# Questions about Hagen-Poiseuille Equation: Non-constant Pressure Drop?

The Hagen–Poiseuille equation states that the fluid flow rate in a tube is proportional to the pressure gradient across that tube. Mathematically, it's stated here: $$\Delta p=\frac{8\mu LQ}{\pi R^{4}}.$$ Imagine however, the following situation: you have two balloons connected by a pipe (with one balloon on each side). The balloon on the right side (Balloon R) has fluid in it, filled up till it's almost going to burst — i.e., the fluid is at a high pressure in that balloon. The left balloon (Balloon L) is empty. Before $$t=0$$, there is a shut valve which separates the fluid in Balloon R from the tube. At time $$t=0$$, you remove the valve.

The H-P equation seems to assume that the pressure drop over the section of the pipe remains constant. But here, the pressure on the left side of the tube will eventually go from 0 to a high value as fluid flows from Balloon R into Balloon L, and similarly the pressure in Balloon R will drop. So the pressures eventually equalize, and flow should stop right?

That is, I'm curious because it seems the fluid flow across the pipe should affect the pressures on each side of the pipe, such that the pressure drop is not constant.

Am I right?

• In your example both $\Delta p$ and $Q$ will be instantaneous, that is a function of time. But the HPE remains valid at every moment in time.
– Gert
Apr 23 at 0:10

## 1 Answer

The equation you give is derived under the following conditions (I quote from Bird et al.'s Transfer Phenomena):

(a) The flow is laminar.

(b) The density is constant ("incompressible flow").

(c) The flow is "steady" (i.e., it does not change with time).

(d) The fluid is Newtonian.

(e) End effects are neglected.

(f) The fluid behaves as a continuum.

(g) There is no slip at the wall.

Your example violates at least (c); thus, you can't depend on this equation to suitably predict the outcome. You would need a transient treatment of the flow.