ElectroMechanical System

ElectroMechanical System - ME375 Handouts...

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Unformatted text preview: ME375 Handouts ElectroElectro-Mechanical Systems • DC Motors – Principles of Operation – Modeling (Derivation of Governing Equations (EOM)) • Block Diagram Representations – Using Block Diagrams to Represent Equations in s - Domain – Block Diagram Representation of DC Motors • Example School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 1 DC Motors Motors are actuation devices (actuators) that generate torque as actuation. torque • Terminology – Rotor : the rotating part of the motor. – Stator : the stationary part of the the stationary part of the motor. motor. – Field System : the part of the motor that provides the magnetic flux. – Armature : the part of the motor which carries current that interacts with the magnetic flux to produce torque. – Brushes : the part of the electrical th th circuit circuit through which the current is supplied to the armature. – Commutator : the part of the rotor that is in contact with the brushes. School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 2 1 ME375 Handouts DC Motors - Principles of Operation • Torque Generation v B v df B v dL i Force will act on a conductor in a magnetic field with current flowing through the conductor. v v v d f = ia ⋅ d L × B Integrate over the entire length: i f= Total torque generated: B τ Coil = School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 3 DC Motors - Principles of Operation Let N be the number of coils in the motor. The total torque generated from the N coils is: τ m = N ⋅ (2 ⋅ ia ⋅ B ⋅ L ⋅ R) = For a given motor, (N, B, L, R) are fixed. We can define KT = 2 ⋅ N ⋅ B ⋅ L ⋅ R Nm / A as the Torque Constant of the motor. Torque The torque generated by DC motor is proportional to the The torque generated by a DC motor is proportional to the armature armature current ia : τ m = KT ⋅ ia Large KT : – Large (N, L, R). (N, L, R) is limited by the size and weight of the motor. For a DC motor, it is desirable to have a large KT . However, size and other physical limitations often limits the achievable KT . School of Mechanical Engineering Purdue University – Large B: Need to understand the methods of generating flux ... ME375 ElectroMechanical - 4 2 ME375 Handouts DC Motors - Principles of Operation • Back-EMF Generation BackElectromotive force (EMF) is generated in a conductor v moving in a magnetic field: vvB v vv v ×B de = ( v × B ) ⋅ dL emf v v Integrate over the entire length L: e em f = Since the N armature coils are in series, the total EMF is: E em f = 2 N ( R ω ) B L = 13 2 v Kb = 2 ⋅ N ⋅ R ⋅ B ⋅ L Define the Back-EMF Constant Kb : Back- V / (rad / sec) The Back-EMF generated due to the rotation of the motor armature is opposing the applied Backvoltage and is proportional to the angular speed ω of the motor: E em f = K b ⋅ ω Note: KT = Kb is true only if consistent SI units are used ! School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 5 DC Motors - Modeling Schematic + eRa − + ei(t) _ RA + eLa − iA LA + Eemf _ Element Laws: Electrical Subsystem θ, ω τm JA τL B Mechanical Subsystem FBD: Interconnection Laws: School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 6 3 ME375 Handouts DC Motors - Modeling Derive I/O Model: I/O Model from ei(t) and τL to angular speed ω : and FG H FG H IJ K IJ K FG H FG H IJ K IJ K b & LA τ L + RA τ L LA J A LA B RJ RA B && & ω+ + A A ω+ + K b ω = ei ( t ) − KT KT KT KT KT I/O Model from ei(t) and τL to angular position θ : and & LA τ L + RA τ L L A J A &&& LA B RJ RA B && & θ+ + A A θ+ + K b θ = ei ( t ) − KT KT KT KT KT School of Mechanical Engineering Purdue University b g g ME375 ElectroMechanical - 7 DC Motors - Modeling Transfer Function: Ω (s) = ⋅ Ei ( s ) − ⋅Τ L ( s ) θ (s) = ⋅ Ei ( s ) − ⋅Τ L ( s ) Q: Q: Let the load torque be zero (No Load), what is the steady state speed (No-Load Speed) of the (Nomotor for a constant input voltage V ? Q: Let the load torque τL = T, what is the steady state speed of the motor for a constant input voltage V ? School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 8 4 ME375 Handouts Block Diagram Representation • Differential Equation → Transfer Transfer Function (System & Signals) Y (s) = G (s) ⋅ U (s) G ( s) Input Signal Y (s) = U1(s) ± U 2 (s) U1(s) Y(s) U(s) • Signal Addition/Subtraction Output Signal Y(s) + ± U2(s) Ex: Draw the block diagram for the following DE: Ex: Draw the block diagram for the following DE: & Jω + Bω = τ & Jω = τ School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 9 Block Diagram Representation • Transfer Function in Series • Multiple Inputs Y ( s ) = G 2 ( s ) ⋅ Y1 ( s ) , Y1 ( s ) = G1 ( s ) ⋅ U ( s ) b g U(s) G1 (s) Y1(s) G2 (s) Input Signal = G1 ( s ) ⋅ U 1 ( s ) ± G 2 ( s ) ⋅ U 2 ( s ) Y (s ) Output Signal • Transfer Function in Parallel X 1 ( s ) = G1 ( s ) ⋅ U ( s ) , X 2 ( s ) = G 2 ( s ) ⋅ U ( s ) Y ( s ) = X1 ( s ) ± X 2 ( s ) b Y1 ( s ) = G 1 ( s ) ⋅ U 1 ( s ) , Y2 ( s ) = G 2 ( s ) ⋅ U 2 ( s ) Y ( s ) = Y1 ( s ) ± Y2 ( s ) Y ( s ) = G 2 ( s ) ⋅ G1 ( s ) ⋅ U ( s ) U1(s) U2(s) Input Signals G1 (s) G2 (s) Y1(s) + Y (s ) ± Output Y2(s) Signal g = G1 ( s ) ± G 2 ( s ) ⋅ U ( s ) U(s) Input Signal G1 (s) G2 (s) X1(s) + Y (s ) ± Output X2(s) Signal School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 10 5 ME375 Handouts Block Diagram Representation • Feedback Loop U(s) + X (s) Y (s ) G (s ) Input − Output Signal Signal H(s) School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 11 Block Diagram Representation of DC Motors Schematic + eRa − + ei(t) _ RA ElectroElectro-Mechanical Coupling: + eLa − iA LA + Eemf _ θ, ω Take Laplace Transform of the Eqs. τm τL JA B Governing Equations: d LA iA + RA iA + Eemf = ei (t ) ( dt & J A ω + Bω = τ m − τ L τ m = KT ⋅ i A ⎫ ⎬ Eemf = Kb ⋅ ω ⎭ ( ⎛ ⎜ ⎝ ) ) ⎞ ⎟ ⎠ School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 12 6 ME375 Handouts Block Diagram Representation of DC Motors Electrical Subsystem: Mechanical Subsystem: Take Laplace Transform of the Eqs. Take Laplace Transform of the Eqs. Q: Now that we generated a block diagram of a voltage driven DC Motor, can we derive the transfer function of this motor from its block diagram ? ( This is the same as asking you to reduce the multiblock diagram to a simpler form just relating inputs ei(t) and τL to the output, either ω or θ .) School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 13 Block Diagram Reduction From Block Diagram to Transfer Function • Label each signal and block + − − 1 LA s + RA KT + 1 JA s + B Kb • Write down the relationships between signals School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 14 7 ME375 Handouts Block Diagram Reduction • Solve for the output signal in terms of the input signals • Substitute the transfer functions’ label with the actual formula and simplify Ω (s) = θ (s) = KT LA s + RA ⋅Τ L ( s ) ⋅ Ei ( s) − LA J A s 2 + ( BLA + RA J A )s + ( RA B + Kb KT ) LA J A s 2 + ( BLA + RA J A )s + ( RA B + Kb KT ) KT LA s + RA ⋅ Ei ( s ) − ⋅Τ L ( s ) s( LA J A s 2 + ( BLA + RA J A )s + ( RA B + K b K T )) s( LA J A s 2 + ( BLA + RA J A )s + ( RA B + K b K T )) School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 15 Model Reduction Q: Physically, what do we mean by neglecting armature inductance ? By neglecting the armature inductance, we are assuming that it takes no time for the current to reach its steady state value when there is a step change in input voltage, i.e., a sudden change in input voltage will result in a sudden change in the armature current, which in turn will result in a sudden change in the motor torque output. This is equivalent to having direct control over the motor current. output. current. From Block Diagram: Mathematically: d L A iA + RA iA + Eemf = ei ( t ) dt & J A ω + Bω = τ m − τ L τ m = K T ⋅ iA Eemf = K b ⋅ ω + − 1 LA s + RA KT U V W School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 16 8 ME375 Handouts Example Media Advance System in InkJet Printers The figure on the right shows the media advance system of a typical inkjet printer. dv sys The objective of the system is to precisely and quickly position the media such that ink droplets can be precisely “dropped” on to the media to form “nice looking” images. The system is driven by a DC motor through two sets of gear trains. You, as the “new kid” on the development team, are given the task of specifying a motor and designing the control system that will achieve the desirable performance. Some time your manager will also walk by your desk and ask you if a certain level of performance is achievable. How would you start your first engineering project? School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 17 Example – DC motor with load Schematic DC Motor + eRa − + ei(t) _ RA + eLa − iA LA + Eemf _ τm θ, ω θL, ωL Assumptions: JL N2 • Gears and shafts are rigid and massless. JA N1 BL B Block Block diagram of the load inertia: Block diagram of the gear train: School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 18 9 ME375 Handouts Example – DC motor with load Block diagram of the DC motor subsystem: + − 1 LA s + RA − KT + 1 JA s + B Kb Reduce the mechanical portion of the block diagram: School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 19 Example – DC motor with load Simplified block diagram: + − 1 LA s + RA KT Kb Transfer Function from input voltage Ei(s) to the angular velocity or position of the load: School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 20 10 ...
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