Question

e connected, 60 Hz, 2300 volt, sik pole, cylindrical rotor synchronous motor is 8 leading, neglect losses, the synchronous reactance is 2.5 sand Fp ohm/phase. Determine: (a) Synchronous speed-n, (b) Developed torque in Horsepower Tmec (c) Active Power-P (d) Apparent Power-S (e) Reactive Power-Q (f) Draw the Power Triangle (g) Armature current-1 (h) Excitation voltage-Et (i) Power Angle-8 U) Draw the phasor diagram

Answer 1

(a) The synchronous speed is given by

, 120 f

Given,

f = 60H

$P=6$

120 × 60 6

1200RPM

(b) As the motor is operating at rated condition and neglecting losses, the developed power would be the rated power which is given as

Pace = 500HP

Therefore, the developed torque in HP is given by

der = 500 H P

(c) The active power is equal to the developed power

So,

P = 500MP

P-500 × 745.711

P- 372850M

P-372,85K

(d) As the power factor is given as

$\cos \phi =0.8$ lead

Also, $\phi =36.86^{\circ}$

The apparent power is given by

$S=\frac{P}{\cos \phi }$

372.85 0,8

$S=466.06KVA$

(e) The reactive power is given by

$Q=S\sin \phi$  

$Q=466.06\times \sin 36.86^{\circ}$

$Q=279.64KVAR$

(f) Power triangle would be as shown in the figure

S-466.0 KVA = 7764 KVAR

(g)  $P=\sqrt{3}VI_a\cos \phi$

So,

$I_a=\frac{P}{\sqrt{3}V\cos \phi }$

$I_a=\frac{372.85\times 10^3}{\sqrt{3} \times 2300\times 0.8 }$

$I_a=116.99A$

$I_a=116.99\angle 36.86^{\circ}A$

(h) The excitation voltage for synchronous motor is given by

$\vec{E_f}=\vec{V}-j\vec{I_a}X_s$

$\vec{E_f}=\frac{2300}{\sqrt{3}}\angle 0-j(116.99\angle 36.86^{\circ})\times2.5$

$\vec{E_f}=\frac{2300}{\sqrt{3}}\angle 0-292.475\angle 126.86^{\circ}$

E, 1742.81しー15.570V

$\left | E_f \right |=1742.81V$

(i) The power angle as calculated above is

$\delta =-15.57^{\circ}$

(j) The phasor diagram is shown in the following figure