Question

A 3 phase, 11 kV, 25 MVA, 60 Hz, salient pole, synchronous motor is running from a 11 KV, 60 Hz, balanced three-phase supply. The machine reactances are Xd=1.2 p.u. Y, = 0.6 p.u (with the machine rating as base). Neglect mechanical losses and armature copper losses. The maximum power input is obtained with no field excitation. Determine the power factor of the motor in this condition .​

Answer 1

Given:

The machine reactance

$X_{d}=1.2\ p.u.$

$X_{q} =0.6 p.u.$

To Find:  

The power factor of the motor.

Solution:

The equation of active power in salient pole synchronous motor

$P=\frac{E_{f}V_{t} }{X_{d} }sin\delta+\frac{1}{2}V_{t}^{2}sin^{2}(2\delta)(\frac{1}{X_{q} }-\frac{1}{X_{d} })$

$E_{f}$ is the excitation emf

$V_{t}$ is the terminal voltage per phase

$X_{d}$ is the d-axis reactance

$X_{q}$ is the q-axis reactance

$\delta$ is the power angle

It is given that field excitation is zero so $E_{f}=0$

$P=\frac{1}{2}V_{t}^{2}sin(2\delta)(\frac{1}{X_{q} }-\frac{1}{X_{d} })$  

Power is maximum so

$sin(2\delta)=1\\\delta=45^{\circ}$

The equation becomes

$P=\frac{1}{2}V_{t}^{2}(\frac{1}{X_{q} }-\frac{1}{X_{d} })$

The equation of reactive power of the synchronous motor work at lagging power factor

$Q=-\frac{E_{f} V_{t} }{X_{d} } cos\delta+\frac{V_{t}^{2} }{X_{d} }+V_{t}^{2}sin\delta(\frac{1}{X_{q} }-\frac{1}{X_{d} })$

Put $E_{f}=0$ in the above equation

$Q=\frac{V_{t}^{2} }{X_{d} }+V_{t}^{2}sin^{2}\delta(\frac{1}{X_{q} }-\frac{1}{X_{d} })$

$\delta=45^{\circ}\\sin45^{\circ}=\dfrac{1}{\sqrt{2} }\\sin^{2}45^{\circ} =\dfrac{1}{2}$

$Q=\dfrac{V_{t}^{2} }{2}(\frac{1}{X_{q} }+\frac{1}{X_{d} })$

Divide Q with P we get

$\dfrac{Q}{P}=\dfrac{X_{q}+X_{q} }{X_{d}-X_{q} }$ 

Put$X_{d}=1.2 \ and \ X_{q}=0.6$

$\dfrac{Q}{P}=\dfrac{1.2+0.6}{1.2-0.6}\\=\dfrac{1.8}{0.6}\\=3$

Assume $\Phi$ is the power factor angle

$\tan \Phi=\dfrac{Q}{P}\\\tan \Phi=3\\\Phi=\tan^{-1}\ (3)\\=71.56^{\circ}$

Power factor

$cos\Phi=cos(71.56^{\circ})\\=0.316 \ lagging$

The power factor of the motor is 0.316.