Electrical phase angle( please help me)?

Question

Electrical phase angle: a) Which significance has the electric phase angle for the generation of electrical energy? b) How does the phase angle (or its cosine) change when in a transmission network a short-circuit occurs?

Answer 1

(A)  When the reactive part of a load impedance causes the current to lead or lag the voltage across it then the real power in watts (I squared R) used by the load is less than the apparent power (V x I) furnished by the generator. This difference in power serves no usefull purpose, can not be stored and is in fact wasted volt amps that cost money to generate.

The angle whose cosine is ( I squared R) / ( V x I) is the phase angle between load current and load voltage. The cosine ( I squared R ) / ( V x I ) of this phase angle is also known as the power factor and is always equal to or less than (1).

Obviously a greater phase angle means a smaller cosine value and thus a smaller power factor. Since real power is equal to (apparent power) x (power factor) it is of great significance to the people who pay for the apparent power to keep the phase angle between load current and load voltage as small as possible.

(B)  When a short circuit occurs in a transmission line the phase angle between voltage and current goes to zero. The reason for this is that the load is no longer in the circuit so it`s power factor no longer has any influence on the circuit. Furthermore when a short circuit occurs, unless a circuit breaker opens the circuit, all the power that the source can deliver is dissipated in the resistance of the transmission line or the internal resistance of the source. Since the load is completely resistive after a short occurs there is no shift in current with respect to voltage.

On top of all that when a short occurs the least of any one`s worrys would be what the phase angle is at the moment between voltage and current.