Test the limits of Poiseuilles law

[Ron Burgundypants]

Homework Statement

I'm not sure if this is introductory or not but here goes anyway...

So we have this experiment to measure the flow rate of viscous fluids by forcing them up a measuring tube using some compressed air in a drum.

The idea is to test the limits of poiseuilles law with various different viscosity fluids. So we first decided to model the system. Using the Hagen-Poiseuille equation we end up with an expression for pressure (constant) and height as a function of time, but we have two height variables and I'm not sure how to deal with them. I'll show my working in a minute but basically my supervisor says we are along the right lines but there is more to be done. I'm not exactly sure how to continue with this problem. Any advice would be most welcome.

p- Pressure
r - radius
h - height of tube
Q - flow rate
μ - Viscosity

The equations come in the next section...I hope that all makes sense, if its a bit confusing its because this is the format we are required to present in.

Homework Equations

Poiseuilles law - ∆p=8μhQ / r^4π

Q (volumetric flow rate) = πr^2h

The Attempt at a Solution

My supervisor said 'h' (there's two h variables here) is not constant. So if I make one 'h' a dh/dt ( so we have a discretized volumetric flow rate) then I can make the other h a function of time right? The rest is constant (pressure is also constant in our experiment as we will make it so!) so we end up with the following

Subbing into poiseuilles equation for Q gives us --> ∆p = 8μ h(t) dh/dt / r^2

Pressure is constant ( I'll call it P from now on. I'll also group and rename all the constant variables on the right hand side to k. So we end up with

P = k * h(t) dh/dt

I'm thinking that if I just isolate h(t) then I have my solution right? But moving k over to the left and the dh/dt would give us

(P/k) / dh/dt = h(t) --> Pdt / kdh = h (t)

Can I even do this? This is the point at which I become confused...

Although I also think if this is correct then its just the exponential function right? I know most of these equations end up as being the exponential function and it would make sense for h(t) to be an exponential decay because; the resistance to flow increases in the tube proportionally to pressure. I'm just not sure I'm explaining/deriving it correctly or how to...

Thanks.

 

[kuruman]

(there's two h variables here)

What are the two h variables? What do they represent in terms of the experiment?

o if I make one 'h' a dh/dt

You just can't do that without physical justification.

Pressure is constant

Why do you say this? There is a pressure drop from one end to the other. How can it be constant in between?

 

[Ron Burgundypants]

I've attached a picture of the experiment below plus the initial workings I did.

In response to your question 'h' is the height of the tube, or length if you will ( it is positioned vertically).

The applied pressure will be kept constant when the experiment runs.

The idea of using a dh/dt was to discretize the volume of the tube and analyse the flow over small sections. Maybe I did this incorrectly.

The response I got from my supervisor was the following ' In the middle of the page you write "h is a constant". Don't write that! Instead realize that h is not a constant, it is a function of t. You have both dh/dt and h(t) in your model.

There are two h's here. Both represent the height/length of the tube. So I'm really confused as to what she wants me to do...

 

[kuruman]

OK, I see the setup now. Can you tell me what you measured as you varied what?

 

[Ron Burgundypants]

Well we haven't actually done anything yet. The setup is being built for us.

Our intention is to try out fluids with different viscosities and measure their flow rate. We did some calculations as to what kind of pressures we would need to get each fluid to flow 50cm within a time period of between 6-12 hours. Then we had a basis for what level of precision in the increments of pressure we would need to apply so the setup could be built. So the plan is to start with water and apply an instantaneous pressure, forcing it up the tube. Then we measure the flow rate and probably the shape of the head of the flow at various time steps. So the only thing that varies in when the pressure is applied will be the distance the fluid travels in the tube, which i guess could be described as the flow rate everything else should be constant.

 

[kuruman]

So the only thing that varies in when the pressure is applied will be the distance the fluid travels in the tube, which i guess could be described as the flow rate everything else should be constant.

I see now what this is about. The flow rate Q is not proportional to the distance the fluid travels in the tube. That's where your confusion is. The flow rate is ##Q= dV/dt## where ##V## is the volume of the column. With ##V = A h##, ##Q=A dh/dt.## That's the physical justification for the derivative. So I would assume that you will record the height of the fluid at regular time intervals [i.e. measure h(t)] and from that figure out its rate of change or the derivative. Be sure to record not only the time interval between height measurements, but also the total time since starting the experiment. That will give the functional form of h(t) which you can check if it is exponential. Good luck with your experiment.

 

[Ron Burgundypants]

I see now what this is about. The flow rate Q is not proportional to the distance the fluid travels in the tube. That's where your confusion is. The flow rate is ##Q= dV/dt## where ##V## is the volume of the column. With ##V = A h##, ##Q=A dh/dt.## That's the physical justification for the derivative. So I would assume that you will record the height of the fluid at regular time intervals [i.e. measure h(t)] and from that figure out its rate of change or the derivative. Be sure to record not only the time interval between height measurements, but also the total time since starting the experiment. That will give the functional form of h(t) which you can check if it is exponential. Good luck with your experiment.

Aha! Eureka as they say. Thank you for your help. Much appreciated.