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Department of Electrical & Communication Engineering CHAPTER 1 1. POWER SYSTEMS Per Unit Calculations Power System Representation Power Component M Symbol Power Component Symbol = Generator = Circuit breaker = Transformer = = Motor = Feeder + load Transmission line = Busbar (substation) Power components and symbols 1 Department of Electrical & Communication Engineering POWER SYSTEMS Interconnections among these components in the power system may be shown is a so-called one-line diagram or single-line diagram. Single-line diagram represents all 3- of balanced system. For the purpose of analysis, the single-line diagram of a particular power system network is represented to its equivalent reactance or impedance diagram. A sample of a interconnected of individual power component is shown in Figure 1.1. This represent a circuit diagram of a power network which is referred to as a single-line diagram. M Figure 1.1 – Single-line diagram 2 Department of Electrical & Communication Engineering POWER SYSTEMS Impedance diagram In power system fault calculations it is often that a single-line diagram representing a typical power network in 3- be converted into its per phase impedance diagram. Some assumptions for converting from single-line diagram into its equivalent impedance diagram needed to be considered. (i) A generator can be represented by a voltage source in series with an inductive reactance. The internal resistance of the generator is assumed to be negligible compared to the reactance. (ii) The loads are usually inductive represented by resistance and inductance. (iii) The transformer core is assumed to be ideal, and the transformer may be represented by a reactance only. (iv) The transmission line is represented by its resistance and inductance, the line-to-ground capacitance is assumed to be negligible. Let us consider the following on how the single-line diagram of Figure 1.2 converted into its impedance diagram counterpart. 3 Department of Electrical & Communication Engineering Generator G1 Station A Transformer T1 Transmission Line TL POWER SYSTEMS Transformer T2 Station B G3 G2 G4 Load L1 Load L2 Figure 1.2 – Single-line diagram of a power network 4 Department of Electrical & Communication Engineering Transformer T1 Transmission Line TL POWER SYSTEMS Transformer T2 Station B Station A j XT1 j X1 j X2 G1 G2 RL1 j XL1 RTL j XTL j XT2 RL2 j XL2 j X3 G3 j X4 G4 Figure 1.3 – Impedance diagram of Figure 1.2 5 Department of Electrical & Communication Engineering POWER SYSTEMS Per-Unit Quantities Per unit quantities are quantities that have been normalized to a base quantity. In general, Z pu Z actual Z base per-unit (p.u) Choice of the base value Zbase is normally a rated value which is often one of the normal full-load operations of power component in a power network. Let us look at two of the most common per unit formula which are widely used when per unit calculations are involved. (i) Base impedance (Zbase) For a given single-line (one-line) diagram of a power network, all component parameters are expressed in 3- quantity whether it is the rating (capacity) expressed as MVA or voltage as kV. Let begin with 3- base quantity of Sbase 3Vbase I base ----- (i) where Vbase = line voltage, Ibase= line or phase current Per phase base impedance, V base Z base I base 3 -----(ii) This is line-to-neutral impedance 6 Department of Electrical & Communication Engineering POWER SYSTEMS Combining (i) and (ii) yields, V base Z base S base 3 Z base 3V base kV base 2 MVAbase where kVbase and MVAbase are 3- qualtities (ii) Changing base impedance (Znew] Sometimes the parameters for two elements in the same circuit (network) are quoted in per-unit on a different base. The changing base impedance is given as, 2 Z NEW kVbase OLD MVAbase NEW pu Z OLD 2 MVAbase OLD kVbase NEW 7 Department of Electrical & Communication Engineering POWER SYSTEMS Example 1 Determine the per-unit values of the following single-line diagram and draw the impedance diagram. 5 MVA Xg = 16% 100 MVA 275 kV/132 kV Transmission line j 3.48 50 MVA 132 kV/66 kV XT2 = 0.04 p.u XT1 = 0.1 p.u Load 40 MW, 0.8 p.f. lagging Solution: Chosen base: Always choose the largest rating, therefore Sbase = 100 MVA, V = 66 kV, 132 kV and 275 kV Per-unit calculations: Transformer T1: Generator G1: 2 Z NEW kVbase OLD MVAbase NEW pu Z OLD 2 MVAbase OLD kVbase NEW X g ( pu ) 0.16 100 0.32 50 p.u. X T 1 ( pu ) 0.1 p.u. 8 Department of Electrical & Communication Engineering Transmission line TL: Z base Transformer T2: kV base 2 MVAbase 3.4 100 X TL ( pu ) POWER SYSTEMS 1322 Z pu Z actual Z base X T 2 ( pu ) 0.04 100 0.08 50 p.u. 0.0195 p.u. Inductive load: 66 103 Z actual Z L ( pu ) 40 10 3 87.1236.87o 6 3 66 103 0.8 87.1236.87o 100 2 66 236.87o or (1.6 j1.2) p.u. 9 Department of Electrical & Communication Engineering POWER SYSTEMS Now, we have all the impedance values in per-unit with a common base and we can now combine all the impedances and determine the overall impedance. 5 MVA Xg = 16% 100 MVA 275 kV/132 kV Transmission line j 3.48 50 MVA 132 kV/66 kV XT2 = 0.04 p.u XT1 = 0.1 p.u Load 40 MW, 0.8 p.f. lagging Transformer T1 j 0.1 p.u. j 0.32 p.u. Transmission Line TL j 0.0195 p.u. Transformer T2 j 0.08 p.u. 1.6 p.u.. Generator G Load j 1.2 p.u. 10 Department of Electrical & Communication Engineering POWER SYSTEMS 11 Department of Electrical & Communication Engineering POWER SYSTEMS Load 12 Department of Electrical & Communication Engineering POWER SYSTEMS Summarise: Load 13 Department of Electrical & Communication Engineering POWER SYSTEMS Load 14