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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