Question

A DC servomotor has a torque constant of 0.075 N-m/A and a voltage constant of 0.12 V/(rad/sec). The armature resistance is 2.5 Ohm. A terminal voltage of 24 V is used to operate the motor. Determine:
a) the starting torque generated by the motor just as the voltage is applied.
b) the maximum speed at a torque of zero.
c) the operating point of the motor when it is connected to a load whose torque characteristic is proportional to speed with a constant of proportionality = 0.0125 N-m/(rad/sec).

Answer 1

Answer:

The answer is below

Explanation:

Given that:

$k_t=torque\ constant=0.075\ Nm/A\\\\k_v=voltage\ constant=0.12\ V/(rad/sec)\\\\R_a=armature \ resistance=2.5 \Omega\\\\V_t=terminal\ voltage=24\ V\\\\a)The \ starting\ current\ I_a\ is\ given \ as:\\\\I_a=\frac{V_t}{R_a} =\frac{24}{2.5} =9.6\ A\\\\The \ starting\ torque(T)\ is:\\\\T=k_tI_a=0.075*9.6=0.72\ N.m$

b) The maximum speed occurs when the terminal voltage and back emf are equal to each other i.e.

$V_t=e_b=k_v\omega\\\\\omega=\frac{V_t}{k_v}=\frac{24}{0.12} =200\ rad/s$

c) The load torque is given as:

$T_L=0.0125\Omega\\\\The\ motor\ torque \ is:\\\\T=k_t(\frac{V_t-k_v\omega}{R_a} )\\\\but\ T = T_L,hence:\\\\0.0125\omega=0.075(\frac{24-0.12\omega}{2.5} )\\\\0.03125\omega=1.8-0.009\omega\\\\0.04025\omega=1.8\\\\\omega=44.72\ rad/sec\\\\N=\frac{60\omega}{2\pi} =\frac{60*44.72}{2\pi} =427\ rpm$