Power factor and VA vs Watts

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Hameedulla-Ekhlas said:
It is ok and p.f is just a ratio and ratio is always unitless.

they also cancel each other, if you want I will do it.
we know that S = P only if p.f = 1 and here we have 0.9 not one.
it means 90% is real power and 10% is reactive power. So, since we multiply it by 0.9 it means we get pure real power without reactive. we know when there is no reactive, V-A = W.
So, 0.9 represent a ratio and I think it is correct. Am not I correct?


The unit cacels each other and we get proper unit ( W, V-A, Var)
Click to expand...
A ratio is unitless only if it is developed from terms of the same unit.

When we discuss voltage drop and use an ohm per kft value. This is also a ratio, but of different units. So when we calculate, can we omit the units of this ratio.

Watts and volt-amperes are similar units but not identical units.

PS: Thread did not automatically close at 400 replies :D
 
Smart $ said:
A ratio is unitless only if it is developed from terms of the same unit.

When we discuss voltage drop and use an ohm per kft value. This is also a ratio, but of different units. So when we calculate, can we omit the units of this ratio.

Watts and volt-amperes are similar units but not identical units.

PS: Thread did not automatically close at 400 replies :D
Click to expand...

yeah, But think and expand the formula you can find the same unit.

even you can cancel unit by unit W/V-A and I am sure you can do it.

Good news. not closed
 
Hameedulla-Ekhlas said:
When we say ratio, it is unitless. we dont have any ratio with unit.
Click to expand...
From wikipedia...
Ratio
A ratio is a quantity that denotes the proportional amount or magnitude of one quantity relative to another.
Ratios are unitless when they relate quantities of the same dimension. When the two quantities being compared are of different types, the units are the first quantity "per" unit of the second ? for example, a speed or velocity can be expressed in "miles per hour". If the second unit is a measure of time, we call this type of ratio a rate.
Click to expand...
 
Besoeker said:
It is power at that instant. No representative about it.
You accepted that a constant 1V times 1A for one second was 1 J.
Rightly so.
You even volunteered that the instantaneous rate was 1W.
Again, rightly so.
So let's break the one second down into a million 1us segments.
It is still 1W in every segment.
Now, any one segment could be part of an ac waveform or continuous DC.
Over that sort of time interval on a system operating at power frequency the difference between ac and dc would not matter.
Still with me?
Click to expand...
Oh, I agree it is power... but in the matter under discussion, it depends greatly on what the power is attributed to: energy consumed, energy transferred, or a combination thereof.
 
Smart $ said:
Sure, it can be done...

P = 120V * 10A * 0.9 * 1W/VA = 1080W = 1.08KW

:D
Click to expand...

no we can not add from ourselves W/VA in equation and you dont need to add either, see



P = 120V * 10A * 0.9 = 1080W =1.08kW

p.f = W/VA ------1 and 0.9 has no unit because

W = V * I * cosx; cosx = 0.9 ------2
S = V * I ------3

now put the Eqs: 2 and 3 into 1

p.f = ( V * I * cosx ) / ( V* I)

after cancellation ampere and voltage we get

p.f = cosx ; if x = 25.84 degree

p.f = 0.9 without any unit and it is a ratio.


P = V * I * cosx ---11
S = V * I -----12

p.f = P/S = (V * I * cosx ) / (V * I ) after cancellation of V, I

p.f = cosx.

P = Watt = V-A * cosx
S = V-A

Note: p.f is not a constant to have a unit and balance our equation, it is just a ratio and unit less
 
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Smart $ said:
From wikipedia...
Click to expand...
Ratio
A ratio is a quantity that denotes the proportional amount or magnitude of one quantity relative to another.
Ratios are unitless when they relate quantities of the same dimension. When the two quantities being compared are of different types, the units are the first quantity "per" unit of the second — for example, a speed or velocity can be expressed in "miles per hour". If the second unit is a measure of time, we call this type of ratio a rate.
Click to expand...

yes, I am agree with wikipedia and we have also same qauntity, see

p.f = cosx is a unitless and a ration lets proof it.

p.f = W/V-A = ( I * V * cosx ) / ( I * V ) = cosx after cancellation of V and I, cosx is a ratio and unitless
 
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Hameedulla-Ekhlas said:
no we can not add from ourselves W/VA in equation and you dont need to add either, see



P = 120V * 10A * 0.9 = 1080W =1.08kW

p.f = W/VA ------1 and 0.9 has no unit because

W = V * I * cosx; cosx = 0.9 ------2
S = V * I ------3

now put the Eqs: 2 and 3 into 1

p.f = ( V * I * cosx ) / ( V* I)

after cancellation ampere and voltage we get

p.f = cosx ; if x = 25.84 degree

p.f = 0.9 without any unit and it is a ratio.


P = V * I * cosx ---11
S = V * I -----12

p.f = P/S = (V * I * cosx ) / (V * I ) after cancellation of V, I

p.f = cosx.

P = Watt = V-A * cosx
S = V-A

Note: p.f is not a constant to have a unit and balance our equation, it is just a ratio and unit less
Click to expand...
I'm not saying pf cannot be unitless. But if you adhere to strict equation protocols on units in equations you cannot write...
P = 120V * 10A * 0.9 = 1080W =1.08kW​
...without any defined parameters.

Now if you change your equation to, say:
P = 120V * 10A * cos(φ) = 1080W =1.08kW​
...and define...
P = V ? I ? cos(φ)
where
P is power in watts
V is voltage in volts
I is current in amperes
φ is the phase angle difference between voltage and current waveforms​
...it is okay. Here cos(φ) is a unitless ratio.

How you went about it is also okay, but you have to define those parameters with the equation for it to be strictly correct.
 
Smart $ said:
I'm not saying pf cannot be unitless. But if you adhere to strict equation protocols on units in equations you cannot write...
P = 120V * 10A * 0.9 = 1080W =1.08kW​
...without any defined parameters.

Now if you change your equation to, say:
P = 120V * 10A * cos(φ) = 1080W =1.08kW​
...and define...
P = V ? I ? cos(φ)
where
P is power in watts
V is voltage in volts
I is current in amperes
φ is the phase angle difference between voltage and current waveforms​
...it is okay. Here cos(φ) is a unitless ratio.

How you went about it is also okay, but you have to define those parameters with the equation for it to be strictly correct.
Click to expand...

yes, in this case I am agree because p.f = cosx and you can replace each other. But in below
Originally Posted by Smart $
Sure, it can be done...

P = 120V * 10A * 0.9 * 1W/VA = 1080W = 1.08KW
Click to expand...

But in above equation which you had written, I was worried.

Ok anyway, how you type phi sign in equation.
 
Hameedulla-Ekhlas said:
... But in below
Originally Posted by Smart $
Sure, it can be done...

P = 120V * 10A * 0.9 * 1W/VA = 1080W = 1.08KW
Click to expand...
But in above equation which you had written, I was worried.
Click to expand...
What I'm saying with that equation is without defining the change of units (either before or after the equation) it has to be included in the equation. Knowledgeable electrical persons take it for granted, and don't define the parameters for others assumed to be knowledgeable.

Ok anyway, how you type phi sign in equation.
Click to expand...
No type. Copy and paste from Windows' Character Map.
 
Smart $ said:
...If your question is limited to our hypothetical one watt-second produced and consumed scenario, then the answer is yes.

If not, the answer is dependent on whether any part of the instantaneous product is VAR (volt-amperes reactive). If there is no VAR present, then the answer is also yes. If there is VAR present, the answer is no.
Click to expand...

Besoeker said:
...The instantaneous value of power is instantaneous voltage times instantaneous current....
Click to expand...

Smart -
Bes is absolutely correct.

These are the same concepts discussed several pages ago. The concept that instantaneous power is either reactive or real does not fit any known math model currently in use. The only thing one can tell is if the instantaneous power is toward the source or toward the load. These are some pretty basic concepts set out 300 years ago.


cf
 
Smart $ said:
What I'm saying with that equation is without defining the change of units (either before or after the equation) it has to be included in the equation. Knowledgeable electrical persons take it for granted, and don't define the parameters for others assumed to be knowledgeable.


No type. Copy and paste from Windows' Character Map.
Click to expand...

Smart$:
Let me clear it from my point of view :
Originally Posted by Smart $
Sure, it can be done...

P = 120V * 10A * 0.9 * 1W/VA = 1080W = 1.08KW
Click to expand...

The above equation which is written is wrong absolutely. because it means that.
cosx = 0.9 * W/V-A -----1;
if you say that this is correct and we can do it then how can you find angle. like this
x = arc(cos(0.9*W/V-A)) Now can you find the angle ?
My objection is how you obtained 0.9 W/V-A and how you replaced that without any proof. If you write such a equation it would be wrong.

If you are thinking 0.9 W/V-A is for cancellation of units but I dont think so. We can not add from ourselves even one unit in a equation without proof.

See below

the difference between Watt and V-A is cosx

V* I * cosx ( represents real power) = V* I ( represents apparent power) ; now we will write it like this

p.f = cosx = ( V * I ) / ( V * I ) = 1 -----------2 no unit
you see we got pure value without any W/V-A. That is why I am saying we can not put W/V-A in our equation like Eq.1 which is written by you.

On the other hand;
cosx = p.f and p.f = 0.9 and even I can easily find angle by
x = arc(cos(0.9)

and Now I can replace any of that in equation : cosx, p.f or 0.9 as below

P = 120V * 10A * 0.9 = 1080W=1.08kW is a correct equation for P even to write it with units.
 
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Hameedulla-Ekhlas said:
Smart$:
Let me clear it from my point of view :


The above equation which is written is wrong absolutely. because it means that.
cosx = 0.9 * W/V-A -----1;
if you say that this is correct and we can do it then how can you find angle. like this
x = arc(cos(0.9*W/V-A)) Now can you find the angle ?
My objection is how you obtained 0.9 W/V-A and how you replaced that without any proof. If you write such a equation it would be wrong.

If you are thinking 0.9 W/V-A is for cancellation of units but I dont think so. We can not add from ourselves even one unit in a equation without proof.

See below

the difference between Watt and V-A is cosx

V* I * cosx ( represents real power) = V* I ( represents apparent power) ; now we will write it like this

p.f = cosx = ( V * I ) / ( V * I ) = 1 -----------2 no unit
you see we got pure value without any W/V-A. That is why I am saying we can not put W/V-A in our equation like Eq.1 which is written by you.

On the other hand;
cosx = p.f and p.f = 0.9 and even I can easily find angle by
x = arc(cos(0.9)

and Now I can replace any of that in equation : cosx, p.f or 0.9 as below

P = 120V * 10A * 0.9 = 1080W=1.08kW is a correct equation for P even to write it with units.
Click to expand...
As you say...
V* I * cosx ( represents real power) = V* I ( represents apparent power)​
...is an incorrect equation because it is incomplete for all possibilities.

The correct equation is...
(V* I * cosx) watts = (V* I) volt-amperes * cosx​
The only time your equation is true, is when cosx equals one (1), and that would make it...
(V* I * 1) watts = (V* I) volt-amperes * 1​
Since the one (1) is a result of the unitless ratio (cosx), we can rewrite the equation as such...
1 * (V* I) watts = (V* I) volt-amperes * 1​
And we thus we can revise as...
(V* I) watts / (V* I) volt-amperes = 1 W/VA​
 
Cold Fusion said:
Smart -
Bes is absolutely correct.
Click to expand...
Your opinion is so noted.

These are the same concepts discussed several pages ago.
Click to expand...
As if it weren't obvious??? ;)

The concept that instantaneous power is either reactive or real does not fit any known math model currently in use.
Click to expand...
I did not say either/or. I said one, the other, or a combination thereof.

The only thing one can tell is if the instantaneous power is toward the source or toward the load.
Click to expand...
Well if you want to limit yourself to this notion, then go right ahead. I'll hold my ground.

These are some pretty basic concepts set out 300 years ago.
Click to expand...
That's exactly right... some pretty basic 300-year-old concepts. With all the knowledge we have accumulated over the past 300 years, don't you think it is time to move on to more advanced concepts... :D
 
Smart $ said:
As you say...
V* I * cosx ( represents real power) = V* I ( represents apparent power)​
...is an incorrect equation because it is incomplete for all possibilities.

The correct equation is...

(V* I * 1) watts = (V* I) volt-amperes * 1​
Since the one (1) is a result of the unitless ratio (cosx), we can rewrite the equation as such...




1 * (V* I) watts = (V* I) volt-amperes * 1​
And we thus we can revise as...
(V* I) watts / (V* I) volt-amperes = 1 W/VA​
Click to expand...
Click to expand...
again you are wrong with 1W/VA
You have not completed what you want to proof continue your equation
cosx = Watt/VA = (I * V) / (I * V) = 1
here all unit cancels each other and nothing remains.

you are thinking it as acceleration formula = v (velocity)/t(time) = v/sec but it is not like this nor it is a constant to have a unit.


V* I * cosx) watts = (V* I) volt-amperes * cosx
The only time your equation is true, is when cosx equals one (1), and that would make it...
Click to expand...

Did you not read it completely that I have proved it in previous post
V* I * cosx ( represents real power) = V* I ( represents apparent power) ; now we will write it like this

p.f = cosx = ( V * I ) / ( V * I ) = 1 -----------2 no unit
you see we got pure value without any W/V-A. That is why I am saying we can not put W/V-A in our equation like Eq.1 which is written by you.
 
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