# Bernoulli's equation in system of coupled tanks

Problem statement. Consider a system of two coupled tanks as the one shown below Several authors such as Bistak and Huba, 2014, Sim et al, 2017, Khalid and Kadri, 2012, Essahafi, 2014 claim that using Bernoulli's principle they can show that

$$\rho A_1 \dot{h}_1 = F_{\mathrm{in}} - c \sqrt{h_1 - h_2},\tag{1a}$$ and $$\rho A_2 \dot{h}_2 = \rho c \sqrt{h_1 - h_2} - c' \sqrt{h_2},\tag{1b}$$

assuming $$h_1>h_2$$.

The underlying assumptions are that:

1. the liquid is incompressible and inviscid,
2. the tube that connects the two tanks is very short,
3. the areas $$a_1$$ and $$a_2$$ are negligible compared to $$A_1$$ and $$A_2$$,
4. the flow is steady.

Background Using Bernoulli's principle I can show that for the case of a single tank the level of liquid is described by

$$\rho A \dot{h} = F_{\mathrm{in}} - c' \sqrt{h},\tag{2}$$

where $$c' = \rho a \sqrt{2g}$$.

My first attempt. I tried to derive equations (1a) and (1b) using Bernoulli's principle (this is what the authors I cited above claim to do).

Let $$A$$ be a point on the surface of tank 1, $$B$$ a point at the entrance of the tube, $$B'$$ is at its exit and and $$C$$ is a point on the surface of tank 2.

Then, by Bernoulli's principle from $$A$$ to $$B$$ we have

$$P_{atm} + \rho g h_1 = P_B + \frac{1}{2}\rho v_{B}^2\tag{3}$$

and from $$B'$$ to $$C$$,

$$P_{B'} + \frac{1}{2}\rho v_{B'}^2 = P_{atm} + \rho g h_2.\tag{4}$$

and from $$B$$ to $$B'$$

$$P_B + \frac{1}{2}\rho v_{B}^2 = P_{B'} + \frac{1}{2}\rho v_{B'}^2\tag{5}.$$

I guess there is something wrong with this approach because it clearly implies that $$h_1 = h_2$$.

My second attempt. If we apply Bernoulli's principle from point $$A$$ to point $$C$$ and assume that the pressure there is $$P_C = \rho g h_2 + P_{atm}$$, then

$$P_{atm} + \rho g h_1 = \tfrac{1}{2}\rho v_C^2 + \rho g h_2 + P_{atm},$$

and it follows that $$v_C = \sqrt{2g(h_1 - h_2)}$$, which leads to equations (1a) and (1b). Yet, I'm not sure I can take $$P_C$$ to be the static pressure at that point (it doesn't seem to follow from Bernoulli's equation).

Question. My question is how one can derive equations (1a) and (1b).

• Your instructor can't answer this question? – David White Apr 26 at 1:33
• @DavidWhite I'm studying by myself. – Pantelis Sopasakis Apr 26 at 2:35

## 1 Answer

Equation 1a is a mass balance for the first tank. The last term is the flow out of the tank.

If you had a free outfall from the first tank, then applying Bernoulli's principle would give you an expression for the rate of flow though the hole at the bottom.

$$p _{atm} + \rho g h_1 = p _{atm} + \frac {1}{2} \rho v_{out} ^2$$

When the tanks are connected, you have the same thing, except now the plume of flow from the first tank is discharging in to the bottom the second tank, where instead of $$p_{atm}$$, the pressure is $$p_{atm} + \rho g h_2$$. This gives you the expression for the rate of flow from the first tank into the second tank.

$$p _{atm} + \rho g h_1 = p _{atm} + \rho g h_2 + \frac {1}{2} \rho v_{out} ^2$$

You can use that to get the last term in equation 1a.

Equation 1b is a mass balance for the second tank. This time the flow into the second tank is just the flow out of the first tank. Also, the flow out of the second tank discrages into $$P_{atm}$$.

• My question is: Why can we assume that the pressure at the bottom of the second tank is $P_{atm} + \rho$? (See "My second attempt"). Is the pressure at the bottom of the second tank unaffected by the flow and the fact that we have connected the first tank to it? The pressure at the end of the second tank should satisfy, by Bernoulli's principle, $P_{B'} + \frac{1}{2}\rho v_{B'}^2 = P_{atm} + \rho g h_2$. Unless you assume that there is a point at the bottom of the tank, away from the point of inflow, where the speed of the liquid is negligible. – Pantelis Sopasakis Apr 26 at 2:41
• So, I guess I should apply Bernoulli's principle on a streamline that connects a point on the surface of tank #1 to such a point in tank #2. We should assume that the liquid enters the second tank gently or that the second tank is large enough for such a point to exist, right? Otherwise we would need to use Euler's equation of flow I suppose. – Pantelis Sopasakis Apr 26 at 3:06
• "Unless you assume that there is a point at the bottom of the tank, away from the point of inflow, where the speed of the liquid is negligible" That is it. The pressure at the bottom is determined by the depth of water in the tank. The water entering the tank has the pressure, plus extra energy. – Daddyo Apr 26 at 3:12