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akch2002, Engineer
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# An 11kV to 415 V transformer has a rating of 100kVA. The winding

### Customer Question

An 11kV to 415 V transformer has a rating of 100kVA. The winding resistance and leakage reactance when referred to the secondry are 0.0142 ohms and 0.0569 ohms respectively. Determine the % regulation of the transformer for a 0.8 lagging power factor.
Submitted: 3 years ago.
Category: Homework
Expert:  akch2002 replied 3 years ago.

Is it single-phase or 3-phase transformer? Also indicate load at which regulation is to be determined?

Customer: replied 3 years ago.

Hi, The transformer is 3 phase "An 11kv to 415v...."
Load for regulation ? Nothing has been omitted from the original question this is all i have. As a load has not been specified i assume the load is the combined resistances of the winding and leakage reactance. Thats my understanding of the question, what do you think ?

Customer: replied 3 years ago.
Relist: Other.
Dont want to wait
Expert:  akch2002 replied 3 years ago.
Do you still want answer for this question?
Customer: replied 3 years ago.

Expert:  akch2002 replied 3 years ago.

word file
Kindly rate the answer. BONUS will be appreciated.

akch2002, Engineer
Category: Homework
Satisfied Customers: 3091
Experience: Home work expert
Customer: replied 3 years ago.

Thanks for your help. To be honest i have calculated an entirely differently way and was unable to follow your workings. The good news is i get the same answer so your reply has still helped. Here are my workings.

Let me know what you think.

Attachment: 2013-03-04_034914_scanned_doc_2.pdf

Expert:  akch2002 replied 3 years ago.

There are two ways to calculate regulation.
1) per unit method
2) Conventional method.
Per unit method is straight and chances of committing mistake are rare, In this method, we should know how resistance and reactance are converted in per unit values. Also it avoids the type of connection, that is, whether it is delta or star connected.
Conventional method is lengthy and chances of committing mistake are quite high. That's why I rarely use this method. In this method, we have to use actual values of current etc.

Don't you see that you also have used per unit method?

Customer: replied 3 years ago.

Not really if i am honest , just happy with an answer, its the last question in a long assignment thats been all maths without much practcal application behind it

Expert:  akch2002 replied 3 years ago.
I will try to clarify later today.
Expert:  akch2002 replied 3 years ago.
You can view this answer by clicking here to Register or Login and paying £3.
Customer: replied 3 years ago.

Now i have to rate you again and pay you a further £11 for an explanation to an answer you have already been paid for ?

I appreciate your help and reckon i paid a fair price for it. May even need your help again at some point and if i do i will pay a fair price again for it

Expert:  akch2002 replied 3 years ago.
You are most welcome for putting up new questions.

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