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ch5sol

# ch5sol - Chapter 5 Problems Solutions 5.1 Based on[3.10 the...

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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 time-varying B -field are given by ( 29 ( 29 1 2 , 0< < a g d v v B t r dt θ θ θ π = = l ( 29 ( 29 1 2 , , g a g a dB t d v v B t r r dt dt θ θ θ θ = = + l l Recognize / d dt θ 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 θ θ ϖ θ = + = + l l The first term on the right-hand 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 π π × = = = 2 5.3 Examine the units for both E k and K . V lines rpm E p m E k n = Φ : V Wb rad /s p m E K ϖ = Φ : 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 π Thus, 10 8 1.6667 10 2 30 10 E pZ pZ k a a π π - = = × × 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

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and 9 8 0.738 1.174 10 2 10 T pZ pZ k a a π - = = × 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 Φ , 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 ϖ ϖ ϖ ϖ Φ = = = = = Φ 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 so-called 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

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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 η = × = = ( 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 cemf is half the value determined in Prob. 5.7. ( 29 1 2 245.89 122.94 V E = = The value of load resistance must be 230 4.409 52.17 tR L LR V R I = = = The new load current is found by Ohm’s law.
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