Reactive Power – why is it so confusing? –Additional notes

This is a follow up article to one we posted before Christmas (2023) which provides some more detail.

In the original article – we introduced geometric algebra (sometimes referred to as Clifford algebra) as a tool that could be applied to elucidate some of the concepts. In order to understand the following – we recommend you first read the first article, which can be found here LINK .

Technical content follows

Products (geometric, wedge, dot and Hadamard)

In order to provide enough algebraic machinery to come up with a robust definition of power and reactive power that works for all circumstances, we will need several different types of products which are briefly described below.

• The dot product represented by a dot “.” is well known in vector calculus. An easy way to think of this product is as the normal matrix product between two (identically dimensioned) vectors arranged so that the first vector is a row vector, and the second a column vector.? The result will always be a scalar ( i.e. a number).

• The wedge product represented by a “wedge” , ?. ?In some ways it is the dual concept to the dot product. In matrix notation it can be defined (for the2 –D case as follows):

• The geometric product combines both of the above concepts together so that (by definition) ab = a.b + a^b , the result being a scalar and a bivector, in matrix notation this is represented by a number and a matrix. ?

Why bother doing this?

It turns out that unique vector division and vector inverses can be defined using the geometric product, which is not possible with either the dot or the wedge products or their own.

• The Hadamard product ( or point by point product) is defined as follows:

In many ways it is the simplest to understand, just multiply corresponding elements together – no need to worry about complicated matrix manipulations.? The circle symbol o is used to distinguish it from the other products. ?It allows us to create results that cannot be obtained with any combination of the dot or wedge products.

Duality between power and reactive power

In our first article – we used the following approach to define the relationships between apparent power, active power and reactive power.

S = Vs Is = Vs.Is + Vs ^ Is = P + Q

However, some things were left out which we will attempt to rectify here.

In reality the apparent power used in electrical engineering is defined using the point by point or Hadamard product.

S = Vs o Is

Which strictly speaking is not equal to Vs.Is + Vs ^ Is

This is because the first term is a scalar value, and the second term is a bivector.?

To make the equation valid we need to change each term into vectors.

Calculating Power

For the first term, this can be achieved as follows:

P =(( Vs.Is )/Is) o Is

In words:

1.?????? Take the dot product of Vs and Is

2.?????? Divide (i.e. multiply using the geometric product and the vector inverse of Is) by Is

These first two steps calculate the projection of the Vs vector onto the Is vector. I.e. the result is a vector pointing in the same direction as the Is vector but scaled to be equal to Vs x Cos(theta), where theta is the angle between the voltage (Vs) and current (Is) vectors.

3. Multiply the result by Is using the Hadamard (i.e. point by point) product.

Some examples may make this clear.

Assume we have a 2-D system so that

Vs = Cos(t) = e1, Is = Sin(t) = e2 (i.e. a 90 degree phase difference between voltage and current ).

Vs . Is = 0 because e1.e2 = 0, so P is zero.

Now assume Vs = Sin(t) = e2, Is = Sin(t) = e2. (i.e. voltage and current are in phase).

Vs. Is = e2.e2 = 1

(Vs.Is )/Is = (e2.e2)/e2 = 1/e2 = e2 = Sin(t)

And finally putting it altogether by including the point by point multiplication

((Vs.Is )/Is) o Is = Sin(t) Sin(t) = 0.5 -0.5 Cos(2 t)

One final example to drive the point home:

Assume Vs = Cos(t)+Sin(t) = e1+e2, Is = Sin(t)=e2 (i.e. voltage and current are out of phase by 45 degrees).

Vs.Is = (e1+e2).e2 = e1.e2+e2.e2=0+1=1

(Vs.Is )/Is = 1/e2 = e2 = Sin(t)

Which leads to the same answer as above:

((Vs.Is )/Is) o Is = Sin(t) Sin(t) = 0.5 -0.5 Cos(2 t)

The reason the answer is the same is that even though the angle between voltage and current vectors is 45 deg, the voltage vector we started with was of larger magnitude – this exactly cancels out the reduction in magnitude caused by the 45 deg phase shift.

Calculating Reactive power

We can do a similar trick with the reactive power term:

Q =( (Vs ^ Is)/Is ) ° Is

In words:

1.?????? Take the wedge product of the voltage and current vectors.

2.?????? Using the geometric product divide out the current vector.

In geometric algebra circles, the output of the first two steps is known as the “rejection” of the vector Is relative to Vs. This is the converse concept of “projection” that is discussed above.

3.?????? Multiply the result by Is using the Hadamard (i.e. point by point) product.

Trying the same examples as above:

Assume we have a 2-D system so that

Vs = Cos(t) = e1, Is = Sin(t) = e2 (i.e. 90 degree phase difference between voltage and current ).

Vs ? Is = e1 ? e2

Vs ? Is/Is = (e1 ? e2)/e2 = e1 = Cos(t)

(Vs ? Is/Is) ° Is = Cos(t) Sin(t) = 0.5 Sin(2 t)

Now assume Vs = Sin(t) = e2, Is = Sin(t) = e2. (i.e. voltage and current are in phase).

Vs. Is = e1 ? e2 =? 0

i.e Q = 0

And finally the 45 deg example:

Assume Vs = Cos(t)+Sin(t) = e1+e2, Is = Sin(t)=e2 (i.e. voltage and current are out of phase by 45 degrees).

Vs ? Is = (e1+e2) ? e2 = e1 ? e2+e2 ? e2 = e1 ? e2 + 0 = e1 ? e2

?(Vs ? Is)/Is = (e1 ? e2)/e2 = e1 = Cos(t)

Which leads to:

Q = (Vs ? Is)/Is ° Is = Cos(t) Sin(t) = 0.5 Sin(2 t)

Representing voltage and current as vectors

In the examples above we represented the voltage and current inputs as Sine and Cosine functions, and arbitrarily assigned e1 = Cos(t) and e2 = Sin(t).? This works because we can identify the dot products of e1 and e2 with the integral - trigonometric identities:

i.e. up to a scaling factor of pi, we have a dot ( more formally an inner product) defined using trigonometric functions. ?

This technique is not confined just to trigonometric functions, we can use any family of functions (e.g. polynomials etc) which integrate to a constant value when the two functions match and integrate to zero when they differ (the technical term is “orthogonal”).? We can then identify each of the functions with a vector ei and use the geometric algebra techniques to calculate power and reactive power.

In particular, suppose we developed amnesia of trigonometry and was reduced to sampling the data one point at a time.

Each sample could be considered a single vector function scaled according to the waveform it is representing.? They are orthogonal (i.e. different sample points integrate to zero) to each other because they don’t overlap, so we could consider each sample to be a different vector ei. The two dimensional vector manipulations considered above are replaced with multidimensional systems which could be hundreds or thousands of vectors ei, and hence hundreds or thousands of dimensions instead of just two.?

If we redo the calculations shown above assuming different dimensions – do we get the same answer?

Hopefully you won’t be surprised to learn that up to rounding errors in the calculations – the answers are identical.

Harmonic distortion is easily incorporated into this scheme, and it also gives the same answer as when we use trigonometric functions as our orthogonal set of vectors .

In fact – because we are using individual samples as our vectors, we can even calculate transient power and transient reactive power, which we can’t do with sine and cosine representations.

Concluding remarks

We hope we have shown that there is a lot more to the simple formula S = Vs o Is than meets the eye. The question we set out to answer “why is reactive power so confusing” has been answered – it’s a bit more complicated than we were taught. However when you break it down into steps – it isn’t difficult to understand why the apparent power S splits naturally into the components power (P) and reactive power (Q).

All of the steps are coded into the formulas.

S = Vs o Is

?= ((Vs .Is)/Is) o Is + ((Vs ^ Is)/Is) o Is

?= P + Q

The P component corresponds to the dot product between Vs and Is, the Q component to the wedge product. ??

In our first article, that was as far as we went. ?

But this overly simplistic picture had some problems, the result of a dot product between vectors is a scalar, i.e. just a number, and the wedge product between vectors is a bivector – which has similar properties to matrices. Power and reactive power are represented as time varying waveforms, i.e. they are like vectors.

The geometric product (which we are using as a division) allows us to convert scalars and bivectors back into vectors – but this still not enough to get the answer we want. ?

The power and reactive power waveforms are typically double the frequency of the voltage and current waveforms. No linear combination of sines and cosines can change their frequency, but multiplying them together point by point can and does. ?This is the final piece in the puzzle.

Using geometric algebra terminology:

The power waveform encodes the projection of the voltage and current waveforms.

The reactive power waveform encodes the rejection of the voltage and current waveforms.

To achieve the answer we wanted, we needed four different types of vector products which may seem to be a bit extreme, but the fact is, vector algebra doesn’t conform to the simple rules of arithmetic, this has been known since the 19th century.?

So what now? Will these additional insights help advance our electrical technology?

Possibly not, but then predicting the future of technology has a very low success rate. Developing tools and making them widely available often creates unexpected synergies. We believe there is more to be discovered here.