Question about calculating voltage drop on 3 phase inductive circuits

[Crohnos01]

Hi folks,

I just found this forum...Very nice and I plan on hanging out a lot!

It's been awhile since I have done any VD calcs, and inductive 3 phase circuits even longer still so pardon my rust. I initially did a calculation using a negligible inductance calc vd = (1.73*K*L*I)/Cm found in Ugly's and then realized that since I was working on a 3 phase circuit with a motor hanging on the end of it, I should really be including the motor PF in the calculation. Using the figures in Table 9 of the NEC code, I came up with a line to line VD of 2.37 volts vs. my original calculation of 2.7 volts. (I pulled a .69 ohm/ft figure from the table). In either case, I am just over a 1% VD and should be good to go....However.... Shouldn't the calculation including the motor PF provide a higher VD rather than a lower one? Granted it's not much, and it matters not in the grand scheme of things, but I am struggling with the "why" Did I go bug-eyed and miss something?

Specifics of the circuit in question:

Motor HP = 15
3 phase
208 volt
Length = 50 ft.
Amps = 39.59 (I rounded to 40)
Wire size = #8 THHW Cu
KW = 11.2 @ FLA

I am planning on running in PVC conduit buried and used the standard 75 deg C. Could someone check my math?

Thanks,

 

[gar]

130301-2020 EST

Crohnos01:

First, assume that the voltage source is ideal to simplify your problem.

What is the voltage drop along one of the hot wires?

How is that voltage drop changed by a change in load power factor if the line current remains constant?

What is the phase difference between the voltage drops on two of the hot wires?

Then what is the sum of the voltage drops on said two hot wires?

.

 

[hurk27]

Hi folks,

I just found this forum...Very nice and I plan on hanging out a lot!

It's been awhile since I have done any VD calcs, and inductive 3 phase circuits even longer still so pardon my rust. I initially did a calculation using a negligible inductance calc vd = (1.73*K*L*I)/Cm found in Ugly's and then realized that since I was working on a 3 phase circuit with a motor hanging on the end of it, I should really be including the motor PF in the calculation. Using the figures in Table 9 of the NEC code, I came up with a line to line VD of 2.37 volts vs. my original calculation of 2.7 volts. (I pulled a .69 ohm/ft figure from the table). In either case, I am just over a 1% VD and should be good to go....However.... Shouldn't the calculation including the motor PF provide a higher VD rather than a lower one? Granted it's not much, and it matters not in the grand scheme of things, but I am struggling with the "why" Did I go bug-eyed and miss something?

Specifics of the circuit in question:

Motor HP = 15
3 phase
208 volt
Length = 50 ft.
Amps = 39.59 (I rounded to 40)
Wire size = #8 THHW Cu
KW = 11.2 @ FLA

I am planning on running in PVC conduit buried and used the standard 75 deg C. Could someone check my math?

Thanks,

The most important thing is to know what the current the load will actually be pulling, motors can be a little frustrating if they don't give you the correct current at the voltage you are supplying, you stated 208 volts but this is not a normal motor voltage rating which can cause the current to be higher or lower depending upon if it is a 230 volt rated motor or a 200 volt rated motor (which is common for 208 rated motors in high torque applications)

If you use the name plate amperage the PF is already figured in, if not you divide the PF into the current to get the actual current unless you use PF correction cap at the load.

Also according to 430.6(A)(1) you must use table 430.250 for sizing your conductors for your 3-phase motor not the name plate, which a 15hp motor at 208 volts has a amperage of 46.2 amps for conductors, but in your case a #8 is still good to go.

As far as your calculations at what input you provided I'm getting 1.3% with a 2.7 VD
If we use the 46.2 amps listed in the NEC then I get a 1.5% drop or a 3.1v VD
If we add in a .8 PF then were looking at 49.49 with a 1.6% drop and a 3.3v VD

So none of the above is bad and its because the run is only 50', on longer runs you will have to be more acurate to what the current is as the margine of error can result in a larger differance of VD.

Here is a online calculator that works well: http://www.electrician2.com/calculators/vd_calculator.html

But again, knowing what a load will pull is very important, a 230 volt rated motor will pull a higher amperage at 208 volts because of pole slippage, and if this is an air compressor application, be ready for problems with hard starting because of the high torque type of application as it is more common for manufactures to use a 200 volt rated motor for this.

 

[Smart $]

... but I am struggling with the "why" Did I go bug-eyed and miss something? ...

Look at the resistance values in Table 9 then compare with the effective Z at 0.85 PF. The effective Z is lower than the resistance values for sizes up to around #1. As the conductor size gets larger the effective Z values are greater than the resistance values...

 

[Besoeker]

Hi folks,

I just found this forum...Very nice and I plan on hanging out a lot!

It's been awhile since I have done any VD calcs, and inductive 3 phase circuits even longer still so pardon my rust. I initially did a calculation using a negligible inductance calc vd = (1.73*K*L*I)/Cm found in Ugly's and then realized that since I was working on a 3 phase circuit with a motor hanging on the end of it, I should really be including the motor PF in the calculation. Using the figures in Table 9 of the NEC code, I came up with a line to line VD of 2.37 volts vs. my original calculation of 2.7 volts. (I pulled a .69 ohm/ft figure from the table). In either case, I am just over a 1% VD and should be good to go....However.... Shouldn't the calculation including the motor PF provide a higher VD rather than a lower one? Granted it's not much, and it matters not in the grand scheme of things, but I am struggling with the "why" Did I go bug-eyed and miss something?

Specifics of the circuit in question:

Motor HP = 15
3 phase
208 volt
Length = 50 ft.
Amps = 39.59 (I rounded to 40)
Wire size = #8 THHW Cu
KW = 11.2 @ FLA

I am planning on running in PVC conduit buried and used the standard 75 deg C. Could someone check my math?

Thanks,

A couple of points.
The motor current, if it's given on the nameplate, is the motor current. It already includes power factor. In any case, 40A is about right for a motor of that power rating and voltage. So no need to include power factor in your calculation.
I think your 0.69 ohm/ft figure just might be a wee bit suspect...........

 

[weressl]

So no need to include power factor in your calculation.

Unless yor calculation includes not only resistance but the more complex impedance formula. It still just approximates the inductance from the resistance as there is very little capacitance, but it helps to be more accurate especially when not dealing with full load, which is what I often use when I am certian that the initial load will not change and dealing with large, single loads and long runs.

 

[Besoeker]

Unless your calculation includes not only resistance but the more complex impedance formula.

That's true. But, for the size of conductor mentioned, it is likely to be insignificant. For multi-core armoured cable, which is what we would normally use for such an application, the difference between R and Z is about 1% with 50mm² conductors. A 50mm² conductor equates to about 1/0 AWG

 

[weressl]

That's true. But, for the size of conductor mentioned, it is likely to be insignificant. For multi-core armoured cable, which is what we would normally use for such an application, the difference between R and Z is about 1% with 50mm² conductors. A 50mm² conductor equates to about 1/0 AWG

We use GRS(Galvanized Rigid Steel) Conduit that is at least as bad as armored cable. We also use Galvanized Steel Interlocked Armor cable, but most of the time the armor is aluminum with PVC overjacket, or corrugated seamless armor - CLX by the OKONITE tradename.

 

[Besoeker]

We use GRS(Galvanized Rigid Steel) Conduit that is at least as bad as armored cable. We also use Galvanized Steel Interlocked Armor cable, but most of the time the armor is aluminum with PVC overjacket, or corrugated seamless armor - CLX by the OKONITE tradename.

You could well be right. I have a notion that the GRS conduit or any conduit might be worse than steel wire armoured (SWA) if worse means higher inductance. With the SWA the conductors are kept in close and controlled proximity that maybe doesn't happen to the same extent in conduit.

Actually, I've just looked at our cable tables (BS7671) and the difference between R and Z is about 4% in conduit compared to 1% that I gave before for 50mm² SWA.

 

[weressl]

You could well be right. I have a notion that the GRS conduit or any conduit might be worse than steel wire armoured (SWA) if worse means higher inductance. With the SWA the conductors are kept in close and controlled proximity that maybe doesn't happen to the same extent in conduit.

Actually, I've just looked at our cable tables (BS7671) and the difference between R and Z is about 4% in conduit compared to 1% that I gave before for 50mm² SWA.

Yep, armored cables are tight and spun with a constant pitch, individual conductors are much looser leaving as much as 50% airpsace and not orderly.

On a related subject, I always wondered about the additonal inductive/heat losst in conduit or armored cables when current unbalance is present. I remember that many of our motors - in Hungary - were wye wound, so you would bring back the unballance on the neutral.

 

[Smart $]

... the difference between R and Z is about 1% with 50mm² conductors. A 50mm² conductor equates to about 1/0 AWG

... the difference between R and Z is about 4% in conduit compared to 1% that I gave before for 50mm² SWA.

According to NEC Table 9, for an effective Z at 0.85PF, the difference between R and Z for copper conductors in conduit of any type is about 11-13% lower for the 8AWG used to calculate V_D in the OP. The difference at the 1/0AWG size is about 1% higher.

 

[Besoeker]

According to NEC Table 9, for an effective Z at 0.85PF, the difference between R and Z for copper conductors in conduit of any type is about 11-13% lower for the 8AWG used

Z lower than R?

 

[Smart $]

Z lower than R?

Yep... from about 1AWG down. At 14AWG the values for Cu in GRSC are R=3.1 and Z(@.85PF)=2.7 ohms/kft... a difference of about 15%.

 

[mivey]

Z lower than R?

Effective Z. Effective Z = RcosΘ + XsinΘ

 

[Besoeker]

Effective Z. Effective Z = RcosΘ + XsinΘ

Z=(R²+X²)^1/2

 

[Haji]

Effective Z. Effective Z = RcosΘ + XsinΘ

He would have accepted it, had you expressed it in the correct complex notation Z = RcosΘ +J.(XsinΘ)

 

[mivey]

Z=(R²+X²)^1/2

Correct, but not the effective Z that Smart$ was talking about. Effective Z is a current multiplier that approximates the line-neutral voltage drop.

 

[mivey]

He would have accepted it, had you expressed it in the correct complex notation Z = RcosΘ +J.(XsinΘ)

Incorrect.

R = |Z|cosΘ
X = |Z|sinΘ

Z = R + jX = |Z|cosΘ +j|Z|sinΘ

 

[mivey]

At 85?F here are the copper R to Z percentages in three conduit types:

CU: PVC, Al, St
18-1: 100.0%, 100.0%, 100.0%
18-7: 100.0%, 100.0%, 100.0%
16-1: 100.0%, 100.0%, 100.0%
16-7: 100.0%, 100.0%, 100.0%
14-1: 100.0%, 100.0%, 100.0%
14-7: 100.0%, 100.0%, 100.0%
12-1: 99.9%, 99.9%, 99.9%
12-7: 99.9%, 99.9%, 99.9%
10-1: 99.9%, 99.9%, 99.8%
10-7: 99.9%, 99.9%, 99.8%
8-1: 99.7%, 99.7%, 99.5%
8-7: 99.7%, 99.7%, 99.5%
6-7: 99.3%, 99.3%, 98.8%
4-7: 98.4%, 98.4%, 97.5%
3-7: 97.7%, 97.7%, 96.4%
2-7: 96.4%, 96.7%, 94.8%
1-19: 94.1%, 94.8%, 92.3%
1/O-19: 91.9%, 92.9%, 88.1%
2/O-19: 89.3%, 89.3%, 84.5%
3/O-19: 84.2%, 85.7%, 79.2%
4/O-19: 79.0%, 81.2%, 72.5%
250-37: 73.4%, 76.4%, 66.3%
300-37: 67.5%, 71.4%, 60.1%
350-37: 63.0%, 67.6%, 55.4%
400-37: 57.5%, 63.0%, 52.0%
500-37: 50.8%, 57.3%, 45.8%
600-61: 44.9%, 52.2%, 40.6%
700-61: 40.9%, 48.8%, 36.7%
750-61: 39.2%, 47.4%, 35.0%
800-61: 37.8%, 45.8%, 34.2%
900-61: 35.0%, 42.7%, 32.8%
1000-61: 32.7%, 40.1%, 31.7%
1250-91: 27.7%, 34.4%, 28.6%
1500-91: 24.5%, 30.7%, 26.8%
1750-127: 22.1%, 27.8%, 25.3%
2000-127: 20.2%, 25.5%, 24.1%

 

[mivey]

At 85?F here are the copper R to effective Z percentages in three conduit types:

CU: PVC, Al, St,
18-1: 117.0%, 117.0%, 116.8%,
18-7: 117.0%, 117.0%, 116.8%,
16-1: 116.7%, 116.7%, 116.4%,
16-7: 116.7%, 116.7%, 116.4%,
14-1: 116.1%, 116.1%, 115.7%,
14-7: 116.1%, 116.1%, 115.7%,
12-1: 115.4%, 115.4%, 114.8%,
12-7: 115.4%, 115.4%, 114.8%,
10-1: 114.2%, 114.2%, 113.3%,
10-7: 114.2%, 114.2%, 113.3%,
8-1: 112.2%, 112.2%, 110.9%,
8-7: 112.2%, 112.2%, 110.9%,
6-7: 109.4%, 109.4%, 107.4%,
4-7: 105.7%, 105.7%, 103.1%,
3-7: 103.5%, 103.5%, 100.4%,
2-7: 100.4%, 101.1%, 97.5%,
1-19: 96.2%, 97.3%, 93.5%,
1/O-19: 92.9%, 94.4%, 88.3%,
2/O-19: 89.6%, 89.6%, 84.5%,
3/O-19: 84.3%, 85.7%, 79.6%,
4/O-19: 79.5%, 81.4%, 74.1%,
250-37: 74.8%, 77.3%, 69.2%,
300-37: 70.1%, 73.2%, 64.5%,
350-37: 66.7%, 70.2%, 60.9%,
400-37: 62.6%, 66.7%, 58.3%,
500-37: 57.4%, 62.4%, 53.4%,
600-61: 52.7%, 58.5%, 49.1%,
700-61: 49.4%, 55.8%, 45.8%,
750-61: 48.0%, 54.7%, 44.2%,
800-61: 46.7%, 53.4%, 43.6%,
900-61: 44.2%, 50.9%, 42.3%,
1000-61: 42.1%, 48.7%, 41.2%,
1250-91: 37.4%, 43.7%, 38.2%,
1500-91: 34.1%, 40.2%, 36.4%,
1750-127: 31.5%, 37.4%, 34.9%,
2000-127: 29.4%, 35.1%, 33.7%,