Arc furnace: kW input, arc voltage, pf, kVA

Interpretation: "resistance/reactance of temperature" means the transformer referred to the secondary, "overall efficiency of steel" means overall efficiency of the furnace, and specific heat is 0.5 kJ/kg°C. Values are per phase.

Given: $m = 4300$ kg/h, $c = 0.5$ kJ/kg°C, $L = 37.2$ kJ/kg, $\theta_m = 1370$°C, $\theta_i = 19.1$°C, $\eta = 0.5$, $I = 5700$ A, $R_t = 0.008\ \Omega$, $X_t = 0.014\ \Omega$.

(i) Average kW input

$$\begin{aligned} H &= m[c(\theta_m - \theta_i) + L] = 4300[0.5(1370 - 19.1) + 37.2] \\ &= 4300 \times 712.65 = 3\,064\,395\ \text{kJ/h} \\ P_{out} &= \frac{3\,064\,395}{3600} = 851.2\ \text{kW} \\ P_{in} &= \frac{851.2}{0.5} = 1702.4\ \text{kW} \end{aligned}$$

(ii) and (iii) Arc resistance and arc voltage

Per phase, all input power is dissipated in $R_t + R_{arc}$:

$$\begin{aligned} R_t + R_{arc} &= \frac{P_{in}}{3I^2} = \frac{1\,702\,442}{3 \times 5700^2} = 0.017466\ \Omega \\ R_{arc} &= 0.017466 - 0.008 = 0.009466\ \Omega \\ V_{arc} &= I R_{arc} = 5700 \times 0.009466 = 53.96\ \text{V} \end{aligned}$$

(iv) Power factor

$$\begin{aligned} Z &= \sqrt{0.017466^2 + 0.014^2} = 0.022385\ \Omega \\ \cos\phi &= \frac{0.017466}{0.022385} = 0.780\ \text{lagging} \end{aligned}$$

(v) Average kVA input

$$S = \frac{P_{in}}{\cos\phi} = \frac{1702.4}{0.7803} = 2181.8\ \text{kVA}$$

Answer: (i) 1702.4 kW, (ii) arc voltage $\approx 53.96$ V, (iii) arc resistance $\approx 0.00947\ \Omega$, (iv) pf $\approx 0.780$ lagging, (v) $\approx 2182$ kVA.