NEC Voltage Drop

[timm333]

For calculating the voltage drop in 3-Phase, we take the actual length of the cable. However for calculating the voltage drop in 1-Phase (120 V, L-N), we take the length of the cable as twice the actual because the current flows from the source to load and then back to source.

What about when we calculate the voltage drop for 1-Phase (240 V, L-L), should we still take the length of the cable as twice the actual? Thanks

 

[xptpcrewx]

For calculating the voltage drop in 3-Phase, we take the actual length of the cable. However for calculating the voltage drop in 1-Phase (120 V, L-N), we take the length of the cable as twice the actual because the current flows from the source to load and then back to source.

What about when we calculate the voltage drop for 1-Phase (240 V, L-L), should we still take the length of the cable as twice the actual? Thanks

Yes

 

[INGMRS]

For calculating the voltage drop in 3-Phase, we take the actual length of the cable. However for calculating the voltage drop in 1-Phase (120 V, L-N), we take the length of the cable as twice the actual because the current flows from the source to load and then back to source.

What about when we calculate the voltage drop for 1-Phase (240 V, L-L), should we still take the length of the cable as twice the actual? Thanks

Yes.

 

[luckylerado]

For calculating the voltage drop in 3-Phase, we take the actual length of the cable.

x 1.73

 

[winnie]

You always have to consider the _entire_ length of the circuit, no matter if you are using single phase l-n, single phase l-l, or three phase l-l-l.

If you look closely at the equations that you are using for the voltage drop calculation, you will find that the _entire_ circuit (there and back again) is being considered.

-Jon

 

[MrJLH]

The nec handbook has a good example calculation you might want to look at for voltage drop.

Somewhere towards the end, cant remember which section.

 

[ggunn]

x 1.73

No.

 

[tersh]

The nec handbook has a good example calculation you might want to look at for voltage drop.

Somewhere towards the end, cant remember which section.

I have been computing for the voltage drop of wire since yesterday so this thread is timely:

ftp://ftp.conagua.gob.mx/OCPBC/aps/...sistencia y Reactancia de cables NEC 2008.pdf

In the NEC list for AWG 1 above, the inductive reactance (Xl) for 1000 feet has value of 0.046 ohm

I'm trying to derive it by manual computations (or using existing conventional formulas).

1000 feet of awg 1 has calculated inductance of 687000 nH or 0.000687 H see: https://www.allaboutcircuits.com/tools/wire-self-inductance-calculator/

The inductive reactance is 2 pi f L... or 2 (3.141) (6) (0.000687) = 0.259 ohm http://www.66pacific.com/calculators/inductive-reactance-calculator.aspx

Why is the 0.259 ohm so far from the NEC listed value of 0.0046 ohm?

How does NEC derive the 0.0046ohm? Can you help derive 0.0046 ohm? I've been trying it for 2 hours already.

Is it for 1000 feet of wire.. or 3 x 1000 feet of wires?

 

[timm333]

The inductive reactance is 2 pi f L... or 2 (3.141) (6) (0.000687) = 0.259 ohm

You should not use "XL=2*pi*f*L" here, because NEC calculations consider several additional factors (600 V, 75 Deg C, PVC conduit)

 

[tersh]

You should not use "XL=2*pi*f*L" here, because NEC calculations consider several additional factors (600 V, 75 Deg C, PVC conduit)

What should you use then that can arrive at the 0.046 ohm value instead of 0.259 ohm?

Could the additional factors decrease the reactance more than 5 times?

Hope you can share your own calculations so have idea how the NEC table is derived. Thanks.

 

[kwired]

x 1.73

No.

? Many publications show using 1.73 factor in three phase three wire circuit VD calculations and 2.0 factor for two wire circuits.

2.0 x length is because you have drop on both conductors, and can only be used if both conductors are same size/type if they are different you must calculate them separately and at a 1.0 factor on each one.

1.73 factor is also assuming all three conductors are same size/type and that current is same in all three conductors.

 

[winnie]

I have been computing for the voltage drop of wire since yesterday so this thread is timely:

This really should be a separate thread https://forums.mikeholt.com/showthread.php?t=197174&p=1986684#post1986684

 

[ggunn]

? Many publications show using 1.73 factor in three phase three wire circuit VD calculations and 2.0 factor for two wire circuits.

2.0 x length is because you have drop on both conductors, and can only be used if both conductors are same size/type if they are different you must calculate them separately and at a 1.0 factor on each one.

1.73 factor is also assuming all three conductors are same size/type and that current is same in all three conductors.

If the current is the same in all three identical conductors, the absolute voltage drop is simply the one way Vd=IR on a single conductor, since the current in the neutral is zero. When the sqrt3 comes into play is calculating percent voltage drop compared to phase to phase voltage, but that's just because P2P voltage is sqrt3 times P2N voltage. There's a sqrt3 in the numerator and the denominator, though, so if you compare Vd to P2N voltage the sqrt3's drop out and you get the same %Vd number.

Single phase is the same. In a balanced 240V circuit the %Vd has a X2 multiplier relative to 240V, but not if you compare Vd to the P2N voltage.

 

[MrJLH]

I have been computing for the voltage drop of wire since yesterday so this thread is timely:

ftp://ftp.conagua.gob.mx/OCPBC/aps/...sistencia y Reactancia de cables NEC 2008.pdf

In the NEC list for AWG 1 above, the inductive reactance (Xl) for 1000 feet has value of 0.046 ohm

I'm trying to derive it by manual computations (or using existing conventional formulas).

1000 feet of awg 1 has calculated inductance of 687000 nH or 0.000687 H see: https://www.allaboutcircuits.com/tools/wire-self-inductance-calculator/

The inductive reactance is 2 pi f L... or 2 (3.141) (6) (0.000687) = 0.259 ohm http://www.66pacific.com/calculators/inductive-reactance-calculator.aspx

Why is the 0.259 ohm so far from the NEC listed value of 0.0046 ohm?

How does NEC derive the 0.0046ohm? Can you help derive 0.0046 ohm? I've been trying it for 2 hours already.

Is it for 1000 feet of wire.. or 3 x 1000 feet of wires?

I would stay away from these online calculators and use the NEC tables. i have found these to be off and when asked how did you calculate voltage drop by a client/superior the lat thing I would want to say is I got it off a online calculator.

Note the .046 is for a aluminum conduit which the online calculator does not seen to take into account.