Question

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?

Answer 1

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.

Step-by-step 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} * \cfrac{1 \, min}{60 \, s}* \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.