Re: Power Query

[Marc_F_Hult <MFHult@xxxxxxxxxxxxxxxxxxxxx>]

On Fri, 02 Feb 2007 09:48:17 -0500, Marc_F_Hult
<MFHult@xxxxxxxxxxxxxxxxxxxxx> wrote in message
<hth6s21mh57ld2ik6jnvj8aefqaav007ik@xxxxxxx>:

>On Thu, 01 Feb 2007 22:54:35 -0500, Beamer Smith <beamer@xxxxxxxxxxxxxx>
>wrote in message  <jcc5s2lj3fi2hu3jek410b0roi234hk8rp@xxxxxxx>:
>
>>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...
>>
>
>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
 XXXX^^^^^^    XXXX^^^^^^^^^^^
     WaterIn     WaterLeaked
>
>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
>www.ECOntrol.org

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
www.ECOntrol.org