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Change in emf to raise power factor to unity

Short answer

Initial emf E1 = √((V cosφ+IR)²+(V sinφ+IX)²) = 7082.87 V per phase. At unity pf, current reduces to 240 A (same power). New emf E2 = √((V+IR)²+(IX)²) = 6043.56 V. Percentage change = (6043.56-7082.87)/7082.87×100 = -14.67%.

A 3 phase, star connected alternator with R = 0.4 Ω and X = 6 Ω per phase delivers 300 A at power factor 0.8 to constant frequency 10 kV busbars. If the steam supply is unchanged, find the percentage change in the induced emf necessary to raise the power factor to unity. Ignore the change in losses.

Given: star, R=0.4 ΩR = 0.4\ \Omega, X=6 ΩX = 6\ \Omega per phase, 300 A at 0.8 pf (taken lagging) on 10 kV constant-frequency bus-bars; steam supply unchanged, so output power is constant.

V=100003=5773.50 VV = \frac{10000}{\sqrt3} = 5773.50\ \text{V}

Initial emf (0.8 pf lagging, 300 A)

Vcos⁡ϕ+IR=4618.80+120=4738.80 VVsin⁡ϕ+IX=3464.10+1800=5264.10 VE1=4738.802+5264.102=7082.87 V\begin{aligned} V\cos\phi + IR &= 4618.80 + 120 = 4738.80\ \text{V} \\ V\sin\phi + IX &= 3464.10 + 1800 = 5264.10\ \text{V} \\ E_1 &= \sqrt{4738.80^2 + 5264.10^2} = 7082.87\ \text{V} \end{aligned}

New current at unity pf (same power)

I2=I1cos⁡ϕ1=300×0.8=240 AI_2 = I_1\cos\phi_1 = 300\times0.8 = 240\ \text{A}

New emf

E2=(V+I2R)2+(I2X)2=(5773.50+96)2+14402=5869.502+14402=6043.56 V\begin{aligned} E_2 &= \sqrt{(V + I_2R)^2 + (I_2X)^2} = \sqrt{(5773.50 + 96)^2 + 1440^2} \\ &= \sqrt{5869.50^2 + 1440^2} = 6043.56\ \text{V} \end{aligned}

Percentage change

E2−E1E1×100=6043.56−7082.877082.87×100=−14.67%\frac{E_2 - E_1}{E_1}\times100 = \frac{6043.56 - 7082.87}{7082.87}\times100 = -14.67\%

Answer: the induced emf must be reduced by about 14.67% (from 7082.9 V to 6043.6 V per phase) to bring the pf to unity.

✓ Answer checked by Claude Sonnet 5.5

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