A 225 kW (≈ 300 hp), 250 V, 1280 r/min dc motor has windage, friction, and iron losses of 8 kW. It drives a large flywheel and the total moment of inertia of the flywheel and armature is 177 kg·m² . The motor is connected to a 210 V dc source, and its speed is 1280 r/min just before the armature is

Question

A 225 kW (≈ 300 hp), 250 V, 1280 r/min dc motor has windage, friction, and iron losses of 8 kW. It drives a large flywheel and the total moment of inertia of the flywheel and armature is 177 kg·m² . The motor is connected to a 210 V dc source, and its speed is 1280 r/min just before the armature is switched across a braking resistor of 0.2 Ω.

Calculate

a. The mechanical time constant [latex]T_{ o }[/latex] of the braking system

b. The time for the motor speed to drop to 20 r/min

c. The time for the speed to drop to 20 r/min if the only braking force is that due to the windage, friction, and iron losses

Accepted Answer

a. We note that the armature voltage is 210 V and the speed is 1280 r/min.
When the armature is switched to the braking resistor, the induced voltage is still very close to 210 V. The initial power delivered to the resistor is

[latex]P_{1}=E^{2} / R=210^{2} / 0.2=220500 W[/latex]

The time constant [latex]T_{ o }[/latex] is

[latex]\begin{aligned}T_{ o } &=J n_{1}^{2} /\left(131.5 P_{1} ight) \&=\frac{177 imes 1280^{2}}{131.5 imes 220500} \&=10 s\end{aligned}[/latex] (Euation 9)

b. The motor speed drops by 50 percent every 10 s. The speed versus time curve follows the sequence given below:

[latex]\begin{array}{rc} ext { speed (r/min) } & ext { time (s) } \1280 & 0 \640 & 10 \320 & 20 \160 & 30 \80 & 40 \40 & 50 \20 & 60\end{array}[/latex]

The speed of the motor drops to 20 r/min afteran interval of 60 s.

c. The initial windage, friction, and iron losses are 8 kW. These losses do not vary with speed in exactly the same way as do the losses in a braking resistor. However, the behavior is comparable, which enables us to make a rough estimate of the braking time. We have

[latex]n_{1}=1280 P_{1}=8000[/latex]

The new time constant is

[latex]\begin{aligned}T_{ o } &=J n_{1}^{2} /\left(131.5 P_{1} ight) \&=\left(177 imes 1280^{2} ight) /(131.5 imes 8000) \&=276 s =4.6 min\end{aligned}[/latex]

The stopping time increases in proportion to the time constant. Consequently, the time to reach 20 r/min is approximately

[latex]\begin{aligned}t &=(276 / 10) imes 60=1656 s \&=28 min\end{aligned}[/latex]

This braking time is 28 times longer than when
dynamic braking is used.

Theoretically, a motor which is dynamically braked never comes to a complete stop. In practice, however, we can assume that the machine stops after an interval equal to 5 [latex]T_{ o }[/latex] seconds.
If the motor is plugged, the stopping time has a definite value given by

[latex]t_{ s }=2 T_{ o }[/latex]

where

[latex]\begin{aligned}t_{ s } &= ext { stopping time using plugging }[ s ] \T_{ o } &= ext { time constant as given in Eq. } 9[ s ]\end{aligned}[/latex]

Question 5.6: A 225 kW (≈ 300 hp), 250 V, 1280 r/min dc motor has windage,...

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