## 4 Transformer and Cable Losses Depending on Motor Power Factor  
An important energy parameter for a motor powered directly by mains is not only the efficiency but also the power factor, because the reactive current feeding a motor flows not only through its winding but also through elements of an electric transmission line from which a motor receives power. This causes additional losses [60]. Since the considered motors have different power factors and total currents (see Figure 3b,c and Table 3), it is necessary to evaluate the influence of the power factor on the cost of electricity for the consumer. The following structure of the electrical system is considered for calculations: three pump units with a capacity of 4 kW connected to the mains via a 100 m cable and a 16 kVA step-down transformer (see Section 5). Losses in the cable sections connecting the individual pump units to the common point of three pump units are not taken into account. Thus, the current creating losses in the cable and transformer can be found as the sum of the currents of individual pump units:  
\[I_{load}=I_{motor\ 1}+I_{motor\ 2}+I_{motor\ 3}=3\cdot I_{motor}, \tag{5}\]  
where \(I_{motor\ 1}\ \ldots\ I_{motor\ 3}\) are the motor currents of pump units of the considered electrical system; \(I_{motor}\) is the phase current of one of these motors. The results of the calculation of the load current of the cable and transformer for various studied pump operating points are presented in Table 5. Single motor current is calculated using polynomial data interpolation Figure 3c. The total load current is then calculated using the Formula (5). In the considered case (see Section 5), it is necessary to calculate the cable and transformer losses, which depend on the load current. For industry, the typical cable length for connecting low-voltage power equipment is about 0.1 km [61]. Many low-voltage feeders operate with loads that exceed those planned for the initial design and are close to the maximum allowable load due to the inevitable increase in consumer energy demand in 10-20 years and the delayed process of the upgrade of transmission lines [62]. Considering this fact, in low-power electrical facilities with a current load of up to 24 A, three-phase cables with PVC insulation and with a cross-section of 2.5 mm\({}^{2}\) are can be used [63]. The specific resistance of one phase of a copper cable with these parameters is approximately \(\rho_{cable}\) = 7.55 Ohm/km. The reactance in the calculation of stranded cables of small cross section is usually neglected. The phase resistance of such a cable will be \(R_{cable}=L_{cable}\rho_{cable}\) = 0.1 7.55 = 0.755 Ohm. Losses in the cable are calculated from (3). The losses in the transformer with the parameters of 400 V and 16 kVA are calculated based on the manufacturer's data as [64]:  
\[P_{T}=A+B\cdot(I_{load}/I_{T\ rate})^{2}, \tag{6}\]  
where \(I_{T\ rate}\) = 23.1 A--nominal phase current of the transformer; \(A\) = 40 W and \(B\) = 440 W are determined based on the value of the transformer losses at \(I_{load}\) = 0.5\(\cdot I_{T\ rate}\) and \(I_{load}\) = \(I_{T\ rate}\) (150 and 480 W, correspondingly, according to [64]). Figure 6 shows the losses in these elements, calculated by the Formulas (3) and (6). Using the above functions, the losses were calculated in various components of the pumping system (motors, cable, and transformer). Figure 7 shows the results of calculating these losses depending on the flowrate and the type of motor. \begin{table}
\begin{tabular}{c c c c c c c} \hline \hline \multirow{2}{*}{**Q, \%**} & \multirow{2}{*}{\(P_{mech}\), W} & \multicolumn{5}{c}{**\(I_{load}\), A**} \\ \cline{3-6}  & & **DOL SynRM** & **DOL PMSM** & **IE2 IM** & **IE3 IM** & **IE4 IM** \\ \hline
110 & 3453 & 22.0 & 19.3 & 22.3 & 22.3 & 20.8 \\
100 & 3335 & 21.4 & 18.8 & 21.7 & 21.7 & 20.3 \\
75 & 2962 & 19.5 & 17.5 & 20.1 & 20.1 & 18.8 \\ \hline \hline \end{tabular}
\end{table} 
Table 5: The current of three motors with various loading conditions of the duty cycle.