RE: Ok.. where (Calorimeter)

To: tesla-at-pupman-dot-com
Subject: RE: Ok.. where (Calorimeter)
From: "Payne, Will E" <will.e.payne-at-lmco-dot-com> (by way of Terry Fritz <terryf-at-verinet-dot-com>)
Date: Wed, 18 Nov 1998 13:06:28 -0700
Approved: terryf-at-verinet-dot-com
I assume my ballast bank thermal mass consists of the 5 gal of water plus
the actual resistance elements, which are mostly steel, plus ceramic,
thermoplastic, and an aluminum rail.  Since the water is the large majority
of the mass, I approximate the thermal mass of my resistance elements as 2kg
of iron, a small adjustment to the total.

It is THE metrological standard for power measurement, I have used it for
microwave power, just dump a roll of lossy coax in a bucket of water and use
it as a dummy load.  Since my meters were freaking out over the ugly
waveform in my TC primary, and I already had known load resistors in a known
water bucket, I only needed to add the thermometer and a simple spreadsheet
to get true power in the ballast.  I'm still assuming the phase angle of my
total power is the phase angle of ballast power.  This aint exactly so,
since my TC is not power factor corrected.  When I get Terry's VI probes
hooked to my scope, I'll know the phase angle as well and can correctly
convert true ballast power to true TC power.

Real "cool"  or real "warm" ;-)

Will

> ----------
> From: 	Tesla List[SMTP:tesla-at-pupman-dot-com]
> Sent: 	Tuesday, November 17, 1998 18:12
> To: 	tesla-at-pupman-dot-com
> Subject: 	Re: Ok.. where am I going wrong
> 
> Original Poster: Bryan Kaufman <bryan-at-apexrad-dot-com> 
> 
> 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
> 
> 
> 
> 
>