1000 kcmil or 750

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chaterpilar said:
Two smaller conductors will produce more heat than one bigger conductor of equivalent cross-sectional area.

as per Table 310.15(B) .2 the effect is pronounced only if number of conductors are more than 3 per conduit.
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Hi,

Did you really mean to say that two smaller conductors will dissipate more heat than one bigger conductor of the equivalent cross-sectional area?

The reason I bring this up is that I'm trying to finalize a software specification to calculate the most efficient arrangement for feeding large motors. For example, using double 250kcmil vs single 500kcmil... from "Table 9", the effective impedance of the 250Copper in steel conduit is 0.073 ohms/1000 feet and that for 500 is 0.050 ohms/1000 feet. Note 2 of table 9 states that "multiplying current by effective impedance gives a good approximation for line-to-neutral voltage drop".

So, after all that mumbo-jumbo, since the current "I" is going to be split across two 250kcmil, the voltage drop is going to be less for the double 250's (vd = 0.073 * I/2) than it will be for the single 500 (vd = 0.050 * I). Less voltage drop across the smaller, doubled-up wiring means less total heat generated in the wiring.

Does this make sense? Let me know what you think.

Thanks,
Nate.
 
iwire said:
That is news to me and I work with them. :smile:

Here is a transformer before we installed the 600s.

It is roughly 4' W x 5' H

WHOITRANS1.jpg



Here it is after.

300KVA3LG.jpg


It was not easy getting all that in and I had a lot of room between the terminals and the bottom.
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Bob what are the two breakers for are there Alternate feeds to this xformer ?? And if there is are they locked out with some of key arangement so both feeds cannot be on at the same time.??
 
chaterpilar said:
I thought "skin effect" comes into play only at high frequencies and is negligible at 60 Hz.
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Actually, both frequency and current are factors. The higher the frequency, the less the current has to be for a given size conductor for skin effect to become significant.

Cable-TV coax uses a copper-clad steel inner conductor. In radio and television transmitter stations, they often use "coax" made of copper tube inside copper tube as transmission lines.
 
quogueelectric said:
Bob what are the two breakers for are there Alternate feeds to this xformer ??
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I was pretty surprised when the unit arrived, the cut sheets said nothing about the fans or the switches.

They are not breakers, they are just shunt trip molded case switches, it is one 800 amp feed with two sets of 600s, the molded case switches are only rated 400 amps so they (GE) split the feed. Take a look at the note to the installer that is on the switch.

The transformer is a 500 kva but is the size of a 300. There are temp probes in the windings, as the temp goes up fans start, if the temp keeps going up a alarm sounds (the noise maker is in the enclosure) and if the temp keeps going up the switches will open and kill the feed to the transformer.

GE calls it a "transformore" or something like that, they say it is a benefit to the customer because it is smaller, I think it is just cheaper to make. :roll:
 
iwire said:
GE calls it a "transformore" or something like that, they say it is a benefit to the customer because it is smaller, I think it is just cheaper to make. :roll:
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It's definitely smaller. The only problem is that they eliminated the space that a
500 KVA needs for wires. :roll:
 
Nateholt,

For example

10 Awg is half the size( diameter) of 4 Awg and ..assuming we use 1 run of 4 Awg instead of 2 runs of 10 Awg,

10Awg = 0.00319 ohms/meter

4 Awg = 0.000793 ohms/meter

so for 1 meter length and 10 amps current

4Awg

Watt loss = I*I*R = 10*10*0.000793= 79 miliwatts

10Awg

Watt loss = ( 5*5*0.00319) * 2 = 159 milliwatts.

More watt loss means more heat.

Cheers.
 
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rconteng said:
I am wondering whether it will make a difference to a contractor if he were asked to pull five parallel runs of 4 conductor 1000 kcmil cable in 5" conduits vs 6 runs of 4 conductor 750 kcmil in 4" conduits. The distance is roughly 600 feet.
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Whatever you choose, DO NOT get 900 kcmil compact aluminum, we've had nothing but trouble with it.
 
Hey

Hey

Iwire,
What is the set bolt used in the multi-lug assemblies, is a special tool required ? It looks like two eye-lets at the end of each lug/bolt?

It also looks like a ceramic block on the underside the breakers?
But in re-examining the photo the aluminum lugs are used on top, it must be the angle and such of the picture!
 
chaterpilar said:
Nateholt,

For example

10 Awg is half the size( diameter) of 4 Awg and ..assuming we use 1 run of 4 Awg instead of 2 runs of 10 Awg,
Click to expand...

If 10 awg is half the diameter of 4 awg, you need 4 runs of 10 awg for the same size (cross sectional area). I don't think the NEC permits parallelling such small conductors though.

chaterpilar said:
10Awg = 0.00319 ohms/meter

4 Awg = 0.000793 ohms/meter

so for 1 meter length and 10 amps current

4Awg

Watt loss = I*I*R = 10*10*0.000793= 79 miliwatts

10Awg

Watt loss = ( 5*5*0.00319) * 2 = 159 milliwatts.

More watt loss means more heat.

Cheers.
Click to expand...

For the 10awg case, the calculation becomes

(2.5*2.5*0.00319)*4 = 79.8 mW
 
cadpoint said:
Iwire,
What is the set bolt used in the multi-lug assemblies, is a special tool required ? It looks like two eye-lets at the end of each lug/bolt?

It also looks like a ceramic block on the underside the breakers?
But in re-examining the photo the aluminum lugs are used on top, it must be the angle and such of the picture!
Click to expand...


No special tool required. You simply remove the two lug assemblies adjacent to the set screws on the first assembly. Install the first set and torque them down. Reinstall the second lug assembly and repeat the process.
 
Hey Bob, not trying to bust your chops, but didn't you exceed the bending raduis on those conductors ?

Reason I'm asking is that I like the way it looks, and I used to make my wires neat like that, but a few years ago I got retrained not to do that. Just wondering what your thoughts are on that.

(also, as a side note, I have received a few sets of main gear where the wires were made neat like that also)
 
i stand corrected,..but still it is more....and multiply it by 200 metres, and remember this loss will be forever....(lifetime of the equipment)


I know it is not done with small wires,...was an example for calculation purposes only.


cheers.
 
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chaterpilar said:
i stand corrected,..but still it is more....and multiply it by 200 metres, and remember this loss will be forever....(lifetime of the equipment)
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I'm not sure which data you were using, but if I use the data from Table 8 in the 2005 NEC, #10 is 5.261 mm^2, #4 is 21.15mm^2, so 4 x #10 conductors in parallel is 21.044mm^2. #4 is 0.5% larger than 4 x #10. The power loss values were 79.3mW for #4 and 79.8mW for 4 x #10, which is a 0.6% difference. So the bulk of that small difference is due to the small difference in conductor cross sectional area between the two cases.

BTW, where did you get the resistance values? Those in Tables 8 and 9 in the 2005 NEC are quite a bit higher than the ones you used.
 
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