Principles of Electric Machines and Power Electronics Chapter 2-3 Transformers Third Edition P. C. Sen
Auto transformer
Per unit system S b = S rate V b1 = V rate1 V b2 = V rate2 S b I b1 = = S rate = I V base1 V rate1 rate1 I b2 = S b = S rate = I V base2 V rate2 rate2 Z b1 = V b1 I b1 = V rate1 I rate1 Z b2 = V b2 I b2 = V rate2 I rate2 Fig_2-15
Chapter 2 Main contents (continued) Main contents Balanced three phase system Three-phase transformer Fig_1-1
Balanced three-phase power system Y-connection with positive phase sequence 120 degree shift between any two lines Neutral connection c a V cn V an n V bn b v v a b 120 2 cos( t) o 120 2 cos( t120 ) v c o 120 2 cos( t120 )
Phase sequence b1 c1 a1 a2 c2 b2 a-b-c-a-b-c c-b-a-c-b-a a b c c b a
Three-phase Y-source Y-connection: line to line variables Line-to line voltages
Deriving line to line voltages V ab = V an V bn = 2V rms cos ωt 2V rms cos(ωt 120 ) = 2V rms (cos ωt cos(ωt 120 )) = 2V rms ( 2sin ωt+(ωt 120 ) sin ωt (ωt 120 ) ) 2 2 = 2V rms ( 2sin(ωt 60 ) sin 60 ) = 2V rms (2 cos(ωt + 30 ) 3 2 ) = 2V rms 3 cos(ωt + 30 )
Three-phase Y-source Phase voltage: voltage per source Line voltage: voltage between output terminals Phase current: current per source Line current: current of one transmission line between source and load Y-source: phase current = line current V cn V bn V an c a n b Z l Z l Z l C A N B I cc I aa I n I bb Z Y Z Y Z Y Three phase load IP IL 0 j30 VL L 3V Pe P 3 VP I P cos 3 VLL I P cos V P IP
Three-phase: Δ -source = V ab + V bc + V ca = V pk cos ωt + V pk cos ωt 120 + V pk cos ωt + 120 = V pk (cos ωt + cos ωt 120 + cos ωt + 120 ) = V pk (cos ωt + 2cos( 2ωt 240 ) cos ) 2 2 = V pk (cos ωt + 2cos(ωt)( 1 2 )) = V pk (cos ωt cos(ωt)) = 0 phase voltage = line voltage 0 j30 IL 3I Pe VL L VP P 3 VP I P cos 3 VP I L cos V P IP Where I L is line current, I P is phase current
Δ-Y transformation of source Use Y source to replace Δ source Maintain same V l-l and I l Vab Va Y 3 I a = 0 ( 30 ) 3I ba e j( 30 ) V ab = 3V an e j30 I ab = I ba 3 ej(30 )
Three-phase load phase current = line current phase voltage=line to line voltage
Δ-Y transformation of load With same V l-l and I l Find the relation between Y load and Δ load Z = V AB I AB = 3e j( 30 ) V AB I aa Z Y Z = 1 3 V AN = V AB 3 ej( 30 ) Z Y = V AN = ej( 30 ) V AB I aa 3 I aa
Balanced three-phase power system Per-phase analysis of balanced three-phase system Balanced voltage sources Balanced line and load impedances Balanced resulting currents For Y-source and Y-load (Y- Y), we have Van Vbn Vcn 0 Ia Ib Ic I n 0 V cn V an c a n Z l Z l C A N I cc I aa Z Y Z Y Va Z l ZY I n n N V bn b Z l B I bb Z Y
Balanced three-phase power system How to analyze other types of three-phase source-power connection (Page 14, reference materials) Y- Δ Δ-Y Δ- Δ Solution: transform to Y-Y connection and apply per-phase analysis During transformation, keep two variables unchanged Line to line voltage Line current As a result: power is conservative
Power of balanced three-phase loads Instantaneous power is constant, and equal to average power
Simulation: Power of balanced three-phase loads and single phase load 1. Show phase voltage and phase current 2. Show line voltage 3. Show neutral voltage add to zero 4. Show single phase power/three phase power
Example (Reference, Page 24) A balanced 120 V-rms three-phase source with positive phase sequence. ZY1 = (30 + j40) Ω, ZΔ2 = (60 - j45)ω. Zline = (2 + j4)ω. Determine (a.) the per phase line current (b.) the total real and reactive power drawn from the source (c.) the per phase line-to-neutral voltage across the parallel load (d.) the per phase load current in the wye and delta loads (e.) the total real and reactive three-phase power delivered to each load and the line.