John120/240
Senior Member
- Location
- Olathe, Kansas
2400 or so post in the other thread Now who's up for a good arguement ?
You would be quoting to a set of specs prepared in advance taking into account present load,required spare capacity,maximum efficiency operating point etc.,and so you may quote confidently expecting the award of work to you.So be ready.If I bid for a project, the overwhelming probability is that I will be in competition. Possibly with competitors and/or other projects for funding.
If I price for larger than required transformer ratings for the application the chances are that Joe Bloggs or John Doe, will be able to undercut my offer and maybe win the business that we could otherwise have had.
Actually, that's not generally the case. Many of specifications, particularly those drawn up by consultants, are pretty vague on ratings and much else.You would be quoting to a set of specs prepared in advance taking into account present load,required spare capacity,maximum efficiency operating point etc.,
Actually, that's not generally the case. Many of specifications, particularly those drawn up by consultants, are pretty vague on ratings and much else.
I really had no such expectation and, anyway, my post wasn't specifically about any particular bid.Then I can not help you to win this bid.
I didn't think there was. I said it was only new to me. I was, and am still, sure it would be old hat to someone such as yourself, that fully understands the physics. The reason I brought it up was no one was discussing the math models.
At my level of understanding, this is not readily apparent to me by inspection. I'm sure it is a simple derivation - its nothing I would ask you to do here. I'll see if I can work this one out.
I don't get this one at all. The efficiency calc you use is for full load - 400kw. But the coil losses you are using are less than that listed for full load. ???.... You will only achieve maximum efficiency when the Cu losses equal the Fe losses. Per example, the total losses = 3960.7 X 2 = 7,921.4 watts; maximum efficiency = 400/(400+7.9214) = 0.98.058 or 98.06%. ...
Again, at my level of understanding, this is not readily apparent to me by inspection. This may be a simple, but I suspect, tedious derivation. Perhaps you could list a reference.
I'm a bit surprised the maximum efficiency is not influenced by the ambient. Is that a third order effect and not part of your model? Or not even relavent?
ice
Surely,higher the efficiency lower is the loss in the transformer.But efficiency isn't the whole story. Efficiency is a ratio. For least cost operation, you need the lowest losses in absolute terms.
Efficiency isn't the whole story. A larger transformer might have a higher efficiency but losses can be greater than those of a smaller transformer depending on loading. As an extreme example to illustrate the point, 50kVA of loading on say, a 75kVA transformer is likely to result in significantly lower losses than if that 50kVA load was fed from a 7,500kVA transformerSurely,higher the efficiency lower is the loss in the transformer.
The trick is to use a higher size transformer so that the net cost to the owner is lower (taking into consideration future load expansion and maximum efficiency point not lower than that for present load) than using a lower size transformer catering to the present load only.Efficiency isn't the whole story. A larger transformer might have a higher efficiency but losses can be greater than those of a smaller transformer depending on loading. As an extreme example to illustrate the point, 50kVA of loading on say, a 75kVA transformer is likely to result in significantly lower losses than if that 50kVA load was fed from a 7,500kVA transformer
What if the transformer is designed for purpose and future expansion isn't a consideration?The trick is to use a higher size transformer so that the net cost to the owner is lower (taking into consideration future load expansion and maximum efficiency point not lower than that for present load) than using a lower size transformer catering to the present load only.
Then it is not of modern electrical design.What if the transformer is designed for purpose and future expansion isn't a consideration?
What percent expansion is included in "modern electrical design"?Then it is not of modern electrical design.
You seem to have completely missed my point:Then it is not of modern electrical design.
Wind mill?? Wot wind mill, guv??Bes:
How did you arrive at maximum KVA of the transformer i.e 471 KVA in the wind mill example of your last post in this thread?Please put up the principles here.
You assumed.Bes:
Thanks for your reply.I thought,since doubly fed induction motor is also used in wind mill application,your post related to it.
I did.But you still did not explain why maximum capacity i.e 471 KVA for the transformer is required?
If you mean what is the lowest operational duty, then it is zero. You can clearly see that from the curves.What about the minimum capacity of the transformer in this case?