Could someone check me out on this calculation of "unused work". not sure?

[Sid Trimmer]

Determine the unused work for a load of 22 amps operating on 240 volts using a # 10 conductor. The length of the circuit is 60 feet.


1.21 ohms/1000= .00121 x 120 feet =.1452 ohms

r = e/I or 240/22=10.9 ohms

w = e x I or 240 x 22 = 5280 watts

add 10.9 + .1452 = 11.04

57,600/11.04 = 5217.39

5280 – 5217.39 = 62.61 watts

62.61 watts x 12 = 751/1000 = .751KWH

.751 x 365 = 274 watts x .11 = 30.14



Try using # 6AWG .122/1000 = .000122 x 120 = .01464 ohms

10.9 + .01464 ohms = 10.91464
57,600/10.91464 = 5677 is more than 5280. There is no work loss due to #6 conductor having not sufficient enough resistance to cause unused work.

 

[GoldDigger]

To me, without seeing your textbook definition, unused work would be exactly the I2R resistive losses in the wire. As long as current flows these will never be zero or negative.
What you calculated is not a useful number, in my opinion.

 

[bob]

Determine the unused work for a load of 22 amps operating on 240 volts using a # 10 conductor. The length of the circuit is 60 feet.

1.21 ohms/1000= .00121 x 120 feet =.1452 ohms

r = e/I or 240/22=10.9 ohms

w = e x I or 240 x 22 = 5280 watts

add 10.9 + .1452 = 11.04

57,600/11.04 = 5217.39

5280 – 5217.39 = 62.61 watts

62.61 watts x 12 = 751/1000 = .751KWH

.751 x 365 = 274 watts x .11 = 30.14

Try using # 6AWG .122/1000 = .000122 x 120 = .01464 ohms

10.9 + .01464 ohms = 10.91464
57,600/10.91464 = 5677 is more than 5280. There is no work loss due to #6 conductor having not sufficient enough resistance to cause unused work.

What you are looking for are the losses in the wire and the cost. The loss is = I²R.
If the load runs 12 hrs per day, the loss would be 22² x the resistance of the wire one way not 2. Resistance = 0.00121 x 60 ft = 0.0726. x 22 x 22 = 35 watts x 12 hrs = .42 kwh per day. 0.42 kwh x 365 = 153 kwh at 10 cents per kwh = $15 per year.
Using #6cu 0.000122 x 60 x x22 x 22 = 3.54 watts x 12 hrs x 365 = 15 kwh @ 10 =$1.50. Now figure the cost difference in the wire and see how low your payback is.

Cost from Lowes #10 $0.89 per ft x 60 =$53.00 #6 $1.69 per ft x 60 = $104.00
About a 2 year return. Not bad. This type of return usually only works when the loads runs long continuous hours. If the load is intermit, such as an AC unit,
the return on investment usually is not worth the added cost.

 

[don_resqcapt19]

What you are looking for are the losses in the wire and the cost. The loss is = I²R.
If the load runs 12 hrs per day, the loss would be 22² x the resistance of the wire one way not 2. Resistance = 0.00121 x 60 ft = 0.0726. x 22 x 22 = 35 watts x 12 hrs = .42 kwh per day. 0.42 kwh x 365 = 153 kwh at 10 cents per kwh = $15 per year.
Using #6cu 0.000122 x 60 x x22 x 22 = 3.54 watts x 12 hrs x 365 = 15 kwh @ 10 =$1.50. Now figure the cost difference in the wire and see how low your payback is.

Cost from Lowes #10 $0.89 per ft x 60 =$53.00 #6 $1.69 per ft x 60 = $104.00
About a 2 year return. Not bad. This type of return usually only works when the loads runs long continuous hours. If the load is intermit, such as an AC unit,
the return on investment usually is not worth the added cost.

So there is only voltage drop on the one of the two conductors?

 

[bob]

YOU ARE ABSOUTLY CORRECT. BRAIN FOG. SORRY. AS I WAS WRITING , I KEPT LOOKING AT IT AS IF IT SEEMED STRANGE. SORRY FOR THE GOOF UP.

 

[Carultch]

Determine the unused work for a load of 22 amps operating on 240 volts using a # 10 conductor. The length of the circuit is 60 feet.


1.21 ohms/1000= .00121 x 120 feet =.1452 ohms

r = e/I or 240/22=10.9 ohms

w = e x I or 240 x 22 = 5280 watts

add 10.9 + .1452 = 11.04

57,600/11.04 = 5217.39

5280 – 5217.39 = 62.61 watts

62.61 watts x 12 = 751/1000 = .751KWH

.751 x 365 = 274 watts x .11 = 30.14



Try using # 6AWG .122/1000 = .000122 x 120 = .01464 ohms

10.9 + .01464 ohms = 10.91464
57,600/10.91464 = 5677 is more than 5280. There is no work loss due to #6 conductor having not sufficient enough resistance to cause unused work.

I've never heard of the term "unused work". I recommend sticking to a more industry standard term, such as resistive power loss, so that it is clear to others what you intend to calculate.

 

[GoldDigger]

I've never heard of the term "unused work". I recommend sticking to a more industry standard term, such as resistive power loss, so that it is clear to others what you intend to calculate.

I strongly suspect that the term comes from standard usage is some other English-speaking country, even though the OP is in the US.
It sounds like a question from a textbook.

 

[JoeStillman]

"Unused work" is a puzzling term. I²R = watts, but "work" is in watt-hours. BTU, foot-pounds, newton-meters and horsepower-hours are other units of work.

Power lost during a specific time would be "unused work", at least, the way I learned my physics. Or work-per-hour could be interpreted as power loss.

:?

 

[gar]

160225-1548 EST

Some comments:

If it is a single phase motor determine the loss in one wire and multiply by 2, if a three phase do the one wire loss and multiply by 3.

It is hard to saturate a motor core because of the large air gap in comparison with what happens in a transformer. Therefore, there is no large inrush current to a motor. In a motor at start there is inductance and resistance in series and no counter EMF. There is no instantaneous current flow when a voltage is applied to an inductor. The motor winding does not look like a resistance only at the instant of start.

In the motor current starting curves I showed in the thread http://forums.mikeholt.com/showthread.php?t=174880
there is no indication of any substantial inrush current. Starting current and inrush current are two different things.

In a prctical DC motor you can encounter field core saturation, and in 1877 - 79. Edison discovered magnetic saturation and as a result he was able to design a dynamo that was not very saturated and had high mechanical to electrical efficiency. Much higher efficiency than anyone else had achieved, 90%. Others were in the 50% or lower range. You can not build a power distribution on low efficiency, Edison understood this. See dynamos and other chapters in "Menlo Park Reminiscences", by Franis Jehl.

.

 

[Smart $]

Sid,

Welcome to the forum. :thumbsup:

Please disregard the immediate bashing. We all get it sooner or later. You just happen to get it out of the gate.

​

 

[Little Bill]

Sid,

Welcome to the forum. :thumbsup:

Please disregard the immediate bashing. We all get it sooner or later. You just happen to get it out of the gate.

​

2nd the welcome!

But Sid, unless you are inviting people to your house you might want to take your address of your location.:happyyes:

 

[Carultch]

I strongly suspect that the term comes from standard usage is some other English-speaking country, even though the OP is in the US.
It sounds like a question from a textbook.

That's a good point too. And I apologize if my comment sounded like it was bashing.

One afterthought I had, is that it could come from a thermodynamics context, where the term work is used in contrast with heat and internal thermal energy.

When I think of the word work, I have a bias to think of it as mechanical "W=F dot d" work. And certainly there is power associated with mechanical work, when expressed as a per unit time concept. Thermodynamically, work is just as ordered of a form of energy as electrical energy. That is to say, they are both the kinds of energy that you can convert into heat with 100% efficiency, while the reverse is physically impossible, unless time were to somehow run backwards.

In rudimentary thermodynamics problems, when you are talking about a component such as pump or a compressor, work is the input energy. It might be mechanical, or electrical. It all depends on whether you draw your system definition to include or exclude the motor.

 

[kwired]

2nd the welcome!

But Sid, unless you are inviting people to your house you might want to take your address of your location.:happyyes:

Maybe it is not his address, but someone he wants to get back at for something