ECN Forum Forums General Topics of the Trade General Discussion Area Watts loss?

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?

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).]

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.

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).]

Good call Bob!.
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.

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.

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

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.

My $ is with Bob on this one. I dont quite figure how using a smaller cable will reduce I2R loss on any given load.

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.