Chap4 PerUnit representation.pdf
- 1. Per Unit Representation Dr Bakary Diarra BIUST Palapye
- 2. Per-Unit representation In system with many generators and transformers Many different currents and voltages Use base quantities to make the system more uniform Quantities: apparent power, voltages, current and impedances Electrical Energy Systems Dr B. Diarra 2
- 3. Example Let’s consider this circuit Electrical Energy Systems Dr B. Diarra 3 𝑉𝑆 𝑍𝑆 𝑉𝐿 DC 𝑍𝐿 𝑉𝑆 = 1000 𝑍𝑆 = 𝑗10 𝑍𝐿 = 60 + 𝑗70 The current in the circuit is 𝐼𝑆 = 𝑉𝑆 𝑍𝑆 + 𝑍𝐿 = 1000 60 + 𝑗80 = 10 < −53° 𝐴 The apparent power of the system 𝑆 = 𝑉𝑆. 𝐼𝑆 ∗ = 1000 ∗ 10 < 53° = 10000 𝑉𝐴 < 53° The overall impedance 𝑍𝑇 = 𝑍𝑆 + 𝑍𝐿 = 60 + 𝑗80 = 100 < 53° The amplitude of these quantities will be used as reference in this circuit
- 4. Example Base quantities are real (modules): 1. Apparent power 𝑆𝑏𝑎𝑠𝑒 = 10 𝑘𝑉𝐴 2. Input voltage 𝑉𝑏𝑎𝑠𝑒 = 1000 𝑉 3. Circuit current 𝐼𝑏𝑎𝑠𝑒 = 10 𝐴 4. Overall impedance 𝑍𝑏𝑎𝑠𝑒 = 100 Ω Electrical Energy Systems Dr B. Diarra 4
- 5. Example Per unit quantities of the circuit, 𝑉𝐿 = 𝑍𝐿 𝑍𝑇 𝑉𝑆 = 60+𝑗70 60+𝑗80 ∗ 1000 = 920 + 𝑗60, so 𝑉𝐿𝑝𝑢 = 𝑉𝐿 𝑉𝑏𝑎𝑠𝑒 = 0.92 + 𝑗0.06𝑝. 𝑢 → 𝑉𝐿𝑝𝑢 = 0.922𝑝. 𝑢 The impedances 𝑍𝐿𝑝𝑢 = 𝑍𝐿 𝑍𝑏𝑎𝑠𝑒 = 60 + 𝑗70 100 = 0.6 + 𝑗0.7𝑝. 𝑢 𝑍𝑆𝑝𝑢 = 𝑍𝑆 𝑍𝑏𝑎𝑠𝑒 = 𝑗10 100 = 𝑗0.1 𝑝. 𝑢 Electrical Energy Systems Dr B. Diarra 5
- 6. Definition The circuit becomes Per definition 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑣𝑎𝑙𝑢𝑒 = 𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑏𝑎𝑠𝑒 𝑣𝑎𝑙𝑢𝑒 On equipment, percent values are used 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑣𝑎𝑙𝑢𝑒 = 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑣𝑎𝑙𝑢𝑒 𝑥 100 Electrical Energy Systems Dr B. Diarra 6 1 𝑝. 𝑢 𝑍𝑆 = 𝑗0.1 𝑝. 𝑢 𝑍𝐿 = 0.922 < 49° 𝑝. 𝑢 DC
- 7. Generators The ratings of the generators are chosen as base In the previous example 𝑆𝑏𝑎𝑠𝑒 = 10 𝑘𝑉𝐴 𝑉𝑏𝑎𝑠𝑒 = 1000 𝑉 𝐼𝑏𝑎𝑠𝑒 = 𝑆𝑏𝑎𝑠𝑒 𝑉𝑏𝑎𝑠𝑒 = 10 𝐴 𝑍𝑏𝑎𝑠𝑒 = 𝑉𝑏𝑎𝑠𝑒 2 𝑆𝑏𝑎𝑠𝑒 = 100 Ω Electrical Energy Systems Dr B. Diarra 8
- 8. Transformers Let’s consider the following transformer The voltage and power ratings are used as base values, e.g the base voltages are different for the primary and secondary windings. Electrical Energy Systems Dr B. Diarra 9 Ratings 𝑆 = 2000 𝑉𝐴 𝑉1/𝑉2 = 200/400 𝑉 𝑍 = 𝑗4 Ω 𝑛 = 2 𝑉1 𝑉2 Low voltage side High voltage side 𝑛 = 2 𝑉1 𝑉2 LV side HV side 𝑍 = 𝑗16 Ω 𝑆1 = 𝑉1𝐼1 ∗ = 𝑉1 2 /𝑍1 ∗ and using 𝑉2 = 𝑛𝑉1 we get 𝑆1 = 𝑉2 2 /𝑛2 𝑍1 ∗ , → 𝑍2 = 𝑛2 𝑍1
- 9. Base values For the transformer Electrical Energy Systems Dr B. Diarra 10 Quantities Low voltage side High voltage side 𝑆𝑏𝑎𝑠𝑒 2000 𝑉𝐴 2000 𝑉𝐴 𝑉𝑏𝑎𝑠𝑒 200 𝑉 400 𝑉 𝐼𝑏𝑎𝑠𝑒 𝑆𝑏𝑎𝑠𝑒 𝑉𝑏𝑎𝑠𝑒 = 10 𝐴 𝑆𝑏𝑎𝑠𝑒 𝑉𝑏𝑎𝑠𝑒 = 5 𝐴 𝑍𝑏𝑎𝑠𝑒 𝑉𝑏𝑎𝑠𝑒 2 𝑆𝑏𝑎𝑠𝑒 = 20 Ω 𝑉𝑏𝑎𝑠𝑒 2 𝑆𝑏𝑎𝑠𝑒 = 80 Ω 𝑍𝑝𝑢 𝑍 𝑍𝑏𝑎𝑠𝑒 = 𝑗4 20 = 𝑗0.2 𝑍 𝑍𝑏𝑎𝑠𝑒 = 𝑗16 80 = 𝑗0.2
- 10. Transformers The p.u of the transformer is the same The per-unit impedance is the same for both sides of the transformer Electrical Energy Systems Dr B. Diarra 11 𝑗0.2 𝑝. 𝑢 Per unit model of the transformer
- 11. Three phase systems The base quantities of three-phase systems are Electrical Energy Systems Dr B. Diarra 12 Quantities Low voltage side Quantities 𝑉𝑏𝑎𝑠𝑒 𝑉𝐿𝐿 = 3𝑉𝐿𝑁 Line-to-line voltage 𝑆𝑏𝑎𝑠𝑒 3𝑉𝑏𝑎𝑠𝑒𝐼𝑏𝑎𝑠𝑒 Apparent power 𝐼𝑏𝑎𝑠𝑒 𝑆𝑏𝑎𝑠𝑒 3𝑉𝑏𝑎𝑠𝑒 Line current 𝑍𝑏𝑎𝑠𝑒 𝑉𝐿𝐿 3𝐼𝑏𝑎𝑠𝑒 = 𝑉𝑏𝑎𝑠𝑒 2 𝑆𝑏𝑎𝑠𝑒 Phase impedance
- 12. Change of base for impedance Let’s consider two bases A and B and Z an impedance 𝑍𝐴𝑝𝑢 = 𝑍 𝑍𝐴𝑏𝑎𝑠𝑒 = 𝑍 𝑆𝐴𝑏𝑎𝑠𝑒 𝑉𝐴𝑏𝑎𝑠𝑒 2 → 𝑍 = 𝑍𝐴𝑝𝑢 𝑉𝐴𝑏𝑎𝑠𝑒 2 𝑆𝐴𝑏𝑎𝑠𝑒 𝑍𝐵𝑝𝑢 = 𝑍 𝑍𝐵𝑏𝑎𝑠𝑒 = 𝑍 𝑆𝐵𝑏𝑎𝑠𝑒 𝑉𝐵𝑏𝑎𝑠𝑒 2 → 𝑍 = 𝑍𝐵𝑝𝑢 𝑉𝐵𝑏𝑎𝑠𝑒 2 𝑆𝐵𝑏𝑎𝑠𝑒 So we find that 𝑍𝐴𝑝𝑢 𝑉𝐴𝑏𝑎𝑠𝑒 2 𝑆𝐴𝑏𝑎𝑠𝑒 = 𝑍𝐵𝑝𝑢 𝑉𝐵𝑏𝑎𝑠𝑒 2 𝑆𝐵𝑏𝑎𝑠𝑒 → 𝑍𝐵𝑝𝑢 = 𝑍𝐴𝑝𝑢 𝑉𝐴𝑏𝑎𝑠𝑒 𝑉𝐵𝑏𝑎𝑠𝑒 2 𝑆𝐵𝑏𝑎𝑠𝑒 𝑆𝐴𝑏𝑎𝑠𝑒 Electrical Energy Systems Dr B. Diarra 13
- 13. Change of base for voltage Let’s consider two bases A and B and 𝑉 a voltage 𝑉𝐴𝑝𝑢 = 𝑉 𝑉𝐴𝑏𝑎𝑠𝑒 → 𝑉 = 𝑉𝐴𝑝𝑢𝑉𝐴𝑏𝑎𝑠𝑒 𝑉𝐵𝑝𝑢 = 𝑉 𝑉𝐵𝑏𝑎𝑠𝑒 → 𝑉 = 𝑉𝐵𝑝𝑢𝑉𝐵𝑏𝑎𝑠𝑒 So we find that 𝑉𝐴𝑝𝑢𝑉𝐴𝑏𝑎𝑠𝑒 = 𝑉𝐵𝑝𝑢𝑉𝐵𝑏𝑎𝑠𝑒 → 𝑉𝐵𝑝𝑢 = 𝑉𝐴𝑝𝑢 𝑉𝐴𝑏𝑎𝑠𝑒 𝑉𝐵𝑏𝑎𝑠𝑒 Electrical Energy Systems Dr B. Diarra 14
- 14. Example A power system consists of one synchronous generator and one synchronous motor connected by two transformers and a transmission line. Create a per-phase, per-unit equivalent circuit of this system using a base apparent power of 100 MVA and the base voltage of G1 of 13.8 kV. G1 ratings: 100 MVA, 13.8 kV, R=0.1 pu, XS=0.9 pu T1 ratings: 100 MVA, 13.8/110 kV, R=0.01 pu, XS=0.05 pu T2 ratings: 50 MVA, 120/14.4 kV, R=0.01 pu, XS=0.05 pu M ratings: 50 MVA, 13.8 kV, R=0.1 pu, XS=1.1 pu L1 ratings: R=15 Ω, X=75 Ω Electrical Energy Systems Dr B. Diarra 15
- 15. Different steps First step: find the base voltages Second step: find the base impedance Third step: convert impedances at the right base Draw the per-unit per phase system Electrical Energy Systems Dr B. Diarra 16
- 16. Example The base quantities fixed by transformers!! Electrical Energy Systems Dr B. Diarra 17 𝑉𝑏𝑎𝑠𝑒1 = 13.8 𝑘𝑉 𝑍𝑏𝑎𝑠𝑒1 = 𝑉𝑏𝑎𝑠𝑒1 2 𝑆𝑏𝑎𝑠𝑒 = 13.8𝑥 103 2 100𝑥106 = 1.9044 Ω 𝑉𝑏𝑎𝑠𝑒2 = 𝑉𝑏𝑎𝑠𝑒1 110 𝑘𝑉 13.8 𝑘𝑉 = 110 𝑘𝑉 𝑍𝑏𝑎𝑠𝑒2 = 𝑉𝑏𝑎𝑠𝑒2 2 𝑆𝑏𝑎𝑠𝑒 = 110𝑥 103 2 100𝑥106 = 121 Ω 𝑉𝑏𝑎𝑠𝑒3 = 𝑉𝑏𝑎𝑠𝑒2 14.4 𝑘𝑉 120 𝑘𝑉 = 13.2 𝑘𝑉 𝑍𝑏𝑎𝑠𝑒3 = 𝑉𝑏𝑎𝑠𝑒3 2 𝑆𝑏𝑎𝑠𝑒 = 13.2𝑥 103 2 100𝑥106 = 1.7424 Ω
- 17. Example For generator G1 𝑅𝑝𝑢 = 0.1𝑝𝑢 and 𝑋𝑆𝑝𝑢 = 0.9 𝑝𝑢 its ratings are equal to the base values For transformer T1 𝑅𝑝𝑢 = 0.01 pu and 𝑋𝑆𝑝𝑢 = 0.05 𝑝𝑢 its ratings are equal to the base values For transmission line L1 𝑅𝑝𝑢 = 𝑅 𝑍𝑏𝑎𝑠𝑒1 = 15 121 = 0.124 𝑝𝑢, 𝑋𝑝𝑢 = 𝑋 𝑍𝑏𝑎𝑠𝑒1 = 75 121 = 0.372 𝑝𝑢 Electrical Energy Systems Dr B. Diarra 18
- 18. Example For transformer T2 𝑅𝑝𝑢𝑛𝑒𝑤 = 𝑅𝑝𝑢𝑜𝑙𝑑 𝑉𝑏𝑎𝑠𝑒𝑜𝑙𝑑 𝑉𝑏𝑎𝑠𝑒𝑛𝑒𝑤 2 𝑆𝑏𝑎𝑠𝑒𝑛𝑒𝑤 𝑆𝑏𝑎𝑠𝑒𝑜𝑙𝑑 = 0.01 14.4 13.2 2 100 50 = 0.024 𝑋𝑝𝑢𝑛𝑒𝑤 = 𝑋𝑝𝑢𝑜𝑙𝑑 𝑉𝑏𝑎𝑠𝑒𝑜𝑙𝑑 𝑉𝑏𝑎𝑠𝑒𝑛𝑒𝑤 2 𝑆𝑏𝑎𝑠𝑒𝑛𝑒𝑤 𝑆𝑏𝑎𝑠𝑒𝑜𝑙𝑑 = 0.05 14.4 13.2 2 100 50 = 0.120 For the machine M 𝑅𝑝𝑢𝑛𝑒𝑤 = 𝑅𝑝𝑢𝑜𝑙𝑑 𝑉𝑏𝑎𝑠𝑒𝑜𝑙𝑑 𝑉𝑏𝑎𝑠𝑒𝑛𝑒𝑤 2 𝑆𝑏𝑎𝑠𝑒𝑛𝑒𝑤 𝑆𝑏𝑎𝑠𝑒𝑜𝑙𝑑 = 0.1 13.8 13.2 2 100 50 = 0.219 𝑋𝑝𝑢𝑛𝑒𝑤 = 𝑋𝑝𝑢𝑜𝑙𝑑 𝑉𝑏𝑎𝑠𝑒𝑜𝑙𝑑 𝑉𝑏𝑎𝑠𝑒𝑛𝑒𝑤 2 𝑆𝑏𝑎𝑠𝑒𝑛𝑒𝑤 𝑆𝑏𝑎𝑠𝑒𝑜𝑙𝑑 = 1.1 13.8 13.2 2 100 50 = 2.405 Electrical Energy Systems Dr B. Diarra 19
- 19. Example Per unit circuit (can be simplified) Electrical Energy Systems Dr B. Diarra 20 𝑗0.9 𝑝𝑢 𝑗0.05 𝑝𝑢 𝑗0.62 𝑝𝑢 𝑗0.119 𝑝𝑢 𝑗2.405 𝑝𝑢 0.01 𝑝𝑢 0.124 𝑝𝑢 0.1 𝑝𝑢 0.0238 𝑝𝑢 0.219 𝑝𝑢 𝐸𝐺 𝐸𝑀 𝐺1 𝑀
- 20. Remarks Apparent power in per unit 𝑆 = 3𝑉𝐼∗ → 𝑆𝑝𝑢 = 𝑆 𝑆𝑏𝑎𝑠𝑒 = 3𝑉𝐼∗ 3𝑉𝑏𝑎𝑠𝑒𝐼𝑏𝑎𝑠𝑒 = 3𝑉𝐿𝐿𝐼∗ 3𝑉𝑏𝑎𝑠𝑒𝐼𝑏𝑎𝑠𝑒 → 𝑆𝑝𝑢 = 𝑉𝐿𝐿𝑝𝑢𝐼𝑝𝑢 ∗ = 𝑉 𝑝𝑢𝐼𝑝𝑢 ∗ Ohm’s law gives 𝑉 = 𝑍𝐼 → 𝑉 𝑝𝑢 = 3𝑉 𝑉𝑏𝑎𝑠𝑒 = 3𝑍𝐼 3𝑍𝑏𝑎𝑠𝑒𝐼𝑏𝑎𝑠𝑒 → 𝑉 𝑝𝑢 = 𝑍𝑝𝑢𝐼𝑝𝑢 Finally the complex power becomes 𝑉 𝑝𝑢 = 𝑍𝑝𝑢𝐼𝑝𝑢 → 𝑆𝑝𝑢 = 𝑉 𝑝𝑢 2 𝑍𝑝𝑢 ∗ Electrical Energy Systems Dr B. Diarra 21
- 21. Example2 • 44 Electrical Energy Systems Dr B. Diarra 22
- 22. Conclusion Per unit representation makes easy the manipulation of systems Only amplitudes are concerned not phases All the quantities can be represented in per unit Useful in fault analysis and power flow studies All electrical laws remains applicable Electrical Energy Systems Dr B. Diarra 23