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The Most Economical Power Factor – Definition, Formula and Examples
Most Economical Power Factor
The value to which the power factor should be improved so as to have maximum net annual saving is called the most economical power factor.
When a consumer improves the power factor, it results in reduction in his maximum kVA demand. Consequently, there will be an annual saving over the maximum demand charges. Although the power factor improvement involves the capital investment on the power factor correction equipment, the consumer will draw expenditure every year in the shape of annual interest and depreciation on the investment made over equipment used for power factor correction. Hence, the net annual saving will be equal to the annual saving in maximum demand charge minus annual expenditure on the power factor correction equipment.
Derivation
Consider a consumer taking a peak load of P kW at a power factor of $\mathrm{cos}\, \phi _{\mathrm{1}}$ and it is charged at a rate of Rs x per kVA of maximum demand per year.
Assume that the consumer improves the power factor to $\mathrm{cos}\, \phi _{\mathrm{2}}$ with the help of a power factor correction device. Consider the expenditure made on the power factor improvement device is Rs y per kVAR per annum.
Refer the figure, the power triangle at the initial power factor ($\mathrm{cos}\, \phi _{\mathrm{1}}$) is ABC and the power triangle for the improved power factor ($\mathrm{cos}\, \phi _{\mathrm{2}}$) is ABD.
Therefore, the maximum kVA demand at original power factor is,
$$\mathrm{\mathit{kV\, A_{\mathrm{1}}\mathrm{\, =\, }\frac{P}{\mathrm{cos}\, \phi _{\mathrm{1}}}\mathrm{\, =\, }P\, \mathrm{sec}\, \phi _{\mathrm{1}}}}$$
And the maximum kVA demand at improved power factor is given by,
$$\mathrm{\mathit{kV\, A_{\mathrm{2}}\mathrm{\, =\, }\frac{P}{\mathrm{cos}\, \phi _{\mathrm{2}}}\mathrm{\, =\, }P\, \mathrm{sec}\, \phi _{\mathrm{2}}}}$$
Then, the saving in maximum demand charges per annum is,
$$\mathrm{Saving\: per \: annum\mathrm{\, =\, } \mathit{Rs.x\left (kV\, A_{\mathrm{1}}-kV\, A_{\mathrm{2}} \right )}}$$
$$\mathrm{\Rightarrow Saving\: per \: annum\mathrm{\, =\, } \mathit{Rs.x\left (\frac{P}{\mathrm{cos}\, \phi _{\mathrm{1}}}-\frac{P}{\mathrm{cos}\, \phi _{\mathrm{2}}} \right )}}$$
$$\mathrm{\Rightarrow Saving\: per \: annum\mathrm{\, =\, } \mathit{Rs.xP\left (\mathrm{sec}\, \phi _{\mathrm{1}}- \mathrm{sec}\, \phi _{\mathrm{2}} \right )}\; \; \cdot \cdot \cdot \left ( 1 \right )}$$
Also, the reactive power at original power factor is,
$$\mathrm{kVAR_{\mathrm{1}}\mathit{\mathrm{\, =\, }P\, \mathrm{tan}\, \phi _{\mathrm{1}}}}$$
And the reactive power at improved power factor is,
$$\mathrm{kVAR_{\mathrm{2}}\mathit{\mathrm{\, =\, }P\, \mathrm{tan}\, \phi _{\mathrm{2}}}}$$
The leading kVAR drawn by the power factor improvement device is given by,
$$\mathrm{\mathrm{\, =\, }kVAR_{\mathrm{1}}-kVAR_{\mathrm{2}} \mathit{\mathrm{\, =\, }P \left ( \mathrm{tan}\, \phi _{\mathrm{1}}-\mathrm{tan}\, \phi _{\mathrm{2}} \right )}}$$
Thus, the cost of power factor improvement device per annum is,
$$\mathrm{Cost \: of\: equipment\: per \: annum\mathrm{\, =\, }Rs\, \mathit{yP \left ( \mathrm{tan}\, \phi _{\mathrm{1}}-\mathrm{tan}\, \phi _{\mathrm{2}} \right )}\: \: \cdot \cdot \cdot \left ( 2 \right )}$$
Therefore, from equations (1) & (2), the net saving per annum is,
$$\mathrm{\mathit{S\mathrm{\, =\, }xP\left (\mathrm{sec}\, \phi _{\mathrm{1}}- \mathrm{sec}\, \phi _{\mathrm{2}}\right )-yP\left ( \mathrm{tan}\, \phi _{\mathrm{1}}-\mathrm{tan}\, \phi _{\mathrm{2}} \right )}}$$
Now, the net saving per annum will be maximum if
$$\mathrm{\mathit{\frac{d\left ( S \right )}{d\, \phi \mathrm{2}}}\mathrm{\, =\, }0}$$
$$\mathrm{\Rightarrow \mathit{\frac{d}{d\, \phi _{\mathrm{2}}}\left [ xP\left (\mathrm{sec}\, \phi _{\mathrm{1}}- \mathrm{sec}\, \phi _{\mathrm{2}}\right )-yP\left ( \mathrm{tan}\, \phi _{\mathrm{1}}-\mathrm{tan}\, \phi _{\mathrm{2}} \right ) \right ]}\mathrm{\, =\, }0}$$
$$\mathrm{\Rightarrow \mathit{P\left [ \frac{d}{d\, \phi _{\mathrm{2}}}\left ( x\, \mathrm{sec}\, \phi _{\mathrm{1}} \right )-\frac{d}{d\, \phi _{\mathrm{2}}}\left (x\, \mathrm{sec}\, \phi _{\mathrm{2}} \right )-\frac{d}{d\, \phi _{\mathrm{2}}}\left ( y\, \mathrm{tan}\, \phi _{\mathrm{1}} \right )\mathrm{\, +\, }\frac{d}{d\, \phi _{\mathrm{2}}}\left ( y\, \mathrm{tan}\, \phi _{\mathrm{2}} \right ) \right ]}\mathrm{\, =\, }0}$$
$$\mathrm{\Rightarrow 0-\mathit{x\, \mathrm{sec}\, \phi _{\mathrm{2}}\, \mathrm{tan}\, \phi _{\mathrm{2}}-\mathrm{0}\mathrm{\, +\, }y\, \mathrm{sec^{2}}\, \phi _{\mathrm{2}}}\mathrm{\, =\, }0}$$
$$\mathrm{\Rightarrow \mathit{\mathrm{tan}\, \phi _{\mathrm{2}}\mathrm{\, =\, }\frac{y}{x}\mathrm{sec}\, \phi _{\mathrm{2}}}}$$
$$\mathrm{\Rightarrow \mathit{\frac{\mathrm{tan}\, \phi _{\mathrm{2}}}{\mathrm{sec}\, \phi _{\mathrm{2}}}\mathrm{\, =\, }\mathrm{sin}\, \phi _{\mathrm{2}}\mathrm{\, =\, }\frac{y}{x}}}$$
Therefore, the most economical power factor is given by,
$$\mathrm{\mathit{\mathrm{cos}\, \phi _{\mathrm{2}}\mathrm{\, =\, } \sqrt{\mathrm{1}-\mathrm{sin^{\mathrm{2}}}\, \phi _{\mathrm{2}}} \mathrm{\, =\, }\sqrt{\mathrm{1}-\left ( \frac{y}{x} \right )^{\mathrm{2}}} }\; \; \cdot \cdot \cdot \left ( 3 \right )}$$
From Equation (3), it is clear that the most economical power factor is independent of the original power factor.
Numerical Example
A paper plant which has a maximum demand of 185 kW at a power factor of 0.8 lagging is charged at Rs 75 per kVA per annum. If the power factor correction equipment costs Rs 130 per kVAR. Then, find the most economical power factor at which the plant should operate. The power factor correction equipment has interest and depreciation charges 10% of its capital investment.
Solution
The original power factor of the paper plant is,
$$\mathrm{\mathit{\mathrm{cos}\, \phi _{\mathrm{1}}}\mathrm{\, =\, }0.8 \, \, lagging}$$
The maximum demand charges are
$$\mathrm{\mathit{x\mathrm{\, =\, }𝑅𝑠.}\,75\: per\: kVA\: per\: annum }$$
And the expenditure on the phase correction equipment is,
$$\mathrm{\mathit{y\mathrm{\, =\, }𝑅𝑠.}\,130\times 10\%\mathrm{\, =\, }\mathit{Rs.}\, 13\: per\: kVA\: per\: annum }$$
Therefore, the most economical power factor of the plant is,
$$\mathrm{\mathit{\mathrm{cos}\, \phi _{\mathrm{2}}\mathrm{\, =\, }\sqrt{\mathrm{1}-\left ( \frac{y}{x} \right )^{\mathrm{2}}}\mathrm{\, =\, }\mathrm{\sqrt{1-\left ( \frac{13}{75} \right )^{2}}}}}$$
$$\mathrm{\mathit{\therefore \mathrm{cos}\, \phi _{\mathrm{2}}\mathrm{\, =\, } }\sqrt{1-0.03}\mathrm{\, =\, }0.984 \: lagging}$$
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