1
$\begingroup$

enter image description here

This bent, flat tube is located on a horizontal plane and ideal fluid is flowing through it, as shown by the arrows.

According to me, pressure at 2 would be higher than pressure at 1. I think so because, for the flow to be sustained as it is depicted, velocity at 1 should be higher since the fluid there would have to cover a larger distance than it has to in 2. So, from Bernoulli's theorem, since heights are same, pressure should be larger at 2.

But my book says just the opposite. But if $v_1$ is lower than $v_2$, the flow wouldn't be straight as depicted in the picture. I'd preferably go with the book's answer but I also want to know WHY that is the case.

$\endgroup$
2
$\begingroup$

This is an ideal fluid so where is the origin of the force which causes the particles to change their speed?

The fluid is moving down the tube and has to turn a corner.

The particles which make up the fluid have to undergo a centripetal acceleration.

For particle 1 touching the wall it is subjected to an "inward" force from the wall and an "outward" force from its fluid neighbours.

The magnitude of the inward force from the wall is larger and so the particle goes around the corner.

Particles within the fluid are subjected to inward and outward forces due to their nearest neighbours and again the inward forces have a larger magnitude.

So as one moves inwards the pressure decreases.

Using Bernoulli this implies that the speed of the outer particles is slower than those on the inside of the bend.
Imagine particle moving along a straight tube encountering a bend.
Those particles on the outside of the bend will encounter a region of greater pressure than on the inside of the bend and being subjected to a larger force will slow down more.

$\endgroup$
0
$\begingroup$

According to Bernoulli, enter image description here $$\rho gh_{2}+P_{2}+\frac{\rho v_{2}^2}{2}=P_{1}+\rho gh_{1}+\frac{\rho v_{1}^2}{2}$$ NOW,I assume the height of the point 2 be $H$(from the given plane) and point 1 be $h$ so we also know that the cross sectional area is same for both the points as they are in same Section. According to the rate of flow being constant non turbulent flow. $A$=area of the section.$v_{1}$ velocity of point 1 and $v_{2}$ $$Av_{1}=Av_{2}$$ From which $$v_{1}=v_{2}$$ Applying bernoulli therom we get $$\rho gh+P_{1}=P_{2}+\rho gH$$ Now $H \gt h$(evident from the figure) Hence rearranging

notice heights are from given plane

so, $$P_{1}-P_{2}=\rho g\left(H-h\right)$$ which proves that $$P_{1}\gt P_{2}$$

$\endgroup$
  • $\begingroup$ I already said it's located on a horizontal plane. $\endgroup$ – Nabaneet Sharma May 6 '17 at 11:20

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.