# Mechanical work to required battery power

I have a very practical question where I've calculated the mechanical work needed by a simple mechanical system by solving the line integral $W = \int_C \ F \ dx$. However, since I have a black spot in my brain for electrical calculations I figured I could just (considering a 1.5V 2700mAh Alkaline battery) calculate the current required by $I = P / V$ and using that calculate how many batteries I need or how long my mechanical system could work. Is this correct or am I missing something crucial here?

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## 2 Answers

You have (1.5V)(2.7A)(3600s) = 14580 Watt-seconds = 14580 Joules of energy in the battery.

BUT you can only get that much energy out if you extract it the same way the manufacturer did it during their optimized tests, which is typically at a much lower amperage than what you want to drive your motor. Also, remember that there are friction and other losses.

If you parallel enough batteries to keep the draw from any one battery to somewhere near a value that would result in total discharge at 100 hours (27mA in your case), and factor in a total system efficiency rate of 50%, my guess is that you will be close.

So, with batteries properly arranged and sized (quantity of batteries) for your system, you should expect to get (14580J)(50%) = 7290J of actual work out of EACH battery. That is about 5368 ft.lb. of work (lift 100 pounds a height of 53.7 feet).

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From the specifications of your battery, that is 1.5V and 2700mAh, you can compute that there is $14580$ Joules of energy stored in your batteries.

The formula $P=U\cdot I$ relates power to voltage and current. You battery specs give voltage and capacity (that is total charge stored). The former is in Volt, the latter in milli-Ampère-hour. The product is therefore not a power but an energy. And that corresponds to the $14580$ Joules I mentioned. Of course, you should also factor in possible losses, so in practice, you'll never draw that much work from the batteries.

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So I can safely go from the Jouls needed to the ones generated by the battery. I factored in losses due to friction, but a general efficiency rule regarding batteries is hard to find. – avanwieringen Jun 15 '11 at 14:37
Problem is that as you use a battery, the current and voltage will not be steady and therefore the power will not be either. The current in particular will depend on the electrical resistance of your system. I don't think much can be said with the little information you gave. – Raskolnikov Jun 15 '11 at 14:41
If the specs are honest (and not marketting lies), and the use is within a reasonable range, the draw current is low enough that internal resistance within the battery isn't import etc. It should be a reasonable approximation. Small sized electric motors are probably not close to 100% efficient, so you need to include a substantial safety factor. – Omega Centauri Jun 15 '11 at 15:54