Page 571 - Handbook of Electrical Engineering

564 HANDBOOK OF ELECTRICAL ENGINEERING

√
The total VA rating for the cable = 3 × Rated line voltage × Rated phase current
√ 6
= 3 × 5000 × 250 = 2.165 × 10 VA
6
The VA rating for the cable per phase = 0.3333 × 2.165 × 10 = 721.67 × 10 3
5000
The 1.0 pu impedance of the cable per phase = V olp = √
3 × 250
= 11.547 ohms per phase.
Hence the per-unit impedance of this particular cable at its own base is:-

0.5145 + j0.1935
R pu + jX pu = = 0.04456 + j0.01676 pu
11.547
Convert this impedance to the system base:-

(base kVA) (cable rated voltage) 2
R cm + jX cm = (R pu + jX pu ) 2
(cable kVA) (system base voltage)
(0.04456 + j0.01676)(3125)(5000) 2
=
6
(2.165 × 10 )(13, 800) 2
= 0.00844 + j0.003175 pu
g) Motor running conditions (suffix ‘r’)

Motor rated voltage = 4000.0 volts
Motor system base voltage = 4181.8 volts
Motor terminal voltage = 4160.0 volts
Rated power output
Input power to each phase =
3 × efficiency
500 3
P mrp = × 10 = 175.44 kW
3 × 0.95
Rated power input
Input VA to each phase =
Power factor
175.44 3
S mrp = × 10 = 199.36 kVA
0.88
√ 2 2
Input VAr to each phase = (S mrp − S mrp )
√ 2 2
Q mrp = 1000.0 × (199.36 − 175.44 ) = 94.68 kVAr

At the motor rating base the phase ohmic resistance R mrp is:-
2
Phase voltage

R olp =
Phase active power
4000 × 4000
= = 30.4 ohms per phase
3 × 175.44 × 10 3