《电路》（英文版）12-3 The delta()connection

§12-3 The delta(△) connection A A--connection three-phase sources a.b. c--line ter min als ∠0 bn Vb=V∠-120 bp ∠+120 P 十 C + b +vbp+vcp=0 2V

§12-3 The delta( ) connection  A --connection three-phase sources a,b,c − −line ter min als  = = 0 • • p Van Vap V  = = −120 • • p Vbn Vbp V  = = +120 • • p Vcn Vcp V V L V p • • = + + = 0 • • • V ap V bp V cp V ap • V bp • V cp • n V ap • V bp • V cp • n V ap V bp • • + V ap V bp V cp V cp • • • • + − = −2 c  +  a + − −  b + − a b c V ap • V bp • V cp • c + a + − − b V bp • + − V cp • V ap •

Y-△ connection Y--source B 十 L bn bn 十 L in=3im∠3P=3∠3v如=3∠-90i=3∠+10 △-- connection load: AB AB BC CA AB BC CA The phase currents P AB,IBC,ICA IAB+IBC+ICA=0 and I,=laB=lBC=lcA are aso balance

Y-- connection p an bn cn L ab bc ca V V V V V V V V = = = = = = VL Vp = 3   = 3 30 = 3 30 • • p Vab Van V  = 3 − 90 • p Vbc V  = 3 +150 • p Vca V  -- connection load: p CA CA p BC BC p AB AB Z V I Z V I Z V I • • • • • • = , = , = The phase currents are also balance. I AB I BC I CA • • • , , AB BC CA p AB BC CA I + I + I = and I = I = I = I • • • 0 Y--source: ( Z Z Z Z ) AB = BC = CA = p    a b c n • V bn • + + + − − −   A B C Z p Z p Z p V an • V cn •

The line currents. aA=aB b B 3IAB∠-30° A b BC √3IBC∠-30° BC 3Ic4∠-30° las+bb+l 0 aA bB L

The line currents: I aA I AB I CA • • • = − I bB I BC I AB • • • = − I cC I CA I BC • • • = − a A b B cC L a A b B cC and I I I I I I I = = = + + = • • • 0 L p I = 3I I aA • I bB • V bn • V cn • V bc • V ca • V ab • V an •  30 I AB • I BC • I CA • I cC •  • = 3 I AB  − 30 • = 3 I BC  − 30 • = 3 I CA  − 30    a b c n • V bn • + + + − − −   A B C Z p Z p Z p V an • V cn •

Example: We are to determine the magnitude of the line current in Vi=300 three-phase system which supplies 12000W to a 4-connected load at a lagging pF of 0.8. Solution: V=300J,PAB=12000/3=4000,c0s9=0.8 4000 16.74 300×0.8 L=√3ln=28.9A 300 ∠cos0.8=18∠369°=144+j10.892 16.7

Example: We are to determine the magnitude of the line current in VL=300V three-phase system which supplies 12000W to a  -connected load at a lagging PF of 0.8. Solution: VL = 300V, PAB = 12000/ 3 = 4000W, cos = 0.8 I p 16.7A 300 0.8 4000 =   =  I L = 3I p = 28.9A =  =  = +  −  cos 0.8 18 36.9 14.4 10.8 16.7 300 1 Z j p

The three-phase power We assume a Y-connected load with a power-factor angle 9 P=V cos 9 √3 L COS The total power P=3×P=3×VI1c0s9=√3 LILcOS9 The power delivered to each phase of a A-connection load Pn= V cos9=。1 √3"hcos9 The total power P=3xP 1 3acos=√3 VI cos9 L COS 9 3丿Icos3

The three-phase power We assume a Y-connected load with a power-factor angle .  cos 3 1 cos p p p L L P =V I = V I The total power cos 3 cos 3 1 3 p 3 L L L L P =  P =  V I = V I The power delivered to each phase of a  - connection load.   cos 3 1 cos 3 1 cos p p p L L L L P =V I =V I = V I The total power cos 3 cos 3 1 3 p 3 L L L L P =  P =  V I = V I P = 3VLI L cos = 3VI cos

Example: If VL=380V, Z,=4.5+j692, ZL=1.5+j2 Q2. Find (1)Ia4, IbB, Icc and IAB, IBC, ICA; (2)PI load Solution. L Vant L n 380 P (1 let van ∠0°=220∠0° 十 ad =44∠-53.1°A Z./3 Z,+Z/3 L Z./3 AB ∠30°=25.4∠-23.1°A 3×2×R Z.13 =3×25.42×4.5=8710W

Example: If VL=380V,Zp =4.5+j6Ω, ZL =1.5+j2 Ω. Find    a b c n • V bn • + + + − − −    A B C Z p Z p Z p V an • V cn • ZL ZL ZL (1) I aA,I bB ,I cC and I AB ,I BC ,I CA; • • • • • • ( ) P . 2 load Solution: ( ) let V an V   0 220 0 3 380 1 =  =  • A Z Z V I L p a n a A  44 53.1 / 3 =  − + = • • − + • V an − + • V bn − + • V cn ZL ZL ZL / 3 Z p n N a A / 3 Z p / 3 Z p − + • V an N a A ZL / 3 Z p A I I aA AB   30 25.4 23.1 3 =  =  − • • . . W ( ) Pload I p Rp 3 25 4 4 5 8710 2 3 2 2 =   = =  

The three-phase circuits analysis 1. a balanced three-phase circuit (1)The balanced three-phase system=>Y-Y (2) Draw a phase circuit(a phase), then find voltages and currents of a phase(a phase); (3)Direct write other two phase(B, c) voltages and currents from the balanced relationship; (4)Find voltages and currents of the original circuit from the y-4 transformation

The three-phase circuits analysis 1. A balanced three-phase circuit (1) The balanced three-phase system => Y-Y; (2) Draw a phase circuit (A phase), then find voltages and currents of a phase(A phase); (3) Direct write other two phase(B,C) voltages and currents from the balanced relationship; (4) Find voltages and currents of the original circuit from the Y-  transformation

Example: If: ZL=j5,Z1 =50, z2=90+j120,V=380V, B b (+sequence. Find the phase voltages and currents of two loads Solution: B A2 Z,/3 z,/3

Example: abc ZL Z1 Z2 A B C If: ZL=j5, Z1=50, Z2=90+j120 , VL=380V, (+)sequence. Find the phase voltages and currents of two loads. abc ZL Z1 Z2 / 3 A B C a ZL Z1 Z2 / 3 A n I A • I A1 • I A2 • Solution:

Solution Al B b 380 n Leta==r=∠0°=220∠0°V 之1x2/372∠-350A Z1+Z,/3 Z,/3 41=1A =4.04∠-8.5°AIA2=IA 4.04∠-61.5°A Z,+Z,/3 Z1+Z,/3 i=1∠30=233∠-3159AVn=|∠1|=50×404=20V △ L 350

a ZL Z1 Z2 / 3 A n I A • I A1 • I A2 • Solution: Let V a n V   0 220 0 3 380 =  =  • A Z Z Z Z Z V I L a n A  7.2 35 / 3 / 3 1 2 1 2 =  − +  + = • • . . A Z Z / Z / I A I A  • • =  − + =  4 04 8 5 3 3 1 2 2 1 A Z Z Z I A I A  4.04 61.5 1 2 / 3 1 2 =  − + =  • • A I I A AB   30 2.33 31.5 3 2 =  =  − • • Vp = VL = 3Vp = 350V Vp = Z1 I A1 = 50 4.04 = 202V • a b c ZL Z1 Z2 A B C

DP:IfZ,=j0.62,Z2=4.8+j3.69,Z1{V1=300V,P1=12000 PF1=0.8(lagging), Find VsL Z,/3 b B A2 C Solution: 12000 LIp coS 16.74 3×300×08 LI 30 18z1=18∠cos-0.8=18∠369°=144+/i0.8 In16.7 300 50 Let y ∠0°VI42=I ∠-36.9° A-IAIT1A2 100∠-36 340 Van=z latva ∠8.13° V,=√3=3401 3

DP: If ZL=j0.6Ω, Z2=4.8+j3.6 Ω, Z1 {VL1=300V, P1=12000W, PF1=0.8(lagging)}, Find VSL. a ZL Z1 / 3 Z2 A n I A • I A1 • I A2 a • b c ZL Z1 A B C Z2 Solution: P VL I p I p 16.7A 3 300 0.8 12000 3 cos 1 1 1 1 1 =   =   = 18 16.7 300 1 1 1 = = = p L I V Z 18 cos 0.8 18 36.9 14.4 10.8 1 1 Z =  =  = + j −  Let V An V  0 3 300 =  •  36.9 3 50 2 = 1 =  − • • I A I A  36.9 3 100 = 1+ 2 =  − • • • I A I A I A  8.13 3 340 = + =  • • • A An L V a n Z I V VsL = 3Van = 340V