A plant has two loadsc The load voltage is 4.6 kVrms. (Note: Since nothing is sa

Question

A plant has two loadsc The load voltage is 4.6 kVrms. (Note: Since nothing is said to the contrary, you must assume that this is the line voltage). Load one has apparent power S, 35kVA and power factor of Pf 0.8, Leading. Load two is consuming 72 kW of real power and has the very bad power factor of Pf,-0.6, Lagging . Assume f = 60 Hz d in parallel. Assume these two are the only loads in the plant. Find the total plant complex power Find the power factor of the total load and specify whether it is leading or lagging. Find the capacitance of the -connected capacitor bank that when installed in parallel to these loads will change the plant power factor to one. To improve the power factor of the plant, instead of the capacitor bank calculated in (c), a Y-connected capacitor bank of rating (4.6 kV rms, 50 kVAR) is ordered and connected in parallel with the load. Calculate the resulting power factor and specify whether it is leading or lagging. Calculate the capacitance of each phase of the capacitor bank of part (d). a. b. C. d.

Explanation / Answer

A. The total complex power of the plant will be the addition of the complex powers of both the load.

S1 = 35KVA , P2=72 KW with power factor(PF) of 0.6,lagging ,

we know , S * PF=P , So S2= P2 / PF2 = 72/0.6 = 120KVA

hence total complex power = S= S1 + S2 = 35 + 120= 155KVA

B. Power factor of the total load can be calculated by the total real power and the total reactive power.

Total real power = P1 + P2 = 35 * (0.8) + 72 = 28 + 72 = 100KW

Total reactive power = Q1 + Q2 = -35 * sin (cos-1(0.8)) + 120 * sin (cos-1(0.6)) = -21 + 96 = 75 KVAR

(we took negative sign while calculating reactive power of load one because the power factor associated with it is leading and we take lagging VARS positive as per standard)

Now the total power factor of the plant = cos (tan-1 Q/P) = 0.8 lagging

(lagging because the total reactive power calculated is positive)

C. Now to make the power factor of the plant to be as one , the ractive power demand of the plant should be zero for which capacitor installed should have a capacity of 75 KVAR .

QC= VL2 * 2* pi * f * C

75 * 1000 = 4.62 * 106 * 2 * 3.14 * 60 * C

C= 9.4 micro farads

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