Combined Delta and Wye Loads

At its worst, a problem of calculating line currents may involve one or more delta loads, one or more wye loads, and one or more single-phase loads connected between one line and any other line, including the neutral line if there is one. Unbalanced wye-connected loads are extremely rare, and their solutions are relatively so complicated that they will be neglected entirely in the general case of unbalance in line currents. Every problem involving unbalanced delta-connected loads and miscellaneous single-phase loadings can be solved by the methods used with the balanced delta-connected load; but, inasmuch as currents at a delta point will almost never be just 60° apart in phase, the factor

cannot be used.

Example 9-1. - As indicated in Fig. 9-4 (a), a factory load consists of the following: One department with seven 10-hp motors, 3-phase, 60-cycle, 230-volt, all of which together draw 190 amp per line lagging their respective line-to-neutral voltages by 58°. One department with a lighting load of 90 amp at 100 per cent power factor between lines 1 and 2. One department with a total load of 50 amp at 70.7 per cent power factor, lagging, connected between lines 2 and 3. One department with a total load of 80 amp at 80 per cent power factor, leading, connected between lines 3 and 1. Calculate the line currents I₁, I₂, and I₃ in lines 1, 2, and 3 with a phase sequence of E₃2, E₂1, E₁3

Solution. - The vector diagram is shown in Fig. 9-4 (b). The currents in line 1 may be combined as follows:

If the phase sequence of the voltages applied to the load is reversed by reversing any two of the leads, then the currents in those lines will be reversed. In assembling the vectors for addition at any corner of the voltage triangle to find the corresponding line current, all current vectors associated with voltage vectors positive to the point must be taken as positive and all current vectors associated with voltage vectors negative to the point must be taken as negative (reversed).