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