Firewood vending- convert to solar

7 Posts
Sep 23, 2009 01:52 pm
Firewood vending- convert to solar

The machine vends prebundled firewood.Plugs into 110v outlet. It uses an electric motor (6.7 amps) that turns a screw drive to push the bundle out into a catch bin. The motor runs approx 15 sec's when activated by the bill validator.The bill validator draws very little and runs all the time.3 exterior 13 watt lights dusk to dawn.  I was hoping to use a 2500 or 3000 watt power inverter with a deep cycle battery and a solar battery charger for now.Or could I connect the power inverter with multiple outlets to a deep cycle battery and plug a battery charger into the inverter then connect the charger to the battery so the battery feeds the inverter, the inverter feeds the battery charger as well as the machine and the charger keeps the battery charged. There is a 10 amp fuse at the electrical panel. Also pure or modified sine. I am not sure how sensitive the bill validator, limit switches or relays are Please advise  Thanks  John     
Sep 26, 2009 10:42 am
Re: Firewood vending- convert to solar

One of the most frustrating things about figuring up an RE system are the amount of variables there can be in calculating the load and supply, as efficient as possible to minimize expenditure.
One device that can help is a simple plug in kiloWatthour meter/recorder. See Alt-E Store.
I would think that, in a campground setting, firewood bundles would be a cool weather (spring/fall) treat as opposed to a hot or even cold weather (summer/winter) treat. But given the nature of PV I always use winter time insolation as the base point for figuring - how many hours of equivalent full rated power from a PV module - when sizing the PV array and battery bank.
A 110 vac motor that "draws" 6.7 amps would of course be 737 watts but at 24 vdc nominal that would be better than 31 amps that the inverter would draw. Roughly 35 amps at 15% inverter efficiency. Now we have to convert that to amphours which would be difficult to do because we don't know the frequency at which wood is purchase, averaged out over, a year??? If we assume 4 times a day/night (24 hrs.) averaged out over a year it makes the math easier but not so accurate. (Please read my next post) You could pull that information from how many bundles are sold in a year. But I will assume 4 times 35 which is 140 amphours.
3 - 13 watt lamps "burning" for 12 hours (winter) at 110 vac would burn 0.36 watts (plus 15%) times 12 hours making that a total of .504 kWh's. At 24 vdc that would be 21 amphours.
Using just those two factors only the total would be 161 amphours. I like to multiply that by a factor of no less than 5. This helps to keep the draw over 24 hours from the battery in the top 20% depth of discharge. Now we have at least 805 amphours at 24 volts nominal battery bank.
Now, to replace the 161 amphours used each 24 hours.
24 vdc times 161 amphours equals 3,864 watt hours or 3.864 kWh's. I would round that up to 4 kWh's. If where this machine is sees the - equivalent of 2 hours of full rated power from a PV module in the winter months - simply divide 4,000 by 2 and you would have a need for a 2,000 PV watt array.

Now I think you can begin to realize how that kiloWatt meter/recorder might be a worthwhile investment. If your best sales for bundles is not in the dead of winter but in the fall, you can size the PV array for those months an so on and so forth.
I find this to be a fair assessment of - the equivalent number of hours of full rated power from a PV module. - Its pretty self explanatory. The nature of PV, if sized right, is to have; a surplus during the summer solstice, just the right amount during the vernal and autumnal equinox and, to just squeak by during the winter solstice. But again I would like to remind you that there are quite a lot of variables to this and "your mileage may vary." Chiefly speaking, the time of year the most firewood bundles is sold.
« Last Edit: Sep 26, 2009 02:42 pm by Thomas Allen Schmidt »
Sep 26, 2009 02:36 pm
Re: Firewood vending- convert to solar

Oops! It appears that I made a huge blunder in one of my calculations. I happen to think to myself while I was at a snack machine a while ago waiting for my Fig Newtons to drop, "who would want to wait 15 minutes at a vending machine for their treat?" Then it hit me, Duh! You stated that the machines motor runs for 15 seconds while delivering the bundle and in my hast to simplify the math I used 15 minutes.
Sorry 'bout that.
So, instead of an 805 amphour battery bank and a 2,000 watt PV array it should have been more like a 110 amphour battery bank and a 254 watt PV array assuming all of the other circumstances used in my last post. The lights would use more power than the motor in terms of kWh's. Getting its power from a battery and inverter assuming a 15% efficiency -
The lights = .504 kWh's per 12 hours of on time during the winter.
The motor = .03 kWh's for every 15 seconds of run time, 4 times in a 24 hour period.
So .507 kWh's or 507 watt hours divided by 24 vdc nominal is 21 amphours times 5 is 106 amphours.
507 watt hours divided by 2 hours of equivalent full rated power from a PV module during the winter months is a 254 watt PV array.
That sounds a lot better huh?
« Last Edit: Sep 26, 2009 02:40 pm by Thomas Allen Schmidt »

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