peak draw will likely be 10kW or more -- 9kW+ more than your fuel cell can supply. Where will that power come from? Typically batteries.

There's also efficiency to consider - the efficiency of charging and discharging batteries, the efficiency of inverting the fuel cell's DC output to AC, etc.

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If Peak/average ratio=1, you need a 1kW fuel cell for 24 hrs. If Pk/Avg= 10 , you need a 10kW fuel cell

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