A factory has an electrical load of 1,800kW at a lagging power factor of 0.6. An

Question

A factory has an electrical load of 1,800kW at a lagging power factor of 0.6. An additional variable power factor load is to be added to the factory. The new loaf will add 600kW to the real power load of the factory. The power factor of the added load is to be adjusted so that the overall power factor of the factory is 0.96 lagging. The factory is fed from a line having an impedance of 0.02 + j0.16 O. The voltage at the factory is maintained at 480 V(rms). Find the average power loss in the line before and after the load is added. Find the magnitude of the voltage at the sending end of the line before and after the loaf is added.

Explanation / Answer

The existing 1800 watts Kw load at 0.6 PF has a phase angle which has a cosine value of 0.6. This angle is about 53 degrees. The tangent of this angle (53 degrees) is 1.327. This is the ratio of the KVAR divided by the Kw or 1.327 = KVAR / 1800 or KVAR = 1800 * 1.327 = 2388.6 KVAR. a) When you add the additional 600 Kw to the 1800 Kw, the cosine of the phase angle for the power factor og 0.96 will be about 16 degrees. The tangent for 16 degrees is 0.287. This is the ratio of the KVAR divided by the Kw or 0.287 = KVAR / 2400 or KVAR = 2400 * 0.287 = 688.8 KVAR b) This means there must be change in reactive power = 688.8-2388.6 = -1699.8 KVAR Thus load delivers reactive power c)Power factor of load Pf = Real Power/v(Real Power^2 + Reactive Power^2) Pf = 600k/ v(600k^2 + (-1699.8k)^2) = 0.1107 leading d) Before adding load Pf = Real Power/ Total Power >>>>>>>Total Power = 1800k/0.6 = 3000 kVA I(rms) = Toatl Power / V(rms) = 3000k / 480 = 6250 A e) After adding load Pf = Real Power/ Total Power >>>>>>>Total Power = 2400k/0.96 = 2500 kVA I(rms) = Toatl Power / V(rms) = 2500k / 480

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