Testing the TK Tar Baby

[TinselKoala]

Now.... does he mean that it took 3 hours for his insulated load cell to increase the temperature of 175 mL of water by 14.4 degrees C?

It takes (14.4 degrees x 175 mL) = 2520 Joules to heat that water that much. 3 hours is 10800 seconds, and a Watt is a Joule per Second, so the average power needed simply to heat the water is 2520 Joules PER 10800 seconds, or 2520/10800 = about a quarter of a Watt average, over the entire period.  (or 0.233333333..... Watt if you prefer.)

The circuit, according to gmeast's measurements, took 0.43885 Watt to do it ( measured to the hundredth of a milliWatt how? ), for a heating efficiency of 0.23333/0.43885 or a little over 50 percent. Most of the losses probably occur in the  mosfet itself, and through the insulation of the load cell.

Where did gmeast's " .9766867Watt to do that" (sic) number come from, can anyone tell me? Is this a calibrated measurement?

[Magluvin]

0.23333/0.43885 or a little over 50 percent.

Where did gmeast's " .9766867Watt to do that" (sic) number come from, can anyone tell me? Is this a calibrated measurement?

Looks like GM knows how to use a calculator. ;]

MaGs

[mrsean2k]

The only way I can see that figure being in some way correct, is if 0.97-blah-blah is the calculated instantaneous power during the on phase of his duty cycle, making his duty cycle ~25%. That would let him arrive at the correct total energy required for that increase in temp.? Very hard to tell from the phrasing used.

[TinselKoala]

http://en.wikipedia.org/wiki/Dolomite

http://en.wikipedia.org/wiki/Diorite

[TinselKoala]

Can't you be more specific, .99? Just where, exactly, is Ainslie wrong?




And three months of careful and patient explanation, poynt by carefully argued poynt, are gone ... like tears in rain.