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DIRECT-CURRENT GENERATORS
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To analyze the performance of a DC generator, it would be useful to obtain an opencircuit characteristic capable of predicting the voltage generated when the machine is driven at a constant speed m by a prime mover The common arrangement is to drive the machine at rated speed by means of a prime mover (or an electric motor) Then, with no load connected to the armature terminals, the armature voltage is recorded as the eld current is increased from zero to some value suf cient to produce an armature voltage greater than the rated voltage Since the load terminals are open-circuited, Ia = 0 and Eb = Va ; and since ka = Eb / m , the magnetization curve makes it possible to determine the value of ka corresponding to a given eld current, If , for the rated speed Figure 1716 depicts a typical magnetization curve Note that the armature voltage is nonzero even when no eld current is present This phenomenon is due to the residual magnetization of the iron core The dashed lines in Figure 1716 are called eld resistance curves and are a plot of the voltage that appears across the eld winding plus rheostat (variable resistor; see Figure 1715) versus the eld current, for various values of eld winding plus rheostat resistance Thus, the slope of the line is equal to the total eld circuit resistance, Rf The operation of a DC generator may be readily understood with reference to the magnetization curve of Figure 1716 As soon as the armature is connected
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Introduction to Electric Machines
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Eb (V) 120 100 80 60 40 20 0 02 04 06 08 10 12 14 If (A)
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Figure 1716 DC machine magnetization curve
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across the shunt circuit consisting of the eld winding and the rheostat, a current will ow through the winding, and this will in turn act to increase the emf across the armature This buildup process continues until the two curves meet, that is, until the current owing through the eld winding is exactly that required to induce the emf By changing the rheostat setting, the operating point at the intersection of the two curves can be displaced, as shown in Figure 1716, and the generator can therefore be made to supply different voltages The following examples illustrate the operation of the separately excited DC generator
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EXAMPLE 174 Separately Excited DC Generator
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Problem
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A separately excited DC generator is characterized by the magnetization curve of Figure 1716 1 2 If the prime mover is driving the generator at 800 rev/min, what is the no-load terminal voltage, Va If a 1- load is connected to the generator, what is the generated voltage
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Solution
Known Quantities: Generator magnetization curve and ratings Find: Terminal voltage with no load and 1-
load
Schematics, Diagrams, Circuits, and Given Data:
Generator ratings: 100 V, 100 A, 1,000 rev/min Circuit parameters: Ra = 014 ; Vf = 100 V; Rf = 100
Analysis:
The eld current in the machine is Vf 100 V =1A = If = Rf 100 From the magnetization curve, it can be seen that this eld current will produce
Part III
Electromechanics
100 V at a speed of 1,000 rev/min Since this generator is actually running at 800 rev/min, the induced emf may be found by assuming a linear relationship between speed and emf This approximation is reasonable, provided that the departure from the nominal operating condition is small Let n0 and Eb0 be the nominal speed and emf, respectively (ie, 1,000 rev/min and 100 V); then, n Eb = Eb0 n0 and therefore Eb = n 800 rev/min 100 V = 80 V Eb0 = n0 1,000 rev/min
The open-circuit (output) terminal voltage of the generator is equal to the emf from the circuit model of Figure 1715; therefore: Va = Eb = 80 V 2 When a load resistance is connected to the circuit (the practical situation), the terminal (or load) voltage is no longer equal to Eb , since there will be a voltage drop across the armature winding resistance The armature (or load) current may be determined from the expression I a = IL = where RL = 1 by Eb 80 V = Ra + R L (014 + 1) = 702 A
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