Why is the unit of hydraulic impedance/resistance kilogram per second and hypermeter? I'm trying to figure out the unit of impedance in the hydraulic analogy of electronic networks. Assuming
$$Z=\frac{P}{Q},$$ 
with $P$ as pressure difference, $Q$ as volumetric flow rate and $Z$ as impedance, the unit would be
$$\rm \frac{Pa}{m^3/s} = \frac{N/m^2}{m^3/s} = \frac{kg/m\cdot s^2}{m^3/s} = \frac{kg}{m^4\cdot s}$$
Intuitively, this unit doesn't make sense. Did I get some part of the unit conversion wrong? If not, could someone help me understand the unit?
 A: If you use a basic forms loss (like the free jet condition) you have:
$$\Delta P = \frac{k}{2} \rho v^2$$
The difference between $\Delta P$ and $P$ doesn't bother me, because I think you might just as well switch that in your question.  But I need to get things in terms of $Q$.
$$Q = \frac{m^3}{s} = v \times A$$
$$\Delta P = \frac{k \rho}{2 A^2} Q^2$$
Now this is close to what you want but unsatisfactory because of the exponent.  Well, let's switch from this to consideration of specifically laminar flow.  In particular, let's use the Darcy friction factor.
$$k = f \frac{L}{D}$$
$$f = \frac{64}{Re}$$
$$Re = \frac{G D}{\mu} = \frac{\rho Q D}{A \mu}$$
$$k = \frac{L}{D} 64 \frac{A \mu}{\rho Q D} = \frac{64 A \mu L}{\rho Q D^2} $$
Back to the head loss equation.
$$\Delta P = \frac{64 A \mu L}{\rho Q D^2} \frac{\rho}{2 A^2} Q^2 = \frac{32 \mu L}{A D^2} Q $$
Okay there, for a specific case of laminar flow through a pipe you can have your resistor analogy with some physical basis.  Now, your question was about units.  Here are the units of the above equation:
$$Pa = {\rm \frac{kg}{m s^2} = \left( \frac{kg}{m^4 s} \right) \left( \frac{m^3}{s} \right)}$$
So yes, right off the bat there doesn't seem to be anything wrong with this.  But I'll take the 2nd approach.
$$Z = \frac{32 \mu L}{A D^2} \rightarrow {\rm \left( \frac{kg}{m s} \right) (m) \left( \frac{1}{m^2} \right) \left( \frac{1}{m^2} \right) = \frac{kg}{m^4 s}} $$
Yep, we're good.
A: Its fine. Never try to extract any sense from higher units.   Does the unit of energy $\mathrm{kgm^2/s^2}$ make sense? Nope, not until you relate it with a formula and break it into bits. Writing your unit as $\mathrm{\frac{Pa}{m^3/s}}$ is the most intuitive thing to do. By no means should you look at this as 'rate of change of mass in hyperspace'. Multiple different quantities can have the same unit. Torque is by no means the same as energy, for example.
A: I find it easier to understand when using mass-flow k$\rm g/s$ , after all electrical current is the flow of mass (electrons):
Then the formula would be
$$\Delta P={\rm\frac{kg}{s}}\cdot R({\rm{mass}})={\rm\frac{kg}{m\cdot s^2}}\iff R({\rm mass})=\frac{1}{{\rm m\cdot s}}$$
This means that this first formula I extracted is independent of the density.
By adding in the density we get the same as the above formula.
$$Z({\rm volume})={\rm Density}\cdot R({\rm mass})={\rm \frac{ kg}{m^3}\frac{1}{m\cdot s}=\frac{kg}{m^4\cdot s}}$$
Propane $1,7~ \rm kg/m^3,$ Water $1000~ \rm kg/m^3$
Let's examine a mass-flow of $2~\rm kg/s$ for each fluid:
For propane:
$${\rm Flow(volume)=\frac{2~\frac{kg}{s}}{1,7~\frac{kg}{m^3}}=1,176~\frac{m^3}{s}}$$
For water:
$${\rm Flow(volume)=\frac{2~\frac{kg}{s}}{1000~\frac{kg}{m^3}}=0,002~\frac{m^3}{s}}$$
So we would expect the speed of the gas to be a lot higher than of the water when using the same pipe.
So now let's assume the same pressure drop for both water and propane: 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 add in the densities:
Propane:
$$Z({\rm volume})=1,7~{\rm \frac{kg}{m^3}}\cdot 25~{\rm \frac{1}{m\cdot s}}=42,5~{\rm \frac{kg}{m^4\cdot s}}$$
And 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:
$${\rm Propaneflow~(volume)}=\frac{\Delta P}{Z{\rm (volume)}}={\rm \frac{50~ Pa}{42,5\frac{kg}{m^4\cdot s}}=1,176~\frac{m^3}{s}}$$
$${\rm Waterflow~(volume)}=\frac{\Delta P}{Z{\rm (volume)}}={\rm \frac{ 50 Pa}{25000~\frac{kg}{m^4\cdot s}}=0,002~\frac{m^3}{s}}$$
As we see we arrive at the right result.
From the calculations above we saw that the unit of the density-independent resistance was: ${\rm \frac{1}{m\cdot s}}$ . When it does not depend on the fluid density it can only depend on other matters. From the units we see the following:
$$\begin{align}R{\rm(mass) \cdot Area} &= {\rm Speed}\\ {\rm \frac{1}{m\cdot s}\cdot m^2} &={\rm \frac{m}{s}}\\\iff ~~~~~ {\rm \frac{1}{m\cdot s}} &={\rm \frac{\frac{m}{s}}{m^2}}\\
\iff  R{\rm (mass)} &={\rm\frac{Speed}{Area}};\end{align}$$
substituting this new formula into the old one we get:
$$Z{\rm (volume)=Density\cdot \frac{Speed}{Area}}$$
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 pressuredrop 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 in [m³/s] and get:
$$DeltaP = Density * Speed^2$$
By dividing this formule with the length in metres we arrive at an interesting result:
$$\frac{DeltaP}{Length} = \frac{Density*Speed^2}{Length}$$
And by multiplying the right side with the area in squaremetres we get:
$$\frac{DeltaP}{Length} = \frac{Density*Area*Speed^2}{Volume}$$
For a circular pipe we get the following:
$$\frac{DeltaP}{Length} = \frac{Density*(PI/4)*Diameter^2*Speed^2}{Length*(PI/4)*Diameter^2}$$
We can remove (PI/4) again. So by separating the right side a bit we get:
$$\frac{DeltaP}{Length} = \frac{Diameter^2}{Diameter^2}* \frac{Density*Speed^2}{Length} = \frac{Diameter}{Length}* \frac{Density*Speed^2}{Diameter}$$
This is actually a simplified form 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:
$$\frac{DeltaP}{Length}=\frac{Frictionfactord}{2} * \frac{Density*Speed^2}{Diameter}$$
<=> 
$${\rm \frac{Pa}{m}=Fd * \frac{\left(\frac{kg}{m^3}\ * (\frac{m}{s})^2\right)}{2*m} =Fd * \frac{kg}{2*m^2 * s^2\cdot}}=\frac{Fd}{2}*\frac{Pa}{m}$$
The Darcy friction factor is equal to ${Fd = \rm \frac{64}{Reynolds-number}}$ and Reynolds number is equal to:
$$Re = \frac{Density*velocity*Length}{dynamic viscosity}$$
If we have constant temperature, both the density and the viscosity are constant. When looking at a specific pipe the length too is constant. This gives us:
$$Re = Factor*velocity (no-unit)$$
Now we can calculate the Darcy friction factor:
$$Fd = \frac{64}{Factor*velocity}$$
Now we have two expressions for a constant without a unit. One is the one we calculated from the "Electric current analogy" and the other one is our Darcy friction factor at constant temperature. Now let's equate them, remembering what our factor consisted of:
$$\frac{Diameter}{Length} = \frac{\frac{64}{Factor*velocity}}{2} = \frac{64}{2*\frac{Density * Length*velocity}{dynamic-viscosity}}= \frac{32*dynamic-viscosity}{Density * Length*velocity}$$
So now we have an equation to solve when we can equate the flow of a fluid in a circular pipe to the rules of resistance in electricity:
$$\frac{Diameter*Density*velocity}{dynamic-viscosity} = 32$$
I would suggest solving it for water in excel for an array of table values and then calculating the percentage we're off from 100% right for each example.
I know your question was about units, but I think I gave some answer to that too.
