- 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 * 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

>

>

> Yahoo! Groups Links

>

>

>

>

>

>

> - --- In new_distillers@yahoogroups.com, "Paul McMillan" <Paul@...>

wrote:>

OK, I think that's where we diverged.

> 10 Gal = 37.854 liters

> 167F - 77F = 90 F = 32.222C

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

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

>

>

> Yahoo! Groups Links

>

>

>

>

>

>

>

>