Skip to main content

Annual saving by pf improvement for 800 kW load

Short answer

Since capacitors are loss-free, kWh cost is unchanged. Improving 800 kW from 0.8 to 0.9 pf cuts kVA demand from 1000 to 888.89, saving Rs 11,111 in demand charge. Against 212.54 kVAR capacitors costing Rs 1,275.30 annually, net saving is about Rs 9,836.

The load on an installation is 800 KW, 0.8 lagging which works for 3000 hours per annum. The tariff is Rs 100 per KVA plus 20 paisa per kwh. If the power factor is improved to 0.9 lagging by means of loss free capacitors costing Rs 60 per KVAR, calculate the annual saving. Allow 10% per annum for interest and depreciation on capacitors.

Capacitors are loss-free, so the energy (kWh) bill does not change; the saving is in the kVA demand charge, minus the annual charges on the capacitors.

Demand before and after

S1=8000.8=1000 kVAS2=8000.9=888.89 kVASaving in demand charge=100×(1000−888.89)=Rs 11,111\begin{aligned} S_1 &= \frac{800}{0.8} = 1000\ \text{kVA} \\ S_2 &= \frac{800}{0.9} = 888.89\ \text{kVA} \\ \text{Saving in demand charge} &= 100 \times (1000 - 888.89) = \text{Rs } 11{,}111 \end{aligned}

Capacitor rating and its annual cost

tan⁡ϕ1=tan⁡(cos⁡−10.8)=0.75,tan⁡ϕ2=tan⁡(cos⁡−10.9)=0.4843QC=800 (0.75−0.4843)=212.54 kVARCapital cost=60×212.54=Rs 12,753Interest and depreciation=0.10×12,753=Rs 1275.3\begin{aligned} \tan\phi_1 &= \tan(\cos^{-1}0.8) = 0.75,\quad \tan\phi_2 = \tan(\cos^{-1}0.9) = 0.4843 \\ Q_C &= 800\,(0.75 - 0.4843) = 212.54\ \text{kVAR} \\ \text{Capital cost} &= 60 \times 212.54 = \text{Rs } 12{,}753 \\ \text{Interest and depreciation} &= 0.10 \times 12{,}753 = \text{Rs } 1275.3 \end{aligned}

Net annual saving

Saving=11,111−1275.3=Rs 9835.9\text{Saving} = 11{,}111 - 1275.3 = \text{Rs } 9835.9

The energy charge (0.20 × 800 × 3000 = Rs 480,000) is the same before and after, so it does not affect the saving.

Answer: Annual saving ≈ Rs 9,836 (capacitors of 212.54 kVAR).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Utilization of Electrical Energy chapter 6 questions →