find the following of questions

Question

The input power to a 220 V lap wound shunt D.C motor runs is 11 kW , the speed is 1150 rpm takes 5 A at no load ,shunt field resistance is 110 Q and armature resistance is 0.5 Q Find a-The speed of motor at this load b- The speed if connected resistance (25 Q) in series with field resistance when the armature torque is constant and the flux is proportional to field current .assume the information in branch a is normal condition.c-Maximum efficiency .before and after adding (R add)

Answer 1

First ... note that e=krpm where is the armature voltage with constant field current ... at No Load the armature current is the Line current minus the field current = 3 amperes. Note that e=220-.5*3 = 218.5 volts at 1150 rpm, thus k=218.5/1150= .19 volt/rpm.

Now with input power at 11,000 watts the line current = 11000/220=50amperes. Note that the field current is still 2 amperes leaving 48 amperes into the armature and e=220-48*.5= 196 volts. Note that the speed is now e/k= 1032 rpm. Part A.

Note that k from above was with constant field current and k=n*flux. If flux is assumed to be proportional to field current then k=nk'I_field when k' is the proportional constant between I_field and flux. If the field current is reduced from 2 amperes to 220/135= 1.63 amperes, k = .19*1.63/2=.155 volts/rpm. If armature current is still 48 amperes then e remains to be 196 volts but the speed is now 196/.155=1266 rpm. Part B

if we assume the constant losses are 220²/110= 440 watts. Setting that equal to I²*.5 we find that the line current for max eff = 29.7amperes with field resistance = 110 ohms. This relates to an armature current of 27.7 amperes and e=220-.5*27.7= 206volts for a speed of 206/.19= 1085rpm