Passive Elements and Phasor Diagrams Resistor Inductor Capacitor v i I V v i I V v i I V Industrial Power Systems Salvador Acevedo

Transformer feeding load: Ideal Transformer Transformer feeding load: V2 = V1/a I2 = V2/Z I1= I2/a V2 V1 I1 I2 Assuming a RL load connected to secondary and ideal source to primary Industrial Power Systems Salvador Acevedo

Two Winding Transformer Model The linear equivalent model of a real transformer consists of an ideal transformer and some passive elements Industrial Power Systems Salvador Acevedo

AC Generators and Motors AC synchronous generator Single-phase equivalent AC synchronous motor AC induction motor (rarely used as generator) Industrial Power Systems Salvador Acevedo

Steady-state Solution In sinusoidal steady-state a circuit may be solved using phasors q vs i 0 p 2p Iy=I sinq Ix=I cosq q I VS I q Rectangular form Polar form Industrial Power Systems Salvador Acevedo

Single-phase Power Definitions Industrial Power Systems Salvador Acevedo

Real Power P = S cos q = V I cos q watts Power Triangle Ssin q=Q S P=Scosq q Real Power P = S cos q = V I cos q watts Reactive Power Q = S sin q = V I sin q vars Industrial Power Systems Salvador Acevedo

Power Consumption by Passive Elements Resistive Load A resistor absorbs P Purely Inductive Load An inductor absorbs Q Purely Capacitive Load A capacitor absorbs negative Q. It supplies Q. Industrial Power Systems Salvador Acevedo

Advantages of Three-phase Systems Creation of the three-phase induction motor Efficient transmission of electric power 3 times the power than a single-phase circuit by adding an extra cable Savings in magnetic core when constructing Transformers Generators Single-phase Load + v - i Three-phase Load va vb vc ia ib ic p = vi p = va ia + vb ib + vc ic Industrial Power Systems Salvador Acevedo

vb(t) = Vm sin (wt - 2p/3) = Vm sin (wt - 120°) volts Three-phase Voltages va vb vc va(t) = Vm sin wt volts vb(t) = Vm sin (wt - 2p/3) = Vm sin (wt - 120°) volts vc(t) = Vm sin (wt - 4p/3) = Vm sin (wt - 240°) volts or vb(t) = Vm sin (wt + 2p/3) = Vm sin (wt + 120°) volts w = 2 p f w: angular frequency in rad/sec f : frequency in Hertz Va Vb Vc 120° Industrial Power Systems Salvador Acevedo

Y-connected Voltage Source Star Connection (Y) Y-connected Voltage Source Industrial Power Systems Salvador Acevedo

D-connected Voltage Source Delta Connection (D ) D-connected Voltage Source Industrial Power Systems Salvador Acevedo

Y-connected Load Industrial Power Systems Salvador Acevedo

D-connected Load Industrial Power Systems Salvador Acevedo

Y-D Equivalence Industrial Power Systems Salvador Acevedo

Power in Three-phase Circuits Industrial Power Systems Salvador Acevedo

Three-phase Power Industrial Power Systems Salvador Acevedo

Three-phase Transformers Use of one three-phase transformer unit Use of 3 single phase transformers to form a “Transformer Bank Industrial Power Systems Salvador Acevedo

Physical Overview Industrial Power Systems Salvador Acevedo

Three-phase Transformers Connections Y-Y; D-D; Y- D; D -Y A single-phase transformer ® Example:10 KVA, 38.1KV/3.81KV Three-phase Transformers Connections Y-Y; D-D; Y- D; D -Y Bank of 3 single-phase transformers Primary terminals Secondary terminals The Bank Transformation Ratio is defined as: Industrial Power Systems Salvador Acevedo

using 3 single-phase transformers: Y-Y connection Ratings for Y-Y bank using 3 single-phase transformers: 3x10KVA = 30 KVA 66 KV / 6.6 KV Industrial Power Systems Salvador Acevedo

D-D connection Ratings for D-D bank: 30 KVA; 38.1 KV / 3.81 KV Industrial Power Systems Salvador Acevedo

Y- D connection Ratings for Y-D bank: 30 KVA; 66 KV / 3.81 KV Industrial Power Systems Salvador Acevedo

D -Y connection Ratings for D-Y bank: 30 KVA; 38.1 KV / 6.6 KV Industrial Power Systems Salvador Acevedo

Power lines operate at kilovolts (KV) Per unit modelling Power lines operate at kilovolts (KV) and kilowatts (KW) or megawatts (MW) To represent a voltage as a percent of a reference value, we first define this BASE VALUE. Example: Base voltage: Vbase = 120 KV ** The percent value and the per unit value help the analyzer visualize how close the operating conditions are to their nominal values. Industrial Power Systems Salvador Acevedo

V: voltage VBASE I: current IBASE S: power SBASE Z: impedance ZBASE Defining bases 4 quantities are needed to model a network in per unit system: V: voltage VBASE I: current IBASE S: power SBASE Z: impedance ZBASE Given two bases, the other two quantities are easily determined. Industrial Power Systems Salvador Acevedo

Three phase bases In three-phase systems it is common to have data for the three-phase power and the line-to-line voltage. With p.u. calculations, three-phase values of voltage, current and power can be used without undue anxiety about the result being a factor of Ö3 incorrect !!! Industrial Power Systems Salvador Acevedo

Sbase=300 MVA (three-phase power) Vbase=100 KV (line-to-line voltage) Example The following data apply to a three-phase case: Sbase=300 MVA (three-phase power) Vbase=100 KV (line-to-line voltage) Three-phase load 270 MW 100 KV pf=0.8 a b c + V=1 p.u. - Single-phase equivalent: I=1.125 p.u. Industrial Power Systems Salvador Acevedo

Transformers in per unit calculations With an ideal transformer High Voltage Bases Low Voltage Bases Sbase1 = 5 KVA Sbase2 = 5KVA Vbase1= 2400 V Vbase2 = 120 V Ibase1 = 5000/2400=2.083 A I base2 = 5000/120=41.667 A Zbase1= 2400/2.083=1152 W Z base2 = 120/41.667=2.88 W From the circuit: V1=2400 V. V2=V1/a=V1/20=120 V. In per unit: V1=1.0 p.u. V2=1.0 p.u. The load in per unit is: Z=(5Ð 30°)/Zbase2 =1.7361 Ð 30° p.u. The current in the circuit is: I=(1.0 Ð0°)/ (1.7361 Ð 30°) =0.576 Ð-30° p.u. The current in amperes is: Primary: I1=0.576 x Ibase1= 1.2 A. Secondary: I2=0.576 x Ibase2= 24 A. Industrial Power Systems Salvador Acevedo

One line diagrams A one line diagram is a simplified representation of a multiphase-phase circuit. Industrial Power Systems Salvador Acevedo

Nodal Analysis Suppose the following diagram represents the single-phase equivalent of a three-phase system Finding Norton equivalents and representing impedances as admittances: I1=y1 V1 + y12(V1-V2) + y13(V1-V3) 0 = y12 (V2-V1) + y2 V2 + y23(V2-V3) I3=y13(V3-V1) + y23(V3-V2) + y3 V3 Industrial Power Systems Salvador Acevedo

General form of the nodal analysis Once the voltages are found, currents and powers are easily evaluated from the circuit. We have solved one of the phases of the three-phase system (e.g. phase ‘a’). Quantities for the other two phases are shifted 120 and 240 degrees under balanced conditions. Actual quantities can be found by multiplying the per unit values by their corresponding bases. Industrial Power Systems Salvador Acevedo