Voltage Drop?

[JasonCo]

I ALMOST understand how to do voltage drop equations. Let me give a quick scenario:

The distance between the panel and a load is 150 feet. The circuit to the load consists of two 12 AWG THHN/THWN solid copper coated conductors that are connected to a 1-pole breaker in a 277/480-volt, 3phase, 4-wire panel and the neutral bar. The load draws 9.8 amps.

So its a 3phase panel so would the equation be 1.732 x K x I x L / CMA? Or because the conductors are on a 1-pole breaker would it be 2 x K x I x L / CMA?

I know L = length will be 300 feet. I = amps will be 9.8. And I know to find K, I need to look at Chapter 9 table 8, but... this is the end of the line... I have no idea how to find K when looking at that table, no one has really explained it in detail when I google it, they just say you can find it on that table... :/ And CMA = Circular mill area so would CMA = 6530? When I look at Chp. 9 Tb. 8.

Thanks for your time and help!

 

[kwired]

I ALMOST understand how to do voltage drop equations. Let me give a quick scenario:

The distance between the panel and a load is 150 feet. The circuit to the load consists of two 12 AWG THHN/THWN solid copper coated conductors that are connected to a 1-pole breaker in a 277/480-volt, 3phase, 4-wire panel and the neutral bar. The load draws 9.8 amps.

So its a 3phase panel so would the equation be 1.732 x K x I x L / CMA? Or because the conductors are on a 1-pole breaker would it be 2 x K x I x L / CMA?

I know L = length will be 300 feet. I = amps will be 9.8. And I know to find K, I need to look at Chapter 9 table 8, but... this is the end of the line... I have no idea how to find K when looking at that table, no one has really explained it in detail when I google it, they just say you can find it on that table... :/ And CMA = Circular mill area so would CMA = 6530? When I look at Chp. 9 Tb. 8.

Thanks for your time and help!

I didn't look it up, but from recollection 6530 CM sounds right for 12 AWG.

The 1.372 factor (square root of three) only applies to three phase loads where all three phases are carrying same current. The 2 factor comes in with a two wire circuit - but remember in your case the voltage between the two wires is 277 and not 480 volts.

The reasons is you are determining the resistance of the conductors - with a two wire circuit you have one conductor out to the load and one returning so you multiply the length times 2 (as long as both conductors are same size and length). If not same size and length then resistance in each segment is not same and you would need to calculate each segment separately and then add them together to get total drop.

Similar with the three phase but the factor is the square root of three because the current going out on phase A returns on B and C, current out on B returns on A and C, current out on C returns on A and B...every line has current in it that also is partly in two other lines, where (two wire) single phase has all the same current in each line at all times. Easiest way I can describe it without drawing vectors that I don't really know how to draw in the first place.

 

[GoldDigger]

The 1.732 factor also only comes in when you use the line-to-line current in your equation.
If you have line to neutral loads and they are roughly balanced you multiply *one* times line current times one-way resistance. And to calculate % VD from that you use line to neutral voltage.
If the line to neutral load is isolated (totally unbalanced) you use a factor of *two*, just like for single phase L to L loads.

 

[GoldDigger]

The 1.732 factor also only comes in when you use the line-to-line current in your equation.
If you have line to neutral loads and they are roughly balanced you multiply *one* times line current times one-way resistance. And to calculate % VD from that you use line to neutral voltage.
If the line to neutral load is isolated (totally unbalanced) you use a factor of *two*, just like for single phase L to L loads.

I take that back, sort of.
If you start with the line current and are looking for the drop in the line to line voltage, you need the 1.732 provided the three phase currents are balanced.
If you have an isolated line-line load, then you use a different factor that I can't calculate in my head.

 

[kwired]

I take that back, sort of.
If you start with the line current and are looking for the drop in the line to line voltage, you need the 1.732 provided the three phase currents are balanced.
If you have an isolated line-line load, then you use a different factor that I can't calculate in my head.

voltage drop on the final segment of the two wire circuit is a factor of 2 isn't it? Assuming both conductors are same size and length half your total conductor resistance is in each conductor. The part you can't calculate in your head is probably the voltage drop ahead of the final 2 wire segment where you do have other currents from other circuits and imbalances factoring into the result.

 

[GoldDigger]

voltage drop on the final segment of the two wire circuit is a factor of 2 isn't it?....

Yes and no.
The voltage drop relative to a separate ground reference (assuming a grounded wye for calculation purposes, with the results being the same for a delta source) will be the current (line current == load current) times the resistance. But the voltage drop will not be in phase with the line voltage. So the magnitude of the dropped voltage will not be the initial magnitude minus the magnitude of the voltage drop.
But when you get down to the difference between the two line voltages, the drop in that voltage should be exactly the load current times twice the line resistance.

 

[jumper]

I know L = length will be 300 feet. I = amps will be 9.8. And I know to find K, I need to look at Chapter 9 table 8, but... this is the end of the line... I have no idea how to find K when looking at that table, no one has really explained it in detail when I google it, they just say you can find it on that table... :/ And CMA = Circular mill area so would CMA = 6530? When I look at Chp. 9 Tb. 8.

Thanks for your time and help!

Here is a thread on the constant K, post #5 explains it well:

http://forums.mikeholt.com/showthread.php?t=143399

 

[Smart $]

Yes and no.
The voltage drop relative to a separate ground reference (assuming a grounded wye for calculation purposes, with the results being the same for a delta source) will be the current (line current == load current) times the resistance. But the voltage drop will not be in phase with the line voltage. So the magnitude of the dropped voltage will not be the initial magnitude minus the magnitude of the voltage drop.
But when you get down to the difference between the two line voltages, the drop in that voltage should be exactly the load current times twice the line resistance.

You are getting way too technical on the matter. I like to hear from anyone that ran across a situation where that mattered on a small conductor circuit at consumer-level voltage.

 

[K8MHZ]

You are getting way too technical on the matter. I like to hear from anyone that ran across a situation where that mattered on a small conductor circuit at consumer-level voltage.

Does 20 amp 120 volt receptacle wiring in a school count?

 

[Smart $]

Does 20 amp 120 volt receptacle wiring in a school count?

Yes, if the voltage phase shift made a significant difference in the voltage drop.

 

[drcampbell]

The voltage drop is equal to current x resistance.
= (9.8 amps) x (300 feet x 0.0016 Ω/foot)
= 4.7 volts

The voltage drop isn't influenced by what kind of a circuit it's connected to. That becomes relevant only when deciding whether the voltage drop is acceptable.

I looked up the resistance for 12 AWG copper wire:
https://en.wikipedia.org/wiki/American_wire_gauge

 

[kwired]

The voltage drop is equal to current x resistance.
= (9.8 amps) x (300 feet x 0.0016 Ω/foot)
= 4.7 volts

The voltage drop isn't influenced by what kind of a circuit it's connected to. That becomes relevant only when deciding whether the voltage drop is acceptable.

I looked up the resistance for 12 AWG copper wire:
https://en.wikipedia.org/wiki/American_wire_gauge

Kind of circuit comes into play as far as whether it is single or three phase. In your calculation you doubled the length which is correct for single phase two wire circuit, had it been balanced three phase three wire circuit you multiply one way circuit length by the square root of three.

 

[K8MHZ]

Yes, if the voltage phase shift made a significant difference in the voltage drop.

I am pretty sure it was the 300+ foot runs. We used #8's for the 20 amp circuits.

 

[Smart $]

I am pretty sure it was the 300+ foot runs. We used #8's for the 20 amp circuits.

You sure you know what I (originally GoldDigger) was referring to...???

Appears you are helping to make my point in a way.

 

[K8MHZ]

You sure you know what I (originally GoldDigger) was referring to...???

Appears you are helping to make my point in a way.

What I was trying to say was that I doubted that voltage phase shift was the reason for the upsize from 12 to 8, rather it was the 300+ foot long runs we were doing.

 

[Smart $]

What I was trying to say was that I doubted that voltage phase shift was the reason for the upsize from 12 to 8, rather it was the 300+ foot long runs we were doing.

Okay... so you are actually emphasizing my point. My query was for anyone having experienced a significant drop as a result of a shift... and you replied nowhere near the detail necessary to support what I thought you were supporting.

 

[quantum]

I ALMOST understand how to do voltage drop equations. Let me give a quick scenario:

The distance between the panel and a load is 150 feet. The circuit to the load consists of two 12 AWG THHN/THWN solid copper coated conductors that are connected to a 1-pole breaker in a 277/480-volt, 3phase, 4-wire panel and the neutral bar. The load draws 9.8 amps.

So its a 3phase panel so would the equation be 1.732 x K x I x L / CMA? Or because the conductors are on a 1-pole breaker would it be 2 x K x I x L / CMA?

I know L = length will be 300 feet. I = amps will be 9.8. And I know to find K, I need to look at Chapter 9 table 8, but... this is the end of the line... I have no idea how to find K when looking at that table, no one has really explained it in detail when I google it, they just say you can find it on that table... :/ And CMA = Circular mill area so would CMA = 6530? When I look at Chp. 9 Tb. 8.

Thanks for your time and help!

Do not use 300 feet, the length is 150 feet for the equation, I.E. the 1-way length. The 2 in the voltage drop equation takes care of doubling the distance already.


Vd = 2 K L I / CM (Use this for 1-Pole and 2-Pole breakers in your panel given in the example, these all produce single phase current)


Now, we can convert this equation to 3-Phase power by multiplying it by sin (120 degrees)

Vd = 2 * sin (120) K L I / CM = 2 * (sqrt(3)/2) K L / CM = sqrt (3) K L I / CM (Use this for 3-Pole breakers in your example)


As for finding K from Chapter 9 Table 8, multiply the resistance column by the circular mills column and divide by 1000. For #12, it would be (1.98 * 6530/1000 ) = 12.93

 

[K8MHZ]

Okay... so you are actually emphasizing my point. My query was for anyone having experienced a significant drop as a result of a shift... and you replied nowhere near the detail necessary to support what I thought you were supporting.

Sorry about that.