# Flow in stems and pipes II

If two sections of a fluid flow past one another with relative velocity, they will pull on one another with a drag force that is proportional to the fluid viscosity, $$\eta$$, as well as to the change in velocity over the change in distance, i.e. the gradient $$\nabla v(r)$$.

Consider the diagram below, which crudely approximates fluid flowing down a pipe as three adjacent cylindrical layers flowing at velocities $$v(r-dr)>v(r)>v(r+dr)$$. The flow of all three layers is out of the screen (as indicated by the white dot at the center).

Focusing on the middle layer, which travels with velocity $$v(r)$$, it feels a forward pull per unit area from the central layer (since $$v(r-dr) > v(r)$$, given by $$\eta\frac{v(r+dr)-v(r)}{dr}$$.

In other words, the central layer is pulling the middle layer through some surface interaction, so that the total pull is given by $$\eta\frac{v(r+dr)-v(r)}{dr}S_A$$ with $$S_A$$ being the surface area of interaction between the two layers.

Similarly, the middle layer feels a backward pull from the top layer given by $$\eta\frac{v(r-dr)-v(r)}{dr}S_A$$.

Using this intuition, we can motivate the solution. Consider the generalization of this picture, where the fluid consists of a continuous series of cylindrical shells each flowing down the pipe and interacting with their neighboring shells. Find a differential relation connecting the velocity of a layer with its radius. Remember the no-slip boundary condition: $$v(R) = 0$$.

Solve the differential equation and find the velocity of the sap at a distance $$r$$ from the center of the tube.

Assumptions

• There is a pressure $$\Delta p=10^6$$ Pa over the pipe
• $$r$$ = 10 $$\mu$$m
• The radius of the pipe is 20 $$\mu$$m
• The viscosity of the sap is $$5\times 10^{-3}$$ Pa s
• The length of the pipe is 35 m
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