332:494:01/599:02 – Smart Grid Homework Assignment – Set 4

Description

Question 1:
For the power system given in figure 1 bus 1 is a slack bus with 𝑉! = 1∠0∘ 𝑝𝑢, bus 2 is
a load bus (PQ) with 𝑆2 = 300𝑀𝑊 + 𝑗1270𝑀𝑉𝐴𝑟 and 𝑆3 = 400𝑀𝑊 + 𝑗220𝑀𝑉𝐴𝑟.
The line impedances are 𝑍12 = 0.01 + 𝑗0.03 𝑝𝑢 and are 𝑍13 = 0.02 + 𝑗0.04 𝑝𝑢. The
base power is 100𝑀𝑉𝐴.
(a) Use the Gauss-Seidel method to write the expression to calculate
𝑉#
(%&!) 𝑎𝑛𝑑 𝑉(
(%&!)
as a function of 𝑉#
(%) , 𝑉(
(%)
`
Figure 1
(b) If after several iterations the voltages at buses B2 and B3 converge to 𝑉# =
0.9046 − 𝑗0.073 and 𝑉( = 0.7618 − 𝑗0.116 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: 𝑆!# =
2. Power flowing from bus 1 to bus 3: 𝑆!( =
3. Power generated by the source: 𝑆! = 𝑆)*+ =
4. Line losses for the line connecting bus 1 to bus 2: 𝑆,-…!# =
5. Line losses for the line connecting bus 1 to bus 3: 𝑆,-…!( =
Question 2:
For the power system given in figure 2 bus 1 is a slack bus with 𝑉! = 1∠0∘ 𝑝𝑢
and bus 2 is a load bus (PQ) with 𝑆2 = 280𝑀𝑊 + 𝑗60𝑀𝑣𝑎𝑟. The line impedance is
0.02 + 𝑗0.04 𝑝𝑢 and the base power is 100𝑀𝑉𝐴
a) Use the Gauss-Seidel method to write the expression to calculate
𝑉#
(%&!) as a function of 𝑉#
(%) and solve for 𝑉#
(!) 𝑖𝑓 𝑉#
(0) = 1∠0∘ 𝑝𝑢
Figure 2
b) If after several iterations the voltage at bus 2 converges to 𝑉# = 0.9 − 𝑗0.1
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: 𝑉#
(%&!)
, 𝑉(
(%&!)
𝑎𝑛𝑑 𝑉1
(%&!) and
given an initial estimate that 𝑉#
(0) = 𝑉(
(0) = 𝑉1
(0) = 1∠0∘ 𝑝𝑢 find
𝑉#
(!)
, 𝑉(
(!)
, 𝑎𝑛𝑑 𝑉1
(!)
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