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Unformatted text preview: 109 Chapter 5 : Synchronous Generators 51. At a location in Europe, it is necessary to supply 300 kW of 60Hz power. The only power sources available operate at 50 Hz. It is decided to generate the power by means of a motorgenerator set consisting of a synchronous motor driving a synchronous generator. How many poles should each of the two machines have in order to convert 50Hz power to 60Hz power? S OLUTION The speed of a synchronous machine is related to its frequency by the equation 120 e m f n P = To make a 50 Hz and a 60 Hz machine have the same mechanical speed so that they can be coupled together, we see that ( ) ( ) sync 1 2 120 50 Hz 120 60 Hz n P P = = 2 1 6 12 5 10 P P = = Therefore, a 10pole synchronous motor must be coupled to a 12pole synchronous generator to accomplish this frequency conversion. 52. A 2300V 1000kVA 0.8PFlagging 60Hz twopole Yconnected synchronous generator has a synchronous reactance of 1.1 and an armature resistance of 0.15 . At 60 Hz, its friction and windage losses are 24 kW, and its core losses are 18 kW. The field circuit has a dc voltage of 200 V, and the maximum I F is 10 A. The resistance of the field circuit is adjustable over the range from 20 to 200 . The OCC of this generator is shown in Figure P51. (a) How much field current is required to make V T equal to 2300 V when the generator is running at no load? (b) What is the internal generated voltage of this machine at rated conditions? (c) How much field current is required to make V T equal to 2300 V when the generator is running at rated conditions? (d) How much power and torque must the generators prime mover be capable of supplying? (e) Construct a capability curve for this generator. Note: An electronic version of this open circuit characteristic can be found in file p51_occ.dat , which can be used with MATLAB programs. Column 1 contains field current in amps, and column 2 contains opencircuit terminal voltage in volts. 110 S OLUTION (a) If the noload terminal voltage is 2300 V, the required field current can be read directly from the opencircuit characteristic. It is 4.25 A. (b) This generator is Yconnected, so A L I I = . At rated conditions, the line and phase current in this generator is ( ) 1000 kVA 251 A 3 3 2300 V A L L P I I V = = = = at an angle of 36.87 The phase voltage of this machine is / 3 1328 V T V V = = . The internal generated voltage of the machine is A A A S A R jX = + + E V I I ( )( ) ( )( ) 1328 0 0.15 251 36.87 A 1.1 251 36.87 A A j = +  +  E 1537 7.4 V A = E (c) The equivalent opencircuit terminal voltage corresponding to an A E of 1537 volts is ( ) ,oc 3 1527 V 2662 V T V = = From the OCC, the required field current is 5.9 A....
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This note was uploaded on 10/15/2009 for the course EE 255 taught by Professor Shaban during the Spring '05 term at Cal Poly.
 Spring '05
 Shaban

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