Something easy but dont know

[rattus]

Doesn't matter, the shortcut provides the correct answer.

True, but I don't see that this is a shortcut.

Help me out here and walk us through this statement.

If you do this calculation with pencil, you are forced to do two calculations, and the units of the intermediate result are correct. However, if you do this with a calculator, you have to press an extra key to see the intermediate result which is unnecessary.

The non-existent voltage is taught as a shortcut in many instructional courses, so it may only be confusing to you..

I am nitpicking the use of the word "shortcut" here.

I monitored voltage on a feeder at a hospital last month and had over 8,000 voltage changes in a 24 hour reading, so how long would one need to wait for the system stabilize to give us more than an approximate?.

My point is that this is not a matter of accuracy. Every measurement you make may be very accurate. The fact that the readings change often does not change the accuracy. I think stability would be a better term.

And where is the wrong one in this thread?.

Nothing truly wrong here, mostly nitpicking, but I still think you are wrong about the 200A line to line current. You have combined two equations, each of which is numerically correct, but not as a combination,

166,200w / 480v = 346.25a / 1.732 = 200 amps phase to phase

Furthermore, your result is wrong, the line current cannot be equal to the line to line current in a balanced system.

Pa = Vline x Ipp x 3

Ipp = Pa/(Vline x 3) = 166,200VA/(480V x 3) = 115.4A, or

Iline = 1.732 x Ipp = 200A

 

[rattus]

I don't recall the poster asking about how you calculate anything.

Iwire, I think it was Tallgirl who mentioned differential equations which we certainly don't need here. In fact we don't even need algebra.

 

[tallgirl]

Iwire, I think it was Tallgirl who mentioned differential equations which we certainly don't need here. In fact we don't even need algebra.

For the sort of precision you seem to want to advance you certainly do. The operation of the system alters the operation of the system -- that's a pretty good clue you're not going to solve the functional parameters of the system with anything short of differential equations. The alternative is to do what all engineers are taught to do -- throw your hands up in disgust and find an approximation that's easier than the really ugly math that's required for an exact answer.

Which is really long winded for "Roger is right."

 

[rattus]

For the sort of precision you seem to want to advance you certainly do. The operation of the system alters the operation of the system -- that's a pretty good clue you're not going to solve the functional parameters of the system with anything short of differential equations. The alternative is to do what all engineers are taught to do -- throw your hands up in disgust and find an approximation that's easier than the really ugly math that's required for an exact answer.

Which is really long winded for "Roger is right."

Tallgirl, I don't know how you got off on precision and DEs. The proven formulas provide good results with elementary arithmetic. And, the fact that the load may change often has nothing to do with the original question, neither does it change the accuracy or precision of any one measurement.

Roger did compute the total power correctly, but he erred when he computed the "phase-to-phase" current. Check it out.

 

[iwire]

And, the fact that the load may change often has nothing to do with the original question,

99% (no math involved ) of this thread has has nothing to do with the original question.

 

[rattus]

Not Exactly:

Not Exactly:

99% (no math involved ) of this thread has has nothing to do with the original question.

Bob, you have to know a little math to understand percentages!

 

[roger]

Rattus, I did not err in the phase to phase current, you are talking a completely different animal.

You are looking at a "system" and current flowing "out of" and back "into" the system. The discussion is current flowing "through" a breaker, and the fact is, 200 amps can flow through all three phases of the breaker period.

VA or apparent power is not in this discussion, maximum current is, so I contend your assertion that I erred is in fact in err. :grin:

Roger

Edited for punctuation

 

[rattus]

Rattus, I did not err in the phase to phase current, you are talking a completely different animal.

You are looking at a "system" and current flowing "out of" and back "into" the system. The discussion is current flowing "through" a breaker, and the fact is, 200 amps can flow through all three phases of the breaker period.

VA or apparent power is not in this discussion, maximum current is, so I contend your assertion that I erred is in fact in err. :grin:

Roger

Edited for punctuation

Roger, when you say "phase-to-phase current", I take this to mean the current through a delta load, and this current is less than the line current. Certainly, the max current can be seen in all three lines at the same time--no argument there.

Now, whether we compute the power in KW of KVA is of no importance here since as you say we are interested in the current.

Now, back to the original question, the answer is simply 200A per phase! No need to compute anything. This is the 1% of the discussion that Iwire mentioned. No algebra, trig, calculus, DEs, slide rules, calculators--no nothing!

 

[roger]

Now, back to the original question, the answer is simply 200A per phase!

Now we're on the same page.

I love concise right to the point threads.:wink:

Roger

 

[George Stolz]

I think math threads should be closed faster than union threads.
I can at least follow a union thread, and they're generally less heated.

 

[Minuteman]

I think math threads should be closed faster than union threads.
I can at least follow a union thread, and they're generally less heated.

I don't dare to debate a math thread. I say, "whoever yells the loudest, wins!"

 

[Timboe]

My brain hurts

My brain hurts

Iwire bob thanks,
My brain was melting.

 

[johnny watt]

Hee hee?I see we finally got a simple answer to the ?something easy?? question.

I just wanted to add that I think that bimetallic breakers do experience more thermal fatigue and ?wear out? quicker when loaded past 80%.

 

[rattus]

I think math threads should be closed faster than union threads.
I can at least follow a union thread, and they're generally less heated.

George, this was not really a math thread; it was much ado about nothing as some old Englishman once said.

 

[tallgirl]

I think math threads should be closed faster than union threads.
I can at least follow a union thread, and they're generally less heated.

If it weren't for the math threads, those of us who aren't in IBEW (I'm a CWA member, so there!) wouldn't have anything really nasty to debate

On the other hand, the math threads wind up being some of the most educational for me, and props to Roger, but I've learned something in this thread. For example, not that I ever expect to work with 277/480, but it hadn't dawned on me how simple things could be if I used Roger's fake voltage shortcut. And understanding that P = E * I, I follow the way he converts to power and all that stuff and why the phase to phase current is what it is -- the amount of power, when using the 277 volt line to neutral voltages has to be the same as when using the 480 volt phase to phase voltages. It's a black box -- 200A @ 277 volts goes in, and that's all there is. However it is used, the answer has to be 166.2kVA. It can be 200A @ 277v per L-N, or it can be 115.4A @ 480v per L-L. Or it can be some weird combination. But the answer has to be 166.2kVA.