X / R RATIO and Equivalent Pf %

Status
Not open for further replies.
J

JEFF MILLAR

Senior Member
I can not find how 15 % PF equals 6.6 X / R. Or how a. PF 0.2 equals an X / R RATIO. 4.9.
These are published values for testing LV. Breakers.The 6.6 X/R and equals 15 % PF .
I have searched the internet and can not find how this conversion is calculated.
Can anyone post the formula for this conversion
Why is it nearly impossible to find the listed X/R ratios of Induction motors.
 
Have mental block understanding how 0.15 % PF is derived from X / R 6.6.

Can you please give the formula or explain how PF 0.15% is derived as 6.6 X / R short circuit value ?
With ExceI i can calculate PF in degrees with the formula = Degrees ( ATAN ( X / R ))
And 0.15 % of what ? is that 360 degrees ?
Thank you if you can clarify.
 
JEFF MILLAR said:
Have mental block understanding how 0.15 % PF is derived from X / R 6.6.

Can you please give the formula or explain how PF 0.15% is derived as 6.6 X / R short circuit value ?
With ExceI i can calculate PF in degrees with the formula = Degrees ( ATAN ( X / R ))
And 0.15 % of what ? is that 360 degrees ?
Thank you if you can clarify.
Click to expand...
PF is the ratio of real power to apparent power. It's not some percentage of something. It is the cosine of the angle between real and apparent power.
So:
PF = cos (theta) = 0.15 or theta = ACOS(PF)
Tan(theta)= X/R
Substituting the value of theta in the above equation:
X/R = TAN(ACOS(PF)) OR X/R OF 0.15 PF = TAN(ACOS(0.15)) =6.59124~ 6.6
 
You may use the formula: tan(φ)=sin(φ)/cos(φ)=sqrt(1-cos(φ)^2)/cos(φ)
 
Thank you for your support. Yes first part of my question has been answered.
And yes how PF is calculated.
Please explain if you can how 15 % PF equals 6.6 X / R ?
And yes X/ R = TAN pf
 
JEFF MILLAR said:
Thank you for your support. Yes first part of my question has been answered.
And yes how PF is calculated.
Please explain if you can how 15 % PF equals 6.6 X / R ?
And yes X/ R = TAN pf
Click to expand...
Not a major deal but, we stop using % with PF. Power factor is a dimensionless quantity as is the X/R ratio. Slang gets us into trouble when everyone in the conversation uses different definitions.
 
JEFF MILLAR said:
Thank you for your support. Yes first part of my question has been answered.
And yes how PF is calculated.
Please explain if you can how 15 % PF equals 6.6 X / R ?
And yes X/ R = TAN pf
Click to expand...
You got it wrong. X/R is not equal to the tangent of the power factor! X/R is = tan(acos(PF))
 
Thank you all very much. Please note, My reference was to circuit breaker testing. What is used to test the breaker short circuit is given as PF = 0.15 % and X/R = 6.6.
Yes we have shown above how the X/R value is calculated. And thanks yo you my confusion reduced.
What remains is to know how this value of PF is calculated. Sure PF = COS theta and KW / KVA. But back to the breaker testing values, where does this PF value of 0.15 come from ?. As far as my confusion is concerned i am of the opinion that not enough data is known to be able to calculate this short circuit power factor of 0.15. Now can you understand the second part of my question. How is 0.15 short circuit PF calculated or associated with or derived. from X/R = 6.6. I do not see how it could be. Something is missing in my opinion. I love this forum and want to learn from you.
 
JEFF MILLAR said:
Thank you all very much. Please note, My reference was to circuit breaker testing. What is used to test the breaker short circuit is given as PF = 0.15 % and X/R = 6.6.
Yes we have shown above how the X/R value is calculated. And thanks yo you my confusion reduced.
What remains is to know how this value of PF is calculated. Sure PF = COS theta and KW / KVA. But back to the breaker testing values, where does this PF value of 0.15 come from ?. As far as my confusion is concerned i am of the opinion that not enough data is known to be able to calculate this short circuit power factor of 0.15. Now can you understand the second part of my question. How is 0.15 short circuit PF calculated or associated with or derived. from X/R = 6.6. I do not see how it could be. Something is missing in my opinion. I love this forum and want to learn from you.
Click to expand...
Testing of breakers/ fuses was taken from thousands of data from experiments. The reason I saw why 0.15 PF lagging was chosen could be that it was the lowest possible power factor that their testing equipment can provide. If you want, you can go lower and test breakers at 10% PF and achieve the same goal as our pioneering engineers had.
The engineers before us did those tests to establish the line A - B in the graph of the fuse instantaneous peak let-through current versus the prospective short-circuit current using the circuit which exhibited a 0.15 PF. The thermal energy developed doing test A was observed to be 1/123 of the thermal energy developed in Test B conditions.
image-10.jpg
 
Thank you. PF = DEGREES (ATAN(X/R )). eg. for X / R 6.6 =
=DEGREES(ATAN(6.6)) = PF in degrees

I find nothing to calculate find X / R from pf % 0.15
 
ANSWER. Short Circuit. PF %. = ( RES ) / SQRT ( RES ^ 2 + REX ^ 2 ) x 100
res = 1.4 %
rex = 19 %. Short circuit PF. = 7.4 %

Mental block.cleared.
 
Status
Not open for further replies.
Top