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.
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.
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
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.
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.
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.
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.
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