Lecture15 - EE 740 Professor Ali Keyhani Lecture#15 Synchronous Machine Sequence Network Sequence Networks for Synchronous Machines jXs If Vf Ra Ra

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Unformatted text preview: EE 740 Professor Ali Keyhani Lecture #15 Synchronous Machine Sequence Network Sequence Networks for Synchronous Machines jXs If Vf Ra Ra Ra jXm a b Va c Vc Vb Ea Ec jXm Eb jXm jXs jXs Zn In=Ia+Ib+Ic Where Zn is the Grounding Impedance. Consider the Steady State Operation with: m Constant Constant Field Current (If) No Saturation Balanced 3windings located 120apart Then we have: Ea = (Ra+jXs+Zn)Ia + (jXm+Zn)Ib + (jXm+Zn)Ic + Va Eb = (Ra+jXs+Zn)Ib + (jXm+Zn)Ia + (jXm+Zn)Ic + Vb Ec = (Ra+jXs+Zn)Ic + (jXm+Zn)Ia + (jXm+Zn)Ib + Vc 1 Which is similar to the transmission line equations in Lecture #14 Balanced Voltage Set: c Ea = E Eb = a2E Ec = aE a 2 a b 2 a E Let Zs = Ra + jXs + Zn Zm = jXm + Zn 3 Then equation (1) can be written as: Ea Eb Ec = Zs Zm Zm Zs Zm Zm Zm Zm Zs Ia Ib Ic + 5 6 7 Va Vb Vc 4 [Eabc] = [Zabc] [Iabc] + [Vabc] Recall that [Eabc] = [Ts] [E012] and [Iabc] = [Ts] [I012] Multiply Eqt. 5 by Ts-1 and replace Iabc with Eqt. 7 [Ts]-1 [Eabc] = [Ts]-1 [Zabc] [ Ts] [I012] + [Ts]-1 [Vabc]8 Observe that: E0 E1 E2 = 1/3 1 1 1 1 a a2 1 a2 a E a E aE 2 = E/3 0 3 0 = 0 E 0 9 Thus for a Balanced Set of Voltages we have: E0 = 0 E1 = E E2 = 0 10 Therefore 0 Z1 0 0 0 Z2 11 Recall that: [T]-1 [Vabc] = [V012] [T] -1 [Zabc] [T] = [Z012] or Z0 0 0 Where Z0 = Z0Gen + 3Zn and Z0Gen = Ra+j(Xs+2Xm) Z0 = Ra + j(Xs+2Xm) Z1 = Zs - Zm = Ra + j(Xs-Xm) Z2 = Zs - Zm = Ra + j(Xs-Xm) 12 Now equation (8) can be written in matrix form as: 0 E 0 Z0 0 0 = 0 Z1 0 0 0 Z2 I0 I1 I2 + V0 V1 V2 And when you multiply this matrix out you get: Z0I0 + V0 = 0 Z1I1 + V1 = E Z2I2 + V2 = 0 Therefore only the positive sequence is excited if the generator is balanced and there is only positive sequence voltage. ...
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This note was uploaded on 07/17/2008 for the course ECE 740 taught by Professor Keyhani during the Fall '07 term at Ohio State.

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