Respuesta :
Answer:
Explanation:
According to the magnetization curve, the field current corresponding to a speed of 1770 r/min = 0.482 A
That is, [tex]I_{shunt, nl} = 0.482 A[/tex]
When the motor is operated at full load, [tex]I_{shunt, fl} = 0.468 A[/tex]
a) The armature Reaction, [tex]R = (I_{shunt, nl} - I_{shunt, fl} )N[/tex]
Number of turns, N = 2400
R = (0.482 - 0.468)*2400
The armature reaction at full load R = 33.6 A turns/pole
b) Electromagnetic torque, [tex]\tau = \frac{E_{a}I_{a} }{w_{m} }[/tex]
The armature emf, [tex]E_{a} = V_{t} - I_{a} (R_{a} + R_{c} )\\[/tex]
[tex]V_{t} = 230 V\\I_{a} = 58.2 A\\R_{a} = 0.18 ohms\\R_{c} = 0.035 ohms[/tex]
[tex]E_{a} = 230 -58.2(0.18 + 0.035 )\\[/tex]
[tex]E_{a} = 217.487 V[/tex]
[tex]w_{m} = (\frac{\pi }{30} ) 1770\\w_{m} = 185.35[/tex]
[tex]\tau = \frac{217.487*58.2 }{185.35 }\\\tau = 68.29 Nm[/tex]
c) Let us calculate the effective field current and get the corresponding motor speed
[tex]I_{eff} = \frac{(\frac{E_{a} }{R_{f} })N_{f} -R }{N_{f} }[/tex]
[tex]I_{eff} = \frac{(\frac{230 }{375 })2400 -175 }{2400 } \\I_{eff} = 0.5404 A[/tex]
The motor speed corresponding to a field current of 0.5404 A is 1310 r/min
The generated voltage will be :
[tex]E_{a} =230 *\frac{1770}{1310} \\E_{a} = 310 .76 V[/tex]
Starting torque, [tex]\tau = \frac{E_{a} I_{a} }{w_{m} }[/tex]
[tex]\tau = \frac{310.76*85 }{1770 * (\frac{\pi }{30} }[/tex]
Torque = 142.5 Nm
