--- title: "Arc furnace power factor and time to melt steel" subject: "Utilization of Electrical Energy" chapter: 5 key: "semester_7/utilization_of_electrical_energy:arc-furnace-pf-melting-time-steel" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/arc-furnace-pf-melting-time-steel site: BCT Engineering --- # Arc furnace power factor and time to melt steel Utilization of Electrical Energy · Chapter 5: Electric Heating - Asked in: 2081 Baisakh, 4+4 marks - Video: [ELECTRICAL HEATING 16 - TWO PROBLEMS IN ELECTRIC FURNACES](https://www.youtube.com/watch?v=UJjplhajYEk) — zenmurali Sachithanandam - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/arc-furnace-pf-melting-time-steel - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-5 ## Question The following data relate to a 3-phase electric arc Furnace: Current drawn = 4000 A Arc Voltage = 60 V Resistance of transformer referred to Secondary = 0.0025 Ω Reactance of transformer referred to Secondary = 0.0050 Ω i) Calculate the power factor and kW drawn from the supply. ii) If the overall efficiency of the furnace is 70%, find the time required to melt 2.5 tonnes of Steel if Latent heat of Steel = 37.2 kJ/kg, Specific heat of Steel = 0.5 kJ/kg-K, Melting Point of Steel = 1370°C Initial temperature of Steel = 15°C ## Short answer Arc resistance is 0.015 ohm, total resistance 0.0175 ohm, Z 0.0182 ohm; pf is 0.9615 lagging and power drawn is 840 kW. Heat needed for 2.5 t steel is 1,786,750 kJ; with 70% efficiency useful power is 588 kW, so time is 3039 s, about 50.6 min. ## Answer 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the arc treated per phase in this calculation? A. A capacitor B. An inductance C. A pure resistance V_arc/I in series with transformer R and X D. A constant voltage source with no resistance Answer: C. R_arc = 60/4000 = 0.015 Ω, in series with the transformer R and X. (Key point: Arc modelled as resistance) 2. What is the power factor? A. 0.866 B. 0.9615 lagging C. 0.75 lagging D. 0.5 Answer: B. R_total = 0.0175 Ω, Z = 0.0182 Ω, so cos(phi) = 0.0175/0.0182 = 0.9615. (Key point: Power factor) 3. Power drawn from supply is 840 kW. Why is this more than the arc power 720 kW? A. The reactance consumes real power B. The extra 120 kW is transformer copper loss C. The supply voltage is higher D. Efficiency is 70% Answer: B. P = 3 I^2 (R_arc + R_t); the part in R_t = 0.0025 Ω is 120 kW copper loss. (Key point: Power drawn) 4. How is the melting time found? A. Heat divided by 840 kW only B. Latent heat only divided by 720 kW C. Heat x efficiency / 840 kW D. Heat for heating plus latent heat divided by 0.7 x 840 kW Answer: D. H = m[c(Δθ) + L] = 1,786,750 kJ; useful power 588 kW; t = 3039 s. 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