# Pressure in an enclosed tank

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, $$P_0$$?

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

$$P + \rho g h = P_0 + \frac{1}{2} \rho v_1^2$$, 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.
$${\rho}gy_1$$ 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.