Electrical load diagram
Electrical load diagram:
The electrical load diagram is shown in Fig. The phasor V represents the constant voltage of the infinite bus-bars. At the extremity of V is drawn an axis showing the direction of the IaZs drops—i.e. the voltage drops for unity-power-factor output currents. This axis must be drawn at the angle θ = arc tan(Xs/r) to V , to scale along the axis is a distance corresponding to, say, full load at unity power factor.
At this point a line is drawn at right angles to the axis. It is the locus of the E values for constant power, or constant-electrical-power line. Other parallel lines are drawn for other loads, one through the extremity of V itself corresponding to zero power output, others on the right-hand side of V corresponding to negative power output, i.e. input to the machine as a motor.
Fig: Electrical load diagram
If the excitation be fixed, the extremity of the e.m.f. vector E, will have a circular locus as indicated by the circular arcs struck with O as centre. Taking 1.0 per unit E as that for which E = V on no load and no current, the per-unit excitation for any other loading condition can be found from the diagram. Thus with 1.5 per unit excitation, the machine will work on full-load power as a generator with a power factor of cos 80 lagging; on half-fun-load power with a power factor of cos 420 lagging; and on zero power output with a power-factor of zero lagging, as shown by the lines pa, pb and pc.
The variation of the power output (controlled by the input from the prime mover in the case of a generator and by the load applied to the shaft for a motor) with constant excitation is thl1S accompanied by changes in the load power factor. If the generator be provided with greater mechanical power with say, 150 per cent (or 1.5 per unit) excitation, then the output power increases with reducing power factor from lagging values until, with an output (for this case) of 1.2 per unit power, the power factor becomes unity. Thereafter the power increases with a reducing power factor-now leading.
Finally the excitation will not include any more constant-power lines, for the circle of its locus becomes tangential to these. If more power is supplied by the prime mover, the generator will be forced to rise out of step, and synchronous running will be lost. The maximum power that can be generated is indicated by intercepts on the limit of stability. The typical point Pemax on the left of the load diagram is for an excitation of 1.5 per unit.
Similarly, if a motor is mechanically overloaded it will fall out of step, because of its limited electrical power intake. The point Pemax in the motor region again corresponds to 1.5 per unit excitation, and all such points again lie on the limiting-stability line. This maximum power input includes I2R loss, and the remainder-the mechanical power output-in fact becomes itself limited before maximum electrical input can be attained.