This article offers a phasor diagram for one phase of a synchronous generator and explains the diagram.
System characteristics:
- Load: vt=240, s=20000VAr; pf=0.85; and
- Generator (s): s=20000VAr; pf=0.9; ra=0.04pu; xs=0.7pu.
.
Above is the phasor diagram for generator #1 in a pair of parallel generators. Green is flux, cyan is armature current, magenta is voltage.
Flux Φ would induces EMF E in the stator. Φar is the armature reaction flux, Φr is the resultant flux which induces Vr in the stator. Vt is the terminal voltage, it is the resultant of phasor E and Ia*(ra+jXs) (note that ra nd Xs are specified pu).
They are not load sharing equally, neither real nor reactive power. Generator 1 supplies 97.4% of the combined power, it is loaded to 16.6kW and has +10.3VAr reactive power. It illustrates the condition where generator #1 has just been paralleled by generator #2.
Some detailed statistics that were plotted to create the phasor diagram:
zs 1: ra: 4.000e-02pu, xs: 7.000e-01pu, rdv: 1.152e-01+2.016e+00johms Ratings 1: rs: 2.000e+04, rpf: 0.900, ri: 8.333e+01, rp: 1.800e+04, rdf: -1.111e-04, rdv: 1.152e-01+2.016e+00j zs 2: ra: 4.000e-02pu, xs: 7.000e-01pu, rdv: 1.152e-01+2.016e+00johms Ratings 2: rs: 2.000e+04, rpf: 0.900, ri: 8.333e+01, rp: 1.800e+04, rdf: -1.111e-04, rdv: 1.152e-01+2.016e+00j Load: v3: 2.400e+02, pf3: 0.850, i3: 7.083e+01-4.390e+01j, s3: 1.700e+04+1.054e+04j, p3: 1.700e+04, q3: 1.054e+04, z3: 2.448e+00+1.517e+00j,v1: 3.367e+02+1.377e+02j,|v1|: 3.637e+02,v2: 2.400e+02,|v2|: 2.400e+02 Synced: f01: 51.889, f02: 50.100, rdf1: -1.111e-04, p3: 1.700e+04 Sharing: f01: 51.889, f02: 50.100, sh1: 0.974, sh2: 0.026, fa: 50.050, fb: 50.050 (68.95833333333324-42.736536669047204j) (334.10085792479913+134.09675097572557j) (94.10085792479916+134.09675097572557j) Gen 1: v: 334.101+134.097j, |v:| 360.007, i: 68.958-42.737j, |i|: 81.127, s: 16550.000+10256.769j, pa: 21.869°, ta: 143.657° Gen 2: v: 242.559+3.646j, |v:| 242.586, i: 1.875-1.162j, |i|: 2.206, s: 450.000+278.885j, pa: 0.861°, ta: 122.650°
Note that the above uses v3 as the phase reference, the diagram uses E as the phase reference.
The diagram shows the meaning of the terms Torque Angle and Power (or Load) Angle. These terms are often used as equivalents. A hint to misuse in a generator context is that Torque Angle cannot be less than 90° and Power Angle must not be more than 90° (otherwise pole slipping occurs and the system is unstable).
rdv is the transformation of ra and Xs to equivalent impedance.
An exercise for the reader
What would the terminal voltage be if load current was reduced to zero?
How would you deal with this if the the generator was:
- operating in island mode as a stand-alone generator;
- synchronised to a grid; or
- operating in island mode, but in parallel with other generators?
Calculate the phasor Ia*(ra+jXs) and add it to the tip of E, did it end at Vr? Tip: use Ia*rdv1.
Conclusions
The phasor diagram for one phase of a three phase 60kVA synchronous generator carrying approximately 97% of a balanced 60kVA PF=0.85 shared load is shown and explained.