7.1 Problems - chapter 7  (Page 2/6)

7.6 The dc machine of Problem 7.4 is to be operated as a motor supplied by a constant armature terminal voltage of 250 V. If saturation effects are ignored, the magnetization curve of Fig. 7.1 becomes a straight line with a constant slope of 150 volts per ampere of field current. For the purposes of this problem, you may assume that saturation effects can be neglected.

a. Assuming that the field current is held constant at 1.67 A, plot the motor speed as a function of motor shaft power as the shaft power varies from 0 to 25 kW.

b. Now assuming that the field current can be adjusted in order to maintain the motor speed constant at 1200 r/min, plot the required field current as a function of motor shaft power as the shaft power varies from 0 to 25 kW.

7.7 Repeat Problem 7.6 including the saturation effects represented by the saturation curve of Fig. 7.1. For part (a), set the field current equal to the value required to produce an open-circuit armature terminal voltage of 250 V at 1200 r/min. (Hint: This problem is most easily solved using MATLAB and its "spline0" function as in Problem 7.5.)

7.8 A 15-kW, 250-V, 1150 r/min shunt generator is driven by a prime mover whose speed is 1195 r/min when the generator delivers no load. The speed falls to 1140 r/min when the generator delivers 15 kW and may be assumed to decrease in proportion to the generator output. The generator is to be changed into a short-shunt compound generator by equipping it with a series field winding which will cause its voltage to rise from 230 V at no load to 250 V for a load of 61.5 A. It is estimated that the series field winding will have a resistance of 0.065 $\Omega$ . The armature resistance (including brushes) is 0.175 $\Omega$ . The shunt field winding has 500 turns per pole. To determine the necessary series-field turns, the machine is run as a separately-excited generator and the following load data are obtained:

Armature terminal voltage = 254 V

Armature current = 62.7 A

Field current = 1.95 A

Speed = 1140 r/min

The magnetization curve at 1195 r/min is as follows:

Determine

a. the armature reaction in equivalent demagnetizing ampere-turns per pole for $I_a$ = 62.7 A and

b. the necessary number of series-field turns per pole. (Hint: This problem can be solved either graphically or by use of the MATLAB "spline()" function to represent the magnetization curve.)

7.9 When operated from a 230-V dc supply, a dc series motor operates at 975 r/min with a line current of 90 A. Its armature-circuit resistance is 0.11 $\Omega$ and its series-field resistance is 0.08 $\Omega$ . Due to saturation effects, the flux produced by an armature current of 30 A is 48 percent of that at an armature current of 90 A. Find the motor speed when the armature voltage is 230 V and the armature current is 30 A.

7.10 A 250-V dc shunt-wound motor is used as an adjustable-speed drive over the range from 0 to 2000 r/min. Speeds from 0 to 1200 r/min are obtained by adjusting the armature terminal voltage from 0 to 250 V with the field current kept constant. Speeds from 1200 r/min to 2000 r/min are obtained by decreasing the field current with the armature terminal voltage remaining at 250 V. Over the entire speed range, the torque required by the load remains constant.