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Re: Ok.. where am I going wrong



Nice calorimeter experiment.
For pure H2O C(T) = 1 cal / g °C (by definition) plug in volume and dT ; the
rest is units conversion. You can actually measure the power deposition and
work backward to see if the voltage and current values match what you expect
them to be . . .

If you want to compensate for the cooling loss in kCal/sec you end up with a
relatively simple relation. Nice, very cool Will.

Where does the 2kg of iron come from?

Bryan Kaufman

> Original Poster: "Payne, Will E" <will.e.payne-at-lmco-dot-com>
>
> Howdy !
>
>         My meters gave very doubtful readings, mainly reading WAAAYYY high
> due to the horrible waveform in the mains circuit.  I have resorted to
> calorimetry to measure my primary current, and it seemd to be pretty
> accurate.
>         The resistance in my ballast bank is electric hot water elements in
> a 5 gal plastic bucket.  I measure the temp before and after a 10 min run
> with a candy thermometer.  Then I measure again 10 min later to
> approximately account for heat lost during the run.  Turns out my bucket
> heats much faster than it cools, so the correction is fairly small.  I stir
> the water to prevent a very hot layer forming at the top and boiling off.
> One could try to measure liquid level before and after to account for
> vaporization, but I dont.
>         Built an Excel spreadsheet to find the Joules and current, etc.  I
> take 5 gal of water, plus 2 kg of iron is my estimated thermal mass.  I find
> the results to be in good agreement with how often by 20 A main breaker
> blows.  My biggest unknown is the phase angle.  This could be estimated with
> a scope.
>
> Will