DC resistance
DC resistance
(OP)
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.
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.





RE: DC resistance
You have both V and I.
The other details don't really matter.
RE: DC resistance
TTFN
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RE: DC resistance
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Entered 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?
RE: DC resistance
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 1.7x10e-8 as the factor? and the length?
RE: DC resistance
TTFN
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RE: DC resistance
RE: DC resistance
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?
RE: DC resistance
Recalculate using meters for length and square meters for cross-section, then the formula given by IRstuff will work using 1.7x10e-8 for p.
Benta.
RE: DC resistance
Rho is what you're after here you go:-
http
desertfox
RE: DC resistance
http:/
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RE: DC resistance
RE: DC resistance
RE: DC resistance
Are you sure?
RE: DC resistance
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RE: DC resistance
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.
RE: DC resistance
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.
RE: DC resistance
=====================================
(2B)+(2B)' ?
RE: DC resistance
RE: DC resistance
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)
RE: DC resistance
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?
RE: DC resistance
RE: DC resistance
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.
RE: DC resistance
Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
RE: DC resistance
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...
RE: DC resistance
The only way I can get 3.5V drop is if the inner diameter is 3.6in and length is halved and the current is halved.
TTFN
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RE: DC resistance
i just re-read 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?
RE: DC resistance
TTFN
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RE: DC resistance
is my terminology not correct by stating the bus is run in parallel?
did you look at the sketch?
RE: DC resistance
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 non-existing 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.
RE: DC resistance
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 true-RMS 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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RE: DC resistance
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=7.6E-07*450*12/20.42/2=0.0001 ohm.
RE: DC resistance
attached is a detailed diagram of the system.
RE: DC resistance
And, finally, what is the problem?
We have been struggling with this ill-defined '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.
RE: DC resistance
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.
RE: DC resistance