Viscosity & Drag: Linear Coeff & Medium Density

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    Drag Viscosity
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SUMMARY

The linear drag coefficient is determined solely by object size and medium viscosity, while the drag force itself is influenced by the medium's density. In low Reynolds number scenarios, such as Stokes flow, drag is primarily viscous, rendering density effects negligible on drag but significant on terminal velocity. The formula for drag force, F = 0.5 * density * speed^2 * drag coefficient * area, illustrates that denser media result in higher drag forces. This understanding aligns with principles outlined in Stokes' Law and classical mechanics, specifically in Taylor's work.

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  • Understanding of Stokes' Law and its application in fluid dynamics
  • Familiarity with Reynolds number and its significance in flow regimes
  • Knowledge of drag force equations and their components
  • Basic principles of viscosity and its role in fluid mechanics
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  • Study the derivation and implications of Stokes' Law in fluid dynamics
  • Explore the concept of Reynolds number and its impact on flow behavior
  • Investigate the relationship between viscosity and drag in various fluids
  • Learn about the nondimensional Navier-Stokes equations and their applications
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Students and professionals in physics, engineering, and fluid dynamics, particularly those interested in the behavior of objects in viscous fluids and the principles governing drag forces.

Vaal
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The linear drag coefficient is a function only of object size and medium viscosity. Multiply sources say viscosity of air is independent of pressure and density. How can the linear drag be independent of density of the medium? It seems like less dense medium should mean fewer collision and in turn less drag force. What am I missing here?
 
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the coefficient may be independent of density but the drag force depends on density. F=0.5 * density * speed^2 * drag coefficient * area. So yes, in denser medium there is higher drag force. ( source: wikipedia, i took the formula from there )
 
It is dependent on density. Stokes' Law shows drag explicitly as a function of terminal velocity, and that terminal velocity is a function of the density. It even shows this in the article you linked.
 
The article I linked may have not been the best example, I think it is referring to the force at terminal velocity (notice that a general v is no were in the expression). Classical Mechanics by Taylor explicitly says drag force is b*v and in problem 2.2 says b is given by 3pi*n*D where d is diamante of a sphere and n is the viscosity.
 
Right well in the case of Stokes flow, the velocity is so low that it is assumed that all drag is viscous in nature, which means that the density has no part in it. It only affects the terminal velocity. With such a low flow velocity, viscous effects are much, much more significant than pressure and density effects.

Consider the nondimensional Navier-Stokes equations:

\frac{\partial u_i^{*}}{\partial t^{*}} + u_j^{*}\frac{\partial u_i^{*}}{\partial x_j^{*}} = -\frac{\partial p^{*}}{\partial x_i^{*}} + \frac{1}{\textrm{Re}}\frac{\partial^2 u_i^{*}}{\partial x_j^{*} \partial x_j^{*}}

Stokes flow assumes that \textrm{Re} \ll 1, so the dissipation term, \frac{\partial^2 u_i^{*}}{\partial x_j^{*} \partial x_j^{*}}, will be an order of magnitude greater than the pressure term. In other words, the forces on the object in question are going to be dominated by viscosity with negligible contribution by pressure/density effects.
 
When you point out the mathematics I guess it does make sense. I don't have a lot of dynamics experience and I guess I am just having trouble intuitively getting used to the idea that the number of interactions (which is proportional to the density) would not change the viscosity.

Thanks for your help.
 

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