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# Dealing with a 10hp motor performance

Dealing with the change in power factor. A 10HP (7.5kW), 400v, IE2, B3, Induction motor at full load will have a power factor of 0.86. At 75% power the power factor is 0.81 and at 50% it is 0.71. Assuming full load then your 10HP is 7.5kW at power factor 0.86 has a reactive power of 4.45kVAr. If the load drops to 75% and the capacitors are still in circuit, the machine only requires 4.07kVAr so it is now 0.38kVAr leading or +0.99. at 50%, 3.82kVAr leading or +0.98.

Dealing with the current. The same motor will draw 14.1A at full load. If corrected with the exact capacitors at the motor for unity power factor as suggested, the motor will still draw 14.1 but the circuit will draw 10.83A. So where do you set the thermal current protection for the motor that is sitting in the starter some distance away? If you set it at 14.1A and the caps are fully functional, the motor can overload by 30% (10.83 -> 14.1) before the overload see the overload. If you set it at 10.83, as the capacitor de-generates, the current will increase and the overload will trip. You will need to protect your circuit so that if a cap does fail, it is removed from the circuit without disruption to the operation. So you need to install fuses. Near the motor this along with the capacitors will require another enclosure. How will you know when your fuse blows? The motor trips on over current? If thermistors, Thermocouples or PT 100s are added into the windings to protect the motor (best as far as I’m concerned) then you require an additional cable and an special control relay associated with that temperature device.

Dealing with the inrush current. You purchase a motor (inductive) contactor with an AC3 rating. This enables you to start the 10hp motor DOL without damage. If you buy a capacitor contactor, it will have an early make set of contacts with inductance added to the circuit to limit the inrush current to that caps to prevent tripping as capacitors are seen as a short circuit to AC when you first close. If you use this contactor for the motor as well, the full motor current will flow through the early make auxiliary contacts which are not designed for this. With both motor and cap inrush current what size contactor do you select?

Dealing with the selection of capacitors, you can get very good capacitors but this question was very specific about a 10HP motor. I can’t see these capacitors being available in the rating required for a single motor. Use of a power factor controller on a single small motor is not cost effective. If you can get them then ROI?

Dealing with the location of pfc. The location of this unit as described as a single load, can only be in the motor starter panel (as far as I’m concerned), added to the circuit by a contactor controlled by a timer from the auxiliary from the motor contact. Value? Dependent on the minimum load of the motor and to give you no more than 0.95. The timer will have to be set so that the caps have time to discharge between starts should the motor be inched or stop and restarted many times. Normally about 10 minutes if the discharge resistors are functional. This should still assist with Maximum demand billing but protect the caps.

Dealing with the harmonics. This particular question was on one motor. If there are devices such as VFDs or LED lighting or electronic ballasts in the installation creating harmonics, these can destroy capacitors. Only a harmonic study would tell you this. The Q factor ringing back and forth between the motor and caps will not happen if you keep the power factor to 0.95. On large installations, to go from 0.95 to 1.00 will double your costs so unless you have a pressing problem, 0.95 is very good. The angle is directly proportional to the kVAr and 0.8pf = 36.87 degrees and 0.95 = 18.87 degrees. This gain of 0.15pf will cost you the same as the next step from 0.95 to 1.00 which is only 0.05.