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Discussion Starter · #1 · (Edited)
I went to an anwser web site this was the anwser i was given...
p=10kw=230v (assumed),v=208v
p=10kw=10,000w
e=rated voltage=230v(assumed)
resistance,r=e^2/p=230^2/10,000=5.29ohms
power conssumed by heater=v^2/r=208^2/5.29=8178.45w
power loss=10,000w - 8178.45w=1821.55w
(does anyone know what the '^ ' means)
I kinda knew there was a way to figure this out without needing any other imfo. I hope this will help others its something needed to know when taking the test.......
I need to go practice this now....
 

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I went to an anwser web site this was the anwser i was given...
p=10kw=230v (assumed),v=208v
p=10kw=10,000w
e=rated voltage=230v(assumed)
resistance,r=e^2/p=230^2/10,000=5.29ohms
power conssumed by heater=v^2/r=208^2/5.29=8178.45w
power loss=10,000w - 8178.45w=1821.55w
(does anyone know what the '^ ' means)
I kinda knew there was a way to figure this out without needing any other imfo. I hope this will help others its something needed to know when taking the test.......
I need to go practice this now....
You did, however, provide additional information.

You are comparing power consumption (watts) at both 230v and 208v, the 230v not mentioned in your original post.

So I think the question you originally referred to is what is the power consumption of a 10kw heater rated at 230v when it is supplied with 208 volts.

Calculate R at 230 volts : R=E²/P : R=230*230/10000 : R=5.29

At 208 volts, P=E²/R : P=208*208/5.29 : P= 43264/5.29 : P=8178.44 watts.

And the carat (^) means 'to the power of", so 'E^2' is the same as 'E²'.
 

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Discussion Starter · #3 ·
that may have been the question 480sparky...sorry for the confusion this was one of my many test questions and the 230v could ave been the part i was missing but the anwser site just filled in te blank for me and so i was just trying to up date te fact that i found the anwser i was looking for and posting this imfo to possibly help those who may have been confused.
thanks for anwsering the ^ question as well...
 

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Discussion Starter · #4 ·
I just want to say that after working through this formula and understanding it. I cant belive how easy it actually is.Just a 6 part formula...
 

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I went to an anwser web site this was the anwser i was given...

power loss=10,000w - 8178.45w=1821.55w
(....
I have to back up 480 on this one. This has nothing to do with power loss. From what you quoted, they were simply looking for the difference in power consumption between the 2 voltages, the rated voltage and the supplied voltage.

all you did is figure the power consumed with each voltage and then calculated the difference between those two wattages.

That is not power loss. An electric resistance heater has no power loss. It is 100% efficient as all power consumed is converted to heat (the intended purpose) so there is no power loss.
 

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Discussion Starter · #7 ·
if its not running at its highest potential wouldnt that be a loss btween its potential and its actual use, sort of one of those 'quote' trick question?
 

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if its not running at its highest potential wouldnt that be a loss btween its potential and its actual use, sort of one of those 'quote' trick question?
You can run your truck at redline all the time too. Do you consider it a loss if you don't?
 

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Discussion Starter · #10 · (Edited)
if your intented use for your vehicle was to run at that level... im just saying the heater was designed to run at a certain nominal value and by use one lower than that there is a loss of the potential that, in this case is power ie power loss ....would your vehicle be considered of a variable speed and the red line an indicator of over load... just like your truck you can over load the heater but by how much dictates how long
You can run your truck at redline all the time too. Do you consider it a loss if you don't?
 

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If you are installing a 240v heater on a 208v system, it should be listed for use at that voltage. Otherwise, you are not installing it according to the manufacturer's listing.

If it is listed for use at 208 volts, then you haven't lost anything. Unless you want to spring for a buck-boost transformer to get the voltage back up to 240.

In many apartments, condos, townhouses, 208 is the operating voltage as it is on a 3-phase system.
 

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Discussion Starter · #12 · (Edited)
so just like you can by a 130v flood light usually used in motion detectors but obviously its installed on a 120v sys. you are not running it at its full potential again the difference can be considered a loss...obviously this is done for the prolonggevity of the fillament ... its not ran at the "redline" running at full capacity per say...
also the only slight down fall is a slightly dimmer light
 

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Discussion Starter · #14 ·
okay i see the book i have says that hysteresis loss also called molecular friction is also found in generators ....caused as the armature rotates in a fixed magnetic field many of the magnetic particles in the armature core remain lined up with the fixed field...this rotation causes internal friction between the magnetic particles
 

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generators-motors; same thing. The only difference is a motor uses electricity to make it move and a generator uses an outside power source to make it move and the electricity produced is used elsewhere but the principles are the same. Only the end result varies.

and no, not using an appliance to full potential is has nothing to do with power loss. If you dim a light, dry your clothes on anything less than high heat, or any other similar situations, you are not causing a power loss. You are using less power.

Power loss is where not all the power consumed is used to provide the desired result/action. Motors have inherent power losses because of the lack of efficiency of a motor. There is slip and other losses, including hysteresis and heat, that mean not all the power consumed will provide rotation (intended work product)

Think of it this way. It is not electricity but it explains it well.

A house furnace;

a 50% efficient furnace will provide 50% of the heat produced by burning the fuel to heat the house (intended purpose). The rest goes up the chimney. That means there is a 50% energy (power) loss.

a 98% efficient furnace will provide 98% of the heat produced by burning the fuel to the intended use of heating the house. The remaining 2% will go out the exhaust. That 2% would be the power loss.

an electric heater utilizes all the electricity consumed to produce heat. That means it is 100% efficient as all power consumed is converted to heat and utilized for the intended purpose. There are no power losses. Before anybody jumps in and says there are losses to the resistance of the wire and such, yes, that is correct but they are negligable and actually, even they are producing heat though.

wacoNot to quibble, but some heaters lose energy as light.
and that light does not produce heat? Ever been burned by IR or UV light? IR grills are some of the hottest grills available and UV light will cook your eyeballs.

ever seen a laser melt metal? all it is is light but it will melt metal.

Light can react with a material to produce heat. Isn;t the sunlight warm? Why is it cooler in the shade?
 

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This theory problem has raised another question in my mind and since I can't understand why the two methods don't give the same result, it makes me crazy! I was never strong in math, so the disconnect is probably somewhere in the math.

My approach to the problem (and the one I use in the real world) when I encounter a specification plate that has, say, 1875 watts at 125 volts. If you set up a ratio, 125 over 1875 equals 120 over X, the answer doesn't come out the same as it does if you solve for R using Ohm's Law, E squared over R.

At least, I don't think it does. Why? :001_huh:
 

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This theory problem has raised another question in my mind and since I can't understand why the two methods don't give the same result, it makes me crazy! I was never strong in math, so the disconnect is probably somewhere in the math.

My approach to the problem (and the one I use in the real world) when I encounter a specification plate that has, say, 1875 watts at 125 volts. If you set up a ratio, 125 over 1875 equals 120 over X, the answer doesn't come out the same as it does if you solve for R using Ohm's Law, E squared over R.

At least, I don't think it does. Why? :001_huh:
Where do you get E squared in ohms law E=IxR
 

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if you solve for R using Ohm's Law, E squared over R.

:001_huh:
Did you mean E²/P = R? That is what the quantities you used represented anyway.

anyway, your formula would have to be 125²/1875=120²/X and it does work out as such. Notice the voltage is squared. You would have 1728 watts with 120 volts. You are not squaring the voltage.

here is a nice Ohm's law calculator

and a nice Ohm's law pie chart
 

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10 kw heater on 208 volts

I have seen this question before. If at 240 volts a single phase heater puts out 10 kw. First I need to find the resistance as this will be constant.

240^/ 10,000 = 5.76 ohms. Then to find what the output would be at 208 volts single phase. Take the new applied voltage. 208^/5.76 = 7,511
 
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