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# Electrical Machines – What is Synchronizing Power Coefficient?

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 (𝑋_{𝑠} >> 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,

𝑋

_{𝑑}is the direct-axis synchronous reactance, and𝑋

_{𝑞}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 =$\frac{6}{2}= 3$

**At no-load,**

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

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

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

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

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

The synchronizing power coefficient per mechanical degree is given by,

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

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

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

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