Voltage drop calculation

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electro7

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Electrician, Solar and Electrical Contractor
Hi,

I was wondering if somebody could follow my voltage drop calculation to see if I am correct? See below:

Total amperage: 421A
Copper (K): 12.9
L = 700ft (one way)
CMA = 503,400 (3 sets of 3/0 copper = 167,800 x 3)
3 phase 480V circuit
1 pf

(1.732x12.9x421x700) ÷ 503,400 = VD
(6,584,423.16) ÷ 503,400 = VD
13.1 Volts
13.1V ÷ 480V = 2.7% VD

3 sets of 3/0 thwn copper OKAY

Thank you in advance
 
Your calculations look correct to me.

Since these conductors are being used below their ampacity it is likely that their temperature will be less than 75C and thus the K value will be less than 12.9, so you will see lower voltage drop than calculated. This doesn't hurt you one bit, but might be useful to know when you are checking the final installation. (Or it might be my inner pedant..... :) :) )

-Jon
 
In your calculation you consider only the d.c. resistance at 90oC.If you'll take the a.c. resistance from NEC Table 9 for cables in steel conduit you'll get for 3/0 awg 0.079 ohm/kft at 75oC.If you'll multiply by (234.5+90)/(234.5+75)=1.0485 you'll get 0.0828 ohm/kft
Total resistance of 3*3/0 cables will be then 0.0828/3*700/1000=0.0193 ohm
If pf=1 then VD=1.73*421*0.0193=14.057 V 14.057/480=2.93%
 
Julius Right said:
In your calculation you consider only the d.c. resistance at 90oC.If you'll take the a.c. resistance from NEC Table 9 for cables in steel conduit you'll get for 3/0 awg 0.079 ohm/kft at 75oC.If you'll multiply by (234.5+90)/(234.5+75)=1.0485 you'll get 0.0828 ohm/kft
Total resistance of 3*3/0 cables will be then 0.0828/3*700/1000=0.0193 ohm
If pf=1 then VD=1.73*421*0.0193=14.057 V 14.057/480=2.93%
Click to expand...


My money is on this method. I think chapter 9 Table 9 is all to often under rated.

I adore your avatar btw 🐈🐱♥
 
You are right, ptonsparky. In the end it's still less than 3%
 
Julius Right said:
In your calculation you consider only the d.c. resistance at 90oC.If you'll take the a.c. resistance from NEC Table 9 for cables in steel conduit you'll get for 3/0 awg 0.079 ohm/kft at 75oC.If you'll multiply by (234.5+90)/(234.5+75)=1.0485 you'll get 0.0828 ohm/kft
Total resistance of 3*3/0 cables will be then 0.0828/3*700/1000=0.0193 ohm
If pf=1 then VD=1.73*421*0.0193=14.057 V 14.057/480=2.93%
Click to expand...
Why 90*C? The installation is more than likely limited to 75*C.
 
electro7 said:
Hi,

I was wondering if somebody could follow my voltage drop calculation to see if I am correct? See below:

Total amperage: 421A
Copper (K): 12.9
L = 700ft (one way)
CMA = 503,400 (3 sets of 3/0 copper = 167,800 x 3)
3 phase 480V circuit
1 pf

(1.732x12.9x421x700) ÷ 503,400 = VD
(6,584,423.16) ÷ 503,400 = VD
13.1 Volts
13.1V ÷ 480V = 2.7% VD

3 sets of 3/0 thwn copper OKAY

Thank you in advance
Click to expand...

Looks accurate to me. In case you had no control of the load pf, the worst case load pf would be 0.835 with a corresponding voltage drop of 3.4% (assuming magnetic raceway).
 
To the OP: in summary your calculations look good. But the equations that you used (like all such equations) are approximations that are good enough to get the job done.

We like geeking out on more accurate equations that make small differences in your results but are not that important at the size of installation you are considering.

I don't think we've yet discussed allowable manufacturing tolerances for the wire nor changes in resistance caused by mechanical working of the conductors during installation. :) :)

Seriously: the AC values described in post #6 are the most significant, and really important if you use larger conductors or have really inductive loads.

But I think your result is close enough for normal humans.

Jon
 
winnie said:
I don't think we've yet discussed allowable manufacturing tolerances for the wire nor changes in resistance caused by mechanical working of the conductors during installation.
Click to expand...

Solid point. Any suggestion on a good multiplying factor to use to account for this variation?

winnie said:
Seriously: the AC values described in post #6 are the most significant, and really important if you use larger conductors or have really inductive loads.
Click to expand...
The cable reactance is arguably more of a significant factor under low power factor conditions.
 
Xptpcrewx said:
Solid point. Any suggestion on a good multiplying factor to use to account for this variation?
Click to expand...

I don't have good information on this.

For the magnet wire that we use to wind the motors that I work with, the manufacturer specifies a diameter tolerance range. I've never seen such for building wiring.

I would expect work hardening to increase resistivity, but I really don't know by how much one would expect.

-Jon
 
winnie said:
I don't have good information on this.

For the magnet wire that we use to wind the motors that I work with, the manufacturer specifies a diameter tolerance range. I've never seen such for building wiring.

I would expect work hardening to increase resistivity, but I really don't know by how much one would expect.

-Jon
Click to expand...

I know if you use a DLRO meter to measure a short section of stranded cable, the variation in resistance can be very significant with bending/routing/training. This variation becomes less and less significant as the total cable length increases (dominating any variations), but it would be interesting to determine at what point it becomes negligible based on empirical data.
 
ptonsparky said:
Not quite what I meant.
I’ve never had to worry about 5.00% or 3.00%. I’ve never measured the VD at the last tower of a pivot irrigation system but I imagine 10% is a dream.
Are there areas that would be concerned about 3.2%?
Click to expand...
I imagine that most the time there is no 10% drop from pivot point to the last tower, maybe factoring in starting current there is. This drop if anything creates somewhat natural soft starting to some extent also. And for no longer than those motors run in normal operation they probably can stand almost 20% voltage drop and still not give much troubles.

With a higher HP well being supplied by same source however you might already be starting with a lower level voltage at the pivot end.

How many roto-phase supplied systems out there with pretty unbalanced voltage to those drive motors? They still take it pretty good, but I think the intermittent duty they experience is main reason why.
 
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