Amperes & Resistance

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tryinghard

tryinghard

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California
I think I?m missing something obvious?

If I have a DC circuit: 240V, 460A I would have .52 ohms resistance (R=E/I)

If this same circuit had undersized conductors that would cause resistance.

How will this same circuit be affected with more resistance?
 
eric -
I don't know - I was thinking constant current. The resistance goes up, the voltage goes up. I * R = E :wink:

carl
 
coulter said:
eric -
I don't know - I was thinking constant current. The resistance goes up, the voltage goes up. I * R = E :wink:

carl
Click to expand...

What a great idea! Perhaps we could apply this and create a perpetual motion machine. :D
 
The circuit in question is a generator exciter DC circuit, it is currently 600A fused protected and averages about 460A. The existing conductors are cu quad 250?s RHW-2 (1000 kcmil), 100 feet long bundled in tray and 20% conduit.

I was originally told the overcurrent protection was 500A, with this I come up with 900 mcm for tri 300?s. The engineer is actually saying parallel 4/0?s can be used; wow only 423,200 CM?s! I don?t even know why he would want to downsize? He did send supporting information for this but this information is referencing table 310-17; we are installing DLO type cable (diesel locomotive) but certainly not in free-air application.

As it unfolded we actually purchased DLO type RHW-2 cable 313,000 CM?s and now that I know it has 600A protection I see I can use quad 313?s to achieve 1,250,000 CM?s (derating 70% for bundling in tray and 7-9 conductors in conduit), any less than this would be erroneously protected according to NEC.

So it is my opinion the circuit conductors are greatly undersized and this caused me to wonder about resistance, writing this I can see I may have voltage drop but I don?t know it is an exciter? I would think with the actual 460A load he would need 750 mcm?s at least but again these would be incorrectly protected.

I got a feeling the current will be felt on the conductor bends as well as heat. With all this said I wonder what the outcome would be if in fact the engineer got his desired conductor size of parallel 4/0?s? or even 600 mcm?
 
Trying,
When you look at the ampacity of parallel conductors, you don't add the circular mils, you add the ampacity shown in the tables. Prallel 4/0s in two conduits would be good for 460 amps and parallel 300s in two conduits would be good for 570 amps. Note I was using Table 310.16. If you have an open air installation the ampacites would be 720 for the parallel 4/0s and 990 for the parallel 300s. Under the conditions show in 392.11(B) you can use the ampacities shown in Table 310.17 for cable tray installations.
Don
 
The application should be 392-11(B)(4) as bundled but it's NOT separated 2.15 times the O.D. of the largest conductor due to the lack of space in the existing tray, so it?s closer to 392-11(B)(2)? The problem is this same circuit is installed the last 20? in conduit.

With all this I?m inclined to follow 310-16 and 310-15(B)(2)(a)
 
600A to a generator DC exciter is some pretty serious generation.

I'd guess it's pretty old. Any I have seen in the last 25 years have had exciter generators with diodes in the shaft. Generators installed in the 60's had brushed fields.

Given that, I wonder if you are working to the NEC.

Couple of curiuosity questions:
1. How old?
2. What is the generator rating?

Now for the VD issues:
I guess I wasn't kidding when I suggested a constant current source. The field current power supply essentially is a current source. Assuming the gen is islanded, the current to the field sets the output voltage. The voltage regulator senses the output and adjusts the field drive (field current) to bring the output voltage to setpoint.

So what happens if there is a change. For example suspose the conductors supplying the field drive heat up and the resistance goes up. If the field supply is a true current source, the field supply current stays the same and the field supply voltage climbs to meet the E = I * R.

Same thing happens if the field copper heats up - and if does - a lot - especially with a lagging pf near full load.

carl
 
Part Two - Field conductor installation:

Generally the NEC isn't very interested in generator field drive installations. The field conductors are part of a generation package originally speced and supplied 40 or 50 years ago. 100" of conductors sounds like a generator (steam turbine?) down on an operating floor with a static exciter up with the rest of the control equipment and maybe even with the main CBs.

I'm guessing:
1. You don't have an AHJ standing over your shoulder demanding this installation comply with the NEC.

2. The engineer works for the company and the prints are not stamped.

Both of these two are okay - just different.

Then again, maybe I'm all wet - equally probable - I'm guessing a lot.

So, going forward with I'm just damp - not all wet:
DLO cable has a lot higher rating than comparable sized cables listed in the NEC. Parallel 4/0 does sound a little small for 600A continuous.

Question for derrick: Assuming there isn't a VD issue (current sources don't have VD issues) how does parallel 4/0 sound for a 600A battery supply?

Summary:
1. The NEC is not a design guide. You are likely in an area where the NEC relies on good engineering practice for installation specs.
2. This is a constant current source, not constant voltage. Pitch out concerns about Voltage Drop (not entirely true, but close).
3. The concern is will the conductors carry the required current with out exceeding their temp rating. This is not a house panel with a 75C rating. Hopefully your engineer paid attention to the specific equipment temp specs.

Give us some feed back on the specific installation. This is interesting.

carl
 
coulter said:
3. The concern is will the conductors carry the required current with out exceeding their temp rating. This is not a house panel with a 75C rating. Hopefully your engineer paid attention to the specific equipment temp specs.
Click to expand...

Carl,

I appreciate your explanation of the field supply current vs. voltage! You make it easy to understand.

Now to answer a few of your questions:
1) The generator was put into service in 1967
2) It is a 43,125 KVA unit propelled by water
3) I?m doing the electrical circuitry related to the unit that is currently being rewound. My portion is design build in that we?ll upgrade chosen circuitry while the unit is down.

The DLO cable we purchased is actually rated as RHW-2

I understand the NEC ?is not intended as a design specification or an instruction manual? (90-1(B)) - so it is not a design guide - BUT it certainly qualifies installations to meet the purposes listed in 90-1(A)(B)

My greatest concern is common with yours in that I want to supply adequate conductors to meet the overcurrent protection. Even though NEC does not apply in a powerhouse the fact remains I have 600A overcurrent protection for circuitry in conduit, so the question is what size conductors are required? And I come up with quad 313 mcm?s. I?ll admit I do get stuck around NEC as it is proven.

Thanks for your input!
 
Even with derating, you would not need 4 sets of 313 kcmil for this application. Three sets of 4/0s would be fine for a 600 amp DC circuit. 4/0 is a 260 amp conductor in the 90?C column. If you put 3 sets, 6 conductors, in the raceway, you would have a 624 amp circuit. Your DLO is rated at about 320 amps in the 90?C column based on it being RHW-2. Three sets would be rated at 768 amps after derating. Four sets would be rated at 896 amps.
Don
 
don_resqcapt19 said:
...Three sets of 4/0s would be fine for a 600 amp DC circuit...
Click to expand...

Okay, with this I should be good with tri 313 mcm's (320A x 3 x .80 = 768A).

Earlier I was looking for the correct ampere rating derated then dividing it's CM for conductor size (665A x .70 = 466 so 1,250,000 CM / 4 = 313,000 CM.

I could actually use parallel 250?s (290A x 2 x .80 = 464) is this correct?
 
Just a thought:

Just a thought:

coulter said:
eric -
I don't know - I was thinking constant current. The resistance goes up, the voltage goes up. I * R = E :wink:

carl
Click to expand...

I too am thinking along these lines. I would guess that the exciter is part of a feedback loop; that is, the exciter current is automatically adjusted to maintain the desired ouput voltage from the alternator.

If that is the case, the voltage drop in the exciter circuit is automatically compensated. Of course, other criteria must be met as have been discussed already in this thread.
 
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