# Power Query

• posted

A bit of a stretch for this group but the experts seem to be here. Our building wants to install seperate meters for our power in part to mitigate the expence of the condo's power bill. The building is 208V but most abbliances are 220. I know for example that my mother can boil a pot of water in about 5 minutes on her 220 service and 22 stove (not new or efficient) but on my 208 service and

220 stove it takes much longer. Q: Am I paying more to boil a pot of water or not? Some say that the meter doesn't work that way and that if I had a 12V service and 24 hours to boil the water it would be the same as 208 for much less time.. Help me out here folks...

Beamer Smith I have spent most of my money on women and beer. The rest I just wasted...

• posted

My thought is that it takes a fixed amount of total energy input to boil a pot of water. Wether a low input is applied over a long time versus a higher input over a shorter time doesn't make much difference.

However, this may not be true, when you think it down to the extremes. In your example, it is likely the 12v input would NEVER boil the water, as the thermal loss would be greater (or at least equal) to the thermal input. So you could run the 12v heater in perpetuity and boil nothing but dollars.

So I suppose a higher temperature is probably more effecient. Measurable in dollars? That would be an interesting experiment.

• posted

No. You pay for power by the Killowatt Hour (KWH). 1 Killowatt Hour = 1 Kilowatt hour regardless of the voltage that it is delivered at or the how fast you use the energy. You pay the same for using: 4.55 Amps @ 220 volts (1 KW) for 1 hour, 4.81 Amps @ 208 Volts (1 KW) for 1 hour, 83.3 Amps @ 12 Volts (1 KW) for 1 hour, .455 Amps @ 220 volts (0.1 KW) for 10 hours, .481 Amps @ 208 Volts (0.1 KW) for 10 hours, .833 Amps @ 12 Volts (0.1 KW) for 10 hours.

Watt Hours are a measure of engergy used, like calories.

1 watt hour = 860.42 calories Heating a cup of water from 72 def F to 212 deg F (but not boiling it) takes 33,122 calories or 0.385 KWH. That ignores any heat that is wasted in the process (heating things near the stove) and does not include turning any water into steam (actually boiling it).

The effect on your stove of running 208 instead of 220 volts is that the heating elements will not draw energy as fast. Looking at the example above, a resistive heater that is designed to run at 1KW @ 220V will have a

48.4 Ohm resistance. Using that same 48.4 Ohm heating element on 208 volts gives us a 4.30 Amps or 0.895 KW. To use the same energy that the 220V heating element uses in an hour, you would need to run the 208V stove for 1 hour and 7 minutes.

The only limitation is that your highest heat is not quite as high. The payback you get for that is that the heating element will last a little longer because it is not running as hot.

• posted

Bill and Lee are correct in the assertion that you pay for watts not volts.

But that's not the whole answer. The Tea-kettle question is an classic example of an energy budget which is parallel/similar to a mass balance problem.).

Think of it first with the close analogy of a kettle with a leak being filled with water (mass balance):

RateWaterIn - RateWaterLeaked = Change in amount of water stored in kettle

Note whether you ever fill the kettle depends on BOTH rates. Too large a leak (hole) and the kettle never fills.

For making a kettle boil, the budget is: For temp < 212F

Heat In - Heat Out = Change in Heat stored in kettle

Heat Out ("lost") is a function of: Time, K, Thermal conductivity of the kettle material Temperature difference between kettle and environment

So if you reduce Time, you reduce heat loss.

So the 12 volt element may never get the water in an uninsulated kettle to boil, the 208 is less efficient than 220 volts and the most efficient is INSTANTLY applying enough heat to boil the water because there is not time to lose any heat.

If you reduce thermal conductivity of the kettle walls (i.e, insulate the kettle as is done with some electric kettles) you reduce heat loss.

Alternatively, you could raise the room temperature to 211 F to minimize the delta T heat loss. Hard on the heating bill among other things...

... Marc Marc_F_Hult

• posted

XXXX^^^^^^ XXXX^^^^^^^^^^^ WaterIn WaterLeaked

I ought to read my messages before I hit the carriage return .... L^3 (amount) not L^3/t (rate), in LHS of the first equation. See correction.

Marc Marc_F_Hult

• posted

It must be entertaining trying to cook with an oven mitt on one hand and an engineering calculator in the other. :^)

• posted

Um...OK...

While my condo is Very hot (being on the top floor contributes to that) I'm not sure I can get the boiler to give me 211F

So the upshot to all this is that I'm not paying any more to boil a

208 pot of water than my mom is moiling a 220 pot of water (all other things being equal.. ) Or at least microscopicly so.

Now all I need to do is solve why I'm usually getting 195 or so volts to that stove outlet (damned aluminum wiring)

Thanks folks

Beamer

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