Recent questions in Fluid Mechanics

Poiseuille's Law
Answered

Adelyn Rodriguez
2022-05-09

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

$\begin{array}{}\text{(1)}& {z}_{1}+\frac{{v}_{1}^{2}}{2g}+\frac{{p}_{1}}{\rho g}={z}_{2}+\frac{{v}_{2}^{2}}{2g}+\frac{{p}_{2}}{\rho g}+{h}_{L}\end{array}$

Where ${h}_{L}$ is the head loss (due to viscosity) calculated using Hagen Poiseuille law.

$\begin{array}{}\text{(2)}& {h}_{L}=\rho g\frac{8\eta L\overline{v}}{{R}^{2}}\end{array}$

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)?

Poiseuille's Law
Answered

Amappyaccon22j7e
2022-05-09

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

$Re=\frac{\rho v2R}{\mu}$

$v=\frac{Q}{A}=\frac{Q}{\pi {R}^{2}}$

From Hagen-Poiseuille:

$Q=\frac{\delta P\pi {R}^{4}}{8\mu L},$

therefore:

$v=\frac{\frac{\delta P\pi {R}^{4}}{8\mu L}}{\pi {R}^{2}}=\frac{\delta P\pi {R}^{4}}{8\mu L\pi {R}^{2}}=\frac{\delta P{R}^{2}}{8\mu L}$

Replacing $v$ on $Re$

$Re=\frac{\frac{\rho \delta P{R}^{2}2R}{8\mu L}}{\mu}=\frac{\rho \delta P{R}^{3}2}{8{\mu}^{2}L}$

From this, if I consider the $\delta P$ as constant with the size scaling, $\rho $ and $\mu $ are material properties that are constant at different scales, I get:

$Re\sim \frac{{R}^{3}}{L}$

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?

Poiseuille's Law
Answered

tuehanhyd8ml
2022-05-09

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).

$R=\frac{8\eta \mathrm{\Delta}x}{\pi {r}^{4}}$

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?

Poiseuille's Law
Answered

Aedan Gonzales
2022-05-08

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:

$\rho vdv+dP=0$

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 $P=\rho RT$ 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 $G={\rho}_{1}{v}_{1}={\rho}_{2}{v}_{2}$, 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 $\rho $ to get

${\rho}^{2}vdv+\frac{1}{RT}PdP=0$

write ${\rho}^{2}v={G}^{2}/v$, 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.

Poiseuille's Law
Answered

Azzalictpdv
2022-05-08

If one measures the pressure drop across any gas flow restriction you can generally fit the relationship to

$\mathrm{\Delta}P={K}_{2}{Q}^{2}+{K}_{1}Q$

where $\mathrm{\Delta}P$ 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?

Poiseuille's Law
Answered

Yaritza Oneill
2022-05-08

I'm looking for a pneumatic formula in order to have the flow. I found many formula, but only for hydraulic :

$\mathrm{\Delta}P={R}_{p}Q$ where ${R}_{p}$ is the resistance, $Q$ the flow, and $\mathrm{\Delta}P$ the pressure potential:

${R}_{p}=\frac{8\eta L}{\pi {R}^{4}}$

with $\eta =$ 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 ; $\mathrm{\Delta}P=$ 2 bars

Thanks !

Poiseuille's Law
Answered

Marissa Singh
2022-05-08

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.

Poiseuille's Law
Answered

Nubydayclellaumvcd
2022-05-07

Express your answer with the appropriate units.

photo

Poiseuille's Law
Answered

Azzalictpdv
2022-05-07

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.

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