.Analyze and Comparison of Voltages at PQ bus and PV bus. 1. At the PQ bus, the value of voltage is decreased or increased with the increase in an iteration? (On basis of zero and 1st Iteration). 2. At PV bus, the value of voltage or angle is decreased or increased with the increase in an iteration? (On basis of zero and 1st Iteration). 3. What is the explanation of the PQ bus and PV bus? (Graph can be made for comparison). 4. What is the effect of line admittances on Voltage values at PQ and PV buses? 5. Any additional comments

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.28P
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Please solve according to the given method only. Thank you


10.Analyze and Comparison of Voltages at PQ bus and PV bus.
1. At the PQ bus, the value of voltage is decreased or increased with the increase in an iteration? (On basis of zero and 1st Iteration).
2. At PV bus, the value of voltage or angle is decreased or increased with the increase in an iteration? (On basis of zero and 1st Iteration).
3. What is the explanation of the PQ bus and PV bus? (Graph can be made for comparison).
4. What is the effect of line admittances on Voltage values at PQ and PV buses?
5. Any additional comments

5
Elements of Jacobian Matrix can be calculated as:
|V2||Vi||Y21| sin(@21 – 8g + ô1) + |V2||V3||Y23|
sin(023 – 82 + 83)
ar = -V||Vs||Y23| sin(@23 – 62 + 63)
[Vi||Y21]cos(021 - 82 + 61) + 2|V2||Y22| cos 022 +
Ale
|Vs||Y23| cos(023 – 82 + 83)
-|Va||Va||Y32| sin(032 – 83 + 82)
= |Va||Vi||Y31|sin(831 – 83 + ổ1) + |V3||V½||Y32}|
sin(@32 – 83 + 82)
|Va||Y32| cos(032 - đ3 + 82)
ƏQ2
|Va||Vi||Y21| cos(021 – đ2 + ôi) + |Va||Va||Y23al
cos(f23 – 62 + 83)
:-|Va||Va||Y23| cos(023 – 82 + 63)
ƏQ2
-|Vi||Y21 |sin(@21 - ô2 + ổi) – 2|Val||Y22| sin ®2 –
Ale
|Va||Y23| sin(823 – 62 + 83)
Scheduled power calculation at bus 2 and bus 3.
Psch - jQsch
Ssch =
100
Mismatch power calculation: (Difference of Power calculated at Step 4 and Step
6).
APO) = P, sah - P0), APO) = P3 sch – P®), AQ) = Q2 sch – Q
(0)
8
Use Jacobian Matrix to find the unknown angles and voltages. Values of Jacobian
Matrix are found in Step 5.
(0)
(0)
= J21 J22 J23|A0
J31 J32 J33!
Lav
Values of Voltage and Angles after 1* Iteration:
s1) = 60 + A80
(0)
(0)
A6 = 80 + As
v4) = v0) + av,0)
Analyze and Comparison of Voltages at PQ bus and PV bus.
1. At PQ bus, value of voltage is decreased or increased with the increase in
10
Transcribed Image Text:5 Elements of Jacobian Matrix can be calculated as: |V2||Vi||Y21| sin(@21 – 8g + ô1) + |V2||V3||Y23| sin(023 – 82 + 83) ar = -V||Vs||Y23| sin(@23 – 62 + 63) [Vi||Y21]cos(021 - 82 + 61) + 2|V2||Y22| cos 022 + Ale |Vs||Y23| cos(023 – 82 + 83) -|Va||Va||Y32| sin(032 – 83 + 82) = |Va||Vi||Y31|sin(831 – 83 + ổ1) + |V3||V½||Y32}| sin(@32 – 83 + 82) |Va||Y32| cos(032 - đ3 + 82) ƏQ2 |Va||Vi||Y21| cos(021 – đ2 + ôi) + |Va||Va||Y23al cos(f23 – 62 + 83) :-|Va||Va||Y23| cos(023 – 82 + 63) ƏQ2 -|Vi||Y21 |sin(@21 - ô2 + ổi) – 2|Val||Y22| sin ®2 – Ale |Va||Y23| sin(823 – 62 + 83) Scheduled power calculation at bus 2 and bus 3. Psch - jQsch Ssch = 100 Mismatch power calculation: (Difference of Power calculated at Step 4 and Step 6). APO) = P, sah - P0), APO) = P3 sch – P®), AQ) = Q2 sch – Q (0) 8 Use Jacobian Matrix to find the unknown angles and voltages. Values of Jacobian Matrix are found in Step 5. (0) (0) = J21 J22 J23|A0 J31 J32 J33! Lav Values of Voltage and Angles after 1* Iteration: s1) = 60 + A80 (0) (0) A6 = 80 + As v4) = v0) + av,0) Analyze and Comparison of Voltages at PQ bus and PV bus. 1. At PQ bus, value of voltage is decreased or increased with the increase in 10
Note: It is expected from student to analyze power flow in three phase power system.
Themal Power Plant and Hydropower plant are interconnected in a way that in case of shut down
of one power plant, the other power plant acts as a backup power supplier. Analyze power flow of
a system as shown in Fig. 1 by the Newton Raphson method to compare voltage values at load
(PQ) bus and Voltage Controlled (PV) bus. Base Values: (100 MVA and 132 kV).
100 MW
0.01 + jo.03
0.0125 +j0.025
Slack Bus
|V3| = 1.05
V1 = 1.0 +j0
100 MW
100 MVAR
Fig. 1. Three Bus Power Network
Hint for Solving Question:
Steps Newton Raphson Method
1
Convert Line Impedances to Admittances
Convert Load and Generation Power (P and Q) into Per Unit Values
Identifying Load Bus (PQ bus) and Voltage Controlled Bus (PV bus).
What are known and unknown values at PQ bus and PV bus?
Identifying/ Designating Slack Bus
Initial Assumption of Voltage Values
V, = 1.0 + 0j per unit, V = 1.0 + 0j per unit, V = 1.05 + 0jper unit
Y-bus Formation by Direct method, or By Building Block Method.
Convert Y-bus values into polar form with angles in radian.
2
3
4
Calculation of Real and Reactive Power at bus 2 and bus 3.
P2 = |Va||Vi||Y21| cos(021 - 82 + ố1) + |VŽ||Y22| cos 02+
|Va||Va||Y23| cos(823 – 62 + b3)
P = |Va||Vi||Y31| cos(f31 - 83 + 81) + |Va||Va||Y32| cos(032 –
83 + ô2) + |V?||Ya3| cos @33
Transcribed Image Text:Note: It is expected from student to analyze power flow in three phase power system. Themal Power Plant and Hydropower plant are interconnected in a way that in case of shut down of one power plant, the other power plant acts as a backup power supplier. Analyze power flow of a system as shown in Fig. 1 by the Newton Raphson method to compare voltage values at load (PQ) bus and Voltage Controlled (PV) bus. Base Values: (100 MVA and 132 kV). 100 MW 0.01 + jo.03 0.0125 +j0.025 Slack Bus |V3| = 1.05 V1 = 1.0 +j0 100 MW 100 MVAR Fig. 1. Three Bus Power Network Hint for Solving Question: Steps Newton Raphson Method 1 Convert Line Impedances to Admittances Convert Load and Generation Power (P and Q) into Per Unit Values Identifying Load Bus (PQ bus) and Voltage Controlled Bus (PV bus). What are known and unknown values at PQ bus and PV bus? Identifying/ Designating Slack Bus Initial Assumption of Voltage Values V, = 1.0 + 0j per unit, V = 1.0 + 0j per unit, V = 1.05 + 0jper unit Y-bus Formation by Direct method, or By Building Block Method. Convert Y-bus values into polar form with angles in radian. 2 3 4 Calculation of Real and Reactive Power at bus 2 and bus 3. P2 = |Va||Vi||Y21| cos(021 - 82 + ố1) + |VŽ||Y22| cos 02+ |Va||Va||Y23| cos(823 – 62 + b3) P = |Va||Vi||Y31| cos(f31 - 83 + 81) + |Va||Va||Y32| cos(032 – 83 + ô2) + |V?||Ya3| cos @33
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