3
$\begingroup$

If you were to measure the viscosity of a fluid you would do the following experiment https://www.wikihow.com/Measure-Viscosity

The formula for dynamic viscosity is the following one:

$n=\frac{2gr^{2}(p_s-p_l)}{9v}=\frac{ms^{-2}m^{2}(kgm^{-3})}{ms^{-1}}=kgm^{-1}s^{-1}= Pa·s$

where n is dynamic viscosity of the liquid in Pa s, g is gravity ($9.81{ms^-}^2$), r the radius of the ball in meters, $p_s$ density of sphere, $p_l$ density of liquid and v the velocity at which the ball travels through the fluid.

This source and many more state that the dynamic viscosity of water is $8.90*{10^-}^4$ Pa s https://www.engineersedge.com/physics/water__density_viscosity_specific_weight_13146.htm

If we plug this value for viscosity and try to find a value for the velocity at witch the sphere travels through the water (in this case). Then we come up with the following.

$8.90*{10^-}^4=\frac{2(9.81)r^{2}(p_s-1000)}{9v}$ (Density of water is 1000kg per meter cube)

Let's say that we drop an aluminum ball of 0.01m radius. Aluminum has a density of 2700kg/$m^3$

$8.90*{10^-}^4=\frac{2(9.81)(0.01)^{2}(2700-1000)}{9v}$

The final value for $v$ is 416.4 ${ms^-}^1$

Is the sphere really traveling through water at that high speed. Is the formula wrong? Or is it my math or what? I'm so confused.

$\endgroup$
  • $\begingroup$ I didn't check on your math. But a ball creates a 3-D wake behind it as it travels through the fluid. Pressure drag plays a role and so does viscous drag. There will be separation behind the ball if it's speed is too large. I don't know how a formula would include these effects to obtain a value for viscosity. $\endgroup$ – jjack Dec 10 '17 at 10:48
  • $\begingroup$ @jjack So would you say that this formula to calculate viscosity is not right at all (not accurate)? $\endgroup$ – Matthew Dec 10 '17 at 10:51
  • $\begingroup$ Sorry, I can't give you a good answer. You'd have to wait for a fluid dynamics guy to answer it. $\endgroup$ – jjack Dec 10 '17 at 10:55
  • $\begingroup$ read this Wikipedia article on viscosity: en.m.wikipedia.org/wiki/Viscosity, be aware that there seems to be difference between dynamic and kinematic viscosity. There's also a short section on measuring kinematic viscosity. $\endgroup$ – jjack Dec 10 '17 at 11:00
  • $\begingroup$ @jjack the one I am talking about is dynamic viscosity, check the units they are Pa s as dynamic, while kinematic is m^2/s. Thanks for the article but I already read it so many times :P $\endgroup$ – Matthew Dec 10 '17 at 11:11
5
$\begingroup$

Your equation comes from an application of Stokes' law whilst equating the viscous drag to the apparent weight of the object when the object is travelling at a terminal speed.

However Stokes' law is not applicable in your example the test for this being to evaluate a parameter called Reynolds' number $R_{\rm e} = \dfrac {\rho_{\rm fluid}R_{\rm shere}v_{\rm terminal}}{\eta_{\rm fluid}}$ which in your example is approximately four million far in excess of the upper limit for Stokes' law to be applicable which is about ten.

In your regime the fluid friction is no longer proportional to the speed but more likely to the speed squared so you will need to do a different analysis.
There is an answer in PSE which might help you and the analysis is done in many textbooks and Internet sites; here is one as an example.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.