EEN320 - Power Systems I ( Συστήματα Ισχύος Ι ) Part 4: The per-unit system Dr Petros Aristidou Department of Electrical Engineering, Computer Engineering & Informatics Last updated: February 10, 2020
Today’s learning objectives After this part of the lecture and additional reading, you should be able to . . . . . . explain the use and advantages of the per unit system in power 1 system computations; . . . convert physical quantities to their corresponding per unit values; 2 . . . calculate stationary network conditions using the per unit system. 3 , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 2/ 33
Outline Principle and advantages 1 Introduction of per unit quantities via an example 2 Conversion between different per unit systems 3 Choice of base values in power systems with several zones 4 , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 3/ 33
1 Outline Principle and advantages 1 Introduction of per unit quantities via an example 2 3 Conversion between different per unit systems Choice of base values in power systems with several zones 4 , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 4/ 33
1 Principle of ”per unit” system Usual representation of physical quantities as product of numerical value and physical unit, e.g. V = 400 kV Alternative: representation of the quantity relative to another (base) quantity value of quantity in physical unit value of quantity in pu = value of corresponding ”base” in same unit Division by ”base” eliminates physical unit → per-unit (pu) system Example: base value for voltage V base = 400 kV (On board) , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 5/ 33
1 Advantages (1) Appropriate choice of base values gives pu-values very useful meaning Example: express bus voltage V relative to nominal grid voltage V base and suppose that v = 0 . 93 pu → We see immediately that value of v is 7% below nominal voltage This is much easier to see than by looking at the absolute value V = 372 . 03 kV Better conditioning of numerical computations Under normal operating conditions, voltage values in pu are close to 1 Networks of different dimensions and voltage levels can be represented in same order of magnitude Example: 100 MVA = 1 pu for large networks and 1 MVA = 1 pu for small networks , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 6/ 33
1 Advantages (2) Easier comparison of components of different power ratings Consider two transformers and suppose that their currents are indicated in pu with respect to their respective maximum currents Suppose that i 1 = 0 . 99 pu and i 2 = 0 . 35 pu We see immediately that transformer 1 is operating much closer to its limit than transformer 2 In general, parameters of similar devices have similar pu values, independently of their power rating (as long as the values are referred to that rating) → Can check quickly if data of a component/machine is within usual range Ideal transformer present in real transformer model is eliminated in the equivalent per-unit circuit (see example later) , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 7/ 33
2 Outline Principle and advantages 1 Introduction of per unit quantities via an example 2 3 Conversion between different per unit systems Choice of base values in power systems with several zones 4 , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 8/ 33
2 Exemplary circuit I It holds that 1 j ω C = − j 1 1 X L = ω L X C = ω C = − jX C R ω C From KVL we have that V jX L V = RI + jX L I − jX C I Goal: Represent variables in pu system − jX C First question: How to choose base quantities? , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 9/ 33
2 Exemplary circuit - Choice of base values Basic relations in stationary power systems S = V I ∗ , V = Z I I → Can choose two independent base quantities R Other base values are obtained using fundamental laws for electric circuits V jX L Typical choice of base quantities 1) Base power S B 2) Base voltage V B − jX C Note 1: base values are always real numbers! Note 2: usual numbers of base values correspond to nominal (power and voltage) ratings of the circuit , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 10/ 33
2 Exemplary circuit - Base voltage Introduce base voltage V B I Then the per unit representation of V is obtained as R v = V V B V jX L Then v = V = RI + jX L I − jX C I V B V B V B V B − jX C Example: rated voltage of circuit is 110 kV → Choose V B ≈ 110 kV , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 11/ 33
2 Exemplary circuit - Base power Introduce (single-phase) base power S B 1 φ S B 1 φ = V B I B = V 2 B Z B I Then the per unit representation of S is obtained as R S s = S B 1 φ V jX L And the base values for currents and impedances follow from the relations V 2 I B = S B 1 φ Z B = V B − jX C B = V B I B S B 1 φ Per unit representation of current and impedance i = I z = Z I B Z B , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 12/ 33
2 Exemplary circuit - Per unit representation (1) For the voltage in per unit we had the relation I v = V = RI + jX L I − jX C I R V B V B V B V B By expressing the current in per unit via V jX L i = I I B we obtain − jX C v = V = RI B i + jX L I B i − jX C I B i V B V B V B V B , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 13/ 33
2 Exemplary circuit - Per unit representation (2) Recall that base impedance is given by Z B = V B I I B Base impedance Z B is base value for both real and R complex impedances V jX L Hence r = R = RI B x L = X L = X L I B x C = X C = X C I B Z B V B Z B V B Z B V B − jX C We obtain v = V = RI B i + jX L I B i − jX C I B i V B V B V B V B = ri + jx L i − jx c i , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 14/ 33
2 Exemplary circuit - Per unit representation (3) I By introducing the overall impedance Z = R + j ( X L − X C ) R and its per unit representation V jX L z = Z = r + j ( x L − x C ) Z B we obtain the compact representation in pu − jX C v = z i , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 15/ 33
2 Per unit quantities - Summary Choose two base quantities, e.g. S B and V B 1 S B can be either single- or three-phase power I S B 3 φ = 3 S B 1 φ R Other values obtained via electrical laws 2 √ S B 1 φ S B 3 φ S B 3 φ Base current I B = = 3 V B = U B = 3 V B √ V B V jX L 3 U B V 2 3 V 2 U 2 Base impedance Z B = V B I B = S B 1 φ = S B 3 φ = B B B S B 3 φ 1 Base admittance Y B = G B = B B = − jX C Z B V B and I B are always RMS values per phase! In non-stationary conditions usually frequency and/or time are also normalised , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 16/ 33
3 Outline Principle and advantages 1 Introduction of per unit quantities via an example 2 3 Conversion between different per unit systems Choice of base values in power systems with several zones 4 , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 17/ 33
3 Conversion between different per unit systems In practice, it is often necessary to convert values from one per unit system to another one Example: machine parameters are given in per unit values with respect to machine rating and we want to convert them into per unit values with respect to base values of power system to which machine is connected This can be done as follows X Per unit value wrt first base: x 1 = X B , 1 X Per unit value wrt second base: x 2 = X B , 2 Hence: X = x 1 X B , 1 = x 2 X B , 2 → Conversion from base 1 to base 2: x 2 = x 1 X B , 1 X B , 2 , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 18/ 33
3 Conversion between different per unit systems - Example Conversion of an impedance Z from base ”old” to base ”new” Per unit value wrt first base: = ZS old = ( V old B ) 2 Z B 1 φ Z old z old = B Z old ( V old S old B ) 2 B B 1 φ Per unit value wrt second base: = ZS new = ( V new ) 2 Z B 1 φ Z new B z new = B Z new ( V new ) 2 S new B B B 1 φ Conversion from base ”old” to base ”new” � V old S new � 2 Z old B 1 φ B B z new = z old = z old Z new V new S old B B 1 φ B , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 19/ 33
3 Example: Conversion between per unit systems (1) Task. A three-phase transformer is rated 400 MVA, 220Y/22 ∆ kV. The Y-equivalent short-circuit impedance measured on the low-voltage side of the transformer is 0.121 Ω . Due to the low resistance, this value can be considered to be equal to the leakage reactance of the transformer. Determine the per-unit reactance of the transformer by taking the secondary voltage as base voltage. Determine the per-unit reactance in a system with base values S B 3 φ = 100 MVA and V B = 230 kV. , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 20/ 33
3 Example: Conversion between per unit systems (2) Solution. (On board) , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: February 10, 2020 21/ 33
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