§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