Ra la A DC permanent-magnet motor has an armature resistance R.-1.45 The motor r
ID: 1766062 • Letter: R
Question
Ra la A DC permanent-magnet motor has an armature resistance R.-1.45 The motor runs at a shaft speed of gem-240 rad/s when the armature voltage is Vs = 44 V and the armature current is le = 2.1 A. Assume that V, remains constant. Notes: The developed power is the mechanical power, not the electrical input power. Km is a constant proportional to the flux (which is itself constant) Part (b) If the armature current is la = 7.3 A 5. What is the shaft speed om? (Unit: rad/s) 6. What is the developed torque T? (Unit: N*m) 7. What is the motor power P? (Unit: W)Explanation / Answer
Ra =1.45 ohms
shaft speed Wm=240 rad/sec=(60/2*pi)*240=2291.83 rpm
armature vlotage Vt=44v
armature current Ia=2.1A
Vt=(I*R)+Ve
Ve=Vt-(I*R)=44-(2.1*1.45)=40.955v=back emf
Ke=Ve/Wm=40.995/240=0.1706=back emf constant of motor
when Ia =7.3A
Wm=Vt-((I*R))/Ke=(44-(7.3*1.45))/0.1706=195.8675 rad/sec
torqe M=(V-Wm*Ke)Km/R=(44-195.8675*0.1706)*Km/1.45=7.3 N/m
Km=motor torque constant
power p= M*Wm=7.3*195.8675=1429.833 watts
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