1 / 30 Lecture 9: Space-Vector Models ELEC-E8405 Electric Drives (5 ECTS) Marko Hinkkanen Autumn 2017
2 / 30 Learning Outcomes After this lecture and exercises you will be able to: Include the number of pole pairs in the machine models Transform phase variables to a space vector (and vice versa) Transform space vectors to different coordinates Express the space-vector model of the synchronous machine in rotor coordinates Calculate steady-state operating points of the synchronous machine
3 / 30 Outline Number of Pole Pairs Space Vectors Synchronous Machine Model in Stator Coordinates Coordinate Transformation Synchronous Machine Model in Rotor Coordinates
ϑ M = ϑ m ϑ M = ϑ m /2 4 / 30 Number of Pole Pairs p 2 poles (p = 1) 4 poles (p = 2) Electrical angular speed ω m = p ω M and electrical angle ϑ m = p ϑ M
Synchronous Rotor Speeds Stator (supply) frequency f (Hz) Electrical angular speed (rad/s) ω m = 2πf Rotor angular speed (rad/s) ω M = ω m p Rotor speed (r/min) n = f p 60 s min Speeds for f = 50 Hz No of pole pairs p Speed n (r/min) 1 3000 2 1500 3 1000 4 750 5 600 6 500 Note that in converter-fed motor drives, the rated supply frequency of the motor does not need to be 50 Hz. 5 / 30
6 / 30 1-Phase Machine Phase voltage i a R L u a = Ri a + dψ a dt Phase flux linkage u a dψ a dt e a ψ a = Li a + ψ fa where ψ fa = ψ f cos(ϑ m ) Back-emf Mechanical power p M = e a i a = T M ω m /p e a = dψ fa dt = ω m ψ f sin(ϑ m ) Torque T M = p i a ψ f sin(ϑ m )
7 / 30 Synchronous Machine: Phase-Variable Model u a = R s i a + dψ a dt u b = R s i b + dψ b dt a b i a i b R s R s L s L s e a e b n u c = R s i c + dψ c dt c i c R s L s e c ψ a = L s i a + ψ f cos(ϑ m ) ψ b = L s i b + ψ f cos(ϑ m 2π/3) ψ c = L s i c + ψ f cos(ϑ m 4π/3) e a = ω m ψ f sin(ϑ m ) e b = ω m ψ f sin(ϑ m 2π/3) e c = ω m ψ f sin(ϑ m 4π/3) T M = pψ f [i a sin(ϑ m ) + i b sin(ϑ m 2π/3) + i c sin(ϑ m 4π/3)]
8 / 30 Outline Number of Pole Pairs Space Vectors Synchronous Machine Model in Stator Coordinates Coordinate Transformation Synchronous Machine Model in Rotor Coordinates
9 / 30 Why Space Vectors? 1. Complex phasor models Simple to use but limited to steady-state conditions 2. Phase-variable models Valid both in transient and steady states Too complicated 3. Space-vector models Phase-variable models can be directly transformed to space-vector models Compact representation, insightful physical interpretations Commonly applied to analysis, modelling, and control of 3-phase systems
10 / 30 About Complex Numbers Complex number z = x + jy Complex conjugate of z z = x jy Magnitude of z z = z = x 2 + y 2 Euler s formula e jϑ = cos ϑ + j sin ϑ Rotating the position vector by 90 jz = j(x + jy) = y + jx Dot product Re{z 1 z 2 } = Re{(x 1 + jy 1 )(x 2 jy 2 )} = x 1 x 2 + y 1 y 2 Cross product Im{z 1 z 2 } = Im{(x 1 + jy 1 )(x 2 jy 2 } = y 1 x 2 y 2 x 1
Magnetic Axes in the Complex Plane e j2π/3 e j0 e j0 Phase a e j4π/3 All 3 phases Windings are sinusoidally distributed along the air gap 11 / 30
Space-Vector Transformation Space vector is a complex variable (signal) i s s = 2 3 (i a + i b e j2π/3 + i c e j4π/3) where i a, i b, and i c are arbitrarily varying instantaneous phase variables Superscript s marks stator coordinates Same transformation applies for voltages and flux linkages Space vector does not include the zero-sequence component (not a problem since the stator winding is delta-connected or the star point is not connected) Peak-value scaling of space vectors will be used in this course. Furthermore, we will use the subscript s to refer to the stator quantities, e.g., the stator current vector i s and the stator voltage vector u s, since this is a very common convention in the literature. 12 / 30
e j2π/3 β e j2π/3 i c e j4π/3 i s s 3 2 is s i c i s s i b e j2π/3 i a α i b i a e j0 e j4π/3 e j4π/3 i a = Re { i s } s i s s = 2 ( ia + i b e j2π/3 + i c e j4π/3) i b = Re { e j2π/3 i s } s 3 i c = Re { e j4π/3 i s } s 13 / 30
14 / 30 Examples: Space Vectors Rotate in Steady State Positive sequence i a = 2I + cos(ω m t + φ + ) i b = 2I + cos(ω m t 2π/3 + φ + ) i c = 2I + cos(ω m t 4π/3 + φ + ) Space vector i s s = 2I + e j(ωmt+φ+) Non-sinusoidal periodic waveform Negative sequence i a = 2I cos(ω m t + φ ) i b = 2I cos(ω m t 4π/3 + φ ) i c = 2I cos(ω m t 2π/3 + φ ) Space vector i s s = 2I e j(ωmt+φ ) i s s = 2I 1 e j(ωmt+φ 1) + 2I 5 e j(5ωmt+φ 5) + 2I 7 e j(7ωmt+φ 7)...
15 / 30 Representation in Component and Polar Forms Component form Polar form i s s = i α + ji β i s s i α β i s s = i s e jθ i = i s cos(θ i ) +j i }{{} s sin(θ i ) }{{} i α Generally, both the magnitude i s and the angle θ i may vary arbitrarily in time i β Positive sequence in steady state: i s = 2I is constant and θ i = ω m t + φ ji β θ i α
Physical Interpretation: Sinusoidal Distribution in Space i s s i α β 3-phase winding creates the current and the mmf, which are sinusoidally distributed along the air gap Space vector represents the instantaneous magnitude and angle of the sinusoidal distribution in space Magnitude and the angle can vary freely in time ji β α Rotating current distribution produced by the 3-phase stator winding 16 / 30
17 / 30 Outline Number of Pole Pairs Space Vectors Synchronous Machine Model in Stator Coordinates Coordinate Transformation Synchronous Machine Model in Rotor Coordinates
Space-Vector Model of the Synchronous Machine Stator voltage u s s = R si s s + dψs s dt Stator flux linkage ψ s s = L si s s + ψ f e jϑm Torque can be expressed in various forms Following form is convenient since it holds for other AC machines as well T M = 3p { } 2 Im i s sψ s s Derive these voltage and flux linkage equations starting from the phase-variable model and the definition of the space vector. Also show that the space-vector and phase-variable formulations for the torque are equal. 18 / 30
19 / 30 Space-Vector Equivalent Circuit Stator voltage can be rewritten as i s s R s L s u s s = R si s s + L s di s s dt + e s s Back-emf e s s = jω mψ f e jϑm is proportional to the speed u s s dψ s s dt e s s
20 / 30 Torque Vectors in the polar form i s s = i s e jθ i ψ s s = ψ se jθ ψ β Instantaneous torque T M = 3p { } 2 Im i s sψ s = 3p 2 i sψ s sin(γ) where γ = θ i θ ψ Nonzero γ is needed for torque production s i s s θ i γ θ ψ ψ s s α
Power Vectors in the component and polar forms u s s = u α + ju β = u s e jθu i s s = i α + ji β = i s e jθ i Instantaneous power fed to the stator p s = 3 2 Re { u s } s is s = 3 2 (u αi α + u β i β ) = 3 2 u si s cos(ϕ) where ϕ = θ u θ i The power calculated using the space vectors naturally agrees with the power p s = u ai a + u b i b + u ci c calculated from the phase variables. Furthermore, in steady state, the rms-valued expression P s = 3U si s cos(ϕ) is obtained, since u s = 2U s and i s = 2U s hold. 21 / 30
22 / 30 Outline Number of Pole Pairs Space Vectors Synchronous Machine Model in Stator Coordinates Coordinate Transformation Synchronous Machine Model in Rotor Coordinates
23 / 30 Example: Stopping the Rotating Vector Positive-sequence space vector in stator coordinates i s s = 2I e j(ωmt+φ) Rotating vector can be stopped by the transformation i s = i s s e jωmt = 2I e jφ In other words, we observe the vector now in a coordinate system rotating at ω m In rotating coordinates, the vector is denoted without a superscript and the components are marked with the subscripts d and q i s = i d + ji q
24 / 30 Coordinate Transformation Previous example assumed the rotor speed ω m to be constant General dq transformation and its inverse are i s = i s se jϑm i s s = i s e jϑm where the rotor angle is ϑ m = ω m dt dq transformation αβ transformation
25 / 30 q β i s q β i s s d ϑ m α ϑ m d α Stator coordinates (αβ) Rotor coordinates (dq) i s = i s s e jϑm
26 / 30 i s s i α q β i s i d q β ji β d ji q ϑ m α ϑ m d α Stator coordinates (αβ) Rotor coordinates (dq) i s = i s s e jϑm
27 / 30 Outline Number of Pole Pairs Space Vectors Synchronous Machine Model in Stator Coordinates Coordinate Transformation Synchronous Machine Model in Rotor Coordinates
28 / 30 Synchronous Machine Model in Rotor Coordinates Substitute ψ s s = ψ s ejϑm, u s s = u s ejϑm, and i s s = i s e jϑm u s e jϑm = R s i s e jϑm + d dt (ψ s e jϑm ) u s = R s i s + dψ s dt + jω m ψ s ψ s e jϑm = L s i s e jϑm + ψ f e jϑm ψ s = L s i s + ψ f Torque is proportional to i q T M = 3p { } 2 Im i s ψ = 3p s 2 ψ fi q while i d does not contribute to the torque
29 / 30 Power Balance Stator voltage can be rewritten as u s = R s i s + L s di s dt + jω m L s i s + e s where e s = jω m ψ f is the back-emf Power balance is obtained from the stator voltage equation p s = 3 2 Re {u s i s} = 3 2 R s i s 2 }{{} Losses Middle term is zero in steady state + 3 L s 2 2 d i s 2 }{{ dt } Rate of change of energy in L s + T M ω m p }{{} Mechanical power
30 / 30 Vector Diagram In steady state, d/dt = 0 holds in rotor coordinates Stator voltage u s q i s u s = R s i s + jω m ψ s = R s i s + jω m (L s i s + ψ f ) Steady-state operating points can be illustrated by means of vector diagrams ψ s Assumption: R s 0 ψ f L s i s d