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Annual charge and pf improvement (Rs 65/kVA tariff)

Short answer

Initial annual charge is Rs 132,000. Improving pf to 0.95 requires about 489.9 kVAR including 5% apparatus loss, giving a new annual charge of Rs 121,664, a saving of about Rs 10,336.

The load on a certain installation may be considered constant at 1200kVA, 0.75 lagging power factor for 3000 hours per annum. The tariff is Rs 65 per kVA of maximum demand plus 2 paisa per kWH. i) Determine the annual charge for electrical energy ii) Power factor improving apparatus is installed to improve the power factor to 0.95 lagging. Determine the kVAR required and the new annual charge if the power factor improving apparatus costs Rs 60 per kVAR, annual interest and depreciation charges are 10% of the capital cost and the losses in the apparatus are 5% of the kVAR rating.

(i) Present annual charge

P=1200×0.75=900 kWDemand charge=65×1200=Rs 78,000Energy charge=0.02×900×3000=Rs 54,000Annual charge=Rs 132,000\begin{aligned} P &= 1200 \times 0.75 = 900\ \text{kW} \\ \text{Demand charge} &= 65 \times 1200 = \text{Rs } 78{,}000 \\ \text{Energy charge} &= 0.02 \times 900 \times 3000 = \text{Rs } 54{,}000 \\ \text{Annual charge} &= \text{Rs } 132{,}000 \end{aligned}

(ii) After improving pf to 0.95

Let the apparatus rating be QCQ_C kVAR. Its loss, 0.05 QC0.05\,Q_C kW, is also drawn from the supply, so the overall pf of 0.95 must include it.

tan⁡ϕ1=tan⁡(cos⁡−10.75)=0.8819,tan⁡ϕ2=tan⁡(cos⁡−10.95)=0.3287Load kVAR=900×0.8819=793.73 kVARCondition: 793.73−QC900+0.05 QC=0.3287793.73−QC=295.82+0.01643 QCQC=497.911.01643=489.9 kVAR\begin{aligned} \tan\phi_1 &= \tan(\cos^{-1}0.75) = 0.8819,\quad \tan\phi_2 = \tan(\cos^{-1}0.95) = 0.3287 \\ \text{Load kVAR} &= 900 \times 0.8819 = 793.73\ \text{kVAR} \\ \text{Condition: } \frac{793.73 - Q_C}{900 + 0.05\,Q_C} &= 0.3287 \\ 793.73 - Q_C &= 295.82 + 0.01643\,Q_C \\ Q_C &= \frac{497.91}{1.01643} = 489.9\ \text{kVAR} \end{aligned}

New annual charge:

Apparatus loss=0.05×489.9=24.5 kWTotal kW=900+24.5=924.5 kWNew max demand=924.50.95=973.15 kVADemand charge=65×973.15=Rs 63,255Energy charge=0.02×924.5×3000=Rs 55,470Capital cost=60×489.9=Rs 29,392Interest and depreciation=0.10×29,392=Rs 2,939New annual charge=63,255+55,470+2,939=Rs 121,664\begin{aligned} \text{Apparatus loss} &= 0.05 \times 489.9 = 24.5\ \text{kW} \\ \text{Total kW} &= 900 + 24.5 = 924.5\ \text{kW} \\ \text{New max demand} &= \frac{924.5}{0.95} = 973.15\ \text{kVA} \\ \text{Demand charge} &= 65 \times 973.15 = \text{Rs } 63{,}255 \\ \text{Energy charge} &= 0.02 \times 924.5 \times 3000 = \text{Rs } 55{,}470 \\ \text{Capital cost} &= 60 \times 489.9 = \text{Rs } 29{,}392 \\ \text{Interest and depreciation} &= 0.10 \times 29{,}392 = \text{Rs } 2{,}939 \\ \text{New annual charge} &= 63{,}255 + 55{,}470 + 2{,}939 = \text{Rs } 121{,}664 \end{aligned}
ItemBefore (Rs)After (Rs)
Demand charge78,00063,255
Energy charge54,00055,470
Apparatus charges02,939
Total132,000121,664

(If the apparatus loss is ignored when sizing, QC=497.9Q_C = 497.9 kVAR; but then the kVA demand including the 24.9 kW loss is about 971 kVA, not 947.4 kVA.)

Answer: (i) Rs 132,000 per year; (ii) kVAR required ≈ 489.9 kVAR, new annual charge ≈ Rs 121,664 (saving about Rs 10,336 per year).

✓ Answer checked by Claude Opus 5.5 (high effort)

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