Page 140 - Solutions Manual to accompany Electric Machinery Fundamentals

(b) The tor que angle  of this machine is –33.7.
(c) The static stability power limit is given by

3VE  3 120 V 144 V 
P max   A   32.4 k W
X S  1.6 

(d) A phasor diagram of the motor operating at a power factor of 0.78 leading is shown below.
 P
}
I
A2
I V 
A1

jX I }  P
S A

E E
A1 A2

Since the power supplied by the motor is constant, the quantity I A cos  , which is directly proportional
to power, must be constant. Therefore,


I 0.8    50 A 1.00
A 2


 I 62.5 36.87 A
A 2
The internal generated voltag required to produce this current would be
e
 E  V R I jX I
A 2  A A 2 S A 2




E 120 0  j  V  1.6   62.50 36.87 A
A 2

E 197  23.9 V
A 2
The internal generated voltage E A is directly proportional to the field flux, and we have assumed in this
problem that the flux is directly proportional to the field current. Therefore, the required field current is
E 197 V
I  A 2 I   2.7 A  3.70 A
2
F
E A 1 F 1 144 V
(e) The new torque angle  of this machine is –23.9.
5 -8. A 4.12 kV, 60 Hz, 3000-hp 0.8-PF-leading, Δ-conn ected, three-phase synchronous motor has a
synchronous reactance of 1.1 per unit and an armature re sistance of 0.1 per unit. If this motor is running
at rated voltage with a line current of 300 A at 0.85 PF leading, what is the internal generated voltage per
phase inside this motor? What is the torque angle δ?
SOLUTION


The output power of the motor is  3000 hp 746 W/hp  2238 kW . If we take this as rated power,
the ratings of this machine are
 
S base  P / PF  2238 kW / 0.8  2798 kVA
V L ,base  4120 V

V  ,base  4120 V

134