Electrical Power Systems Technology

304 Electrical Power Systems Technology

20,000 watts
= ——————————
480 volts × 1.73 × 0.75

= 32.11 amperes

Then, for the power per phase (PP), divide the 20,000 watts of the sys-
tem by 3 (the number of phases), for a value of 6,666.7 watts (6.66 kW).

Unbalanced Three-phase Loads
Often, thr ee-phase systems are used to supply power to both three-
phase and single-phase loads. If three, identical, single-phase loads were
connected across each set of power lines, the three-phase system would
still be balanced. However, this situation is usually difficult to accomplish,
particularly when the loads are lights. Unbalanced loads exist when the
individual power lines supply loads that are not of equal resistances or
impedances.
The total power converted by the loads of an unbalanced system
must be calculated by looking at each phase individually. Total power of a
three-phase unbalanced system is:

PT = PP–A + PP–B + PP–C

where the power-per-phase (PP) values are added. Power per phase is
found in the same way as when dealing with balanced loads:

PP = VP × IP × power factor

The current flow in each phase may be found if we know the power
per phase and the phase voltage of the system. The phase currents are
found in the following manner:

Sample Problem:
Given: the following 120-volt single-phase loads are connected to a
120/208-volt wye system. Phase A has 2000 watts at a 0.75 power factor,
Phase B has 1000 watts at a 0.85 power factor, and Phase C has 3000 watts
at a 1.0 power factor.
Find: the total power of the three-phase system, and the current flow
through each line.
Solution: To find the phase currents, use the formula PP = VP × IP ×