Download Advanced Electric Drives: Analysis, Control, and Modeling by Ned Mohan PDF
By Ned Mohan
Complex electrical Drives makes use of a physics-based method of clarify the basic options of contemporary electrical force regulate and its operation below dynamic conditions.
• Gives readers a “physical” photograph of electrical machines and drives with no resorting to mathematical modifications for simple visualization
• Confirms the physics-based research of electrical drives mathematically
• Provides readers with an research of electrical machines in a manner that may be simply interfaced to universal strength digital converters and regulated utilizing any keep watch over scheme
• Makes the MATLAB/Simulink documents utilized in examples on hand to an individual in an accompanying website
• Reinforces basics with various dialogue questions, proposal quizzes, and homework difficulties
Read Online or Download Advanced Electric Drives: Analysis, Control, and Modeling Using MATLAB / Simulink PDF
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Extra info for Advanced Electric Drives: Analysis, Control, and Modeling Using MATLAB / Simulink
2-18) and 14 INDUCTION MACHINE EQUATIONS IN PHASE QUANTITIES b-axis B-axis ωm ib A-axis θm iB a-axis iA ia iC ic c-axis C-axis (a) iB ib + + vb − − − Vc VB = 0 Rs Va ia Rr + − − + iA − VA = 0 VC = 0 + + ic Stator circuit iC Rotor circuit (b) Fig. 2-5 Rotor circuit represented by three-phase windings. Note that with the choice of the same number of turns in the equivalent three-phase rotor windings as in the stator windings, the rotor leakage inductance Lℓr in Eq. (2-18) is the same as L′r in the perphase, steady-state equivalent circuit of an induction motor.
3-40) into Eq. (3-41), 2 Td , rotor p µ 3 / 2 N s = π 0 r p 2 g isq + Lr irq ird. Lm (3-42) Rewriting Eq. (3-42) below, we can recognize Lm from Eq. (2-13) N p3 µ = π 0 r s p 2 2 g 2 Td ,rotor isq + Lr irq ird . Lm Lm Hence, Td , rotor = p p (Lm isq + Lr irq ) ird = λrq ird . 2 2 λ (3-43) rq 3-5-2 Torque on the Rotor q-Axis Winding On the rotor q-axis winding, the torque produced is due to the flux density produced by the d-axis windings in Fig.
3-6) and The factor 2 / 3 , reciprocal of the factor 3 / 2 used in choosing the number of turns for the dq windings, ensures that the dq-winding currents produce the same mmf distribution as the three-phase winding currents. In Fig. 3-1b, the d and the q windings are mutually decoupled magnetically due to their orthogonal orientation. Choosing 3 / 2 N s turns for each of these windings results in their magnetizing inductance to be Lm (same as the per-phase magnetizing inductance in Chapter 2 for three-phase windings with ia + ib + ic = 0) for the following reason: the inductance of a winding is proportional to the square of the number of turns and therefore, the magnetizing inductance of any dq winding (noting that there is no mutual inductance between the two orthogonal windings) is dq winding magnetizing inductance = ( 3 / 2 )2 Lm,1-phase = (3 / 2)Lm,1-phase (3-7) = Lm (using Eq.