This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 109 Chapter 5 : Synchronous Generators 5-1. At a location in Europe, it is necessary to supply 300 kW of 60-Hz power. The only power sources available operate at 50 Hz. It is decided to generate the power by means of a motor-generator set consisting of a synchronous motor driving a synchronous generator. How many poles should each of the two machines have in order to convert 50-Hz power to 60-Hz 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 10-pole synchronous motor must be coupled to a 12-pole synchronous generator to accomplish this frequency conversion. 5-2. A 2300-V 1000-kVA 0.8-PF-lagging 60-Hz two-pole Y-connected 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 P5-1. (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 open-circuit terminal voltage in volts. 110 S OLUTION (a) If the no-load terminal voltage is 2300 V, the required field current can be read directly from the open-circuit characteristic. It is 4.25 A. (b) This generator is Y-connected, 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 open-circuit 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....
View Full Document
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