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For a 75 Hp ,380 line voltage PF 0.85 motor when i calculated the rated current P = 1.73*VIPF i get Rated Current = 98 A While what is printed on the nameplate is 124 ampere enter image description here

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  • \$\begingroup\$ i just added the picture \$\endgroup\$ – Mahmoud Maged May 12 '17 at 16:19
  • \$\begingroup\$ May have something to do with VFD motor with normal operation at 50Hz and range going from 6Hz to 60Hz. \$\endgroup\$ – StainlessSteelRat May 12 '17 at 16:28
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    \$\begingroup\$ 75 HP is output. You need to consider efficiency. Peak output power, peak PF, and peak current may not come together in the same operating condition. E.g. peak PF comes under heavy load, so does peak current, but peak output power is at a certain optimal operating point. \$\endgroup\$ – user3528438 May 12 '17 at 16:38
  • \$\begingroup\$ I get 105.7A at nameplate values. \$\endgroup\$ – StainlessSteelRat May 12 '17 at 17:28
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$$P_m = 75hp \times 746 W/hp = 55,950W$$ $$P_e = \frac {P_m} {\eta} = \frac {55,950W} {0.936} = 59,775W$$ $$I_{Line} = \frac {P_e} {\sqrt {3} V_{Line} cos \theta} = \frac {59,775W} {\sqrt {3} \times 380V \times 0.859} = 105.7A$$

106A is not 124A. Typically all the performance data (voltages, currents, efficiency, pf, torque) apply to full load conditions. The current does not.

Stated frequency is at 50Hz, but the motor is a VFD with 6Hz to 60Hz operation.

At 60Hz, motor will spin 20% faster (n = 1800rpm). Torque should drop off if motor was designed for 50Hz, but it could be constant to 60Hz. If we assume torque (T) is constant, increase should provide 20% more hp. $$P_m = \frac {nT}{9.55}$$ Do the current calc with 20% more hp, you get 126.9A, which is much closer to 124A. Difference is probably due to changes in pf or efficiency.

To be honest, without input from manufacturer, this extrapolation is just a guess.

You must also factor in the 209A and 104A currents, off to the right of the 6Hz-60Hz. Do these currents apply to 190V/380V and at what frequency.

You can probably learn more by looking at how the VFD is set for this motor, but very hard to guess from nameplate.

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The calculation including the nominal efficiency marked on the nameplate gives 105.7 amps. The current marked on the nameplate may include an additional allowance for the harmonic content due to inverter operation. However, the inverter duty data seems to be marked below the words "inverter Duty." That data shows lower currents. However, the torque and speed marked indicate that the motor can only produce 60 Hp at the maximum operating speed. Without an explanation from the manufacturer, we can only guess why the nameplate is marked the way it is.

The NEMA standard that was presumably the guide for the information marked on the nameplate requires the rated output mechanical power to be marked on the nameplate. IEC standards have a similar requirement. Therefore:

P = SQRT(3) X Current X Voltage X Efficiency X Power Factor

1 Horsepower = 746 Watts

Current = (75 X 746) / (1.732 X 380 X 0.936 X 0.859 = 105.7 amps

I would be inclined to use that value for motor the protection and VFD tuning parameters. The current limit setting that limits the sort-time overload limit can be set higher, perhaps as high as 150% if necessary.

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  • \$\begingroup\$ Sorry but how did you get 105.7 A from the above formula what i used is P = SQRT(3)*380*PF and I'm using drive altivar 61 want to adjust rated current and thermal overcurrent which valves should i pick \$\endgroup\$ – Mahmoud Maged May 13 '17 at 0:19

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