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Induction furnace at 10 V, half-full hearth

Short answer

Full hearth gives R = 0.5×10⁻⁴ ohm and X = 0.866×10⁻⁴ ohm. Half full gives Z' = 0.9014×10⁻⁴ ohm and I' = 110,940 A, so power is about 307.7 kW and pf is 0.277 lagging.

A low frequency induction furnace operating at 10V in the secondary circuit takes 500kW at 0.5 p.f when the hearth is full. If the secondary voltage be maintained at 10V, estimate the power absorbed and the p.f when the hearth is half full. Assume the resistance of the secondary circuit to be thereby halved and the reactance to remain the same.

In a low-frequency core-type induction furnace the charge forms a single-turn secondary, so its current depends on the secondary resistance RR and reactance XX.

Given: V2=10V_2 = 10 V, P=500P = 500 kW at pf 0.5 (hearth full); half-full: R′=R/2R' = R/2, XX unchanged.

Hearth full

S=5000.5=1000 kVA,I2=1000×10310=100 000 AZ=10100 000=1.0×10−4 ΩR=Zcos⁡ϕ=0.5×10−4 Ω,X=Zsin⁡ϕ=0.866×10−4 Ω\begin{aligned} S &= \frac{500}{0.5} = 1000\ \text{kVA}, \quad I_2 = \frac{1000\times10^3}{10} = 100\,000\ \text{A} \\ Z &= \frac{10}{100\,000} = 1.0 \times 10^{-4}\ \Omega \\ R &= Z\cos\phi = 0.5\times10^{-4}\ \Omega, \quad X = Z\sin\phi = 0.866 \times 10^{-4}\ \Omega \end{aligned}

Hearth half full

R′=0.25×10−4 ΩZ′=0.252+0.8662×10−4=0.9014×10−4 ΩI2′=100.9014×10−4=110 940 AP′=I2′2R′=110 9402×0.25×10−4=307.7 kWcos⁡ϕ′=R′Z′=0.250.9014=0.277 lagging\begin{aligned} R' &= 0.25 \times 10^{-4}\ \Omega \\ Z' &= \sqrt{0.25^2 + 0.866^2} \times 10^{-4} = 0.9014 \times 10^{-4}\ \Omega \\ I_2' &= \frac{10}{0.9014 \times 10^{-4}} = 110\,940\ \text{A} \\ P' &= I_2'^2 R' = 110\,940^2 \times 0.25 \times 10^{-4} = 307.7\ \text{kW} \\ \cos\phi' &= \frac{R'}{Z'} = \frac{0.25}{0.9014} = 0.277\ \text{lagging} \end{aligned}

Answer: Power absorbed ≈307.7\approx 307.7 kW at power factor ≈0.277\approx 0.277 lagging. Halving the resistance raises the current but lowers both the power and the power factor, because the circuit becomes mainly reactive.

✓ Answer checked by Claude Opus 5.5 (high effort)

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