3 phase Voltage Drop square root 3 factor !?

[AWinston]

I understand the concept of voltage drop but I am worried about the sqrt 3 factor. To get voltage drop you have to know the current. In a 3 phase system that would be I = P/(sqrt(3)*V). In order to determine the voltage drop it's Vdrop = sqrt(3)*I*Z*L.

My question is, when figuring out the current, do you calculate it with the sqrt (3) prior to entering the I value into the Vdrop equation? If that's the case, the voltage drop equation would then become
Vdrop = 3*I*Z*L. The 3 coming from sqrt(3)*sqrt(3).

 

[GoldDigger]

Before reaching a specific conclusion you need to be clear what current you are using and what voltage you are looking at too. Let's start with a wye.
The most accessible current value to work with is the line current in one of the ungrounded conductors. If you have a line to neutral load and can assume balanced loads the current I will result from that one load and the voltage drop will be in phase with the applied voltage (for a resistive load).
If R is the end to end resistance of one ungrounded line, VD is simply IR.
If you look instead at an unbalanced line to line load the voltage on each of the two lines is IR and those two drops are in phase, so line to line VD will be 2IR.
But if you have a balanced delta load with each line current being I, the line to line current will be 2I/sqrt(3). And although the line to ground voltage drop will be IR the drops on two lines will be out of phase. That makes the line to line voltage drop only sqrt(3)IR.

Now, what current value are you using and what VD are you trying to find?

 

[AWinston]

Before reaching a specific conclusion you need to be clear what current you are using and what voltage you are looking at too. Let's start with a wye.
The most accessible current value to work with is the line current in one of the ungrounded conductors. If you have a line to neutral load and can assume balanced loads the current I will result from that one load and the voltage drop will be in phase with the applied voltage (for a resistive load).
If R is the end to end resistance of one ungrounded line, VD is simply IR.
If you look instead at an unbalanced line to line load the voltage on each of the two lines is IR and those two drops are in phase, so line to line VD will be 2IR.
But if you have a balanced delta load with each line current being I, the line to line current will be 2I/sqrt(3). And although the line to ground voltage drop will be IR the drops on two lines will be out of phase. That makes the line to line voltage drop only sqrt(3)IR.

Now, what current value are you using and what VD are you trying to find?

It's a 3PH 208 Wye system. I am trying to find the voltage drop in all 3 phases. I just want to make sure I'm not using an extra sqrt(3).

 

[GoldDigger]

It's a 3PH 208 Wye system. I am trying to find the voltage drop in all 3 phases. I just want to make sure I'm not using an extra sqrt(3).

"The voltage drop in all three phases" is not an unambiguous (or even agreed upon) term. You can use more precise words or draw a diagram and indicate the two points between which the VD is being measured.
In any case I believe you have at least one extra factor of sqrt(3).

 

[AWinston]

"The voltage drop in all three phases" is not an unambiguous (or even agreed upon) term. You can use more precise words or draw a diagram and indicate the two points between which the VD is being measured.
In any case I believe you have at least one extra factor of sqrt(3).

I am trying to determine the voltage drop from a switchboard to a main panel. Total load draws 241A (I = 86400VA/(sqrt(3)*208V). The cable is being ran 170ft in PVC. The cable run I am considering is 4#4/0. Based on the customer's electric bill, I determined that the phase angle is 22.38 degrees which gives me a resistance of 0.072 ohms/1000 ft (Total resistance of 0.01 ohms). Is that enough detail?

 

[kwired]

I am trying to determine the voltage drop from a switchboard to a main panel. Total load draws 241A (I = 86400VA/(sqrt(3)*208V). The cable is being ran 170ft in PVC. The cable run I am considering is 4#4/0. Based on the customer's electric bill, I determined that the phase angle is 22.38 degrees which gives me a resistance of 0.072 ohms/1000 ft (Total resistance of 0.01 ohms). Is that enough detail?

What is this 22.38 degree angle? There should always be 120 degrees between two phases of a wye system.

 

[AWinston]

What is this 22.38 degree angle? There should always be 120 degrees between two phases of a wye system.

The angle is from the power triangle based on the reactive power given on the electric bill.

 

[kwired]

The angle is from the power triangle based on the reactive power given on the electric bill.

That is power factor information, voltage drop in the conductors is still going to be dependent on actual current in the conductor whether it is true power or reactive power.

If you are given watts and need to convert to VA - then you need to know what power factor is to do the conversion.

 

[AWinston]

That is power factor information, voltage drop in the conductors is still going to be dependent on actual current in the conductor whether it is true power or reactive power.

If you are given watts and need to convert to VA - then you need to know what power factor is to do the conversion.

Sorry, In my calculations, I did the conversion. Back to my original question....am I using an extra factor of sqrt(3)?

 

[kwired]

Sorry, In my calculations, I did the conversion. Back to my original question....am I using an extra factor of sqrt(3)?

extra factor of sqrt3? Likely yes.

Looking back at what you said in OP, you are trying to combine two portions of two formulas when you add the second sqrt3 in there.

VD is current based, yes you used sqrt3 when calculating current from VA, but once you determine current, that is the current and you use current later in the VD calculation.

Where the sqrt3 factor comes in for the VD formula is the fact that current is dividing across three paths. If you were calculating VD for a two wire circuit you use a factor of 2 instead of sqrt3 because you have length of supply conductor and a return conductor. With three phase (same current all three lines) you have length of supply conductor and length of two return conductors some current is going to each other phase (at a sqrt3 factor).

 

[AWinston]

extra factor of sqrt3? Likely yes.

Looking back at what you said in OP, you are trying to combine two portions of two formulas when you add the second sqrt3 in there.

VD is current based, yes you used sqrt3 when calculating current from VA, but once you determine current, that is the current and you use current later in the VD calculation.

Where the sqrt3 factor comes in for the VD formula is the fact that current is dividing across three paths. If you were calculating VD for a two wire circuit you use a factor of 2 instead of sqrt3 because you have length of supply conductor and a return conductor. With three phase (same current all three lines) you have length of supply conductor and length of two return conductors some current is going to each other phase (at a sqrt3 factor).

Thank you so much for the help.

 

[Carultch]

I understand the concept of voltage drop but I am worried about the sqrt 3 factor. To get voltage drop you have to know the current. In a 3 phase system that would be I = P/(sqrt(3)*V). In order to determine the voltage drop it's Vdrop = sqrt(3)*I*Z*L.

My question is, when figuring out the current, do you calculate it with the sqrt (3) prior to entering the I value into the Vdrop equation? If that's the case, the voltage drop equation would then become
Vdrop = 3*I*Z*L. The 3 coming from sqrt(3)*sqrt(3).

It depends on if you think of the phase-to-phase, or phase-to-neutral voltage in this calculation. The ratio between the two, in a standard WYE system, is sqrt(3).

When you use the phase-to-neutral voltage only (call it lowercase v), you don't even need to think about the square root of 3. In this example, I = P/(3*v), and Vdropratio = I*R/v, where R is the one-way resistance and v is the phase-to-neutral voltage. The power is carried by three circuits at voltage V, and the current travels out on the line and back on the neutral. Because the currents on the neutral add up to zero, the only current that generates a drop in voltage is the outgoing current on the line.

Note that combining these two equations to eliminate I, we get:
I = P/(3*v)
Vdropratio = I*R/v
Vdropratio = P*R/(3*v)

Now when we translate the formulas for capital V = phase-to-phase voltage instead, we introduce:
V = v*sqrt(3)

And substitute for v:
v = V/sqrt(3)

And then:
I = P/(3*(V/sqrt(3)))
Vdropratio = I*R/(V/sqrt(3))

Simplify:
I = P/(V*sqrt(3))
Vdropratio = I*R*sqrt(3)/V

If you eliminate current from the formula, you get:
Vdropratio = (P/(V*sqrt(3)))*R*sqrt(3)/V

Simplify:
Vdropratio = P*R/V^2

There isn't an "extra square root of 3" factor, because it ends up cancelling out of the equation.

 

[kwired]

It depends on if you think of the phase-to-phase, or phase-to-neutral voltage in this calculation. The ratio between the two, in a standard WYE system, is sqrt(3).

When you use the phase-to-neutral voltage only (call it lowercase v), you don't even need to think about the square root of 3.

It doesn't matter if it is phase to neutral or single phase to phase circuit. With balanced load across all three phases you use the sqrt3 factor in your formula, if it is a single phase two wire circuit regardless of what two points it is supplied from - you use a factor of 2 because there is a supply and a return conductor, all you are doing is doubling the one way length of the conductor to account for the return path. In the three phase circuit the return path is through two other conductors at a sqrt(3) factor.

 

[Smart $]

It doesn't matter if it is phase to neutral or single phase to phase circuit...

Ahhh, but it does.

Assuming we are talking 3Ø wye system only...

...the factor can be 1 or 2 for line to neutral, depending on whether the loads are coincident and balanced MWBC wye circuit or not, respectively.

...line to line can actually vary from 1.732 to 2, depending on whether the load is balanced or not.

In texts, the preceding is usually oversimplified to just 2 and 1.732 respectively because they either do not understand the variations or just don't want to get into the specifics.

 

[kwired]

Ahhh, but it does.

Assuming we are talking 3Ø wye system only...

...the factor can be 1 or 2 for line to neutral, depending on whether the loads are coincident and balanced MWBC wye circuit or not, respectively.

...line to line can actually vary from 1.732 to 2, depending on whether the load is balanced or not.

In texts, the preceding is usually oversimplified to just 2 and 1.732 respectively because they either do not understand the variations or just don't want to get into the specifics.

I was talking about a two wire circuit needing a factor of 2 regardless of what points of the system it connects to.

 

[Smart $]

I was talking about a two wire circuit needing a factor of 2 regardless of what points of the system it connects to.

That's a fact!