Fluid dynamics: Reynolds Number, Drag Constant

  • #1
I completed 4a successfully, and with 4b, i have 2 queries:
a)why can't I let Reynolds # equal to 2.19 x 10^5 (from part a) then simply sub v=4 instead of 5m/s and rearrange for viscosity? I tried it this way first and got a very wrong answer. Why do we, essentially, need to work backwards to get the dynamic viscosity?
b)How do they find the reynolds number from the drag constant? I know they used the table but how?

Below is proof of my workings, and i have screen shotted the question and the answer workings as well.


Relevant equations are

F (drag)= drag constant (Cd) x pi/4 x d^2 x density fluid x velocity^2 x 1/2

Reynolds number= (density of fluid x diameter of spehere x velocity)/viscosity

F (drag) = Weight- Buoyancy
= pi/6 x d^3 x gravity x (density of sphere-density of fluid)


kinematic viscosity= dynamic viscoity/ density
 

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Answers and Replies

  • #2
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I completed 4a successfully, and with 4b, i have 2 queries:
a)why can't I let Reynolds # equal to 2.19 x 10^5 (from part a) then simply sub v=4 instead of 5m/s and rearrange for viscosity?
I tried it this way first and got a very wrong answer. Why do we, essentially, need to work backwards to get the dynamic viscosity?

Because the velocity also appears in the ##\rho v^2/2## term. So here, it does not appear in combination with the dynamic viscosity.
b)How do they find the reynolds number from the drag constant? I know they used the table but how?
You either plot the relationship on a graph, and use the graph (preferrably a log-log plot) to get the reynolds number, or you interpolate (preferrably logarithmically) in the table.

Chet
 
  • #3
Thanks Chet, I also realized that Reynolds number will not be the same due to a variation in the drag constant. So we cannot use the same Reynolds number as the first part
 

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