Hi

I am trying to solve this i really help .anyone i would be thankful.

 

 

Fluid of specfic gravilty 0.06 flow in a long vertical 2.56 ^10-2 m diameter pipe with average velocity of 0.15m/s .if the preasure is constant through the fluids,what is viscosity of fluid? assume laminar flow/

 

i) Determine the shear stress

ii) Viscosity of the fluid

iiI) Dertermine Reynolds numbers

 

i tried that i used my best to put reymonds number which is shown in following link.

http://en.wikipedia.org/wiki/Shear_stress

 

but i couldt . i hope you would be able to help me thx.

http://en.wikipedia.org/wiki/Reynolds_number

thx for your kind reply i have seen the link you sent me. i have seen the three formulas givens for the reynolds numbers but i dont understand whihc one to put as you see the 1st one is (density x mean fluid velocity x length of the object / dynamic viscosity ) if we see we have all the information we need to find reynolds numbers but we are also asked to find the dynamic viscosity . so that does not fit here all are to me used if the dynamic viscosity of is given .

So i m still confused .

waiting for your kind reply

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Ok this is what I have tried. I tried using the two formulas which come close to solving the problem.

 

 

[math] u® = Vc [1-(\frac{2r}{D})^2] [/math]

 

and [math] V = \frac{\pi*R^2*Vc}{2*\pi*R^2} =\frac{Vc}{2}=\frac{\Delta p*D^2}{32*\mu*l}[/math]

 

if i use the 2nd formula, then I can calculate [math] \mu [/math] by making it the subject and doing the arithematics but the question, as I am sure you would have noticed by now, doesnt give the length or the pressure difference so I have more than 2 parameters missing rather than 1. Does anyone think that we can assume the pressuse as 1 atm (as it is constant) and the length as 1 m to get [math] \mu [/math]? If I do that I get [math] \mu = 7440 [/math]

 

any help now please.