Shows a Wound-Rotor Induction Motor with Controller (Rheostat)

Wound motor operates in the same manner as squirrel-cage motor when the external controller is set at zero resistance as the rotor will be short-circuited. Imagine now that the torque is carrying a given load because of the rotating of rotor at a constant speed. The rotor resistance will be doubled per phase as the controller is moved.

Reactions

Following are some reaction that will be observed for the mentioned example.

The current flowing in the rotor will switch to half of the previous value neglecting the transient effect

  • When the rotor current is reduced, it reduces the developed torque
  • The motor slows down when the load on the motor is constant which develops greater torque than the developed torque
  • The slip also increases in this condition which results into increased rate of stator flux cutting the rotor conductors.
  • The rotor current will increase subsequently in order to develop the required value of torque which is necessary to carry the load at the constant speed.
  • The rotor induced voltage increases.

Rotor resistance along with the reactance increases in the same portion even on the condition when rotor factor remains same. The induction motor will develop the same value of torque which is rotating at a lower speed but carrying the same current as before. It happens because of the slid before the increase in the rotor circuit resistance.

So if you place more load on the induction motor, I will always deliver higher slip and rotor developing the required torque in the complete process. Talking about the situation when pullout torque will be attained” it will happen when slip will be twice as great as the shortened rotor. This is because of rotor resistance which is twice the value of the rotor alone. The value of the pull out torque, thus, will be same as of short circuited rotor.

Fig. 18 shows the torque-slip characteristic of the shorted rotor, together with that for a total resistance of, Rt = 2R.

It will not be worth if the torque is developed at the standstill position which is higher along with increase in rotor resistance. Why, if you ask? The fact is that the rotor frequency is equal to the line frequency which results into maximum rotor resistance.

In order to conclude the best results, the additional resistance in the rotor circuit will result into improved power factor in the circuit when the rotor current is reducing. The develop torque will be greater with the added resistance because of the standstill rotor resistance. The standstill rotor resistance is greater than its town resistance; it helps in increasing the power factor as compared to decrease in the rotor current resulting into great torque.

Examples:

Example 1. A 3-phase, 4-pole, 60-Hz induction motor is running at 1280 r.p.m. Determine the slip speed and slip.

Solution. Number of poles, p = 4

Frequency, f = 60 Hz

Actual speed of rotor, N = 1280 r.p.m.

Slip speed = ? 

 

Example 2:

A 3-phase induction motor has 2 poles and is connected to 500V, 60Hz supply. Calculate the actual rotor speed and rotor frequency when the slip is 5%.

Solution:

Given, p = 2;

S = 5% = 0.05; f = 60Hz

 

Links of Next Electrical Engineering Topics:-

Trustpilot

We’re rated Excellent

4.7 out of 5 based on 39 reviews

Order Homework Help
Click or drag files to this area to upload. You can upload up to 5 files.
We do not share numbers with 3rd party.

Join over 200,000+ happy students served our tutors

Take a look at independent reviews to see what users have to say about our homework help service.

4.5/5

overall rating
on LiveChat

96%

of users
recommend service

How It Works

1. Get Price Estimate.
1. Get Price Estimate.

Submit.

Send us your homework. Mention deadline & upload relevant files.

2. Pay Securely.
2. Pay Securely.

Pay.

We accept all major credit cards. Partial payments also accepted.

3. Get Solutions.
3. Get Solutions.

Delivered

Receive error-free, step-by-step solutions in your email.

FAQs

Why use Myhomeworkhelp?

We provide you with the homework help from top experienced experts. Your satisfaction is a priority to us. Get better grades or money-back. It’s that risk-free! Furthermore, everything about you is kept confidential.

How quickly can you help me with my homework?

We take pride in our 24×7 homework help services. Which means, student can approach us anytime, to get help even on short notices. As short as few hours! And yes, we provide complementary plagiarism-free report.

How much will it cost?

We do not have a monthly fees or minimum payments. Each homework task is quoted a unique price. It is based on tutor’s estimate of the time it will take them to complete your task. There are no obligations, and you are free to discuss the price quote with the tutor.
We do accept partial payment (subject to our policy) to start working on your assignment help. You can pay the remaining amount when your task gets completed. No pressure of up-front payment. No hidden costs.

Who are your tutors?

All our experts are highly qualified professionals – holding at least Master’s degree of their respective fields of expertise. We are very selective and choose only the best qualified tutors for each academic subject. All tutor applicants must provide academic transcripts for each degree they hold, and are tested and screened carefully by our staff.

I have privacy concerns. Can I trust you?

Myhomeworkhelp has completed over 250,000 homework help tasks over the last 12 years. But do not take our word for it! See our here and on Trustpilot. Furthermore, everything about you is kept confidential. Your credit card information is not stored anywhere, and use of PayPal relies on their secure payment networks. Your identity, payment and homework help are in safe hands.

Check our video

Our Reviews
Ratings