Low frequency induction furnace with half-full hearth

In a low-frequency (core-type) induction furnace the charge forms the short-circuited secondary, so the secondary current is set by the secondary resistance $R$ and reactance $X$.

Given: $V_2 = 20$ V, $P = 500$ kW at pf 0.5 (hearth full). Half-full: $R' = 2R$, $X$ unchanged.

Secondary impedance when hearth is full

$$\begin{aligned} S &= \frac{P}{\cos\phi} = \frac{500}{0.5} = 1000\ \text{kVA} \\ I_2 &= \frac{S}{V_2} = \frac{1000 \times 10^3}{20} = 50\,000\ \text{A} \\ Z &= \frac{V_2}{I_2} = \frac{20}{50\,000} = 4 \times 10^{-4}\ \Omega \\ R &= Z\cos\phi = 2 \times 10^{-4}\ \Omega \\ X &= Z\sin\phi = 4\times10^{-4} \times 0.866 = 3.464 \times 10^{-4}\ \Omega \end{aligned}$$

Hearth half full

$$\begin{aligned} R' &= 2R = 4 \times 10^{-4}\ \Omega, \quad X = 3.464 \times 10^{-4}\ \Omega \\ Z' &= \sqrt{(4)^2 + (3.464)^2} \times 10^{-4} = 5.2915 \times 10^{-4}\ \Omega \\ I_2' &= \frac{20}{5.2915 \times 10^{-4}} = 37\,796\ \text{A} \\ P' &= I_2'^2 R' = 37\,796^2 \times 4 \times 10^{-4} = 571.4\ \text{kW} \\ \cos\phi' &= \frac{R'}{Z'} = \frac{4}{5.2915} = 0.756\ \text{lagging} \end{aligned}$$

Answer: With the hearth half full, power absorbed $\approx 571.4$ kW at a power factor of $0.756$ lagging.