Assuming you do mean sin, it is the sine of an angle, the ratio of the opposite side to the hypotenuse in a right angle triangle with that angle.As the angle rotates through a full turn the sine oscillates between plus and minus one. (negative angles are the same as those that are more than half a turn.)

 That is useful in electrical calculations when considering the resultant of pairs of quantities that are orthogonal in time, such as the total current composed of both resistive and reactive current (in inductors and capacitors) One quantity can be represented as a sin wave and the other as a cosine wave , and the resultant complex quantity is the vector sum.

Apart from a few angles wtith simple geometry like 60 and 45 degrees, there is generally no simple method of calculation, but the value can be approximated by truncating the sum of an infinite series that converges, if the angle is expressed in radians,

Pre-calculator sin, cos. and tan were normally either a scale on the slide rule or available to higher accuracy in printed tabules, but nowadays electronic calculators approximate the values using Chebychev polynomials as these converge much faster than the 'correct' McLaurin series.

IF this is not quite what you wanted, and I fear it may not be, then rephrase the question and we can have another go.

Mike

Thank you for your explanation it's definitely explained a few questions but I was wondering where do we get these values in order to calculate the sin angel? Is it through appendix 4 of bs7671, where ratings for resistance, reactance and impedance are provided? 

Or is it normally provided on a motor plate such as the cos value? 

Thank you again for your previous explanation 

The real answer is, it's a combination of:

 - The supply impedance (resistance and reactance)
 - The cable impedance to the load (resistance and reactance)
 - The load impedance (resistance and reactance).

This will allow you to calculate currents and voltages at different points in the system ... and the phase angle difference between them.

So, what does a motor 'cos value' (power factor) tell you? It tells you the phase angle between the voltage and current, which you can use to determine the resistance and reactance of the motor, so you can include that in your calculations.

You might see some books use complex number notation to carry out resistance/reactance calculations. In this case, z=(r+jx) where j represents the imaginary number of the square root of negative 1 (mathematicians use i for this, but we use i for current).

There is nothing especially magical about cables greater or less than 16mm2, except that for larger cable sizes, the resistance drops but the inductive effects (the magnetic fields created around the wire ) tend to stay the same or get worse as the separation of flow and return current increases. In effect energy goes into creating that magnetic field inside the long thin coil formed by the live and neutral conductors, and then destroying and reversing it every half cycle, but unlike resistance, where the energy is lost to heat, the inductive energy is returned to the supply as the field collapses - so there is a maximum of inductively generated current flowing when the voltage crosses zero.
The tables of voltage drop represent this as two parts - the resistive bit, and the reactive bit - for thin cables, carrying lower currents, at least at 50Hz there is an assumption the inductive stuff can be safely neglected - indeed a quick squint at the tables shows that even when both are in the tables, it starts of with an R to X ratio that makes you think it was scarcely worth it. However, by cables of hundreds of sq mm it really is worth it, as the inductive part dominate.s

The real answer is, it's a combination of:

 - The supply impedance (resistance and reactance)
 - The cable impedance to the load (resistance and reactance)
 - The load impedance (resistance and reactance).

I suspect that's the crux of the OP's question.. The load PF can generally be had from rating plates (either directly or via comparison of W and A (or VA) ratings), but supply and wiring system values? Or are they often insignificant compared with the load?

   - Andy.

The real answer is, it's a combination of:

 - The supply impedance (resistance and reactance)
 - The cable impedance to the load (resistance and reactance)
 - The load impedance (resistance and reactance).

I suspect that's the crux of the OP's question.. The load PF can generally be had from rating plates (either directly or via comparison of W and A (or VA) ratings), but supply and wiring system values? Or are they often insignificant compared with the load?

   - Andy.

Really you are asking 'what is my supply impedance' and generally the answer is a mix of L and R, - for low current circuits, mostly R, and as you get fatter cables, so the magnetic loop opens up and the resistance falls, it becomes more L and less R.

As a rule of thumb, not in the regs, expect around 1nH per mm, or 1microhenry per metre for isolated wires of sensible diameters, lower for fatter or flatter ones, and a lot lower if you can cancel some of the close-in magnetic field with a current going the other way. Higher if coiled by square of no of turns.

Also, hopefully unsurprisingly, given it is a coil inside, at the 11kV/400V transformer secondary, it is mostly L.

If you have cables in bundles of trefoils or quadrafoils  miswired so there is a bundle of one phase, then a gap and bundle of another, it is a lot worse than when each bundle has one wire of each phase or one of each phase and a neutral - a lot more energy goes into filling the space between conductors with magnetic field, as there is more space to fill  - so the reactive voltage drop is greater.

As an aside, you can make special  low impedance transmission lines with almost no inductance using interleaved wide flat conductors - think ribbons of alternating Live and Neutral  separated by thin insulation, or bundles of wires with many L and many N interwoven.

The only folk who have the budget and the need to do this are the pulse power mob and for those sort of events,  the trade off between competing effects of line capacitance versus line inductance is very carefully thought out. This defines the surge impedance, the ratio of volts to amps in the time between the load changing and the information arriving at the source that it has - that time is more or less the length of the cable divided by the speed of light, times a single digit correction factor for the geometry and dielectric choices. All irrelevant at 50Hz unless your cables are hundreds of km long, but background to get the mental image right. And explains the appearance of oddly made braided cables draped around nuclear physics and laser labs.

(spot the low inductance earth bond - same CSA, less inductance)

Mike.

Thank you for the indepth reply definitely explains a few things and gives a lot to think about and a good starting point to go and research things a little more In depth 

AJJewsbury said:

but supply and wiring system values? Or are they often insignificant compared with the load?

Typically, supply impedance is ignored for installations up to 100 A single-phase. For larger installations, reactance should be taken into account, and either:

 - Typical values will be provided by DNO or via other industry guidance (ENA Engineering Recommendations); or
 - The designer will determine from the transformers/generators used.

Typically cable reactance is ignored for cables up to and including 16 sq mm (or the very short lengths of 'tails' used in 100 A installations). Greater csa, it is advised to take cable reactance into account.
An exception is SWA cables, where even though the copper csa may be below 16 sq mm, the armour csa is typically above 16 sq mm, and has a more significant effect, and reactance should be taken into account for earth faults in SWA at least, even for copper csa up to 16 sq mm.