Armature Reaction, The Effect Of Reactive Load 1

[cdol]

Am I correct in thinking that a capacitive load on a generator has the effect of magnetizing the feild, and inductive loads demagnetize the field and that this is due to armature reaction. What is happening here and why do these type of loads cause this to happen.

Many thanks.

 

[electricpete]

I think that inductive load will require more excitation, capacitive will require less excitation (for constant terminal voltage). For inductive load the generator must have higher internal generated voltage since voltage drop accross generator internal inductance is vectorially significant when feeding inductive load. Opposite effect when feeding capacitive load.

 

[jbartos]

Suggestions:
1. There are some good postings pertaining to your posting. They may be retrieved over the key word search.
2. Reference:
a) Slemon G. R. "Magnetoelectric Devices, Transducers, Transformers, and Machines," John Wiley and Sons, Inc., New York, 1966
Reference a) 5.5 Polyphase Synchronous Machines includes Figure 5.38 "The terminal voltage |Es| as a function of field current "if", with no-load and zero-power factor loads at constant stator current Is."
This Figure shows the dependence of a capacitive load curve on field current "if", inductive load curve dependent on field current "if" and open-circuit curve dependent on field current "if". Essentially, at the stator terminal voltage |Es|=constant, a capacitive load (curve) will necessitate the smallest "if", open-circuit curve (no-load) will have "if" somewhere between the capacitive load curve point and inductive load curve point, and a inductive load (curve) will necessitate the highest "if". However, the best way to understand the phenomena is to study Reference a) or similar ones.

 

[electricpete]

The result posted by jbartos is the same one I explained by circuit analysis.

I'm using a simple model of a generator as an internal ideal voltage source (dependent on excitation current), and an internal impedance (indctive).

If it's an inductive load, then the internal impedance (inductive) of the generator has a similar angle to the imepedance of the load. Therefore voltage drop accross the internal impedance is maximum (compared to voltage drop for load at any angle different than internal impedance), and you have to boost the internal voltage (more field current) to achieve the same terminal voltage.

If it's a pure resistive load, then internal impedance of generator is approx at 90 degrees away from impedance of the load. Terminal voltage magnitude does not drop below nearly as much below internal voltage as it did in the inductive load case. Lower internal voltage is required for same terminal voltage, and less excitation current.

If it's capacitive load, then the inductive internal impedance is more than 90 degrees away from the load impedance and there will likely be a voltage magnitude increase accross the internal impedance (terminal voltage is higher than internal voltage). In this case even lower internal voltage is required and even less field excitation than for resistive.

Try drawing the vector diagrams for each of these three cases.