Power(2)-EPM320-L01-16 Lectures of Power System Analysis.pdf
- 1. Power System Analysis EPM320 3rd Year Electrical Power and Machines Dept. Ahmed A. Daoud, Ph.D.
- 2. Course Contents Lecture: 2h/week Tutorial: 2h/week 1- Introduction(Unified Egyptian Network, Power system analysis, Electrical power systems construction) 2-Electrical Power Transformers(Single phase & three phase transformers, Equivalent circuit, open and short circuit test, three winding and Auto transformer, Parallel operation of transformers) 3- Synchronous Generator(synchronous machine, Three phase generator, Synchronous reactance and equivalent circuit, Real and reactive power control, Loading capability diagram) 4-Power Systems Economics(Thermal power station, Characteristics of power generation units, Economic dispatch, Lagrange relaxation method) 5- Load Flow Calculations(Necessity for Power Flow Studies, Conditions for Successful Operation of a Power System, The Power Flow Equations, Gauss-Seidel Iterative Method, Newton-Raphson Method, Sparsity of Network Admittance Matrices)
- 3. Introduction The Egyptian Electrical Unified Network (EEUN)in Egypt is divided into six geographical regions, namely, Cairo, Canal, Delta, Alexandria/ west delta, Middle Egypt, and Upper Egypt. The Egyptian Electricity transmission system is composed of 500 kV, 400 kV, 220 kV, 132 kV, and 66 kV levels.
- 4. Introduction
- 5. Introduction
- 6. Introduction Power system analysis An electric power system is made up of electrical components to generate, transmit and use electric power. This could be the elaborate network that supplies power to a region’s home and industry through an electrical grid transmission system from generating plants located faraway. Or, this could even be a captive power plant/micro-grid which generates and consumes power within the same premises itself. The majority of these systems rely upon three-phase AC power - the standard for large-scale power transmission and distribution across the modern world. Specialized power systems are also found in aircraft, electric rail systems, ocean liners and automobiles that do not always rely upon three-phase AC power. The planning, design, and operation of these commercial and industrial power systems requires in-depth engineering studies to evaluate existing and proposed system performance, reliability, safety, and economics.
- 7. Introduction Electrical power systems construction Generation Transmission Load
- 8. Per Unit Quantities There are several reasons for using a per-unit system: •Similar apparatus (generators, transformers, lines) will have similar per- unit impedances and losses expressed on their own rating, regardless of their absolute size. •Use of the constant is reduced in three-phase calculations. •Per-unit quantities are the same on either side of a transformer, independent of voltage level. •By normalizing quantities to a common base, both hand and automatic calculations are simplified. 𝐼𝑏𝑎𝑠𝑒 = 𝑆𝑏𝑎𝑠𝑒 𝑉𝑏𝑎𝑠𝑒 For Single phase 𝑍𝑏𝑎𝑠𝑒 = 𝑉𝑏𝑎𝑠𝑒 𝐼𝑏𝑎𝑠𝑒 = 𝑉𝑏𝑎𝑠𝑒 2 𝐼𝑏𝑎𝑠𝑒𝑉𝑏𝑎𝑠𝑒 = 𝑉𝑏𝑎𝑠𝑒 2 𝑆𝑏𝑎𝑠𝑒 For Three phase 𝑆𝑏𝑎𝑠𝑒 = 3𝑉𝑏𝑎𝑠𝑒𝐼𝑏𝑎𝑠𝑒 𝐼𝑏𝑎𝑠𝑒 = 𝑆𝑏𝑎𝑠𝑒 3𝑉𝑏𝑎𝑠𝑒 𝑍𝑏𝑎𝑠𝑒 = 𝑉𝑏𝑎𝑠𝑒 𝐼𝑏𝑎𝑠𝑒 = 3𝑉𝑏𝑎𝑠𝑒 2 𝑆𝑏𝑎𝑠𝑒
- 10. Per Unit Quantities G1 G2 M 30 MVA, 15 kV XG1= j 0.1 pu 25 MVA, 13.8 kV XG2= j 0.15 pu 25 MVA, 18/138 kV XT1= j 0.1 pu 25 MVA, 138/13.8 kV XT2= j 0.1 pu 10 MVA, 12 kV XM= j 0.1 pu 20 + j60 W Take 30 MVA, 15 kV as base on generating bus bar draw the equivalent circuit of the system using per unit quantities.
- 11. Per Unit Quantities The terminl voltage of a Y connected load consisting of three equal impedances of 2030 is 4.4 kV line to line. The impedance of each of the three lines connecting the load to a bus at a bus station is 1.475W. (a)Find the line-to-line voltage at the substation bus. (b)Find the solution by working in per unit on a base of 4.4 kV, 127 A so that both voltage and current magnitudes will be 1 .0 per unit. Current rather than KVA is specified here since the latter quantity does not enter the problem .
- 12. Bus Admittance Matrix Nodal Analysis
- 13. Bus Admittance Matrix Node (3) 𝑉3𝑌𝑎 + 𝑉3 − 𝑉2 𝑌𝑏 + 𝑉3 − 𝑉1 𝑌𝑐 = 𝐼3 Rearranging Node (1) 𝑉1 𝑌𝑐 + 𝑌𝑑 + 𝑌 𝑓 − 𝑉2𝑌𝑑 − 𝑉3𝑌𝑐 − 𝑉4𝑌 𝑓 = 0 Node (3) -𝑉1𝑌𝑐 − 𝑉2𝑌𝑏 + 𝑉3(𝑌𝑎 + 𝑌𝑏 + 𝑌𝑐) = 𝐼3 Similarly for nodes (2) and (4) We obtain: 𝑌11 𝑌12 𝑌13 𝑌14 𝑌21 𝑌22 𝑌23 𝑌24 𝑌31 𝑌41 𝑌32 𝑌42 𝑌33 𝑌34 𝑌43 𝑌44 𝑉1 𝑉2 𝑉3 𝑉4 = 𝐼1 𝐼2 𝐼3 𝐼4
- 15. Exercises