# How do I calculate the pressure $p_0$ on top of the water reservoir?

Can anyone help me with this question, I can't get the right answer :-)

A water reservoir with depth D=200m, which is placed H=500m above on a mountain. A pipe is placed from the bottom of the reservoir to the bottom of the mountain. This pipe has two cross-section areas $$A_1=10m^2$$ and $$A_2=5m^2$$. The atmospheric pressure in the bottom is $$p_2=100kPa$$ and the air density down here is $$\rho_2=1 kg/m^3$$ and waters density is $$\rho_w=1000 kg/m^3$$.

What is the atmospheric pressure $$p_0$$ at the surface of the reservoir?
It has to be 93 kPa

My attempt on a solution:

I know I need to use Bernoulli's equations:

$$p_{0}+\rho_w g(H+D)+\frac{1}{2} \rho_w v_{0}^{2}=k$$ $$p_{1}+\rho_w g H+\frac{1}{2} \rho_w v_{1}^{2}=k$$ $$p_{2}+\rho_w g(0)+\frac{1}{2} \rho_w v_{2}^{2}=k$$

And the continuity equation. $$A_{1} v_{1}=A_{2} v_{2}$$

Usually I would assume $$v_{0} \simeq 0$$, but they don't mention that the surface area is much larger than the two cross sections. So can I still assume this, or should I forget about the first equation?

But so far (assuming v0=0), I got 4 equations and 5 unknowns.

We are assuming the water got the same density. And if the water is a standing liquid (v0=0), we could calculate $$p_{1}=p_{0}+\rho_{w} g D$$

But solving this system of equations gives me: $$p_{0}=p_{2}-\rho_{w} g(H+D),$$ which is completely wrong, since I get a negative pressure and the cross sections areas becomes irrelevant. How do I solve this exercise? And why do I even need to know the density of the air at the bottom of the mountain?

• Static water pressure only depends on depth below the surface, so you don't need most of the data in your problem. Also, at the bottom of a 700 m column of water, it's not possible to have a pressure of 100 kPa. The pressure should be at least 70 times that value. Also, note that the pressure $P_0$ should be equal to the local atmospheric pressure. Aug 17, 2019 at 21:25
• @DavidWhite But my final result also gives me, that the pressure p2 only depends on the atmospheric pressure p0 and the depth below the surface (and water density + g). But why is the correct result of the pressure $P_0=93kPa$? But the water is not static, so that's probably where the issue are.
– mhj
Aug 17, 2019 at 21:31
• $P_0$ is a boundary condition, not something that you calculate. You start at the value of $P_0$, and calculate pressures going down the mountain from there. Aug 17, 2019 at 21:33
• @DavidWhite: "it's not possible to have a pressure of 100 kPa". I think it's meant to be the atmospheric pressure at $P2$.
– Gert
Aug 17, 2019 at 21:34
• @DavidWhite - Well, it's an exam question and I know the correct answer is $P_0=93kPa$ - So it has to be possible to calculate.
– mhj
Aug 17, 2019 at 21:36