Question

A drive has following parameters J-10 kg-m', T-15 OSN, N-m, and T-5 0.06N, N-m where N is speed in rpm Initially the drive is working in steady state. Now the drive is braked by electric braking Torque of the motor in braking is given by = -10-0.04N, N-m. Calculate time taken by drive to stop

Answer 1

Explanation:

drive to stop the km is N and knmn kshbsjs ugsnvs jdgs sjb oh KBC if JC kvw

Answer 2

Answer:

The correct answer will be -

It will take 21.33 seconds to stop.

Explanation:

Correction in the question -

The given parameters are -

J = 10 kg-m²

T = 15 + 0.05N, N-m

$T_l$ = 5 + 0.06N, N-m

The torque of the motor in braking is T = -10 - 0.04N, N-m

Initially, the drive is working in steady-state so-

$T -T_l = 0\\\\15+0.05N-5-0.06N = 0\\\\10-0.01N= 0\\\\N = \frac{10}{0.01} = 1000$

So the speed of the drive in the steady-state is 1000 rpm.

The net torque while braking -

$T_b = -10-0.04N-(5+0.06N)\\\\T_b = -15-0.1N = -(15+0.1N)$

As we know,

$T_b = J\frac{dN}{dt} \\\\J\frac{dN}{dt} = -(15+0.1N)$

$\frac{dN}{15+0.1N} = -\frac{3}{\pi} dt$

At t = 0, N = 1000 rpm

At t = t, N = 0 rpm

Integrating with proper limits -

$\int\limits^0_1000 {\frac{dN}{15+0.1N}} \, dN = \int\limits^t_0 {-\frac{3}{\pi}} \, dt \\\\10(ln15-ln115) = -\frac{3}{\pi} t$

$t = \frac{10\pi}{3}ln(\frac{115}{15} ) = 21.33\ s$

So the time taken by the drive to stop is 21.33 seconds.

#SPJ3