barcode reader code in c# net DC Series Motor Analysis in Software

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EXAMPLE 178 DC Series Motor Analysis
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Determine the torque developed by a DC series motor when the current supplied to the motor is 60 A
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Electromechanics
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Known Quantities: Motor ratings; operating conditions Find: T60 , torque delivered at 60-A series current Schematics, Diagrams, Circuits, and Given Data: Motor ratings: 10 hp; 115 V; full
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load speed = 1,800 rev/min Operating conditions: motor draws 40 A
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Assumptions: The motor operates in the linear region of the magnetization curve Analysis: Within the linear region of operation, the ux per pole is directly proportional to the current in the eld winding That is,
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= k S Ia The full-load speed is n = 1,800 rev/min or 2 n = 60 rad/s 60 Rated output power is m = Prated = 10 hp 746 W/hp = 7,460 W and full-load torque is T40A = Prated 7,460 = 3958 N-m = m 60
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Thus, the machine constant may be computed from the torque equation for the series motor:
2 2 T = ka kS Ia = KIa
At full load, 3958 N-m N-m = 00247 2 2 A2 40 A and we can compute the torque developed for a 60-A supply current to be K = ka kS =
2 T60A = KIa = 00247 602 = 8892 N-m
EXAMPLE 179 Dynamic Response of PM DC Motor
Problem
Develop a set of differential equations and a transfer function describing the dynamic response of the motor angular velocity of a PM DC motor connected to a mechanical load
Solution
Known Quantities: PM DC motor circuit model; mechanical load model Find: Differential equations and transfer functions of electromechanical system
17
Introduction to Electric Machines
Analysis: The dynamic response of the electromechanical system can be determined by
applying KVL to the electrical circuit (Figure 1720), and Newton s second law to the mechanical system These equations will be coupled to one another, as you shall see, because of the nature of the motor back emf and torque equations Applying KVL and equation 1747 to the electrical circuit we obtain: VL (t) Ra Ia (t) La or dIa (t) + Ra Ia (t) + KaPM m (t) = VL (t) dt Applying Newton s second law and equation 1746 to the load inertia, we obtain: La J or d (t) + b (t) = Tload (t) dt These two differential equations are coupled because the rst depends on m and the second on Ia Thus, they need to be solved simultaneously To derive the transfer function, we Laplace-transform the two equations to obtain: KT PM Ia (t) + J (sLa + Ra )Ia (s) + KaPM (s) = VL (s) KT PM Ia (s) + (sJ + b) (s) = Tload (s) We can write the above equations in matrix form and resort to Cramer s rule to solve for m (s) as a function of VL (s) and Tload (s) (sLa + Ra ) KT PM with solution det
m (s)
dIa (t) Eb (t) = 0 dt
d (t) = T (t) Tload (t) b dt
KaPM (sL + b) (sLa + Ra ) KT PM (sLa + Ra ) KT PM
Ia (s)
m (s)
VL (s) Tload (s)
VL (s) Tload (s) Ka PM (sJ + b)
= det
m (s)
(sLa + Ra ) Tload (s) (sLa + Ra )(sJ + b) + Ka PM KT PM + KT PM VL (s) (sLa + Ra )(sJ + b) + Ka PM KT PM
Comments: Note that the dynamic response of the motor angular velocity depends on
both the input voltage and on the load torque This problem is explored further in the homework problems
DC Drives and DC Motor Speed Control The advances made in power semiconductors have made it possible to realize lowcost speed control systems for DC motors The basic operation of controlled recti er and chopper drives for DC motors was described in 11 In the present section we describe some of the considerations that are behind the choice of a speci c drive type, and of some of the loads that are likely to be encountered
Part III
Electromechanics
Constant-torque loads are quite common, and are characterized by a need for constant torque over the entire speed range This need is usually due to friction; the load will demand increasing horsepower at higher speeds, since power is the product of speed and torque Thus, the power required will increase linearly with speed This type of loading is characteristic of conveyors, extruders, and surface winders Another type of load is one that requires constant horsepower over the speed range of the motor Since torque is inversely proportional to speed with constant horsepower, this type of load will require higher torque at low speeds Examples of constant-horsepower loads are machine tool spindles (eg, lathes) This type of application requires very high starting torques Variable-torque loads are also common In this case, the load torque is related to the speed in some fashion, either linearly or geometrically For some loads, for example, torque is proportional to the speed (and thus horsepower is proportional to speed squared); examples of loads of this type are positive displacement pumps More common than the linear relationship is the squared-speed dependence of inertial loads such as centrifugal pumps, some fans, and all loads in which a ywheel is used for energy storage To select the appropriate motor and adjustable speed drive for a given application, we need to examine how each method for speed adjustment operates on a DC motor Armature voltage control serves to smoothly adjust speed from 0 to 100 percent of the nameplate rated value (ie, base speed), provided that the eld excitation is also equal to the rated value Within this range, it is possible to fully control motor speed for a constant-torque load, thus providing a linear increase in horsepower, as shown in Figure 1724 Field weakening allows for increases in speed of up to several times the base speed; however, eld control changes the characteristics of the DC motor from constant torque to constant horsepower, and therefore the torque output drops with speed, as shown in Figure 1724 Operation above base speed requires special provision for eld control, in addition to the circuitry required for armature voltage control, and is therefore more complex and costly
Torque (% rated) 100 Field control Armature control 200 100 Base speed 300 400 500 Speed (% rated)
Horsepower (% rated) 100 Field control Armature control 200 100 Base speed 300 400 500 Speed (% rated)
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