Arc furnace power factor and time to melt steel

Each phase of the arc furnace is the arc (a pure resistance) in series with the transformer resistance and reactance.

Given (per phase): $I = 4000$ A, $V_{arc} = 60$ V, $R_t = 0.0025\ \Omega$, $X_t = 0.0050\ \Omega$.

i) Power factor and kW drawn

$$\begin{aligned} R_{arc} &= \frac{V_{arc}}{I} = \frac{60}{4000} = 0.015\ \Omega \\ R_{total} &= 0.015 + 0.0025 = 0.0175\ \Omega \\ Z &= \sqrt{0.0175^2 + 0.005^2} = 0.01820\ \Omega \\ \cos\phi &= \frac{R_{total}}{Z} = \frac{0.0175}{0.01820} = 0.9615\ \text{lagging} \\ P &= 3I^2R_{total} = 3 \times 4000^2 \times 0.0175 = 840\ \text{kW} \end{aligned}$$

(Arc power alone $= 3 \times 60 \times 4000 = 720$ kW; the other 120 kW is transformer copper loss.)

ii) Time to melt 2.5 tonnes of steel

Heat required:

$$\begin{aligned} H &= m\left[c(\theta_m - \theta_i) + L\right] \\ &= 2500\left[0.5(1370 - 15) + 37.2\right] \\ &= 2500 \times 714.7 = 1\,786\,750\ \text{kJ} \end{aligned}$$

Useful power with 70% overall efficiency:

$$P_{useful} = 0.7 \times 840 = 588\ \text{kW}$$

Time:

$$t = \frac{1\,786\,750}{588} = 3038.7\ \text{s} = 50.6\ \text{min}$$

Answer: pf $= 0.9615$ lagging, power drawn $= 840$ kW; time to melt 2.5 t of steel $\approx 3039$ s $\approx 50.6$ minutes.