## Electricity/Temperature conversion

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• If someone could check my math, I d be thankful If I want to raise 10 gallons of water from 77 degrees F to 167 F (in an insulated container) I think it takes
Message 1 of 4 , Jun 30, 2006
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If someone could check my math, I'd be thankful

If I want to raise 10 gallons of water from 77 degrees F to 167 F (in
an insulated container) I think it takes 2197 watt hours.

In metric, 37.85 liters raised 50 C.

So,
37850 grams, times 50 calories each, or 1892500 calories, or 7910650
joules. One joule is a watt second. Divide by 3600 seconds in an
hour, and I get 2197 watt hours.

Did I miss somewhere?
• 10 Gal = 37.854 liters 167F - 77F = 90 F = 32.222C Assuming 1atm in your vessel, 1kcal energy raises 1 liter of water 1 degree C so, 37.854 liters * 32.222C *
Message 2 of 4 , Jul 1, 2006
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10 Gal = 37.854 liters
167F - 77F = 90 F = 32.222C

1kcal energy raises 1 liter of water 1 degree C so,
37.854 liters * 32.222C * 1kcal/(1L*1C) = 1219.7kcal
1 Joule ~ 2.390x10^-4kcal
1219.7kcal * 1J/2.390x10^-4 = 5,103,300 J
1 Joule = 1 Watt / second
1 Joule = (1 Watt / second) * (3600 seconds/ 1 hour)
5103,300 J * (1 Watt/second) * (3600 seconds/ 1 hour) = 1417.6 Watt/Hours

Did I screw that up? Not sure where you got your 50C, but that's the
point at which my calculations diverged.

Paul

On 6/30/06, richardt2005 <richardt2005@...> wrote:
> If someone could check my math, I'd be thankful
>
> If I want to raise 10 gallons of water from 77 degrees F to 167 F (in
> an insulated container) I think it takes 2197 watt hours.
>
> In metric, 37.85 liters raised 50 C.
>
> So,
> 37850 grams, times 50 calories each, or 1892500 calories, or 7910650
> joules. One joule is a watt second. Divide by 3600 seconds in an
> hour, and I get 2197 watt hours.
>
> Did I miss somewhere?
>
>
>
>
>
>
>
>
> New Distillers group archives are at http://archive.nnytech.net/
> FAQ and other information available at http://homedistiller.org
>
>
>
>
>
>
>
>
>
• ... OK, I think that s where we diverged. 77F = 25C 167F = 75C 75C-25C = 50C So, for a 20 gallon boiler, which I expect to fill half full of wash, 4kw in
Message 3 of 4 , Jul 1, 2006
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--- In new_distillers@yahoogroups.com, "Paul McMillan" <Paul@...>
wrote:
>
> 10 Gal = 37.854 liters
> 167F - 77F = 90 F = 32.222C

OK, I think that's where we diverged.

77F = 25C
167F = 75C
75C-25C = 50C

So, for a 20 gallon boiler, which I expect to fill half full of wash,
4kw in heating elements to get up to runable temerature (and then
expet to run at 1 kw) isn't outlandish?
• Right. the issue is that the subtraction should be done in kelvin, making your 50C number correct. F is such a screwy scale. My bad... Paul
Message 4 of 4 , Jul 2, 2006
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Right. the issue is that the subtraction should be done in kelvin,
making your 50C number correct. F is such a screwy scale. My bad...

Paul

On 7/1/06, richardt2005 <richardt2005@...> wrote:
> --- In new_distillers@yahoogroups.com, "Paul McMillan" <Paul@...>
> wrote:
> >
> > 10 Gal = 37.854 liters
> > 167F - 77F = 90 F = 32.222C
>
> OK, I think that's where we diverged.
>
> 77F = 25C
> 167F = 75C
> 75C-25C = 50C
>
> So, for a 20 gallon boiler, which I expect to fill half full of wash,
> 4kw in heating elements to get up to runable temerature (and then
> expet to run at 1 kw) isn't outlandish?
>
>
>
>
>
>
>
>
> New Distillers group archives are at http://archive.nnytech.net/
> FAQ and other information available at http://homedistiller.org
>
>