# Does this make any sense? Ohms law applied to laminar liquid fluid flow in pipes

Originally the below post was an answer to another post, but I thought my findings were interesting enough to post on their own. I would like to have confirmation that I found the right formula.

My question is about the analogy between (laminar) flow of liquid fluid and flow of electric current through an ohmic resistor.

Can Ohms law be applied to water? And if yes, when?

I began by substituting mass-flow k$\rm g/s$ as the current, after all electrical current is the flow of (a very small) mass:

Changing the voltage to delta pressure and with $R$(mass) being the resistance calculated from the mass-flow the formula would be:

\begin{align} U=IR\Longleftrightarrow\Delta P&={\rm\frac{kg}{s}}\cdot R({\rm{mass}})\\ &={\rm\frac{kg}{m\cdot s^2}}\iff R({\rm mass})=\frac{1}{{\rm m\cdot s}} \end{align}

This means that this first formula I extracted is independent of the density.

By multiplying with the density we we reach the standard unit for volumetric resistance (I will refer to this as Z(volume) from this point):

$$Z({\rm volume})={\rm Density}\cdot R({\rm mass})={\rm \frac{ kg}{m^3}\frac{1}{m\cdot s}=\frac{kg}{m^4\cdot s}}\qquad(1)$$

I will try to calculate a small example using these units:

Water: Density= $1000~ \rm kg/m^3$

Let's examine a mass-flow of $2~\rm kg/s$:

Water:

$${\rm Flow(volume)=\frac{2~\frac{kg}{s}}{1000~\frac{kg}{m^3}}=0.002~\frac{m^3}{s}}$$

So now let's assume a pressure drop of $50 Pa$

Which resistance could cause such a pressure drop?

$$R({\rm mass})=\frac{\Delta P}{{\rm \frac{kg}{s}}}=\frac{{\rm 50~ Pa}}{{\rm 2~\frac{kg}{s}}}=25~{\rm \frac{1}{m\cdot s}}$$

Now let's multiply with the density:

Water:

$$Z({\rm volume})=1000~{\rm \frac{kg}{m^3}}\cdot 25~{\rm {\frac{1}{m\cdot s}}}=25000~{\rm \frac{kg}{m^4\cdot s}}$$

Now let's calculate the flow:

$${\rm Waterflow~(volume)}=\frac{\Delta P}{Z{\rm (volume)}}={\rm \frac{ 50 Pa}{25000~\frac{kg}{m^4\cdot s}}=0.002~\frac{m^3}{s}}$$

As we see we arrive at the right result.

From the calculations above we saw that the unit of the density-independent resistance was: $\displaystyle{{\rm \frac{1}{m\cdot s}}}$ . When it does not depend on the fluid density it can only depend on other matters. From the units we see the following:

\begin{align} &R(mass)\times Area=Speed\Longleftrightarrow \frac{1}{m\times s}\times m^2=\frac{m}{s}\\ &\Longleftrightarrow\frac{1}{m\times s} =\frac{\frac{m}{s}}{m^2}\\ &\Longleftrightarrow R(mass)=\frac{Speed}{Area} \end{align}

substituting this new formula into Formula (1) we get:

$$Z{\rm (volume)=Density\cdot \frac{Speed}{Area}}\qquad(2)$$

And in the unit analysis:

$${\rm \frac{kg}{m^4\cdot s}=\frac{\left(\frac{kg}{m^3}\cdot \frac{m}{s}\right)}{m^2} =\frac{kg}{m^4\cdot s}}$$

As we see this rings quite true. So, using this logic, with constant density, the resistance only depends on the speed and the size of the pipe. If the size of the pipe is also a constant, we see that the pressure drop increases exponentially to the speed. We see this from the following analysis:

$$Z{\rm (volume)=Density\cdot \frac{Speed}{Area}}$$

We multiply this with the flow and get:

$$\Delta P = \frac{\text{Density} \times \text{Speed} \times \text{Speed} \times \text{Area}}{Area}=\text{Density} \times \text{Speed}^2$$

By dividing this formula with the length in meters we arrive at an interesting result:

$$\frac{\Delta P}{\text{Length}} = \frac{\text{Density}\times \text{Speed}^2}{\text{Length}}$$

And by multiplying the right side with the area in square meters we get:

$$\frac{\Delta P}{\text{Length}} = \frac{\text{Density}\times \text{Area}\times \text{Speed}^2}{\text{Volume}}$$

For a circular pipe we get the following:

$$\frac{\Delta P}{\text{Length}} = \frac{\text{Density}\times (\pi/4)\times \text{Diameter}^2\times \text{Speed^2}}{\text{Length}\times (\pi/4)\times\text{Diameter^2}}$$

We can remove $(\pi/4)$ again. So by separating the right side a bit we get:

$$\frac{\Delta P}{\text{Length}} = \frac{\text{Diameter}^2}{\text{Diameter}^2}\times \frac{\text{Density}\times \text{Speed}^2}{\text{Length}} = \frac{\text{Diameter}}{\text{Length}}\times \frac{\text{Density}\times\text{Speed}^2}{\text{Diameter}}\qquad(3)$$

This is actually a special case of the Darcy-Weisbach formula, but note that I derived it solely from using unit analysis. The only difference is that they add a friction factor. The Darcy-Weisbach formula reads:

$$\frac{\Delta P}{\text{Length}}=\frac{\text{Friction factor} d}{2} \times \frac{\text{Density}\times\text{Speed}^2}{\text{Diameter}}\qquad(4)$$

$${\rm \frac{Pa}{m}=Fd * \frac{\left(\frac{kg}{m^3}\ * (\frac{m}{s})^2\right)}{2*m} =Fd * \frac{kg}{2*m^2 * s^2\cdot}}=\frac{Fd}{2}*\frac{Pa}{m}$$

The Darcy friction factor is equal to ${Fd = \rm \frac{64}{Re}}$ for a laminar flow where Re is Reynolds number equal to:

$$Re = \frac{Density*velocity*Diameter}{dynamic viscosity}$$

At this point let's remember Formula (3) and (4)

Formula (3) $$\frac{DeltaP}{Length} = \frac{Diameter}{Length}* \frac{Density*Speed^2}{Diameter}$$

Formula (4) $$\frac{DeltaP}{Length}=\frac{Frictionfactord}{2} * \frac{Density*Speed^2}{Diameter}$$

Now we have two similar expressions describing the pressuredrop. One is the one we calculated using unitconversion from the "Electric current analogy" and the other one is our Darcy-Weisbach at constant temperature and a circular pipe. Now let's equate them:

$$\frac{DeltaP}{Length} =\frac{DeltaP}{Length}$$ <=> $$\frac{Diameter}{Length}* \frac{Density*Speed^2}{Diameter}=\frac{Frictionfactord}{2} * \frac{Density*Speed^2}{Diameter}$$ <=>

Formula (5)$$\frac{Diameter}{Length}=\frac{Frictionfactord}{2}$$

Formula five is a general formula describing the conditions that need to be present in order for Ohms law to apply to liquid fluid this being both laminar and turbulent. By inserting the frictionfactor for laminar flows we get:

$$\frac{Diameter}{Length} = \frac{\frac{64}{Re}}{2}$$ <=> $$\frac{Diameter}{Length} =\frac{64}{2*\frac{Density * Diameter * velocity}{dynamic-viscosity}}= \frac{32*dynamic-viscosity}{Density * Diameter *velocity}$$

By isolating the constant we get an equation that describes the relation between a set of conditions. If these conditions are met, meaning that the equation equals 32, then Ohms law is a perfect analogy between the laminar pipe flow and an electric current.

$$\frac{Diameter²*Density*velocity}{Length*dynamic-viscosity} = 32$$ <=> $$\frac{Diameter}{Length}*\frac{Diameter*Density*velocity}{dynamic-viscosity} = 32$$ From this we get: $$\frac{Diameter}{Length}*Re = 32$$ But let's remember the situation it can be used in: laminar flow and circular pipe.

Keeping the length in the calculation constant at 1 meter (Equal to pressure drop pr meter) I tried to draw some curves in excel to see how it related to 32.   As we see in the pictures the first two lines are quite parallel at very low velocities and cross-sectional area. I think what this means is that Ohms law can with some accuracy be used here. But in order for it to be precise the two lines should have been identical. We also se that a change in diameter makes a bigger difference than a similar change in velocity.

Now let's look at the turbulent flow for 3000 < Re < 10^5

Then the frictionfactor is equal to:

$$Frictionfactor=\frac{0,3164}{fourth-root-of-Re}$$

By substituting this into formula (5) we get

$$\frac{Diameter}{Length}=\frac{\frac{0,3164}{fourth-root-of-Re}}{2}=\frac{0,3164}{2*fourth-root-of-Re}$$ <=> $${\frac{1}{\frac{2*Diameter}{Length*0,3164}}}^4=Re$$ <=> $$4*log(\frac{Length*0,3164}{2*Diameter})=Re$$ <=> $$\frac{Re}{log(\frac{Length*0,3164}{2*Diameter})}=4$$

We have now once again arrived at an expression that needs to be give a constant result. In this case we're talking about turbulent flow in smooth tubes. Let us once again use the 1 meter distance for the length so that we can compare. I made a curve in excel for this equation too: Clearly the turbulent flow never gets close to ohms law. The only situation where this formula gets nearly parallel to the constant are at very low velocities, which would then be equal to laminar flow anyway.

My question now is: Does all of this make any sense or did I miscalculate???

• The similarities are startling. I relearned my electricity after knowing my fluid dynamics. Diode like a flow safety valve, current like pressure, transformer like a compressor, resistors like restrictive orifice Etc. – william deets Oct 15 '16 at 1:55

If voltage drop is analogous to pressure drop $\Delta p$, and current is analogous to flow rate $Q$, then for Ohm's law to hold we must have $\Delta p~\alpha ~Q$, with constant of proportionality being independent of $Q$. This does indeed happen for $\textit{laminar}$ flow through a pipe, so if you had a network of pipes in which the flow was laminar, you could apply the analogy of electrical circuits. Unfortunately, the flow in real world applications is almost always turbulent, in which case $\Delta p~\alpha ~Q^n,n>1$.
Of course the flow is laminar if Reynolds number of the flow is small (for pipes, less than about 2000). Reynolds number is defined as $\frac{Ud}{\nu}$, where $U$ is mean flow speed, $d$ is pipe diameter, and $\nu$ is kinematic viscosity of fluid. So the flow speed and cross-sectional area need not be separately small, but only the combination $\frac{Ud}{\nu}$.