Excel input for 7671 calculation

Not at desk but think

=sqrt(1-PF^2)

(Where PF is the cell containing cos(th))

will do the trick.

Note that excel’s trig functions take radians as their inputs, just to catch you out (c;

So to get the phase angle from PF=cos(th), you need =degrees(acos(PF))

**Edited to correct formula!**

Ps should have said first that the reverse for PF from phase angle is =cos(radians(th))

Whoops, forgot a square root there. Edited.

as above pythagoras is your friend. you know the cos, that is adjacent over hyp,

or exact adjacent if  hyp is 1

and  diff of squared hyp and squared adj gets you the squared opp.

no need to think about rads or degrees

Not sure if that is helpful or not. I may come back when the beer has worn off..

Mike

Jam said:

=sqrt(1-PF^2)

DEFINITELY - mathematicians use the following identity to remember it by:

sin²θ + cos²θ = 1

As Mike says, it falls straight out of Pythagoras. If we have A as the side of a right-angled triangle adjacent to the angle θ, B as the side opposite, and H as the hypotenuse, then according to Pythagoras (I'm sure someone will correct me that this ought to be the 47^th Proposition of Euclid though), we have:

A² + B² = H²

Dividing both sides by H², we have:

(A/H)² + (B/H)² = 1 ... and if we remember that A/H is defined as cosθ, and B/H is defined as sinθ, the mathematical identity pops out.

I think both Euclid and Pythagoras we’re in my trig class, but they sat at the front!
Many thanks all. That’s a neat way of remembering the function Graham. Works perfectly on my calculator but haven’t tried on a cell input on Excel. 
I am actually setting it up for the IS10101 2020 voltage drop formula. That standard, unlike BS7671 has no helpful voltage drop tables and one is forced to use the rather unwieldy method. However, it’s not that different to the methodology adopted in 7671 in that it takes account of load power factor, although not Ct.

I will post it as a photo below. Note that the three-phase calculation is incorrect as it gives only the core voltage drop. 

Thanks Graham and Jam, works a treat!

Good to see that we all agree on pythag.

But could you explain your 'core drop' comment please ?

Do we need to consider neutral drop ? surely only for an out of balance term ?

or is it a star/delta thing ?

Say if we have a 230/400V 3 phase supply,  with a balanced 10V drop,

I would read that to be 10V off each phase,(so a drop of 1/23) so it becomes 220/ 382 .

Do others  understand it to be   a 10V drop phase-phase (so a drop of 1/40) so 226 /390?

M

If you run through the formula it gives the voltage drop in volts along one straight length of single core cable. Thus the coefficient b that they refer to needs to be multiplied by 2 for single phase. Now they say it only needs to be multiplied by 1 to get the three phase voltage drop. That would only be the voltage drop along the single core and the line voltage would be root 3 that value.