Power Systems Test 6

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Power Systems Test 6

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Question 1
1 / -0

A power flow problem is solved using Newton-Raphson power flow in polar coordinate. The Jacobian size is 50 × 50. There is one slack bus in the system. The number of PV buses is 4. The number of PQ buses is

A
23-23

B
234-231

C
237-236

D
239-232

Solution

Concept:
In the Newton-Raphson method,
The size of a Jacobian matrix = (2n – m – 2) × (2n – m – 2)
Where n = number of buses
m = number of PV buses or generator buses
Calculation:
Given that Slack bus = 1
PV buses or Generator buses = 4
Size of Jacobian matrix = 50 × 50
2n – m – 2 = 50
2n – 4 – 2 = 50
⇒ n = 28
Number of PQ buses = 28 – 1 – 4 = 23

Question 2

1 / -0

Which of the following has a problem in convergence for a system with long radial lines?

Newton-Raphson

Gauss-Seidel

Both Newton-Raphson and Gauss Siedel

None of the two methods

Solution

Comparison of two load flow methods is given below.

Type of problem	Gauss-Seidel Method	Newton-Raphson Method
Heavily loaded system	Usually cannot solve systems with more than 70° phase shift	Solve systems with shifts up to 90°
Systems containing negative reactance such as three-winding transformers or series capacitors	Unable to solve	Solves with ease
Systems with slack bus at a desired location	Often requires trail and error to find a slack bus location that will yield a solution	More tolerant of slack bus location
Long and short lines terminating the same bus	Usually cannot solve if long to short ratio is above 1000	Can solve a system with a long to short ratio at any bus of 1,000,000
Long radial type of system	Difficulty in solving	Solves a wide range of such problems
Acceleration factor	Number of iterations depends on choice of factor	Not required

Question 3
1 / -0

A 500 × 500 bus admittance matrix for an electric power system 4000 non-zero elements the minimum number of branches in this system are –

A
1750-1750

B
17504-17501

C
17507-17506

D
17509-17502

Solution

No. of non-zero elements = no. of diagonal element + 2 (no. of line)
4000 = 500 + 2 (lines)
\(lines = \frac{{4000 - 500}}{2} = 1750\)
Question 4
1 / -0

A 3ϕ SC occurs at BUS – 4. The SC current flowing between buses 1 and 2 is___
Assume that the pre fault voltages at all the buses 1.0 Pu.
\({Z_{bus}} = j\left[ {\begin{array}{*{20}{c}} {0.15}&{0.08}&{\begin{array}{*{20}{c}} {0.04}&{0.07} \end{array}}\\ {\begin{array}{*{20}{c}} {0.08}\\ {\begin{array}{*{20}{c}} {0.04}\\ {0.07} \end{array}} \end{array}}&{\begin{array}{*{20}{c}} {0.15}\\ {\begin{array}{*{20}{c}} {0.06}\\ {0.09} \end{array}} \end{array}}&{\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {0.06}\\ {\begin{array}{*{20}{c}} {0.13}\\ {0.05} \end{array}} \end{array}}&{\begin{array}{*{20}{c}} {0.09}\\ {\begin{array}{*{20}{c}} {0.05}\\ {0.12} \end{array}} \end{array}} \end{array}} \end{array}} \right]\)

A
-2.13∠90°

B
2.13∠90°

C
4.08∠-90°

D
-4.08∠-90°

Solution

before fault,
\(V_1' = 1.0,\ V_2' = 1.0,\ V_3' = 1.0,\ V_4' = 1.0\\ {I_f} = {I_4} = \frac{{V_4'}}{{{Z_{44}}}} = \frac{{1.0}}{{j\ 0.12}} = - j\ 8.33\)
During fault,
\(V_1^"= {I_f}{Z_{14}} = {I_4}{Z_{14}} = 0.58\\ V_2^" = {I_f}{Z_{24}} = {I_4}{Z_{24}} = 0.75\)
The SC current flowing between buses 1 and 2 is
\({I_{12}} = \frac{{V_1^"- V_2^"}}{{{Z_{12}}}} = \frac{{0.58 - 0.75}}{{j0.08}} = 2.13\angle 90^\circ\)
Question 5
1 / -0

The bus impedance matrix of a 4 – bus power system is given by
\({Z_{bus}} = \left[ {\begin{array}{*{20}{c}}{j0.3435}&{j0.2860}&{j0.2723}&{j0.2277}\\{j0.2860}&{j0.3408}&{j0.2586}&{j0.2414}\\{j0.2723}&{j0.2586}&{j0.2791}&{j0.2209}\\{j0.2277}&{j0.2414}&{j0.2209}&{j0.2791}\end{array}} \right]\)
A branch having an impedance of j 0.2 Ω is connected between bus 2 and the reference Then the values of Z₂₂,new and Z₂₃,new of the bus impedance matrix of the modified network are respectively.

A
j0.5408 Ω and j0.4586 Ω

B
j0.1260 Ω and j0.0956 Ω

C
j0.5408 Ω and j0.0956 Ω

D
j0.1260 Ω and j0.1630 Ω

Solution

\({Z_{B\left( {new} \right)}} = {Z_{B\left( {old} \right)}} - \frac{1}{{{Z_{ij}} + {Z_b}}}\left[ {\begin{array}{*{20}{c}}{{Z_{ij}}}\\ \vdots \\{{Z_{nj}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{Z_{ji}}}& \ldots &{{Z_{jn}}}\end{array}} \right]\)
New element Z_b= j0.2Ω is connected in j^th and reference bus j = 2, n = 4 so
\(\frac{1}{{{Z_{ij}} + {Z_b}}}\left[ {\begin{array}{*{20}{c}}{{Z_{12}}}\\{{Z_{22}}}\\{{Z_{23}}}\\{{Z_{24}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{Z_{21}}}&{{Z_{22}}}&{{Z_{23}}}&{{Z_{24}}}\end{array}} \right]\)
\(= \frac{1}{{\left[ {j\left( {0.3408} \right) + j\left( {0.2} \right)} \right]}}\left[ {\begin{array}{*{20}{c}}{j0.2860}\\{j0.3408}\\{j0.2586}\\{j0.2414}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{j.0.2860}&{j0.3408}&{j0.2586}&{j0.2414}\end{array}} \right]\)
Given that we are required to change only Z₂₂, Z₂₃
\(\begin{array}{l}Z_{22}^1 = \frac{{{j^2}{{\left( {0.3408} \right)}^2}}}{{j\left( {0.5408} \right)}} = j0.2147\\Z_{23}^1 = \frac{{{j^2}\left( {0.3408} \right)\left( {0.2586} \right)}}{{j\left( {0.5408} \right)}} = j0.16296\\{Z_{22\left( {new} \right)}} = {Z_{22\left( {old} \right)}} - Z_{22}^1 = j0.3409 - j0.2147\end{array}\)
= j 0.1260
\({Z_{23\left( {new} \right)}} = {Z_{23\left( {old} \right)}} - Z_{23}^1 = j0.2586 - j0.16296\)
= j 0.0956