Not sure what you mean. Typically if a SYSTEM voltage is 600V, the MOTOR design voltage is 575V.
But cable size would be based on the motor size and depending on the code standards that you use, may be tied to the nameplate power rating of the motor.
" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
A 575 V motor could be run with a cable rated for 600V.
[b said:
NEC:[/b] a) The size of the conductors should be at least 125% of the motor FLC [430.22(A)]b) The size the overloads no more than 115% to 125% of the motor nameplate current rating, depending on the conditions [430.32(A)(1)].
c) The short-circuit ground-fault protection device from 150% to 300% of the motor FLC [Table 430.52].]
I'm sorry, yes, I'm assuming the system voltage as 600V, but in order to calculate the current that will draw this motor to determine voltage regulation (applying some fancy formulas), which current should I consider as input to that formula? HP/[Rated kV x 0.746 x sqrt(3)] or HP/[Nominal kV x 0.746 x sqrt(3)], being Rated kV = 0.575 kV and Nominal kV = 0.6 kV, which corresponds to the up-stream busbar that is supplying this motor.
or no?
Cable size is a code matter.
Depending on the code, the cable sizing may be based on the motor nameplate rated current.
Some codes may determine the ampacity of the motor feeders by HP, as specified by a Current per HP table in the code.
Sizing by HP and the table in the code considers that in the future the motor may be replaced with a motor of the same HP but higher current.
You don't calculate the motor current for sizing supply conductors.
You use either the nameplate current value or the code table, depending on the local code.
Also your formula will not be acceptable in the real world. You must also consider the efficiency of the motor and the power factor.
In Canada, refer to the Canadian Electrical Code: 28-106 Conductors — Individual motors
(1) The conductors of a branch circuit supplying a motor for use on continuous duty service shall have an
ampacity not less than 125% of the full load current rating of the motor.
28-704 Horsepower rated equipment
.......
(2) Where the full load current rating is not marked, an equivalent full load current rating shall be determined
EXAMPLE:
Consider a 15 HP motor without a current rating.
By your formula, 15 HP x 746 / (1.73 x 575 V) = 11.24 Amps
11.24 Amps x 125% = 14 Amps. That would be a #14 AWG wire
By code. 15 HP per table 44 = 17 Amps.
17 Amps x 125% = 21.25 Amps. That would be a # 10 AWG wire.
from the horsepower rating by referring to Table 44 or 45 as applicable.
Bill
--------------------
"Why not the best?"
Jimmy Carter
That 25% is the Service Factor and should be used, according to Nema MG1:
To accommodate inaccuracy in predicting intermittent system horsepower needs.
To lengthen insulation life by lowering the winding temperature at rated load.
To handle intermittent or occasional overloads.
To allow occasionally for ambient above 40°C.
To compensate for low or unbalanced supply voltages.
However, to make sure that a 600V motor will have 3% voltage regulation (for example if the design criteria requests that 3% as maximum voltage drop) at 100, 500 or 1000 meters from the 600V CCM or Switchgear, etc, we need to calculate the current that this motor will draw running at 100%. Obviously considering motor power factor, efficiency, correction factor by altitude, etc. What I'm having trouble to find out is the current input to this formula.
I understand that FLA is defined as the Full Load Ampere at Rated Voltage which the manufacturer of the motor measured at 575V. So, do I need to calculate this current considering 575V or 600V in order to calculate voltage regulation at the cable that will feed this motor (sized according to NEC tables, of course).
I suppose that if the motor is at full capacity ideally running under 600V (with an ideal feeder) then the current consumption will be lower than 575V.
Since you say you're using the NEC, the current you should be using for all calculations (conductor size, OCPD size, disconnect size, started size, etc.) except the overload trip setting and possibly the voltage drop calculation (though to be conservative you could use the table value for voltage drop also) should come from the current in Table 430.250 that corresponds with your motor voltage and horsepower ratings. The overload trip setting and possibly the voltage drop calculation use the nameplate current value. Do not try to correct the current values for the difference between nominal system voltage of 600 V and nominal utilization voltage of 575 V.
xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
I don't think he said he was using the NEC, which is why I said it is "depending on the code standards that you use". We don't really know because the rules may be different depending on the code involved.
But yes, if it was the NEC (and I believe the CEC is the same, because given the use of 600V, there's a 90% chance this involves Canadian standards), you don't "calculate" the motor amps at all, you read it from a table for that HP and nominal voltage and then use that value x 1.25.
" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
READ MY LIPS.
LOOK AT THE MOTOR NAMEPLATE.
The manufacturer has already calculated the FLA.
The manufacturer's value is the value that we use to determine conductor sizing.
Alternately you can use Table 430-250 in the NEC or table 44 in the CEC.
Nema MG1 is a standard for motor manufacture.
For cable sizing you are not manufacturing. You are installing.
Use your local installation code.
Voltage drop. Use the cable impedance and the NAMEPLATE current. Calculate the voltage drop and then express the voltage drop as a percentage of 600 Volts.
Stop worrying about the voltage. Use the nameplate current the same as everyone else. The nameplate current is based on the worst case situation when the voltage at the motor has dropped to 575 Volts. If the voltage is a little higher and the current is a little less, great, but that doesn't change the code provisions which are based on the worst case.
Bill
--------------------
"Why not the best?"
Jimmy Carter
So that 125% factor assures that I'll have the correct voltage regulation at Motor Terminals? I feel that considering only that factor then the distance between busbar and motor won't matter, which I believe is wrong.
Thanks waross, that was exactly what I told my boss but he kept insisting that I should use 600V in order to calculate the current that will flow through the cable. But if the manufacturer says that the rated voltage is 575 then that is the voltage that I'll have to use to calculate the current consumption (in a hypothetical case that the FLA is erased from the nameplate)
davidbeach, and at design stage what should I do? I only have information on service voltage (600V)... should I lookup for catalogues of motors at 600V (can't find yet) or 575V?
Horsepower. System voltage, not motor voltage, and horsepower is all you need. The Code takes it from there. That you probably can’t find a motor at that voltage and horsepower combination that draws that much current is immaterial.
You have to stop trying to calculate the current.
The Canadian you use the nameplate current.
In the event that the nameplate current is not available, look up the current in table 44.
If you are using the US NEC, look up the current in table 430-250.
Table 44 is based on motor voltage.
Table 430-250 is based on system voltage.
It doesn't matter. You are not doing any calculations at this point, your are looking at a nameplate or looking up a value in a table.
The voltage just gets you into the correct column of the table.
Step #1 Find the current from a nameplate or from a table.
Step #2 Multiply this current by 125% and select a cable with this ampacity or greater.
Step #3 Use the nameplate current or the table current to calculate the voltage drop in volts.
It is advisable to determine the impedance of the conductors from a table. The impedance will be greater than the DC resistance.
The CEC allows a total voltage drop from supply to motor of 5% but only allows a maximum of 3% drop in the motor branch circuit.
Express the voltage drop as a percentage of the supply voltage.
For most motors the conductor ampacity is based on 75 degree rated insulation.
If the voltage drop is excessive, increase the conductor size and check the voltage drop again.
Jeff; The Canadian code first uses the nameplate current. If that is not available then we go to the table.
The NEC allows for the possibility that the motor may be replaced with a same HP motor drawing more current then the original motor.
The CEC does not. I like the NEC method better.
Bill
--------------------
"Why not the best?"
Jimmy Carter
Dvhez. North American motor voltages are based on multiples of 115 Volts.
You aren't going to find a standard North American motor rated at 600 Volts.
Look for 575 Volt motors.
Where are you located. I have done some work out of the country but I have not yet seen 600 Volts outside of Canada.
Bill
--------------------
"Why not the best?"
Jimmy Carter
I'm located in Chile but the specific project I'm seeing considers a 600V voltage level. I understand that I have to check those tables but I feel that with this method, for a motor located at a distance of 1 meter the cable will be the same as this motor being located at 500 meters. In other words, I don't know if I can assure 3% voltage drop at motor terminals just considering this x1.25 factor applied to the motor's FLA.
No, you can't. The FLA x 1.25 factor is used to pick the MINIMUM cable size allowed. You MUST consider the voltage drop for a long run of cable, which might result in a further increase in the cable size.
