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A couple of things come to mind
1) Challenge FZs post if you feel it needs to be.
2) Its just an Internet forum, anyone with a lick of commonsense is going to go to a number of places to verify what they read on it.
THE PHYSICS OF... POWER
- Thread starter Phil Corso
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A couple of things come to mind
1) Challenge FZs post if you feel it needs to be.
2) Its just an Internet forum, anyone with a lick of commonsense is going to go to a number of places to verify what they read on it.
One is power. The other isn't. No matter what you call it.
FionaZuppa
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- AZ
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- Part Time Electrician (semi retired, old) - EE retired.
very true. so lets bring it back to VA
real world, if the load is 1kW + 1kVAr, what size gen and what size input do you need, and why?
also explain the kickback the gen sees from kVAr
i understand the urge to think that real R changes, it does not
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FionaZuppa
Senior Member
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- AZ
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- Part Time Electrician (semi retired, old) - EE retired.
how about
T = HP * 5252 / RPM w/ and w/o load
we know w/ load means torque when load reduces rpm slightly, hmmm, looks like HP remains constant
HPfree-wheel=HPloaded
but thats so confusing because we think of HP as doing work because it has the word "power" in it, but in free-wheel there is no torque thus no work being done, so where's the "power" in "HP" ?? perhaps thats imaginary power
and btw, R'/s term (it represents a virtual resistor in the model), R' is constant, only the s changes, you are not removing R from R' to make that term go down, s in increasing during the very simple principle of counter reaction to shaft resistance.
T = HP * 5252 / RPM w/ and w/o load
we know w/ load means torque when load reduces rpm slightly, hmmm, looks like HP remains constant
HPfree-wheel=HPloaded
but thats so confusing because we think of HP as doing work because it has the word "power" in it, but in free-wheel there is no torque thus no work being done, so where's the "power" in "HP" ?? perhaps thats imaginary power
and btw, R'/s term (it represents a virtual resistor in the model), R' is constant, only the s changes, you are not removing R from R' to make that term go down, s in increasing during the very simple principle of counter reaction to shaft resistance.
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VA is not a measure of power.
Phil Corso
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The HP rating of a motor is a motor rating for what it can do.
Not what it is doing.
Proves that you can get it right - sometimes..................
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And the toruque "rating" of a motor also reflects the torque it can produce at a given nominal speed rather than the torque it is producing. An unloaded motor is not producing any torque, except that necessary to overcome frictional losses.
active
reactive
total/complex/apparent
all the same units technically
by convention
P watts
Q var
S va
va = volt x amp = joule/coulomb x coulomb/sec = joule/sec = watt
forms
electrical
nuclear
mechanical
chemical
all governed by different physics (with overlap)
maxwell et al
Bohr et el
Newton's mechanics etc
thermodynamics, etc
That's just plain wrong. You ought to know better.
Prove it
those are equivilent units
not MY physics, but EVERYONE's physics
S = P + jQ
is this true?
if so to be additive and for the unit analysis to conform don't P and Q have to be the same to equal S?
apples + apples = apples
The voltage V in volts (V) is equal to the energy E in joules (J), divided by the charge Q in coulombs (C):
V(V) = E(J) / Q(C)
So volt = joule / coulomb
or V = J / C
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There are none so blind as those who will not see.
VA is volt amperes. You can have VA and no power. Can't you see that??
you ignored my questions as usual
is S = P + jQ valid?
if so must not all 3 be the same unit?
Prove this to be false
va = volt x amp = joule/coulomb x coulomb/sec = joule/sec = watt
you can repeat yourself all day
you should know better
power is power
that is absurd
that is just plain wrong
ad nauseum
but unless you give an explanation with support/back-up it's all imho hooey
until then I'm done with you
readers will form their own opinions
no shortage of that around here lol
you have a pf of 0 and you have reactive POWER Q and total POWER S but no active power P
Power is watts. VA is not watts. It's VA. Why can't you see that?
S = P + jQ, j is a vector rotation operator, by 90 deg. j is an independent dimension, its variables have a different effect or meaning compared to variables in the real dimension.
In the instantaneous case in the j dimension, work is performed either charging energy into the E and M fields or recovering that energy back out. In the time averaged case, the net power is zero.
Variables in the j dimension have very different effect, function, and meaning compared to the variable in the real dimension.
Using the definition of power, the time averaged case is in the definition and the instantaneous case is not. 6c, http://www.merriam-webster.com/dictionary/power
Watt is a unit of power.
Distance and altitude have the same unit dimension, but would you say travelling 10 meters in altitude is the same as travelling 10 meters in distance? The effect and meaning is completely different.
If P then Q.
If two parameters have the same function or meaning, they have the same unit dimension.
If Q then P.
If two parameters have the same unit dimension, they have the same function or meaning. Converse logical error.
In the instantaneous case in the j dimension, work is performed either charging energy into the E and M fields or recovering that energy back out. In the time averaged case, the net power is zero.
Variables in the j dimension have very different effect, function, and meaning compared to the variable in the real dimension.
Using the definition of power, the time averaged case is in the definition and the instantaneous case is not. 6c, http://www.merriam-webster.com/dictionary/power
Watt is a unit of power.
Distance and altitude have the same unit dimension, but would you say travelling 10 meters in altitude is the same as travelling 10 meters in distance? The effect and meaning is completely different.
If P then Q.
If two parameters have the same function or meaning, they have the same unit dimension.
If Q then P.
If two parameters have the same unit dimension, they have the same function or meaning. Converse logical error.
From the perspective of instructing someone who is going to use or employ your information, for example loading on a UPS rated 100 kVA 80 kW where the VA rating is bolded on the equipment labels and marketing promotions but the kW rating is in fine print on the nameplate or specs ...
Would you say VA = watts ?
Is it true or is it false?
True or false?
Would you say VA = watts ?
Is it true or is it false?
True or false?
FionaZuppa
Senior Member
- Location
- AZ
- Occupation
- Part Time Electrician (semi retired, old) - EE retired.
it cannot be wrong.
all of the units are the same, otherwise the math cant jive
S=P+Q
S = units of power (watt)
P = units of power (watt)
Q = must have units of power (watt)
except Q is more like I2 x jX
the imaginary part comes from j=√-1 = not real
in the application of reactance in the load, the load itself can do no work with this power.
if the gen and transmission was lossless then all of the Q would avg to be zero, however, the input to the gen would need to be able to handle instantaneous watts = VA, because to charge jX the gen has to do work, which it gets back later. the avg work done by gen would = kW of load, but the power curve of the gen is not a flat kW, it will oscillate above and below kW of load. you cannot build gen using avg, just as we do not provide insulation to wires based on the RMS voltage value.
but again, lossless and pure jX is not real world. the Q power will be lost by gen and transmission. 1MW motor with PF=0.833, wow, sucky, poco gen has to have input of 1.2MW and that 0.2MW the cust wont pay for.
there are two views from poco. gen/transmission and customer. the Q on cust side is √-1 to customer as cust gets no work from it, but that Q is very real for gen/transmission.
this is nothing more than impedance mismatch because poco buy a gen design, but the load varies in what it's impedance looks like. so when a poco feed has poor PF due to inductive reactance, they add XC to "re-tune" the source to better match the load.
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yes and no
by fact yes
by convention no
P = S x pf
pf is unitless, hence P and S have the same units
by convention after applying the pf we call P watts to indicate such
Read the para after equation 2.3.2 S has the 'same units a P and Q'
http://forums.mikeholt.com/attachment.php?attachmentid=15644&d=1473429532
no one has refuted
va = volt x amp = joule/coloumb x coulomb/sec = joule/sec = watt
NO ONE
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