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Well if you're interested I have a tank (actually a PVC tube with an end cap) that has an interior diameter of 86 mm. It stands on its circular face. Coming out near the bottom of the tank at a right angle is a metal pipe 11 mm in diameter (interior) and 312 mm long (in total that is, note that some of the pipe is protruding within the tank and not just out of the tank to stabilise it). Here is a picture similar to what I did on page 3 http://seniorphysics.com/physics/ejp_sianoudis.pdf (I’m not doing the same experiment).

I fill the tank with water until enough water is in the tank so that it starts to flow out of the pipe (there is some space beneath the exit pipe's hole). I do this so then I know if I add another litre of water now, 1 litre should come out (no space inside the tank beneath the pipe hole now). My experiment is aimed at investigating the effect of viscosity on the time the tank takes to empty (not really empty because when the water level is beneath the top of the pipe the theory does not stand, so about there). First I would like a formula that can predict the volume of fluid in the tank at any time. I know that according to the Hagen–Poiseuille equation

ΔP=8μLQ/(πr^4 )

ΔP is the pressure drop

L is the length of pipe

μ is the dynamic viscosity

Q is the volumetric flow rate

r is the radius

π is the mathematical constant

This can be arranged for the flowrate:

ΔP(πr^4 )/8μL = Q

The pressure difference across the ends of the pipe is due solely because of the column of water in the tank right? This pressure equals:

P=hρg

P is the hydrostatic pressure (Pa),

h is the height of the water (m)

ρ is the fluid density (kg/m3),

g is gravitational acceleration (m/s2),

Thus substituting this in the flow rate equation:

Q = ΔP(πr^4 )/8μL and P=hρg

-> Q = hρg(πr^4 )/8μL

Q = hρgπr^4/8μL

The height of the water column in the tank can be related to the tank volume. Note the radius of the tank will be notated with [r2] as the other plain "r" is for the pipe radius:

Volume of a cylinder = hπr^2

V/(π[r2]^2)=h

Thus

Q = hρgπr^4/8μL

Q = ρgπr^4/8μL x V/(π[r2]^2)

Q = Vρgπr^4/8μLπ[r2]^2

Q = Vρgr^4/8μL[r2]^2

Or Q = kV where k = ρgr^4/8μL[r2]^2 <- These are all constant

Since Q = dV/dt

Thus dV/dt = kV <- Flipping this...

Or dt/dV = 1/k x 1/V Integrating with respect to V

Thus t=1/k x ln V + c

t-c=1/k x ln V

k(t-c)= ln V

e^[k(t-c)]=V

V=e^(kt-kc)

V=e^kt x e^-kc

when t=0 V =Vo

Thus Vo=e^-kc sub this back in to V=e^kt x e^-kc

V=e^kt x Vo

Thus V=Vo x e^kt

where k = ρgr^4/8μL[r2]^2

So there we go I got a formula for volume at any time.

Taking the dynamic viscosity out of k we get:

k = ρgr^4/8L[r2]^2

and

V = Vo x e^(kt/μ)

So according to this if I doubled μ the graph of "V" would be stretched in the x direction away from the y-axis by a factor of 2. So if it originally took 20 secs to half empty the tank, doubling the viscosity should make it take 40 seconds to empty the tank right? Yet rapeseed (canola) oil and most things are hundreds of times as viscous as water even at room temperature, yet just from experience we know they do not flow 100 times slower nor would drain from a tank taking 100 times longer.

Using my tank setup I tested rapeseed oil (μ = 160 mPa•s ) and water at 20 degrees Celsius (μ - 1mPa•s) and rapeseed oil (just roughly looking at it now) took maybe 6 times as long. I thought it might be because when viscosity changes so does the density values, but they are fairly similar so that cant be it. What’s worse the equation V=Vo x e^(kt/μ) doesn’t seem to fit my data or even appear realistic, although I haven't checked this last part. So what’s going on? Do I lack understanding? Am I just doing it wrong? Can anyone explain this?

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# How come a liquids dynamic viscosity is not indicative of flowrate?

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