# Electrical Machines – What is Synchronizing Power Coefficient?

Electronics & ElectricalElectronDigital Electronics

When a synchronous machine is synchronised to infinite busbars, it has an inherent tendency to remain in synchronism.

• Rotor is accelerated – Consider an alternator delivering a steady power P at a steady load angle δ. Suppose, due to some transient disturbance, the rotor is accelerated so that the load angle increases by an angle $dδ$. The operating point of the machine shifts to a new constant power line and the load on the machine increases to (𝑃 + 𝑑𝑃)). As the power input to the machine remains unchanged, this additional load decreases the speed of the machine and brings it back to the synchronism.

• Rotor is decelerated – Similarly, if due to some transient disturbance, the rotor of the machine decelerates so that the load angle decreases by an angle dδ. Thus, the operating point of the machine shifts to a new constant power line and the load on the machine decreases to (𝑃 − 𝑑𝑃). Since the power input to the machine remains unchanged, the reduction in load accelerates the rotor, i.e., increases the speed of the machine. As a result, the machine again comes back to synchronism.

From the above discussion, it can be concluded that the effectiveness of restoring action depends upon the change in power transfer for a given change in load angle. The measure of this effectiveness is given by synchronising power coefficient.

Therefore,the synchronising power coefficient is defined as the rate at which the synchronous power (P) varies with the load angle (δ) of the synchronous machine. It is denoted by $𝑃_{syn}$

$$\mathrm{𝑃_{syn} =\frac{𝑑𝑃}{𝑑𝛿}… (1)}$$

The synchronising power coefficient is also known as stability factor, rigidity factor or stiffness of coupling.

The power output per phase of the cylindrical rotor synchronous generator is given by,

$$\mathrm{𝑃 =\frac{𝑉}{𝑍_{𝑠}}\left [𝐸_{𝑓}\:cos(𝜃_{𝑧} − 𝛿) − 𝑉\:cos\:𝜃_{𝑧}\right ]}$$

From Eqns. (1) & (2), we get,

$$\mathrm{𝑃_{𝑠𝑦𝑛}=\frac{𝑑𝑃}{𝑑𝛿}=\frac{𝑉𝐸_{𝑓}}{𝑍_{𝑠}}sin(𝜃_{𝑧} − 𝛿) … (3)}$$

In many synchronous machines (𝑋𝑠 >> 𝑅). Therefore,

• For a cylindrical rotor synchronous machine, the synchronizing power coefficient is given by,

$$\mathrm{𝑃_{𝑠𝑦𝑛} =\frac{𝑉𝐸_{𝑓}}{𝑋_{𝑠}}cos\:𝛿 … (4)}$$

• For a salient-pole synchronous machine, the power output is given by,

$$\mathrm{𝑃 =\frac{𝑉𝐸_{𝑓}}{𝑋_{𝑠}}sin\:𝛿 +\frac{1}{2}{𝑉^{2}}\left ( \frac{1}{𝑋_{𝑑}}-\frac{1}{𝑋_{𝑞}}\right )sin\:2𝛿 … (5)}$$

Where,

• 𝑋𝑑 is the direct-axis synchronous reactance, and

• 𝑋𝑞 is the quadrature-axis synchronous reactance.

Therefore, the synchronising power coefficient will be,

$$\mathrm{𝑃_{𝑠𝑦𝑛} =\frac{𝑉𝐸_{𝑓}}{𝑋_{𝑠}}cos\:𝛿 + {𝑉^{2}}\left ( \frac{1}{𝑋_{𝑑}}-\frac{1}{𝑋_{𝑞}}\right )cos\:2𝛿 … (6)}$$

## Numerical Example

A 3 MVA, 3-phase, 6-pole alternator is connected to 11 kV, 50 Hz busbars and has a synchronous reactance of 5 per phase. Calculate the synchronizing power coefficient per mechanical degree of rotor displacement at no-load. Assume normal excitation.

Solution

Total number of poles, 𝑃 = 6

∴ Number of pole pairs,𝑝 =$\frac{6}{2}= 3$

$$\mathrm{Load\:angle,\:𝛿 = 0°\:and\:𝐸_{𝑓} = 𝑉_{𝑝ℎ}}$$

$$\mathrm{Line\:voltage\:at\:busbars,\:𝑉_{𝐿} = 11000\:V}$$

$$\mathrm{Phase\:voltage,\:𝑉_{𝑝ℎ} =\frac{11000}{\sqrt{3}}V = 6351\:V}$$

$$\mathrm{∴ \:𝐸_{𝑓} = 𝑉_{𝑝ℎ} = 6351\:V}$$

$$\mathrm{𝑁_{𝑠} =\frac{120𝑓}{𝑃}=\frac{120 × 50}{6}= 1000\:RPM}$$

The synchronizing power coefficient per mechanical degree is given by,

$$\mathrm{𝑃_{syn} =\left (\frac{3𝑉_{𝑝ℎ}𝐸_{𝑓}}{𝑋_{𝑠}}cos\:𝛿\right)\left(\frac{𝑝𝜋}{180°}\right)}$$

$$\mathrm{\Rightarrow\:𝑃_{syn} =\frac{3 × 6351 × 6351}{5}× cos\:0° ×\frac{3 × 𝜋}{180°}}$$

$$\mathrm{\Rightarrow\:𝑃_{syn} = 1266525\:Watts\: per\:mech\:degree}$$