Question

A 3-phase induction motor is being driven at 75 Hz. Initially, the motor is connected to a mechanical load that requires 10Nm of torque while spinning at a speed of 950rpm. Assume rotational losses are negligible. a) How many poles does the motor probably have? b) What is the slip? c) What is the power supplied to the load? d) If a new load with different torque is connected that requires 1400 W of power, what is the new speed of the motor in rpm and slip assuming that the motor is operating in its normal operating range?

Answer 1

a)Therefore, the motor probably has approximately 10 poles. b)The slip is approximately 0.2083, or 20.83%.

c)The power supplied to the load is approximately 994.7 watts.

d)the new slip is approximately -0.1125, or -11.25%.

Let's break down each part of the problem step by step:

a) How many poles does the motor probably have?

The synchronous speed (Ns) of a 3-phase induction motor is given by the formula:

Ns = (120 * f) / P

Where:

- Ns is the synchronous speed in RPM

- f is the frequency in Hz

- P is the number of poles

Given:

- Frequency (f) = 75 Hz

- Initial speed (N) = 950 RPM

We can rearrange the formula to find the number of poles (P):

P = (120 * f) / N

P = (120 * 75) / 950

P ≈ 9.47

Since the number of poles must be a whole number, we can round to the nearest whole number. Therefore, the motor probably has approximately 10 poles.

b) What is the slip?

The slip (S) of an induction motor can be calculated using the formula:

S = (Ns - N) / Ns

Where:

- Ns is the synchronous speed (calculated in part a)

- N is the actual speed (given as 950 RPM)

S = (Ns - 950) / Ns

S = (1200 - 950) / 1200

S ≈ 0.2083

So, the slip is approximately 0.2083, or 20.83%.

c) What is the power supplied to the load?

The power supplied to the load can be calculated using the formula:

Power = Torque * Angular Speed

Given:

- Torque (T) = 10 Nm

- Speed (N) = 950 RPM

First, we need to convert the speed from RPM to radians per second (rad/s):

ω = (N * 2π) / 60

ω = (950 * 2π) / 60 ≈ 99.47 rad/s

Now, we can calculate the power:

Power = T * ω

Power = 10 Nm * 99.47 rad/s ≈ 994.7 W

So, the power supplied to the load is approximately 994.7 watts.

d) New Speed of the Motor and Slip with Different Load:

Given:

- New power required by the load (P_new) = 1400 W

We can calculate the new speed (N_new) using the power formula:

P_new = T * ω_new

Solving for ω_new:

ω_new = P_new / T

ω_new = 1400 W / 10 Nm ≈ 140 rad/s

Now, we can find the new speed in RPM:

N_new = (ω_new * 60) / (2π)

N_new ≈ (140 * 60) / (2π) ≈ 1334 RPM

To find the new slip, we'll use the formula from part b:

S_new = (Ns - N_new) / Ns

S_new = (1200 - 1334) / 1200 ≈ -0.1125

The negative slip indicates that the motor is now running faster than its synchronous speed. Therefore, the new slip is approximately -0.1125, or -11.25%.