《电路》（英文版）12-2 Three-phase Y-Y connection

8 12-2 Three-phase Y-Y connection Three-phase sources have three terminals called the line terminals(a, b,c) Neutral connection--n n Balanced three-phase sources may be defined as having 十 C Van =vbn vcn and van+y bn+ cn=0 These three voltages, each existing between one line and the neutral, are called phase voltages(van, Vbn, V cn

§12-2 Three-phase Y-Y connection Three-phase sources have three terminals called the line terminals(a,b,c). Neutral connection--n Balanced three-phase sources may be defined as having = = + + = 0 • • • a n b n cn Va n Vb n Vcn and V V V These three voltages, each existing between one line and the neutral, are called phase voltages ( V an V bn V cn ) . • • • , ,    a b c n • V bn • + + + − − − V an • V cn •

Let bn bn=V∠-120 cn=V∠+120° C Vp-rms magnitude of any of the phase voltages. an=V,∠0 =V,∠120°,Vcm=Vn∠240°

Let  = 0 • p Van V  = −120 • p Vbn V  = +120 • p Vcn V Vp --rms magnitude of any of the phase voltages.  •  •  • or V an =Vp 0 , V bn =Vp 120 , V cn =Vp 240    a b c n • V bn • + + + − − − V an • V cn •

Phase sequence(相序) a→b→c- positive phase sequence c→b→a- negative phase sequence ch n Van+v br bn anty bn (+ phase sequence C phase sequence anty butyn =0 anty btv cn =0

Phase sequence(相序) (-) phase sequence (+) phase sequence + + = 0 • • • V an V bn V cn V bn • V cn • V an • V an V bn • • + V an • V bn • V cn • V an V bn • • + a  b  c -- positive phase sequence c  b  a -- negative phase sequence + + = 0 • • • V an V bn V cn

The line voltage--the line to line voltage Vab y=Vm-Vlmn=√3an∠30°=√3∠30 30 b Jbn∠30°=√3∠-909 a=in-Vm=3∠30°=V3Vn∠-2l0° b 3y and ab leads v bc by 120 be leads v ca by120° ca leads vab by120° ab+ybctvca=0(balanced source

The line voltage--the line to line voltage. V bc • V ca • V ab • V bn • V cn • V an •   = − = 3 30 = 3 30 • • • • p Vab Van Vbn Van V   = − = 3 30 = 3 − 90 • • • • p Vbc Vbn Vcn Vbn V   = − = 3 30 = 3  − 210 • • • • p V ca V cn V an V cn V    120 120 3 120 V leadsV by V leadsV by V V and V leadsV by ca a b b c ca a b b c L p • • • • • •  = V a b +V b c +V ca = 0 ( balanced source ) • • •  30

Y-Y connected Z bB ad 十 bn a4- bB bn⊥Fam∠-120=ia∠-1200 P an∠+120 aA∠+120 Nn A+bB+ 0

Y-Y connected p an aA Z V I • • =   120 120 =  −  − = = • • • • a A p a n p b n b B I Z V Z V I   120 120 =  +  + = = • • • • a A p a n p cn cC I Z V Z V I = + + = 0 • • • • I Nn I aA I bB I cC I aA • I bB • I cC • V an • V bn • V cn •    a b c n • V bn • + + + − − −    A B C • Z p Z p Z p N V cn • V an •

If an impedance Zi is inserted in series with each of the three lines and N is inserted in the neutral B VNn-y Nn bn Z,+ P Nn 十 0 十 L 3 +ontv Nn L 十 十 P P

If an impedance ZL is inserted in series with each of the three lines and ZN is inserted in the neutral. + = 0 + − + + − + + − • • • • • • • N Nn L p Nn cn L p Nn bn L p Nn an Z V Z Z V V Z Z V V Z Z V V ) 0 3 1 ( = + + + + = + • • • • L p a n b n cn L p N N n Z Z V V V Z Z Z V  = 0 • V Nn    a b c n • V an • + + + − − −    A B C • Z p Z p Z p N V bn • V cn • ZL ZL ZN ZL

十 P 十 bn b bB aA∠-120 Z+Z P 十 ∠+120 十 P N aA+lbb+lcc =0 VN=0 NI 0 If we have balanced sources. balanced loads and balanced line impedance, a neutral wire of any impedance may be replaced by other impedance including a short circuit and an open circuit

L p a n aA Z Z V I + = • •  =  − 120 + = • • • a A L p b n b B I Z Z V I  =  + 120 + = • • • a A L p c n cC I Z Z V I = + + = 0 • • • • I Nn I aA I bB I cC = = 0 • • N Nn Nn Z V or I If we have balanced sources, balanced loads, and balanced line impedance, a neutral wire of any impedance may be replaced by other impedance, including a short circuit and an open circuit.    a b c n • V an • + + + − − −    A B C • Z p Z p Z p N V bn • V cn • ZL ZL ZN ZL  = 0 • V Nn

Example: Find Z, and IA 400W Solution: A〖 0. 8 PLeading A2, 300/√3∠0°V/,4004 b 0. 8 PLeading 200 300/√3 300 200W Let van ∠0°p 200 300 =115540°4400=12×0.8∴IA2=289∠3690A 300/√3 3 IA=IA1+IA2=3.87∠26.6°A 300/√3 ∠-36.9°=60∠-36.9°A 2.89

Example: Find Z and I A . • 2 Solution: Let V an V  0 3 300 =  • . A / I A  1 155 0 300 3 200 1 = =  • 0.8 3 300 400 = I 2  I A I A I A A  = 1+ 2 = 3.8726.6 • • • Z A   36.9 60 36.9 2.89 300 / 3 2 =  − =  − I A1 • I A2 • a − + I A • . PFleading W 0 8 400 / V  300 30 200W n Z2 a − + I A • − + − + b c 200W . PFleading W 0 8 400 300 / 3V Z2 I A A   2 = 2.8936.9 •