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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.

Qmechanic
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Étienne Bézout
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  • 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). –  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
  • Related : Pressure just before the hole in a draining tank, http://physics.stackexchange.com/q/263458 – sammy gerbil Jul 13 '16 at 18:38
  • @DavidWhite Yes, you're right about the second equation, and I have edited it accordingly. However, why is pB = p0 incorrect? We're talking about an ideal flow, and we can assume it is steady if the container is large. Under such assumptions, Bernouilli's principle holds, and for a stream line in a horizontal pipe of constant cross-sectional area there is not going to be any pressure drop. – Étienne Bézout Jul 13 '16 at 19:30
  • @Lelouch I don't have that book available at the moment. Why can we not consider a point just at the end of the pipe? – Étienne Bézout Jul 13 '16 at 19:32
  • The pressure has to change from inside to outside. For a continuos vhange, a finite length of water flowing is needed. – Lelouch Jul 14 '16 at 08:34
  • You could add in displacement terms from the density parameters, then expand them in a vector and take the rotation of pressure. –  Jul 14 '16 at 11:27

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Suppose the hole is small compared to dimensions of the container. Then motion of water inside the container will be small (read up "sink flow"). So net vertical momentum of water inside the container will be small as well. Now if the tube is horizontal so that water jet issuing out of it is horizontal, then a control volume analysis (in which you draw a control volume enclosing container+tube) tells you that weight reading shown by the balance at any instant is equal to sum of container's weight and weight of water present inside container+tube. So as water is emptied out, weight reading decreases gradually and not abruptly.

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.

Deep
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