ELEC0014 - Introduction to power and energy systems The per unit system Thierry Van Cutsem t.vancutsem@ulg.ac.be www.montefiore.ulg.ac.be/~vct October 2018 1 / 12
Principle The per unit system Principle value of a quantity in physical unit = value of quantity in per unit (pu) value of corresponding base in same unit Advantages: the parameters of devices with similar design have close values in per unit, whatever the power of the device, provided they are referred to that power check of data validity is easier default values can be substituted to unavailable parameters in normal operating conditions, voltages in per unit are close to one better conditioning of numerical computations the ideal transformer present in the model of a real transformer disappears after converting the parameters in per unit. 2 / 12
Example The per unit system Example It is known that the internal reactance of a synchronous machine lies typically in the range [1.5 2.5] pu (on the machine base) A machine with the characteristics (20 kv, 300 MVA) has a reactance of 2.667 Ω. Is this a normal value? We will see that the base impedance is 20 2 /300 = 1.333 Ω value of reactance in per unit = 2.667/1.333 = 2 pu quite normal value! Same question for a machine with the characteristics (15 kv, 30 MVA) The base impedance is now 15 2 /30 = 7.5 Ω value of reactance in per unit = 2.667/7.5 = 0.356 pu abnormal small value! 3 / 12
Converting a simple circuit in per unit Converting a simple circuit in per unit Converting an electric circuit in per unit = choosing 3 base quantities for instance: power S B voltage V B time t B The other base values are obtained using fundamental laws of Electricity: base current: = S B V B base impedance: Z B = V B = V 2 B S B base magnetic flux: ψ B = V B t B base inductance: L B = ψ B = V 2 B t B S B base angular frequency: ω B = Z B L B = 1 t B V B, : RMS values. 4 / 12
Converting a simple circuit in per unit Variant (adopted in this course) Consider an AC circuit operating at frequency f N. Choosing a base angular frequency instead of a base time: ω B = ω N = 2πf N rad/s from which one derives: t B = 1 ω B = 1 ω N = 1 2πf N s With this choice: X pu = X Z B = ω N L ω B L B = L L B = L pu At frequency f N, the reactance and the inductance have the same per unit value! 5 / 12
Converting a simple circuit in per unit Converting in per unit a typical sinusoidal relation In MVA, MW and Mvar: S = V I cos(θ ψ) + j V I sin(θ ψ) In per unit: S pu = S = V I V I cos(θ ψ) + j sin(θ ψ) S B V B V B = V pu I pu cos(θ ψ) + j V pu I pu sin(θ ψ) Same equation in physical units and in per unit! In the above steady-state equation, time does not appear explicitly. Hence, only S B and V B are used. 6 / 12
Converting a simple circuit in per unit Converting in per unit a typical dynamic equation In volts: In per unit: v = R i + L d i d t v pu = v V B = R i Z B + L d i ω B L B d t = R 1 d i pu pui pu + L pu ω B d t = R pu i pu + L pu d i pu d t pu In the above equation, time appears explicitly. There are two options: all variables, including time, are converted in per unit identical equations in physical units and in per unit time is kept in seconds there appears a factor 1/ω B in front of the derivation operator. 7 / 12
Converting in per unit two magnetically coupled circuits Converting in per unit two magnetically coupled circuits Flux-current relations: ψ 1 = L 11 i 1 + L 12 i 2 ψ 2 = L 21 i 1 + L 22 i 2 Identical times. For reasons of simplicity we take: t 1B = t 2B Symmetry of inductance matrices. In Henry, one has: L 12 = L 21. We want that this property still holds true in per unit. ψ 1pu = ψ 1 = L 11 i 1 + L 12 i 2 = L 11pu i 1pu + L 12I 2B ψ 1B L 1B I 1B L 1B I 1B L 1B I }{{ 1B } = L 12pu i 2pu Similarly for the second circuit: L 21pu = L 21I 1B L 2B I 2B L 12pu = L 21pu I 2B = I 1B S 1B t 1B = S 2B t 2B S 1B = S 2B L 1B I 1B L 2B I 2B A per unit system preserving symmetry of inductance matrices is called reciprocal. 8 / 12
Converting in per unit two magnetically coupled circuits Summary circuit # 1 circuit # 2 S B S B one degree of freedom t B t B = 1/ω N V 1B V 2B one degree of freedom in each circuit Application to a network including several voltage levels connected through transformers: the same base power S B is taken at all voltage levels. Usual value in transmission systems: 100 MVA the same base time t B is taken everywhere (not used in steady state) at each voltage level, the base voltage is chosen in relation with the nominal voltage of the equipment. 9 / 12
Converting in per unit a three-phase circuit Converting in per unit a three-phase circuit Same base time t B and same base power S B everywhere at each voltage level, a base V B is chosen for all phase-to-neutral voltages. 1st case. Unbalanced operation - analysis of the three phases Convenient choice: S B = single-phase power base current: = S B V B base impedance: Z B = V B = V 2 B S B etc. Example. Convert in per unit the expression of the three-phase complex power: S pu = S S B = V a V B Ī a + V b V B S = V a Ī a + V b Ī b + V c Ī c Ī b + V c V B Ī c = V a pu Ī a pu + V b pu Ī b pu + V c pu Ī c pu Identical expressions in physical units and in per unit! 10 / 12
Converting in per unit a three-phase circuit 2nd case. Balanced operation - per phase analysis Convenient choice: S B = three-phase power base current: = S B = S B 3V B 3UB U B = 3V B base impedance: Z B = V B = 3V 2 B S B = U2 B S B etc. Example. Convert in per unit the expression of the three-phase complex power: S = V a Ī a + V b Ī b + V c Ī c = 3 V a Ī a S pu = S S B = 3 V a Ī a 3 V B = V a pu Ī a pu 1 All calculations are performed in a single phase, and in per unit 2 at the end, the power in all three phases is obtained by multiplying the power in per unit by S B. 11 / 12
Change of base The per unit system Change of base In a three-phase circuit, an impedance Z (in ohm) becomes in per unit in the first base: Z pu1 = Z Z B1 = Z S B1 3V 2 B1 in the second base: Z pu2 = Z Z B2 = Z S B2 3V 2 B2 Hence the formula to change from the first to the second base is: Z pu2 = Z pu1 S B2 S B1 ( ) 2 VB1 V B2 12 / 12