# Modeling of Voltage Fluctuations Generated by Arc Furnaces
## 3 Modeling of Voltage Fluctuations in Parallel Operation of Arc Furnaces  
Due to the non-linear nature of the electric arc, it is difficult to determine the voltage fluctuations in the parallel operation of arc devices. Figure 11 shows the power supply diagram for two parallel arc furnaces. The same power value of furnace transformers was assumed. Resistances \(\mathrm{R_{L}}\) and \(\mathrm{R_{T}}\) in the electric arc supply circuit were omitted. The voltages on the \"replacement electric arc furnace\" for each of these conditions can be determined:  
\[\mathrm{U_{Arc}}=\frac{\frac{\mathrm{U_{Arc1}}}{\mathrm{X_{T1}}}+\frac{ \mathrm{U_{Arc2}}}{\mathrm{X_{T2}}}}{\frac{1}{\mathrm{X_{T1}}}+\frac{1}{ \mathrm{X_{T2}}}}=\frac{\mathrm{U_{Arc1}}+\mathrm{U_{Arc2}}}{2} \tag{12}\]  
which is the same as its replacement reactance:  
\[\mathrm{X_{TE}}=\frac{1}{\frac{1}{\mathrm{X_{T1}}}+\frac{1}{\mathrm{X_{T2}}}} =\frac{\mathrm{X_{T}}}{2} \tag{13}\]  
In case of different arc voltages in each furnace (e.g., \(\mathrm{U_{ArcA}}\) in one furnace and \(\mathrm{U_{ArcB}}\) in the other), the matter is more complicated. Changes of phase angles of voltages \(\mathrm{U_{ArcA}}\) and \(\mathrm{U_{ArcB}}\) must be adjusted to the unchanged value of the source voltage of the supply network (\(\mathrm{U_{L}}=100\%\)) and to maintain the same values of \(\mathrm{U_{S}}\) modules on each of the furnaces operating in parallel and on the \"replacement\" furnace, i.e.,  
\[\mathrm{U_{ArcA}^{2}+I_{A}^{2}X_{T}^{2}}=\mathrm{U_{ArcB}^{2}+I_{B}^{2}X_{T}^ {2}}=\mathrm{U_{ArcE}^{2}+I^{2}\bigg{(}\frac{\mathrm{X_{T}}}{2}\bigg{)}^{2}}= \mathrm{U_{S}} \tag{14}\]  
Based on the vector diagram, as shown in Figure 12 the dependence can be written:  
\[\mathrm{U_{L}^{2}}=(\mathrm{U_{S}}+\mathrm{Dsin}\ \alpha)^{2}+(\mathrm{Dcos}\ \alpha )^{2} \tag{15}\]  
Figure 10: Voltage change as a function of current changes the furnace with the resistance in the power supply circuit of an electric arc. Substituting:  
\[ \begin{array}{l}\text{D}\text{sin}\alpha=\text{U}_{\text{T}1}\frac{\text{X}_{ \text{t}}}{\text{X}_{\text{T}}}\text{sin}\ \alpha_{1}+\text{U}_{\text{T}2}\frac{\text{X}_{\text{t}}}{\text{X}_{\text{T}}} \text{sin}\ \alpha_{2}=\text{U}_{\text{S}}\text{sin}\ \alpha_{1}\frac{\text{X}_{\text{t}}}{\text{X}_{\text{T}}}\text{sin}\ \alpha_{1}+\text{U}_{\text{S}}\text{sin}\ \alpha_{2}\frac{\text{X}_{\text{t}}}{\text{X}_{\text{T}}}\text{sin}\ \alpha_{2}=\\ =\text{U}_{\text{S}}\frac{\text{X}_{\text{t}}}{\text{X}_{\text{T}}}\text{sin}\ \alpha_{1}+\text{sin}\ \alpha_{2}\text{cos}\ \alpha_{2}\end{array}  \tag{16}\]  
\[ \begin{array}{l}\text{D}\text{cos}\alpha=\text{U}_{\text{T}1}\frac{\text{X} _{\text{t}}}{\text{X}_{\text{T}}}\text{cos}\ \alpha_{1}+\text{U}_{\text{T}2}\frac{\text{X}_{\text{t}}}{\text{X}_{\text{T}}} \text{cos}\ \alpha_{2}=\text{U}_{\text{S}}\text{sin}\ \alpha_{1}\frac{\text{X}_{\text{t}}}{\text{X}_{\text{T}}}\text{cos}\ \alpha_{1}+\text{U}_{\text{S}}\text{sin}\ \alpha_{2}\frac{\text{X}_{\text{t}}}{\text{X}_{\text{T}}}\text{cos}\ \alpha_{2}=\\ =\text{U}_{\text{S}}\frac{\text{X}_{\text{t}}}{\text{X}_{\text{T}}}\text{sin} \ \alpha_{1}\text{cos}\ \alpha_{1}+\text{sin}\ \alpha_{2}\text{cos}\ \alpha_{2}\end{array}  \tag{17}\]  
Assuming the dependencies:  
\[\cos\ \alpha_{1}=\frac{\text{U}_{\text{Arcl}}}{\text{U}_{\text{S}}}\text{sin} \ \ \alpha_{1}=\sqrt{1-\left(\frac{\text{U}_{\text{Arcl}}}{\text{U}_{\text{S}}} \right)^{2}} \tag{18}\]  
Figure 11: Scheme of the supply system for the steel plant, (**a**) for two identical furnaces arched, (**b**) for a replacement electric arc furnace. Figure 12: Vector diagram of the currents and voltages of the arc furnace. \[\cos\ \alpha_{2}=\frac{\mathrm{U}_{\mathrm{Arc2}}}{\mathrm{U}_{\mathrm{S}}}\mathrm{ sin}\ \ \alpha_{2}=\sqrt{1-\left(\frac{\mathrm{U}_{\mathrm{Arc2}}}{\mathrm{U}_{\mathrm{S}}} \right)^{2}} \tag{19}\]  
and substituting them to Equations (18) and (19), we obtain Formula (19) in the form:  
\[