Page 145 - Solutions Manual to accompany Electric Machinery Fundamentals


Q  S 2 sin  2  1 cos PF   80 kW sin cos  1 0.8  48 kvar

The real, reactive, and apparent power supplied to the plant is


P TOT  P  1 P  2 100 kW 64 kW 164 kW

Q TOT  Q  1 Q  2 48.4 kvar 48 kvar  96.4 kvar

S TOT  P TOT 2  TOT 2   Q 164 kW 2    96.4 kVA 2  190 kVA

The line current is given by
S 190 kVA
I  TOT   228.5 A
L
3 L  V V  3 480

Note: In the following sections, we will treat the synchronous motor as though it
were Y connected. The problem doesn’t specify the connection, and the
answers are the same whether we assume a  or a Y connection, so we will
just choose one for convenience.


(b) The rated power of the motor is P motor  100 hp 746 W/hp  74.6 kW . Assuming that the
motor is Y-connected, the base quantities for the synchronous motor are
 
S  P / PF  74.6 kW / 0.8  93.2 kVA
base
V  460 V
L ,base
V  ,base  V L,base / 3  460 V / 3   266 V

S 93.2 kVA
I  base   117 A
L ,base
3V L ,base  V  3 460
I  ,base  I L ,base  117 A

3 V 2   3 266 V 2
Z   ,base   2.278 

,base
S base 93,200 VA
Therefore the impedances can be expressed in ohms per phase as

R  A Z  ,base R A ,pu  2.278   0.01  0.023 

X  S Z  ,base S ,pu   X 2.278    2.51 1.1 

The rated real and reactive power of the motor is
  P 100 hp 746 W/hp    74.6 kW
motor
 
Q motor  P tan  motor  1 cos PF  74.6 kW tan cos  1 0.8    56 kvar

Note that the motor is supplying reactive power, not consuming it.

The phase voltage would be V   V L / 3  480 V / 3   277 V , and magnitude of the armature
current would be

139