## Terminology (power)

Hello all. I recently read something that has me a bit confused. Consider a single-phase 100kVA transformer that supplies a 40kW heating load at unity power factor. What is meant by "how much additional kW of inductive load at 0.8 power factor can the transformer carry" mean? Is it asking how much more real power will the supply be transferring to the load given it is no longer unity?
Seems odd talking about a transformer "carrying" kiloWatts.
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 Mentor 40 kW at 1 PF is 40 kVA. 100-40 is 60. So how many kW is 60 kVA at .8 PF? Yes, that's real power.
 (60 kVA)0.8=48 W. Essentially they are simply wanting to know how much more real power the source can supply to a load with 0.8 PF given it is still supplying the unity PF heating load?

## Terminology (power)

Thats the way I read it. It's asking how much additional load can be added to the transformer assuming all additional loads have a PF of 0.8.

So yeah, 48KW of true power will give you an apparent power of 60KVA at a PF of 0.8.
 Recognitions: Gold Member Science Advisor Word problems are best parsed carefully. Word problem authors dont always give the facts in a logical sequence. I don't know if that's because English is such a mishmash of Olde English, Latin and Old Germanic, but English works well enough if one is careful about transcribing it into equations.... First impulse is to leap to the answer -there's 60 kva of room left at .8 pf so answer is 48. That was my first answer but i think it's wrong. Too much internet has made me impatient. Going back to ninth grade algebra and geometry, We started with just a line - no imaginary component. Adding some kva at an angle changed our line into a triangle that represents our real(kw) and imaginary(kvar) and total(kva) components. specifically, a right triangle whose hypotenuse is 100 kva and whose two sides represent real and imaginary power,, kilowatts and kilovars respectively. so : let x = real kw to be added. PF of .8 defines a 3-4-5 right triangle, so if x kw of real power are added, then (3/4)x kvar of imaginary power are added and (5/4)x of total kva are added ; the resulting right triangle has real side (40 + x) kw imaginary side (3/4)x kvar hypotenuse 100 kva so pythagoras says (40+x)^2 + (0.75 x)^2 = (100)^2 which by quadratic equation gave me x = 52.06183 kw real and (3/4)x = 39.0467 kvar imaginary and a quick check by windows calculator: sqrt( (40 + 52.0618)^2 + (39.0467)^2 ) = 100 so answer could be 52.0618 kw added instead of 48 .......... (and 65.0773 kva instead of 60) . would somebody check my thinking and algebra? And is phrase "kW inductive load" an oxymoron, or am i nitpicking? old jim
 Yeah, this is exactly why I asked. I felt as though it was worded awkwardly; in a way that none of my textbooks have ever worded a problem. I see where you are coming from, though. I drew a phasor diagram, ended up with two triangles, then applied Sine Law and ultimately ended up with the same result as you! Regards!
 Recognitions: Gold Member Science Advisor Thanks Sandy - diverse methods came to same number, so i am convinced it's right. But what a wakeup it was for me to see how easily i leapt to wrong answer ! my self discipline needed a tuneup.. I wonder if that was a subtle choice of words, or did author make same mistake? thanks old jim

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