0
$\begingroup$

I'm trying to solve this task:

enter image description here

Where I have to find v1, the velocity at the exit point of the hole.

I know I can use Bernoulli's equation. My question is what should I consider to be the pressures. I was thikning of writing the equation like this:

$$P_0 + \frac12 \rho v_1^2 = P + \frac12 \rho v_2^2$$

Yet in my textbook it is written: enter image description here

Why do they add the terms with $y_2$ and $y_1$? I would have thought that $\rho g y_2$ is the pressure in the bottom, not in the surface. And doesn't the pressure in point 1 have to be simply the atmospheric pressure, $P_0$?

$\endgroup$
  • $\begingroup$ Bernoulli is an energy conservation equation. The potential energies $\rho gh$ along the flow line must be accounted for. $\endgroup$ – Gert Sep 26 '19 at 21:38
  • $\begingroup$ the tank seems closed, I do not think P is the atmospheric pressure $\endgroup$ – Wolphram jonny Sep 26 '19 at 21:43
0
$\begingroup$

The author based his tank pressure reference point on the bottom of the tank, which in my opinion is mathematically correct, but more difficult to interpret from a physical standpoint. Another way to formulate the Bernoulli equation for this problem is to only consider the height of the liquid above the hole, which gives:

$P + \rho g h = P_0 + \frac{1}{2} \rho v_1^2$, where $P_0$ is atmospheric pressure. Note that there is no $\frac{1}{2}v_2^2$ term in the left hand side of this equation because the continuity equation must be followed, and the assumption for this type of problem is that $A_2$ is so much larger than $A_1$ that $v_2$ approaches zero, and can be ignored.

Since $h = y_2 - y_1$, a substitution yields

$P + \rho g (y_2 - y_1) = P_0 + \frac{1}{2} \rho v_1^2$, which is equivalent to what the author wrote.

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

${\rho}gy_1$ and ${\rho}gy_2$ denote the potential energy of the fluid. This is important because the two points are not at the same level. Point 2 is at a higher potential than point 1(that is why fluid is flowing from 2 to 1). Also,yes the external pressure at both point 1 and point 2 will be equal to the atmospheric pressure and they will cancel out.

| 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.