Fluid Mechanics Newtons law of viscosity

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Homework Help Overview

The discussion revolves around the application of Newton's law of viscosity in fluid mechanics, specifically focusing on differentiating a velocity profile given by the equation for wall shear stress.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the differentiation of the velocity profile equation and question the interpretation of variables, particularly the definition of 'r' in relation to 'R'. There is an exploration of how to correctly apply the differentiation to find the shear stress.

Discussion Status

Some participants have provided guidance on the differentiation process and clarified the relationship between the variables involved. There is an acknowledgment of the need to differentiate correctly before applying the wall shear stress equation.

Contextual Notes

There appears to be confusion regarding the definition of 'u' and the variable 'r', with some participants questioning whether 'r' refers to the radius of a cylindrical pipe. The original poster's attempts at differentiation led to discrepancies in expected results.

wozzers
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Homework Statement


having problems differentiating


Homework Equations


Tw ( wall shear stress)= -U(viscosity)*du/dr
been given u as 2V(1-(r/r)^2)


The Attempt at a Solution


i substituted u in and got

d/dr (2V(1-(r/r)^2)
i tried to multiply out the minus sign and 2V and got a very different answer i got -2V+2V (r/r)^2

when the answer is supposed to be -4VR/R^2

please help
 
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What do you mean here:

been given u as
2V(1-(r/r)^2) = 2V(1 - 1) = 0

?
 
wozzers said:

Homework Statement


having problems differentiating


Homework Equations


Tw ( wall shear stress)= -U(viscosity)*du/dr
been given u as 2V(1-(r/r)^2)

Do you mean
<br /> u = 2V\left(1 - \left(\frac{r}{r_0}\right)^2\right)<br />
where r_0 is the radius of what I assume to be a cylindrical pipe?


The Attempt at a Solution


i substituted u in and got

d/dr (2V(1-(r/r)^2)
i tried to multiply out the minus sign and 2V and got a very different answer i got -2V+2V (r/r)^2

when the answer is supposed to be -4VR/R^2

please help

The given answer is for du/dr; to get T_w you have to multiply that by -\mu.

So far all you've done is multiply u by -1. Try differentiating with respect to r first.
 
r=R and its (r/R)^2
 
okay now i have differentiated it with respect to r i get 2V2r/R^2) or -(4Vr/R^2) thanks again i was then required to use the wall shear stress to determine the frictional force but it is very straight forward from there
 

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