Mechanical Power Developed by a Synchronous Motor

Electronics & ElectricalElectronDigital Electronics

Consider an under-excited (i.e., Ef < V), 3-phase cylindrical rotor synchronous motor driving a mechanical load. The figure shows the per phase phasor diagram of the motor. Since the motor is under-excited, it will be operating at a lagging power factor cos φ.

In practice, for a synchronous motor, XS>>Ra, then the armature resistance (Ra) of the motor can be neglected. Since Ra is neglected, the armature copper loss will be zero. Therefore, the mechanical power developed (Pm) by the synchronous motor is equal to the input power (Pi) to the motor.

Also, Ra= 0, Er=IaXS, thus the armature current (Ia) lags the resultant voltage (Er) by 90°.

Input power per phase,

$$\mathrm{P_{i}=V\:I_{a}\:cosφ}$$

Therefore, the mechanical power developed per phase is,

$$\mathrm{P_{m}=P_{i}=V\:I_{a}\:Cosφ\:\:\:\:\:\:...(1)}$$

Referring to the phasor diagram of the motor, we get,

In triangle ABO,

$$\mathrm{AB=E_{r}\:Cosφ=I_{a}X_{S}\:Cosφ}$$

And, in triangle ABC,

$$\mathrm{AB=E_{f}\:Sin(180°-δ)=E_{f}\:Sinδ}$$

$$\mathrm{\therefore\:I_{a}X_{S}\:Cosφ=E_{f}\:Sinδ}$$

$$\mathrm{\Longrightarrow\:I_{a}Cosφ=\frac{E_{f}}{X_{S}}\:Sinδ\:\:\:\:\:\:...(2)}$$

Substituting the value of IaCosφ in Equation (1), we get,

$$\mathrm{P_{m}=\frac{VE_{f}}{X_{S}}\:Sinδ\:\:\:\:\:\:...(3)}$$

Equation (3) gives the mechanical power developed per phase of the synchronous motor.

The total mechanical power developed for the three phases is,

$$\mathrm{P_{m}=\frac{3VE_{f}}{X_{S}}\:Sinδ\:\:\:\:\:\:...(4)}$$

Condition for Maximum Mechanical Power Developed

For maximum mechanical power developed of the synchronous motor,

$$\mathrm{\frac{dp_{m}}{dδ}=0\:and\:\frac{d^2p_{m}}{dδ^2}<0}$$

$$\mathrm{\Longrightarrow\:\frac{d}{dδ}\left [\frac{VE_{f}}{X_{S}}Sinδ\right ]=0}$$

$$\mathrm{\Longrightarrow\:\frac{VE_{f}}{X_{S}}\:Cosδ=0}$$

$$\mathrm{\Longrightarrow\:Cosδ=Cos90°}$$

$$\mathrm{\therefore\:δ=90°}$$

Therefore, the mechanical power developed will be maximum when torque angle is equal to 90° (electrical). The maximum power developed is given by,

$$\mathrm{P_{m(max)}=\frac{VE_{f}}{X_{S}}...per phase}$$

$$\mathrm{P_{m(max)}=\frac{3VE_{f}}{X_{S}}...per phase}$$

Numerical Example

A 3-phase, 5000 kW, 11 kV, 200 RPM, 50 Hz synchronous motor has per phase synchronous reactance of 1.5 Ω. At full-load, the torque angle of the motor is 23° electrical. If the excitation EMF is 3.4 kV. Calculate the mechanical power developed and maximum mechanical power developed by the motor.

Solution

Terminal voltage per phase,

$$\mathrm{V=\frac{11000}{\sqrt{3}}=6351.04V}$$

Mechanical power developed,

$$\mathrm{P_{m}=\frac{3VE_{f}}{X_{S}}Sinδ=\frac{3\times6351.04\times3400}{1.5}\times\:Sin23°}$$

$$\mathrm{\Longrightarrow\:P_{m}=16.875 MW}$$

The maximum developed mechanical power is,

$$\mathrm{P_{m(max)}=\frac{3VE_{f}}{X_{S}}=\frac{3\times6351.04\times3400}{1.5}= 43.187 MW}$$

raja
Updated on 28-Oct-2021 11:35:20

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