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

  1. Aug 14, 2011 #1
    REWRITTEN:

    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?
     
    Last edited: Aug 14, 2011
  2. jcsd
  3. Aug 14, 2011 #2
    Good morning, Zaxo.

    It is a good idea to use breaks and paragraphing in your text. It makes it considerably easier to read.

    First to clear up some misconceptions.

    Dynamic viscosity has nothing to do with density. That property is called kinematic viscocity, which equals dynamic viscocity divided by density.

    Poiselles formula does not contain a density factor, unless it is presented in terms of kinematic viscocity and is applicable to pipes.

    For a pipe of length L and radius r with a pressure difference between the ends the discharge Q is

    [tex]Q = \frac{{\pi ({p_1} - {p_2}){r^4}}}{{\eta L}}[/tex]

    where [tex]\eta [/tex] is the dynamic viscocity.

    The kinematic viscocity,[tex]\nu [/tex] is

    [tex]\nu = \frac{\eta }{\rho }[/tex]

    Now I do not have figures for rape seed oil but here is a comparison between water and castor oil

    Water Dynamic viscosity 1.79; 1.00; 0.65 centipoise
    Castor oil dyn viscosity 6400; 973; 227 centipoise

    Taken at 0oC; 20oC; 40oC

    You can see how dramatically the oil viscosity reduces with temperature, relative to the water.

    So when you do this experiment it is vital that you record the temperature to obtain an accurate viscosity value.
     
  4. Aug 14, 2011 #3
    Thanks for the response. I've corrected my question now, i actually meant the density that was substituted into Poiselles formula with water colum height and gravity in place of pressure difference. Ill insert some paragraph breaks too. Also even your figures show the castor oil is hundreds times more viscous even at room temperature of testing. I would still appreciate an explanation as to why the time taken for the different fluids to (for instance) reach half the tank volume is not hundreds of times greater for the thicker oil, even though viscosity is.
     
  5. Aug 14, 2011 #4
    If you were to post some details of your experiment I might be able to comment more.
     
  6. Aug 14, 2011 #5
    I added a lot of additional information into the original question. You now know as much as I do about this. Can you figure it out?
     
  7. Aug 14, 2011 #6
    Experimentally your method should 'work'.

    Hoever I wonder if your outlet pipe bore (r=5.5mm) was not too large?

    Your reference article notes extreme sensitivity to pipe radius (r4) and the max radius in their tests was less than half yours at 2.5mm

    This experiment is normally conducted with small bores pipes (capilliaries).

    Incidentally you can use the X2 and X2 icons on the main reply box to obtain powers and indices.
     
  8. Aug 14, 2011 #7
    Thanks a lot for going to the trouble of reading all my work and helping me. At current what you say seems like the most logical explanation. And thanks for the tip.
     
  9. Aug 14, 2011 #8
  10. Aug 15, 2011 #9
    Hey so since Reynolds Number = pvD/μ
    and in this case
    p=Density of water = 1000 kg/m³
    the pipe diameter D = 0.011 m
    The dynamic viscosity of water is 1.002 mPa•s 20 °C (the temperature of testing)
    and the velocity that was initially measured of the water exiting the pipe was 0.856 m/s this goes to:

    Reynolds Number = 1000 * 0.856 * 0.011 / 0.001 = 9416

    This is greater than 4000 and 2000 so is my experiment not laminar flow but actually turbulent flow? That is until v = .364 when the flow will become transitional and then when v = 0.182 when the flow becomes laminar?

    Also is there any way to calculate the theoretical volume at any time either which does take into account viscosity and pipe length?!?
     
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