# Synchronous generator – phasor diagram and explanation

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.