Symmetrical Components Fall 2007

Transcription

1 0.1 Variables STEADYSTATE ANALYSIS OF SALIENT-POLESYNCHRONOUS GENERATORS This paper is intended to provide a procedure for calculating the internal voltage of a salientpole synchronous generator given the terminal voltage and the complex power delivered. First the notation, terms, parameters, and variables used in the procedure will be defined. Assumptions will be listed. An equations for the power delivered by a salient-pole synchronous generator is given. The actual procedure will be presented followed by an development of the procedure. A modified procedure is given for a generator connected through a reactance to an infinite bus. 1 Definitions 1.1 Notation Phasors will be designated by putting a tilde over the top of a variable; e.g., I a = I a e jθ VI. Variables without a tilde over the top are magnitudes. The subscripts on angles specify between what phasors the angles span; e.g., the angle θ VI is the angle of the voltage with respect to the current. A superscript * denotes the complex conjugate of a phasor or complex number; e.g., I aλ = I a e jθ VI. 1.2 Terms The direct axis is defined as the direction along the rotor that the field winding current causes magnetic flux to flow. The quadrature axis is defined as the axis located π 2 electrical radians behind the direct axis of the rotor. The stator internal voltage is that portion of the terminal voltage due to the flux caused exclusively by the field current. 1.3 Parameters and Variables X d direct axis reactance X q quadrature axis reactance Page 1/7

2 S the total three phase complex power E q a phasor representing the a phase line-to-neutral stator internal voltage V a a phasor representing the a phase line-to-neutral terminal voltage I a a phasor representing the a phase line current a phasor representing the quadrature axis component of the a phase line current The quadrature component is in phase with the stator internal voltage Ẽ q. a phasor representing the direct axis component of the a phase line current The direct component leads or lags the quadrature component by π=2 radians. θ VI the angle of V a with repect to I a θ EI the angle of E q with repect to I a δ the angle of the E q with repect to V a 2 Assumptions It is assumed that the voltage due to the stator winding resistance is negligible; i.e., the stator winding resistance is zero. Magnetic saturation will be ignored. The a phase line-to-neutral voltage, V a is the reference for angles; i.e., V a = V a. 3 Power Delivered by a Salient-Pole Synchronous Generator P = 3» Eq V a X d sin(δ)+ V 2 a 2 Xd X q X d X q sin(2δ) (1) Remove the 3 in Eq. 1 if calculations are being performed in per unit. Page 2/7

3 4 Procedure for Calculating E q Given V a and S d-axis E xq j(x d -X q ) E q q-axis jx q jx q I a delta jx d theta VI Va I a Ĩ a = SΛ 3Ṽ Λ a ) I a = I a e jθ VI (2) θ VI = angle(ĩ s ) (3) Ẽ xq = Ṽ a + jx q Ĩ a ) Ẽ xq = E xq e jδ (4) δ = angle(ẽ xq ) (5) = I a sin(θ VI + δ) (6) Ĩ d = e j(δ+ π 2 ) ) Ĩ d = e j(δ+ π 2 ) (7) Ẽ q = E xq + j(x d X q )Ĩ d ) Ẽ q = E q e jδ (8) Remove the 3 in Eq. 2 if calculations are being performed in per unit. Page 3/7

4 5 Development of the Procedure The components of the stator current acting along the direct and quadrature axes result in different magnetic flux per ampere due to the non-uniform airgap of a salient-pole synchronous generator. The salient-pole steady state model accounts for this effect by decomposing the stator current in to a direct,, and quadrature,, component. The direct and quadrature components act through two different reactances: X d and X q. The quadrature component is in phase with the stator internal voltage, Ẽ q. The direct component leads or lags the quadrature component by π=2 radians. Equations 9 and 10 are used to determine I a and Ẽ q given Ĩ q, Ĩ d, and the terminal voltage, Ṽ a. Ĩ a = Ĩ q + Ĩ d (9) Ẽ q = Ṽ a + jx d Ĩ d + jx q Ĩ q (10) The relationships between the phasors described by Eq. 9 and 10 are displayed in the phasor diagram shown below. d-axis E q q-axis jx q delta jx d theta VI Va I a Page 4/7

5 Typically Ĩ q and Ĩ d are not initially known. Usually the line current, Ĩ a, and the terminal voltage, Ṽ a, are known. To decompose the stator current, Ĩ a,intoĩ q and Ĩ d the phase of the stator internal voltage, Ẽ q, must be know. To solve the problem of needing the phase of Ẽ q to calculate Ẽ q, Eq. 10 can be modified as shown in the following three equations. First zero in the form of jx d Ĩ d jx d Ĩ d is added to the right hand side of Eq. 10. Ẽ q = Ṽ a + jx d Ĩ d + jx q Ĩ q + jx q Ĩ d jx q Ĩ d Λ (11) Rearranging terms yields: Ẽ q = Ṽ a + jx q (Ĩ q + Ĩ d )+ j(x d X q )Ĩ d (12) Recognizing that Ĩ a = Ĩ q + Ĩ d yields: Ẽ q = Ṽ a + jx q Ĩ a + j(x d X q )Ĩ d (13) A new voltage, Ẽ xq,isdefined as the portion of Eq. 13 that is in terms of only Ṽ a and Ĩ a. Ẽ xq Ṽ a + jx q Ĩ a ) δ = angle(ẽ xq ) (14) Equation 13 can be rewritten in terms of the newly defined voltage Ẽ xq. Ẽ q = Ẽ xq + j(x d X q )Ĩ d (15) Observing that the voltage j(x d X q )Ĩ d is in phase with Ẽ q means that the voltage Ẽ xq is also in phase with Ẽ q. That is, the voltages j(x d X q )Ĩ d, Ẽ xq and Ẽ q are all in phase. The result of the above observation is that Ẽ xq can be calculated with Eq. 14 using only Ṽ a and Ĩ a. By calculating Ẽ xq the angle of Ẽ q with respect to Ṽ a is known. The angle of Ẽ q with respect to Ṽ a is defined as δ. With δ known, Ĩ a can be decomposed in to Ĩ q and Ĩ d using Eq. 16 and 17. Ĩ q = I a cos(δ+θ VI )e jδ (16) Ĩ d = I a sin(δ+θ VI )e j(δ+ π 2 ) (17) Page 5/7

6 6 Procedure for Calculating E q Given Ṽ sys and S sys This section presents a procedure for calculating Ẽ q given the system voltage, Ṽ sys, and the complex power delivered to the system, S sys, through a reactance X th. 6.1 Development Solving Eq. 10 for Ṽ a yields: Ṽ a = Ẽ q jx d Ĩ d jx q Ĩ q (18) Kirkhoff s voltage law yields Ṽ a = Ṽ sys + jx th Ĩ a (19) Equating Eq. 18 and 19 yields Ẽ q jx d Ĩ d jx q Ĩ q = Ṽ sys + jx th (Ĩ q + Ĩ d ) (20) Rearranging terms yields: Ẽ q = Ṽ sys + j(x q + X th )Ĩ q + j(x d + X th )Ĩ d (21) Ẽ q = Ṽ sys + jx qeq Ĩ q + jx deq Ĩ d (22) where X deq X d + X th (23) X qeq X q + X th (24) Page 6/7

7 6.2 Procedure I a = SΛ sys 3Ṽ Λ sys ) I a = I a e jθ VI (25) θ VI = angle(ĩ a ) (26) E xq = Ṽ sys + jx qeq I a ) E xq = E xq e jδ sys (27) δ sys = angle(ẽ xq ) (28) = I a sin(θ VI + δ sys ) (29) I d = e j(δ sys+ π) 2 ) I d = e j(δ sys+ π) 2 (30) E q = E xq + j(x deq X qeq ) I d ) E q = E q e jδ sys (31) P sys = 3 " E q V sys sys sin(δ sys )+ V 2 X deq 2 Xdeq X qeq X deq X qeq # sin(2δ sys ) (32) Remove the 3 in Eq. 25 and 32 if calculations are being performed in per unit. Page 7/7