barcode reader code in c# net is the load resistance The terminal (load) voltage is therefore given in Software

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is the load resistance The terminal (load) voltage is therefore given
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VL = IL RL = 702 1 = 702 V
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EXAMPLE 175 Separately Excited DC Generator
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Determine the following quantities for a separately excited DC: 1 2 3 Induced voltage Machine constant Torque developed at rated conditions
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Known Quantities: Generator ratings and machine parameters Find: Eb , ka , T Schematics, Diagrams, Circuits, and Given Data:
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Generator ratings: 1,000 kW; 2,000 V; 3,600 rev/min Circuit parameters: Ra = 01 ; ux per pole = = 05 Wb
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The armature current may be found by observing that the rated power is equal to the product of the terminal (load) voltage and the armature (load) current; thus, Ia = Prated 1,000 103 = 500 A = VL 2,000
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17
Introduction to Electric Machines
The generated voltage is equal to the sum of the terminal voltage and the voltage drop across the armature resistance (see Figure 1614): Eb = Va + Ia Ra = 2,000 + 500 01 = 2,050 V 2 The speed of rotation of the machine in units of rad/s is 2 3,600 rev/min 2 n = = 377 rad/s 60 60 s/min Thus, the machine constant is found to be m = ka = 3 V-s Eb 2,050 V = 10876 = m 05 Wb 377 rad/s Wb-rad
The torque developed is found from equation 166: T = ka Ia = 10875 V-s/Wb-rad 05 Wb 500 A = 2,7189 N-m
Comments: In many practical cases, it is not actually necessary to know the armature
constant and the ux separately, but it is suf cient to know the value of the product ka For example, suppose that the armature resistance of a DC machine is known and that, given a known eld excitation, the armature current, load voltage, and speed of the machine can be measured Then, the product ka may be determined from equation 1620, as follows: ka = Eb VL + Ia (Ra + RS ) = m m
where VL , Ia , and m are measured quantities for given operating conditions
Since the compound-connected generator contains both a shunt and a series eld winding, it is the most general con guration, and the most useful for developing a circuit model that is as general as possible In the following discussion, we shall consider the so-called short-shunt, compound-connected generator, in which the ux produced by the series winding adds to that of the shunt winding Figure 1717 depicts the equivalent circuit for the compound generator; circuit models for the shunt generator and for the rarely used series generator can be obtained by removing the shunt or series eld winding element, respectively In the circuit of Figure 1717, the generator armature has been replaced by a voltage source corresponding to the induced emf and a series resistance, Ra , corresponding to the resistance of the armature windings The equations describing the DC generator at steady state (ie, with the inductors acting as short circuits) are:
DC Generator Steady-State Equations Eb = ka m V T = P Eb Ia = = ka Ia N-m m m (1720) (1721) (1722) (1723)
VL = Eb Ia Ra IS RS Ia = IS + If
Part III
Electromechanics
If Ra Rf Lf La + Eb Rx Va m Prime mover + Ia LS RS
IL +
Figure 1717 Compound generator circuit model
Note that in the circuit of Figure 1717, the load and armature voltages are not equal, in general, because of the presence of a series eld winding, represented by the resistor RS and by the inductor LS where the subscript S stands for series The expression for the armature emf is dependent on the air-gap ux, , to which the series and shunt windings in the compound generator both contribute, according to the expression = sh S = sh kS Ia (1724)
Check Your Understanding
175 A 24-coil, 2-pole DC generator has 16 turns per coil in its armature winding The eld excitation is 005 Wb per pole, and the armature angular velocity is 180 rad/s Find the machine constant and the total induced voltage 176 A 1,000-kW, 1,000-V, 2,400-rev/min separately excited DC generator has an armature circuit resistance of 004 The ux per pole is 04 Wb Find: (a) the induced voltage; (b) the machine constant; and (c) the torque developed at the rated conditions 177 A 100-kW, 250-V shunt generator has a eld circuit resistance of 50 and an armature circuit resistance of 005 Find: (a) the full-load line current owing to the load; (b) the eld current; (c) the armature current; and (d) the full-load generator voltage
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