Recent questions in Poiseuille's Law

Fluid MechanicsAnswered question

Selena Hardin 2023-02-25

The density of seawater is $1024kg/{m}^{3}$. The pressure at a depth of 5 m below the surface of the sea is ___ $N/{m}^{2}$.

Fluid MechanicsOpen question

J Ashkanani2022-12-20

The radius of the aorta is about 1.1 cm , and the blood passing through it has a speed of about 34 cm/s . Coefficient of viscosity for the whole blood (37∘) is η = 4×10−3Pa⋅s Calculate the pressure drop per cm along the aorta.

Fluid MechanicsAnswered question

Gharib4Pe 2022-12-04

What is the pressure in atmospheres at $11$ km depth?

Fluid MechanicsAnswered question

ramirezherePYM 2022-11-28

Theoretical question from the book: The pressure exerted by a liquid depends on...?

Fluid MechanicsAnswered question

Abdii Diroo2022-07-25

A sample of ideal gas has an internal energy U and is then compressed to one half of its original volume while the temperature stays the same. What is the new internal energy of the ideal gas?

Fluid MechanicsAnswered question

Osmarq5ltp 2022-05-20

For the electrical resistance of a conductor, we have

$R=\rho \frac{l}{A}$

Noting the structural similarity between the Hagen-Poiseuille law and Ohm's law, we can define a similar quantity for laminar flow through a long cylindrical pipe:

${R}_{V}=8\eta \frac{l}{A{r}^{2}}$

So there's a structural difference of a factor of ${r}^{2}$ between the two. What's the intuition behind this?

$R=\rho \frac{l}{A}$

Noting the structural similarity between the Hagen-Poiseuille law and Ohm's law, we can define a similar quantity for laminar flow through a long cylindrical pipe:

${R}_{V}=8\eta \frac{l}{A{r}^{2}}$

So there's a structural difference of a factor of ${r}^{2}$ between the two. What's the intuition behind this?

Fluid MechanicsAnswered question

Carley Haley 2022-05-20

Theoretical vs Experimental flow rate of fluid coming out of a water bottle with a hole in it

Say there is a water bottle that is filled with 300 mL of water and has a circular hole with a radius of 2 mm. In this bottle, the water sits 7.8cm above the top of the hole (which has been drilled 1.5cm above the bottom of the bottle).

According to Bernoulli's law the velocity $v$ of the water flowing out is equal to $\sqrt{2gh}$

Using this, the flow rate can be calculated as $Q\text{}=\text{}Av\text{}=\text{}\pi (0.002\text{}m{)}^{2}\ast 1.24\text{}m/s=0.000016\text{}{m}^{3}/s=16\text{}mL/s$

This doesn't seem accurate considering that the experimental flow rate is equal to 8 mL/s (40 mL over 5 seconds). However I understand that it ignores viscosity (and other things?)

I'm wondering a few things, firstly, does the theoretical math here apply to the situation I'm describing? The hole in the bottle isn't exactly a pipe and the only examples I've seen with water flow involve pipes.

Secondly, can Poiseuille's Law be used to determine the flow rate instead, with a more accurate result? (From what I understand Q=πPR^4/8nl, however I don't understand what P is, seeing as in Bernoulli's law pressure cancels and as aforementioned this isn't a typical pipe example.)

Thirdly, I assume the theoretical flow rate will still be different from the experimental flow rate, what factors cause this?

Say there is a water bottle that is filled with 300 mL of water and has a circular hole with a radius of 2 mm. In this bottle, the water sits 7.8cm above the top of the hole (which has been drilled 1.5cm above the bottom of the bottle).

According to Bernoulli's law the velocity $v$ of the water flowing out is equal to $\sqrt{2gh}$

Using this, the flow rate can be calculated as $Q\text{}=\text{}Av\text{}=\text{}\pi (0.002\text{}m{)}^{2}\ast 1.24\text{}m/s=0.000016\text{}{m}^{3}/s=16\text{}mL/s$

This doesn't seem accurate considering that the experimental flow rate is equal to 8 mL/s (40 mL over 5 seconds). However I understand that it ignores viscosity (and other things?)

I'm wondering a few things, firstly, does the theoretical math here apply to the situation I'm describing? The hole in the bottle isn't exactly a pipe and the only examples I've seen with water flow involve pipes.

Secondly, can Poiseuille's Law be used to determine the flow rate instead, with a more accurate result? (From what I understand Q=πPR^4/8nl, however I don't understand what P is, seeing as in Bernoulli's law pressure cancels and as aforementioned this isn't a typical pipe example.)

Thirdly, I assume the theoretical flow rate will still be different from the experimental flow rate, what factors cause this?

Fluid MechanicsAnswered question

dresu9dnjn 2022-05-20

How does Newtonian viscosity not depend on depth?

I am currently trying to self-study basic Classical Mechanics and right now I am trying to understand fluid dynamics. I have read about Newtonian viscosity: as my book says, the force exerted on a viscous fluid depends on the coefficient of viscosity, the cross-section and the velocity gradient, but, to my amazement, not on the fluid's height, or in any other way on its total amount.

Usually I'd be more than willing to trust such a result even if not proven, and the way it is used in deriving something like Poiseuille's law, for example, almost makes sense to me. But I still, to my best efforts, fail to understand or even get an intuition on what should be the most basic case of application, i.e. a fluid that can be approximated as many rectangular plates flowing on top of each other. If I try to apply the same method I would use for a cylindrical tube, I get that considering a section of any height $h$, if the velocity gradient in the rectangle is constant at every depth, the force I need to apply to that section is the same no matter the value of $h$. In particular, dividing the rectangle in an arbitrarily large number of sections of arbitrarily small value of $h$, keeping the gradient the same I get that the total force I need to apply to maintain it is arbitrarily large.

Also, it sounds unintuitive that keeping the same gradient and halving the amount of fluid I would need the same force (and double the pressure). In addition to this, even with the tube, if instead of applying the viscosity law to a full cylinder I try applying it to a cylindrical ring, I get the same problems.

Can someone explain? Am I doing this wrong (probably)? Does the viscosity law somehow only work for the "top" section of the liquid? If so, what does top mean? How can the fluid "know" what is the top and what isn't? Are different pressures really needed depending on the amount of fluid to maintain the same gradient? What's an intuitive explanantion of this? Is there something else I'm missing that would make the explanation work?

I am currently trying to self-study basic Classical Mechanics and right now I am trying to understand fluid dynamics. I have read about Newtonian viscosity: as my book says, the force exerted on a viscous fluid depends on the coefficient of viscosity, the cross-section and the velocity gradient, but, to my amazement, not on the fluid's height, or in any other way on its total amount.

Usually I'd be more than willing to trust such a result even if not proven, and the way it is used in deriving something like Poiseuille's law, for example, almost makes sense to me. But I still, to my best efforts, fail to understand or even get an intuition on what should be the most basic case of application, i.e. a fluid that can be approximated as many rectangular plates flowing on top of each other. If I try to apply the same method I would use for a cylindrical tube, I get that considering a section of any height $h$, if the velocity gradient in the rectangle is constant at every depth, the force I need to apply to that section is the same no matter the value of $h$. In particular, dividing the rectangle in an arbitrarily large number of sections of arbitrarily small value of $h$, keeping the gradient the same I get that the total force I need to apply to maintain it is arbitrarily large.

Also, it sounds unintuitive that keeping the same gradient and halving the amount of fluid I would need the same force (and double the pressure). In addition to this, even with the tube, if instead of applying the viscosity law to a full cylinder I try applying it to a cylindrical ring, I get the same problems.

Can someone explain? Am I doing this wrong (probably)? Does the viscosity law somehow only work for the "top" section of the liquid? If so, what does top mean? How can the fluid "know" what is the top and what isn't? Are different pressures really needed depending on the amount of fluid to maintain the same gradient? What's an intuitive explanantion of this? Is there something else I'm missing that would make the explanation work?

Fluid MechanicsAnswered question

Merati4tmjn 2022-05-20

Why is pressure gradient assumed to be constant with respect to radius in the derivation of Poiseuille's Law?

Poiseuille's Law relies on the fact that velocity is not constant throughout a cross-section of the pipe (it is zero at the boundary due to the no-slip condition and maximum in the center). By Bernoulli's Law, this means that pressure is maximum at the boundary and minimum at the center. But in the book I have it is assumed that the pressure gradient is independent of radius (distance from the center of the pipe), and the pressure gradient is thus extricated from a radius-integral. Can anyone justify this?

Poiseuille's Law relies on the fact that velocity is not constant throughout a cross-section of the pipe (it is zero at the boundary due to the no-slip condition and maximum in the center). By Bernoulli's Law, this means that pressure is maximum at the boundary and minimum at the center. But in the book I have it is assumed that the pressure gradient is independent of radius (distance from the center of the pipe), and the pressure gradient is thus extricated from a radius-integral. Can anyone justify this?

Fluid MechanicsAnswered question

Laila Andrews 2022-05-19

How to convert this derivation of Poiseuille's law into the standard one?

I am trying to derive Poiseuille's law. I have reached a point in the derivation where I have:

$V=\frac{(p1-p2)({R}^{2})}{4lu}$

Where $l$ is length, u is viscosity, p is pressure, v is flow velocity and R is radius. What I am stuck on is shifting this to the volumetric flow rate:

$V\pi {R}^{2}=Q=\frac{(p1-p2)({R}^{4})}{4lu}.$

$V\pi {R}^{2}=Q=\frac{(p1-p2)({R}^{4})}{4lu}.$

However this is incorrect as Poiseuille's law is divided by 8. I know that I am probably missing something obvious, but I can't think of a reason as to why the whole equation needs to be halved. Any help understanding why (or whether my original derivation for V was inaccurate) would be appreciated.

I am trying to derive Poiseuille's law. I have reached a point in the derivation where I have:

$V=\frac{(p1-p2)({R}^{2})}{4lu}$

Where $l$ is length, u is viscosity, p is pressure, v is flow velocity and R is radius. What I am stuck on is shifting this to the volumetric flow rate:

$V\pi {R}^{2}=Q=\frac{(p1-p2)({R}^{4})}{4lu}.$

$V\pi {R}^{2}=Q=\frac{(p1-p2)({R}^{4})}{4lu}.$

However this is incorrect as Poiseuille's law is divided by 8. I know that I am probably missing something obvious, but I can't think of a reason as to why the whole equation needs to be halved. Any help understanding why (or whether my original derivation for V was inaccurate) would be appreciated.

Fluid MechanicsAnswered question

Brooklynn Hubbard 2022-05-18

Calculate flow rate of air through a pressurized hole

I was wondering about this:

If there is a pressurized container, like a tank of compressed air at some pressure that is greater than the ambient air pressure, and this tank of air has a hole in it, what is the velocity of the escaping air through the hole? Is there a formula for this?

I was wondering about this:

If there is a pressurized container, like a tank of compressed air at some pressure that is greater than the ambient air pressure, and this tank of air has a hole in it, what is the velocity of the escaping air through the hole? Is there a formula for this?

Fluid MechanicsAnswered question

othereyeshmt4l 2022-05-18

Hydraulic dynamics question

We know the length of a pipe, say L, and the starting point pressure p1, ending point p2, cross-sectional area A. What else do we need to compute the mean velocity of flow in this pipe?

My full question actually runs as follows. Given the mean cardiac output(volume flux) be 5.5L per min, the radius of the aorta is about 1.1cm. In the systemic circulatory system, the mean radius of a capillary is about 3µm. The mean pressure at the arterial end of the capillary bed (beginning of the capillary system) is estimated to be about 30mmHg, and about 15mmHg at the venous end (end of the capillary system). The length of capillary is 0.75mm.

Calculate the mean velocity in the aorta and in a capillary.

My idea is to use the volume flux 5.5L/min and the radius of aorta, divide the flux by the crossectional area to find the mean velocity in the aorta. However, I have no idea how to deal with the velocity in a capillary. I tried to use Poiseulle's law, but given that we only know the difference of pressure and length of the capillary, it is still unsolvable. Could anyone tell me how to deal with it?

We know the length of a pipe, say L, and the starting point pressure p1, ending point p2, cross-sectional area A. What else do we need to compute the mean velocity of flow in this pipe?

My full question actually runs as follows. Given the mean cardiac output(volume flux) be 5.5L per min, the radius of the aorta is about 1.1cm. In the systemic circulatory system, the mean radius of a capillary is about 3µm. The mean pressure at the arterial end of the capillary bed (beginning of the capillary system) is estimated to be about 30mmHg, and about 15mmHg at the venous end (end of the capillary system). The length of capillary is 0.75mm.

Calculate the mean velocity in the aorta and in a capillary.

My idea is to use the volume flux 5.5L/min and the radius of aorta, divide the flux by the crossectional area to find the mean velocity in the aorta. However, I have no idea how to deal with the velocity in a capillary. I tried to use Poiseulle's law, but given that we only know the difference of pressure and length of the capillary, it is still unsolvable. Could anyone tell me how to deal with it?

Fluid MechanicsAnswered question

Jordon Haley 2022-05-18

How does a hole's size affect the distance that water will squirt

I took a bucket, drilled 2 different sized holes on the side near the bottom and filled it with water. The stream of water the proceeded from the larger hole traveled further than the stream from the smaller one. How does the size of the hole affect the distance that the water travels?

I took a bucket, drilled 2 different sized holes on the side near the bottom and filled it with water. The stream of water the proceeded from the larger hole traveled further than the stream from the smaller one. How does the size of the hole affect the distance that the water travels?

Fluid MechanicsAnswered question

hetriamhageh6k20 2022-05-18

Proving that the water leaving a vertical pipe is exponential (decay)

How can I prove that the rate of which water leaves a vertical cylindrical container (through a hole at the bottom) is exponential of the form :

$A{e}^{kx}$

I know that Torricelli's law is:

$\sqrt{2gh}$

But this only proves a square root relationship. I have data points every 10 seconds and graphed it suggests a decay function. I know the distance between the pipe is 1.5M and the internal diameter is 5cm. The hole diameter is 0.25cm, if this helps. I need to prove that the water leaving the pipe is exponentially decaying.

How can I prove that the rate of which water leaves a vertical cylindrical container (through a hole at the bottom) is exponential of the form :

$A{e}^{kx}$

I know that Torricelli's law is:

$\sqrt{2gh}$

But this only proves a square root relationship. I have data points every 10 seconds and graphed it suggests a decay function. I know the distance between the pipe is 1.5M and the internal diameter is 5cm. The hole diameter is 0.25cm, if this helps. I need to prove that the water leaving the pipe is exponentially decaying.

Fluid MechanicsAnswered question

Iyana Macdonald 2022-05-17

Flow through a cylinder with pores

I'm building a model to study the flow of fluid through a cylinder with pores on its surface.

For flow through a cylinder, the velocity of the fluid flow is given by the Hagen-Poiseuille equation.

I would like to ask for suggestions on references from which I can look at derivations for a cylinder with porous walls. By porous, I mean openings on the surface of the cylinder. Example, the pores present in the fenestrated capillary.

I'm building a model to study the flow of fluid through a cylinder with pores on its surface.

For flow through a cylinder, the velocity of the fluid flow is given by the Hagen-Poiseuille equation.

I would like to ask for suggestions on references from which I can look at derivations for a cylinder with porous walls. By porous, I mean openings on the surface of the cylinder. Example, the pores present in the fenestrated capillary.

Fluid MechanicsAnswered question

hovudverkocym6 2022-05-17

What happens to pipe length when the pipe diameter changes?

Intuitively, when the diameter of pipe is decreased, there will be more friction loss, more water pressure, and a higher flow rate. Is there a direct relationship/equation derived to see the affect of the pipe lengths?

Deriving from Hagen-Poiseuille's equation, we get:

$\text{}Q=\frac{\mathrm{\Delta}P\pi {r}^{4}}{8\mu L}$

where :

$Q=$ flow rate

$\mathrm{\Delta}\text{}P=$ change in fluid pressure

$r=$ radius of pipe,

$\mu =$ dynamic viscosity of fluid,

$L=$ length of pipe

To keep similar flow rates, are we able to use Poiseuille's derived formula to find the new lengths of pipe with a change in $r$ (pipe radius)?

Intuitively, when the diameter of pipe is decreased, there will be more friction loss, more water pressure, and a higher flow rate. Is there a direct relationship/equation derived to see the affect of the pipe lengths?

Deriving from Hagen-Poiseuille's equation, we get:

$\text{}Q=\frac{\mathrm{\Delta}P\pi {r}^{4}}{8\mu L}$

where :

$Q=$ flow rate

$\mathrm{\Delta}\text{}P=$ change in fluid pressure

$r=$ radius of pipe,

$\mu =$ dynamic viscosity of fluid,

$L=$ length of pipe

To keep similar flow rates, are we able to use Poiseuille's derived formula to find the new lengths of pipe with a change in $r$ (pipe radius)?

Fluid MechanicsAnswered question

Edith Mayer 2022-05-17

Why exactly is the resistance of a conductor inversely proportional to the area of its cross-section?

Before I explain my query, I would like to clarify that I am a ninth-grader who got this question while studying the formula $R\propto \frac{1}{A}$ where $A$ is the area of cross-section.

I have often asked this question to my teachers and they always give me the classic "corridor and field example". They told me that if 5 people walk in a corridor, they will find it harder to get across than if they were to be walking through a field- the same goes for electrons passing through a conductor. My counter-argument would be that if the width of the conductor increases, so will the number of positive ions (my textbook says that positive ions in conductors hinder the flow of current) and hence, more the resistance.

I would really appreciate it if the answer could be explained to me in simple terms as I'm not well versed with the more complex formulae involved in this concept. If not, do let me know of the concepts I should read about (preferably the specific books) to understand the solution better.

Before I explain my query, I would like to clarify that I am a ninth-grader who got this question while studying the formula $R\propto \frac{1}{A}$ where $A$ is the area of cross-section.

I have often asked this question to my teachers and they always give me the classic "corridor and field example". They told me that if 5 people walk in a corridor, they will find it harder to get across than if they were to be walking through a field- the same goes for electrons passing through a conductor. My counter-argument would be that if the width of the conductor increases, so will the number of positive ions (my textbook says that positive ions in conductors hinder the flow of current) and hence, more the resistance.

I would really appreciate it if the answer could be explained to me in simple terms as I'm not well versed with the more complex formulae involved in this concept. If not, do let me know of the concepts I should read about (preferably the specific books) to understand the solution better.

Fluid MechanicsAnswered question

Jayden Mckay 2022-05-15

Flow rate of a syringe

Suppose a syringe (placed horizontally) contains a liquid with the density of water, composed of a barrel and a needle component. The barrel of the syringe has a cross-sectional area of $\alpha \text{}{m}^{2}$, and the pressure everywhere is $\beta $ atm, when no force is applied.

The needle has a pressure which remains equal to $\beta $ atm (regardless of force applied). If we push on the needle, applying a force of magnitude $\mu \text{}N$, is it possible to determine the medicine's flow speed through the needle?

Suppose a syringe (placed horizontally) contains a liquid with the density of water, composed of a barrel and a needle component. The barrel of the syringe has a cross-sectional area of $\alpha \text{}{m}^{2}$, and the pressure everywhere is $\beta $ atm, when no force is applied.

The needle has a pressure which remains equal to $\beta $ atm (regardless of force applied). If we push on the needle, applying a force of magnitude $\mu \text{}N$, is it possible to determine the medicine's flow speed through the needle?

Fluid MechanicsAnswered question

encamineu2cki 2022-05-15

How does the radius of a pipe affect the rate of flow of fluid?

Poiseuille's law states that the rate of flow of water is proportionate to ${r}^{4}$ where $r$ is the radius of the pipe. I don't see why.

Intuitively I would expect rate of flow of fluid to vary with ${r}^{2}$ as the volume of a cylinder varies with ${r}^{2}$ for a constant length. The volume that flows past a point is equal to rate of flow fluid, and thus it should vary with ${r}^{2}$

I cannot understand the mathematical proof completely. Is there an intuitive explanation for this?

Poiseuille's law states that the rate of flow of water is proportionate to ${r}^{4}$ where $r$ is the radius of the pipe. I don't see why.

Intuitively I would expect rate of flow of fluid to vary with ${r}^{2}$ as the volume of a cylinder varies with ${r}^{2}$ for a constant length. The volume that flows past a point is equal to rate of flow fluid, and thus it should vary with ${r}^{2}$

I cannot understand the mathematical proof completely. Is there an intuitive explanation for this?

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As you proceed with your studies of fluid mechanics or need to apply Physics for some mechanical experiment, Poiseuille's law example problems will be one of the questions that you will have to solve. In basic terms, it is a law related to hemodynamics based on the volumetric flow rate sensitivity. For example, this law is used to explain why constricted capillaries lead to higher blood pressure as one of the examples. This is where Poiseuille's law equation can help you explain the fluid mechanics principles. See how such problems are addressed both verbally and graphically in the presented problems.