• Most Economical Power Factor when kW demand is constant

    If a consumer improves the power factor, there is reduction in his maximum kVA demand and hence there will be annual saving over the maximum demand charges. However, when power factor is improved, it involves capital investment on the power factor correction equipment. The consumer will incur expenditure every year in the shape of annual interest and depreciation on the investment made over the p.f. correction equipment. Therefore, the net annual saving will be equal to the annual saving in maximum demand charges minus annual expenditure incurred on p.f. correction equipment. The value to which the power factor should be improved so as to have maximum net annual saving is known as the most economical power factor.

    Consider a consumer taking a peak load of P kW at a power factor of cos φ1 and charged at a rate of Rs x per kVA of maximum demand per annum. Suppose the consumer improves the power factor to cos φ2 by installing p.f. correction equipment. Let expenditure incurred on the p.f. correction equipment be Rs y per kVAR per annum. The power triangle at the original p.f. cos φ1 is OAB and for the improved p.f. cos φ2, it is OAC [See Fig.]. 


    kVA max. demand at cos φ1, kVA1 = P/cos φ1 = Psec φ1

    kVA max. demand at cos φ2, kVA2 = P/cos φ2 = Psec φ2

    Annual saving in maximum demand charges

         = Rs x.(kVA1 − kVA2)

         = Rs x.(Psec φ1 – Psec φ2) = Rs x.P (sec φ1 – sec φ2)  ...(i)

    Reactive power at cos φ1, kVAR1 = P tan φ1

    Reactive power at cos φ2, kVAR2 = P tan φ2

    Leading kVAR taken by p.f. correction equipment = P (tan φ1 − tan φ2)

    Annual cost of p.f. correction equipment = Rs P.y (tan φ1 – tan φ2)                             ...(ii)

    Net annual saving, S = exp. (i) − exp. (ii)

            = x.P (sec φ1 – sec φ2)  − y.P (tan φ1 – tan φ2)

    In this expression, only φis variable while all other quantities are fixed. Therefore, the net annual saving will be maximum if differentiation of above expression w.r.t. φ2 is zero i.e.

    dS/dφ2 = 0

    0 – x.P sec φ2 tan φ2 - 0 + y.P sec2φ2 = 0

    -  x.tan φ2 + y. sec φ2 = 0

    sin φ2 = y/x

    Hence, Most economical power factor,

    It may be noted that the most economical power factor (cos φ2) depends upon the relative costs of supply and is independent of the original p.f. cos φ1. It is of the order of 0.95.