# Hagen Poiseuille's Law for water flowing through a bent tube

Recently, in a chemical engineering class, I was introduced the derivation for the form of the Hagen-Poiseuille equation. We started from the differential equation $$- A \frac{dh}{dt} = \frac{\rho g h}{R}$$, which is based on the conceptual idea that Flow = Driving force / Resistance. In the class, we were basically supposed to find out that for an outlet fitted with a bent tube, is the "h" we are supposed to use "h" (in black) or "h' " (in blue), as shown in the diagram below.

We did this by measuring the rate of change of the height of the water surface in the cylindrical tank. We then plotted two graphs of $$\ \frac{h}{h_0}$$ against time, one based on using the height "h" (in black) as the height and the other based on using "h' " (in black) as the height. We then plotted a similar graph based on the Hagen-Poiseuille equation using $$\ R = \frac{128 \mu L}{\pi d^4}$$ and then visually compared which experimental plot best matched the plot based on the theoretical model.

Intuitively, I thought that the correct height is "h" (in black) since the driving force for the flow should be the pressure difference between the inlet and outlet of the small tube and since the inlet pressure = $$\ P = P_0 + \rho g h$$, then the pressure difference should be $$\ \rho g h$$. However, the data suggests otherwise. It suggests that the correct height is "h' " (in blue) and based on the theory, that must mean that the pressure difference that causes the flow is $$\ \rho g h'$$. Could someone help to explain to me why this should be the pressure difference?

It suggests that the correct height is "$$h'$$" (in blue) and based on the theory, that must mean that the pressure difference that causes the flow is $$ρgh′$$. Could someone help to explain to me why this should be the pressure difference?

The mistake being made can be shown using Bernoulli's Principle.

With points $$1$$, $$2$$ and $$3$$ on the same flowline:

For points on the same flow line (and assuming inviscid flow), Bernoulli states:

$$\frac{v^2}{2}+gh+\frac{P}{\rho}=\text{Constant}$$ where $$v$$ is the flowspeed. Assume that the bent bit (down to point $$(3)$$) just isn't there for now, then for points $$1$$ and $$2$$ we can write: $$\frac{v_2^2}{2}+0+\frac{P_0}{\rho}=\frac{v_1^2}{2}+gh+\frac{P_0}{\rho}\tag{1}$$ because the pressures in $$1$$ and $$2$$ are both $$P_0$$ (and the initial level in the pipe is $$h=0$$)

Assuming the pipe's diameter is much larger than the outlet's diameter, then:

$$v_2 \gg v_1$$ so that from $$(1)$$ we can glean:

$$\frac{v_2^2}{2}\approx gh$$ By the same reasoning for point $$(3)$$:

$$\frac{v_3^2}{2}\approx gh'$$

So the flowspeed $$v_3$$ (and associated volumetric throughput) is larger than $$v_2$$ because the fluid falls over a larger height, thus acquiring more kinetic energy, rather than any pressure difference.

One can see this intuitively even more clearly by strongly increasing $$h'-h$$: flow speed will then increase strongly because of the fluid's increased kinetic energy.

In my opinion the classroom experiment in question is a very poor choice for deriving the Hagen-Poiseuille equation and seems more suited for [Torricelli's Law][3].

For instance:

$$-A \frac{dh}{dt} = \frac{\rho g h}{R}$$

is a wrong starting point. Solved, it would suggest exponential 'decay' of $$h(t)$$ in time, when really the starting DE should be:

$$-A\frac{dh}{dt}=a\sqrt{2gh}$$

where $$A$$ is the cross-section of the pipe and $$a$$ the cross-section of the outlet.

A relatively simple (crude even) set up for quantifying the Hagen-Poiseuille equation using pressure differences could be as follows:

Note that HP states:

$$\boxed{\Delta P=\frac{8 \mu L}{\pi R^4}Q}\tag{2}$$

and is only strictly valid for laminar flow through straight cylindrical pipes. Here $$\Delta P$$ is the pressure drop over $$L$$ length of pipe.

The pressure points are:

$$P_1=P_0$$ $$P_2=P_0+\rho gh$$ $$P_3=P_0$$ Thus:

$$\Delta P=P_3-P_2=\rho g h$$

By measuring $$Q$$ for various laminar combinations of $$h$$, $$L$$ and $$R (D)$$, $$(2)$$ can then be verified.

• I agree with you. This is a typical example for the application of Bernoulli theorem, under the assumption of negligible vorticity and viscosity effects, typically related with flow regimes with high Reynolds number (assumptions needed here to derive Bernoulli theorem from the very fundamental principles of fluid dynamics). This could be very misleading if used for deriving Poiseuille velocity profile, that is an exact solution for straight pipes in flow regimes characterized by very low Reynolds number (typically very small pipe diameter, small velocity, high viscosity) +1 Aug 27, 2022 at 13:07
• Thanks for the upvote.
– Gert
Aug 27, 2022 at 14:17
• So for the bent outlet scenario, HP would be completely inappropriate for modelling it? Additionally, you are also saying that the flow rate out of a bent outlet would also be higher since it travels through an additional vertical distance (during which, it experiences gravitational acceleration, and hence gains KE)? Aug 27, 2022 at 15:20
• Hi! Correct on both. However, one complication: water flow experiences a pressure drop in right angled bends (see eg. nvlpubs.nist.gov/nistpubs/jres/21/jresv21n1p1_A1b.pdf) This loss might offset the gain you get due to the higher falling height. All to play for. For simple set up to quantify HP, avoid all bends, valves, pipe roughness etc etc.
– Gert
Aug 27, 2022 at 15:41

I added labels to your figure for clarity.

Imagine if the tank were filled up to Point 2. Water would still be spilling out the downspout, no? Because the downspout mouth (at height 3) is still lower than height 2.

The difficulty you may be seeing is that in real life, as soon as some water flowed out, the level would dip below height 2, and flow would stop. But what matters is what is happening over a single instant when the water is at Height 2. Or if you like, imagine the diameter of the tank is very large. It would take quite a long time for the downspout flow to cause any change in the water level. So water level would sit at point 2 and spill out the downspout, under a pressure difference of $$(h'-h)$$