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Locked Rotor Current Calculation when i dont have NEMA code 1
- Thread starter valsgum
- Start date
Hello Valsgum,
The locked rotor current of a motor is depenadant purely upon rotor design. The fact that you are starting a 20hp load with a 25hp motor will not reduce starting current at all..assuming of-course you are referring to DOL or Full VOltage Starting.
As indicated by DougMSOE the typical range is between 5 & 7 times rated current. Having said that it is not uncommon to find motors that draw less than that (4.5 x FLC) and more than that (up to 9 x FLC) in general industry.
Regards,
GGOSS
The locked rotor current of a motor is depenadant purely upon rotor design. The fact that you are starting a 20hp load with a 25hp motor will not reduce starting current at all..assuming of-course you are referring to DOL or Full VOltage Starting.
As indicated by DougMSOE the typical range is between 5 & 7 times rated current. Having said that it is not uncommon to find motors that draw less than that (4.5 x FLC) and more than that (up to 9 x FLC) in general industry.
Regards,
GGOSS
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1
- #4
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: Reference:
Donald G. Fink, H. Wayne Beaty "Standard Handbook for Electrical Engineers," 14th Edition, McGraw-Hill, 2000,
page 20-40 Equation (20-30)
Locked rotor current=kVA/hp x hp x 1000/(k x line volts)
where
k = 1 for single-phase motors
k = sqrt3 for 3-phase motors
kVA/hp corresponds to NEMA motor Code Letter
It means that information posted in the original posting is inadequate to determine the motor locked rotor current.
Donald G. Fink, H. Wayne Beaty "Standard Handbook for Electrical Engineers," 14th Edition, McGraw-Hill, 2000,
page 20-40 Equation (20-30)
Locked rotor current=kVA/hp x hp x 1000/(k x line volts)
where
k = 1 for single-phase motors
k = sqrt3 for 3-phase motors
kVA/hp corresponds to NEMA motor Code Letter
It means that information posted in the original posting is inadequate to determine the motor locked rotor current.
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: Reference:
NFPA 70-2002 National Electrical Code
Table 430.151(B) Conversion Table of Polyphase Design B, C, D, and E Maximum Locked-Rotor Currents for Selection
of Disconnecting Means and Controllers as Determined from Horsepower and Voltage Rating and Design Letter
Since the 370V and most likely 50Hz motor is closely related, by approximation (prorating) the motor locked-rotor current could be a value between 183A and 365A, say about 274A.
NFPA 70-2002 National Electrical Code
Table 430.151(B) Conversion Table of Polyphase Design B, C, D, and E Maximum Locked-Rotor Currents for Selection
of Disconnecting Means and Controllers as Determined from Horsepower and Voltage Rating and Design Letter
Since the 370V and most likely 50Hz motor is closely related, by approximation (prorating) the motor locked-rotor current could be a value between 183A and 365A, say about 274A.
Dears,
Please note that in the first small period of one tenth of the motor voltage period, a current of 20 times the FLC is drawn which many people are not considering and may be don't know. I have the experience that this very high impulse of current on the begining could cause tripping the main feeder of the motor if not provided with a suitable time delay to avoid this impulse.After this very small period, the inrush current is 4-7 times the FLC as all you know and mentioned.
Regards
Please note that in the first small period of one tenth of the motor voltage period, a current of 20 times the FLC is drawn which many people are not considering and may be don't know. I have the experience that this very high impulse of current on the begining could cause tripping the main feeder of the motor if not provided with a suitable time delay to avoid this impulse.After this very small period, the inrush current is 4-7 times the FLC as all you know and mentioned.
Regards
tharwat,
In jB's defence, the distributed capacitances are present in any cable or busbar system, and between winding turns within the motor. They don't have to be present as an actual capacitor. Closing a contactor onto a de-energised cable at the voltage peak will indeed cause a current transient, although you would need a fast current probe to catch it (e.g. a Tektronix AM503B). How long were the cables feeding the motors in question, and what was the duration of the transient you observed?
-----------------------------------
Start each new day with a smile.
Get it over with.
In jB's defence, the distributed capacitances are present in any cable or busbar system, and between winding turns within the motor. They don't have to be present as an actual capacitor. Closing a contactor onto a de-energised cable at the voltage peak will indeed cause a current transient, although you would need a fast current probe to catch it (e.g. a Tektronix AM503B). How long were the cables feeding the motors in question, and what was the duration of the transient you observed?
-----------------------------------
Start each new day with a smile.
Get it over with.
tharwat:
The internal parallel capacitance in an induction motor at the line frequency results in a high impedance that will draw a few amperes or milli-amperes from the line.
If you have measured such a peak current lasting T/10 ( around 1.7 milliseconds?) please give me more details like; frequency, peak, phase angle and rms values.
That time seems well into the subtransient reactance of a machine, does that peak will be because the motor performance or due to the power supply?
The internal parallel capacitance in an induction motor at the line frequency results in a high impedance that will draw a few amperes or milli-amperes from the line.
If you have measured such a peak current lasting T/10 ( around 1.7 milliseconds?) please give me more details like; frequency, peak, phase angle and rms values.
That time seems well into the subtransient reactance of a machine, does that peak will be because the motor performance or due to the power supply?
jbartos
Electrical
- Jan 15, 2000
- 6,440
Comment on tharwat (Electrical) Mar 6, 2004 marked ///\\Please note that in the first small period of one tenth of the motor voltage period, a current of 20 times the FLC is drawn which many people are not considering and may be don't know. I have the experience that this very high impulse of current on the begining could cause tripping the main feeder of the motor if not provided with a suitable time delay to avoid this impulse.After this very small period,
///The capacitance cause of the peak is:
i(t)=Cxdv(t)/dt
relationship. Now, if v(t)=Vmax sinwt
then dv/dt=w x Vmax x coswt
where
w=2pif=2pi/T
When sinwt crosses zero it will be equal to zero and the dv/dt will be equal to one. If the applied voltage is equal to Vmax at time T/4, implying dv/dt large, then dv/dt will be large number, implying i(T/4) will be a high current spike.\\ the inrush current is 4-7 times the FLC as all you know and mentioned.
///The capacitance cause of the peak is:
i(t)=Cxdv(t)/dt
relationship. Now, if v(t)=Vmax sinwt
then dv/dt=w x Vmax x coswt
where
w=2pif=2pi/T
When sinwt crosses zero it will be equal to zero and the dv/dt will be equal to one. If the applied voltage is equal to Vmax at time T/4, implying dv/dt large, then dv/dt will be large number, implying i(T/4) will be a high current spike.\\ the inrush current is 4-7 times the FLC as all you know and mentioned.
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