Double experiment: "Falling ball viscometry" and viscous flow in an horizontal pipe

[HighPhy]

TL;DR Summary: Analyze, with the help of @Chestermiller, the importance of viscous interactions in a vertical geometry ("Falling ball viscometry") and in an horizontal viscous flow.

The following double experiment was conducted by me a few weeks ago. @Chestermiller and I discussed this experiment privately, and @Chestermiller advised me to open a thread on this Forum so that other users can also benefit from the discussion he and I will have on this topic.

Consider a fluid-dynamic resistance force of intensity $F$ ($\vec F = - \alpha \vec V$) opposing motion at velocity $V$ of a sphere (of radius $R$, homogeneous, density $\rho$) inside an incompressible fluid (water, with density $\rho_W$, contained in a tube of radius $A \gg R$).1) In the first experiment, the fluid is contained in a tube placed vertically and closed
below, and the sphere is placed inside the tube at various distances below the upper surface of the liquid and left free to move. The sphere left free in the liquid rises to the upper surface; at equilibrium it floats leaving a spherical cap outside the liquid.

I measured the ascent times with a precision of one hundredth of a second (times) as a function of the distance below the surface of the liquid at which the sphere is left (distances). The uncertainty on the distances traveled by the sphere can be considered to be 1 cm. I have reported these values in a table:

$$\small \begin{array}{l|l}
\text{distance (cm)} & \text{time (s)}\\
\hline 100 & 1.77\\
200 & 2.85\\
300 & 3.87\\
400 & 4.86\\
500 & 5.87\\
\end{array}$$

Values are: $\rho_W = 1.000 \ \mathrm{g/cm^3}$; volume of the sphere $V_R = 500 \pm 15 \ \mathrm{cm^3}$; volume of the sphere outside water $V_C = 50 \pm 2 \ \mathrm{cm^3}$

My goal is to calculate (with some relative uncertainty) the viscosity coefficient $\alpha$. @Chestermiller addressed me to the literature on "Falling ball viscometry".

So, I wrote the equation of motion given by the presence of the buoyancy force, the weight of the sphere, and the viscous resistance $F = \alpha v$. Noting that the sphere reaches its terminal velocity after a few tenths of a second (the time it takes the sphere to accelerate), I assumed that the acceleration was zero and calculated the velocity as an approximation of the terminal velocity $v \approx \frac{F_{\text{buoyancy}}-F_{\text{weight}}}{\alpha}$, or alternatively, $v = \frac{2(\rho_{\text{sphere}}-\rho_{\text{liquid}})r^2 g}{9 \mu}$

But, as @Chestermiller noted, this works only assuming the pipe is horizontal, and that there is no buoyancy. Assuming the pipe is vertical only makes the problem harder. So, we would like to approach the problem from this point of view.

2) Subsequently, I varied the experiment by considering a reserve of the fluid to be contained in a container of cross-section
much greater than that of the tube considered above. The fluid discharges outwards through the tube (therefore, with an L shape) exiting from a circular opening of radius $a < A$ located at a distance $h$ below the level of the fluid in the container. (So, the total thickness of the tube is $2A$ and the total length of the opening is $2a$).

I considered the two hypotheses in which there is or is not the sphere held stationary inside the tube (in the middle). Obviously, there is a difference in velocity in the two cases. In addition to describing the difference from a qualitative point of view, I would like an estimate of the two velocities.
At the very beginning, I calculated the efflux velocity with Bernoulli's law in the case where there is no sphere, and used the continuity equation in the case where there is a sphere, adopting the theorem of mechanical energy and work of non-conservative forces (work of viscous friction) to find the efflux velocity.

But, as @Chestermiller pointed out to me, Bernoulli's equation works only for steady-state inviscid fluids. Admitting the presence of viscous resistance makes the fluid viscous. So, Bernoulli's equation is no longer applicable. It needs to be approached in another way, and @Chestermiller suggested lubrication theory. Besides that, we would like to look at different cases: creeping flow with low Reynolds number, turbulent flow with high Reynolds number, treat Reynolds number as a parameter.

 

[Chestermiller]

TL;DR Summary: Analyze, with the help of @Chestermiller, the importance of viscous interactions in a vertical geometry ("Falling ball viscometry") and in an horizontal viscous flow.

The following double experiment was conducted by me a few weeks ago. @Chestermiller and I discussed this experiment privately, and @Chestermiller advised me to open a thread on this Forum so that other users can also benefit from the discussion he and I will have on this topic.

Consider a fluid-dynamic resistance force of intensity $F$ ($\vec F = - \alpha \vec V$) opposing motion at velocity $V$ of a sphere (of radius $R$, homogeneous, density $\rho$) inside an incompressible fluid (water, with density $\rho_W$, contained in a tube of radius $A \gg R$).1) In the first experiment, the fluid is contained in a tube placed vertically and closed
below, and the sphere is placed inside the tube at various distances below the upper surface of the liquid and left free to move. The sphere left free in the liquid rises to the upper surface; at equilibrium it floats leaving a spherical cap outside the liquid.

I measured the ascent times with a precision of one hundredth of a second (times) as a function of the distance below the surface of the liquid at which the sphere is left (distances). The uncertainty on the distances traveled by the sphere can be considered to be 1 cm. I have reported these values in a table:

$$\small \begin{array}{l|l}
\text{distance (cm)} & \text{time (s)}\\
\hline 100 & 1.77\\
200 & 2.85\\
300 & 3.87\\
400 & 4.86\\
500 & 5.87\\
\end{array}$$

Values are: $\rho_W = 1.000 \ \mathrm{g/cm^3}$; volume of the sphere $V_R = 500 \pm 15 \ \mathrm{cm^3}$; volume of the sphere outside water $V_C = 50 \pm 2 \ \mathrm{cm^3}$

My goal is to calculate (with some relative uncertainty) the viscosity coefficient $\alpha$. @Chestermiller addressed me to the literature on "Falling ball viscometry".

So, I wrote the equation of motion given by the presence of the buoyancy force, the weight of the sphere, and the viscous resistance $F = \alpha v$. Noting that the sphere reaches its terminal velocity after a few tenths of a second (the time it takes the sphere to accelerate), I assumed that the acceleration was zero and calculated the velocity as an approximation of the terminal velocity $v \approx \frac{F_{\text{buoyancy}}-F_{\text{weight}}}{\alpha}$, or alternatively, $v = \frac{2(\rho_{\text{sphere}}-\rho_{\text{liquid}})r^2 g}{9 \mu}$

But, as @Chestermiller noted, this works only assuming the pipe is horizontal, and that there is no buoyancy. Assuming the pipe is vertical only makes the problem harder. So, we would like to approach the problem from this point of view.

I never said any of this. You must have mis-interpreted me for selection 1.

The equation you wrote is correct if the sphere radius is much smaller than the pipe radius, and the Reynolds number for the flow is very low (so that you essentially nave creeping flow). The following oar gives correction factors if these two conditions are not satisfied: https://www.hindawi.com/journals/itees/2022/3897120/ That's all that needs to be said about situation 1 for now.

 

[Chestermiller]

2) Subsequently, I varied the experiment by considering a reserve of the fluid to be contained in a container of cross-section
much greater than that of the tube considered above. The fluid discharges outwards through the tube (therefore, with an L shape) exiting from a circular opening of radius $a < A$ located at a distance $h$ below the level of the fluid in the container. (So, the total thickness of the tube is $2A$ and the total length of the opening is $2a$).

View attachment 337973

I considered the two hypotheses in which there is or is not the sphere held stationary inside the tube (in the middle). Obviously, there is a difference in velocity in the two cases. In addition to describing the difference from a qualitative point of view, I would like an estimate of the two velocities.
At the very beginning, I calculated the efflux velocity with Bernoulli's law in the case where there is no sphere, and used the continuity equation in the case where there is a sphere, adopting the theorem of mechanical energy and work of non-conservative forces (work of viscous friction) to find the efflux velocity.

But, as @Chestermiller pointed out to me, Bernoulli's equation works only for steady-state inviscid fluids. Admitting the presence of viscous resistance makes the fluid viscous. So, Bernoulli's equation is no longer applicable. It needs to be approached in another way, and @Chestermiller suggested lubrication theory. Besides that, we would like to look at different cases: creeping flow with low Reynolds number, turbulent flow with high Reynolds number, treat Reynolds number as a parameter.

Situation 2 is very different from situation 1. Assuming that the final converging section and sphere are not present, let's see your analysis of situation 2 for a highly viscous fluid (negligible inertia).

 

[HighPhy]

I never said any of this. You must have mis-interpreted me for selection 1.

The equation you wrote is correct if the sphere radius is much smaller than the pipe radius, and the Reynolds number for the flow is very low (so that you essentially nave creeping flow). The following oar gives correction factors if these two conditions are not satisfied: https://www.hindawi.com/journals/itees/2022/3897120/ That's all that needs to be said about situation 1 for now.

You are quite correct. I misinterpreted your statements. Let me offer my apologies for that.

 

[HighPhy]

Situation 2 is very different from situation 1. Assuming that the final converging section and sphere are not present, let's see your analysis of situation 2 for a highly viscous fluid (negligible inertia).

I will try the following. Please take what I write with a grain of salt, as I have no expertise on the subject.

Consider the steady flow of fluid of constant density in fully developed flow through a horizontal pipe. Visualize a disk shaped element, concentric with the access of the tube, of radius $r$ and length $\mathrm{d}L$.

The fluid pressure acting on upward stream is $P_1 = P$, the fluid pressure acting on downward stream is $P_2 = -(P + \mathrm{d}P)$. Since the fluid is possessing viscosity, we have to assume that shear stress will exist on the rim of element $\sigma$. Since pipe is horizontal, $x_1 = x_2$. Since flow is fully developed, we have $V_1 \ \text{upstream} = V_2 \ \text{downstream}$

The resultant of all the forces acting on the fluid element would be zero, i.e. $\sum F = 0$, from which: $$P_1 \ s_1 - P_2 \ s_2 + F_{\text{stress}} - F_{\text{weight}} = 0 \tag 1$$

where $P_1 \ s_1 = P \pi r^2$ is the force acting upward the stream, $P_2 \ s_2 = (P + \mathrm{d}P) \pi r^2$ is the force acting downward the stream, $F_{\text{stress}}$ is the force acting on the stream due to wall, i.e. shear stress on the rim of the element. I calculated it as $$F_{\text{stress}} = - (2 \pi r \mathrm{d}L)\sigma$$

Now substituting the values and dividing by $\mathrm{dL}\pi r^2$ eq. $(1)$ we get

$$\frac{\mathrm{d}P}{\mathrm{d}L} + \frac{2 \sigma}{r} = 0 \tag 2$$

Pressure drop will occur due to skin friction, therefore $\mathrm{dp} = \Delta P$ and $\mathrm{dL} = L$. Eq. $(2)$ becomes: $$\Delta P = - \frac{2L \sigma}{r} \tag 3$$

Now to calculate friction we use Bernoulli's equation, combining it to the fact that the pipe is horizontal and the flow is fully developed, I get:

$$\frac{P_1}{\rho} = \frac{P_2}{\rho} + h \tag 4$$

Since $\Delta P = P_1 - P_2$, we have $P_2 = P_1 - \Delta P \qquad (5)$

Substituting $(5)$ in $(4)$, I get: $$\frac{P_1}{\rho} = -\frac{\Delta P}{\rho} + \frac{P_1}{\rho} + h \tag 6$$ so, putting $(3)$: $$\frac{\Delta P}{\rho} = h = - \frac{2L \sigma}{\rho r}$$ from which $$\sigma = - \frac{\Delta P r}{2L} \tag 7$$

The fluid will possess some velocity, hence considering a thin ring with radius $R$ and width $\mathrm{d}R$, area is given by $\mathrm{d}s = 2 \pi R \mathrm{d}R$

Using the definition of viscosity:

$$\frac{\mathrm{d}u}{\mathrm{d}R} = - \frac{\sigma}{\mu r}R \tag 8$$

Integrating from $0$ to $u$ and from $r$ to $R$: $$u = \frac{\sigma (r^2-R^2)}{2 r \mu} \tag 9$$

The average velocity is given by: $$v = \frac{1}{s} \int u \mathrm{d}s \tag {10}$$

Substituting $(8)$ and $(9)$ in $(10)$, I get: $$v = \frac{\sigma r}{4 \mu} \tag {11}$$

Substituting $(7)$ in $(11)$: $$v = - \frac{\Delta P r^2}{8 L \mu}$$

Setting $r = \frac{D}{2}$, we have: $$v = - \frac{\Delta P D^2}{32 L \mu}$$

This equation corresponds to the one you sent me privately, but the proof does not seem very precise to me. I would therefore ask you to rectify my incorrectness and propose your (certainly more accurate) solution. I would also like literature reference where this is shown in a more thorough manner.

If your request meant another solution or derivation, feel free to correct me and begin over.

 

[Chestermiller]

I will try the following. Please take what I write with a grain of salt, as I have no expertise on the subject.

Consider the steady flow of fluid of constant density in fully developed flow through a horizontal pipe. Visualize a disk shaped element, concentric with the access of the tube, of radius $r$ and length $\mathrm{d}L$.

The fluid pressure acting on upward stream is $P_1 = P$, the fluid pressure acting on downward stream is $P_2 = -(P + \mathrm{d}P)$. Since the fluid is possessing viscosity, we have to assume that shear stress will exist on the rim of element $\sigma$. Since pipe is horizontal, $x_1 = x_2$. Since flow is fully developed, we have $V_1 \ \text{upstream} = V_2 \ \text{downstream}$

The resultant of all the forces acting on the fluid element would be zero, i.e. $\sum F = 0$, from which: $$P_1 \ s_1 - P_2 \ s_2 + F_{\text{stress}} - F_{\text{weight}} = 0 \tag 1$$

where $P_1 \ s_1 = P \pi r^2$ is the force acting upward the stream, $P_2 \ s_2 = (P + \mathrm{d}P) \pi r^2$ is the force acting downward the stream, $F_{\text{stress}}$ is the force acting on the stream due to wall, i.e. shear stress on the rim of the element. I calculated it as $$F_{\text{stress}} = - (2 \pi r \mathrm{d}L)\sigma$$

Now substituting the values and dividing by $\mathrm{dL}\pi r^2$ eq. $(1)$ we get

$$\frac{\mathrm{d}P}{\mathrm{d}L} + \frac{2 \sigma}{r} = 0 \tag 2$$

Pressure drop will occur due to skin friction, therefore $\mathrm{dp} = \Delta P$ and $\mathrm{dL} = L$. Eq. $(2)$ becomes: $$\Delta P = - \frac{2L \sigma}{r} \tag 3$$

Now to calculate friction we use Bernoulli's equation, combining it to the fact that the pipe is horizontal and the flow is fully developed, I get:

$$\frac{P_1}{\rho} = \frac{P_2}{\rho} + h \tag 4$$

Since $\Delta P = P_1 - P_2$, we have $P_2 = P_1 - \Delta P \qquad (5)$

Substituting $(5)$ in $(4)$, I get: $$\frac{P_1}{\rho} = -\frac{\Delta P}{\rho} + \frac{P_1}{\rho} + h \tag 6$$ so, putting $(3)$: $$\frac{\Delta P}{\rho} = h = - \frac{2L \sigma}{\rho r}$$ from which $$\sigma = - \frac{\Delta P r}{2L} \tag 7$$

The fluid will possess some velocity, hence considering a thin ring with radius $R$ and width $\mathrm{d}R$, area is given by $\mathrm{d}s = 2 \pi R \mathrm{d}R$

Using the definition of viscosity:

$$\frac{\mathrm{d}u}{\mathrm{d}R} = - \frac{\sigma}{\mu r}R \tag 8$$

Integrating from $0$ to $u$ and from $r$ to $R$: $$u = \frac{\sigma (r^2-R^2)}{2 r \mu} \tag 9$$

The average velocity is given by: $$v = \frac{1}{s} \int u \mathrm{d}s \tag {10}$$

Substituting $(8)$ and $(9)$ in $(10)$, I get: $$v = \frac{\sigma r}{4 \mu} \tag {11}$$

Substituting $(7)$ in $(11)$: $$v = - \frac{\Delta P r^2}{8 L \mu}$$

Setting $r = \frac{D}{2}$, we have: $$v = - \frac{\Delta P D^2}{32 L \mu}$$

This equation corresponds to the one you sent me privately, but the proof does not seem very precise to me. I would therefore ask you to rectify my incorrectness and propose your (certainly more accurate) solution. I would also like literature reference where this is shown in a more thorough manner.

If your request meant another solution or derivation, feel free to correct me and begin over.

This derivation looks pretty good. You can find the detailed derivation in Transport phenomena based on an annular shell balance. Your final answer is correct. The other relation for your diagram is $-\Delta P=mgh$

 

[HighPhy]

This derivation looks pretty good. You can find the detailed derivation in Transport phenomena based on an annular shell balance. Your final answer is correct. The other relation for your diagram is $-\Delta P=mgh$

Good. I'm glad to hear that.

If I understand correctly, I could rewrite the relationship as $v = \dfrac{mghD^2}{32L \mu}$, where $m$ is the mass of fluid involved and $D=2A$ in my scenario (?). Just for confirmation.
A few days ago you privately sent me the equation $-gh+\dfrac{v^2}{2}=-\dfrac{32Lv\mu}{\rho D^2}$, corresponding to the one I wrote. This is a quadratic in $v$. How does it fit with the previous expressions?

I guess this works for creeping laminar flow at low Reynolds numbers. But how to continue?

 

[Chestermiller]

Good. I'm glad to hear that.

If I understand correctly, I could rewrite the relationship as $v = \dfrac{mghD^2}{32L \mu}$, where $m$ is the mass of fluid involved and $D=2A$ in my scenario (?). Just for confirmation.
A few days ago you privately sent me the equation $-gh+\dfrac{v^2}{2}=-\dfrac{32Lv\mu}{\rho D^2}$, corresponding to the one I wrote. This is a quadratic in $v$. How does it fit with the previous expressions?

I guess this works for creeping laminar flow at low Reynolds numbers. But how to continue?

First let me correct myself. I meant that $$\Delta P=-\rho g h $$where $\rho$ is the density of the fluid. D is the diameter of the tube.

The equation I wrote with $$\frac{v^2}{2}$$ includes the inertial contribution to accelerate the fluid down the tube. The equation I wrote without the term is for negligible inertia.

The point is that, even for a constant diameter tube without an end contraction, the pressure drop is not zero (as predicted by Bernoulli.)

You say, "how to continue?" What would you want to include next?

 

[HighPhy]

First let me correct myself. I meant that $$\Delta P=-\rho g h $$where $\rho$ is the density of the fluid. D is the diameter of the tube.

The equation I wrote with $$\frac{v^2}{2}$$ includes the inertial contribution to accelerate the fluid down the tube. The equation I wrote without the term is for negligible inertia.

The point is that, even for a constant diameter tube without an end contraction, the pressure drop is not zero (as predicted by Bernoulli.)

You say, "how to continue?" What would you want to include next?

I would like to continue with the selections we talked about: turbulent flow with high Reynolds number and treat Reynolds number as a parameter.

Of course, we should also examine the case where there is the sphere. Situation which, as you said, is more difficult than the one we solved.

 

[Chestermiller]

I would like to continue with the selections we talked about: turbulent flow with high Reynolds number and treat Reynolds number as a parameter.

For selection 2, without the sphere present, the analysis for turbulent flow can be derived from the development in Transport Phenomena, Chapter 7. Let's see what you come up with.

 

[HighPhy]

For selection 2, without the sphere present, the analysis for turbulent flow can be derived from the development in Transport Phenomena, Chapter 7. Let's see what you come up with.

I went through Chapter 7 of 'Transport Phenomena', not before reading parts of Chapter 5. This book is so detailed and full of information that I really struggled. I could not digest all the information that is written in the whole chapter. I am not very satisfied with what I have been able to produce, but I will try.

The mechanical energy balance is

$$\frac{v^2}{2}+ \frac{\Delta P}{\rho} + h_f = 0 \tag 1$$

where $h_f$ is the term contributing to the head loss due to friction from the pipe wall and $\Delta P = - \rho g h \qquad (1.1)$

The Darcy equation gives the friction head loss for flow in a straight pipe: $$h_f = 2 \frac{fL}{D} v^2 \tag 2$$ where $f$ is a dimensionless friction factor and $D$ is the diameter of the pipe.

Substituting $(1.1)$ and $(2)$ into equation $(1)$: $$-gh + \dfrac{v^2}{2} + 2 \dfrac{fL}{D} v^2 = 0 \tag 3$$

We can obtain an estimate of the friction factor $f$ using the empirical relation known as Blasius equation, applicable to smooth pipe and turbulent flow with Reynolds numbers in the range of $4,000 < Re < 100,000$

$$f = \dfrac{791}{100 \ Re^{\frac{1}{4}}} \tag 4$$

where $Re = \frac{\rho v D}{\mu} \qquad (4.1)$

So, substituting $(4.1)$ in $(4)$, and $(4)$ in $(3)$, we have:

$$-gh + \dfrac{v^2}{2} + \dfrac{791 \ L \mu^{\frac{1}{4}}}{50 \ D^{\frac{5}{4}} \rho^{\frac{1}{4}}} v^{\frac{7}{2}} = 0$$

I do not believe that this derivation and the final result are correct. I therefore explicitly request your assistance.

 

[Chestermiller]

I went through Chapter 7 of 'Transport Phenomena', not before reading parts of Chapter 5. This book is so detailed and full of information that I really struggled. I could not digest all the information that is written in the whole chapter. I am not very satisfied with what I have been able to produce, but I will try.

The mechanical energy balance is

$$\frac{v^2}{2}+ \frac{\Delta P}{\rho} + h_f = 0 \tag 1$$

where $h_f$ is the term contributing to the head loss due to friction from the pipe wall and $\Delta P = - \rho g h \qquad (1.1)$

The Darcy equation gives the friction head loss for flow in a straight pipe: $$h_f = 2 \frac{fL}{D} v^2 \tag 2$$ where $f$ is a dimensionless friction factor and $D$ is the diameter of the pipe.

Substituting $(1.1)$ and $(2)$ into equation $(1)$: $$-gh + \dfrac{v^2}{2} + 2 \dfrac{fL}{D} v^2 = 0 \tag 3$$

We can obtain an estimate of the friction factor $f$ using the empirical relation known as Blasius equation, applicable to smooth pipe and turbulent flow with Reynolds numbers in the range of $4,000 < Re < 100,000$

$$f = \dfrac{791}{100 \ Re^{\frac{1}{4}}} \tag 4$$

where $Re = \frac{\rho v D}{\mu} \qquad (4.1)$

So, substituting $(4.1)$ in $(4)$, and $(4)$ in $(3)$, we have:

$$-gh + \dfrac{v^2}{2} + \dfrac{791 \ L \mu^{\frac{1}{4}}}{50 \ D^{\frac{5}{4}} \rho^{\frac{1}{4}}} v^{\frac{7}{2}} = 0$$

I do not believe that this derivation and the final result are correct. I therefore explicitly request your assistance.

In Eqn. 4, the 791/100 should be 791/1000. Otherwise, if you did the rest of the algebra correctly, the remainder of the development is correct. Very nicely done. Congrats. Cudos on mastering the information in Chapters 6 and 7. You seem to have a knack for fluid dynamics.

You can now see how, under many circumstances (particularly turbulent flow), the inviscid Bernoulli equation can give highly incorrect results.

 

[HighPhy]

In Eqn. 4, the 791/100 should be 791/1000. Otherwise, if you did the rest of the algebra correctly, the remainder of the development is correct. Very nicely done. Congrats. Cudos on mastering the information in Chapters 6 and 7. You seem to have a knack for fluid dynamics.

You can now see how, under many circumstances (particularly turbulent flow), the inviscid Bernoulli equation can give highly incorrect results.

You are making me blush. I didn't think I would do this selection correctly as well.

You are absolutely right about the numerical correction. Good catch. I am going to correct my mistake.

I think now the good stuff will come. Still missing to analyze the selection in a way that treats the Reynolds number based on the average velocity of the fluid relative to the sphere and based on the sphere diameter as a parameter (as you said).
More importantly, the case in the presence of the sphere, for which you promised to show lubrication theory. I foresee something quite difficult.

 

[HighPhy]

I went through Chapter 7 of 'Transport Phenomena', not before reading parts of Chapter 5. This book is so detailed and full of information that I really struggled. I could not digest all the information that is written in the whole chapter. I am not very satisfied with what I have been able to produce, but I will try.

The mechanical energy balance is

$$\frac{v^2}{2}+ \frac{\Delta P}{\rho} + h_f = 0 \tag 1$$

where $h_f$ is the term contributing to the head loss due to friction from the pipe wall and $\Delta P = - \rho g h \qquad (1.1)$

The Darcy equation gives the friction head loss for flow in a straight pipe: $$h_f = 2 \frac{fL}{D} v^2 \tag 2$$ where $f$ is a dimensionless friction factor and $D$ is the diameter of the pipe.

Substituting $(1.1)$ and $(2)$ into equation $(1)$: $$-gh + \dfrac{v^2}{2} + 2 \dfrac{fL}{D} v^2 = 0 \tag 3$$

We can obtain an estimate of the friction factor $f$ using the empirical relation known as Blasius equation, applicable to smooth pipe and turbulent flow with Reynolds numbers in the range of $4,000 < Re < 100,000$

$$f = \dfrac{791}{1000 \ Re^{\frac{1}{4}}} \tag 4$$

where $Re = \frac{\rho v D}{\mu} \qquad (4.1)$

So, substituting $(4.1)$ in $(4)$, and $(4)$ in $(3)$, we have:

$$-gh + \dfrac{v^2}{2} + \dfrac{791 \ L \mu^{\frac{1}{4}}}{500 \ D^{\frac{5}{4}} \rho^{\frac{1}{4}}} v^{\frac{7}{2}} = 0$$

This is the correct result.

 

[Chestermiller]

You are making me blush. I didn't think I would do this selection correctly as well.

You are absolutely right about the numerical correction. Good catch. I am going to correct my mistake.

I think now the good stuff will come. Still missing to analyze the selection in a way that treats the Reynolds number based on the average velocity of the fluid relative to the sphere and based on the sphere diameter as a parameter (as you said).
More importantly, the case in the presence of the sphere, for which you promised to show lubrication theory. I foresee something quite difficult.

Yes. I'm going to help you work through the lubrication approach by gradually leading you through it.

The analysis starts, surprisingly, with pressure driven flow between two infinite parallel plates. The development is very similar to what you did for viscous flow in a tube, based on the information in the first couple of chapters in Transport Phenomena. Suppose you have a fluid of viscosity $\mu$ sandwiched between two infinite parallel plates, and you impose a pressure gradient in the z direction, parallel to the plates. Let h be the distance between the plates. Express the average fluid velocity in the z direction as a function of the pressure gradient, the viscosity, and the distance h.

 

[Fred Wright]

Many years ago when I was an undergraduate student there was a commercial on tv advertising a shampoo called "Prell". On the commercial a lady dropped a pearl in to a tube of Prell and watched it slowly descend. She called it "The Prell pearl test". One day I came to lab class and the instructor asked us, "Are you guys ready to take the Prell test?" Although we hadn't yet taken fluid dynamics we were challenged to compute the viscosity of Prell. Sadly my analysis fell rather short but thankyou for reminding me of simpler days.

 

[Chestermiller]

Answer to my question in post #15: $$\frac{dp}{dx}=-\frac{12\mu v}{h^2}$$

 

[HighPhy]

Answer to my question in post #15: $$\frac{dp}{dx}=-\frac{12\mu v}{h^2}$$

Sorry for not giving more news. I've been quite busy. I will try to work out a derivation these days to submit it for your consideration.

 

[HighPhy]

Yes. I'm going to help you work through the lubrication approach by gradually leading you through it.

The analysis starts, surprisingly, with pressure driven flow between two infinite parallel plates. The development is very similar to what you did for viscous flow in a tube, based on the information in the first couple of chapters in Transport Phenomena. Suppose you have a fluid of viscosity $\mu$ sandwiched between two infinite parallel plates, and you impose a pressure gradient in the z direction, parallel to the plates. Let h be the distance between the plates. Express the average fluid velocity in the z direction as a function of the pressure gradient, the viscosity, and the distance h.

Let's try. The following derivation is the result of my reading of the first two chapters of Transport PhenomenaBoth plates are at rest and the flow is caused by a pressure gradient $\dfrac{\mathrm{d}p}{\mathrm{d}z}$ in the direction $z$ parallel to the plates. Assuming that the only non-zero component of the velocity is $u_z$ and that the velocity and pressure are independent of time the resulting continuity equation for an incompressible fluid yields
$$\frac{\partial u_z}{\partial z} = 0 \tag 1$$
so that $u_z(y)$ is a function only of $y$, the coordinate perpendicular to the plates.

Using this the planar Navier-Stokes equations for an incompressible fluid of constant and uniform viscosity reduce to
$$\frac{\partial u_z}{\partial z} = \mu \frac{\partial^2}{\partial y^2} u_z \tag 2$$ $$\frac{\partial p}{\partial y} = 0 \tag 3$$
Eq. $(3)$ shows that the pressure $p(z)$ is a function only of $z$ and hence the gradient $\dfrac{\mathrm{d}p}{\mathrm{d}z}$ allows eq. $(2)$ to be integrated so that the velocity $u_z$, can be written as:
$$u_z = \frac{1}{\mu}\frac{\mathrm{d}p}{\mathrm{d}z} \frac{y^2}{2}+ C_1y + C_2 \tag 4$$
where $C_1$ and $C_2$ are integration constants to be determined by the application of the boundary conditions at the two plates.

Applying the no-slip conditions at the lower and upper walls, namely $$(u_z)_{y=0} = 0 \ \text{and} \ (u_z)_{y=h} = 0 \tag 5$$
yields
$$C_2 = 0 \ \text{and} \ C_1 = −\frac{h}{2\mu} \frac{\mathrm{d}p}{\mathrm{d}z} \tag 6$$
and so we would have
$$u_z = -\frac{1}{\mu}\frac{\mathrm{d}p}{\mathrm{d}z} \frac{y}{2} (h-y) \tag 7$$
The average velocity $\left\langle v \right\rangle$ of the flow is
$$\left\langle v \right\rangle = \frac{1}{h} \underbrace{\int_{0}^{h} u_z \mathrm{d}y}_{\dot{Q}} \tag 8$$
where $\int_{0}^{h} u_z \mathrm{d}y$ is the volume flow rate $\dot {Q}$

Computing the integral, we'll have $$\int_{0}^{h} u_z \mathrm{d}y = - \frac{1}{\mu}\frac{\mathrm{d}p}{\mathrm{d}z}\int_{0}^{h} \frac{y}{2} (h-y) \mathrm{d}y = - \frac{h^3}{12\mu}\frac{\mathrm{d}p}{\mathrm{d}z} \tag 9$$
Plugging $(9)$ in $(8)$:
$$\left\langle v \right\rangle = - \frac{1}{\cancel{h}} \frac{h^{\cancel{3} 2}}{12\mu}\frac{\mathrm{d}p}{\mathrm{d}z}$$
So: $$\left\langle v \right\rangle = - \frac{h^2}{12\mu}\frac{\mathrm{d}p}{\mathrm{d}z}$$

This result is the same as yours, but I am not sure of my derivation. As usual, I would like bibliographical references where I can find a more accurate derivation.

I await your reply on my results and your tips as to how to proceed going forward.

 

[Chestermiller]

Let's try. The following derivation is the result of my reading of the first two chapters of Transport PhenomenaBoth plates are at rest and the flow is caused by a pressure gradient $\dfrac{\mathrm{d}p}{\mathrm{d}z}$ in the direction $z$ parallel to the plates. Assuming that the only non-zero component of the velocity is $u_z$ and that the velocity and pressure are independent of time the resulting continuity equation for an incompressible fluid yields
$$\frac{\partial u_z}{\partial z} = 0 \tag 1$$
so that $u_z(y)$ is a function only of $y$, the coordinate perpendicular to the plates.

Using this the planar Navier-Stokes equations for an incompressible fluid of constant and uniform viscosity reduce to
$$\frac{\partial u_z}{\partial z} = \mu \frac{\partial^2}{\partial y^2} u_z \tag 2$$ $$\frac{\partial p}{\partial y} = 0 \tag 3$$
Eq. $(3)$ shows that the pressure $p(z)$ is a function only of $z$ and hence the gradient $\dfrac{\mathrm{d}p}{\mathrm{d}z}$ allows eq. $(2)$ to be integrated so that the velocity $u_z$, can be written as:
$$u_z = \frac{1}{\mu}\frac{\mathrm{d}p}{\mathrm{d}z} \frac{y^2}{2}+ C_1y + C_2 \tag 4$$
where $C_1$ and $C_2$ are integration constants to be determined by the application of the boundary conditions at the two plates.

Applying the no-slip conditions at the lower and upper walls, namely $$(u_z)_{y=0} = 0 \ \text{and} \ (u_z)_{y=h} = 0 \tag 5$$
yields
$$C_2 = 0 \ \text{and} \ C_1 = −\frac{h}{2\mu} \frac{\mathrm{d}p}{\mathrm{d}z} \tag 6$$
and so we would have
$$u_z = -\frac{1}{\mu}\frac{\mathrm{d}p}{\mathrm{d}z} \frac{y}{2} (h-y) \tag 7$$
The average velocity $\left\langle v \right\rangle$ of the flow is
$$\left\langle v \right\rangle = \frac{1}{h} \underbrace{\int_{0}^{h} u_z \mathrm{d}y}_{\dot{Q}} \tag 8$$
where $\int_{0}^{h} u_z \mathrm{d}y$ is the volume flow rate $\dot {Q}$

Computing the integral, we'll have $$\int_{0}^{h} u_z \mathrm{d}y = - \frac{1}{\mu}\frac{\mathrm{d}p}{\mathrm{d}z}\int_{0}^{h} \frac{y}{2} (h-y) \mathrm{d}y = - \frac{h^3}{12\mu}\frac{\mathrm{d}p}{\mathrm{d}z} \tag 9$$
Plugging $(9)$ in $(8)$:
$$\left\langle v \right\rangle = - \frac{1}{\cancel{h}} \frac{h^{\cancel{3} 2}}{12\mu}\frac{\mathrm{d}p}{\mathrm{d}z}$$
So: $$\left\langle v \right\rangle = - \frac{h^2}{12\mu}\frac{\mathrm{d}p}{\mathrm{d}z}$$

This result is the same as yours, but I am not sure of my derivation. As usual, I would like bibliographical references where I can find a more accurate derivation.

I await your reply on my results and your tips as to how to proceed going forward.

This looks correct, except for Eqn. 2, which has a partial with respect to z of p rather than $u_z$ on the left hand side. Nice, nice, job.

Next, suppose that the volumetric flow rate per unit width of channel is uniform at q=vh. In terms of q, what does this equation become/

Suppose, rather than parallel plates, we have two concentric cylinders of radii $\bar{r}-\frac{h}{2}$ and $\bar{r}+\frac{h}{2}$, where $h<<\bar{r}$, so that we can neglect the curvature of the cylinders. Based on the previous results, what is the pressure gradient along the cylinder axis in terms of the total volumetric flow rate Q, the mean radius $\bar{r}$, the gap h, and the viscosity $\mu$.

 

[HighPhy]

This looks correct, except for Eqn. 2, which has a partial with respect to z of p rather than $u_z$ on the left hand side. Nice, nice, job.

Next, suppose that the volumetric flow rate per unit width of channel is uniform at q=vh. In terms of q, what does this equation become/

Suppose, rather than parallel plates, we have two concentric cylinders of radii $\bar{r}-\frac{h}{2}$ and $\bar{r}+\frac{h}{2}$, where $h<<\bar{r}$, so that we can neglect the curvature of the cylinders. Based on the previous results, what is the pressure gradient along the cylinder axis in terms of the total volumetric flow rate Q, the mean radius $\bar{r}$, the gap h, and the viscosity $\mu$.

Could you formulate slightly better?
Sorry, I didn't understand it properly...

 

[Chestermiller]

The volume flow rate per unit width of parallel plate channel is $q=\left \langle v\right\rangle h$, so
$$q = - \frac{h^3}{12\mu}\frac{\mathrm{d}p}{\mathrm{d}z}$$

For axial annular flow in which the radii of the two cylinders comprising the annulus are nearly equal, the volumetric flow rate axially is $Q=2\pi Rq$, where R is the outer radius (nearly equal to the inner radius) and q is the axial flow per unit (width) of circumference . So, for such axial annular flow,
$$Q = -2\pi R \frac{h^3}{12\mu}\frac{\mathrm{d}p}{\mathrm{d}z}$$

OK so far?

 

[HighPhy]

The volume flow rate per unit width of parallel plate channel is $q=\left \langle v\right\rangle h$, so
$$q = - \frac{h^3}{12\mu}\frac{\mathrm{d}p}{\mathrm{d}z}$$

For axial annular flow in which the radii of the two cylinders comprising the annulus are nearly equal, the volumetric flow rate axially is $Q=2\pi Rq$, where R is the outer radius (nearly equal to the inner radius) and q is the axial flow per unit (width) of circumference . So, for such axial annular flow,
$$Q = -2\pi R \frac{h^3}{12\mu}\frac{\mathrm{d}p}{\mathrm{d}z}$$

OK so far?

OK.

 

[Chestermiller]

OK.

Suppose, rather than an inner cylinder, we have an axisymmetric body having a radius nearly equal to that of the outer cylinder, but for which the clearance between the body and the outer cylinder is h=h(z) (rather than a constant h in the case of the annulus). Then we can write $$Q = -2\pi R \frac{h(z)^3}{12\mu}\frac{\mathrm{d}p}{\mathrm{d}z}$$Still OK?

 

[Chestermiller]

Let $h_0$ represent the small minimum clearance between the sphere (in our problem) and the cylinder. Let z = 0 represent the axial coordinate of the center of the sphere. Show that, for axial locations on either side of the sphere close to this center location, the clearance can be approximated by the parabolic shape $$h(z)=h_0+\frac{z^2}{2R}$$

 

[Chestermiller]

Incidentally, if you are interested in falling ball viscometry, shouldn't the sphere be stationary, the wall of the cylinder moving axially, and the fluid doing whatever it needs to in order to match these boundary conditions?

 

[HighPhy]

Incidentally, if you are interested in falling ball viscometry, shouldn't the sphere be stationary, the wall of the cylinder moving axially, and the fluid doing whatever it needs to in order to match these boundary conditions?

Yes. I think that's part of the lubrication approach you're leading me to, right?

 

[Chestermiller]

Yes. I think that's part of the lubrication approach you're leading me to, right?

You posed the problem of a stationary sphere in a stationary pipe, with flow past the sphere. My understanding was that I was helping you to solve that. I was not thinking in terms of a moving pipe wall, but your title for the thread suggests a moving wall.. I'm not sure my entry into the problem (i.e., starting with simpler models) would have been the same. How do you want to proceed (moving pipe wall or continue with stationary pipe wall and sphere)?

 

[HighPhy]

You posed the problem of a stationary sphere in a stationary pipe, with flow past the sphere. My understanding was that I was helping you to solve that. I was not thinking in terms of a moving pipe wall, but your title for the thread suggests a moving wall.. I'm not sure my entry into the problem (i.e., starting with simpler models) would have been the same. How do you want to proceed (moving pipe wall or continue with stationary pipe wall and sphere)?

I would say to continue with stationary pipe wall and sphere and arrive at a moving pipe wall after finishing the full analysis of the previous selection. What do you think?

 

[Chestermiller]

I would say to continue with stationary pipe wall and sphere and arrive at a moving pipe wall after finishing the full analysis of the previous selection. What do you think?

Do you know what to do next?

 

[HighPhy]

Do you know what to do next?

I have to show the following, right?

Let $h_0$ represent the small minimum clearance between the sphere (in our problem) and the cylinder. Let z = 0 represent the axial coordinate of the center of the sphere. Show that, for axial locations on either side of the sphere close to this center location, the clearance can be approximated by the parabolic shape $$h(z)=h_0+\frac{z^2}{2R}$$

 

[Chestermiller]

I have to show the following, right?

yes.