-1
$\begingroup$

I was looking up at the derivation of Torricelli's law in my textbook:

$$P_a + \frac{1}{2}{\rho}v_1^2 + \rho gy_1=P + \rho gy_2$$

This is the equation representing one of the steps for its derivation. Before I start the main part of my question I would like to create an imaginary model with which the question is related.

The model Consider an open tank containing a liquid of density $\rho$ with a small hole in one side at a height $y_1$ from bottom. There is air above the surface of the liquid whose height is $y_2$. This air is at pressure P. The cross-sectional area of the tank is much larger than that of the hole which is the reason why I neglected the kinetic energy term on the right hand side of my equation above for velocity here as it would be negligible.

The main question Why have we used the pressure energy term on the left hand side of the equation i.e. pressure energy of water at hole equal to atmospheric pressure? This pressure according to me must be greater than atmospheric pressure for it is at depth that the pressure of water increases. My textbook says:

Since both the cross sections are open to the atmosphere, the pressure there equals to atmospheric pressure.

Improve this question
$\endgroup$
3
  • $\begingroup$ If somebody can give me a hint through comment,to my question, it will be really helpful.I don,t want to get stuck on it with wrong concept. $\endgroup$
    – Sikander
    Jul 16, 2016 at 18:15
  • $\begingroup$ With the velocity in there, surely you mean Bernoulli's principle rather than Torricelli's. $\endgroup$
    – dmckee --- ex-moderator kitten
    Jul 16, 2016 at 20:13
  • $\begingroup$ @dmckee basically I am focussing on Bernoulli's principle but since it was a part to derive formula for the speed of efflux so I wrote down Torricelli's law as is in my textbook to give just an idea what I was going to do but it may look superfluos over here. $\endgroup$
    – Sikander
    Jul 16, 2016 at 20:23

1 Answer 1

Reset to default
4
$\begingroup$

The pressure at any point open to the atmosphere is the atmospheric pressure.

1) You are using concepts of fluid statics and applying it to fluid dynamics. When an entire fluid is in complete equilibrium, pressure must increase proportionally with depth, since the mass of the fluid above a certain depth must be completely balanced by the pressure forces acting from below. For example, if you are standing in a stationary elevator, then the force exerted by the floor on you must be exactly equal to your weight, since you are in equilibrium. But if the elevator is descending with some acceleration, the force must be lesser than your weight. Similarly, in this case the fluid is flowing and is not in equilibrium.

2)If the hole were absent, the wall of the container would exert a force on the liquid, which prevents it from just flowing out through the wall. Now, by making a hole, there IS no wall. The air will exert a force on the water having magnitude of the product of atmospheric pressure and the area of cross section of the hole. And this force would not be sufficient to hold back the liquid, causing it to flow out.

Also, when a fluid accelerates due to gravity through a hole, the pressure variation in the fluid is quite complicated and does not linearly vary with depth. The value of pressure gets closer and closer to the atmospheric pressure as we get closer to the hole, inside the fluid.

This may also be explained in the basis of continuity. If the pressure JUST outside the hole is equal to the atmospheric pressure, one can expect the pressure at any point in the near vicinity to be approaching that value.

Improve this answer
$\endgroup$
3
  • $\begingroup$ In simple words, pressure at a point is DEFINED as the perpendicular force acting per differential area at the point. The source of pressure in this case is the air in the atmosphere. My first point was just to clear the misconception you had that pressure always increases with depth $\endgroup$
    – Newton
    Jul 16, 2016 at 18:49
  • $\begingroup$ For what I have understood I would like to add to your answer through comment that the extra pressure energy of the liquid at depth has been converted to kinetic energy so that now the pressure of water is equal to atmospheric pressure. $\endgroup$
    – Sikander
    Jul 16, 2016 at 18:50
  • $\begingroup$ For more details you should look at lagrangian fluid dynamics. $\endgroup$
    – Newton
    Jul 16, 2016 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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