# POLYPHASE CIRCUITS LEARNING GOALS Three Phase Circuits Advantages

POLYPHASE CIRCUITS LEARNING GOALS Three Phase Circuits Advantages of polyphase circuits Three Phase Connections Basic configurations for three phase circuits Source/Load Connections Delta-Y connections Power Relationships Study power delivered by three phase circuits Power Factor Correction Improving power factor for three phase circuits THREE PHASE CIRCUITS V an 3 phase v o lta g e V bn V 0 120 cn 240 Instantane ousPhaseVoltages

van (t ) Vm cos( t )(V ) vbn (t ) Vm cos( t 120 )(V ) vc (t ) Vm cos( t 240 )(V ) Vm 120 2 a a + _ W y e C o n n e c te d S ource V0 n V -2 4 0 + c _ _ V -1 2 0 + b b

c Delta Source a a _ D e lta S o urce + + _ b c _ b + c Vab = | Vab | 0 Vbc = Vab -120 Vca = Vab -240 Wye Wye System a

A Z l ZL n c N ZL b B Z l Z l ZL C Delta Delta System a A Z _

l + c L ZL Z + _ _ + b B Z l Z l C ZL Delta Wye System

a A Z _ ZL + + c l _ _ + ZL b B Z l Z l

ZL C a a IaA + _ A V0 n V -2 4 0 + _ _ IC Z Z IB V -1 2 0 +

IAB b b c C C B Z c A V cn - V bn V ab 30 o

V V bn V V an ab = V ab = 3 V an - V an bn 30o Vab Van Vbn | V p | 0 | V p | 120 | V p | 1 (cos120 j sin 120) 1 3 | V | p | V p | j 2 2

3 | V p | 30 Vbc 3 | V p | 90 Vca 3 | V p | 210 VL 3 | V p | Line Voltage IC 3 I A B -3 0 o IaA = A IAB IaA IB C - IC A INSTANTANEOUS POWER Instantane ousPhaseVoltages van (t ) Vm cos( t )(V ) BalancedPhaseCurrents ia (t ) I m cos( t ) vbn (t ) Vm cos( t 120 )(V )

ib ( t ) I m cos( t 120 ) vc (t ) Vm cos( t 240 )(V ) ic (t ) I m cos( t 240 ) Instantane ous power p(t ) van (t )ia (t ) vbn (t )ib (t ) vcn (t )ic (t ) Theorem For a balanced three phase circuit the instantaneous power is constant p(t ) 3 Vm I m cos (W ) 2 REVIEWOF Y Transforma tions Ra R1R2 R1 R2 R3 Rb R2 R3 R1 R2 R3 Rc

R3 R1 R1 R2 R3 Y R1 Ra Rb Rb Rc Rc Ra Rb R2 Ra Rb Rb Rc Rc Ra Rc R3 Ra Rb Rb Rc Rc Ra Ra Y Y Rab Ra Rb Y Rab R2 || ( R1 R3 ) REVIEWOF Y Transforma tions Ra R1 Rb R1 Rb R2 Rb R1

R1R2 R R 3 2 Ra Rc R1 Rc R2 ( R1 R3 ) Ra R1 R2 R3 Rb R3 Ra Rb REPLACE IN THE THIRD AND SOLVE FOR R1 R1 R2 R3 R2 R3 Rb R1 R2 R3 Ra Rb Rb Rc Rc Ra R3 ( R1 R2 ) R 1 Rb Rc Rb R R 3 1 R1 R2 R3 Rc R1 R2 R3 R R Rb Rc Rc Ra R2 a b Rc

Y R (R R ) Rc Ra 1 2 3 R1 R2 R3 SUBTRACT THE FIRST TWO THEN ADD TO THE THIRD TO GET Ra R3 Ra Rb Rb Rc Rc Ra Ra Y POWER FACTOR CORRECTION Similar to single phase case. Use capacitors to increase the power factor Balanced load Low pf lagging Keep clear about total/phase power, line/phase voltages Q Qnew Qold

ReactivePowerto be added To use capacitors this value should be negative pf cos f sin f 1 pf 2 tan pf f 2 1 pf Q P tan f lagging Q 0 f 60 Hz , | Vline |34.5kV rms. Required: pf 0.94 leading LEARNING EXAMPLE Pold 18.72 MW S P jQ P | S | cos f Q | S | sin f pf cos f Q P tan f tan f pf 1 pf 2 lagging Qold 0 pf cos f sin f 1 pf 2 0.626 | Qold |15.02 MVA Pold 18.72 MW

Qnew 6.8 MVA pf new 0.94 leading Q 6.8 15.02 21.82 MVA Qper capacitor 7.273MVA 34.5 Y connection Vcapacitor kV rms 3 34.5 103 6 7.273 10 2 60 C 3 C 48.6 F 2 LEARNING BY DESIGN Proposed new store #4ACSRwireratedat 170A rms S1 70036.9 S2 100060 kVA S3 80025.8 kVA 560 j 420 kVA 500 j866 kVA 720 j 349 kVA Stotal 1780 j1635 kVA 241742.57 kVA

| S total | 2.417 106 | I line | 101.1A rms 3 3 Vline 3 13.8 10 Wire is OK Pold Qnew P tan f ( new ) 758.28kVA pf new Q Qnew Qold 876.72kVA |Q per capacitor | CV 2 876.72 10 / 3 C 2 60 13.8 10 3 3 2 12.2 F /3

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