# Get help with fluid mechanics questions

Recent questions in Fluid Mechanics

### Matching Bernoulli equation and Hagen Poiseuille law for viscous fluid motionI 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 particularIf the flow is not laminar, I cannot use $\left(2\right)$, but can I still write $\left(1\right)$ in that way?Is $\left(1\right)$ valid only along the singular streamline or between different ones (assuming the fluid irrotational)?

Amappyaccon22j7e 2022-05-09

### Deriving the scaling law for the Reynolds numberI'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?

tuehanhyd8ml 2022-05-09

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

Aedan Gonzales 2022-05-08

### Steady isothermal flow of an ideal gasSo 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\left(\frac{1}{{P}_{1}^{2}}-\frac{1}{{P}_{2}^{2}}\right)}$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.

Azzalictpdv 2022-05-08

### 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$\mathrm{\Delta }P={K}_{2}{Q}^{2}+{K}_{1}Q$where $\mathrm{\Delta }P$ is the pressure drop and $Q$ is the volumetric flowand 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?

Yaritza Oneill 2022-05-08

### Pneumatic flow formulaI'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 pipeI 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 barsThanks !

Marissa Singh 2022-05-08