Most Economical Power Factor

Most Economical Power Factor

It is known that power factor improvement of the system load has number of advantages. The main advantage of the power factor improvement is reduction in maximum kVA demand. Due to this, consumer saves annually over the maximum demand charges. But for improvement in the power factor, consumer requires power factor improvement apparatus. Thus consumer has to make capital investment for such an apparatus. According to depreciation and interest rate, consumer has to spend certain cost annually. Hence net annual saving of consumer is difference between maximum demand charge and annual cost on power factor improvement apparatus.

There exists a certain value of power factor for which the net annual saving of the consumer is maximum such a power factor is called most economical power factor.

Let consumer is operating at a power factor of cos ϕ₁ with a peak load of P kW and power factor is to be improved to cos ϕ₂.

Let x = Charge in  per maximum kVA demand per annum

while y = Expenditure in kVAR per annum on p.f. improvement

Consider the power triangle for the load as shown in the Fig. 7.23.1.

Now from the power triangle we can write,

KVA₁ = P / cos ϕ ₁

and

KVA₂ = P / cos ϕ ₂

Reduction in KVA demand = P / cos ϕ _{1 -} P / cos ϕ ₂ kVA

As charge per maximum kVA demand per annum is ₹ x we can write,

Annual saving in maximum demand charges,

Now size of the power factor improvement device will be according to the reduction in kVAR required. 

kVAR₁ = P tan ϕ₁

and    kVAR₂ = P tan ϕ₂

Hence leading kVAR to be supplied by the power factor improvement apparatus must be difference between kVAR₁ and kVAR₂

Leading kVAR supplied = P (P tan ϕ₁ - P tan ϕ₂)'

The expenditure per kVAR per annum is y hence,

Annual cost on power factor improvement

= yP( tan ϕ₁-  tan ϕ₂)

Hence net annual saving is,

The variable ϕ₂ variable which decides the new power factor. Hence for saving S to be maximum,

It can be observed that most economical power factor depends on costs of maximum kVA demand and equipment but it is not dependent on the original power factor cos ϕ₁.

Example 7.23.1 A certain industrial load has maximum demand of 150 kW at 0.7 lagging. It is charged at a rate of ? 50 per kVA per annum. The power factor improvement device costs ? 100 per kVAR and total interest and depreciation rate is 10 % per annum. Find the maximum economic limit for the power factor improvement.

Solution : Original p.f. cos ϕ₁  = 0.7 lagging

Maximum demand charges = x = 50 per kVA/annum

Capital cost for p.f. improvement device

= 100 per kVAR

Expenditure on p.f. improvement device per annum is,

y = Capital cost x Interest and depreciation rate

= 100 × 10 / 100  = 10 per kVAR/annum

Most economical power factor is,

= 0.9797 lagging

Hence maximum economic limit for the p.f. improvement is 0.9797 lagging.

Review Question

1. Derive the expression for the most economical power factor.