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A 3-phase, 500-V motor load has a power factor of 0.4 Two watt meters connected to measure the power show the input to be 30 kW. Find the reading on each instrument. |
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Answer» tanφ = (√3(W1 - W2))/(W1 + W2) ....(i) Now, cosφ = 0.4; φ cos-1(0.4) = 66.6°; tan 66.6° = 2.311 W1 + W2 = 30 ....(ii) Substituting these values in equation (i) above, we get 2.311 = (√3(W1 - W2))/30 ∴ W1 - W2 = 40 .... (iii) From Eq. (ii) and (iii), we have W1 = 45 kW and W2 = – 5 kW Since W2 comes out to be negative, second watt meter reads ‘down scale’. Even otherwise it is obvious that p.f. being less than 0.5, W2 must be negative. |
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