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#1
October 28th 06, 01:30 AM
posted to rec.radio.amateur.antenna,rec.radio.amateur.policy
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ans calculations/ fuel cell battery buffer or cap?
Roy Lewallen wrote in
: Tom Ring wrote: . . . 1 week is 168 hours which gives us 403 ampere hours. . . . which at 13.6 volts is about 5480 watt-hours. This is the energy storage requirement. The OP asked about using a capacitor. The energy stored in a capacitor is C * V^2 / 2. For capacitance in farads and voltage in volts, the result is joules, or watt-seconds -- you need 5480 * 3600 ~ 20,000,000 joules in round numbers. So suppose you wanted to store this same amount of energy in a capacitor, and you had a switching regulator which would handle 50 volts maximum input voltage. Solving for C, assuming it's charged to 50 volts, C = 20,000,000 * 2 / (50^2) = 16,000 farads. This assumes you can get all the energy out of the capacitor, which requires your regulator to work down to zero volts. But you'll get 3/4 of the energy out of it if your regulator cuts off at 25 volts, 7/8 if it cuts off at 12.5 volts, etc. How does that sound? Roy Lewallen, W7EL Send that to me in CW and I'll have a look at it. SC |
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#2
November 2nd 06, 04:13 AM
posted to rec.radio.amateur.antenna,rec.radio.amateur.policy
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ans calculations/ fuel cell battery buffer or cap?
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