Velocity of water across depth

[fonseh]

Homework Statement

Here's the velocity profile of the water across the depth . I don't understand why The max velocity occur at the top surface[/B]

Homework Equations

The Attempt at a Solution

Is it because the friction on the top surface of water only comes from the below , whereas for the water below the top surafce , it has higher friction from the moving on the top , and moving water beneath it ... Is my concept correct ? [/B]

 

[BvU]

Hi.
What's holding back the water ? The viscous friction with something that doesn't move, namely the bottom. So all transport of horizontal momentum takes place in a vertical direction - as you can see from the variation in velocity going from 0 at the bottom to max at the surface.

 

[Chestermiller]

Homework Statement

Here's the velocity profile of the water across the depth . I don't understand why The max velocity occur at the top surface[/B]

Homework Equations

The Attempt at a Solution

Is it because the friction on the top surface of water only comes from the below , whereas for the water below the top surafce , it has higher friction from the moving on the top , and moving water beneath it ... Is my concept correct ? [/B]

The friction at the top surface (i.e., the shear stress) is zero. There is friction at the bottom surface because of the non-slip boundary condition.

 

[fonseh]

The friction at the top surface (i.e., the shear stress) is zero. There is friction at the bottom surface because of the non-slip boundary condition.

Do you mean as the depth increases , the shear stress increases, causing the velocity to decreases parabolically from top surface to the bottom >

 

[Chestermiller]

Do you mean as the depth increases , the shear stress increases, causing the velocity to decreases parabolically from top surface to the bottom >

Yes

 

[fonseh]

Yes

can you explain further why the shear stress will increase with depth ?

 

[Chestermiller]

can you explain further why the shear stress will increase with depth ?

You do understand that the shear stress is zero at the upper surface, correct? And you do understand that the fluid does not slip at the wall, correct.

 

[fonseh]

y

You do understand that the shear stress is zero at the upper surface, correct? And you do understand that the fluid does not slip at the wall, correct.

yes , i understand both ... Why the shear stress will incresae down the depth ?

 

[Chestermiller]

y

yes , i understand both ... Why the shear stress will incresae down the depth ?

Since the fluid at the wall is stationary (the velocity at the wall is zero), and you have a volumetric flow rate, the velocity has to be higher away from the wall. This means that the velocity gradient at the wall must be positive. From Newton's law of viscosity, that means that there is a shear stress at the wall. So you have shear stress at the wall and no shear stress at the interface with the air. So the shear stress must be increasing with distance from the wall.

 

[fonseh]

Since the fluid at the wall is stationary (the velocity at the wall is zero), and you have a volumetric flow rate, the velocity has to be higher away from the wall. This means that the velocity gradient at the wall must be positive. From Newton's law of viscosity, that means that there is a shear stress at the wall. So you have shear stress at the wall and no shear stress at the interface with the air. So the shear stress must be increasing with distance from the wall.

why ? ?

 

[Chestermiller]

why ? ?

If the velocity were zero everywhere, there would be no flow.

 

[fonseh]

If the velocity were zero everywhere, there would be no flow.

why it has to increase parabolically ? Why the shape of graph can't be others ? I mean the graph has still positive gradient ...

 

[Chestermiller]

why it has to increase parabolically ? Why the shape of graph can't be others ? I mean the graph has still positive gradient ...

To obtain that result, you need to actually solve the flow problem involving the differential force balance and Newton's law of viscosity.

 

[fonseh]

To obtain that result, you need to actually solve the flow problem involving the differential force balance and Newton's law of viscosity.

I still doesn't get it . Can you show the problem involving the differential force balance and Newton's law of viscosity?

 

[Chestermiller]

I still doesn't get it . Can you show the problem involving the differential force balance and Newton's law of viscosity?

It's too extensive to show here. Are you familiar with the derivation of the Hagen-Poiseuille law for laminar viscous flow in a pipe.

 

[fonseh]

It's too extensive to show here. Are you familiar with the derivation of the Hagen-Poiseuille law for laminar viscous flow in a pipe.

Can you provide some link and explain briefly here ?

 

[Chestermiller]

Can you provide some link and explain briefly here ?

Google is your friend.

 

[BvU]

Bird Stewart and Lightfoot are even better friends. Whole derivation in 2.2 in my edition.

 

[fonseh]

Bird Stewart and Lightfoot are even better friends. Whole derivation in 2.2 in my edition.

What do you mean ?

 

[fonseh]

Google is your friend.

https://en.wikipedia.org/wiki/Hagen–Poiseuille_equation
Which part of the equation relate to differential force balance and Newton's law of viscosity ?

 

[fonseh]

Since the fluid at the wall is stationary (the velocity at the wall is zero), and you have a volumetric flow rate, the velocity has to be higher away from the wall. This means that the velocity gradient at the wall must be positive. From Newton's law of viscosity, that means that there is a shear stress at the wall. So you have shear stress at the wall and no shear stress at the interface with the air. So the shear stress must be increasing with distance from the wall.

Do you mean the shear stress increase to max from the wall to the center ? Which means the shear stresss is max at the center at particular depth ?

 

[Chestermiller]

https://en.wikipedia.org/wiki/Hagen–Poiseuille_equation
Which part of the equation relate to differential force balance and Newton's law of viscosity ?

The part about "Putting it Together".

 

[Chestermiller]

Do you mean the shear stress increase to max from the wall to the center ? Which means the shear stresss is max at the center at particular depth ?

No. It is zero at the top and maximum at the wall. It decreases linearly with distance from the wall.

 

[fonseh]

No. It is zero at the top and maximum at the wall. It decreases linearly with distance from the wall.

can you show the graph of shear stress ? I'm very confused now ... Do you mean the shear stress increases from zero at the top down to the bottom ?

The red part means for a particular depth , the shear stress decreases from the wall to the center ?

 

[fonseh]

It's too extensive to show here. Are you familiar with the derivation of the Hagen-Poiseuille law for laminar viscous flow in a pipe.

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/RiverViscosity.pdf

I found this , is it suitable to apply the same concept here based on the the link ?

 

[BvU]

https://en.wikipedia.org/wiki/Navier–Stokes_equations#Representations
https://en.wikipedia.org/wiki/Hagen–Poiseuille_flow_from_the_Navier–Stokes_equations

Google is your friend.

What do you mean ?

This is textbook material. We don't feel inclined to completely re-type it. But specific questions for elucidation are always welcome.

 

[Chestermiller]

can you show the graph of shear stress ? I'm very confused now ... Do you mean the shear stress increases from zero at the top down to the bottom ?

Yes.

 

[Chestermiller]

In the Wiki article on viscous laminar flow in a tube, they do an axial force balance on the shell of fluid between radial locations r and r+Δr. An easier way to analyze the problem is to do an axial force balance on the plug of fluid between r = 0 and arbitrary radial location r:
$$\pi r^2\Delta p+2\pi r L\tau_{rz}(r)=0$$where $\Delta p$ is the pressure drop between the inlet and outlet of the the section of pipe under consideration, L is the axial length of the section of pipe, and $\tau_{rz}(r)$ is the shear stress the radial surface of the plug (i.e., at constant r in the z direction). From this equation, we get that:
$$\tau_{rz}=-\frac{\Delta p}{2L}r$$
This tells us that, for steady flow in a pipe, the shear stress varies linearly with radial distance from the axis.

 

[fonseh]

In the Wiki article on viscous laminar flow in a tube, they do an axial force balance on the shell of fluid between radial locations r and r+Δr. An easier way to analyze the problem is to do an axial force balance on the plug of fluid between r = 0 and arbitrary radial location r:
$$\pi r^2\Delta p+2\pi r L\tau_{rz}(r)=0$$where $\Delta p$ is the pressure drop between the inlet and outlet of the the section of pipe under consideration, L is the axial length of the section of pipe, and $\tau_{rz}(r)$ is the shear stress the radial surface of the plug (i.e., at constant r in the z direction). From this equation, we get that:
$$\tau_{rz}=-\frac{\Delta p}{2L}r$$
This tells us that, for steady flow in a pipe, the shear stress varies linearly with radial distance from the axis.

Can you point out which part of the wiki article shows that the velocity profile of water across depth is parabolic shape ?

 

[Chestermiller]

Can you point out which part of the wiki article shows that the velocity profile of water across depth is parabolic shape ?

The entire section Putting It Together, culminating with the next-to-last equation which gives the parabolic velocity profile.

 

[fonseh]

In the Wiki article on viscous laminar flow in a tube, they do an axial force balance on the shell of fluid between radial locations r and r+Δr. An easier way to analyze the problem is to do an axial force balance on the plug of fluid between r = 0 and arbitrary radial location r:
$$\pi r^2\Delta p+2\pi r L\tau_{rz}(r)=0$$where $\Delta p$ is the pressure drop between the inlet and outlet of the the section of pipe under consideration, L is the axial length of the section of pipe, and $\tau_{rz}(r)$ is the shear stress the radial surface of the plug (i.e., at constant r in the z direction). From this equation, we get that:
$$\tau_{rz}=-\frac{\Delta p}{2L}r$$
This tells us that, for steady flow in a pipe, the shear stress varies linearly with radial distance from the axis.

Where do you get this equation actually ?

 

[Chestermiller]

Where do you get this equation actually ?

It's just a force balance (you remember, like in freshman physics). Please show us what you think the free body diagram looks like.

 

[fonseh]

It's just a force balance (you remember, like in freshman physics). Please show us what you think the free body diagram looks like.

i know that A = pi(r^2) , so dA = 2pi(r)dr , so in $$\pi r^2\Delta p+2\pi r L\tau_{rz}(r)=0$$ , why there is L ? shouldn't it $$= 2\pi r \tau_{rz}(r)=0$$ only ?

 

[Chestermiller]

i know that A = pi(r^2) , so dA = 2pi(r)dr , so in $$\pi r^2\Delta p+2\pi r L\tau_{rz}(r)=0$$ , why there is L ? shouldn't it $$= 2\pi r \tau_{rz}(r)=0$$ only ?

There are pressure forces on the two ends that have to be balanced by the viscous frictional (tangential) force.

 

[Chestermiller]

i know that A = pi(r^2) , so dA = 2pi(r)dr , so in $$\pi r^2\Delta p+2\pi r L\tau_{rz}(r)=0$$ , why there is L ? shouldn't it $$= 2\pi r \tau_{rz}(r)=0$$ only ?

There are pressure forces on the two ends that have to be balanced by the viscous frictional (tangential) force.