How do I correctly integrate the formula for deriving Posseuille's Law?

[JohnnyGui]

Hello,

I’ve been trying to derive Posseuille’s Law and I’m very close. However there seems to be a small difference between my concluded formula and the real one but I don’t know why.

Here’s what I get:

Picture a cilindrical tube with a radius $R$ in which a fluid is flowing. From what I know, there are 2 forces that play a role here. There’s the force $F_P$ that causes the flow itself and is proportional to the pressure difference $ΔP$; it’s equal to $F_P = ΔP \cdot π \cdot R^2$. And there’s a force $F_V$ that counteracts $F_P$ because of the fluid’s viscosity $ɳ$ which causes friction at the walls of the tube; it’s equal to $F_V = ɳ \cdot 2 \cdot R \cdot L \cdot \frac{v}{r}$. The $r$ is a chosen radius where you want to know $F_V$.

From what I understand, the $F_P$ is equal to $F_V$ at every chosen radius $r$ within radius $R$ of the tube. So we can say that:

$$ \frac{ΔP \cdot π \cdot R^2}{ɳ \cdot 2 \cdot \cdot π \cdot R \cdot L} = \frac{v}{r} $$

To calculate $v$ here, one would have to rewrite this as:

$$ \frac{ΔP \cdot R \cdot r}{ɳ \cdot 2 \cdot L} = v $$

We here now have written $v$ as a function of a chosen $r$ within the tube. If we now choose very small $r$’s, we would get more or less the velocity of the portion of the fluid within that chosen $r$. That velocity times the surface $π \cdot r^2$ would give the flow rate $Q$ through that chosen $r$. We’d thus have to integrate the formula:

$$ \frac{ΔP \cdot R \cdot r}{ɳ \cdot 2 \cdot L} \cdot π \cdot r^2 = \frac{ΔP \cdot R \cdot r^3 \cdot π}{ɳ \cdot 2 \cdot L} $$

Integrating this would finally give the formula:

$$ \frac{ΔP \cdot R \cdot π \cdot r^4}{8 \cdot L \cdot ɳ} = Q $$

The correct formula is missing the $R$ parameter in the numerator, which is the radius of the tube. I don’t get why that is missing since that’s the determining factor in the formulae for the $F_P$ and $F_V$.

 

[Chestermiller]

Where did you get your equation for Fv from? To me, it makes no sense.

 

[JohnnyGui]

Where did you get your equation for Fv from? To me, it makes no sense.

Got it from here: http://www.insula.com.au/physics/1250/poiseuille.html

Looking at it now, I noticed it doesn't fit since a smaller $r$ should give a smaller $F_V$. How should it be noted then?

 

[Chestermiller]

Got it from here: http://www.insula.com.au/physics/1250/poiseuille.html

Looking at it now, I noticed it doesn't fit since a smaller $r$ should give a smaller $F_V$. How should it be noted then?

If should be $$F_v=(2\pi R L)\left(-\eta \frac{dv}{dr}\right)_{r=R}$$And dv/dr is not constant. v is maximum at the center of the tube and zero at the wall.

 

[JohnnyGui]

If should be $$F_v=(2\pi R L)\left(-\eta \frac{dv}{dr}\right)_{r=R}$$And dv/dr is not constant. v is maximum at the center of the tube and zero at the wall.

I don't really get what $dr$ is. Is it $R - r$? And how about $dv$? How would one get the actual velocity at a particular $r$ with that formula?

 

[Chestermiller]

I don't really get what $dr$ is. Is it $R - r$? And how about $dv$? How would one get the actual velocity at a particular $r$ with that formula?

It's hard to know where to start without knowing more about your background. Have you had calculus yet?

 

[JohnnyGui]

It's hard to know where to start without knowing more about your background. Have you had calculus yet?

It was a long time ago for me during high school. However, after watching a video, I conclude that $dv$ is the difference of velocity between a radius of 0 and a chosen $r$ ( = $dr$), inside the tube. This $dv$ should be negative since increasing the $dr$ lowers the velocity because it approaches the walls of the tube, hence the minus sign in the formula. Because the velocity at the walls is 0, I can calculate the velocity at the middle of the tube if I use $dr$ as $R$, thus:

$$ v_(max) = \frac{ΔP \cdot R^2}{ɳ \cdot 2 \cdot L} $$

Now that I have the maximum velocity at the middle of the tube ($dr = 0$), I can calculate the velocity at any $dr$ away from $dr = 0$ by substracting your mentioned formula with any chosen $dr$, from my concluded formula. Thus:

$$ v = \frac{ΔP \cdot R^2}{ɳ \cdot 2 \cdot L} - \frac{ΔP \cdot dr^2}{ɳ \cdot 2 \cdot L} = \frac{ΔP}{ɳ \cdot 2 \cdot L}(R^2 - dr^2) $$

Am I going the right way here? If I am, then this is the formula of $v(r)$. Now, if I choose a very small $r$, calculate $v$ out of that and then multiply this $v$ with the surface of that very small $r$ which is $π \cdot r^2$, I'd get the volume rateflow within that small surface. I'd have to add all these small surfaces until I reach a radius of $R$. Thus I'd have to integrate:

$$ \frac{ΔP}{ɳ \cdot 2 \cdot L}(R^2 - dr^2) \cdot π \cdot dr^2 $$

Which still doesn't give me the correct equation for volume rate flow. What am I doing wrong here? Should I substract each small surface from the previous surface in the formula?

 

[Chestermiller]

It was a long time ago for me during high school. However, after watching a video, I conclude that $dv$ is the difference of velocity between a radius of 0 and a chosen $r$ ( = $dr$), inside the tube. This $dv$ should be negative since increasing the $dr$ lowers the velocity because it approaches the walls of the tube, hence the minus sign in the formula. Because the velocity at the walls is 0, I can calculate the velocity at the middle of the tube if I use $dr$ as $R$, thus:

$$ v_(max) = \frac{ΔP \cdot R^2}{ɳ \cdot 2 \cdot L} $$

Now that I have the maximum velocity at the middle of the tube ($dr = 0$), I can calculate the velocity at any $dr$ away from $dr = 0$ by substracting your mentioned formula with any chosen $dr$, from my concluded formula. Thus:

$$ v = \frac{ΔP \cdot R^2}{ɳ \cdot 2 \cdot L} - \frac{ΔP \cdot dr^2}{ɳ \cdot 2 \cdot L} = \frac{ΔP}{ɳ \cdot 2 \cdot L}(R^2 - dr^2) $$

Am I going the right way here? If I am, then this is the formula of $v(r)$. Now, if I choose a very small $r$, calculate $v$ out of that and then multiply this $v$ with the surface of that very small $r$ which is $π \cdot r^2$, I'd get the volume rateflow within that small surface. I'd have to add all these small surfaces until I reach a radius of $R$. Thus I'd have to integrate:

$$ \frac{ΔP}{ɳ \cdot 2 \cdot L}(R^2 - dr^2) \cdot π \cdot dr^2 $$

Which still doesn't give me the correct equation for volume rate flow. What am I doing wrong here? Should I substract each small surface from the previous surface in the formula?

None of this is correct. You are not going to be able to do this without using calculus, and it looks like your calculus skills are currently not adequate.

For example: dv is the difference between the velocity at locations r and r + dr, and dr is not the radial location; it is the tiny incremental difference between two closely neighboring radial locations.

 

[JohnnyGui]

For example: dv is the difference between the velocity at locations r and r + dr, and dr is not the radial location; it is the tiny incremental difference between two closely neighboring radial locations.

Ok, I do remember some calculus but I thought that it could be interpreted in another way. I think I have 2 questions that would make things more clear for me.

1. If $\frac{dv}{dr}$ is the derivative of a function $v(r)$, then doesn't this mean that $\frac{ΔP \cdot r}{ɳ \cdot 2 \cdot L}$ is the derivative as well of the function $v(r)$? So I'd have to integrate that formula to get this function?

2. You say that $dr$ is the difference in velocity between r and r + dr. If I set $r$ to 0 and choose $dr$ to be $R$, would the formula give me the difference of $v$ between a radius of $R$ and a radius of 0?

 

[Chestermiller]

Ok, I do remember some calculus but I thought that it could be interpreted in another way. I think I have 2 questions that would make things more clear for me.

1. If $\frac{dv}{dr}$ is the derivative of a function $v(r)$, then doesn't this mean that $\frac{ΔP \cdot r}{ɳ \cdot 2 \cdot L}$ is the derivative as well of the function $v(r)$? So I'd have to integrate that formula to get this function?

2. You say that $dr$ is the difference in velocity between r and r + dr. If I set $r$ to 0 and choose $dr$ to be $R$, would the formula give me the difference of $v$ between a radius of $R$ and a radius of 0?

No, no, and no. If you want to see the correct derivation using a "shell" force balance, see Transport Phenomena by Bird, Stewart, and Lightfoot.

 

[JohnnyGui]

No, no, and no. If you want to see the correct derivation using a "shell" force balance, see Transport Phenomena by Bird, Stewart, and Lightfoot.

Then why are there people who are integrating that formula as part of deriving Posseuille's Law?

Source: http://www.insula.com.au/physics/1250/poiseuille.html

Also, this lecturer does the same at 5:15:

 

[Chestermiller]

Then why are there people who are integrating that formula as part of deriving Posseuille's Law?

View attachment 111235
Source: http://www.insula.com.au/physics/1250/poiseuille.html

Also, this lecturer does the same at 5:15:

These derivations look OK.

 

[JohnnyGui]

These derivations look OK.

But that's what I proposed in my first question, integrating $\frac{ΔP \cdot r}{ɳ \cdot 2 \cdot L}$ to get the function of $v(r)$. Why wasn't it correct?

 

[Chestermiller]

But that's what I proposed in my first question, integrating $\frac{ΔP \cdot r}{ɳ \cdot 2 \cdot L}$ to get the function of $v(r)$. Why wasn't it correct?

You are aware that there is a difference between dv/dr and v/r, correct?

 

[JohnnyGui]

You are aware that there is a difference between dv/dr and v/r, correct?

Yes, I am after your explanation in your post #8. The only thing that bothers me about it though is why, when you choose r = 0 and r + dr = R (thus $dr = R$), it doesn't give the difference between the velocity ($dv$) in the middle of the tube ($r = 0$) and the velocity at radius of the tube ($R$). Since the velocity at $R$ is 0, the difference should be equal to the velocity at the middle of the tube, which is the maximum velocity in the tube. This may be all out of the subject of deriving Poisseulle's law but I'd really like to understand this first. Is there a way to explain this?

 

[Chestermiller]

Yes, I am after your explanation in your post #8. The only thing that bothers me about it though is why, when you choose r = 0 and r + dr = R (thus $dr = R$), it doesn't give the difference between the velocity ($dv$) in the middle of the tube ($r = 0$) and the velocity at radius of the tube ($R$). Since the velocity at $R$ is 0, the difference should be equal to the velocity at the middle of the tube, which is the maximum velocity in the tube. This may be all out of the subject of deriving Poisseulle's law but I'd really like to understand this first. Is there a way to explain this?

There is a key difference between dr and $\Delta r$, and between dv and $\Delta v$. dr is strictly reserved for tiny differences in r, and dv is reserved for tiny differences in v. It is improper to write dr = R - 0.

dv/dr can only be set equal to v/r if we know for a fact that v is directly proportional to r. But for flow in a tube, v is not directly proportional to r. For flow in a tube, v varies quadratically with r. Look at the final equation for v as a function of r. So you can't just set $dv/dr =-v_{max}/R$. It is incorrect mathematically.

 

[boneh3ad]

Perhaps it is fruitful here to discuss the definition of a derivative in this case?

$$\dfrac{dv}{dr} = \lim_{\Delta r \to 0}\dfrac{v(r+\Delta r) - v(r)}{\Delta r}$$

So if you just pick arbitrary values for $r$ for $\Delta r$, you aren't satisfying the limit. Sometimes you can approximate a derivative by doing something similar to that (see: finite differences), but it isn't even approximately correct when you take your $\Delta r$ to be the entire domain.

 

[JohnnyGui]

There is a key difference between dr and ΔrΔr\Delta r, and between dv and ΔvΔv\Delta v. dr is strictly reserved for tiny differences in r, and dv is reserved for tiny differences in v. It is improper to write dr = R - 0.

dv/dr can only be set equal to v/r if we know for a fact that v is directly proportional to r. But for flow in a tube, v is not directly proportional to r. For flow in a tube, v varies quadratically with r. Look at the final equation for v as a function of r. So you can't just set dv/dr=−vmax/Rdv/dr=−vmax/Rdv/dr =-v_{max}/R. It is incorrect mathematically.

That's where my confusion was! I didn’t know there was a difference between $Δr$ and $dr$. I do understand that choosing a fixed $Δr$ would give a different $Δv$ since the relation is a parabola and thus $Δv$ would depend on which r to which r you want to take that fixed $Δr$ from. But apparently $\frac{dv}{dr}$ really is about a very local steepness of a curve instead of a larger range.

So, after integration I’d get the following function of $v(r)$:

$$v(r) = \frac{ΔP}{4ηL} \cdot (R^2-r^2)$$

Now, to get the flow per time unit I’d have to multiply that formula $v(r)$ with the area of the tube $π \cdot r^2$. But since the velocity differs with $r$ I’d have to integrate the whole formula over small $dr$’s.

Here’s where I’m stuck, because if I choose to integrate the circle area $π \cdot dr^2$, then the integration would just keep integrating a fixed circle area. That’s incorrect because substracting a circle area of $π \cdot (n \cdot dr)^2$ from a circle area of $π \cdot ((n-1) \cdot dr)^2)$ with $n$ being whole numbers, gives circle areas of $πr^2$, $3 \cdot πr^2$, $5 \cdot πr^2$, etc. I don’t know how to put this in a formula before starting the integration.

Watching the previous video I gave a link to shows that I’d have to integrate $2 \cdot π \cdot r \cdot dr$ multiplied by $v(r)$. But I don’t have the slightest idea how he came to $2 \cdot π \cdot r \cdot dr$ based on the following picture:

 

[JohnnyGui]

Anyone who could please help me with my question in the above post?

 

[Chestermiller]

That's where my confusion was! I didn’t know there was a difference between $Δr$ and $dr$. I do understand that choosing a fixed $Δr$ would give a different $Δv$ since the relation is a parabola and thus $Δv$ would depend on which r to which r you want to take that fixed $Δr$ from. But apparently $\frac{dv}{dr}$ really is about a very local steepness of a curve instead of a larger range.

So, after integration I’d get the following function of $v(r)$:

$$v(r) = \frac{ΔP}{4ηL} \cdot (R^2-r^2)$$

Now, to get the flow per time unit I’d have to multiply that formula $v(r)$ with the area of the tube $π \cdot r^2$. But since the velocity differs with $r$ I’d have to integrate the whole formula over small $dr$’s.

Here’s where I’m stuck, because if I choose to integrate the circle area $π \cdot dr^2$, then the integration would just keep integrating a fixed circle area. That’s incorrect because substracting a circle area of $π \cdot (n \cdot dr)^2$ from a circle area of $π \cdot ((n-1) \cdot dr)^2)$ with $n$ being whole numbers, gives circle areas of $πr^2$, $3 \cdot πr^2$, $5 \cdot πr^2$, etc. I don’t know how to put this in a formula before starting the integration.

Watching the previous video I gave a link to shows that I’d have to integrate $2 \cdot π \cdot r \cdot dr$ multiplied by $v(r)$. But I don’t have the slightest idea how he came to $2 \cdot π \cdot r \cdot dr$ based on the following picture:

View attachment 111690

The area over which v(r) applies is the anulus of area $\pi(r+\frac{dr}{2})^2-\pi (r-\frac{dr}{2})^2$, where r is the radial location at the mid-point of the anulus. But this is just equal to $2\pi r dr$.

Also, from calculus, $d(r^2)=2rdr$, so, in your own development, $\pi d(r^2)=2\pi r dr$

 

[JohnnyGui]

The area over which v(r) applies is the anulus of area $\pi(r+\frac{dr}{2})^2-\pi (r-\frac{dr}{2})^2$, where r is the radial location at the mid-point of the anulus. But this is just equal to $2\pi r dr$.

Also, from calculus, $d(r^2)=2rdr$, so, in your own development, $\pi d(r^2)=2\pi r dr$

Ah, this clears it up, thanks a lot! I got confused from the drawing because he's showing the length $r$ not extending into the mid-point of the annulus but being a separate length away from the width $dr$. From that drawing I'd have therefore concluded that the area of the annulus should be $π(r + dr)^2 - πr^2$. But why can't one describe $r$ like this and give this formula instead before integrating?

 

[Chestermiller]

Ah, this clears it up, thanks a lot! I got confused from the drawing because he's showing the length $r$ not extending into the mid-point of the annulus but being a separate length away from the width $dr$. From that drawing I'd have therefore concluded that the area of the annulus should be $π(r + dr)^2 - πr^2$. But why can't one describe $r$ like this and give this formula instead before integrating?

Doing it this way is fine too. So, to terms of first order in dr , what do you get for this?

 

[JohnnyGui]

Doing it this way is fine too. So, to terms of first order in dr , what do you get for this?

I'm sorry for my late reply. The derivation of my mentioned formula would be $2πdr$ I think, which seems to be close to the one that should be integrated. I have a bit of a struggle to derive an equation when the initial formula already has a $dx$ variable in it.

Also, I thought that my mentioned formula $π(r + dr)^2 - πr^2$ should be integrated instead of taking its first order derivative since my formula shows that I have to add up every annulus with width of $dr$. Might be missing something obvious here.

 

[Chestermiller]

//..

I'm sorry for my late reply. The derivation of my mentioned formula would be $2πdr$ I think, which seems to be close to the one that should be integrated. I have a bit of a struggle to derive an equation when the initial formula already has a $dx$ variable in it.

Also, I thought that my mentioned formula $π(r + dr)^2 - πr^2$ should be integrated instead of taking its first order derivative since my formula shows that I have to add up every annulus with width of $dr$. Might be missing something obvious here.

You are multiplying it by a function of r (the velocity), so you can't integrate it separately. Another option is to write it as $\pi d(r^2)$ and integrating with respect to $r^2$

 

[JohnnyGui]

//..
You are multiplying it by a function of r (the velocity), so you can't integrate it separately. Another option is to write it as $\pi d(r^2)$ and integrating with respect to $r^2$

Apologies, I wasn't clear enough. I meant integrating that formula after multiplying it with $v(r)$. So I'd have to integrate:

$$ \frac{ΔP}{4ηL} \cdot (R^2-r^2) \cdot (π(r + dr)^2 - πr^2)) $$

Isn't it possible like that? Since you said $π(r + dr)^2 - πr^2$ is fine too, I expected that it should be part of the integration because it will then give you every bit of annulus area over the whole circle times the velocity at each $r$ which would then give you the total flow rate. That's why I don't get why I should take the first derivative of $π(r + dr)^2 - πr^2$ instead. At least that's what I thought you meant when you asked about the terms of first order in dr in your post #22.

 

[Chestermiller]

Apologies, I wasn't clear enough. I meant integrating that formula after multiplying it with $v(r)$. So I'd have to integrate:

$$ \frac{ΔP}{4ηL} \cdot (R^2-r^2) \cdot (π(r + dr)^2 - πr^2)) $$

Isn't it possible like that? Since you said $π(r + dr)^2 - πr^2$ is fine too, I expected that it should be part of the integration because it will then give you every bit of annulus area over the whole circle times the velocity at each $r$ which would then give you the total flow rate. That's why I don't get why I should take the first derivative of $π(r + dr)^2 - πr^2$ instead. At least that's what I thought you meant when you asked about the terms of first order in dr.

You need to go back and review calculus. What I meant was that $$\pi (r+dr)^2-\pi r^2=2\pi rdr(1+\frac{dr}{2r})$$But, in calculus, the second term in parenthesis is negligible compared to the 1, so it is dropped. So, we're still left with $2\pi r dr$. It is not possible, in any calculus I've ever seen, to integrate the relationship the way you have it written.

 

[JohnnyGui]

You need to go back and review calculus. What I meant was that $$\pi (r+dr)^2-\pi r^2=2\pi rdr(1+\frac{dr}{2r})$$But, in calculus, the second term in parenthesis is negligible compared to the 1, so it is dropped. So, we're still left with $2\pi r dr$. It is not possible, in any calculus I've ever seen, to integrate the relationship the way you have it written.

I will review calculus. After reading this through again, I think the culprit of my confusion is the meaning of how the term $dr$ is put in the formula that has to be integrated. I thought that I could integrate a formula where the $dr$ term can be substracted or added, while (I think) integration should only be possible if the formula is formulated in such a way that I'd have to multiply it with $dx$.

I also noticed that one could also just take the derivative of $πr^2$ which is $2πr$ without putting $dr$ in it. If I then multiply $2πr$ with $v(r)$ and then integrate that from 0 to $R$, one would also get the same correct formula. All this time I thought you should integrate the $dr$ term itself as well, which got me putting it in the formula in any other way, like in the form of substraction/addition as in my post #25.