# Velocity of water across depth

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Chestermiller
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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
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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.

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
Mentor
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
Mentor
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.

Last edited:
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 ? shouldnt it $$= 2\pi r \tau_{rz}(r)=0$$ only ?

Chestermiller
Mentor
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 ? shouldnt 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.

• fonseh
Chestermiller
Mentor
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 ? shouldnt 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.

There are pressure forces on the two ends that have to be balanced by the viscous frictional (tangential) force.
Can you explain further how to get 2\pi r L\tau_{rz}(r) ??

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

Chestermiller
Mentor • fonseh