Transformer engineering problem

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Homework Help Overview

The discussion revolves around a transformer engineering problem involving the calculation of input power based on output power and efficiency. The transformer steps down voltage from 3900V to 220V and is intended to deliver 1180kW to a factory at 91% efficiency.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between input and output power in transformers, particularly focusing on the impact of efficiency. There are attempts to apply formulas related to power, current, and voltage, but confusion arises regarding the correct application of efficiency in the calculations.

Discussion Status

Some participants have provided insights into the need to consider efficiency when calculating input power, suggesting that the original poster may be misunderstanding the relationship between power in and power out for nonideal transformers. There is acknowledgment of the original poster's efforts, but also recognition of potential missteps in their calculations.

Contextual Notes

Participants note that the original poster's homework submission system is indicating incorrect answers, which adds a layer of frustration to the discussion. There is also mention of a previous post that provided a formula, indicating that some guidance has been shared, though it may not have been fully understood by the original poster.

neraketam
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Homework problem:
A transformer on a pole near a factory steps the voltage down from 3900V to 220V. The transformer is to deliver 1180kW to the factory at 91% efficiency. Find the power delivered to the primary in kW.

What I've done:
Well, I thought that power in equals power out. So I worked all the math to prove that and used P=IsVs, solved for Is then used that to find Ip (Ip=VsIs/Vp). Then found P=IpVp. I got 1180kW! I then calculated in the efficiency--not correct. And then in the follow up question, I found out that my Ip isn't even correct. What's wrong here?
 
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Originally posted by neraketam
Well, I thought that power in equals power out. So I worked all the math to prove that and used P=IsVs, solved for Is then used that to find Ip (Ip=VsIs/Vp). Then found P=IpVp. I got 1180kW!
Sounds like you are going in a circle.

True, power in = power out for an ideal transformer. A nonideal one will waste some of that power. If I read the question correctly, the first part just asks for what input power is needed to get the output power given, given the efficiency:

Pin X 0.91 = Pout
 
transformers

I did that but my silly internet homework submission page is saying it's not correct. That is not the first time it's said that so...at least I know I'm doing the problem right and thanks for getting back to me so quickly!
 
You don't need to know anything about voltage and current to do this. All you need to know is the power out and the efficiency. It is supposed to be 1180 Kwatts out. So, figure the efficiency from that. Can't help you with your internet problem though. Sorry.

P.S. Why did I just post this when Doc Al did it so nicely with just a formula?
 

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