[Phys-L] AC power calculations

[John Denker <jsd@av8n.com>]

On 2/22/21 7:14 AM, Arlyn DeBruyckere wrote:

> To use P=V*V/R you REALLY need (delta)V. 

That is excellent advice, in this context and every other.

Forsooth, in any context whatsoever, the physics depends on ΔV, never
directly on V.  The fancy name for this is /gauge invariance/.  It is
a fundamental property of the EM field.  It is built into the Maxwell
equations.

This is a really simple, really fundamental, really powerful idea.

I know a guy who wrote a PhD thesis that revolved around exploiting
gauge invariance in clever ways.
  https://ieeexplore.ieee.org/document/97537

>  The Voltage drop (delta V) for
> the power company will only be 1 Volt (0.01 A * 100 Ohms) so the power for
> the power company will be 1 V * 1 V / 100 Ohms = 0.01 Watts for the power
> lines.

That's true in the original context ... but in other situations the
AC power calculation is more complicated than that.

Let I, ΔV, and P represent the *instantaneous* current, voltage drop,
and power, as functions of time.  So I(t) is the current at a particular
time t.  We assume the resistance R and the impedance Z are constants.

Then let ⟨I⟩ and ⟨ΔV⟩ represent the RMS averages.  For simplicity,
let's imagine these to be constants, although they could be slowly
varying functions of time.  An ordinary AC voltmeter indicates ⟨I⟩
and ⟨ΔV⟩.

Let [P] represent the algebraic average of P.  We never care about
the RMS average of P.  The energy transfer is ΔE = [P] Δt.

Since Ohm's law is linear (!) we can average both sides, so:
	ΔV = I R
becomes:
	⟨ΔV⟩ = ⟨I⟩⟨R⟩

The instantaneous power is P = ΔV I.  This couldn't possibly be any
simpler.  This is analogous to mechanics, where the formula is
force dot velocity.

However, beware: The RHS is nonlinear, so you cannot reliably calculate:
	☠ [P] = ⟨ΔV⟩⟨I⟩ ☠
Really not.

The same warning applies to any other nonlinear expression:
    *The average of the product is not the product of the averages.*

For example, if you connect an AC signal to a capacitor, ⟨ΔV⟩ will be
nonzero, ⟨I⟩ will be nonzero, but [P] will be zero. Energy will flow
into and out of the capacitor, but the average [P] is zero.  If we
integrate P over a cycle, we find the net energy transfer is zero.

Power companies know all about this, and they have a name for it:
	PF ≡ *power factor*

It depends on the phase angle between the voltage and the current.
If you apply a voltage to a purely resistive load, then the phase
angle is zero and the power factor is 100%.  This is the special
case in which you can get away with writing [P] = ⟨ΔV⟩² / R and
[P] = ⟨I⟩² R.

Resistive loads include things like electric ovens and old-fashioned
incandescent light bulbs.

We now segue to /reactive/ loads including capacitors and inductors.

For a reactive load, you need to use the impedance Z, and you need
to throw in a factor of PF on the RHS here:
	[P] = ⟨ΔV⟩⟨I⟩ PF

For a purely reactive load such as an ideal capacitor or inductor,
the power factor is zero.  Assuming I and V are sinusoidal, you
can understand this mathematically using trig identities such as:
	P(t) = cos(ωt) sin(ωt) = ½ sin(2ωt)
where the algebraic average of the RHS is obviously zero.

A typical municipal power grid presents a markedly reactive load to
the generating station.  That's because:
 -- Electric motors have a terrible power factor when they are
  lightly loaded. They look like inductors.
 -- Traditional fluorescent light installations have a /ballast/
  which is an inductor.
 -- The distribution lines themselves are somewhat inductive,
  although they take steps to alleviate this.
 -- Et cetera, or worse.

And (!) it's even worse than that.  The grid makes heavy use of
/saturable transformers/ to provide a steady voltage to customers
even when the voltage on the distribution lines changes under load.
This creates a tremendous amount of /third harmonic/.  That is, you
get something like this:
	V = cos(ωt)
	I = cos(ωt) + 0.3 cos(3ωt) + higher harmonics
which also contributes to a bad power factor.

And (!!) your typical dimmer circuit draws current during some
small part of the circuit, which means the current waveform is
really nasty.  Compact fluorescent lights have a "electronic"
ballast which may have a highly distorted current waveform ...
although you can get special ones that are better behaved.

This is a really big deal because many of their costs (including
dissipation in the distribution network) depend on ⟨I⟩² and/or ⟨ΔV⟩
...  whereas in most cases they can only charge customers for the
actual power ⟨I V⟩ delivered, which will be less by a factor of PF.
In *some* cases they can charge customers for the reactive load.
Note that reactive loads are measured in VA (volts times amperes)
or kVA, never in W or kW; the latter are reserved for actual power.

Grid operators sometimes install eeeenormous capacitors in an effort
to improve the power factor.  Customers who are being charged by the
kVA may install their own capacitors.

Some useful diagrams (and discussion) are here:
https://www.pecanstreet.org/2018/10/power-factor-a-little-known-feature-of-a-reliable-grid/

===================

For simplicity I have avoided mentioning phasors, and have instead
just used trig functions.

The same caveats apply: Phasors are great for *linear* calculations
and not otherwise.  I've seen lots of people who ought to know
better get this wrong.

If there is any nonlinearity involved, be sure to take the real
part of your phasor before plugging it into the nonlinear formula.