Flow rate experiment not producing same results as the Hagen-Poiseuille law?

[mahdis]

Homework Statement

Hello I was asked to design an experiment which showcases the Hagen-Poiseuille law. The law states that a shorter length of a tube should increase my flow rate. Why is it when I conduct my experiment comparing the length of 150 cm tube to 90 cm and 30 cm it is the longer length that has the fastest flow rate? The experiment design is a bag filled with 1 liter of water on the top, connected to a 4mm radius pipe which is hung vertically the length being compared is of the radius pipe. My experiment shows me that my 150 cm pipe has the fastest flow rate under gravity flowing at 313ml/min and my 90 cm flows at 238 ml/min and the 30 cm which is the slowest runs at 120 m/min

Relevant Equations

why is the shorter length in the experiment producing a faster flow rate despite the Hagen-Poiseuille law claiming otherwise

I have repeated the experiment several times and it always shows my longer length has the fastest flow rate. the same results occur when I apply 300mmH pressure to the fluid bag on top, the flow rate is still faster in the longer pipe

 

[haruspex]

"pipe which is hung vertically"

I can think of a reason that might affect matters.

 

[mahdis]

"pipe which is hung vertically"

I can think of a reason that might affect matters.

Please explain why flow down a vertical pipe would be slower if the length is shorter

 

[Orodruin]

Please explain why flow down a vertical pipe would be slower if the length is shorter

Consider the case of a zero viscosity liquid. What forces would act on a liquid element in the pipe?

 

[mahdis]

Consider the case of a zero viscosity liquid. What forces would act on a liquid element in the pipe?

if we say resistance then it should be increased with a longer length of a pipe (R=8nL/piR^4) but here the results are showing that the flow is faster with a longer pipe

 

[Orodruin]

if we say resistance then it should be increased with a longer length of a pipe (R=8nL/piR^4) but here the results are showing that the flow is faster with a longer pipe

I just told you the liquid had zero viscosity …

 

[mahdis]

gravity, radius, pressure and length?

 

[Chestermiller]

For flow in a vertical tube, the Hagen Poiseuille law becomes $$P_{top}+\rho g L=\frac{128 Q\eta}{\pi D^4}L$$where ##P_{top}## is the gauge pressure at the top of the tube (bottom of bag) and ##\eta## is the fluid viscosity. How does the pressure at the bottom of the bag change when you increase the length?

 

[Chestermiller]

Are the velocities you measured the average over the time it took to empty the bag?

 

[Chestermiller]

If you calculate the Reynolds number for all three of your flows, you find that it is in the range 30,000 to 80,000. The transition from laminar flow to turbulent flow occurs at a Reynolds number of 2100, so your 3 flows are all turbulent. The Hagen Poiseuille equation applies only to laminar flow.

Try using the Blasius equation to determine the turbulent friction factor and then the pressure drop for turbulent flow at the observed flow rates.

 

[kuruman]

If you calculate the Reynolds number for all three of your flows, you find that it is in the range 30,000 to 80,000. The transition from laminar flow to turbulent flow occurs at a Reynolds number of 2100, so your 3 flows are all turbulent. The Hagen Poiseuille equation applies only to laminar flow.

If that's the case and OP says

Hello I was asked to design an experiment which showcases the Hagen-Poiseuille law.

then doesn't it follow that a different design is needed for this experiment?

 

[erobz]

It's not clear that the OP understands what is happening in this experiment with the vertical tube, or the EQ provided (because the haven't indicated so). If they are studying this equation, it stands to reason they would have examined Bernoulli's, and/or The First Law of Thermo. as it pertains to incompressible flows with viscous losses?

It should be of some help to just look at a simplified steady state version of the experiment (approximately) to see why the longer pipe is experiencing the higher flowrate.

In engineering we write these equations is terms of head. In going from (1) to (2), with ##z_1## being zero elevation datum:

$$ \frac{P_1}{\rho g} + z_1 + \alpha_1 \frac{V_1^2}{2g} =\frac{P_2}{\rho g} + z_2 + \alpha_2 \frac{V_2^2}{2g} + \beta L \frac{V_2^2}{2g}$$

Simplifying via assumptions:

$$ \cancel{\frac{P_1}{\rho g}}^0 + \cancel{z_1}^0 + \cancel{ \alpha_1 \frac{V_1^2}{2g}}^{V_1 \ll V_2} =\cancel{\frac{P_2}{\rho g}}^0 + z_2 + \alpha_2\frac{V_2^2}{2g} + \beta L \frac{V_2^2}{2g}$$

Here ##z_2 = - ( z+L)##, moving to the LHS:

$$ z+ L = \alpha_2 \frac{V_2^2}{2g} + \beta L \frac{V_2^2}{2g}$$

Where ##\beta## is a constant for a particular pipe geometry.

If the total volumetric flowrate is ##Q = V_2 A ##, solving for ##Q^2##:

EDIT: there is supposed to be a kinetic head correction factor ## \alpha_2 = 2 ## applied to ##\frac{V_2^2}{2g}## for laminar flow, thus( factoring the 2 into the numerator):

$$Q^2 = g A^2 z \frac{2+\frac{2}{z}L}{2 + \beta L}$$

From your experimental results it would seem that ## \frac{2}{z} > \beta## in the setup.I hope that elaborates on why your experimental set up is not quite working out like you had hoped, as all the others have suggested...the vertical pipe is an issue. Also, as @Chestermiller notes you've exceeded the Reynolds number where the linear proportionality (Hagen-Poiseuille) ##f = \beta L ## holds.

I would parrot what @kuruman suggests and design a new experiment, so that things fall into line.

 

[Chestermiller]

It's not clear that the OP understands what is happening in this experiment with the vertical tube, or the EQ provided (because the haven't indicated so). If they are studying this equation, it stands to reason they would have examined Bernoulli's, and/or The First Law of Thermo. as it pertains to incompressible flows with viscous losses?

It should be of some help to just look at a simplified steady state version of the experiment (approximately) to see why the longer pipe is experiencing the higher flowrate.

View attachment 336575In engineering we write these equations is terms of head. In going from (1) to (2), with ##z_1## being zero elevation datum:

$$ \frac{P_1}{\rho g} + z_1 + \alpha_1 \frac{V_1^2}{2g} =\frac{P_2}{\rho g} + z_2 + \alpha_2 \frac{V_2^2}{2g} + \beta L \frac{V_2^2}{2g}$$

Simplifying via assumptions:

$$ \cancel{\frac{P_1}{\rho g}}^0 + \cancel{z_1}^0 + \cancel{ \alpha_1 \frac{V_1^2}{2g}}^{V_1 \ll V_2} =\cancel{\frac{P_2}{\rho g}}^0 + z_2 + \alpha_2\frac{V_2^2}{2g} + \beta L \frac{V_2^2}{2g}$$

Here ##z_2 = - ( z+L)##, moving to the LHS:

$$ z+ L = \frac{V_2^2}{2g} + \beta L \frac{V_2^2}{2g}$$

Where ##\beta## is a constant for a particular pipe geometry.

If the total volumetric flowrate is ##Q = V_2 A ##, solving for ##Q^2##:

EDIT: there is supposed to be a kinetic head correction factor ## \alpha_2 = 2 ## applied to ##\frac{V_2^2}{2g}## for laminar flow, thus( factoring the 2 into the numerator):

$$Q^2 = g A^2 z \frac{2+\frac{2}{z}L}{2 + \beta L}$$

From your experimental results it would seem that ## \frac{2}{z} > \beta## in the setup.I hope that elaborates on why your experimental set up is not quite working out like you had hoped, as all the others have suggested...the vertical pipe is an issue. Also, as @Chestermiller notes you've exceeded the Reynolds number where the linear proportionality (Hagen-Poiseuille) ##f = \beta L ## holds.

I would parrot what @kuruman suggests and design a new experiment, so that things fall into line.

What happened to the term for the viscous frictional pressure drop in the tube?

 

[erobz]

What happened to the term for the viscous frictional pressure drop in the tube?

The head loss is the last term on the right, ( ## \beta L \frac{V_2^2}{2g}##). I just lumped all the other constants in to ##\beta## and left proportionality to ## \frac{V_2^2}{2g}##

 

[Chestermiller]

The head loss is the last term on the right, ( ## \beta L \frac{V_2^2}{2g}##). I just lumped all the other constants in to ##\beta## and left proportionality to ## \frac{V_2^2}{2g}##

So this is $$\Delta P_{frictional}=\frac{1}{2}\rho v^2f\left(\frac{4L}{D}\right)$$wehre f is the fanning friction factor?

 

[erobz]

So this is $$\Delta P_{frictional}=\frac{1}{2}\rho v^2f\left(\frac{4L}{D}\right)$$wehre f is the fanning friction factor?

I'm using Darcy-Weisbach

$$ h_f = f \frac{L}{D} \frac{V^2}{2g} $$

For laminar flow ##f = \frac{64}{Re}##

Maybe its a bit abusive of me to say ##\beta## is a constant because it depends on ##V##. Is that the issue?

EDIT: In actuality using the equation I've outlined the OP would need to iterate ##f##, by guessing an initial flow velocity, calculate ##V##, and checking/update ##f##. My point wasn't to show the whole process, I was leaving that up to them. I was just trying to illuminate the experimental issue with the vertical pipe "cleanly", so it could be factorized.

If it has theoretical issues, I apologize.

 

[erobz]

Being precise about the friction factor in the laminar regime:

$$ z + L = \alpha_2\frac{V_2^2}{2g} + f \frac{L}{D}\frac{V_2^2}{2g} $$

With ##\alpha_2 = 2 ##, ## f = 64/Re ##, and ## Re = \frac{V_2 D}{ \nu }## ( ##\nu## is kinematic viscosity)

$$ z + L = \frac{V_2^2}{g} + \frac{32 \nu}{g D^2} L V_2 $$

And now solve the quadratic for ##V_2## and see what happens to ##V_2## as you increase ##L##.

 

[Chestermiller]

I think that,since we already have data on this, we should fit it with the turbulent flow calculation.