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#48352 02/09/05 08:58 PM
Joined: Nov 2004
Posts: 138
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My textbook says:

watts loss = load * load / resistance

although i can easily enough plug in numbers, i wonder what it all means.


if watts loss = 100

does that mean

power consumption = load + watts loss?

#48353 02/09/05 10:33 PM
Joined: Mar 2002
Posts: 582
R
Ron Offline
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I'm not sure what result your looking for. Checkout this website for some power (wattage) formula http://www.tpub.com/neets/book1/chapter3/1-7.htm

[This message has been edited by Ron (edited 02-09-2005).]


Ron
#48354 02/09/05 11:26 PM
Joined: Nov 2004
Posts: 138
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Member
thanks, i'll check it out. i'm just trying to figure out what exactly watts loss is. my boss looked at me like i was crazy when i told him about it.

#48355 02/10/05 12:29 AM
Joined: Feb 2002
Posts: 182
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Bob Offline
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watts loss = amps x amps x resistance of the conductor in your example. This is the energy that heats the conductor. The amps used is the load amps.I often see ads that show cost saving benefits by increasing wire size above what is normally required by the NEC. Depending on the load and length of time the plant is in operation, it can be shown that the added wire size can return the extra costs over a short period of time.



[This message has been edited by Bob (edited 02-10-2005).]

#48356 02/10/05 05:57 AM
Joined: Jul 2002
Posts: 8,443
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Good call Bob!. [Linked Image]
Yes, any loss within an Electrical system can be put down to I2R, provided that it is a resistive load.
And it is usually a Heating effect.
However, should the load involve things like Capacitive and Inductive Reactance, things become a tad more difficult, with respect to the total impedance offered to the Supply Voltage.

#48357 02/11/05 12:31 AM
Joined: Nov 2004
Posts: 138
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thanks, Bob. so the eccentric homeowner how we just ran 12 wire for everything in his home (except the range, dryer, ect) was actually on to something.

#48358 02/11/05 04:23 PM
Joined: Oct 2003
Posts: 156
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clydesdale, it might help you if you were to draw out a simple circuit and do the math using ohms law to see how it works. Just use a source like 120 VDC, load resistor of 144 ohms (equivalent to 100 W light bulb), and say something like 2-2 ohm resistors representing each leg of the wire and see what you come up with.

I will give you a hint to one of the numbers you should come up with 96.6

#48359 02/11/05 04:28 PM
Joined: Oct 2003
Posts: 156
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Member
Bob, I have to disagree with your statement with respect to resistive loads. If you were to use a larger wire with less resistance on a resistive load, you would increase the load current, thereby cost more to operate, not save.

#48360 02/11/05 06:01 PM
Joined: May 2004
Posts: 186
A
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My $ is with Bob on this one. I dont quite figure how using a smaller cable will reduce I2R loss on any given load.

#48361 02/11/05 07:27 PM
Joined: Feb 2002
Posts: 182
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Bob Offline
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Clydesdale
"thanks, Bob. so the eccentric homeowner how we just ran 12 wire for everything in his home (except the range, dryer, ect) was actually on to something."
In a residence it would take forever to get a reasonable return on the investment because the load is not on that long.

dereckbc
For a resistive load that is online for a long time I think you would be correct.
My point was
"Depending on the load and length of time the plant is in operation, it can be shown that the added wire size can return the extra costs over a short period of time."
This idea is for large loads that run all day and night.

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