RadioBanter

RadioBanter (https://www.radiobanter.com/)
-   Policy (https://www.radiobanter.com/policy/)
-   -   ans calculations/ fuel cell battery buffer or cap? (https://www.radiobanter.com/policy/108088-re-ans-calculations-fuel-cell-battery-buffer-cap.html)

Slow Code October 28th 06 01:30 AM

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

RadioGuy November 2nd 06 04:13 AM

ans calculations/ fuel cell battery buffer or cap?
 
In article t,
says...
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

You can't multiply so how could you help?


All times are GMT +1. The time now is 02:54 AM.

Powered by vBulletin® Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
RadioBanter.com