Description
Question 1:
For the power system given in figure 1 bus 1 is a slack bus with �! = 1∠0∘ ��, bus 2 is
a load bus (PQ) with �2 = 300�� + �1270���� and �3 = 400�� + �220����.
The line impedances are �12 = 0.01 + �0.03 �� and are �13 = 0.02 + �0.04 ��. The
base power is 100���.
(a) Use the Gauss-Seidel method to write the expression to calculate
�#
(%&!) ��� �(
(%&!)
as a function of �#
(%) , �(
(%)
`
Figure 1
(b) If after several iterations the voltages at buses B2 and B3 converge to �# =
0.9046 − �0.073 and �( = 0.7618 − �0.116 determine the following:
G V1 =1∠0!
Z13 = 0.02+ j0.04 pu P3 = 400MW
Q3 = 220MVAr
B1
S3
B3
P2 = 300MW
Q2 =170MVAr
S2
B2
Z12 = 0.01+ j0.03pu
332:494/599– HW Set 4 2
1. Power flowing from bus 1 out to bus 2: �!# =
2. Power flowing from bus 1 to bus 3: �!( =
3. Power generated by the source: �! = �)*+ =
4. Line losses for the line connecting bus 1 to bus 2: �,-…!# =
5. Line losses for the line connecting bus 1 to bus 3: �,-…!( =
Question 2:
For the power system given in figure 2 bus 1 is a slack bus with �! = 1∠0∘ ��
and bus 2 is a load bus (PQ) with �2 = 280�� + �60����. The line impedance is
0.02 + �0.04 �� and the base power is 100���
a) Use the Gauss-Seidel method to write the expression to calculate
�#
(%&!) as a function of �#
(%) and solve for �#
(!) �� �#
(0) = 1∠0∘ ��
Figure 2
b) If after several iterations the voltage at bus 2 converges to �# = 0.9 − �0.1
determine the poser S1
c) For part (b) determine the line losses for the line connecting bus 1 and bus 2
Question 3:
For the power system in figure 3, assume a base power of 100MVA
(a) Find the admittance matrix Y
(b) Write the equations for the Gauss-Seidel iteration: �#
(%&!)
, �(
(%&!)
��� �1
(%&!) and
given an initial estimate that �#
(0) = �(
(0) = �1
(0) = 1∠0∘ �� find
�#
(!)
, �(
(!)
, ��� �1
(!)
G V1 =1∠0!
Z12 = 0.02+ j0.04 pu
P2 = 280MW
B2
Q2 = 60M var
B1
S1
332:494/599– HW Set 4 3
Figure 3
G
B2
Z13 = j0.0125pu
B1
B4
Z14 = j0.02 pu
Z24 = j0.005pu Z2 = j0.5pu
Z4 = j1pu
V1 =1∠0! pu Z3 = j0.5pu
B3
Z23 = j0.025pu
S3 S = 210MW + j80MVAr 2 =130MW + j25MVAr
S4 =175MW + j55MVAr
