40HP 3Phase 208 Volt Motor AMP Draw

I am hooking up a new 40HP motor 3Phase 208V motor. I am concerned
about the amps required for initial startup. I believe the running
load will be around 100AMPs but how can I determine what the startup
draw will be. I don't want to spend $$$ on a soft start but also don't
have 200AMPs to dedicate to this one motor.
Thanks for your input.
Reply to
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The motor nameplate likely has Locked Rotor Amps (LRA) listed. That is the peak starting current.
Why would you "dedicate" 200 amps to the motor? The starting current lasts for a few to several seconds depending on the load being driven. It is not a continuous draw.
Charles Perry P.E.
Reply to
Charles Perry
Hello, You are correct your motor will run around 100A ( 106 actually) and full load amps will be 132 FLA .
here is a website with a calculator for you to use its pretty straight forward
formatting link

Reply to
trike Mike
The 2005 NEC Table 430.250 lists the full load current for a 208 volt 3 phase squirrel cage induction motor as 114 amperes. The name plate current is probably slightly less than this. As a rule of thumb the starting current for the first half second or so will be about 6 times this. This current is of such a short duration that an inverse time circuit breaker according to the NEC Table 430.52 should be 250 per cent of 114 amperes or 285 amperes or rounded up to 300 amperes. This circuit breaker provides short circuit and ground fault protection but does not provide overload protection. Overload protection is provided by a separate overload protective device sized at about 125 per cent of the Name Plate current rating. Your 200 ampere service is too small. Note that Design B motors have a much higher starting current.
Reply to
I should be more accurate. If a 200 ampere inverse time circuit breaker is used for the service disconnect then it probably needs to be increased to 300 amperes. The service conductors sized for a 200 ampere load may still be sufficient.
Ref: Areticle 230 Part VII. Service Equipment - Overcurrent Protection 230.90 Where Required. Each ungrounded service conductor shall have overload protection. (A) Ungrounded Conductor. Such protection shall be provided by an overcurrent device in series with each ungrounded service conductor that has a rating or setting not higher than the allowable ampacity of the conductor. A set of fuses shall be considered all the fuses required to protect all the ungrounded conductors of a circuit. Single-pole circuit breakers, grouped in accordance with 230.71(B), shall be considered as one protective device. Exception No. 1: For motor-starting currents, ratings that conform with 430.52, 430.62, and 430.63 shall be permitted.
Article 430 Part V. Motor Feeder Short-Circuit and Ground-Fault Protection 430.61 General. Part V specifies protective devices intended to protect feeder conductors supplying motors against overcurrents due to short circuits or grounds. FPN: See Annex D, Example D8. 430.62 Rating or Setting - Motor Load. (A) Specific Load. A feeder supplying a specific fixed motor load(s) and consisting of conductor sizes based on 430.24 shall be provided with a protective device having a rating or setting not greater than the largest rating or setting of the branch-circuit short-circuit and ground-fault protective device for any motor supplied by the feeder [based on the maximum permitted value for the specific type of a protective device in accordance with 430.52, or 440.22(A) for hermetic refrigerant motor-compressors], plus the sum of the full-load currents of the other motors of the group. Where the same rating or setting of the branch-circuit short-circuit and ground-fault protective device is used on two or more of the branch circuits supplied by the feeder, one of the protective devices shall be considered the largest for the above calculations. Exception No. 1: Where one or more instantaneous trip circuit breakers or motor short-circuit protectors are used for motor branch-circuit short-circuit and ground-fault protection as permitted in 430.52(C), the procedure provided above for determining the maximum rating of the feeder protective device shall apply with the following provision: For the purpose of the calculation, each instantaneous trip circuit breaker or motor short-circuit protector shall be assumed to have a rating not exceeding the maximum percentage of motor full-load current permitted by Table 430.52 for the type of feeder protective device employed. Exception No. 2: Where the feeder overcurrent protective device also provides overcurrent protection for a motor control center, the provisions of 430.94 shall apply. FPN: See Annex D, Example D8.
Reply to
You failed to notice it is 208V. 50A would be at 460V.
For the OP, 40HP @ 208V will be around 114A. Starting current if started Across-the-Line will be around 680A, albeit for a very short time. The MAXIMUM inverse time/current (Thermal Magnetic) breaker size that you can put ahead of that motor is going to be 250% of FLA then the next size up if necessary, per NEC table 430.52. That is the table that matters in the case of a motor branch circuit.
In your case 114 x 2.5 = 285A, so use a 300A breaker. A 200A will likely trip on you unless you use a soft starter. So if you use only a 300A breaker, the wire must be rated for 300A.
However, if you use a motor starter with a Thermal Overload Relay, the breaker then becomes only the short circuit brotective device and the OL relay will protect the wires, so you can revert back to the rules for conductor sizing for motor circuits, 125% of FLA then up to the next available size, adjusted as necessary for voltage drop over distance.
All that said, if you do not have at least 230A available (200% FLA), that motor will never fully accelerate even with a soft starter. You either need more available power or you will need to use a VFD to accelerate the motor.
Reply to
Bob Ferapples
Check the nameplate for a locked rotor current or code letter (see NEC Table 430.7(B) to interpret code letters).
I don't understand what you mean by 'don't have 200A to dedicate'. Service and feeder size doesn't have to be sized to the locked rotor current. However, if you have a limited power source (like a generator), you are going to have to take this into account somehow. A soft start motor is probably cheaper than increasing the source's transient capacity.
Reply to
Paul Hovnanian P.E.

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