Calculating Shear Stress: Solving for μ = ρv in 8-41 Equation

[hotjohn]

Homework Statement

how to make μ = ρ v as in 8-41 ?

Homework Equations

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 ?

 

[SteamKing]

Homework Statement

how to make μ = ρ v as in 8-41 ?

Homework Equations

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 ?

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 $\nu = \frac{\mu}{\rho}$

See this article for a discussion of the units:

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

 

[hotjohn]

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 $\nu = \frac{\mu}{\rho}$

See this article for a discussion of the units:

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

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 ?

 

[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 $u$.

 

[hotjohn]

Note that $\sqrt{\tau_w/\rho}$ is defined to be the "fictitious velocity" $u$_-. (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 $u$.

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

 

[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.

 

[hotjohn]

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.

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

 

[haruspex]

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

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