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Power-Angle Characteristics of Salient Pole Synchronous Machine
The resistance $R_{a}$ of the armature can be neglected since it has negligible effect on the relationship between the power output of a synchronous machine and its torque angle $\delta$. The phasor diagram at lagging power factor for a salient pole synchronous machine, neglecting $R_{a}$ is shown in Figure-1. The power-angle characteristics of a salient-pole machine may be derived from the phasor diagram.
The complex power output per phase of the alternator is,
$$\mathrm{π_{1π} =π{πΌ^{*}_{π}}… (1)}$$
Taking excitation voltage ($E_{f}$) as the reference phasor, then,
$$\mathrm{π = π\angle − \delta = π\:cos\:\delta − ππ\:sin\:\delta … (2)}$$
$$\mathrm{πΌ_{π} = πΌ_{π} − ππΌ_{π}}$$
$$\mathrm{∴\:{πΌ^{*}_{π}}= πΌ_{π} + ππΌ_{π} … (3)}$$
Hence, from Eqns. (1), (2) & (3), we get,
$$\mathrm{π_{1π} = (π\:cos\:\delta − ππ\:sin\:\delta)(πΌ_{π} + ππΌ_{π}) … (4)}$$
From the phasor diagram, we obtain,
$$\mathrm{πΆπ· = π΄π = πΌ_{π}π_{π} = π\:sin\:\delta}$$
$$\mathrm{∴\:πΌ_{π} =\frac{π\:sin\:\delta}{π_{π}}… (5)}$$
$$\mathrm{π΄πΆ = ππ· = ππ· − ππ = πΌ_{π}π_{π} = πΈ_{π} − π\:cos\:\delta}$$
$$\mathrm{∴\:πΌ_{π} =\frac{πΈ_{π} − π\:cos :\delta}{π_{π}}… (6)}$$
Substituting the values of $πΌ_{π}$ and $πΌ_{π}$ in Eqn. (4), we have,
$$\mathrm{π_{1π} = (π\:cos\:\delta − ππ\:sin\:\delta)\left(\frac{π\:sin\:\delta}{π_{π}}+ π\frac{πΈ_{π} − π\:cos\:\delta}{π_{π}}\right)}$$
$$\mathrm{\Rightarrow\:π_{1π} =\left(\frac{π^{2}}{π_{π}}sin\:\delta\:cos\:\delta +\frac{ππΈ_{π}}{π_{π}}sin\:\delta −\frac{π^{2}}{π_{π}}sin\:\delta\:cos\:\delta\right)}$$
$$\mathrm{+ π\left(\frac{ππΈ_{π}}{π_{π}}cos\:\delta −\frac{π^{2}}{π_{π}}cos^{2}\:πΏ −\frac{π^{2}}{π_{π}}sin^{2}\:\delta\right)}$$
$$\mathrm{\Rightarrow\:π_{1π} =\left[\frac{ππΈ_{π}}{π_{π}}sin\:\delta+\frac{π^{2}}{2} \left(\frac{1}{{π_{π}}}-\frac{1}{{π_{π}}}\right)sin \:2\delta\right]}$$
$$\mathrm{+ π\left[\frac{ππΈ_{π}}{π_{π}}cos\:\delta -\frac{π^{2}}{2π_{π}}(1 + cos\:2\delta) −\frac{π^{2}}{2π_{π}}(1 − cos\:2\delta)\right]}$$
$$\mathrm{\Rightarrow\:π_{1π} =\left[\frac{ππΈ_{π}}{π_{π}}sin\:\delta +\frac{π^{2}}{2} \left(\frac{1}{{π_{π}}}-\frac{1}{{π_{π}}}\right)sin\: 2\delta\right]}$$
$$\mathrm{+ π\left[\frac{ππΈ_{π}}{π_{π}}cos\:\delta-\frac{π^{2}}{2π_{π}π_{π}}\{(π_{π} + π_{π} ) − (π_{π} − π_{π})\:cos\:2\delta\} \right]… (7)}$$
Also,
$$\mathrm{π_{1π} = π_{1π} + ππ_{1π} … (8)}$$
Comparing Eqns. (7) & (8), we obtain,
For 3-phase system,
$$\mathrm{π_{1π} =\frac{ππΈ_{π}}{π_{π}}sin\:\delta +\frac{π^{2}}{2} \left(\frac{1}{{π_{π}}}-\frac{1}{{π_{π}}}\right)sin\: 2\delta … (9)}$$
Reactive power per phase,
$$\mathrm{π_{1π} =\frac{ππΈ_{π}}{π_{π}}cos\:\delta-\frac{π^{2}}{2π_{π}π_{π}}\left[(π_{π} + π_{π} ) − (π_{π} − π_{π})\:cos\:2\delta \right]… (10)}$$
For 3-phase system,
$$\mathrm{π_{3π} = 3π_{1π} =\frac{3ππΈ_{π}}{π_{π}}sin\:\delta +\frac{3π^{2}}{2} \left(\frac{1}{{π_{π}}}-\frac{1}{{π_{π}}}\right)sin\: 2\delta… (11)}$$
$$\mathrm{π_{3π} = 3π_{1π} =\frac{3ππΈ_{π}}{π_{π}}cos\:\delta −\frac{3π^{2}}{2π_{π}π_{π}}\left[(π_{π} + π_{π} ) − (π_{π} − π_{π})\:cos\:2\delta \right]… (12)}$$
Equations (11) & (12) are applicable to both salient-pole synchronous generator and motor. The torque angle (\delta) is positive for the synchronous generator and negative for the synchronous motor.
Figure-2 and Figure-3 show the (P- $\delta$) and (Q-$\delta$) curves respectively for a salient pole synchronous machine. Also, Figure-4 shows the power-angle curve for a salient pole synchronous machine. From the Figure-4, it is to be noted that peak-power or steady-state limit occurs at a value of δ less than 90°. The maximum value of load angle ($\delta_{max}$) depends upon the relative magnitudes of $V, E_{f}$ and saliency.
Also, for a cylindrical rotor synchronous machine, the 3-phase real power is given by,
$$\mathrm{π_{3π} =\frac{3ππΈ_{π}}{π_{π}}sin\:\delta… (13)}$$
Again, the electromagnetic torque developed by the synchronous machine is given by,
$$\mathrm{τ_{π} =\frac{3π_{1π}}{π_{π}}= \frac{3}{2ππ_{π }}\left[\frac{ππΈ_{π}}{π_{π}}sin\:\delta +\frac{π^{2}}{2} \left(\frac{1}{{π_{π}}}-\frac{1}{{π_{π}}}\right)sin\: 2\delta \right]… (14)}$$
From eqn. (13), it can be seen that the developed torque has two components – excitation torque and reluctance torque. The first term represents the torque due to field excitation and is given by,
$$\mathrm{π_{ππ₯π} =\frac{3ππΈ_{π}}{2ππ_{π }π_{π}}sin\:\delta… (15)}$$
And the second term is known as reluctance torque and is given by,
$$\mathrm{τ_{πππ} =\frac{3π^{2}}{4ππ_{π }}\left(\frac{1}{{π_{π}}}-\frac{1}{{π_{π}}}\right)sin\:2\delta… (16)}$$
The reluctance torque of the machine is independent of the field excitation and exists only if the synchronous machine is connected to a system receiving reactive power from other synchronous machines operating in parallel with the terminal voltage V. Actually, the reluctance torque is due to the saliency of the field poles which tend to align the direct axis with the axis of the armature MMF.