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Is there a contradiction between the continuity equation and Poiseuilles Law?
The continuity equation states that flow rate should be conserved in different areas of a pipe:
$Q=v_1{A}_1=v_2{A}_2=v\mathrm{\pi <!--\; \pi \; -->}{r}^2$
We can see from this equation that velocity and pipe radius are inversely proportional. If radius is doubled, velocity of flow is quartered.
Another way I was taught to describe flow rate is through Poiseuilles Law:
$Q=\frac{\mathrm{\pi <!--\; \pi \; -->}{r}^4\mathrm{\Delta <!--\; \Delta \; -->}P}{8\mathrm{\eta <!--\; \eta \; -->}L}$
So if I were to plug in the continuity equations definition of flow rate into Poiseuilles Law:
$vA=v\mathrm{\pi <!--\; \pi \; -->}{r}^2=\frac{\mathrm{\pi <!--\; \pi \; -->}{r}^4\mathrm{\Delta <!--\; \Delta \; -->}P}{8\mathrm{\eta <!--\; \eta \; -->}L}$
Therefore:
$v=\frac{r^2\mathrm{\Delta <!--\; \Delta \; -->}P}{8\mathrm{\eta <!--\; \eta \; -->}L}$
Now in this case, the velocity is proportional to the radius of the pipe. If the radius is doubled, then velocity is qaudrupled.
What am I misunderstanding here? I would prefer a conceptual explanation because I feel that these equations are probably used with different assumptions/in different contexts.

Do you need to increase pressure to pump a fluid from a large pipe through a small pipe?
The application of this question is towards clogged arteries and blood flow. I am wondering why people with clogged arteries have high blood pressure if a smaller artery (assuming it is smooth and cylindrical) means less pressure exerted by the blood according to Bernoulli's Equation. Is this because the heart has to pump the blood harder to be able to travel through a smaller arterial volume?

Power generation with a long, horizontal pipe, using atmospheric pressure differences
Straight to the point.
If I float a pipe starting from a point in the ocean that has the atmospheric pressure of 101 kPa (Call it Point A) all the way to another geographic position on the ocean with an atmospheric pressure of 96 kPa (point B). Lets say the distance between the two points is 1000 km. The air temperature at point A is 28 °C while the air temp at point is 18 °C.
If you opened both ends of the pipe,would the atmospheric pressure cause air to flow from Point A to Point B? (factoring in reasonable friction co-eff for a pipe.)
Can you please provide factors that would hinder the flow of air!
Could such a pipe, practically be used to generate a flow of air to generate electrical power?

Pressure changes in Continuity equations and Poiseuille's Law?
Continuity says Q=AV, and we know that velocity and pressure are inversely related. So if we are in a closed system, like vasculature for example, Q is constant and any decrease in vessel radius would be expected to raise velocity, which would result in lower pressure.
If we look at Poiseuille's Law, on the other hand, we see the opposite! If Q is constant, then a decrease in radius/cross sectional area we should expect pressure to be raised!
What's going on?

Physical meaning of $\frac{\pi }{8}$ in Poiseuille's equation
Recently I read about the Poiseuille's equation which relates the flow rate of a viscous fluid to coefficient of viscosity ($\mathrm{\nu <!--\; \nu \; -->}$), pressure per unit length($\frac{P}{l}$) and radius of the tube ($r$) in which the fluid is flowing. The equation is
$\frac{V}{t}=\frac{\pi P{r}^4}{8\mathrm{\nu <!--\; \nu \; -->}l},$
where $V$ denotes volume of the fluid.
One thing which I don't understand in this equation is that why do we have the constant term as $\frac{\pi }{8}$ . Is there some physical significance of this specific number here or is it just a mathematical convention?

Blood as Newtonian fluid
In some of the literature I read that blood can be considered as Newtonian fluid when a larger vesses with high shear stress is considered... How is the shear stress calculated for aorta and how do they claim that shear stress is more for larger vessel when compared to the smaller ones . What is the relationship between the shear stress and the size of the vessel

Matching Bernoulli equation and Hagen Poiseuille law for viscous fluid motion
I thought that Bernoulli equation could be used only in the case of non viscous fluid. But doing exercises on 2500 Solved Problems In Fluid Mechanics and Hydraulics (Schaum's Solved Problems Series) I found that this procedure is followed.
In the case of viscous laminar flow, Bernoulli equation is written as
$\begin{array}{}\text{(1)}& z_1+\frac{v_1^2}{2g}+\frac{p_1}{\mathrm{\rho <!--\; \rho \; -->}g}=z_2+\frac{v_2^2}{2g}+\frac{p_2}{\mathrm{\rho <!--\; \rho \; -->}g}+h_L\end{array}$
Where $h_L$ is the head loss (due to viscosity) calculated using Hagen Poiseuille law.
$\begin{array}{}\text{(2)}& h_L=\mathrm{\rho <!--\; \rho \; -->}g\frac{8\mathrm{\eta <!--\; \eta \; -->}L\stackrel{¯<!--\; ¯\; -->}{v}}{R^2}\end{array}$
Is this a correct way to solve exercises involving viscosity?
Furthermore are there limitation to this use of Bernoulli equation (in case of viscosity)?
In particular
If the flow is not laminar, I cannot use $(2)$, but can I still write $(1)$ in that way?
Is $(1)$ valid only along the singular streamline or between different ones (assuming the fluid irrotational)?

Deriving the scaling law for the Reynolds number
I'm trying to derive a scaling law for the Reynolds number, to get a better understanding of how it changes for microsystem applications, but I'm getting stuck. From textbook tables I should end up with $l^2$, but can't get there. The goal is to find a simplistic relation of Reynolds number and the system dimension.
This is my reasoning so far:
Considering a tube
$Re=\frac{\mathrm{\rho <!--\; \rho \; -->}v2R}{\mathrm{\mu <!--\; \mu \; -->}}$
$v=\frac{Q}{A}=\frac{Q}{\mathrm{\pi <!--\; \pi \; -->}{R}^2}$
From Hagen-Poiseuille:
$Q=\frac{\mathrm{\delta <!--\; \delta \; -->}P\mathrm{\pi <!--\; \pi \; -->}{R}^4}{8\mathrm{\mu <!--\; \mu \; -->}L},$
therefore:
$v=\frac{\frac{\mathrm{\delta <!--\; \delta \; -->}P\mathrm{\pi <!--\; \pi \; -->}{R}^4}{8\mathrm{\mu <!--\; \mu \; -->}L}}{\mathrm{\pi <!--\; \pi \; -->}{R}^2}=\frac{\mathrm{\delta <!--\; \delta \; -->}P\mathrm{\pi <!--\; \pi \; -->}{R}^4}{8\mathrm{\mu <!--\; \mu \; -->}L\mathrm{\pi <!--\; \pi \; -->}{R}^2}=\frac{\mathrm{\delta <!--\; \delta \; -->}P{R}^2}{8\mathrm{\mu <!--\; \mu \; -->}L}$
Replacing $v$ on $Re$
$Re=\frac{\frac{\mathrm{\rho <!--\; \rho \; -->}\mathrm{\delta <!--\; \delta \; -->}P{R}^{2}2R}{8\mathrm{\mu <!--\; \mu \; -->}L}}{\mathrm{\mu <!--\; \mu \; -->}}=\frac{\mathrm{\rho <!--\; \rho \; -->}\mathrm{\delta <!--\; \delta \; -->}P{R}^{3}2}{8{\mathrm{\mu <!--\; \mu \; -->}}^{2}L}$
From this, if I consider the $\mathrm{\delta <!--\; \delta \; -->}P$ as constant with the size scaling, $\mathrm{\rho <!--\; \rho \; -->}$ and $\mathrm{\mu <!--\; \mu \; -->}$ are material properties that are constant at different scales, I get:
$Re\sim <!--\; \sim \; -->\frac{R^3}{L}$
I can only think that I could consider it as $\frac{l^3}{l}=l^2$ but doesn't seem right.
What am I missing here?

Intuitive explanation of Poiseuille's law -- why $r^4$?
According to Poiseuille's law, the effective resistance of a tube is inversely proportional to the fourth power of its radius (as given by the following equation).
$R=\frac{8\mathrm{\eta <!--\; \eta \; -->}\mathrm{\Delta <!--\; \Delta \; -->}x}{\mathrm{\pi <!--\; \pi \; -->}{r}^4}$
I can go through the derivation of the law but is there an intuitive explanation for why it is the fourth power of radius, or a thought experiment that might make it more obvious?

Steady isothermal flow of an ideal gas
So I have a steady isothermal flow of an ideal gas through a smooth duct (no frictional losses) and need to compute the mass flow rate (per unit area) as a function of pressures at any two different arbitrary points, say 1 and 2. I have the following momentum equation in differential form:
$\mathrm{\rho <!--\; \rho \; -->}vdv+dP=0$
where $v$ is the gas the flow velocity and $P$ is static pressure. The mass flow rate per unit cross section $G$, can be calculated by integrating this equation between points 1 and 2. This is where it gets confusing. I do the integration by two ways:
1) Use the ideal gas equation $P=\mathrm{\rho <!--\; \rho \; -->}RT$ right away and restructure the momentum equation:
$vdv+RT\frac{dP}{P}=0$
, integrate it between points 1 and 2 and arrive at:
$v_1^2-<!--\; -\; -->v_2^2+2RTln\frac{P_1}{P_2}=0$
Since the flow is steady, I can write $G={\mathrm{\rho <!--\; \rho \; -->}}_1{v}_1={\mathrm{\rho <!--\; \rho \; -->}}_2{v}_2$, again use the ideal gas law to write density in terms of gas pressure and finally arrive at the mass flow rate expression:
$G^2=\frac{2ln\frac{P_2}{P_1}}{RT(\frac{1}{P_1^2}-<!--\; -\; -->\frac{1}{P_2^2})}$
2) In another way of integrating (which is mathematically correct), I start by multiplying the original momentum equation by $\mathrm{\rho <!--\; \rho \; -->}$ to get
${\mathrm{\rho <!--\; \rho \; -->}}^{2}vdv+\frac{1}{RT}PdP=0$
write ${\mathrm{\rho <!--\; \rho \; -->}}^{2}v=G^2/v$, integrate between points 1 and 2 to arrive at
$G^{2}ln\frac{v_2}{v_1}=\frac{P_1^2-<!--\; -\; -->P_2^2}{2RT}$
Using the ideal gas law the velocity ratio can be written as the pressure ratio to finally arrive at the mass flow rate equation
$G^2=\frac{P_1^2-<!--\; -\; -->P_2^2}{2RTln\frac{P_1}{P_2}}$
Both the expressions are dimensionaly sound and I know that the second expression is the correct one. My question is, whats wrong with first expression.

What determines the dominant pressure-flow relationship for a gas across a flow restriction?
If one measures the pressure drop across any gas flow restriction you can generally fit the relationship to
$\mathrm{\Delta <!--\; \Delta \; -->}P=K_2{Q}^2+K_{1}Q$
where $\mathrm{\Delta <!--\; \Delta \; -->}P$ is the pressure drop and $Q$ is the volumetric flow
and what I've observed is that if the restriction is orifice-like, $K_2>>K_1$ and if the restriction is somewhat more of a complex, tortuous path, $K_1>>K_2$ and $K_2$ tends towards zero.
I get that the Bernoulli equation will dominate when velocities are large and so the square relationship component. But what's determining the $K_1$ component behavior? Is this due to viscosity effrects becoming dominant? Does the Pouiselle relationship become dominant?

Pneumatic flow formula
I'm looking for a pneumatic formula in order to have the flow. I found many formula, but only for hydraulic :
$\mathrm{\Delta <!--\; \Delta \; -->}P=R_{p}Q$ where $R_p$ is the resistance, $Q$ the flow, and $\mathrm{\Delta <!--\; \Delta \; -->}P$ the pressure potential:
$R_p=\frac{8\mathrm{\eta <!--\; \eta \; -->}L}{\mathrm{\pi <!--\; \pi \; -->}{R}^4}$
with $\mathrm{\eta <!--\; \eta \; -->}=$ dynamic viscosity of the liquid ; $L=$ Length of the pipe ; $R=$ radius of the pipe
I don't know if I can use them, since I'm in pneumatic ! After doing research on the internet, I found some other variables like : sonic conductance, critical pressure coefficient, but no formula...
I think that I have all the information in order the calculate the flow : Pipe length = 1m ; Pipe diameter = 10mm ; $\mathrm{\Delta <!--\; \Delta \; -->}P=$ 2 bars
Thanks !

Hydraulics, flow rate, and tubing size questions
Here are a couple assumptions / things I believe to be true:
1.Pumps create a certain pressure in a piping circuit.
2.Pressure is like voltage and constant across the piping circuit, with differences in piping size increasing flow rate by reducing friction loss ie resistance
3.Given constant pressure, flow rate is variable as a fuction of pipe radius per the Poiseuille law.
This is where I am confused:
Imagine the case that a 2" pump outlet, runs for 1 foot of 2" piping and then connects in with a 4" pipe for reduced friction.
So if the pressure across both pipes (the system) are the same, the gpm for the larger pipe is something like 16x bigger for the larger pipe than the smaller pipe. The larger pipe is fed by the smaller pipe, so where does this water come from?
Understand what I am getting at?
One of my assumptions seems erroneous.

The figure shows four charges at the corners of a square of side L. What magnitude and sign of charge Q will make the force on charge q zero?
Express your answer with the appropriate units.
photo

Difference between Pressure and Pressure Energy in fluids
For a fluid with viscosity to flow through a pipe that has the same cross-sectional area at both ends, at a constant velocity, there has to be a pressure difference according to Poiseuille's Law. Why exactly is there a change in pressure required to keep the velocity constant?
Is it because according to Bernoulli's principle that Pressure or Pressure-Energy gets converted to Kinetic Energy to speed up the fluid so the mass flow rate at both ends of the pipe stays the same?
So if that's the case in light of Bernoulli's principle, that means the change in pressure in Poiseullie's Law is there so that pressure energy gets converted to Kinetic Energy to fight off the viscosity of the fluid to keep the velocity of the fluid constant?
And
What exactly is Pressure? I know it is the Force divided by the area. I understand that concept, but I've seen the terms Pressure and Pressure Energy used interchangeably when talking about fluids, which creates for some amount of confusion. Aren't Pressure and Pressure Energy different? But when we talk about fluids in light of Bernoulli's principles, it seems as if Pressure and Pressure Energy are the same, which is pretty confusing.