How does reactive power effect system voltage

[rockman7892]

I understand the general concepts that a decrease in reactive power will cause system voltage to decrease and that an increase in reactive power will cause an increase in system voltage but for some reason I am not able to see this analytically. For instance I know that when capacitors are added to a system or that when synchronous generators are overexcited there is an increase in reactive power (MVARS) which in turn can increase the system voltage. Conversely I know that when there is not enough reactive power in the system to supply inductive loads such as motors, transformers, etc... then the system voltage will drop.

Is there a way that someone can explain this to me analytically? Perhaps I am just missing something simple. Am I able to somehow see it by drawing vectors that include voltage vectors with leading or lagging current vectors?

Thanks

 

[waross]

Reactive power causes reactive voltage drops, or voltage drops at 90 degrees to the applied voltage.
regardless of the phase angle of a current, it causes I²R losses.
Depending on the type of reactive current or power (capacitive or inductive) and the dirrection of the reactive current, the effect may be a voltage drop or a voltage rise. There are instances when the flow of reactive power may be the opposite direction to the flow of real power.
Vectors for reactive currents are 90 degrees out of phase with the applied voltage. However the phase angle of the load current has an effect on the total voltage drop.
The addition of capacitors may cause a reactive voltage drop.
The addition of capacitors may cause a cancellation of part or all of an inductive voltage drop with a net voltage rise.
The addition of capacitors may cause increased excitation of the supply generator with an attending rise in voltage.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[fpietryga]

Consider a simple series circuit consisting of an AC voltage source, resistor, inductor and a load (either purely resistive or purely capacitive). Select the component values such that in Case 1 (resistive load), the current lags the voltage (lagging power factor), and for Case 2 (capacitive load), the current leads the voltage (leading power factor). Now assume the voltage across the load to be 1 per unit and solve for the voltage required by the AC voltage source using vector math and KVL. For Case 1, the inductance will draw lagging vars from the source and more than 1 per unit voltage will be required from the source to overcome the voltage drop to maintain 1 per unit voltage across the load. For Case 2, the capacitor will provide leading vars to the circuit and less than 1 per unit voltage will be required from the source since the capacitor will cause a voltage rise at the load. Hope this helps some!

 

[jghrist]

Changing the current angle by increasing the load power factor with capacitors will reduce the amount of source voltage VS needed to get the load voltage VR. This can be seen by drawing the voltage and voltage drop vectors. See attached diagram. In the first vector diagram, the current lags the load voltage by the angle theta. In the second vector diagram, the power factor is improved to unity and theta = 0. Graphically, you can see that for the same VR, the second diagram has a smaller VS, meaning less voltage drop. If the IR-IX voltage drop vectors are rotated even more counterclockwise, representing the load current leading the load voltage, VS will get smaller and after a lot of rotation, it will be smaller than VR (voltage rise from VS to VR).

The voltage rise can be approximated by:

% volt rise = (kvar·XL)/(10·kV²)

Where: XL = the inductive reactance of the line (ohms)
kvar = three-phase capacitor size
kV = line-to-line voltage

This voltage rise is above the voltage as reduced by load voltage drop. That is, if there is 5% voltage drop with no capacitors, and 6% voltage rise caused by the capacitors, then the net voltage rise will be 1%.

 

[constantlylearning]

Without going into phasor diagrams, I'll give you the 10-cent version of this. If you have a high-current load that is primarily inductive, the high-magnitude current is going to cause the voltage to drop between the source and the load due to the fact that the lines supplying the load have impedance. If you can supply the reactive current very close to the load (using a cap bank, for instance), the reactive part of the current does not have to travel the length of the line between the source and the load, thus it does not contribute to the voltage drop.

 

[waross]

I believe that rockman7892 has generating capability. With more than one source of power simple solutions may not be adequate. It is possible and not uncommon for reactive power to flow in the opposite direction to real power. Reactive power may have an effect on the excitation and resulting voltage.
When a generator is connected to a relatively large grid:
The throttle setting determines the output of real power.
The voltage setting determines the amount and direction of reactive power produced.
The grid determines the voltage.
In one instance the addition of capacitors or the production of leading reactive power may cancel reactive voltage drops and improve the voltage.
In another instance and under light load conditions, the addition of capacitors may and the induction of the supply transformer may act as a series LC circuit and cause voltages high enough to damage equipment.
Many years ago power factor was measured and penalties assessed based on the ratio of a months reactive power consumption to real power consumption. It was common to correct plant power factors by connecting an unswitched bank of capacitors 24/7.
During the night when the major load was just a part of the normal lighting the voltage may rise high enough to cause rapid failure of the lights.
Note: These are two distinct and separate effects.
I have seen an instance where diesel generators were run unloaded as synchronous condensers to cancel the reactive voltage drop of the transmission line feeding a city so as to improve the supply voltage regulation for the city.

Bill
--------------------
"Why not the best?"
Jimmy Carter