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Physics Lab Experiment- Confused on what we are measuring

  1. Mar 10, 2012 #1
    In my physics class for an assessment we designed our own lab.

    In our lab we had plastic cups and cut small holes in the bottom ranging from 0.3 cm to 1.3 cm diameters. We then measured the amount of time it took for different volumes of water, detergent, syrup and rubbing alcohol. I understand that each of these liquids have a different density and that the amount of time for each substance to flow out of the cup will vary.

    Here is my question:
    In this lab, are we measuring flow rate or another value? How exactly can i construct a lab write up based on these variables:

    the density of the liquids vs the time it took to flow out
    the diameter of the cups vs the time it took to flow out
    the density of the liquids vs the diameter of the cups?
     
  2. jcsd
  3. Mar 10, 2012 #2
    When a fluid flows through a constriction, another important characteristic of the fluid is the so called viscosity. It enters in Newton's Law of viscous force. Namely, if you have two parallel sheets of the fluid a distance [itex]\Delta l[/itex] apart, flowing with a difference in velocities [itex]\Delta v[/itex], then the tangential force felt by each one is proportional to the common area [itex]A[/itex] of the two sheets, the above mentioned velocity gradient [itex]\Delta v/\Delta l[/itex]. The coefficient of proportionality is the dynamical viscosity [itex]\eta[/itex] of the fluid:
    [tex]
    F = \eta \, A \, \frac{\Delta v}{\Delta l}
    [/tex]
    Of course the direction of this force is opposite on each sheet. It tends to slow down the faster moving sheet, and accelerate the more slowly moving sheet.

    From the above law, you can find the dimensions of viscosity:
    [tex]
    \left[ \eta \right] = \frac{[ F ] \, [ l ]}{ [ A ] \, [ v ] } = \frac{\mathrm{T}^{-2} \, \mathrm{L} \, \mathrm{M} \, \mathrm{L} } { \mathrm{L}^2 \, \mathrm{T}^{-1} \mathrm{L} } = \mathrm{T}^{-1} \, \mathrm{L}^{-1} \, \mathrm{M}
    [/tex]

    Because [itex]F/A[/itex] has the meaning of a strain, with the same dimension as pressure, and the velocity gradient has the dimensions of [itex][v]/[l] = \mathrm{T}^{-1} \mathrm{L}/\mathrm{L} = \mathrm{T}^{-1}[/itex] inverse time, the dynamical viscosity is usually reported in units of [itex] [\mathrm{pressure}] \cdot [\mathrm{time} ][/itex].

    Now, your problem is a pretty complicated hydrodynamics problem. Nevertheless, you may draw a series of conclusions by employing arguments from dimensional analysis. The "fundamental constants" that govern this process are:
    1) acceleration due to Earth's gravity [itex]g, \ [g] = \mathrm{T}^{-2} \mathrm{L}[/itex]

    2) density of the liquid, [itex]\rho, \ [\rho] = \mathrm{L}^{-3} \mathrm{M}[/itex]

    3) viscosity of the liquid, [itex]\eta, \ [\eta] = \mathrm{T}^{-1} \mathrm{L}^{-1} \mathrm{M}[/itex]
    ----------------------------------
    Then, there is the parameter of each cup, namely the diameter of its hole [itex]D, \ [D] = \mathrm{L}[/itex]

    A quantity which you experimentally measure is the flow time [itex]T, [ T ] = \mathrm{T}[/itex]

    By dimensional analysis, you can construct a combination of the first three constants with the dimensions of length [itex]L_0[/itex] and time [itex]T_0[/itex], respectively. Find these!

    Then, you can form the dimensionless variables [itex]D/L_0[/itex], and [itex]T/T_0[/itex]. It is a consequence of the scaling laws in hydrodynamics that there is a general functional relationship:

    [tex]
    \frac{T}{T_0} = f \left( \frac{D}{L_0} \right)
    [/tex]

    You can use it, in principle to measure the viscosity of an unknown fluid in terms of the viscosity of a known fluid, say water. I will let you figure out what viscosity may be measured in this way.
     
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