1. 1 Department of Electrical & Communication Engineering CHAPTER 1 Per Unit Calculations 1.Power System Representation Power Component SymbolPower ComponentSymbol = Generator = Circuit breaker M = Transformer = Transmission line = Motor =Feeder + load = Busbar (substation) Power components and symbols POWER SYSTEMS

2. 2 Department of Electrical & Communication Engineering 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 POWER SYSTEMS

3. 3 Department of Electrical & Communication Engineering 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. POWER SYSTEMS

4. 4 Department of Electrical & Communication Engineering Generator G 1 G2G2 Station A Station B G3G3 G4G4 Load L 1 Load L 2 Transformer T 1 Transmission Line T L Transformer T 2 Figure 1.2 – Single-line diagram of a power network POWER SYSTEMS

5. 5 Department of Electrical & Communication Engineering G2G2 G1G1 G3G3 G4G4 j X 1 j X 2 j X 3 j X 4 j X T1 j X T2 R L1 j X L1 R L2 j X L2 R TL j X TL Transformer T 1 Transformer T 2 Transmission Line T L Station A Station B Figure 1.3 – Impedance diagram of Figure 1.2 POWER SYSTEMS

6. 6 Department of Electrical & Communication Engineering Per-Unit Quantities Per unit quantities are quantities that have been normalized to a base quantity. In general, per-unit (p.u) Choice of the base value Z base 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 (Z base ) 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 ----- (i) where Vbase = line voltage, Ibase= line or phase current Per phase base impedance, -----(ii) This is line-to-neutral impedance POWER SYSTEMS

7. 7 Department of Electrical & Communication Engineering Combining (i) and (ii) yields, where kV base and MVA base 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, POWER SYSTEMS

8. 8 Department of Electrical & Communication Engineering Determine the per-unit values of the following single-line diagram and draw the impedance diagram. Example 1 X T1 = 0.1 p.u 5 MVA X g = 16% 100 MVA 275 kV/132 kV 50 MVA 132 kV/66 kV Transmission line j 3.48  X T2 = 0.04 p.uLoad 40 MW, 0.8 p.f. lagging Solution: Chosen base: Always choose the largest rating, therefore S base = 100 MVA, V = 66 kV, 132 kV and 275 kV Per-unit calculations: Generator G1: p.u. Transformer T1: p.u. POWER SYSTEMS

9. 9 Department of Electrical & Communication Engineering Transmission line TL: p.u. Transformer T2: p.u. Inductive load: p.u.  POWER SYSTEMS

10. 10 Department of Electrical & Communication Engineering 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. Load G j 0.32 p.u. j 0.1 p.u. j 0.0195 p.u. Transformer T 1 Transformer T 2 Transmission Line T L j 0.08 p.u. 1.6 p.u.. j 1.2 p.u. Generator POWER SYSTEMS X T1 = 0.1 p.u 5 MVA X g = 16% 100 MVA 275 kV/132 kV 50 MVA 132 kV/66 kV Transmission line j 3.48  X T2 = 0.04 p.uLoad 40 MW, 0.8 p.f. lagging

11. 11 Department of Electrical & Communication Engineering POWER SYSTEMS

12. 12 Department of Electrical & Communication Engineering Load POWER SYSTEMS

13. 13 Department of Electrical & Communication Engineering Load POWER SYSTEMS Summarise:

14. 14 Department of Electrical & Communication Engineering Load POWER SYSTEMS