Voltage Drop "M" Multiplier

[gh0st]

Why is it that online VD drop calculators don't use the "M" multiplier M=1 for a 3PH/4W?

These calculators all seem to use M=1.732 regardless of 3PH/3W or 3PH/4W voltage configuration.

For example....for a 250A, 480V, 3PH, 4W panelboard feed
Distance = 600 ft. from distribution board
Load = 250 A
Wire = 250kcmil, cu
M=1 yields 1.61 %VD vs M=1.732 yields 2.79%VD

In what application would we use M=1? I would think that even in situations where we are feeding a cluster of office furniture (multiwire branch circuit), the load would not truly be balanced.

 

[xptpcrewx]

Why is it that online VD drop calculators don't use the "M" multiplier M=1 for a 3PH/4W?

These calculators all seem to use M=1.732 regardless of 3PH/3W or 3PH/4W voltage configuration.

For example....for a 250A, 480V, 3PH, 4W panelboard feed
Distance = 600 ft. from distribution board
Load = 250 A
Wire = 250kcmil, cu
M=1 yields 1.61 %VD vs M=1.732 yields 2.79%VD

In what application would we use M=1? I would think that even in situations where we are feeding a cluster of office furniture (multiwire branch circuit), the load would not truly be balanced.

Can you provide a link to the calculator you are using?

In general 1.732 (sqrt 3) is used for 3PH 3w and 4w. This is a “per phase equivalent” factor. A multiplier of 1 would never be used. For single phase circuits, the multiplier would be 2.

 

[Hv&Lv]

“M” = phase multiplier: Use 2 for a single phase or DC circuit and 3, or 1.732, for a three-phase circuit.

Voltage Drop Calculator

Calculate voltage drop in an AC or DC circuit, find the minimum wire gauge required when planning a circuit, or determine the maximum length of conductors in a circuit.

www.inchcalculator.com

M=1 would be a useless step in any calculation.

 

[gh0st]

This website mentions the use of M=1: http://www.electrician2.com/voltdroppractice/vdr03.htm

The calculator from Southwire uses M=1.732 at 3 or 4 wire configurations: https://www.southwire.com/calculator-vdrop

We have an old program which generally aligns with southwire's results:

 

[Carultch]

Why is it that online VD drop calculators don't use the "M" multiplier M=1 for a 3PH/4W?

These calculators all seem to use M=1.732 regardless of 3PH/3W or 3PH/4W voltage configuration.

For example....for a 250A, 480V, 3PH, 4W panelboard feed
Distance = 600 ft. from distribution board
Load = 250 A
Wire = 250kcmil, cu
M=1 yields 1.61 %VD vs M=1.732 yields 2.79%VD

In what application would we use M=1? I would think that even in situations where we are feeding a cluster of office furniture (multiwire branch circuit), the load would not truly be balanced.

The M multiplier is a factor that determines the effective round trip length of the circuit, as it relates to the 1-way length of the circuit.

In a DC circuit, the round trip distance is twice the 1-way length of the circuit. Current leaves the positive terminal, and returns on the negative terminal. It has to travel the 1-way length, and then return to the source through another round of the 1-way length.

A single phase circuit would work just like a DC circuit, in how you determine the round trip length. And this is true for both line-to-neutral single phase circuits (where you'd use the 120V value), and line-to-line single phase circuits (where you'd use the 240V value). Worst case scenario for the voltage drop calculation in a 1-phase/3-wire system, the full current is on both the black and the red conductors. If there is any imbalance, the voltage drop will be slightly less on the conductor carrying less of the load.

In a three phase circuit, when you consider line-to-line voltage for your calculation, the M multiplier is the square root of 3. A mathematician would insist on maintaining its identity as sqrt(3), so that it is always possible to evaluate it to greater precision. An engineer in the days before calculators, would memorize that it equals 1.73, and call that good enough. The worst case scenario voltage drop, just like the 1-ph/3-wire system, will occur when all three conductors carry the full current. An imbalance will mean less voltage drop on the conductor carrying less of the current. There will be a shift in the neutral-to-ground voltage at the load if the ohmic losses and imbalance among the phases are both significant.

"In what application would we use M=1?"

Good question. And the answer will give you some understanding as to why it is sqrt(3), in the 3-phase calculation in the first place.

Suppose we chose not to use the line-to-line voltage for our calculation, but instead to use the line-to-neutral voltage. Given balanced currents, the current among all three phases adds up to zero, just by using the three phase conductors alone. The three phase conductors do the full job of carrying the current in a complete path. The neutral is idle, until there is an imbalance. Suppose phase A is a cosine curve, starting at Imax at time t=0. Phase B and phase C, will both be at -1/2*Imax at this instant in time. One will be rising back to zero, and the other will be falling to negative Imax. The currents of phases B and C will add up to the opposite of phase A's current. This is the principle advantage of 3-phase systems, because the current of the other two phases' waveforms, will simultaneously bring the current of the first phase's waveform back to the source. No conductor has to be dedicated to carrying return current, in a balanced 3-phase system It is as if the current only had to take the 1-way trip, driven by the phase-to-neutral voltage.

This means using the line-to-neutral voltage (Vpn), the voltage drop calculation becomes %Vd = 100% * I*r*L/Vpn, where I is current, r is the resistance per unit length, L is the 1-way length. And you multiply by 100 to make it a percent. If r has units Ohms/kft, then you'd have to divide by 1000 to account for the number of ft in 1 kft. The M-multiplier would be 1.

Now to convert this formula for using Vpp (phase-to-phase) instead of Vpn, we can recall that Vpp = sqrt(3)*Vpn, which you can derive through the geometry of how we represent phases as vectors. Take the above formula, and multiply by 1 in a fancy way.

1 = sqrt(3)/sqrt(3)

%Vd = 100% * (I*r*L/Vph) * (sqrt(3)/sqrt(3))
%Vd = 100% * (I*r*L*sqrt(3)/(Vpn*sqrt(3))
%Vd = 100% * (I*r*L*sqrt(3)/(Vpp)

This also gives you some insight as to why the kVA (or kW with unity power factor) relates to voltage and current in a 3-phase system with a sqrt(3) factor. You can either calculate kVA = 3*I*Vpn, or kVA =sqrt(3)*I*Vpp. Each of the phases carries its third of the kVA on the current driven by the phase-to-neutral voltage.

 

[gh0st]

The M multiplier is a factor that determines the effective round trip length of the circuit, as it relates to the 1-way length of the circuit.

In a DC circuit, the round trip distance is twice the 1-way length of the circuit. Current leaves the positive terminal, and returns on the negative terminal. It has to travel the 1-way length, and then return to the source through another round of the 1-way length.

A single phase circuit would work just like a DC circuit, in how you determine the round trip length. And this is true for both line-to-neutral single phase circuits (where you'd use the 120V value), and line-to-line single phase circuits (where you'd use the 240V value). Worst case scenario for the voltage drop calculation in a 1-phase/3-wire system, the full current is on both the black and the red conductors. If there is any imbalance, the voltage drop will be slightly less on the conductor carrying less of the load.

In a three phase circuit, when you consider line-to-line voltage for your calculation, the M multiplier is the square root of 3. A mathematician would insist on maintaining its identity as sqrt(3), so that it is always possible to evaluate it to greater precision. An engineer in the days before calculators, would memorize that it equals 1.73, and call that good enough. The worst case scenario voltage drop, just like the 1-ph/3-wire system, will occur when all three conductors carry the full current. An imbalance will mean less voltage drop on the conductor carrying less of the current. There will be a shift in the neutral-to-ground voltage at the load if the ohmic losses and imbalance among the phases are both significant.

"In what application would we use M=1?"

Good question. And the answer will give you some understanding as to why it is sqrt(3), in the 3-phase calculation in the first place.

Suppose we chose not to use the line-to-line voltage for our calculation, but instead to use the line-to-neutral voltage. Given balanced currents, the current among all three phases adds up to zero, just by using the three phase conductors alone. The three phase conductors do the full job of carrying the current in a complete path. The neutral is idle, until there is an imbalance. Suppose phase A is a cosine curve, starting at Imax at time t=0. Phase B and phase C, will both be at -1/2*Imax at this instant in time. One will be rising back to zero, and the other will be falling to negative Imax. The currents of phases B and C will add up to the opposite of phase A's current. This is the principle advantage of 3-phase systems, because the current of the other two phases' waveforms, will simultaneously bring the current of the first phase's waveform back to the source. No conductor has to be dedicated to carrying return current, in a balanced 3-phase system It is as if the current only had to take the 1-way trip, driven by the phase-to-neutral voltage.

This means using the line-to-neutral voltage (Vpn), the voltage drop calculation becomes %Vd = 100% * I*r*L/Vpn, where I is current, r is the resistance per unit length, L is the 1-way length. And you multiply by 100 to make it a percent. If r has units Ohms/kft, then you'd have to divide by 1000 to account for the number of ft in 1 kft. The M-multiplier would be 1.

Now to convert this formula for using Vpp (phase-to-phase) instead of Vpn, we can recall that Vpp = sqrt(3)*Vpn, which you can derive through the geometry of how we represent phases as vectors. Take the above formula, and multiply by 1 in a fancy way.

1 = sqrt(3)/sqrt(3)

%Vd = 100% * (I*r*L/Vph) * (sqrt(3)/sqrt(3))
%Vd = 100% * (I*r*L*sqrt(3)/(Vpn*sqrt(3))
%Vd = 100% * (I*r*L*sqrt(3)/(Vpp)

This also gives you some insight as to why the kVA (or kW with unity power factor) relates to voltage and current in a 3-phase system with a sqrt(3) factor. You can either calculate kVA = 3*I*Vpn, or kVA =sqrt(3)*I*Vpp. Each of the phases carries its third of the kVA on the current driven by the phase-to-neutral voltage.

Carultch, I really appreciate the detailed response! I think I've pieced together why M=1 is never used, but I'm afraid the origins of sqrt(3) is still lost on me. I suppose that I've just memorized that 3ph wye systems have a LL to LN voltage ratio of sqrt(3).

So, it would seem that there aren't any real-life applications where M=1 would be used in a 3ph system since all single phase connections must use M=2... LN (neutral as a 'return' path) and LL (one of the two phases as a 'return' path). M=sqrt(3) must be used for all 3 phase connections by shear design that each of the three phase's waveforms 'share' the load.

 

[Carultch]

Carultch, I really appreciate the detailed response! I think I've pieced together why M=1 is never used, but I'm afraid the origins of sqrt(3) is still lost on me. I suppose that I've just memorized that 3ph wye systems have a LL to LN voltage ratio of sqrt(3).

So, it would seem that there aren't any real-life applications where M=1 would be used in a 3ph system since all single phase connections must use M=2... LN (neutral as a 'return' path) and LL (one of the two phases as a 'return' path). M=sqrt(3) must be used for all 3 phase connections by shear design that each of the three phase's waveforms 'share' the load.

It's more of a calculation preference for why we seldom (if ever) use a multiplier of M=1. You could opt to use the phase to neutral voltage, and use M=1 all the time. It is just a less common practice.

As for the origins of sqrt(3), it helps to draw it to scale. Get a compass and a protractor, and draw the following diagram, with the length of vectors a, b, and c each equal to 12 cm, and 120 degrees between each of them. Measure length L, between any two corners, and get 20.8 cm. You can also prove it with trig, by cutting triangle BCN in half. You see the 60 degree angle between vector c and the horizontal. Take its sine, and equate it to (L/2) over c. Sine of 60 degrees is our special case, where it equals sqrt(3)/2. Cancel the 2 from both sides, and get: L/c = sqrt(3).

The way this represents the phase of electricity waveforms, is that the system of vectors is rotating about point N (assume clockwise rotation), and the projection onto the horizontal axis (or real number axis) is what the voltage is in real time. This is a real world application of imaginary numbers; keeping track of phase. Phase A will start at V = Vmax, and proceed along the cosine curve, slowly decreasing in value from where it started at its upper turning point. Phase B starts at V = -1/2*Vmax, and continues descending toward the bottom turning point where V=-Vmax. Phase C starts at V = -1/2*Vmax, and climb back to zero. If you form the time-domain equations for voltage, you get the following. The +120 degrees, and -120 degrees, is the phase shift. The w indicates the angular frequency. Using degrees as the units for the trig functions, means that it is 360 degrees * frequency in Hertz. Mathematicians prefer radians as the angle unit, so that is why it is a lot more common for you to see w=2*pi*f.

Va = Vmax*cos(w*t)
Vb = Vmax*cos(w*t + 120 degrees)
Vc = Vmax*cos(w*t - 120 degrees)

You can also rearrange the cosine with a phase shift, such that it is a sum of sine and cosine, with no phase shift in either term.
Va = Ax*cos(w*t) + Ay*sin(w*t)
Vb = Bx*cos(w*t) + By*sin(w*t)
Vc = Cx*cos(w*t) + Cy*sin(w*t)

Solving for Ax, Bx, By, Cx, and Cy, we get:
Ax = Vmax
Ay = 0
Bx = -1/2*Vmax
By = -sqrt(3/2) * Vmax
Cx = -1/2*Vmax
Cy = +sqrt(3/2) * Vmax

These are also the components of each of the three vectors, at the snapshot when time = 0. The X-component of each vector in phase-space is the coefficient of cosine, and the Y-component is the coefficient of sine.

The amplitude of the waveforms is not simply the phase-to-neutral voltage, but rather there is a correction term of sqrt(2). So if it is 120 V nominal to neutral, it is really 170V from neutral to the peak of the waveform. This has to do with a special kind of time averaging called root-mean-square or RMS. So it would've been more accurate to draw the lengths a, b, and c to equal 17 cm instead of 12 cm, as they represent 120/208V 3-phase systems. However, that distracts from the point of how 208V = 120V * sqrt(3), because both nominal voltages are multiplied by sqrt(2) to get the amplitudes.

 

[gh0st]

It's more of a calculation preference for why we seldom (if ever) use a multiplier of M=1. You could opt to use the phase to neutral voltage, and use M=1 all the time. It is just a less common practice.

As for the origins of sqrt(3), it helps to draw it to scale. Get a compass and a protractor, and draw the following diagram, with the length of vectors a, b, and c each equal to 12 cm, and 120 degrees between each of them. Measure length L, between any two corners, and get 20.8 cm. You can also prove it with trig, by cutting triangle BCN in half. You see the 60 degree angle between vector c and the horizontal. Take its sine, and equate it to (L/2) over c. Sine of 60 degrees is our special case, where it equals sqrt(3)/2. Cancel the 2 from both sides, and get: L/c = sqrt(3).
View attachment 2557105
The way this represents the phase of electricity waveforms, is that the system of vectors is rotating about point N (assume clockwise rotation), and the projection onto the horizontal axis (or real number axis) is what the voltage is in real time. This is a real world application of imaginary numbers; keeping track of phase. Phase A will start at V = Vmax, and proceed along the cosine curve, slowly decreasing in value from where it started at its upper turning point. Phase B starts at V = -1/2*Vmax, and continues descending toward the bottom turning point where V=-Vmax. Phase C starts at V = -1/2*Vmax, and climb back to zero. If you form the time-domain equations for voltage, you get the following. The +120 degrees, and -120 degrees, is the phase shift. The w indicates the angular frequency. Using degrees as the units for the trig functions, means that it is 360 degrees * frequency in Hertz. Mathematicians prefer radians as the angle unit, so that is why it is a lot more common for you to see w=2*pi*f.

Va = Vmax*cos(w*t)
Vb = Vmax*cos(w*t + 120 degrees)
Vc = Vmax*cos(w*t - 120 degrees)

You can also rearrange the cosine with a phase shift, such that it is a sum of sine and cosine, with no phase shift in either term.
Va = Ax*cos(w*t) + Ay*sin(w*t)
Vb = Bx*cos(w*t) + By*sin(w*t)
Vc = Cx*cos(w*t) + Cy*sin(w*t)

Solving for Ax, Bx, By, Cx, and Cy, we get:
Ax = Vmax
Ay = 0
Bx = -1/2*Vmax
By = -sqrt(3/2) * Vmax
Cx = -1/2*Vmax
Cy = +sqrt(3/2) * Vmax

These are also the components of each of the three vectors, at the snapshot when time = 0. The X-component of each vector in phase-space is the coefficient of cosine, and the Y-component is the coefficient of sine.

The amplitude of the waveforms is not simply the phase-to-neutral voltage, but rather there is a correction term of sqrt(2). So if it is 120 V nominal to neutral, it is really 170V from neutral to the peak of the waveform. This has to do with a special kind of time averaging called root-mean-square or RMS. So it would've been more accurate to draw the lengths a, b, and c to equal 17 cm instead of 12 cm, as they represent 120/208V 3-phase systems. However, that distracts from the point of how 208V = 120V * sqrt(3), because both nominal voltages are multiplied by sqrt(2) to get the amplitudes.

Carultch... Its going to take me reading your posts a few times for me to absorb all of this but again, many thanks for the the detailed breakdown.

 

[Hv&Lv]

See if I can simplify this…

draw the triangle above. Equilateral triangle with all sides equal to “2”
now bisect the triangle. You have a long side =2, a short side =1, and a side that’s somewhere in the middle shorter than one but longer than the other. So now you have a 30,60,90 triangle.
Label the sides
”2” side is “c”
”1” side is ”a”
”x” side is b

Pythagorean Theorum. a^2 + b^2 = c^2
so ”1”^2 + b^2 = ”2”^2
manipulation gives us
b^2= “2”^2 - “1”^2
b^2= 4-1
b^2 =3
b=(sq rt) 3, or 1.73205

We use it because the phases are 120° apart. So the points on a circle (generator) are 120 degrees apart and are separated by (sq rt)3 times the radius

sorry so “elementary” with symbology. I’m on an iPhone cruising on a boat…

 

[gh0st]

See if I can simplify this…

draw the triangle above. Equilateral triangle with all sides equal to “2”
now bisect the triangle. You have a long side =2, a short side =1, and a side that’s somewhere in the middle shorter than one but longer than the other. So now you have a 30,60,90 triangle.
Label the sides
”2” side is “c”
”1” side is ”a”
”x” side is b

Pythagorean Theorum. a^2 + b^2 = c^2
so ”1”^2 + b^2 = ”2”^2
manipulation gives us
b^2= “2”^2 - “1”^2
b^2= 4-1
b^2 =3
b=(sq rt) 3, or 1.73205

We use it because the phases are 120° apart. So the points on a circle (generator) are 120 degrees apart and are separated by (sq rt)3 times the radius

sorry so “elementary” with symbology. I’m on an iPhone cruising on a boat…

See if I can simplify this…

draw the triangle above. Equilateral triangle with all sides equal to “2”
now bisect the triangle. You have a long side =2, a short side =1, and a side that’s somewhere in the middle shorter than one but longer than the other. So now you have a 30,60,90 triangle.
Label the sides
”2” side is “c”
”1” side is ”a”
”x” side is b

Pythagorean Theorum. a^2 + b^2 = c^2
so ”1”^2 + b^2 = ”2”^2
manipulation gives us
b^2= “2”^2 - “1”^2
b^2= 4-1
b^2 =3
b=(sq rt) 3, or 1.73205

We use it because the phases are 120° apart. So the points on a circle (generator) are 120 degrees apart and are separated by (sq rt)3 times the radius

sorry so “elementary” with symbology. I’m on an iPhone cruising on a boat…

Hv&Lv, I think I just had my 'ah-ha' moment! Your simplified version was just what I needed to have everything posted in this thread click! I truly am in awe with the community here...so helpful and patient. I don't have anyone in my office that I can discuss these things with as most conversations end with 'that's how its always been done'. Much appreciated!

 

[Carultch]

Hv&Lv, I think I just had my 'ah-ha' moment! Your simplified version was just what I needed to have everything posted in this thread click! I truly am in awe with the community here...so helpful and patient. I don't have anyone in my office that I can discuss these things with as most conversations end with 'that's how its always been done'. Much appreciated!

The simplification explains the first portion of my post, up through the picture insert.

After the picture insert, I explain why the triangle and vectors represent phases of electricity, and why it works to solve a simple triangle geometry problem to solve for nominal voltage between any two phases given the phase-to-neutral voltage for each of them. Smart$ at one point had a profile picture that animates the concept, but the website no longer supports animated profile pictures.

 

[xptpcrewx]

This website mentions the use of M=1: http://www.electrician2.com/voltdroppractice/vdr03.htm

The calculator from Southwire uses M=1.732 at 3 or 4 wire configurations: https://www.southwire.com/calculator-vdrop

We have an old program which generally aligns with southwire's results:
View attachment 2557102

Looking through the explanation for this calculator, the assumption for M=1 is they are using the "line-to-neutral" voltage for 3ph, 4w circuits. Traditionally, the way this concept is taught/used, is with using "line-to-line" values (which is why M=1.732 or M=2 pops up). M=1 makes sense as long as you understand this is only a one-way voltage drop using line-to-neutral voltages and ignoring the neutral conductor (100% balanced conditions only).

Some members have stated the M=1.732 factor can be thought of as the effective length for a 3ph circuit, but I personally see this as being fictitious and not very useful to visualize. The best way to think about the multiplier is by dividing both sides of the equation VD = M x [I x (Rdc/1000) x L] by M. The result is (VD/M) = [I x (Rdc/1000) x L]; where (VD/M) is nothing other than the "line-to-neutral" voltage and VD is the "line-to-line" voltage. (No more mysterious effective 3ph lengths"

 

[Carultch]

The best way to think about the multiplier is by dividing both sides of the equation VD = M x [I x (Rdc/1000) x L] by M. The result is (VD/M) = [I x (Rdc/1000) x L]; where (VD/M) is nothing other than the "line-to-neutral" voltage and VD is the "line-to-line" voltage. (No more mysterious effective 3ph lengths"

This reasoning works, until you consider how to calculate voltage drop on single phase to neutral circuits. The multiplier is 2 in such examples, despite it being line-to-neutral voltage.

 

[xptpcrewx]

This reasoning works, until you consider how to calculate voltage drop on single phase to neutral circuits. The multiplier is 2 in such examples, despite it being line-to-neutral voltage.

You’d have to use the line-to-line voltage to make it work.


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