Voltage Drop

[tobee]

Could any one explain the 0.75 multiplying factor in the following equation. According to the American Electricians' Handbook, when a three-wire system is tapped from a four-wire three-phase system the voltage drop is calculated as follow provided that the effect of inductance is neglected:

Voltage Drop = 0.75*dc VD,

Where:

dc VD= 2R*L*I

When:

dc VD = drop in volts in the circuit.
R = resistance of wire in ohms/foot.
I = current (load) in amperes.
L = One-Way length of circuit in feet.

Any clarification on this matter is appreciated. Thank You.

 

[buzzp]

That is kinda bizzare. If it is true I can only offer one explanation that I would even question (in other words I am shooting an explanation from the hip and not thinking about it too much).

Perhaps it is related to the return current going back on one of the three wires rather than the neutral. Some current cancels due too the phase shift between the two wires. This in turn would reduce the voltage dropped. Oooh what an ugly thought. Do not attempt to hold water with my quick theory.

 

[busbar]

I am not familiar with the cited reference, but two 1.0 per-unit-current loads served phase-to-neutral on two of three phases in a 4-wire wye circuit will manifest 1.0 per-unit “neutral” current. That would produce voltage drop in the grounded-circuit “neutral” conductor as well as in the two ungrounded “phase” conductors. That would seem to only increase voltage drop at the load terminals.

 

[buzzp]

What do they say to use for 4 wire to 4 wire? How about 3 wire to 3 wire?

 

[tobee]

buzzp

Since you asked, you got it. I am sure this will bring more questions. The reference formula in the Hand Book that is used to calculate balanced load for the 4-wire 3-phase system is given as follows:

Voltage Drop for Lamps = (dc VD)/2
Voltage Drop for motors= 0.866*(dc VD)

And for 3-wire 3-phase system, the Hand book gives the following formula:

Voltage Drop = 0.866*(dc VD)

Any other questions?!

 

[fourleaf]

After finding the formulas in the American Electricians Handbook, I have a theory. If you look at the formula right above it, it reads Voltage Drop = drop factor X dc voltage drop. This is the formula for a four wire two-phase system and each phase is considered as a separate two-wire single-phase system. Then there is the formula in question and I think it is figured .75 which is three of the four wires or 3/4 (.75) of the wires, which means you would only have .75 of the voltage drop.
As I said this is just a theory, let me know what you think.

 

[buzzp]

I believe on the others (motor) they are using the fact that the pf is not one. They are estimating it to be .86. Below is a link that shows some of the details of calculations based on NEC.

http://members.aol.com/ocsee/el_calc.htm

I think the bottom line is they are making assumptions to simplify the calculations. One would have to review how these formulas were derived, mathmatically, to see what assumptions were made. If I have time, I will try to find a better source of info in making these calculations. If you have a copy of the NEC you might try there. A misconception is that the NEC percent voltage drop threshold is a requirement. It is not. It is only recommended (fine print note).

 

[rbulsara]

I am at home this week and do not have access to the hand book. But probale explantion for the later post,saying

Voltage Drop for Lamps = (dc VD)/2
Voltage Drop for motors= 0.866*(dc VD)
is as follows: I am not saying it is not debatable. Drawing a picture will help.

In a 3-ph, 4-w balanced load consisting of all 'single phase' lamps the neutral current will be zero. The voltage drop in the neutral conductor is assumed to be zero (this is true on paper but near imppossible in practice), there for the voltage drop at the single phase lamp 'terminals' will be Vdc/2.

Now for a 3 balanced phase motor (or any other fundamental frequency or line voltage lamps) load the any two wires involved will be line voltage wires, in which the voltage drop will be Vdc. Since the VD per phase will be VD line to line divided by sqrt(3), the VD per phase will be 0.866 Vdc as 1/SQRT (3) =.866. %VD per phase=%VD L-L.

I have not thought through .75 factor given in the original post. (may be someone is multiplying (.866)* (.866)=0.75)

 

[rbulsara]

I suspect as follows:

In a balanced 3-wire system tapped off a 3ph-4W system, the magnitude of current in the neutral will be numerically same as the any of the line current (two equal currents 120 deg apart are added). This is gives you the same (load) current in 3 wires compared to two separate single phase circuits containing 4 wires. Therefore the total voltage drop in such a 3 wire system will be 3/4 or 0.75 times of dc voltage drop (in two separate circuits).

Feel free to throw darts at this.

 

[skiier]

rbulsara check your math.

1/3sqrt= 0.577 last time I checked

 

[skiier]

I think I figured this one out.
By calculating the ratio of the voltage drop Vd ( the line voltage drop) in the 3 wire polyphase system to the 2 wire system I came up with a ratio of 2/7sqrt = 0.7559.
By drawing some vectors of voltage drops in each line of the polyphase system I determined the magnitude of the total voltage drop to the load was equal to 7sqrt*the Vd in any line.
The current(s) in the common conductor will produce a Vd in that common conductor that has a magnitude of 3sqrt of the Vd in either of the other "hot wires" at an angle of 30 degrees.
So in effect you have 2 vectors to add to find the total voltage drop.
One at 30 degrees with a magnitude 3sqrt*Vd and the other at a 60 degrees with a magnitude of Vd. The one at 60 degrees is the Vd of either hot wire.
When you sum these together to get the total voltage drop you get angle of 150 degrees between the 2. Let Vd represent the absolute magnitude of the voltage drop and let Vp represent the magnitude of the voltage drop in either hot wire:

So using the cosine law;

a^2 = b^2 + c^2 + 2bc*cosA
or Vd^2 = Vp^2 + (3sqrt*Vp)^2 - 2*Vp*(3sqrt*Vp)* cos150

then Vd^2 = 4Vp^2 - (2*Vp*(3sqrt*Vp)* (-3sqrt/2)

and Vd^2 = 4Vp^2 + 3Vp^2 = 7Vp^2

or Vd = 7sqrt*Vp in a polyphase system

while in a 2 wire system
Vd = 2 Vp
So it follows then the ratio of voltage drop of a poly phase system to a 2 wire system is 7sqrt/2 = 1.322 or that the voltage drop will be 132% greater in a poly phase system than the same 2 wire system. Or conversely you can say that for a given 2 wire system a similiar polyphase system will have 2/sqrt7=0.7559 or 75% of the voltage drop capacity of the 2wire system.

hope I helped

 

[skiier]

For a true 3 phase aplication such as a 3 phase 3wire motor. The same as above applies.

The voltage seen at the load will be the supply line voltage minus the vector sum of the voltage drops in any 2 lines. The two line voltage drops are out of phase with the line voltage but they do have a vector sum that just nicely adds up to have a magnitude of - you guessed it - 3sqrt/2*Vp.

or Vd=3sqrt/2*Vp = 0.866Vp

have fun

 

[gsimson]

For lamps, the drop in voltage is half of the total voltage drop in the conductor from the source to the lamp and back. Hence the voltage drop at lamp terminal id (dc VD/2). Lighting loads can not be operated with 3 wire system.

For motor loads the line current ( or the line voltage ) is Sq.Root of 3 phase current ( or phase voltage ) for star( or delta ) system. As the return path is the other line in a 3 phase system the voltage drop is half of sq.root of 3 ie .866 (dc VD)

 

[skiier]

umm gsimson lighting loads can be operated on 3 wire systems just fine thank you or were you referring to 3 phase 3wire systems?
The person that posted the question was asking what the ratio 0.75 meant. Your Vd/2 anology works just fine for 2 wire systems or "balanced" 3 phase 3 wire systems but do not attempt to apply this analogy to a polyphase 3 wire system containing 2 "hots" and a "common" or "wht wire".
You have to "vectorally sum" the "unbalanced" voltage drops produced in the line conductors to get the correct voltage drop. This is not simply the sum of numbers but the "vector sum" of numbers or did you miss that class?