0
$\begingroup$

In Bernoulli equation, the only term which corresponds to what one usually calls "pressure" is $p$, the other are still pressures dimensionally but their meaning is linked to kinetic and potential energy per unit volume. Nevertheless I get confused especially for the term $\rho g h$. Can it be seen as a pressure due to weight, similarly to Stevin law, but for hydrodynamics?

My guess would be a clear no because Stevin law in only valid in hydrostatics, but consider, for istance this example, where a liquid (say water) is flowing in the venturi meter.

enter image description here

To read the manometer and calculate pressure difference I have to take into account not only the height $h$, but also the difference $z_2-z_1$ between the two points in the tube.

But, since in the manometer the consideration are about hydrostatics that means to consider a further pressure difference given by $\rho g (z_2-z_1)$, which reminds Stevin law for the height difference $z_2-z_1$. But Stevin law is valid iff the fluid is static, which is not here.

So does a fluid in motion have the same "effect" in terms of pressure caused by weight, than a situation where the fluid is static?

The question is also: is there a pressure difference between point $1$ in the center of the tube and a point on the same section but very close to the tube boundary? (Does the weight of the fluid between $1$ and the boundary create a pressure difference between these two points?)

| cite | improve this question | |
$\endgroup$
1
$\begingroup$

The question is also: is there a pressure difference between point 11 in the center of the tube and a point on the same section but very close to the tube boundary? (Does the weight of the fluid between 11 and the boundary create a pressure difference between these two points?)

The answer to this is Yes. Even with a horizontal tube, the pressure at the center of the tube is lower than at the bottom of the cross section.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the reply! My guess would be that this pressure difference is given by $\rho g H$, where $H$ is the difference in height between the point in the center of the tube and the one on the boundary. This would explain why in the manometer the difference in height between $1$ and $2$ is taken into account as in a hydrostatic problem. But that would be incorrect since the fluid is moving, right? So I'm really failing in understanding why a pressure difference $\rho g (z_2-z_1)$ is considered in the manometer. $\endgroup$ – Sørën Jun 19 '16 at 7:58
  • 1
    $\begingroup$ If you assume that the pipe diameter is small compared to the distances z1 and z2, then the pressure at level z1 on the right exceeds the pressure at z1 on the left by $(\rho_B-\rho_W) g h$, where the subscript B refers to the black fluid and the subscript W refers to the right fluid. So, $P_1-P_2=(\rho_B-\rho_W) g h+\rho_W g (z_2-z_1)=\rho_Bgh+\rho_W(z_2-z_1-h)$. If the density of the white fluid is negligible, then $P_1-P_2=\rho_Bgh$ $\endgroup$ – Chet Miller Jun 19 '16 at 11:40
  • $\begingroup$ Ok thanks! Just to be sure that I got that right I added the points $3$ and $4$ in the picture. If the density of water $\rho_W$ is not neglegible, then I can say that $p_1=p_3-\rho_W g (z_1-z_3)$, right? And this comes from Bernoulli equation for irrotational flow between point $3$ and $1$ (in fact $v_1=v_3$) , correct? In the same way $p_2=p_4-\rho_W g (z_1-z_3)$. Then I used $p_3$ and $p_4$ to do the balancing of pressures in the manometer and I get the correct result, so it looks right. $\endgroup$ – Sørën Jun 20 '16 at 13:21
  • 1
    $\begingroup$ Right. Incidentally, I would have put the manometer connections at the locations here the plane normal to the axis intersects the wall to determine the pressure difference along the axis, and I would have oriented to connection normal to the wall. $\endgroup$ – Chet Miller Jun 20 '16 at 13:57
  • 1
    $\begingroup$ Sure. If the flow is horizontal, then in the vertical direction from point 1 to point 3, the pressure variation is hydrostatic. For laminar flow in a tube, the pressure gradient in the horizontal direction is constant, and independent of radial position. $\endgroup$ – Chet Miller Jun 23 '16 at 23:41
0
$\begingroup$

In Bernoulli equation, the only term which corresponds to what one usually calls "pressure" is $p$, the other are still pressures dimensionally but their meaning is linked to kinetic and potential energy per unit volume. Nevertheless I get confused especially for the term $ρgh$.

That is corect. The term $\rho gh$, better written as $\rho g(z_2-z_1)$, is the potential energy (per unit of volume) increase of a mass element travelling along the flow line.

In a gravitational field (not too far from the Earth's surface) that potential energy change only depends on any change in $z$ (and not on path e.g.), where $z$ is the vertical spatial coordinate. So that change is $\rho gz_2-\rho gz_1=\rho g(z_2-z_1)=\rho gh$ (if we set $h=z_2-z_1$).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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