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dangina [55]
3 years ago
12

A 3-phase induction motor with 4 poles is being driven at 45 Hz and is running in its normal operating range. When connected to

a certain load, the motor rotates at a speed of 1070 rpm and outputs 800W a. What is the slip and developed torque? b. What is the new slip and motor speed if the developed torque is reduced to 4 Nm?
Engineering
1 answer:
Dennis_Churaev [7]3 years ago
5 0

Answer:

a) The slip for the given conditions is 0.2074 or 20.74% and the developed torque is 7.14 Nm.

b) The new slip after reducing the torque to 4 Nm is 0.1162 or 11.62% and the new motor speed is 1193.13 rpm.

Explanation:

In order to find the slip we can use the definition as the relative difference between Synchronous speed and the rotor speed and that the developed torque is the ratio between output power and rotor angular velocity.

Slip.

We can find the synchronous speed u sing the following formula

N_s =\cfrac{120f}p

Where f stands for the frequency and p the number of poles, so we have

N_s = \cfrac{120(45)}{4}\\N_s =1350 \,rpm

Replacing on the slip formula

S = \cfrac{N_s-N_r}{N_s}

we get

S = \cfrac{1350-1070}{1350}\\ S = 0.2074

Thus the slip for the given conditions is 0.2074 or 20.74%.

Developed torque.

We can find the angular velocity in radians per second

\omega_r =1070  \cfrac{rev}{min} \times \cfrac{1 \, min}{60 \, s}\times \cfrac{2\pi rad}{1 \, rev}\\\omega_r =112.05 \, \cfrac{rad}{s}

Thus we can replace on the torque formula

\tau = \cfrac{P}{\omega_r}\\\tau=\cfrac{800 W}{112.05 \, \cfrac{rad}{s}}

We get

\tau = 7.14\,  N m

The developed torque is 7.14 Nm.

Slip for the reduced torque.

The torque is proportional to the slip, so we can write

\cfrac{\tau_1}{\tau_2}= \cfrac{S_1}{S_2}

Thus solving for the new slip S_2 we have:

S_2 = S_1 \cfrac{\tau_2}{\tau_1}

Replacing the values obtained on the previous part we have

S_2 = 0.2074 \cfrac{4 Nm}{7.14 Nm}\\ S_2=0.1162

So the new slip after reducing the torque to 4 Nm is 0.1162 or 11.62%

Motor speed.

We can use the slip definition

S_2 =\cfrac{N_s-N_{r_2}}{N_s}

Solving for the motor speed we have

S_2N_s =N_s-N_{r_2}

N_{r_2}=N_s-S_2N_s \\N_{r_2}=N_s(1-S_2)

Replacing values we have

N_{r_2}=1350 \, rpm(1-0.1162)\\N_{r_2}=1193.13 rpm

The new motor speed is 1193.13 rpm.

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