From: "wombat" 
Subject: Re: crock pot question
Date: 1999/03/12
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On Thu, 11 Mar 1999 22:10:45 -0600, "TC Dufresne" 
wrote:

>I think you may be a little off your calculations.
>Assume the crockpot runs at  300 watts and 120 VAC. Thats 2.5 amps, right?

1==> Purely resistive load, so that is right.    And 300 watts = 0.3
kilowatts.

>(Watts= Volts x amps) If it takes 8 hours to cook, then you are using 8 x
>2.5 amps, or 20 amp/hours of battery "juice", correct?

2==>  Um, well, an "amp hour" is actually an old measure of battery
capacity.   Perhaps it will be less confusing if we use conventional
volts, amps, ohms, watt hours, etc.  

The crockpot *doesn't* , in fact,  run   off "battery juice" of 12
volts,  it runs off 120 volts AC  drawing 2.5 amps.

As is a resistive load, can use Ohms Law to find resistance.  R = E /
I  where R= Resistance in ohms,  E = Voltage in volts, I = Current in
Amperes [ "Amps"]

120 / 2.5 =  48 ohms

3==> If we  try to run the crockpot  off 12 volts, can predict the
result by using a different form of Ohms Law   I = E / R.     
12 / 48 =  0. 25 amps, or 250 ma  [milliampere]

Watts = volts times amps, so 12 * 0.25 = 4 watts.  
Since it is designed to run on 300 watts, all this 4 watts would
do is slowly drain the battery.

4==> OK, now lets run the crockpot  off an inverter [ or "converter",
or  whatever we want to call it]  to change that 12 volts to 120
volts.   

We would need an inverter of at least 300 watts capacity, and this
would pull about 27 amps from the battery when drawing a full
300 watts.  Assume that we start out with fully charged battery at
12.7 volts terminal voltage.  

12.7 * 27 =  342.9  watts.  

5==> Inverters aren't 100% efficient, so although the input is 342.9
watts, the output is only 300 watts.  About a 10.8% conversion loss.

6==> Anyway, the first consideration is how long the battery will
supply the required 27 amps.     You've specified 8 hours cooking
time.   

7==> I don't have a current Sears catalog here, but do have one from
1992.  Largest super duty deep discharge marine battery at that time
claims to supply 25 amps continuous for 6.5 hours.  Now that isn't
quite 27 amps.   Moreover, that is at 80F operating temp before the
voltage drops under 10.5 volts.   So lets assume that 2 of these
batteries in parallel will supply 25 amps for 13 hours.   

That will provide you with your 8 hours of cooking time without
completely discharging the battery.  

One hours operation  will require 27 amps at 12.7 volts, or  342.9
watt, or  .3429 KW .   For 8 hours,  .3429 * 8 =  2.742 KWH [kilowatt
hours].

8==> OK, how long to put that energy back into the battery and bring
it back up to its original full charge?   Storage batteries aren't
100% efficient either.   A conservative figure is 20% more energy than
took out.   We've used 2.742 KWH, so 1.2 * 2.742 =  3.29184 KWH

9==> Now lets look at the solar panels:      

Lets assume that  each  50 watt panel will actually each produce
3 amps at 16 volts or ~48 watts, and we can effectively parallel them
to produce 15 amps at 16 volts.     15 * 16 = 240 watts.   If we run
them for an hour, they will have produced 240 watt hours, or 0.24 KWh.

We have to replace  3.29184  KWH, so 
3.29184  / .24 = 13.716  hours of operation, or about 13 hours and 45
minutes of sunshine.   

10==> OK, suppose we use a generator instead.  I don't have the
charging figures for these batteries, but generally deep discharge
batteries aren't as tolerant of fast charges as ordinary car batteries
if want the maximum life from them. 

Assume a small generating set will put out 13 volts at 40 amps, and we
can safely charge the batteries at 40 amps.   13 * 40 =  520  watt or
0.520 KWH. 3.29184 / .520 =   6.33  hrs or a bit over 6  hours 
30 minute charge time.

11==> Anyone see any mistake in these calculations?   It might be
worth checking Sears for current specs on their deep cycle marine
batteries.   Car batteries won't stand up as well on this type of
service as the deep discharge will.

12==> As to "amp hour":  Defined this in an earlier post, but will
repeat here for those who may have missed it.  The amp/ hour is
fairly rough measure, but basically it was a rating of
how much current could be supplied continuously over a
specified period of time without the voltage falling
below a minimum level.  The time was usually taken to be
eight hours.  So the formula was amp hour rating / 8  =
current in amps.

As an example, a 120 ah battery is supposed to be able
to supply 120 / 8 = 15  amps  continuously for 8 hours. 
It will supply more current for less time or less
current for more time.  Theoretically, if will supply 15
amps for 8 hours it should supply 30 amps for
4 hours or 7.5 amps for 16 hours.     

However, the curve isn't linear, especially at higher
discharge rates.   At 60 amps, won't last two hours, and
certainly won't deliver 120 amps for more than a few
minutes.   On the other hand, if only drawing a couple
of amps from it, will last longer than the curve would
indicate.   

Temperature has a pretty marked effect, though.  The amp
hour rating was usually calculated for 80 deg F.  [27 C] 
Higher temperatures increase the chemical reaction,
hence the output, but operation above 110 F [ 43.3 C]
shortens battery life.   

The amp hr capacity was said to reduce about 0.75% for
each one degree F  [ ~ 0.55 C] below the rated temp.   
 At 0 deg F.  [ minus 17.77 C] the available output
would only be 60 percent of the rated output.  

The measure isn't used as often today as used to be,
possibly because of its imprecision.  As the battery
ages, its capacity can decrease for various reasons.  

14==> If someone has other figures for inverter effeciency,  battery
discharge rates against time, please post on group or e-mail me.
My data may not be completely up to date.

~ larryn











 Now, are you going to
>say that your system only gets 20 amp/hours a day? With 5 panels? Wow, I
>think you got ripped off! Again, assume you have 5 50 watt panels, at 3.0
>amps. and 16 volts. Assume too that you have 4 hours of sun (its winter)
>Thats 12 amp hours (per panel) x 5 panels  = 60 amp/hours a day. Now, that
>leaves you with a nice reserve of 40 amp/hours still in your battery. Am I
>wrong? Please let me know if my math or thinking is incorrect.
>TCD
>
>>Any crockpot I've seen wanted about 300 watts of 117vac power. Having said
>>that I could use a crockpot but it would use up all the power my 5 panel
>>system ($4000 worth of stuff)would make that day if used 8 hours. A solar
>>cooker would be much cheaper and work just fine when the sun shines....DCK
>>
>