Units and Significance of Synchronizing Power Coefficient

Units of Synchronizing Power Coefficient (?_???)

Generally, the synchronizing power coefficient is expressed in Watts per electrical radian, i.e.,

$$\mathrm{?_{syn} =\frac{? ?_{?}}{?_{?}}cos\:? \:\:Watts/electrical\:radian …(1)}$$

$$\mathrm{? \:?\:radians = 180\:degrees}$$

$$\mathrm{\Rightarrow\:1\:radian =\frac{180}{?}\:degrees}$$

$$\mathrm{? \:?_{syn}=\frac{??}{??}\:\:Watts/ \left(\frac{180}{?}\:degrees \right)}$$

$$\mathrm{\Rightarrow\:?_{syn}=\left( \frac{??}{??}\right)\left(\frac{?}{180}\right)\:\:Watt/electrical\:degree …(2)}$$

If p is the total number of pole pairs in the machine, then

$$\mathrm{?_{electrical} = ? \cdot ?_{mechanical}}$$

Therefore, the synchronizing power coefficient per mechanical radian is given by,

$$\mathrm{?_{syn} = ? \cdot\left( \frac{??}{??}\right)\:\:Watts/mech. radian …(3)}$$

And, the synchronizing power coefficient per mechanical degree is given by,

$$\mathrm{?_{syn} =\left( \frac{??}{??}\right)\left(\frac{?\:?}{180}\right)\:Watts/mech.degree …(4)}$$

Significance of Synchronizing Power Coefficient

The synchronizing power coefficient ($?_{syn}$) is the measure of the stiffness of the electromagnetic coupling between the stator and the rotor. A large value of the synchronizing coefficient ($?_{syn}$) shows that the electromagnetic coupling is rigid. The synchronising power coefficient is given by,

$$\mathrm{?_{syn} =\frac{3\:?\:?_{?}}{?_{?}}cos\:? … (5)}$$

Equation (5) indicates that $?_{syn}$ is inversely proportional to the synchronous reactance of the machine. The synchronous machines with large air gaps have relatively small reactances. Thus, the coupling in a synchronous machine with large air gap is more rigid than a machine with smaller air gap.

Also, the $?_{syn}$ is directly proportional to $?_{?}$, hence, an over-excited synchronous machine is more stiff than an under-excited synchronous machine.

From Eqn. (5), it can also be seen that the restoring action is maximum when the load angle $\delta$ = 0°,, i.e., at no-load while the restoring action is zero when the load angle $\delta$ = ±90°. At these values of load angle ($\delta$), the machine would be at the steady state limit of stability and in the condition of unstable equilibrium. Thus, it is impossible to run a synchronous machine at the steady-state limit of stability because its ability to resist small changes is zero.