Voltage Drop

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Maybe you should calculate it yourself.

How to calculate voltage dropIn situations where the circuit conductors span large distances, the voltage drop is calculated. If the voltage drop is too great, the circuit conductor must be increased to maintain the current between the points. The calculations for a single-phase circuit and a three-phase circuit differ slightly. Single-phase voltage drop calculation:

Three-phase voltage drop calculation:

VD = Voltage drop (conductor temp of 75?C) in volts
VD% = Percentage of voltage drop (VD ? source voltage x 100). It is this value that is commonly called "voltage drop" and is cited in the NEC 215.2(A)(4) and throughout the NEC.
L = One-way length of the circuit's feeder (in feet)
R = Resistance factor per NEC Chapter 9, Table 8, in ohm/kft
I = Load current (in amperes)
Source voltage = The voltage of the branch circuit at the source of power. Typically the source voltage is either 120, 208, 240, 277, or 480 V.
Important Note: According to NEC 215.2(A)(4) informational note No. 2, the voltage drop for feeders should not exceed 3% and the voltage drop for branch circuits should not exceed 5%, for efficient operation.[4]
 
bob said:
Maybe you should calculate it yourself.

How to calculate voltage dropIn situations where the circuit conductors span large distances, the voltage drop is calculated. If the voltage drop is too great, the circuit conductor must be increased to maintain the current between the points. The calculations for a single-phase circuit and a three-phase circuit differ slightly. Single-phase voltage drop calculation:

Three-phase voltage drop calculation:

VD = Voltage drop (conductor temp of 75?C) in volts
VD% = Percentage of voltage drop (VD ? source voltage x 100). It is this value that is commonly called "voltage drop" and is cited in the NEC 215.2(A)(4) and throughout the NEC.
L = One-way length of the circuit's feeder (in feet)
R = Resistance factor per NEC Chapter 9, Table 8, in ohm/kft
I = Load current (in amperes)
Source voltage = The voltage of the branch circuit at the source of power. Typically the source voltage is either 120, 208, 240, 277, or 480 V.
Important Note: According to NEC 215.2(A)(4) informational note No. 2, the voltage drop for feeders should not exceed 3% and the voltage drop for branch circuits should not exceed 5%, for efficient operation.[4]
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Ray C. Mullin had a slightly different equation using CMs and gave different answers, than, for example, by assuming #10 has 1 ohm/1000' (and #20 having 10 ohms/1000', and so on, up).
In any case impedance at some average PF, rather than DC resistance, should probably be used.
 
My program gave the following L to N
4.81 V @ 1.0 PF
4.96 V @ 0.9 pf
4.74 v @ 0.8 pf

Is Rays equation R cos A + X sin A where A is the
PF angle?
 
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bob said:
My program gave the following L to N
4.81 V @ 1.0 PF
4.96 V @ 0.9 pf
4.74 v @ 0.8 pf

Is Rays equation R cos A + X sin A where A is the
PF angle?
Click to expand...
I don't know, the derivation wasn't shown.

BTW, in the models below you need to know the PF of the load to figure the Vd.
-You can model a power line as a transmission line as a series resistor.
-The next more sophisticated model would be a resistor in series with an inductor.
-The next would be by adding a shunt capacitor to ground.
-You could also model skin effect but then you need to know the conductor diameter.
The load is across the cap, and the source voltage comes into the resistor.

The models above use "lumped elements", but actually a transmission line has distributed R, L and C and the equations become way more complicated. But 60 Hz is so low that some of these terms may not be important.
The NFPA may use a distributed model, they certainly have the resources and computational power to do this.
 
The way I do it is find the resistance of the wire I need to hit the VD I want to maintain, kind of a back words way but I find it much easer:

Lets say you want to maintain a 3% VD @ 208 and 80 amp load, so that would be a VD of 6.24volts/80 amps would require a wire that would give you a resistance of .078 ohms @640 feet find a wire size that is at or below this resistance.

1/0 having a resistance of .122 per K will have a resistance of .07808 ohms at 640 feet total circuit length. bingo there is you answer if you wish to maintain a 3% VD @80 amps
 
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And I get
0.09827


instead of 0.122 because metals have a temperature coefficient of resistance and of course if they are carrying current the conductor temperature will increase.
This is 20% less than the 0.122. If this is close enough then you don't have to worry.

I have tried to find out what is "close enough" for the NEC, but have not been able to.
From #14 to #12 is 100x1.6/2.5 = 64% but from 4/0 to 3/0 is 100x49/62= 79%. If more precision was needed I guess they'd make conductors of intermediate size.

Also important is how you can tell if you are not close enough. The insulation only lasts 25 years instead of 50? The motor lasts 7 years instead of 10?
 
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infinity said:
I'm guessing that he used 100 amps but Eric mentioned 80% so maybe the load is 80 amps.
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That's just it isn't it. 100 amps, eighty amps, twenty? Who knows!!??

One thing is for sure, a hundred amp panel does not equal a hundred amps of load. And until you know what the load is what good is a voltage drop calculation?
 
ActionDave said:
That's just it isn't it. 100 amps, eighty amps, twenty? Who knows!!??

One thing is for sure, a hundred amp panel does not equal a hundred amps of load. And until you know what the load is what good is a voltage drop calculation?
Click to expand...

Well if you use the maximum possible load for the calculation and it's still good then anything less will be good too. :)
 
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