# Shear stress

1. Jan 21, 2016

### hotjohn

1. The problem statement, all variables and given/known data
how to make μ = ρ v as in 8-41 ?

2. Relevant equations

3. The attempt at a solution
since μ = viscocity , it has unit =( kgms^-2)s m^-2 , = kg(m^-1)(s^-1)
ρ v = (kgm^-3)(ms^-1 ) = kg(m^-2)(s^-1) , the unit for LHS and RHS are not the same , can someone help pls ?

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2. Jan 21, 2016

### SteamKing

Staff Emeritus
You've mixed up lower case v for velocity with the lower case Greek letter nu ($\nu$), which is used to represent the kinematic viscosity of a fluid.

The kinematic viscosity is defined $\nu = \frac{\mu}{\rho}$

https://en.wikipedia.org/wiki/Viscosity

3. Jan 21, 2016

### hotjohn

ok , can you pls explain how the author get the formula of viscous sublayer ? after substitute u= sqrt rt (τ / ρ ) into 8-41 , I have u = ( v / y ) sqrt rt (τ / ρ ) , hwo to get the same form as in 8-42 ?

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4. Jan 21, 2016

### TSny

Note that $\sqrt{\tau_w/\rho}$ is defined to be a fictitious velocity $u$- (called "friction velocity"). (I'm not sure I got the notation right. It looks like there is a subscript "-" on the $u$ for the fictitious velocity. But it is hard for me to read.) See discussion below equation 8-41.
$u$- is not the same as the actual velocity $u$.

Last edited: Jan 21, 2016
5. Jan 21, 2016

### hotjohn

u / $u$- , i haver u [ sqrt rt ( p / т ) ] , how to get the same form as in the 8-42 ?

6. Jan 21, 2016

### TSny

Solve $u$- = $\sqrt{\tau/\rho}$ for $\tau$. Then sustitute this expression for $\tau$ into equation 8-41. You should then be able to rearrange it to get 8-42.

7. Jan 22, 2016

### hotjohn

hwo to do that ? I have tried , but didn't get the same form ,

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8. Jan 22, 2016

### haruspex

Some of your cancellations are wrong. $\frac x{\sqrt x}$ is not $\frac 1{\sqrt x}$