Electrical Power Systems

166 Electrical Power Systems
\ Q 31 = 30.0629 – 31.0098 = –0.9469
\ Q 23 = (1.00099) 2 × 35.77 sin (116.6° ) – 1.00099 × 0.98265 × 35.77
sin (116.6° + 2.68° – 3.048°)
\ Q 23 = 32.0472 – 31.5606 = 0.4866 pu MVAr.
Reactive power loss in line 1-2, 1-3 and 2-3.
\ QLoss 12 = Q 12 + Q 21 = 0.8948 – 0.746 = 0.1488 = 14.88 MVAr
\ QLoss 13 = Q 13 + Q 31 = 1.088 – 0.9469 = 0.1411 = 14.11 MVAr
\ QLoss 23 = Q 23 + Q 32 = –0.4746 + 0.4866 = 0.012 = 1.2 MVAr
Note that all the results are computed after 2nd iteration and details calculation are given
for the purpose of understanding.
Example 7.3: Solve problem Ex-7.2 considering bus 2 is P-|V| bus. Details are given in
ig. 7.8. Use same line and load data as given in Table 7.3 and 7.2.
Solution
ig. 7.8: Sample power system.
P2 = –2.556, P3 = –1.386, Q3 = –0.452
(0) V1 = (1.05 + j0), V2 = 1.0 + j0.0 Þ P-|V| bus.
( 0)
V3 = 1.0 + j0.0 Þ P- Q bus.
Bus-2 is a regulated bus where voltage magnitude and real power are specified. or the
voltage controlled bus, first the reactive power is computed.
Using eqn. (7.16),
3
Q2 = - V2VkY2k q2k - d2+ dk
k = 1
å| || | |sin( )