1_Intro + Per Unit.pdf
- 1. BEF 43303 BEF 43303 POWER SYSTEMS ANALYSIS POWER SYSTEMS ANALYSIS POWER SYSTEMS ANALYSIS POWER SYSTEMS ANALYSIS AND PROTECTION AND PROTECTION 1 SEM I 2013/2014 SEM I 2013/2014
- 2. Course Overview Course Overview Synopsis: The subject deals with topics related to the power system analysis and protection: per-unit power system analysis and protection: per unit system, power flow analysis, analysis of balanced and unbalanced faults, power system stability and application power system control overcurrent application, power system control, overcurrent protection, differential protection and application, distance protection and application. Overall, this course focuses on analysis of power system and course focuses on analysis of power system and the protection schemes for power system network. 2
- 3. Course Outcome At the end of this course the student At the end of this course the student should be able to: 1. Analyze power system faults based on balance 1. Analyze power system faults based on balance and unbalanced faults techniques. (PLO4- CTPS-C4) 2. Demonstrate power flow analysis using related software. (PLO3-CS-P4) 3 Express the suitable protection schemes based 3. Express the suitable protection schemes based on power system requirement. (PLO11-SD-A3) 3
- 4. Lecture Plan WEEKS CONTENTS 1 Per-Unit System 2 Power Flow Analysis 3 Analysis of Balanced Fault 4 Analysis of Unbalanced Fault 5 Power System Stability 5 Power System Stability 6 Application of Power System Stability 7 Load Frequency and Automatic Generation Controls 8 Reactive Power and Voltage Controls 9 Non-Directional Overcurrent and Earth Fault Protection 4
- 5. L t Pl Lecture Plan WEEK CONTENTS 10 Directional Overcurrent and Earth Fault Relay 11 Differential Protection Scheme 12 Differential Protection Application 13 Distance Protection Scheme 14 Distance Protection Application Syllabus details 5
- 6. A t Assessments Oral 5 % Oral 5 % Test 1 15 % (week 4) Test 2 15 % (week 8) Test 2 15 % (week 8) Assignment 15 % Final exam 50 % Final exam 50 % Total 100% WITH WISDOM WE EXPLORE 6
- 7. This Week Lecture Plan This Week Lecture Plan CHAPTER CONTENTS 1 Per-Unit System (3 Hours) • Introduction • Vectors • Operators • Convention Used for Voltage Direction B Q titi d P U it S t • Base Quantities and Per-Unit System • Transferring Per-Unit Quantities from One Set of Base Values 7
- 8. Representation of Electric Power System One-Line Diagram (OLD) Definition: ◦ A diagram showing the interconnection of various components of a balanced three phase various components of a balanced three- phase power system by standard symbols on a single phase basis. R 3-phase system One line diagram R Y B = 8 3-phase system One-line diagram
- 9. S Advantages of OLD Simplicity 1-Φ represents all 3-Φs of the balanced system The equivalent circuits of the components are replaced The equivalent circuits of the components are replaced by their standard symbols The completion of the circuit through the neutral is p g omitted 9
- 10. Symbols for One-Line Diagram Machine or rotating armature Or Machine or rotating armature 2-winding power transformer Or Or 3-winding power transformer Or Power Circuit Breaker (CB) Load Or Power Circuit Breaker (CB) (oil/liquid) (OCB) Air CB (ACB) 10 skip next page
- 11. Busbar 3-phase, 3-wire delta connection Transmission line delta connection 3-phase wye, neutral ungrounded Fuse neutral ungrounded 3-phase wye, neutral grounded A Current transformer (CT) neutral grounded Ammeter V or Potential transformer (PT or VT) Ammeter Voltmeter 11 ( )
- 12. Impedance (Z) and Reactance (X) Diagram Impedance (Z = R + jX) diagram is converted from OLD showing the equivalent circuit of each t f th t It i d d t l l t th component of the system. It is needed to calculate the performance of a system under normal and abnormal conditions i.e. load conditions (Load Flow (LF) studies) th f f lt/ h t i it (f lt or upon the occurrence of a fault/short circuit (fault analysis studies). Reactance (jX) diagram is further simplified from Z diagram by omitting all static loads, all Rs, the magnetizing I (Im) of each transformer, and the capacitance (C) of the transmission line. It is applied only to fault calculations, and not to LF studies. Z and X diagrams sometimes called the Positive- sequence diagram. sequence diagram. 12
- 13. Z and X Diagrams Example: OLD of an EPS Load B T2 T1 Load B Load A WITH WISDOM WE EXPLORE 13
- 14. Z diagram corresponding to the OLD g p g E1 E2 E3 Gen. 3 Load B Transformer T2 Transmission Line Transformer T1 Load A Generators 1 and 2 14
- 15. X diagram corresponding to the OLD E1 E2 E1 Generators 1 and 2 Transmission Line Transformer T2 Gen. 3 Transformer T1 15
- 16. Per - unit (P.U) Representation Common quantities used in power system analysis (PSA) are voltage (V) (in kV), current (I) (in kA), voltamperes (in kVA or MVA), and impedance (in Ω). It is very cumbersome to convert I t diff t lt l l i PS h i t V Is to different voltage levels in a PS having two or more V levels. P.U. representation is introduced in such a way that the various h i l titi d d i l f ti physical quantities are expressed as a decimal fraction or multiples of base quantities and is defined as: actual quantity Quantity in per-unit base value quantity Example: For instance, if a base voltage of 275 kV is chosen, actual voltages of 247 5 kV 275 kV and 288 75 kV becomes 0 90 16 voltages of 247.5 kV, 275 kV, and 288.75 kV becomes 0.90, 1.00, and 1.05 per-unit.
- 17. For 1-Φ systems: The formula relates the various quantities for 1-Φ system: (1- ) ( ) base kVA (in kVA) Base I (in A) base V (in kV) LN LN 1φ V voltage base kV voltage, base kVA base A current, Base ( ) 2 ( ) base V (in V) Base Z (in ohms) base I (in A) (base V ) (in kV) B Z (i h ) LN LN LN A current, base V voltage, base impedance Base 2 LN ) kV voltage, (base impedance Base ( ) (1- ) (1- ) (1- ) Base Z (in ohms) base MVA (in MVA) Base power (in kW) base kVA ( LN in kVA) Base power (in MW) base MVA 1φ MVA base impedance Base Base P, MW1φ = Base MVA1φ B Q MVAR1 B MVA1 (3- ) (1- ) Base power (in MW) base MVA ) ( l Base Q, MVAR1φ = Base MVA1φ 2 . ) ( base base base u p V VA Z Z actual Z Z
- 18. F 3 Φ t For 3-Φ systems: The formula relates the various quantities for 3-Φ system: (3- ) ( ) base kVA Base I (in A) 3 X base V (in kV) LL LL 3φ kV voltage, base X 3 kVA base A current, Base ( ) 2 ( ) (3- ) (base V ) (in kV) Base Z (in ohms) base MVA Base power (in kW) base kVA LL LL 3φ 2 LL MVA base ) kV voltage, (base impedance Base (3- ) (3- ) (3- ) (3- ) Base power (in kW) base kVA Base power (in MW) base MVA 3φ 3φ 3φ 3φ MVA base MW power, Base kVA base kW power, Base 2 ) ( base u p VA Z actual Z Z 18 2 . base base u p V Z
- 19. Example: The base impedance and base voltage for a given power system are 10Ω and 400V, respectively. Calculate the base kVA and the base current. Calculate the base kVA and the base current. Solution: F m Ohm’ l A 40 400 From Ohm’s law, Base current = A 40 10 kVA X 16 400 40 Base kVA = Base current 19 kVA 16 1000 Base kVA 19
- 20. Example: The base current and the base voltage of a The base current and the base voltage of a 345kV system are chosen to be 3000A and 300 kV, respectively. Determine the base impedance and the per unit voltage for the system and the per-unit voltage for the system. Solution: Solution: Base impedance = 100 10 300 3 345 Base impedance 100 3000 1.15pu 300 345 Per-unit voltage = 20
- 21. Example: A 3-Φ, Y-connected system is rated at 100 MVA and 132 kV. Express 80 MVA of 3-Φ apparent power (S) as a p.u. value referred to: (a) the 3-Φ system MVA as base and (b) the 1-Φ system MVA as base. (a) For the 3-Φ base, Base MVA = 100 MVA = 1 p.u. and Base kVLL = 132 kV = 1 p.u. LL p so p.u. MVA = 80/100 = 0.8 p.u. (b) For the 1-Φ base, Base MVA = 100/3 MVA = 33.33 MVA = 1 p.u. and Base kV = 132/√3 = 76 21 kV = 1 p u 21 and Base kV = 132/√3 = 76.21 kV = 1 p.u. so p.u. MVA = (1/3)*(80/33.333) = 0.8 p.u.
- 22. Changing the Base of P.U. Quantities The Z of individual generators and transformers are generally in terms of % or p.u. quantities based on their ratings given by manufacturer. their ratings given by manufacturer. For PSA, all Zs must be expressed in p.u. on a common system base. Thus, it is necessary to f convert the p.u. Zs from one base to another (common base, for example: 100 MVA). P.U. Z of a circuit element 2 (actual Z in ) * (base MVA) (base V) in kV The equation shows that p.u. Z is directly proportional to the base MVA and inversely proportional to the 22 square of the base V.
- 23. Therefore, to change from old base p.u. Z to new 2 , g p base p.u. Z, the following equation applies: 2 old new new old new old base kV base MVA P.U. Z P.U. Z base kV base MVA Example 1: The reactance X” of a generator is given as 0.20 p.u. based on the generator’s nameplate rating of 13.2 kV, 30 MVA. The base for calculations is 13.8 kV, 50 MVA. Find X” on this calculations is 13.8 kV, 50 MVA. Find X on this new base. 2 13 2 50 23 13.2 50 x" 0.20 0.306 p.u. 13.8 30
- 24. Example 2: Calculate the p.u. impedance of a synchronous motor rated 200 kVA, 13.2 kV and having reactance of 50 ohm of 50 ohm Example 3: The primary and secondary sides of a single phase 1 MVA, 4kV/2kV transformer have a p , leakage reactance of 2 ohm each. Find p.u. X if the transformer referred to primary and secondary side side. 24
- 25. A l i f PS bl l i lifi d b Analysis of PS problems are greatly simplified by using single-line Z diagram in which system parameters are expressed in p.u. The steps to compute p.u. values are summarized as follows: compute p.u. values are summarized as follows: Step 1: Select a common volt-ampere base for the entire power system and a voltage base for one part of the system part of the system. Step 2: Compute voltage bases for all parts of the PS by correlating the transformation ratios of the y g transformer banks. 25
- 26. Step 3: Convert p.u. values (which is provided by the Step 3: Convert p.u. values (which is provided by the nameplate of the equipment) to the common system volt-ampere base and the applicable voltage base. In case the parameters are provided in actual ohmic case the parameters are provided in actual ohmic values, compute base Z for the part of the PS in which the equipment is connected and calculate the p.u. values. values. Step 4: Draw a single-line diagram of the PS indicating values of all parameters in p u Proceed to analyze the values of all parameters in p.u. Proceed to analyze the PS. Step 5: Convert to actual values wherever required Step 5: Convert to actual values wherever required. 26
- 27. Example 4: Draw the reactance diagram of system shown below. Assume reactance for the transmission line is 60 ohm and select the generator rating as base in 60 ohm and select the generator rating as base in the generator circuit 6.6/66 kV 27
- 28. Example 5: Example 5: A 30 MVA 13.8 kV 3-Φ generator has a sub-transient reactance (Xd’’) of 15%. The generator supplies two motors over a tr. line having transformers at both motors over a tr. line having transformers at both ends, as shown in OLD below. The motors have rated inputs of 20 MVA and 10 MVA, both 12.5 kV with x” = 20%. The 3-Φ transformer T1 is rated 35 MVA, 13.2/116 , (∆/Y) kV with leakage reactance (Xl) of 10%. 3-Φ transformer T2 is rated at 10 MVA, 116/12.5 (Y/∆) kV with Xl of 10%. Series X of the tr. line is 80 Ω. Draw the X diagram with all Xs marked in p.u. Select the generator rating as base in the generator circuit. 28