# Unexpected result: flux using Poiseuille greater than using Bernoulli

I have a simple, easy to solve, Poiseuille type of problem, to which I added a second question, in which viscosity is zero, so I can solve it using Bernoulli. I am arriving a greater flux using Pouseuille than Bernoulli, which is to me counterintuitive. Here is the problem and a summary of my solution:

From a jar, and through a circular tube, water flows that is poured into the atmosphere through a needle 3 cm in length and with an internal radius of 0.36 mm. When the jar is at a height of 1.5 m, a) what is the flow rate of fluid that comes out through the needle? (density of the fluid, 1.05 g cm–3; viscosity of the fluid, 1.3 cp). b) what would be the flow rate from the needle if we did not take into account the viscosity of the plasma?

*a) I solve it using Poiseuille's equation, considering $$\Delta P=\rho gh_A$$, in which $$h_A=1.5m$$: $$Q=\frac{\pi R^4}{8\mu L}\Delta P=\frac{\pi (0,36\cdot 10^-3)^4}{8\cdot 1,3\cdot 10^{-3}\cdot 3\cdot 10^-2)}1050\cdot9,81\cdot1,5=2,61\cdot 10^{-6}\ m^3/s$$

b) If there's no viscosity, we can use Bernoulli's equation between the top of the jar and the exit of the needle, assuming the velocity of the fluid on top of the jar $$v_A=0\ m/s$$ since it is much wider than the needle: $$P_{atm}+\rho gh_A=P_{atm}+\frac{1}{2}\rho g v_B^2$$, which results in $$v_B=\sqrt{2gh_A}=5,42\ m/s$$, which then I use to find then for finding the flux, $$Q=vS=v\pi R^2=2,21\cdot10^{-6}\ m^3/s$$

How can the flux when I use Poiseuille's equation be greater than when I use Bernouilli? Shouldn't it be the other way around? Thank you

• Bernoulli ignore viscosity and ignores boundary conditions. Completely different domain of applicability. Commented Oct 27, 2023 at 23:11

If you are going to include Poiseuille, you are going to have to recognize that the Bernoulli and Poiseuille resistances are in series. If v is the velocity in the capillary, then, from Bernoulli, the pressure p at the entrance to the capillary is $$p_{atm}+\rho g h = p + \frac{1}{2}\rho v^2$$or $$p=p_{atm}+\rho gh-\frac{1}{2}\rho v^2$$From Poiseuille, $$p-p_atm=\frac{2L}{R}\mu\frac{4v}{R}=\mu\frac{8Lv}{R^2}$$So we have $$\rho g h=\frac{1}{2}\rho v^2+\mu\frac{8Lv}{R^2}$$This is a quadratic equation for v.