bass1mj
Electrical
- Aug 15, 2002
- 11
I am doing the design of a new 6.6kV Substation consisting of two Incomers, a Bussection,two motor feeders and two spare panels in an industrial plant. The supply for the substation come from a main substation which is fed from the supply authority.
The above is just background info and my question is on the PFCorrection on the motors.
The following options exist
1. Use Static power factor correction i.e. connect capacitors in parallel with the motor in which case you have to consider taking it from either after or before the protection CT's. Before and you need to install protection for the caps or after and you need to compensate for it in the motor protection.
All other factors like resonance will be taken into consideration.
2. Increase the size of the existing capacitance at the feeder sub to cater for the additional inductive load on the new substation downstream. Upgrade control and protection on old system.
I would like to hear what option other people would prefer and why?
More info: New Sub rating = 6MVA
Motor1 data : 620kW :Ifl = 100A: pf=0.57 @FL: Inl = 70A
Motor2 data : 500kW :ifl = 74A : pf=0.63@FL: Inl = 49A
Thank you
Thank you
The above is just background info and my question is on the PFCorrection on the motors.
The following options exist
1. Use Static power factor correction i.e. connect capacitors in parallel with the motor in which case you have to consider taking it from either after or before the protection CT's. Before and you need to install protection for the caps or after and you need to compensate for it in the motor protection.
All other factors like resonance will be taken into consideration.
2. Increase the size of the existing capacitance at the feeder sub to cater for the additional inductive load on the new substation downstream. Upgrade control and protection on old system.
I would like to hear what option other people would prefer and why?
More info: New Sub rating = 6MVA
Motor1 data : 620kW :Ifl = 100A: pf=0.57 @FL: Inl = 70A
Motor2 data : 500kW :ifl = 74A : pf=0.63@FL: Inl = 49A
Thank you
Thank you