Electrical Water – Dave McGlone

The flow of water in a hose is often used as an analogy for electrical current. Let’s see about that, first considering the electrical parameters.

Define a few values:

Avogadro Number nA = 6.02214e23 particles/mol
Molar mass of copper mM = 63.546 g/mol
Copper density $\rho$ cu = 8.960 g/cm³
Electric charge q = 1.60218e-19 C
Electron mass Me = 9.10938e-31 kg
Reduced Planck’s Constant $\hslash$ = 1.05457e-34 J s
Speed of light c = 299792458 m/s → 300e6 m/s
m c² = 510998.950 eV

The number of copper atoms per cubic cm is found from:

$\begin{displaymath}nCu\,=\,\frac{\,\mathcal{N}_A\,\rho_{Cu}\,}{M}\,=\,\frac{\,6.02214\textnormal{e23}\,\times\,8.96\,}{63.55} \;=\;8.491\textnormal{e22}/cm^3\end{displaymath}$

where $\mathcal{N}_A$ is Avogadro’s Number, $\rho_{Cu}$ is the atomic density of copper, and $M$ is the molar mass of copper.

Since copper has only one conduction electron per atom, this is the same number of free electrons per $cm^3$ . Multiply by 10⁶ for quantity per m³.

The charge density is found from:

$\begin{displaymath}\rho_q\,=\,n\;q \;\; \rightarrow\;\;13.604\textnormal{e3}\;C/cm^3\;\; \rightarrow\;\;1.3604\textnormal{e10}\;C/m^3\end{displaymath}$

or, inverting: there is 1C of charge contained within a 73.505e-6 cm³ volume of copper. This assumes all conduction electrons are free … which is not fully valid but we’ll use it here.

The diameter of AWG#20 wire is 0.0812 cm. The cross-section area is therefore:

$\begin{displaymath}\textnormal{area20}\,=\,\pi\,\left(\frac{\,0.0812\,}{2}\right)^2\;=\; 5.1785\textnormal{e-3}\,cm^2\end{displaymath}$

So the length of AWG20 wire containing 1C of free charge is found from:

$\begin{displaymath}\textnormal{area20}\,cm^2\times\,\textnormal{length}\,cm\,=\,73.5053\textnormal{e-6}\,cm^3\;\rightarrow\;\textnormal{length}\,=\,0.0141944\,cm\;=\;141.9\,\mu m\end{displaymath}$

Might be a bit difficult to trim with a pair of nippers …

Assume electron motion is fully random with net velocity of zero; convert to units of electronVolts (eV)

The Fermi energy of copper is given as 7.00 eV. Using $E\;=\;m\,c^2$ (see? it really is used) where $m\,c^2\;=\;510999 eV$ . With this conversion of units, the Fermi velocity is found to be:

$\begin{displaymath}\nu_F\,=\,c\,\sqrt{\,\frac{\,2\,eF_{Cu}\,}{m\,c^2}\,}\;=\;1.5692\textnormal{e6}\;m/s\end{displaymath}$

or about 0.5% the speed of light.

Oh, what the heck … I couldn’t find more than a tabular entry for the Fermi energy of copper … but it can be easily calculated. Note that Fermi energy is not temperature dependent (this discussion is pertinent because all those free electrons form a Fermi gas within the conductor … and anyway, I was once an incomplete undergrad nuclear physics major)

The expression used is:

$\begin{displaymath}eF\,=\,\frac{\hslash^2}{\,2\,Me\,}\,\left(\frac{\,3\,\pi^2\,nCu\,}{V}\,\right)^{2/3}\;\Rightarrow\;eF\,=\,1.12865\textnormal{e-18}\,J \;=\;7.0445\,eV\end{displaymath}$

The mean free path of an electron in copper is estimated from:

$\begin{displaymath}mfp\,=\,\frac{\,\sigma_{Cu}\,\mathcal{M}_e\,\nu_F\,}{n\,q^2}\;=\; 38.69\,nm\end{displaymath}$

where $\sigma_{Cu}$ is the conductivity of copper (5.9e7 S/m)

The diameter of a copper atom is about 250 pm so the mean free path of a copper electron is about 160 atoms

The mean time between collisions is found from:

$\begin{displaymath}\tau\,=\,\frac{\,\sigma_{Cu}\,\mathcal{M}_e\,}{n\,q^2}\;=\; 24.66\,fs\end{displaymath}$

The resistance of a 1m length of copper wire:

$\begin{displaymath}R\,=\,\frac{\,\textnormal{length}}{\,\sigma_{Cu}\,\textnormal{area}\,}\;=\; 32.67\,m\Omega\end{displaymath}$

A current of 1A (1C/s) through this segment causes a voltage of $32.67\,mV$ across it.

The drift velocity is found from the charge density and current density:

$\begin{displaymath}\nu_D\,=\,\frac{\textnormal{J}}{\,n\,q\,}\,}\;=\;141.7\textnormal{e-6}\;m/s\end{displaymath}$

Which makes sense as that defines the length of wire containing 1C of free charge moving in 1 sec … 1A

Electrons don’t move very fast, eh? About 1.7 ft/hr. And here we think of electrons going zippity-doo-dah all over the place. I move that fast – if I’m in a hurry. On the other hand, consider a tube full of marbles. I can push a marble into the tube really slow … but one comes out the other end instantaneously.

A bit of a detour:

This analysis assumes that all free electrons (1 per atom in copper) contribute to “current”. These conduction electrons are more or less at the Fermi energy level on the valence band. Since this level shifts with an applied electric field, these “free” electrons are not always in the conduction band. A decrease in the number of free electrons causes an increase in the velocity and the free mean path length for those electrons that are free.

Electrons have quantum spin – they may be in only one of two states. In the event of a collision, a corresponding state must be available before the collision; the total energy before needs to equal that after or a collision can not occur. Lower energy electrons are locked into their respective bands and are not part of “current”.

The drift velocity as calculated above assumes all electrons in the valence band (only 1 for copper) are available for conduction. Thermal energy is sometimes sufficient to move an electron from the valence band to the conduction band.

Think of a copper conductor as a neighborhood lattice of homes with 16yo boys and much younger siblings inside. Normally these boys are bouncing off the walls with anxiety to be set free but they remain within the home (as do the siblings) – their net movement is zero even though they bounce off the walls at Fermi velocity. The boys are in the valence band; the siblings have lower energy in lower bands and are content to stay put. However, when the boys are given sufficient stimulus – an electric field or temperature or a neighborhood girl – it ends up giving him an open door to the conduction band; off he goes but he has to walk to the girl. The speed of bouncing off the walls is much greater than walking down the street. He keeps vibrating at Fermi velocity as he slowly walks though. But let’s not carry the analogy too far, eh?

If 1 mole of particles passes by a point in 1 second, this may be equated to an equivalent current of:

$\begin{displaymath}iEq\,=\,\frac{nA}{\,1/q\,}\,}\;=\;96485.3\;C/mol \quad\textnormal{or}\quad 10.3643\textnormal{e-6}\,mol/C\end{displaymath}$

1 mole of water is contained within a volume of 18 cm³. The equivalent volume of water containing “1C” of particles is:

$\begin{displaymath}volW\;=\;10.3643\textnormal{e-6}\,mol/C\;\times \; 18 \,cm^3/mol\;=\;186.557\textnormal{e-6} \,cm^3/C\end{displaymath}$

Assume a section of 1/16″ ID tubing. This is equivalent to a diameter of 0.15875 cm. The cross-sectional area of the pipe is therefore:

$\begin{displaymath}\textnormal{areaP}\;=\;\pi\,\left(\frac{\,0.15875\,}{2}\right)^2\;=\;0.0197933\, cm^2\end{displaymath}$

The length of pipe necessary to contain 1C-equivalent of water molecules is found from:

$\begin{displaymath}0.0197933\, cm^2\,\times\,\textnormal{lengthP}\,cm\;=\;\;186.557\textnormal{e-6} \,cm^3\;\Rightarrow\;\textnormal{lengthP}\,=\,94.2527\textnormal{e-6}\,cm\end{displaymath}$

Noting that the volume unit “mL” is equivalent to “ $cm^3$ ” (or “cc”), the volume of water equivalent to 1C has been found to be 186.56e-6 mL

Consider an unscientific, semi-official unit of volume known as the “drip”. 1 drip is equivalent to ¼cc. For an equivalent current of 1A, there are this many drips per second:

$\begin{displaymath}\textnormal{dps}\,=\,\frac{186.56\textnormal{e-6}}{\,0.25\,}\;=\;746.23\textnormal{e-6}\;d/s\end{displaymath}$

Better to invert; in seconds per drip:

$\begin{displaymath}\textnormal{spd}\,=\,\frac{\,0.25\,}{186.56\textnormal{e-6}}\;=\;1340.1\;s/d\end{displaymath}$

Converting to minutes:

$\begin{displaymath}\textnormal{mpd}\,=\,\frac{spd}{\,60\,}\;=\;22.3\;m/d\end{displaymath}$

The length of pipe containing 1 drip:

$\begin{displaymath}areaP\,cm^2\,\times\,lengthD\,cm\,=\,0.25 \,cm^3\,\Rightarrow\,lengthD\,=\,12.631\,cm\end{displaymath}$

So the water needs to flow at a rate of 9.4253e-3 cm/s (1.11 ft/hr) to equal 1A of electrical current in a AWG20 wire. For that matter, this is the flow rate of water which compares to 1A of current regardless of wire size.

When using the water = electricity analogy, one can note that 1A of electrical current is equivalent to about 3 drips/hour out of a 1/16″ piece of tubing.

So why bother with all the electrical equivalency stuff if the water-to-current conversion is independent of the electrical?

For the fun of it. What else?

Leave a Comment Cancel Reply