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 φ_{1 }and charged at a rate of Rs x per kVA of maximum demand per annum. Suppose the consumer improves the power factor to cos φ_{2 }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 φ_{1 }is OAB and for the improved p.f. cos φ_{2}, it is OAC [See Fig.].
kVA max. demand at cos φ_{1}, kVA_{1 }= P/cos φ_{1 }= Psec φ_{1 }
kVA max. demand at cos φ_{2}, kVA_{2 }= P/cos φ_{2 }= Psec φ_{2}
Annual saving in maximum demand charges
= Rs x.(kVA_{1 }− kVA_{2})
= Rs x.(Psec φ_{1 }– Psec φ_{2}) = Rs x.P (sec φ_{1 }– sec φ_{2}) ...(i)
Reactive power at cos φ_{1}, kVAR_{1 }= P tan φ_{1 }
Reactive power at cos φ_{2}, kVAR_{2 }= P tan φ_{2 }
Leading kVAR taken by p.f. correction equipment = P (tan φ_{1 }− tan φ_{2})
Annual cost of p.f. correction equipment = Rs P.y (tan φ_{1 }– tan φ_{2}) ...(ii)
Net annual saving, S = exp. (i) − exp. (ii)
= x.P (sec φ_{1 }– sec φ_{2}) − y.P (tan φ_{1 }– tan φ_{2})
In this expression, only φ_{2 }is variable while all other quantities are fixed. Therefore, the net annual saving will be maximum if differentiation of above expression w.r.t. φ_{2 }is zero i.e.
dS/dφ_{2} = 0
0 – x.P sec φ_{2 }tan φ_{2 }- 0 + y.P sec^{2}φ_{2} = 0
- x.tan φ_{2 }+ y. sec φ_{2 }= 0
sin φ_{2 }= y/x
Hence, Most economical power factor,
It may be noted that the most economical power factor (cos φ_{2}) depends upon the relative costs of supply and is independent of the original p.f. cos φ_{1}. It is of the order of 0.95.