Transformer efficiency
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rg1
Well-known member
I knew. except that I jumped to full load while calculating the eff. at 25% there was nothing wrong. I did it by actually finding the resistance (the value is somewhere 3.58 Ohms) too when Omer said two mistakes. I was sure there was no other mistake. That was mistake of, being out of touch, loss of concentration. Lol.
Dont know why this hasn't come to me till now, maybe a cold beer on a Friday evening does the trick, but looking from the output back makes sense. To evaluate any equipment, you need to know how it responds to a certain condition. The only way to do this is to expose it to said condition and record its behavior.
I feel like a dipshidiot!!!
I feel like a dipshidiot!!!
I did some research and generally in transformer efficiency problems they give you the LOAD at which you are required to find the efficiency at. Thus, it is the output not the input as said by @rg1.
One more thing to consider, sometimes they specify the VA rating of the load and the PF. in this case VA rating is used to calculate currents and losses , however, efficiency calculations is done on the Watts ratings only in this case. I thought it is worth mentioning.
One more thing to consider, sometimes they specify the VA rating of the load and the PF. in this case VA rating is used to calculate currents and losses , however, efficiency calculations is done on the Watts ratings only in this case. I thought it is worth mentioning.
rg1
Well-known member
@Omer Some misunderstanding there. I never meant the power ratings of a xmer are input ratings. In my solution I took it as output and in my posts too I mention that for all machines, the Power ratings are output of that machine. The second part of your post is really worth mentioning, I agree. Good point. For losses we have too keep in mind what is the current through the Xmer because they purely depend on Input Voltage and current , I will just extend it a little further. If the pf improvement is done on feeding side (Primary side) of Xmer we take the current from VA of load and Power in W but if pf improvement is done on Load side of the Xmer (Secondary side) then current with improved pf needs to taken flowing through the Xmer. Just think on it. It is interesting.
rg1
Well-known member
LOl. My bad. Sorry.I got your English wrong. it is the output not the input as said by.........
The "anomaly" with the math is simply that the problem statement explicitly says the XFMR is loaded at 25%. It doesn't say the output is loaded at any percent. Loaded, to me, means input. I get it. I know the math, and I understand it clearly and can see why it's done that way, but the language and math do not jibe. Loads are depleted by series impedance, looking toward the load, not the other way around. Hence, why I adamantly maintain it's not necessarily intuitive. But yes, if you run across this type of problem, solve it by the approach used in academia. I'm not saying academia is wrong. Hell, I've been there and done it and have gone through this whole process. I'm merely saying the verbiage and approach do not mesh. That's all.
