Question

A 440-V, 3-phase, delta-connected induction motor has an output of 14.92 kW at a p.f. of 0.82 and efficiency 85%. Calculate the readings on each of the two watt meters connected to measure the input. Prove any formula used. If another star-connected load of 10 kW at 0.85 p.f. lagging is added in parallel to the motor, what will be the current draw from the line and the power taken from the line?

Answer 1

Motor input = 14,920/0.85 = 17,600 W 

∴ W₁ +W₂ = 17.6 kW ... (i)

cosφ = 0.82, φ = 34.9°, tan34.9° = 0.6976; 0.6976 = (√3(W₁ - W₂)/17.6))

∴ W₁ - W₂ = 7.09 kW ..... (ii)

From (i) and (ii) above, we get W₁ = 12.35 kW and W₂ = 5.26 kW

Motor kVA, S_m = motor kW/cosφ_m = 17.6/0.82 = 21.46

∴ S_m = 21.46 ∠ - 34.9° = (17.6 - j12.28)KVA

Load p.f. = 0.85

∴ φ = cos^-1(0.85) = 31.8° ; Load KVA, S_Y = 10/0.85 = 11.76

∴ S_Y = 11.76 ∠- 31.8° = (10 - j6.2) kVA 

Combined kVA, S = S_m + S_Y = (27.6 – j 18.48) = 32.2 ∠ – 33.8° kVA

I = S/√3.V = (33.2 x 10³)/(√3 x 440) = 43.56A

Power taken = 27.6 kW