kissfan4 Posted April 8, 2012 Share Posted April 8, 2012 this is off topic but i dont know where else to ask this, and i know there are some on here with electrical expertise here. this is a question on my final and i am lost,"The nameplate of a transformer indicates 4160Y/2400- 480/277, 50KVA." the question is... ".) if the transformer is loaded to 80%, how much power is being delivered to the load?" i have no clue Link to comment Share on other sites More sharing options...
Dr. Jones Posted April 8, 2012 Share Posted April 8, 2012 power(watts) = Volts x Amps Link to comment Share on other sites More sharing options...
Ken Benedict Posted April 8, 2012 Share Posted April 8, 2012 Just guessing that 50KVA is the maximum rating, so 80% of that would be 40KVA, ignoring the voltage conversion from 4K to 480. Link to comment Share on other sites More sharing options...
-klb- Posted April 8, 2012 Share Posted April 8, 2012 If it is linear load, with power factor of 1, it is 40,000 watts.The voltages don't matter. 50KVA into resistive, or linear, 1.0 power factor loads is 50,000 watts. Then take 80% of that..Hopefully this is not related to any LOR loads, as even 480 delta, 277 is not controller friendly, and neither is 3 phase generally. Link to comment Share on other sites More sharing options...
Al in Raleigh Posted April 12, 2012 Share Posted April 12, 2012 As stated above, with purely resistive loads, 40,000 watts as long as 3 pancakes are attached with 6 AWG wire. Link to comment Share on other sites More sharing options...
Max-Paul Posted April 12, 2012 Share Posted April 12, 2012 I'll see your 3 pancakes with 6 hot dogs (resistive heating). Did anyone ever buy one of those hot dog cookers that you put the hot dog between two spikes and current ran through the dog to cook it? Link to comment Share on other sites More sharing options...
jldavis1969 Posted April 12, 2012 Share Posted April 12, 2012 Max-Paul wrote: I'll see your 3 pancakes with 6 hot dogs (resistive heating). Did anyone ever buy one of those hot dog cookers that you put the hot dog between two spikes and current ran through the dog to cook it?Sadly, YES. The flavor ws terrible too. Link to comment Share on other sites More sharing options...
Al in Raleigh Posted April 13, 2012 Share Posted April 13, 2012 Max-Paul wrote: .. Did anyone ever buy one of those hot dog cookers that you put the hot dog between two spikes and current ran through the dog to cook it?Yes I call them cheap LED strings. Link to comment Share on other sites More sharing options...
Tim Fischer Posted April 14, 2012 Share Posted April 14, 2012 Max-Paul wrote: I'll see your 3 pancakes with 6 hot dogs (resistive heating). Did anyone ever buy one of those hot dog cookers that you put the hot dog between two spikes and current ran through the dog to cook it?When I was a kid, I had a kid's science project book that suggested you bang two nails in a board, hook an electrical wire with plug to the bottom of them (one wire on each nail),l and you've created an instant hot-dog cooker. They did tell you to write a huge "WARNING Don't touch this when plugged in!" on the board, but I can't imagine anyone would dare suggest kids do this sort of thing these days.I never actually did it though. I kind of wish I would have. Link to comment Share on other sites More sharing options...
bobschm Posted April 14, 2012 Share Posted April 14, 2012 Max-Paul wrote:Did anyone ever buy one of those hot dog cookers that you put the hot dog between two spikes and current ran through the dog to cook it?We used to plug a "widowmaker" (made of a couple of nails and an extension cord with alligator clips on the end) through a dill pickle. Made it glow yellow-green!:shock:Hey...I could run that with my LOR! Link to comment Share on other sites More sharing options...
Max-Paul Posted April 14, 2012 Share Posted April 14, 2012 Oh oh! I think I might be the cause that a few villages are going to be short of a few idiots. :shock: Link to comment Share on other sites More sharing options...
Randy Posted April 14, 2012 Share Posted April 14, 2012 kissfan4 wrote:this is off topic but i dont know where else to ask this, and i know there are some on here with electrical expertise here. this is a question on my final and i am lost,"The nameplate of a transformer indicates 4160Y/2400- 480/277, 50KVA." the question is... ".) if the transformer is loaded to 80%, how much power is being delivered to the load?" i have no clueAh, a word problem... My favorite The easy answer would be to say the power delivered to the load would be 80% of 50 kVA or 40 kVA. If the power factor is 1.0, that would be 40 kW. But that would be an assumption that would need to be stated and a unity power factor is not always achieved in real life (of course, it depends on the type of load).Also, every transformer incurs some losses in the form of heat and these losses consume a small amount of power.If I were answering this question on an exam, I would base my answer on the 80% loading given information, assume a reasonable power factor and work that in, assume some reasonable copper losses and work those factors into the final answer. It will show the professor that you are thinking about different factors that can affect transformer loading and efficiency...Of course, your question was a week ago, so maybe it's too late....How did it go? Link to comment Share on other sites More sharing options...
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