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hello,
i'm having a hard time figuring out the actual voltage drop of a DC circuit and i think it's due to a bad calculation with the resistance. amps=70kA VDC=100 L=450 ft
Conductor is 7.50" o.d x 5.5" i.d. water cooled bus. sufficient water flow through out circuit on multiple inlet connections.
cross section is 20.42 sq. in.
can someone help me with this? I can figure out the voltage drop once i get the right resistance of copper for this size bus.
thanks. 

magoo2 (Electrical) 
4 Oct 11 15:35 
R = V/I
You have both V and I.
The other details don't really matter. 

VE1BLL (Military) 
4 Oct 11 22:23 
Here's an online calculator that can (perhaps) be pressed into service: http://circuitcalculator.com/wordpress/2006/01/24/traceresistancecalculatorEntered 20.42 by 1 inches and 450 feet, and the result was 0.000177 ohms. You certainly shouldn't trust this result without quadruple checking it in different ways. I suggest this primarily as yet another easy datapoint to add to the mix. PS: What was the "VDC=100" that you mentioned? 

100 v(dc)total output from rect/trans
i need to figure out the total resistance of the conductor across 450 ft. to calculate the total voltage drop of the circuit.
there are are 2 conductors in parallel carrying 70kA @ 100vdc.
i'm having problems figuring the length into the equation.
i know to use r=v/i=.00142, but where do figure in the resistivity of copper of 1.7x10e8 as the factor? and the length?


Yeah, IR, it seems that you did. 

The drop should be I*ρ*len/area
I=70000 ρ=??? (does this factor include the mat'l? this is where i get confused...i think) L=450 A=in sq. in? cir mil?


benta (Electrical) 
5 Oct 11 3:53 
You are driving yourself crazy by insisting on using antique units (how about throwing in a "furlong" just to make things funnier).
Recalculate using meters for length and square meters for crosssection, then the formula given by IRstuff will work using 1.7x10e8 for p.
Benta. 

VE1BLL (Military) 
5 Oct 11 10:07 
I just did the manual calculation using the above and it matched the online 'trace resistance calculator' to four digits (higher apparent accuracy than the significant digits of the input variables).


iop95 (Electrical) 
5 Oct 11 14:39 
If temperature in cable is about 20C, drop voltage is aprox. 0.146V, or 0.146%. 

VE1BLL (Military) 
5 Oct 11 15:25 
Quote:...0.146V..."
Are you sure? 

iop95 (Electrical) 
6 Oct 11 4:08 
If I undestood corectly, area A = (pi*7.5^2/4  pi*5.5^2/4) = 20.41sq.in = 131.67 cm2. Cooper resistivity at 20C Rcu = 2*10^8 ohm.m Lenght l = 2*450ft = 297m R = Rcu*l/A = 2*10^8 * 297/131.67/10^4 = 4.51*10^4 ohm DV=R*i=4.51*10^4 * 70*10^3 = 31.57V... in first calculation I used a wrong cooper resistivity value... And power lose in cable is Pcu = 2.2MW... sure it's not a real system wuth such power lose. 

VE1BLL (Military) 
6 Oct 11 7:44 
I had found 12.4 volts (the IR drop) oneway  both from the online calculator (I linked above) and by way of handraulic calculation. This matches what IRstuff found. iop95 is using a slightly different value for ρ. Regarding the 450 feet  I wonder why they don't move the transformer a bit closer. It would pay for itself from the recycled copper, power savings, and waste heat. 

Quote:sure it's not a real system wuth such power lose.
70kA is a little unusual, unless it is a a fault current. ===================================== (2B)+(2B)' ? 

I think your conductor is a steel of 0.1 ohm.mm^2/m resistivity and 0.005 1/degreeC temperature coefficient.The conductor temperature has to be 90 degrees C. 

70kA is fully load current 24/7. Customer shows 3.5v drop.
i used so many different k factors and never got close to 3.5.
maybe i should clarify just in case it's not clear. 2 water cooled conductors running in parallel for a total of 450' (450'/conductor)


iop95 (Electrical) 
6 Oct 11 9:32 
I think is much better to offer all details from begining. With this 3.5V drop we close to a real system form loses point of view. At 7MW installed power,with 3.5V drop there are about 245kW lose, which mean 3.5% lose or 96.5% efficiency of power line, much acceptable than 2.2MW... Even now, with your clarification, something is not ok. If there are 2 such conductors in parallel, total resistance is half and drop also, about 15.7V from my calculations, or 1.1MW lose which is still too much. May you give a detailed schematic an parameters of this strange instalation? 

HERE IS A GENERAL OVER VIEW OF THE SYSTEM 

12.8 V one way here. Hand calculation with slide ruler and assuming that current flows in one conductor to and the other conductor fro. If current flows in both conductors (OP says something about parallel  but does he really mean that?) then the voltage drop drops (yes, pun) to half, or 6.4 V. Since this is DC, it may very well be some kind of electrolysis plant where 70 kA isn't at all uncommon. On the low side, actually, if it is an Aluminum plant and standard if it is some halogen or sodium plant. I fail to see what the problem is. Gunnar Englund www.gke.org  Half full  Half empty? I don't mind. It's what in it that counts. 

You should've posted that before. You have 6.4 V drop in one of those conductors. Gunnar Englund www.gke.org  Half full  Half empty? I don't mind. It's what in it that counts. 

VE1BLL (Military) 
6 Oct 11 9:58 
Quote:...maybe i should clarify just in case it's not clear. 2 water cooled conductors running in parallel for a total of 450' (450'/conductor)...
If there are two (not four) conductors, and assuming it's not a unipolar current source (< joke), then I would assume the two conductors are one for source and one for return. If my assumption is correct, then the conductors may be parallel, but are not "[i]in parallel[/]". One isn't going to use a ground return for 70kA... 

sorry about not making this clear from the start. i just reread my first post and i failed to mention anything about parallel. i also cannot give too much info on this, but yes, this is an electrolysis type plant.
can you show me how you came up with that figure? 

the rectifiers are run in parallel to output the 70kA @ 100vdc.
is my terminology not correct by stating the bus is run in parallel?
did you look at the sketch? 

Cunfusion, confusion. How can you get 100 V out of two 50 V rectifiers in parallel? Are you sure they are not in series? And that sketch of yours doesn't exactly make things any clearer. As I said before, I fail to see the problem. Are you trying to tell a supplier that the equipment is faulty? Or are you trying to find a nonexisting fault? Or what is it you are trying to do? Gunnar Englund www.gke.org  Half full  Half empty? I don't mind. It's what in it that counts. 

What exactly is "Rect" putting out? Is it actually DC or rectified AC? kVA is normally used with AC systems, since DC systems ostensibly have no phasing differences between voltage and current. And where is the load? And what is the load? Your "sketch" essentially has a dead short across the outputs of each "Rect" If your outputs are halfwave rectified a trueRMS meter might measure about 7V, and if it was measured with a crappy meter, 3.5V might not be unexpected. Obviously, if your kVA/V calculation isn't what its supposed, say, only 35kA, then the RMS value of a halfwave rectified ouput would be 3.5V, almost exactly. TTFN
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You cannot have, in the same time, 100 V and 70 kA but, you could have 100V and 35 kA or 50 V and 70 kA. I think you have 50 V and 2*35 KA. The resistance of the loop [from "+"of the rectifier to "" of the same rectifier could be 3.5V/35000A=0.0001 ohm.] Let's say from "+" to "" are laid 2 conductors parallel [copper] of 20.42 sqr.inch each one. The total loop resistance will be: R=ro*length/scopper/2 [for 2 parallel conductors]. Where ro =ohm.inch length =inches scopper =sqr.inch For copper ro= 0.0000006793 ohm.inch at 20oC. I think the actual temperature could be 50oC.So ro=ro20*(50+234)/(20+234)=7.6/10^7 ohm.inch .Then the total loop has: R=7.6E07*450*12/20.42/2=0.0001 ohm. 

Ok...This is the diagram with more detail. I apologize for the confusion thus far. the red dashed conductors are water cooled copper tubes. We calculate that water cooled bus can pass 3800amps per sq. in. in DC. attached is a detailed diagram of the system. 

Sorry. I cannot understand how you can get 70 kA out of two series connected 35 kA rectifiers. Or are you counting the sum of the two circuits to get 70 kA? And, please, where is your load? You do not apply 100 V to a dead short? Do you? And, is the complete length of the two circuits in series equal to 450 feet? Or are they 450 feet each? And, finally, what is the problem? We have been struggling with this illdefined 'problem' for quite a while now. Either just drop it  or present the facts and the problem in a way that doesn't cause confusion. Please. Gunnar Englund www.gke.org  Half full  Half empty? I don't mind. It's what in it that counts. 

If "TRANS.RECT." means "Transformer for Rectifier Supply" and "RECT.35 KA" is a rectifying bridge then through these conductors connecting transformers to rectifiers is only Alternative Current [A.C.] and there it is not Direct Current. The Direct Current will flow to the receivers [electrolysis devices or whatever is] from the rectifiers and no direct connection to the transformers, I guess. 

i think i figured it out. thanks for the help. 



