Consider the linear dc machine shown below. Its main parameters are: VB = 240 V,
ID: 1715423 • Letter: C
Question
Consider the linear dc machine shown below. Its main parameters are: VB = 240 V, R = 0.5 W, B = 0.25 T and a l = 4 m.
A) What would be the speed (magnitude and direction) of the bar at no-load?
B) What external force (magnitude and direction) would be required to charge the battery with 5 kW.
C) What would be the speed (magnitude and direction) of the bar if the linear dc machine operates as a motor with 2.5 kW being drawn from the battery? Note: Consider the direction of the current as defined in the figure.
Explanation / Answer
A)
Let E be the voltage across the motor.
Then E=Blv.
Since no load, Vb=E=240 V.
v=(E/(B*l))=240/(0.25*4)
=>v=240 m/s
B)
To charge the battery with 5 kW, I=5000/Vb=5000/240=20.833 A has to flow from motor to battery.
So E=Vb+R*I=250.415 V
Force=BIL=0.25*20.833*4=20.833 N
c)
2.5 kW has to be provided by the battery.
So current flowing from battery to the motor,I=2500/240=10.42 A.
So E=Vb-IR=234.79 V
E=Blv=>v=E/(Bl)=234 m/s.