This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 5 Problems Solutions 5.1 Based on [3.10], the indicated voltages across the conductor sides of Fig. 5.8 for the case of a timevarying Bfield are given by ( 29 ( 29 1 2 , 0&lt; &lt; a g d v v B t r dt = = l ( 29 ( 29 1 2 , , g a g a dB t d v v B t r r dt dt = = + l l Recognize / d dt as the angular mechanical speed and apply KVL to yield ( 29 ( 29 1 2 , 2 , 2 g c a g m a dB t v v v B t r dt = + = + l l The first term on the righthand side of the equation is known as a speed voltage while the second term is known as a transformer voltage . 5.2 Based on [5.6], and recalling from Sec. 5.3.2 that 2 a = for a wave winding, ( 29 ( 29 4 45 2 V s 28.648 2 2 Wb rad pZ K a = = = 2 5.3 Examine the units for both E k and K . V lines rpm E p m E k n = : V Wb rad /s p m E K = : Unit analysis gives Wb rad/s lines rpm E k K : The required conversion factor is 8 1 Wb rad /s 30 rpm 10 lines Thus, 10 8 1.6667 10 2 30 10 E pZ pZ k a a  = = Likewise, Wb ft lbs lines N m T k K : The conversion factor is 8 1 Wb 0.738 ft lb N m 10 lines 67 and 9 8 0.738 1.174 10 2 10 T pZ pZ k a a  = = 5.4 The graph below is the required plot. For the 1000 rpm condition, the graph vertical height is 2/3 the value for 1500 rpm. Since constant values of f I correspond to constant values of field mmf, thus constant p , 2 1000 2 2 1500 1 1 1 1000 2 1500 3 p m m m p m m m K E n E K n = = = = = 5.5 With f a I R neglected, f f E I R 2245 so that the point of operation is determined by the point that the socalled field resistance line described by t f f f V V I R = = crosses the open cir cuit characteristic ( 29 f E f I = . From the figure below, it is seen that 615 V t V = for a 68 speed of 1500 rpm and 445 V t V = for a speed of 1250 rpm. The 1250 rpm curve was de termined by plot of ( 29 1250/1500 E from the table of Prob. 5.4. 5.6 Construct a field resistance line on the above graph that passes through the origin and the value of 500 V E = for a speed of 1500 rpm. Choosing a convenient set of values along the field resistance line, this line represents a total field circuit resistance of 600 480 1.25 f rh R R + = = Thus, 480 480 375 105 rh f R R = = = 69 5.7 ( a ) Let subscript R denote rated conditions. Rated line current is 12,000 52.17 A 230 LR I = = By KCL, aR LR fR I I I = + 52.17 1.5 53.67 A = + = ( b ) By KVL, ( 29 230 53.67 0.296 245.89 V R tR aR a E V I R = + = + = ( c ) The mechanical input power is ( 29 245.89 53.67 231 13.428 kW in R aR FW P E I P = + = + = With the output power known to be nameplate rated value, the efficiency follows as ( 29 12 100 100 89.36% 13.428 out in P P = = = ( d ) The total resistance of the field circuit is 230 153.33 1.5 tR f rh fR V R R I + = = = Hence, 153.33 100 53.33 rh R = = 5.8 ( a ) Since the field current, and thus field flux, is unchanged but the speed is halved, the...
View
Full
Document
This document was uploaded on 01/20/2012.
 Spring '09
 Volt

Click to edit the document details