0195136047.pdf

10.3 POWER TRANSMISSION AND DISTRIBUTION 465

of the system is underutilized. Utilities often define load periods in terms ofon-peakandoff-peak
hours. In order to level the demand by diverting a portion of the energy usage from the on-peak
to the off-peak periods, economic incentives (such as lower electric rates for the sale of off-
peak energy) are generally offered.

EXAMPLE 10.3.1

A three-phase, 34.5-kV, 60-Hz, 40-km transmission line has a per-phase series impedance of
0. 2 +j 0. 5 /km. The load at the receiving end absorbs 10 MVA at 33 kV. Calculate the following:

(a) Sending-end voltage at 0.9 PF lagging.

(b) Sending-end voltage at 0.9 PF leading.

(c) Transmission system efficiency and transmission-line voltage regulation corresponding

to cases (a) and (b).

Solution

The per-phase model of the transmission line is shown in Figure E10.3.1.

VS LN

IS = IR IR

R + jX

8 + j 20 Ω Load

+

−

+

−

VRLN =^33

√− 3

∠ 0 ° kV

Figure E10.3.1Per-phase model of transmission line (with only series impedance).

V ̄R=^33 √,^000

3

 0°= 19 , 052  0° V

(a) I ̄R=

10 × 106

√

3 × 33 × 103

 −cos−^10. 9 = 175  − 25 .8° A

V ̄SLN= 19 , 052 0°+( 175  − 25 .8°)( 8 +j 20 )= 21 , 983  6 .6°VLN

VSLL= 21. 983

√

3 = 38 .1 kV (line - to - line)

(b) I ̄R=

10 × 106

√

3 × 33 × 103

 +cos−^10. 9 = 175  + 25 .8° A

V ̄SLN= 19 , 052  0°+( 175  + 25 .8°)( 8 +j 20 )= 19 , 162  11 .3° VLN

VSLL= 19. 162

√

3 = 33 .2 kV (line - to - line)

(c) (i) 0.9 PF lagging:

PR= 10 × 0. 9 =9MW

PS=

√

3 × 38. 1 × 0 .175 cos( 25. 8 + 6. 6 )°= 9 .75 MW