Pressure in an enclosed tank

Question

I'm trying to solve this task:

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:

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, $P_0$?

Answer 1

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.

Answer 2

${\rho}gy_1$ 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.