# Units and Significance of Synchronizing Power Coefficient

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## Units of Synchronizing Power Coefficient (𝑷𝐬𝐲𝐧)

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

$$\mathrm{𝑃_{syn} =\frac{𝑉 𝐸_{𝑓}}{𝑋_{𝑠}}cos\:𝛿 \:\:Watts/electrical\:radian …(1)}$$

$$\mathrm{∵ \:𝜋\:radians = 180\:degrees}$$

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

$$\mathrm{∵ \:𝑃_{syn}=\frac{𝑑𝑃}{𝑑𝛿}\:\:Watts/ \left(\frac{180}{𝜋}\:degrees \right)}$$

$$\mathrm{\Rightarrow\:𝑃_{syn}=\left( \frac{𝑑𝑃}{𝑑𝛿}\right)\left(\frac{𝜋}{180}\right)\:\:Watt/electrical\:degree …(2)}$$

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

$$\mathrm{𝜃_{electrical} = 𝑝 \cdot 𝜃_{mechanical}}$$

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

$$\mathrm{𝑃_{syn} = 𝑝 \cdot\left( \frac{𝑑𝑃}{𝑑𝛿}\right)\:\:Watts/mech. radian …(3)}$$

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

$$\mathrm{𝑃_{syn} =\left( \frac{𝑑𝑃}{𝑑𝛿}\right)\left(\frac{𝑝\:𝜋}{180}\right)\:Watts/mech.degree …(4)}$$

## Significance of Synchronizing Power Coefficient

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

$$\mathrm{𝑃_{syn} =\frac{3\:𝑉\:𝐸_{𝑓}}{𝑋_{𝑠}}cos\:𝛿 … (5)}$$

Equation (5) indicates that $𝑃_{syn}$ 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 $𝑃_{syn}$ is directly proportional to $𝐸_{𝑓}$, 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.

Updated on 19-Oct-2021 13:02:51