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ʴǷɱڱǷ..نظمقدرة.ٳ
- 2. • Power-flow studies are of great importance in planning and designing the future expansion of power systems as well as in determining the best operation of existing systems. • The principal information obtained from a power-flow study is the magnitude and phase angle of the voltage at each bus and the real and reactive power flowing in each line.
- 3. Successful power system operation under normal balanced three-phase steady-state conditions requires the following: 1. Generation supplies the demand (load) plus losses. 2. Bus voltage magnitudes remain close to rated values. 3. Generators operate within specified real and reactive power limits. 4. Transmission lines and transformers are not overloaded.
- 4. • The power-flow computer program (sometimes called load flow) is the basic tool for investigating these requirements. • This program computes the voltage magnitude and angle at each bus in a power system under balanced three-phase steady-state conditions. • It also computes real and reactive power flows for all equipment interconnecting the buses, as well as equipment losses.
- 5. •The power-flow problem is therefore formulated as a set of nonlinear algebraic equations suitable for computer solution. • Two commonly used iterative techniques:
- 6. The system buses are generally classified into three categories: 1- Slack bus (reference bus) 2- Load buses 3- Voltage controlled buses
- 7. Steps of solution Step -1 : Initial computation Step -2 : Formation of 𝑌𝐵𝑢𝑠 Matrix Step -3 : Iterative computation of bus voltage Step -4 : Computation of slack Bus power Step -5 : Computation of line flows
- 8. Step -1 : Initial computation •Convert all Power in per unit •Compute net-injected power at buses
- 9. Step -2 : Formation of 𝑌𝐵𝑢𝑠 Matrix 𝑌𝐵𝑢𝑠 = 𝑌11 𝑌12 𝑌13 𝑌21 𝑌22 𝑌23 𝑌31 𝑌23 𝑌33
- 10. Step -3 : Iterative computation of bus voltage 𝑉𝑖 = 1 𝑌𝑖𝑖 𝑃𝑖 − 𝑗𝑄𝑖 𝑉𝑖 ∗ − 𝑘 = 1 𝑘 ≠𝑖 𝑛 𝑌𝑖𝑘𝑉𝑘 𝑉 2 (𝑝+1) = 1 𝑌22 𝑃2 − 𝑗𝑄2 𝑉 2 (𝑝) ∗ − 𝑌21𝑉1 − 𝑌23 𝑉 3 (𝑃) 𝑉 3 (𝑝+1) = 1 𝑌33 𝑃3 − 𝑗𝑄3 𝑉 3 (𝑝) ∗ − 𝑌31𝑉1 − 𝑌32 𝑉 2 (𝑃+1)
- 11. Step -4 : Computation of slack Bus power 𝑃1 = 𝑘=1 𝑛 𝑉1𝑉𝑘𝑌𝑖𝑘 cos(𝜃1𝑘 − 𝛿𝑖 + 𝛿𝑘) 𝑄1 = − 𝑘=1 𝑛 𝑉1𝑉𝑘𝑌1𝑘 sin(𝜃1𝑘 − 𝛿1 + 𝛿𝑘)
- 12. Step -5 : Computation of line flows 𝑃𝑖𝑘 = −𝑉𝑖 2 𝑌𝑖𝑘 cos 𝜃𝑖𝑘 + 𝑉𝑖𝑉𝑘𝑌𝑖𝑘 cos(𝜃𝑖𝑘 − 𝛿𝑖 + 𝛿𝑘) 𝑄𝑖𝑘 = 𝑉𝑖 2 𝑌𝑖𝑘 sin 𝜃𝑖𝑘 − 𝑉𝑖𝑉𝑘𝑌𝑖𝑘 sin (𝜃𝑖𝑘 − 𝛿𝑖 + 𝛿𝑘) 𝑃𝑙𝑜𝑠𝑠𝑖𝑘 = 𝑃𝑖𝑘 + 𝑃𝑘𝑖 𝑄𝑙𝑜𝑠𝑠𝑖𝑘 = 𝑄𝑖𝑘 + 𝑄𝑘𝑖
- 13. Example Single line diagram of a sample 3- bus power system , data for this system are given in table.
- 15. 1- Using the Gauss Seidel Method, determine the phasor values of the voltage at buses 2 and 3 (perform only two iterations). 2- find the slack bus real and reactive power after second iteration. 3- determine the line flows and line losses after second iteration (neglect line charging admittance).
- 16. Solution Step -1 : initial computation 𝑃𝐿2 = 305.6 100 = 3.056 𝑝𝑢 , 𝑄𝐿2 = 140.2 100 = 1.402 𝑝𝑢 𝑃𝐿3 = 138.6 100 = 1.386 𝑝𝑢 , 𝑄𝐿3 = 45.2 100 = 0.452 𝑝𝑢 𝑃𝑔2 = 50 100 = 0.50 𝑝𝑢 , 𝑄𝑔2 = 30 100 = 0.30 𝑝𝑢
- 17. 𝑃2 = 𝑃𝑔2 − 𝑃𝐿2 = 0.5 − 3.056 = −2.556 pu 𝑄2 = 𝑄𝑔2 − 𝑄𝐿2 = 0.3 − 1.402 = −1.102 pu 𝑃3 = 𝑃𝑔3 − 𝑃𝐿3 = 0 − 1.386 = −1.386 pu 𝑄3 = 𝑄𝑔3 − 𝑄𝐿3 = 0 − 0.452 = −0.452 pu
- 18. Step -2 : Formation of 𝑌𝐵𝑢𝑠 Matrix 𝑦12 = 𝑦21 = 1 𝑧12 = 1 0.02 + 𝑗 0.04 = 10 − 𝑗20 𝑦13 = 𝑦31 = 1 𝑧13 = 1 0.01 + 𝑗 0.03 = 10 − 𝑗30 𝑦23 = 𝑦23 = 1 𝑧23 = 1 0.0125 + 𝑗 0.025 = 16 − 𝑗32
- 19. 𝑌11 = 𝑦12 + 𝑦13 = 53.85/−68.2 𝑌22 = 𝑦21 + 𝑦23 = 58.13/−63.4 𝑌33 = 𝑦31 + 𝑦32 = 67.23/−67.2
- 20. 𝑌12 = 𝑌21 = − 𝑦12 = 22.36/116.6 𝑌13 = 𝑌31 = − 𝑦13 = 31.62/108.4 𝑌23 = 𝑌32 = − 𝑦23 = 35.77/116.6
- 21. 𝑌𝑏𝑢𝑠 = 53.85/−68.2 22.36/116.6 31.62/108.4 22.36/116.6 58.13/−63.4 35.77/116.6 31.62/108.4 35.77/116.6 67.23/−67.2
- 22. Step -3 : Iterative computation of bus voltage 𝑉𝑖 = 1 𝑌𝑖𝑖 𝑃𝑖 − 𝑗𝑄𝑖 𝑉𝑖 ∗ − 𝑘 = 1 𝑘 ≠𝑖 𝑛 𝑌𝑖𝑘𝑉𝑘 𝑉 2 (𝑝+1) = 1 𝑌22 𝑃2 − 𝑗𝑄2 𝑉 2 (𝑝) ∗ − 𝑌21𝑉1 − 𝑌23 𝑉 3 (𝑃) 𝑉 3 (𝑝+1) = 1 𝑌33 𝑃3 − 𝑗𝑄3 𝑉 3 (𝑝) ∗ − 𝑌31𝑉1 − 𝑌32 𝑉 2 (𝑃+1)
- 23. Slacks bus voltage : 𝑉1 = (1.05 + 𝑗 0.0) Starting voltage : 𝑉 2 (0) = (1 + 𝑗 0.0) 𝑉 3 (0) = (1 + 𝑗 0.0)
- 24. 𝑃2 − 𝑗𝑄2 𝑌22 = 0.0478/−139.9 𝑌21 𝑌22 = −0.3846 𝑌23 𝑌22 = −0.6153
- 25. 𝑉 2 (𝑝+1) = 0.0478/−139.9 𝑉 2 (𝑝) ∗ + 0.3846 𝑉1 + 0.6153 𝑉 3 (𝑃) P = 0 𝑉 2 (1) = 0.98305 /−1.8
- 26. 𝑃3 − 𝑗𝑄3 𝑌33 = 0.0217/−130.86 𝑌31 𝑌33 = 0.47/175.6 𝑌32 𝑌33 = 0.532/183.8
- 27. 𝑉 3 (𝑝+1) = 0.0217/−130.86 𝑉 3 (𝑝) ∗ − 0.47/175.6𝑉1 − 0.532/183.8𝑉 2 (𝑃+1) P = 0 𝑉 3 (1) = 1.0011 /−2.06
- 28. P = 1 𝑉 2 (2) = 0.98265 /−3.048 P = 1 𝑉 3 (2) = 1.00099 /−2.68
- 30. Step -4 : Computation of slack Bus power 𝑃1 = 𝑉1 2 𝑌11 cos 𝜃11 + 𝑉1𝑉2𝑌12 cos 𝜃12 − 𝛿1 + 𝛿2 + 𝑉1𝑉3𝑌13 cos 𝜃13 − 𝛿1 + 𝛿3 𝑃1 = (1.05)2 × 53.85 × cos(−68.2) + 1.05 × 0.98265 × 22.36 cos 116.56 − 0 − 3.048 + 1.05 × 1.00099 × 31.62 cos 108.4 − 0 − 2.68 𝑃1 = 𝑘=1 𝑛 𝑉1𝑉𝑘𝑌1𝑘 cos(𝜃1𝑘 − 𝛿𝑖 + 𝛿𝑘)
- 31. 𝑃1 = 3.84 𝑃𝑢 𝐵𝑎𝑠𝑒 = 100 𝑃1 = 384 𝑀𝑊
- 32. 𝑄1 = − 𝑘=1 𝑛 𝑉1𝑉𝑘𝑌1𝑘 sin(𝜃1𝑘 − 𝛿1 + 𝛿𝑘) 𝑄1 = −𝑉1 2 𝑌11 sin 𝜃11 − 𝑉1𝑉2𝑌12 sin 𝜃12 − 𝛿1 + 𝛿2 − 𝑉1𝑉3𝑌13 sin 𝜃13 − 𝛿1 + 𝛿3 𝑄1 = −(1.05)2 × 53.85 × sin (−68.2) − 1.05 × 0.98265 × 22.36 sin 116.56 − 0 − 3.048 − 1.05 × 1.00099 × 31.62 sin 108.4 − 0 − 2.68
- 33. 𝑄1 = 1.9786 𝑃𝑢 𝐵𝑎𝑠𝑒 = 100 𝑄1 = 197.86 𝑀𝑊
- 34. Step -5 : Computation of line flows 𝑃𝑖𝑘 = −𝑉𝑖 2 𝑌𝑖𝑘 cos 𝜃𝑖𝑘 + 𝑉𝑖𝑉𝑘𝑌𝑖𝑘 cos(𝜃𝑖𝑘 − 𝛿𝑖 + 𝛿𝑘) 𝑃12 = 1.8189 𝑃𝑢 𝑃13 = 2 𝑃𝑢 𝑃23 = − 0.4903 𝑃𝑢 𝑃21 = − 1.744 𝑃𝑢 𝑃31 = −1.95 𝑃𝑢 𝑃32 = 0.496 𝑃𝑢
- 35. Real power losses 𝑃𝑙𝑜𝑠𝑠𝑖𝑘 = 𝑃𝑖𝑘 + 𝑃𝑘𝑖 𝑃𝑙𝑜𝑠𝑠12 = 0.0749 𝑃𝑢 𝑃𝑙𝑜𝑠𝑠13 = 0.05 𝑃𝑢 𝑃𝑙𝑜𝑠𝑠23 = 0.0057 𝑃𝑢
- 36. 𝑄𝑖𝑘 = 𝑉𝑖 2 𝑌𝑖𝑘 sin 𝜃𝑖𝑘 − 𝑉𝑖𝑉𝑘𝑌𝑖𝑘 sin (𝜃𝑖𝑘 − 𝛿𝑖 + 𝛿𝑘) 𝑄12 = 0.8948 𝑃𝑢 𝑄13 = 1.088 𝑃𝑢 𝑄23 = − 0.4746 𝑃𝑢 𝑄21 = − 0.746 𝑃𝑢 𝑄31 = − 0.9469 𝑃𝑢 𝑄32 = 0.4866 𝑃𝑢
- 37. Reactive power losses 𝑄𝑙𝑜𝑠𝑠𝑖𝑘 = 𝑄𝑖𝑘 + 𝑄𝑘𝑖 𝑄𝑙𝑜𝑠𝑠12 = 0.1488 𝑃𝑢 𝑄𝑙𝑜𝑠𝑠13 = 0.1411 𝑃𝑢 𝑄𝑙𝑜𝑠𝑠23 = 0.012 𝑃𝑢