Motor voltage drop
- Thread starter olc
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JonD
New member
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- Saskatoon, Canada
A motor load will yield a lagging power factor, and the more lagging a power factor, the higher your voltage drop will be.
An example of this is Table 3-12 from IEEE 141-1993.
The formula for voltage drop is % Vd = ? 3 x I x (R COS ? + X SIN ?) x L x 100/ N X V
Where:
Your power factor will vary by load. For a 75 HP TECO Westinghouse motor:
50% load => 75.5 % P.F.
75% load => 83.5 % P.F.
100% load => 86.5 % P.F.
An example of this is Table 3-12 from IEEE 141-1993.
The formula for voltage drop is % Vd = ? 3 x I x (R COS ? + X SIN ?) x L x 100/ N X V
Where:
| Vd = Voltage drop |
| I = Current in amperes |
| R = Conductive resistance in ohms/ 1000 ft. |
| V = System Voltage in volts |
| Cos ? = Power Factor |
| X = Conductor inductive reactance in ohms/1000 ft. |
| L = One way length of circuit ( source to load) in thousands of feet. |
| N = No of cable runs per circuit. |
| ? = Phase angle of load |
Your power factor will vary by load. For a 75 HP TECO Westinghouse motor:
50% load => 75.5 % P.F.
75% load => 83.5 % P.F.
100% load => 86.5 % P.F.
- Location
- San Francisco Bay Area, CA, USA
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- Electrical Engineer
But as your PF gets worse, it is BECAUSE your actual load is dropping, which means LESS current to begin with, hence LESS voltage drop.
The NEC wire sizing using the HP and FLC charts, not the nameplate data, will require you to BEGIN with 125% of the value in the chart, and THEN begin to adjust for voltage drop over distance. So that already factors the effects of power factor on the current draw as one of the reasons for the 125% starting point.
The NEC wire sizing using the HP and FLC charts, not the nameplate data, will require you to BEGIN with 125% of the value in the chart, and THEN begin to adjust for voltage drop over distance. So that already factors the effects of power factor on the current draw as one of the reasons for the 125% starting point.
Julius Right
Senior Member
- Occupation
- Electrical Engineer Power Station Physical Design Retired
NEC does not deal with voltage drop [except the informational note no.2 in art.215.2 Minimum Rating and Size. (A) Feeders Not More Than 600 Volts:
?Informational Note No. 2: Conductors for feeders as defined in Article 100, sized to prevent a voltage drop exceeding 3 percent at the farthest outlet of power, heating, and
lighting loads, or combinations of such loads, and where the maximum total voltage drop on both feeders and branch circuits to the farthest outlet does not exceed 5 percent, will
provide reasonable efficiency of operation.?
No reference to current as in art.430 for cable ampacity.
The voltage drop formula presented by is good enough. For an more accurate formula
see IEEE Std 141 fig.3-11.
Vd=V+I*R*cos(fi)+I*X*sin(fi)-sqrt(V^2-(I*R*cos(fi)-I*R*sin(fi))^2) [Volt]
By the way, in connection with current lagging angle (fi) pf=cos(fi) then fi=acos(pf)
for pf=0.85 fi=31o47' 18"[31degrees,47min.18sec].
?Informational Note No. 2: Conductors for feeders as defined in Article 100, sized to prevent a voltage drop exceeding 3 percent at the farthest outlet of power, heating, and
lighting loads, or combinations of such loads, and where the maximum total voltage drop on both feeders and branch circuits to the farthest outlet does not exceed 5 percent, will
provide reasonable efficiency of operation.?
No reference to current as in art.430 for cable ampacity.
The voltage drop formula presented by is good enough. For an more accurate formula
see IEEE Std 141 fig.3-11.
Vd=V+I*R*cos(fi)+I*X*sin(fi)-sqrt(V^2-(I*R*cos(fi)-I*R*sin(fi))^2) [Volt]
By the way, in connection with current lagging angle (fi) pf=cos(fi) then fi=acos(pf)
for pf=0.85 fi=31o47' 18"[31degrees,47min.18sec].
Julius Right
Senior Member
- Occupation
- Electrical Engineer Power Station Physical Design Retired
Correction:
Vd=es+I*R*cos(fi)+I*X*sin(fi)-sqrt(es^2-(I*X*cos(fi)-I*R*sin(fi))^2) [Volt]:ashamed1:
?The voltage drop V obtained from this formula is the voltage drop in one conductor, one way,
commonly called the line-to-neutral voltage drop. The reason for using the line-to-neutral
voltage is to permit the line-to-line voltage to be computed by multiplying by the following
constants: for Three-phase= sqrt(3)~ 1.732.?
Vd=es+I*R*cos(fi)+I*X*sin(fi)-sqrt(es^2-(I*X*cos(fi)-I*R*sin(fi))^2) [Volt]:ashamed1:
?The voltage drop V obtained from this formula is the voltage drop in one conductor, one way,
commonly called the line-to-neutral voltage drop. The reason for using the line-to-neutral
voltage is to permit the line-to-line voltage to be computed by multiplying by the following
constants: for Three-phase= sqrt(3)~ 1.732.?
jumper
Senior Member
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- 3 Hr 2 Min from Winged Horses
:happyno:
You might want to look at articles 647 Sensitive electronic equipment and 695 Fire pumps.
Julius Right
Senior Member
- Occupation
- Electrical Engineer Power Station Physical Design Retired
Thank you, jumper.However,I meant this only for usually motor operation.
kwired
Electron manager
- Location
- NE Nebraska
- Occupation
- EC
Always been my understanding as well, but does not agree with this information:
kwired
Electron manager
- Location
- NE Nebraska
- Occupation
- EC
If all conductors of the circuit are in same raceway or cable reactance should be fairly minimal, resistance of conductors is the main factor contributing to voltage drop. Intentionally introducing reactance like with a line reactor will be more significant. Not much for naturally occuring reactors in a typical raceway with all conductors of the circuit pulled through it though.
Originally Posted by JonD View Post
Your power factor will vary by load. For a 75 HP TECO Westinghouse motor:
50% load => 75.5 % P.F.
75% load => 83.5 % P.F.
100% load => 86.5 % P.F.
[/quote]
I think it does - 75HP is 56kW
50% is 28kW. At 0.755pf that equates to 37kVA
75% at 0.836pf is 50kVA
100% at 0.865pf is 65kVA
So, in this case at least, as the load reduces, the supply current reduces even though the power factor gets worse.
kwired
Electron manager
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- NE Nebraska
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- EC
I think it does - 75HP is 56kW
50% is 28kW. At 0.755pf that equates to 37kVA
75% at 0.836pf is 50kVA
100% at 0.865pf is 65kVA
So, in this case at least, as the load reduces, the supply current reduces even though the power factor gets worse.[/QUOTE]
Ok you are on to something here, but I think we need to factor in motor efficiency as well. If a motor is 93% efficient - that is efficiency at full load, rated voltage and frequency being applied. What is efficiency for same motor if load is changed to 50%?
Fair point and ought to be taken into account.
- Location
- San Francisco Bay Area, CA, USA
- Occupation
- Electrical Engineer
That is a better illustration of the point I was trying to make earlier.
Still, not likely significant as it relates to voltage drop and cable sizing. The worst case motor efficiency I have ever seen, a 12 pole motor built in 1913 (bullet proof, but nobody worried about efficiency then), was 65% eff. at half load. So even if we apply that somewhat extreme value to Besoeker's numbers above, 37kVA /.65 = 57kVA. Cable size (if done properly) is not going to be affected.
Agreed.
Some figures I calculated for one of our drive projects - as it happens, figures that we were contractually obliged to provide and meet.
Calculated Performance LRT "2/4" "3/4" FLC
Efficiency (%) 0.0% 96.7% 97.2% 97.3%
Torque (kNM) 0.41 2.00 3.00 4.00
PU Torque (pu) 0.102 0.499 0.749 1.000
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