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AC Motor Control and Electrical Vehicle Applications by Kwang Hee Nam

By Kwang Hee Nam

Motor keep watch over expertise maintains to play a necessary position within the initiative to cast off or not less than reduce petroleum dependency and greenhouse fuel emissions world wide. elevated motor potency is a vital element of this technology within the worldwide transition to wash strength use in components equivalent to commercial functions and residential appliances―but quite within the layout of vehicles.

Summarizes the evolution of motor using devices towards excessive potency, within your budget, excessive energy density, and versatile interface with different components

AC Motor keep an eye on and electrical motor vehicle Applications addresses the themes pointed out in its identify but in addition elaborates on motor layout standpoint, corresponding to again EMF harmonics, loss, flux saturation, and reluctance torque, and so on. conserving theoretical integrity in AC motor modeling and regulate all through, the writer makes a speciality of the advantages and straightforwardness of the rotor field-oriented keep an eye on, describing the fundamentals of PWM, inverter, and sensors. He additionally clarifies the basics of electrical autos and their linked dynamics, motor matters, and battery limits. a strong compendium of functional info, this booklet serves as an total useful gizmo for the layout and keep watch over of high-efficiency motors.

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Extra resources for AC Motor Control and Electrical Vehicle Applications

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A) Calculate the back EMF coefficient, Kb . b) Calculate current, torque, and efficiency at 950rpm. 4Vsec/rad. Assume that the maximum armature voltage is vamax = 120V . a) Determine the motor speed when the load torque is TL = 8Nm. 3Vsec/rad under the same load. Recalculate the speed. 1kg · m2 . Then the current loop transfer function with a PI controller, Kp + Ki /s is K Ki p 1 (Kp + Ksi ) La s+r Ia (s) La s + La a = = r +K 1 Ia∗ (s) 1 + (Kp + Ksi ) La s+r s2 + aLa p s + a Ki La = Kp La s + ωn2 s2 + 2ζωn s + ωn2 a) Determine the proportional gain, Kp such that ζωn = 1000.

23) Note that the positive sequence components are identical, whereas the negative sequence components are summed to be zero. 23) shows that the injection of the three-phase AC current [cos ωt, 4π jωt in the complex plane. cos(ωt − 2π 3 ), cos(ωt − 3 )] creates a rotating vector e 44 AC Motor Control and Electric Vehicle Applications Fig. 9 shows a schematic representation of the mapping into the stationary frame. However, it is worthwhile to note that the direction of the current sum vector is identical to the direction of fs in the complex plane.

4 Transformation of Impedance Matrices Resistance Consider a simple three-phase resistor load: vabc = riabc , where iabc = [ia , ib , ic ]T and vabc = [va , vb , vc ]T . Let iedq = Tiabc e vdq = Tvabc , where idq0 = [id , iq , i0 ]T and vdq0 = [vd , vq , v0 ]T . Then, e vdq = Tvabc = rTiabc = riedq . The resistor, being a scalar, is invariant under the coordinate change. Inductance Consider flux linkage λabc = Labc iabc , where Labc ∈ R3×3 and λabc = [λa , λb , λc ]T . Let λdq0 = [λd , λq , λ0 ]T .

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