Math Check

I picked up something called a Chill-Chaser on eBay a while ago. It's a heat stick with a temperature control on it. I'm having trouble finding any specs on it, but would like to figure it out.

I metered the element at 15.6 ohms
and my outlet at 125 volts.

V = I * R
125 = I * 15.6
I = 8 amps

P = I * I * R
P = 1000 Watts

Did I use the formulas correctly?

The reason I ask is that when I use the Time to Boil Calculator at 1000 watts, I have to set the efficiency at 60% to match the time it take me to raise the temperature. This is even at lower temperatures increases (say 140 to 150 degrees) with an insulation wrap around the pot. Is this typical or am I missing something?

Your math is right.

Can't help you with the rest, but 60% seems low. At least I can tell you your math is right . (for me, P=VI is easier to remember than P=I^2R, but whatever works for you!)

shortyjacobs said:

. . . for me, P=VI is easier to remember than P=I^2R, but whatever works for you!

Click to expand...

Oh yeah. Now I remember algebra.

Thanks for the confirmation on the math. Running some more tests to see if my original efficiency numbers were correct.

Well remember that resistance changes with temperature. At what temperature did you measure the 15.6 ohms?

I mean, we could easily be looking at 20+ ohms of resistance at higher temperatures, which means your amps delivered is going to be more like 5-6, giving you more like 600-700W which seems in line with your results.

hmmm. Did not know that. I took the initial reading at room temperature.

Made me curious, so I just unplugged it and took another reading. I'm sure that the element cools down quickly, but sitting in 195 degree water, it still read 15.6 ohms.

Getting to that temperature was interesting. At the lower end of the scale, around room temperature, the efficiency was in the 90’s. By the time it got close to 200 degrees it dropped to 30%. I’m sure I’m loosing a lot of heat to surface area, but this supports what you’re saying about getting less wattage to the liquid at higher temperatures.

I don’t expect to be able to boil with this thing, but it’s an interesting experiment that probably relates to the way that all electric systems work.

ressistance does increase with temperature, but its by a percentage.

if you were dealing with something with thousands of ohms, or huge temperature ranges, then temperature is going to matter. but the difference between a short, thick wire resistor (the element) at 70 degrees and 200 degrees is going to be negligable, as you found.

so for example if the resistance increases by, say, 0.5% per 100 degrees, (15.6 ohms x 100.7%) is only 15.7 which you might not even pick up without a really good meter.

I used a 1000W bucket heater for years to heat up my Strike water. It takes several hours to heat up 10 gallons. I would let it run overnight on a timer so it would be ready in the morning. I don't know the efficiency, but it's not great.

yea, 1000w is only 3400 BTU/hr. it takes almost 12,000BTUs to boil 10 gallons, so it would take about 4 hours to heat it even if the pot was completely insulated and lost no heat to the atmosphere in that amount of time. and immersion heaters like that heat stick are almost 100% efficient (in converting electricity to heat).

realistically it would take 6-8 hours to boil 10 gallons with only a 1kW element... if you could keep the boil going. the rate of heat loss increases as the temperature difference between the inside and outside increases. so the closer to boil you got, the faster you would loose heat.

science!