# What is Synchronizing Torque Coefficient?

The synchronising torque coefficient is defined as the torque at which the synchronous speed gives the synchronising power. If represents the synchronising torque coefficient, then

$$\mathrm{𝜏_{𝑠𝑦𝑛} =\frac{1}{𝜔_{𝑠}}𝑚\frac{𝑑𝑃}{𝑑𝛿}\:Nm/electrical\:radian …(1)}$$

Also,

$$\mathrm{𝜏_{𝑠𝑦𝑛} =\left (\frac{1}{𝜔_{𝑠}}𝑚\frac{𝑑𝑃}{𝑑𝛿} \right)\cdot \frac{𝜋𝑝}{180°}\:Nm/mechanical \:degree …(2)}$$

Where,

• m is the number of phase of the machine,

• 𝜔𝑠 = 2𝜋𝑛𝑠 is the angular synchronous speed,

• 𝑛𝑠 is the synchronous speed in r.p.s.

• 𝑝 is the total number of pair of poles of the machine.

The synchronising torque coefficient may also be given by,

$$\mathrm{𝜏_{𝑠𝑦𝑛} =\frac{𝑑𝜏}{𝑑𝛿}=\frac{1}{2𝜋𝑛_{𝑠}}\frac{𝑑𝑃}{𝑑𝛿}… (3)}$$

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

$$\mathrm{∴\:𝜏_{𝑠𝑦𝑛} =\frac{𝑉𝐸_{𝑓}}{2𝜋𝑛_{𝑠 }\cdot 𝑍_{𝑠}}sin(𝜃_{𝑧} − 𝛿) … (4)}$$

In many synchronous machines, Xs >>R. Therefore, for a cylindrical rotor machine, neglecting the saturation and stator resistances, the eqn. (4) becomes,

$$\mathrm{𝜏_{𝑠𝑦𝑛} =\frac{𝑉𝐸_{𝑓}}{2𝜋𝑛_{𝑠} \cdot 𝑋_{𝑠}}cos \:𝛿 … (5)}$$

Since,

$$\mathrm{𝑃_{𝑠𝑦𝑛} =\frac{𝑉𝐸_{𝑓}}{𝑍_{𝑠}}sin(𝜃_{𝑧 }− 𝛿)\:\:\:and\:\:\:𝜔_{𝑠} = 2𝜋𝑛_{𝑠}}$$

Hence, Eqn. (4) may also be written as,

$$\mathrm{𝜏_{𝑠𝑦𝑛} =\frac{𝑃_{𝑠𝑦𝑛}}{2𝜋𝑛_{𝑠}}=\frac{𝑃_{𝑠𝑦𝑛}}{𝜔_{𝑠}}… (6)}$$

## Numerical Example

A 4-pole, 50 Hz, 3-phase, turbo alternator is excited to generate the busbar voltage of 11 kV on no-load. The machine is star-connected and the synchronising power of the machine is 332.48 kW per mechanical degree. Calculate the synchronising torque of the machine.

Solution

The synchronous speed of the machine in r.p.s. is,

$$\mathrm{𝑛_{𝑠} =\frac{2𝑓}{𝑃}=\frac{2 × 50}{4}= 25 r. p. s.}$$

Synchronising torque,

$$\mathrm{𝜏_{𝑠𝑦𝑛} =\frac{𝑃_{𝑠𝑦𝑛}}{2𝜋𝑛_{𝑠}}=\frac{332.48 × 1000}{2𝜋 × 25}= 2117.71\:Nm}$$

Updated on: 13-Oct-2021

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