6
$\begingroup$

I'm having trouble understanding Bernoulli's principle, in particular, why an increase in the section area of a hose increases the pressure?

All the answers I read say: " In order for the energy to be conserved" or "when the tube is compressed, the fluid has to speed up so that the same amount of the fluid gets out of the tube in the same interval of time (continuity law)".

I'm not looking for the finality : energy conservation. I'm trying to understand the mechanism, what happens at the molecular level?

$\endgroup$
  • $\begingroup$ At he molecular level number of particles coliding with the unit area of the surface decreases as the area increases. $\endgroup$ – Žarko Tomičić Feb 7 '17 at 16:29
  • $\begingroup$ So far none of the answers below have dealt with your original misconception. According to Bernoulli's principle, the pressure increases in the section of hose with the larger cross section. $\endgroup$ – D. Ennis Feb 7 '17 at 18:09
  • $\begingroup$ I fixed the OP's mistake (probably just a typo) to eliminate some of the confusion. $\endgroup$ – Pirx Feb 7 '17 at 21:37
  • $\begingroup$ Corrected the title $\endgroup$ – Bill N Feb 7 '17 at 23:13
  • $\begingroup$ Hello! I think I've been asking the wrong question. I wanted to know what happens at the "transition region" and Bernoulli's law describes what happens when the fluid is already flowing. Also the principle is applied in continuum mechanics where things are not explained at the molecular level. $\endgroup$ – user142405 Feb 8 '17 at 10:30
6
$\begingroup$

I'm trying to understand the mechanism, what happens at the molecular level?

Bernoulli's Principle belongs to continuum mechanics, so it's not well suited to make pronouncements about action at the molecular level (but I'll get back to that further down).

Between two points on the same flowline Bernoulli states:

$$P_1+\frac12 \rho v_1^2+\rho gz_1=P_2+\frac12 \rho v_2^2+\rho gz_2$$

The case below is for an inviscid, incompressible fluid, with no potential energy changes ($z=\mathrm{constant}$) and regularly shaped conduits (the cross sections $A_1$ and $A_2$ are well defined):

Bernoulli equation

Then:

$$P_1+\frac12 \rho v_1^2=P_2+\frac12 \rho v_2^2$$

The relationship between the flow speeds is given by incompressible continuity:

$Q_v=A_1v_1=A_2v_2=\mathrm{constant}\implies v_2=\frac{A_1}{A_2}v_1$

So that:

$P_2=P_1+\frac12 \rho (v_1^2-v_2^2)$

$P_2=P_1+\frac12 \rho \Big(1-\frac{A_1^2}{A_2^2}\Big)v_1^2$

Thus: $\boxed{A_2>A_1\implies v_2<v_1\implies P_2>P_1}$

So what you call the 'mechanism' is entirely due to satisfying the continuity requirement. Counter-intuitively perhaps, here a decrease in flow speed $v$ results in an increase in pressure $P$.

At the molecular level, lower bulk fluid pressure is caused by increased average distances between molecules, as these increases decrease Coulombic repulsions between the electron clouds that make up the fluid's molecules. This results in lower numbers of collisions with the counduit's wall and thus lower pressure.

$\endgroup$
  • $\begingroup$ I was trying to think of a satisfying way to explain it closer to a molecular level and kept ending up at the unsatisfying solution you gave in your last paragraph. Your point on continuity is great and was what I was struggling to realize. Continuity laws by definition aren't applicable to systems close to the molecular level. $\endgroup$ – JMac Feb 7 '17 at 16:54
  • $\begingroup$ @Gert you show that the pressure increases with increasing cross-sectional area, but the OP begins with the opposite assumption. Also, the "molecular level" explanation seems to contradict the finding in the derivation by explaining lower pressure. $\endgroup$ – D. Ennis Feb 7 '17 at 19:05
  • $\begingroup$ @D.Ennis: that's because the OP's assertion was wrong, see e.g. hyperphysics.phy-astr.gsu.edu/hbase/pber.html . Also, the molecular level explanation of pressure stands alone from Bernoulli (which is continuum mechanics). $\endgroup$ – Gert Feb 7 '17 at 19:58
  • $\begingroup$ @Gert, Yes, I know that his assertion was incorrect, as I expressed in my comment following his question. I just found it remarkable that no one told him so explicitly in their answers. Also, are you saying that the molecular level explanation is entitled to contradict Bernoulli's principle? $\endgroup$ – D. Ennis Feb 7 '17 at 20:22
  • $\begingroup$ @D.Ennis how does it contradict it? $\endgroup$ – JMac Feb 7 '17 at 20:50
2
$\begingroup$

Your quest for a "mechanism" is almost sure to fail as you are talking about how an Avogadro number of molecules interact with each other. That is precisely why macroscopic quantities like energy and density (in the continuity equation) are used.

However, we can still try to get some intuition more than just saying energy is conserved (I will still use it though) for the case of ideal gas.

The point is pressure come from force per unit area from random motion of molecules. When the fluid has to (continuity equation) pass through a narrow tube, it has to increase its velocity in a certain direction (i.e. the tube direction) and then since energy has to be conserved, assuming 'thermalization' happens at time scales much faster than it takes to pass through the tube, the fraction of energy available for 'random' kinetic energy and thus pressure, is less.

If you prefer mathematical expressions, for $N \gg 1$ molecules of unit mass, the total energy (ignoring potential energies from bound states and walls of container) is

$$ 2 E = \sum_{i=1}^N \vec v_i.\vec v_i = \sum_{i=1}^N (\vec v_i- \vec v_0 + \vec v_0 ).(\vec v_i- \vec v_0 + \vec v_0 ) \\ =\sum_{i=1}^N (\vec v_i- \vec v_0).(\vec v_i- \vec v_0) + N~\vec v_0. \vec v_0 $$ where $\vec v_0 = \frac{1}{N} \sum_{i=1}^N \vec v_i$ is the mean velocity of the $N$ molecules and the cross terms vanishes in the limit $N \to \infty$ because the 'random flucatuations' about the mean are isotropic. The first part contributes to pressure and the second is the velocity through the tube. You can clearly see the effect you mentioned.

$\endgroup$
  • $\begingroup$ My answer should be taken with a grain of salt as I use ideal gas kind of arguments and yet take it to be in-compressible. However, I think as a hand-wavey argument it does 'explain' how lower preassure comes from a lack of equi-partition of energy. $\endgroup$ – Borun Chowdhury Feb 9 '17 at 14:15
-1
$\begingroup$

When The Section Area Increases,the Fluid Molecules Are Spread Widely Thus The Collissions Per Unit Area Decreases .This Makes The Corresponding Exerted Pressure Decrease.

$\endgroup$
  • $\begingroup$ the density of the fluid remains, it is not in number of collisions/area $\endgroup$ – jaromrax Feb 7 '17 at 17:28

protected by Qmechanic Feb 9 '17 at 11:35

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.