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Re: [new_distillers] Electricity/Temperature conversion

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  • Paul McMillan
    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 1 of 4 , Jul 1, 2006
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      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
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    • richardt2005
      ... 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 2 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?
      • Paul McMillan
        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 3 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
          >
          >
          > Yahoo! Groups Links
          >
          >
          >
          >
          >
          >
          >
          >
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