SUBJECT Energy Systems and Power Electronics Please answer q

SUBJECT: Energy Systems and Power Electronics

Please answer question and show all work.

4. HW 1 O4 Problem 3.12 An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power factor of 0.75 lagging from the substation bus. The supply voltage is 12.47 kV. The power factor can be improved by connected a capacitor in parallel with the supply or by using a synchronous motor, which generates reactive power. Analyze both of these cases independently: (a) Find the required kVAR rating of a capacitor connected across the load to raise the power factor to 0.95 lagging. (b) Assuming a synchronous motor rated at 250 hp, with an 80 efficiency is operated from the same bus at rated conditions and a power factor of 0.85 leading, calculate the resultant supply power factor.

Solution

n 41.41=529.15 KVAR

a) For pf=cos phi=0.95

phi=18.19 deg

So net Qt=PL*tan 18.19=197.21 KVAR

So KVAR rating required by capacitor is QL-Qt=529.15-197.21=331.94 KVAR

So required KVAR is 331.94 KVAR

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b) Given Synchronous motor output Po=250 hp=186.425 kW

Input Pi=Po/efficeincy=186.425/0.8=233.03 kW

Total Real power Pt=PL+Pi=600+233.03=833.03 kW

cos phi=0.85 so phi=31.79 deg

Syncronous motor reactive power Qi=Pi*tan phi=233.03*tan 31.79

Qi=-144.43 KVAR (negative because leading pf)

so net reactive power Qt=QL-Qi=529.15-144.43=384.72 KVAR

so Resultant power factor angle phi=tan^-1(Qt/Pt)=tan^-1(384.72/833.03)=24.789 deg

so Resultant power factor Pf=cos phi=0.9078 lag