Calculating Water Viscosity from Flow Rate

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SUMMARY

The viscosity of water can be calculated using the flow rate and height from a tube. Given a flow rate of 2.0 cm³/s, a height of 0.1 m, a length of 0.2 m, and a radius of 1 mm, the viscosity is determined using the equation η = (πρgr⁴)/(8l) where ρ is the density (1000 kg/m³) and g is the gravitational acceleration (9.81 m/s²). After converting the flow rate to m³/s, the calculated viscosity is 9.6 x 10⁻⁴ Pa·s. This calculation was verified by multiple participants in the discussion.

PREREQUISITES
  • Understanding of fluid dynamics principles
  • Familiarity with the equation of motion for viscous flow
  • Basic knowledge of unit conversions (cm³/s to m³/s)
  • Proficiency in algebra for rearranging equations
NEXT STEPS
  • Study the Navier-Stokes equations for fluid dynamics
  • Learn about the Hagen-Poiseuille equation for laminar flow
  • Explore viscosity measurement techniques in laboratory settings
  • Investigate the effects of temperature on water viscosity
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Students and professionals in physics, engineering, and fluid mechanics who are involved in calculating and understanding fluid viscosity and flow characteristics.

JosephL
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Homework Statement


Find the viscosity of water given that;

flow rate ([tex]cm^3 s^{-1}[/tex]) = 2.0
height from tube (m) = 0.1

l (m) = 0.2
r (mm) = 1
[tex]\rho[/tex] = 1000
g = 9.81

Using the equations;

x = [tex]\frac{Flow Rate}{Height}[/tex]

x = [tex]\frac{\pi \rho gr^4}{8l\eta}[/tex]

The Attempt at a Solution



Okay so first, and the most important bit I don't know if I'm doing right I convert the flow rate from [tex]cm^3 s^{-1}[/tex] to [tex]m^3 s^{-1}[/tex] by multiplying by [tex]10^{-6}[/tex]

So flow rate is now [tex]2 * 10^{-6} m^3 s^{-1}[/tex] ***I don't know if that bit is correct.***

Then x = [tex]\frac{2 * 10^{-6}}{0.1} = 2 * 10^{-5}[/tex]

and [tex]2 * 10^{-5} = \frac{\pi \rho gr^4}{8l\eta}[/tex]

After re-arranging and substituting I get;

[tex]\eta = 9.6*10^{-4}[/tex]------

If anyone could check my answer it would be greatly appreciated I haven't done physics for so long I've pretty much forgotten everything.

Thanks in advance :smile:
 
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I got the exact same answer you got, using your data.
 

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