When electric power is being supplied to large industrial consumers by a power company, the company will frequently include a PF clause in its rate schedules. Under this clause, an additional charge is made to the consumer whenever the PF drops below a certain specified value, usually about 0/85 lagging. Very little industrial power is consumed at leading PFs, because of the nature of typical industrial loads. There are several reasons that force the power company to make this additional charge for low PFs. In the first place, it is evident that larger current-carrying capacity must be built into its generators in order to provide the larger currents that go with lower-PF operation at constant power and constant voltage. Another reason is found in the increased losses in its transmission and distribution system. In an effort to recoup losses and encourage its consumers to operate at high PF, a certain utility charges a penalty of Rs 0.22 kVAr for each kVAr above a benchmark value computed as 0.62 times the average power demand:

S = P + jQ = P + j0.62P = P (1+j0.62) = P (1.177∠31.8°)

This benchmark targets a PF of 0.85 lagging, as cos 31.8° = 0.85 and Q is positive. Customers with a PF angle larger than this are subjected to financial penalties.

The reactive power requirement is commonly adjusted through the installation of compensation capacitors placed in parallel with load (typically at the substation outside the customer’s facility). The value of the required capacitance can be shown to be

where ω is the frequency, θ_old is the present PF angle, and θ_new is the target PF angle. For convenience, however, compensation capacitor banks are manufactured in specific increments rated in units of kVAr capacity.

Now let us consider a specific example. A particular industrial machine plant has a monthly peak demand of 5000 kW and a monthly reactive requirement of 6000 kVAr. Using the rate schedule above, what is the annual cost to this utility customer associated with PF penalties? If compensation is available through the utility company at a cost of Rs 2390 per 1000 kVAr increment and Rs 3130 per 2000 kVAr increment, what is the most cost-effective solution for the customer?

The PF of the installation is the angle of the complex power S, which in this case is 5000 + j 6000 kVA. Thus, the angle is tan^–1 (5000/6000) = 50.19° and the PF is 0.64 lagging. The benchmark reactive power value, computed as 0.62 times the peak demand, is 0.62 (5000) = 3100 kVAr. So, the plant is drawing 6000 – 3100 = 2900 kVAr more reactive power than the utility company is willing to allow without penalty. This represents an annual assessment of 12 (2900) (0.22) = Rs 7656 in addition to regular electricity costs.

If the customer chooses to have a single 1000 kVAr increment installed (at a cost of Rs 2390), the excess reactive power draw is reduced to 2900 – 1000 = 1900 kVAr, so that the annual penalty is now 12 (1900) (0.22) = Rs 5016. The total cost this year is then Rs 5016 + Rs 2390 = Rs 7406, for a savings of Rs 250. If the customer chooses to have a single 2000 kVAr increment installed (at a cost of Rs 3130), the excess reactive power draw is reduced to 2900 – 2000 = 900 kVAr, so that the annual penalty is now 12 (900) (0.22) = Rs 2376. The total cost if this year is then Rs 2376 + Rs 3130 = Rs 5506, for a first-year savings of Rs 2150. If, however, the consumer goes overhead and installs 3000 kVAr of compensation capacitors so that no penalty is assessed, it will actually cost Rs 14 more in the first year than if only 2000 kVAr were installed.