# Water container on a balance and Bernouilli's principle

Consider a cylindrical water container on a balance. Suppose the water surface is initially at level $h$ above the balance. The pressure at the bottom of the container is then $$p_i = p_0+\rho g h,$$ where $p_0$ is the atmospheric pressure. Because the air pressure also acts on the opposite side of the balance, the balance reading is $$m_cg + a\rho g h,$$ where $m_c$ is the mass of the container and $a$ is the cross-sectional area of the container.

Now suppose we create a hole in the container at some level $h_t$ and insert a tube trough the hole. Let $A$ be a point at the water surface, $B$ be a point at the end of the tube inside the container and let $C$ be a point at the other end of the tube, outside of the container.

With obvious notation, we then have, by Bernouilli's principle, $$p_0+\frac{1}{2}\rho v_A^2+\rho g h = p_0 + \frac{1}{2}\rho v_C^2 + \rho g h_t$$ and $$p_B + \frac{1}{2}\rho v_B^2 + \rho g h_t = p_0 + \frac{1}{2}\rho v_C^2 + \rho g h_t.$$ Since the flow is incompressible, we must have $v_B = v_C$, and consequently, $p_B = p_0$. This all seems well, there shouldn't be any pressure drop for an ideal flow in a horiontal pipe of constant cross-sectional area.

However, what does this imply for the balance reading, since we now have $p \neq p_i$ at the bottom of the container? Is it going to drop dramatically? Certainly, at other points on the streamline we are not required to have $v = v_C$, so the pressure can be greater there. At the same time, it would surprise me if we could have a pressure $p \approx p_i$ near the bottom, as this would imply, for large $h$, a large pressure gradient along the bottom towards the tube.

Perhaps my question isn't very well phrased. Although I haven't yet done the experiment, it seems that simply drilling a hole in the container would dramatically decrease the balance reading, which seems rather counter intuitive to me.

• The second equation, if correct, implies that a zero balance reading is impossible, with the minimum varying with the weather... – DJohnM Jul 13 '16 at 14:28
• your assumption that the balance reads $m_c g + ap_i$ is only valid if your vase has vertical walls, but assuming that is the case in your example, it is true that the reading will diminish if you make a hole in the vase and create a pressure drop at the bottom. However, I believe is not straightforward to estimate how much, as it depends on the size of the hole (which creates a pattern of drop pressure across the bottom). – Wolphram jonny Jul 13 '16 at 15:55
• The second equation is definitely incorrect. There is atmospheric pressure on top of the liquid in question, but there is ALSO atmospheric pressure on the under-side of the balance, pushing back up. Thus, the atmospheric pressure term should NOT be inside the parentheses on the right hand side of the second equation. – David White Jul 13 '16 at 16:09
• The conclusion that pB = p0 is also incorrect. The driving force for fluid flow is a pressure difference. If the pressures on both ends of the tube are the same, the flow must be zero! – David White Jul 13 '16 at 16:24
• vb = vc is incorrect as well. You have to vonsider the diameter of the water jet just outside, which is smallrer than the hole diameter. Read fluid dynamics chapter from feynman vol2. – Lelouch Jul 13 '16 at 16:51

Now pressure difference between bottom of container and tube entrance, $p_{bottom}-p_B$, can be large if total height $h$ is large. There is nothing surprising about it. Consider a streamline from bottom to tube entrance. As a blob of water rises along that streamline, part of the pressure difference ($p_{bottom}-p_B$) available to it is utilized in rising to the height where tube entrance is located. The left over is exactly what is required to provide speed $v_B$ (just inside the tube) to water.