# 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

$$\mathrm{π_{π π¦π} =\frac{1}{π_{π }}π\frac{ππ}{ππΏ}\:Nm/electrical\:radian …(1)}$$

Also,

$$\mathrm{π_{π π¦π} =\left (\frac{1}{π_{π }}π\frac{ππ}{ππΏ} \right)\cdot \frac{ππ}{180°}\:Nm/mechanical \:degree …(2)}$$

Where,

• m is the number of phase of the machine,

• ππ  = 2πππ  is the angular synchronous speed,

• ππ  is the synchronous speed in r.p.s.

• π is the total number of pair of poles of the machine.

The synchronising torque coefficient may also be given by,

$$\mathrm{π_{π π¦π} =\frac{ππ}{ππΏ}=\frac{1}{2ππ_{π }}\frac{ππ}{ππΏ}… (3)}$$

$$\mathrm{β΅\:\frac{ππ}{ππΏ}=π_{π π¦π} =\frac{ππΈ_{π}}{π_{π }}sin(π_{π§} − πΏ)}$$

$$\mathrm{∴\:π_{π π¦π} =\frac{ππΈ_{π}}{2ππ_{π }\cdot π_{π }}sin(π_{π§} − πΏ) … (4)}$$

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

$$\mathrm{π_{π π¦π} =\frac{ππΈ_{π}}{2ππ_{π } \cdot π_{π }}cos \:πΏ … (5)}$$

Since,

$$\mathrm{π_{π π¦π} =\frac{ππΈ_{π}}{π_{π }}sin(π_{π§ }− πΏ)\:\:\:and\:\:\:π_{π } = 2ππ_{π }}$$

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

$$\mathrm{π_{π π¦π} =\frac{π_{π π¦π}}{2ππ_{π }}=\frac{π_{π π¦π}}{π_{π }}… (6)}$$

## 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,

$$\mathrm{π_{π } =\frac{2π}{π}=\frac{2 × 50}{4}= 25 r. p. s.}$$

Synchronising torque,

$$\mathrm{π_{π π¦π} =\frac{π_{π π¦π}}{2ππ_{π }}=\frac{332.48 × 1000}{2π × 25}= 2117.71\:Nm}$$

Updated on: 13-Oct-2021

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