Password avoidance

On Sun, 14 Mar 2010 22:57:37 -0500, "Dave" <>
wrote:

>
>
>"Char Jackson" <> wrote in message
>news:...
>> On Sun, 14 Mar 2010 22:03:16 -0500, "Dave" <>
>> wrote:
>>
>>>
>>>
>>>"Char Jackson" <> wrote in message
>>>news:...
>>>> On Sun, 14 Mar 2010 16:28:40 -0500, "Dave" <>
>>>> wrote:
>>>>
>>>>>
>>>>>
>>>>>"Peter Foldes" <> wrote in message
>>>>>news:hnjj9t$d8b$...
>>>>>> Dave
>>>>>>
>>>>>> You do know the difference between a household 100v or a 347v
>>>>>> commercial
>>>>>> in the amount of their usage
>>>>>>
>>>>>> --
>>>>>> Peter
>>>>>>
>>>>>
>>>>>I can't give you a scientific example of the differences between those
>>>>>two.
>>>>>Off the top of my head I think you might be comparing single phase to a
>>>>>multiple, like delta or y. I can say the usage, no matter what the
>>>>>supply,
>>>>>is going to be dependant on the demand.
>>>>>One final attempt at trying to not look stupid is that for the same
>>>>>device,
>>>>>if it's capable of handling the higher voltage, usually the higher the
>>>>>voltage, the lower the current so the lower the cost to use.
>>>>
>>>> No, the cost will be the same regardless of the voltage. Usage is
>>>> measured in Watts. Watts are the product of voltage times current, so
>>>> if voltage goes up by a certain factor then current comes down by the
>>>> same factor. In the end, the Watts are the same, therefore the usage
>>>> and the cost are the same.
>>>>
>>>
>>>I agree with you as far a theory goes, P=I*E. But in application, if you
>>>have a dryer for instance. Hook it to 120V and it might use 15 Amps, hook
>>>it
>>>to 240V and it will probably be less than half the Amperage. So, the total
>>>power used will be less for the 240V than for the 120V.
>>
>> Come on, you know that's not true.
>>
>
>Actually, it is and IIRC it has to do with wires ability to carry power more
>efficiently at higher voltages. As I understand it, that's also one of the
>reasons utilities transmit power at kilovolts instead of 120 volts.
>But, since I can't quote any statistics and am too lazy to google any more,
>I'm going to say you win. I think the difference is minor anyway so let's
>revert to theory and say you're right.
>Dave

You're confusing Ohm's Law with transmission theory, two totally
different things. You correctly stated one form of Ohm's Law above,
P=IE. That darn equals sign gets in the way of your current position.

On Sun, 14 Mar 2010 23:03:10 -0500, "Dave" <>
wrote:

>My bad, you are correct. The acronym I should have use is KVA
>(KiloVoltAmps). IIRC KWH is true power and KVA is apparent power. I don't
>know that much about the difference, I know I used to rent generators rated
>in KWH and now they are in KVA and there is a slight difference in the
>output ability. 2) There is a formula for this one. :-D
>Dave

Oy. Glad we got to the bottom of it.

"Char Jackson" <> wrote in message
news:...
> On Sun, 14 Mar 2010 22:57:37 -0500, "Dave" <>
> wrote:
>
>>
>>
>>"Char Jackson" <> wrote in message
>>news:...
>>> On Sun, 14 Mar 2010 22:03:16 -0500, "Dave" <>
>>> wrote:
>>>
>>>>
>>>>
>>>>"Char Jackson" <> wrote in message
>>>>news:...
>>>>> On Sun, 14 Mar 2010 16:28:40 -0500, "Dave" <>
>>>>> wrote:
>>>>>
>>>>>>
>>>>>>
>>>>>>"Peter Foldes" <> wrote in message
>>>>>>news:hnjj9t$d8b$...
>>>>>>> Dave
>>>>>>>
>>>>>>> You do know the difference between a household 100v or a 347v
>>>>>>> commercial
>>>>>>> in the amount of their usage
>>>>>>>
>>>>>>> --
>>>>>>> Peter
>>>>>>>
>>>>>>
>>>>>>I can't give you a scientific example of the differences between those
>>>>>>two.
>>>>>>Off the top of my head I think you might be comparing single phase to
>>>>>>a
>>>>>>multiple, like delta or y. I can say the usage, no matter what the
>>>>>>supply,
>>>>>>is going to be dependant on the demand.
>>>>>>One final attempt at trying to not look stupid is that for the same
>>>>>>device,
>>>>>>if it's capable of handling the higher voltage, usually the higher the
>>>>>>voltage, the lower the current so the lower the cost to use.
>>>>>
>>>>> No, the cost will be the same regardless of the voltage. Usage is
>>>>> measured in Watts. Watts are the product of voltage times current, so
>>>>> if voltage goes up by a certain factor then current comes down by the
>>>>> same factor. In the end, the Watts are the same, therefore the usage
>>>>> and the cost are the same.
>>>>>
>>>>
>>>>I agree with you as far a theory goes, P=I*E. But in application, if you
>>>>have a dryer for instance. Hook it to 120V and it might use 15 Amps,
>>>>hook
>>>>it
>>>>to 240V and it will probably be less than half the Amperage. So, the
>>>>total
>>>>power used will be less for the 240V than for the 120V.
>>>
>>> Come on, you know that's not true.
>>>
>>
>>Actually, it is and IIRC it has to do with wires ability to carry power
>>more
>>efficiently at higher voltages. As I understand it, that's also one of the
>>reasons utilities transmit power at kilovolts instead of 120 volts.
>>But, since I can't quote any statistics and am too lazy to google any
>>more,
>>I'm going to say you win. I think the difference is minor anyway so let's
>>revert to theory and say you're right.
>>Dave
>
> You're confusing Ohm's Law with transmission theory, two totally
> different things. You correctly stated one form of Ohm's Law above,
> P=IE. That darn equals sign gets in the way of your current position.
>
>

So, when you transmit power it acts one way, but when you apply it to an
appliance it acts another? I was never taught that and never found it in
application either. YRMV.
I actually did not state Ohm's law, it is I=E/R. What I stated was Watt's
law, which is P=I*E. You do realize one is relevant to power and one is
relevant to resistance?
I would agree, except you are assuming the power is the same in both
instances, or more correctly, the current draw, which I am saying is not.
So, in my case, you can't solve for P in both 120 and 240 by assuming I is
automatically going to be half if E is doubled.
I hope you're not going to have us Thevonize a circuit, I hated that more
than anything. Plus, I'd have to dig out some reference books just to get
all the steps correct. :-D
Dave

"TOM" <> wrote in message
news:mfenn.252620$...
> Dave wrote:
>>
>>
>> "Peter Foldes" <> wrote in message
>> news:hngtic$srd$...
>>> Ken
>>>
>>> 35 yrs being an Electrician. The draw when you open a light switch with
>>> a single 100 w bulb will be approximately +- 0.03 kvh at start up where
>>> as the light staying open will draw 0.01 per kvh per every 10 hrs, Now
>>> open and close the switch 10 times per day which will cause 0.30kvh
>>> registration on the meter as compared to a bulb continuously burning for
>>> a 24hr period as approximately .025. Same applies to any electrical
>>> apparatus be it a light bulb or the computer plus adding the opening
>>> surge and then the burning ( running) time
>>> But I will try and find the documentation on this Ken and will get over
>>> to you (right now the documentation is in my head and I do not have my
>>> CA book at hand here so as to show a hard copy
>>>
>>> --
>>> Peter
>>>
>>
>> Not to seem irrational or argumentative, but 24 years as an electronic
>> tech along with two side businesses dealing in electronics repair, plus
>> google showed me that the surge when starting a bulb lasts appx 1/2
>> cycle, or 1/120th of a second, plus only 10-15% of the energy consumed by
>> an incandescent bulb is turned into light, the rest is heat. So, if you'd
>> have to cycle your bulbs off and on an unbelievable amount of times a day
>> and it would only be equivalent to a few seconds of burn time at most.
>> Fluorescents are a little more economical, rule of thumb is if you are
>> going to turn one back on within 15 minutes, it's usually cheaper to
>> leave it on, unless it's in the high-usage part of the day when many
>> utilities charge more for KWH of consumption, then the rule of thumb is 5
>> minutes. Another offset in the flourescent is the fact the bulbs,
>> transformers and fixtures are quite a bit more expensive, so shortening
>> their life can account for a bit more monetary loss by cycling.
>> Information I found concerning computers was pretty much the same, if
>> it's going to be idle for more than 5 minutes, put it to sleep or 10
>> minutes then turn it off and you will save energy.
>>
>> http://www.energysavers.gov/your_hom.../mytopic=12280
>> http://www.scientificamerican.com/ar...you-leave-room
>> http://green.yahoo.com/blog/the_cons...ur-lights.html
>> Original google was: turn lights off or leave on
>>
>> To be fair, I did find articles to support your claim, but the number was
>> probably 10 to 1 against leaving them on. One article stated turning an
>> incandescent light on used as much power as leaving it on for 5 minutes.
>> This may sound reasonable, but I turned a cold bulb on and in less than a
>> minute it was too hot to touch so I doubt this data. Most articles, for
>> and against, agreed cycling electrical devices cuts down of their life,
>> but the savings in electricity outweigh the loss in life (which is
>> minimal at best).
>> Finally, this subject came up when I was in college (for electronics) and
>> worked with a guy who was pursuing his Master Electrician's License. He
>> stated pretty much the same as you, so we worked it out in lab one day.
>> Don't have my references, but the data I remember is the same as the
>> articles I found this time, the surge duration is so short it takes a lot
>> of cycles to add up to a second of burn time.
>> Just my 2cents,
>> Dave
>
> I got to thinking about the lamps used on theater marques. I think they
> last as long as the do because the switching cycle is short enough the
> filament doesn't have to go through the expansion/contraction of a
> household-type on/of cycle.
>
> Probably wrong, but it sounds sort of logical to me... :>))
> --
> TOM - Vista, CA - USA

Long ago, in the 50s and before I got my BSEE from Purdue, I
worked part-time at a Westinghouse Sales facility. We stocked
"ruggedized" light bulbs for traffic lights and such. They were
marked "130v" and had a heavy duty filament structure. They
cost 2 times as much as standard bulbs, but I was told
that they lasted 5 times as long!

Chris

On Mon, 15 Mar 2010 09:26:32 -0500, "Dave" <>
wrote:

>
>
>"Char Jackson" <> wrote in message
>news:...
>> On Sun, 14 Mar 2010 22:57:37 -0500, "Dave" <>
>> wrote:
>>
>>>
>>>
>>>"Char Jackson" <> wrote in message
>>>news:...
>>>> On Sun, 14 Mar 2010 22:03:16 -0500, "Dave" <>
>>>> wrote:
>>>>
>>>>>
>>>>>
>>>>>"Char Jackson" <> wrote in message
>>>>>news:...
>>>>>> On Sun, 14 Mar 2010 16:28:40 -0500, "Dave" <>
>>>>>> wrote:
>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>"Peter Foldes" <> wrote in message
>>>>>>>news:hnjj9t$d8b$...
>>>>>>>> Dave
>>>>>>>>
>>>>>>>> You do know the difference between a household 100v or a 347v
>>>>>>>> commercial
>>>>>>>> in the amount of their usage
>>>>>>>>
>>>>>>>> --
>>>>>>>> Peter
>>>>>>>>
>>>>>>>
>>>>>>>I can't give you a scientific example of the differences between those
>>>>>>>two.
>>>>>>>Off the top of my head I think you might be comparing single phase to
>>>>>>>a
>>>>>>>multiple, like delta or y. I can say the usage, no matter what the
>>>>>>>supply,
>>>>>>>is going to be dependant on the demand.
>>>>>>>One final attempt at trying to not look stupid is that for the same
>>>>>>>device,
>>>>>>>if it's capable of handling the higher voltage, usually the higher the
>>>>>>>voltage, the lower the current so the lower the cost to use.
>>>>>>
>>>>>> No, the cost will be the same regardless of the voltage. Usage is
>>>>>> measured in Watts. Watts are the product of voltage times current, so
>>>>>> if voltage goes up by a certain factor then current comes down by the
>>>>>> same factor. In the end, the Watts are the same, therefore the usage
>>>>>> and the cost are the same.
>>>>>>
>>>>>
>>>>>I agree with you as far a theory goes, P=I*E. But in application, if you
>>>>>have a dryer for instance. Hook it to 120V and it might use 15 Amps,
>>>>>hook
>>>>>it
>>>>>to 240V and it will probably be less than half the Amperage. So, the
>>>>>total
>>>>>power used will be less for the 240V than for the 120V.
>>>>
>>>> Come on, you know that's not true.
>>>>
>>>
>>>Actually, it is and IIRC it has to do with wires ability to carry power
>>>more
>>>efficiently at higher voltages. As I understand it, that's also one of the
>>>reasons utilities transmit power at kilovolts instead of 120 volts.
>>>But, since I can't quote any statistics and am too lazy to google any
>>>more,
>>>I'm going to say you win. I think the difference is minor anyway so let's
>>>revert to theory and say you're right.
>>>Dave
>>
>> You're confusing Ohm's Law with transmission theory, two totally
>> different things. You correctly stated one form of Ohm's Law above,
>> P=IE. That darn equals sign gets in the way of your current position.
>>
>>
>
>So, when you transmit power it acts one way, but when you apply it to an
>appliance it acts another? I was never taught that and never found it in
>application either. YRMV.

Yes, transmission lines have known losses which need to be accounted
for and factored into their designs, but that has nothing to do with
the current discussion. There are no long distance transmission lines
in a residential or even commercial property, so the losses from such
lines are irrelevant to the current discussion. I suspect the reason
you were never taught that is because this is far outside your area of
expertise, which I believe is what you said in a previous post.

Take your example of the electric dryer. If you have two dryers, one
designed for 120v and the other designed for 240v, the 120v model will
draw twice as many amps as the 240v model. Both dryers will use the
same number of Watts, (because P=IE), assuming the only differences
between the two models are the voltages they are designed to operate
with.

>I actually did not state Ohm's law, it is I=E/R. What I stated was Watt's
>law, which is P=I*E. You do realize one is relevant to power and one is
>relevant to resistance?
>I would agree, except you are assuming the power is the same in both
>instances, or more correctly, the current draw, which I am saying is not.

If we're having a technical discussion, then P (power) is expressed in
Watts, not amps. Current draw refers to amps, but power refers to
Watts. Watts are the product of amps times voltage. (P=IE)

>So, in my case, you can't solve for P in both 120 and 240 by assuming I is
>automatically going to be half if E is doubled.

If you're saying that "P equals IE" is wrong, and should be restated
as "P is approximately equal to IE", well we better get the textbooks
updated with this new formula. In the meantime, yes, if E is doubled
then I will be halved and P will remain unchanged. It doesn't just
work that way in the textbooks, it works that way in the field, too.

>I hope you're not going to have us Thevonize a circuit, I hated that more
>than anything. Plus, I'd have to dig out some reference books just to get
>all the steps correct. :-D

Not a bad idea to dust off the books now and then.

On Fri, 12 Mar 2010 21:00:31 -0500, Peter Foldes wrote:

> Gene
>
>>> Once a day. Jeeeze.....
>>> If you start up your computer LOTS of times a day a) it uses lots more
>>> electricity than if you just keep it running and b) places excessive wear on
>>> parts like the disk heads...
>
> Actually Gordon is 100% correct with that statement. I agree . Same scenario if you
> open and close a 100w light in the room on a 110v 60hz circuit it will take up twice
> that amount of power than if you would leave the light on for a 24hr period without
> shutting it off. Same exact principle for the computer

It could hardly be the same exact principle for the computer, since the
light bulb is a hot filament - a resistive load - with its resistance
changing as it heats up, whereas a computer is a reactive load.

I think you guys need to plug a Kill A Watt in and look what (watt?)
happens. You should also measure the cold resistance of a filament bulb to
get some idea how much current it might draw before it reaches its
operating temperature - and note that it takes well under a second to get
to that temperature.

Here's a hint: 100W bulb = 90 ohms. At 120V, that's 1.33A, or 160W at
startup. For under 1 second...

So tell me - I *really* want to know - how can this bulb use up twice as
much power when I turn it on, compared to its steady state consumption for
a day?

--
Gene E. Bloch letters0x40blochg0x2Ecom

On Sat, 13 Mar 2010 15:29:36 -0800, TOM wrote:

> Peter Foldes wrote:
>> Ken
>>
>> 35 yrs being an Electrician. The draw when you open a light switch with
>> a single 100 w bulb will be approximately +- 0.03 kvh at start up where
>> as the light staying open will draw 0.01 per kvh per every 10 hrs, Now
>> open and close the switch 10 times per day which will cause 0.30kvh
>> registration on the meter as compared to a bulb continuously burning for
>> a 24hr period as approximately .025. Same applies to any electrical
>> apparatus be it a light bulb or the computer plus adding the opening
>> surge and then the burning ( running) time
>> But I will try and find the documentation on this Ken and will get over
>> to you (right now the documentation is in my head and I do not have my
>> CA book at hand here so as to show a hard copy
>
>
> I was told, by the journeyman electrician I worked for, that the terms
> "open" and "close" originated "way back when." He assumed that when
> candles were used to provide light, they had shutters; you open the
> shutters to light the room and close the shutters to darken the room.
>
> Another idea was the "barn door" shutters on Klieg lights:
>
> http://en.wikipedia.org/wiki/Klieg_light
>
> Barn doors:
> http://en.wikipedia.org/wiki/Barn_doors
>
> Closer to the original off topic:
> http://michaelbluejay.com/electricity/myths.html

Have you ever seen an old-fashioned knife switch? One glance at that (or
for that matter at the interior of a modern wall switch) would tell you
where those terms come from. They are quite literal...

--
Gene E. Bloch letters0x40blochg0x2Ecom

On Mon, 15 Mar 2010 12:47:24 -0500, Char Jackson wrote:

> On Mon, 15 Mar 2010 09:26:32 -0500, "Dave" <>
> wrote:
>
>>
>>
>>"Char Jackson" <> wrote in message
>>news:...
>>> On Sun, 14 Mar 2010 22:57:37 -0500, "Dave" <>
>>> wrote:
>>>
>>>>
>>>>
>>>>"Char Jackson" <> wrote in message
>>>>news:...
>>>>> On Sun, 14 Mar 2010 22:03:16 -0500, "Dave" <>
>>>>> wrote:
>>>>>
>>>>>>
>>>>>>
>>>>>>"Char Jackson" <> wrote in message
>>>>>>news:...
>>>>>>> On Sun, 14 Mar 2010 16:28:40 -0500, "Dave" <>
>>>>>>> wrote:
>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>"Peter Foldes" <> wrote in message
>>>>>>>>news:hnjj9t$d8b$...
>>>>>>>>> Dave
>>>>>>>>>
>>>>>>>>> You do know the difference between a household 100v or a 347v
>>>>>>>>> commercial
>>>>>>>>> in the amount of their usage
>>>>>>>>>
>>>>>>>>> --
>>>>>>>>> Peter
>>>>>>>>>
>>>>>>>>
>>>>>>>>I can't give you a scientific example of the differences between those
>>>>>>>>two.
>>>>>>>>Off the top of my head I think you might be comparing single phase to
>>>>>>>>a
>>>>>>>>multiple, like delta or y. I can say the usage, no matter what the
>>>>>>>>supply,
>>>>>>>>is going to be dependant on the demand.
>>>>>>>>One final attempt at trying to not look stupid is that for the same
>>>>>>>>device,
>>>>>>>>if it's capable of handling the higher voltage, usually the higher the
>>>>>>>>voltage, the lower the current so the lower the cost to use.
>>>>>>>
>>>>>>> No, the cost will be the same regardless of the voltage. Usage is
>>>>>>> measured in Watts. Watts are the product of voltage times current, so
>>>>>>> if voltage goes up by a certain factor then current comes down by the
>>>>>>> same factor. In the end, the Watts are the same, therefore the usage
>>>>>>> and the cost are the same.
>>>>>>>
>>>>>>
>>>>>>I agree with you as far a theory goes, P=I*E. But in application, if you
>>>>>>have a dryer for instance. Hook it to 120V and it might use 15 Amps,
>>>>>>hook
>>>>>>it
>>>>>>to 240V and it will probably be less than half the Amperage. So, the
>>>>>>total
>>>>>>power used will be less for the 240V than for the 120V.
>>>>>
>>>>> Come on, you know that's not true.
>>>>>
>>>>
>>>>Actually, it is and IIRC it has to do with wires ability to carry power
>>>>more
>>>>efficiently at higher voltages. As I understand it, that's also one of the
>>>>reasons utilities transmit power at kilovolts instead of 120 volts.
>>>>But, since I can't quote any statistics and am too lazy to google any
>>>>more,
>>>>I'm going to say you win. I think the difference is minor anyway so let's
>>>>revert to theory and say you're right.
>>>>Dave
>>>
>>> You're confusing Ohm's Law with transmission theory, two totally
>>> different things. You correctly stated one form of Ohm's Law above,
>>> P=IE. That darn equals sign gets in the way of your current position.
>>>
>>>
>>
>>So, when you transmit power it acts one way, but when you apply it to an
>>appliance it acts another? I was never taught that and never found it in
>>application either. YRMV.
>
> Yes, transmission lines have known losses which need to be accounted
> for and factored into their designs, but that has nothing to do with
> the current discussion. There are no long distance transmission lines
> in a residential or even commercial property, so the losses from such
> lines are irrelevant to the current discussion. I suspect the reason
> you were never taught that is because this is far outside your area of
> expertise, which I believe is what you said in a previous post.
>
> Take your example of the electric dryer. If you have two dryers, one
> designed for 120v and the other designed for 240v, the 120v model will
> draw twice as many amps as the 240v model. Both dryers will use the
> same number of Watts, (because P=IE), assuming the only differences
> between the two models are the voltages they are designed to operate
> with.
>
>>I actually did not state Ohm's law, it is I=E/R. What I stated was Watt's
>>law, which is P=I*E. You do realize one is relevant to power and one is
>>relevant to resistance?
>>I would agree, except you are assuming the power is the same in both
>>instances, or more correctly, the current draw, which I am saying is not.
>
> If we're having a technical discussion, then P (power) is expressed in
> Watts, not amps. Current draw refers to amps, but power refers to
> Watts. Watts are the product of amps times voltage. (P=IE)
>
>>So, in my case, you can't solve for P in both 120 and 240 by assuming I is
>>automatically going to be half if E is doubled.
>
> If you're saying that "P equals IE" is wrong, and should be restated
> as "P is approximately equal to IE", well we better get the textbooks
> updated with this new formula. In the meantime, yes, if E is doubled
> then I will be halved and P will remain unchanged. It doesn't just
> work that way in the textbooks, it works that way in the field, too.
>
>>I hope you're not going to have us Thevonize a circuit, I hated that more
>>than anything. Plus, I'd have to dig out some reference books just to get
>>all the steps correct. :-D
>
> Not a bad idea to dust off the books now and then.

It seems that people are ignoring Ohm's law.

If you double the voltage into a fixed device, you will not halve the
current but in fact double it. Ohm's law says E = IR (that's for DC; it's
more complicated in AC with capacitance and inductance to consider - i.e.,
reactive components, but that wouldn't affect this argument).

The result is that the putative dryer will consume four times the power (in
the short amount of time before it bursts into flame).

The case where the power is the same at 120 vs 140 volts is where we have
two entirely separate devices, each one properly designed for its voltage.

--
Gene E. Bloch letters0x40blochg0x2Ecom

On Mon, 15 Mar 2010 00:41:34 -0500, Char Jackson wrote:

> On Sun, 14 Mar 2010 23:03:10 -0500, "Dave" <>
> wrote:
>
>>My bad, you are correct. The acronym I should have use is KVA
>>(KiloVoltAmps). IIRC KWH is true power and KVA is apparent power. I don't
>>know that much about the difference, I know I used to rent generators rated
>>in KWH and now they are in KVA and there is a slight difference in the
>>output ability. 2) There is a formula for this one. :-D
>>Dave
>
> Oy. Glad we got to the bottom of it.

We didn't. KVA is a measure of power (when the power factor is included),
KWH is a measure of energy consumed.

Energy consumed is power times time, or equivalently, power is energy per
unit time.

--
Gene E. Bloch letters0x40blochg0x2Ecom

On Mon, 15 Mar 2010 18:07:25 -0700, "Gene E. Bloch"
<not-> wrote:

>On Mon, 15 Mar 2010 12:47:24 -0500, Char Jackson wrote:
>
>> Take your example of the electric dryer. If you have two dryers, one
>> designed for 120v and the other designed for 240v, the 120v model will
>> draw twice as many amps as the 240v model. Both dryers will use the
>> same number of Watts, (because P=IE), assuming the only differences
>> between the two models are the voltages they are designed to operate
>> with.
>>
>It seems that people are ignoring Ohm's law.
>
>If you double the voltage into a fixed device, you will not halve the
>current but in fact double it. Ohm's law says E = IR (that's for DC; it's
>more complicated in AC with capacitance and inductance to consider - i.e.,
>reactive components, but that wouldn't affect this argument).
>
>The result is that the putative dryer will consume four times the power (in
>the short amount of time before it bursts into flame).
>
>The case where the power is the same at 120 vs 140 volts is where we have
>two entirely separate devices, each one properly designed for its voltage.

I totally agree with your assessment. I may have written poorly or
clumsily above, but at least my dryer example had two dryers. ;-)