So I understand that if we have a system that involves a container of water the pressure will equal atmospheric pressure at the top and as we go further down the container the pressure will increase with depth until it is at its maximum pressure at the bottom of the container. In terms of a 1-D fluid equation (neglecting viscosity) we have

$$\frac{\partial v}{\partial t} = \frac{1}{\rho} \frac{\partial p}{\partial z} - \frac{\partial v}{\partial z} + g$$

$$0 = \frac{1}{\rho} \frac{\partial p}{\partial z} - 0 + g$$

$$\frac{\partial p}{\partial z} = \rho g$$

$$p = \rho g z + const$$

$$p = \rho g z $$

where we have taking the constant to be $0$.

Now say you removed the bottom of this container. Obviously the water starts to flow out the bottom. So the velocity changes with time. As the fluid is incompressible we still have $\frac{\partial v}{\partial z} = 0$.


$$\frac{\partial v}{\partial t} = \frac{1}{\rho} \frac{\partial p}{\partial z} + g$$

But if pressure still balances with gravity we have no fluid flow! But the fluid does flow so the pressure must not balance with gravity anymore. So what is the pressure in the system now that the bottom of the container has been removed?


2 Answers 2


Take a 1-cm square tube and place it vertically in the container from top to bottom, touching the bottom so that the bottom of the container is the bottom of the tube.

The pressure at the bottom of the tube is nothing but the weight of water it is supporting - the water in the tube. Supporting means to keep from falling. (Forget the air pressure - that's just confusing the issue.)

If you take away the bottom of the container, you are taking away the bottom of the tube, so it is no longer supporting the column of water, so the weight of water above goes to zero because it is not being supported. So the pressure goes to zero because it's nothing more than the weight of water being supported.

You don't need equations to understand this.

  • 2
    $\begingroup$ Maybe a good analogy is weightlessness in free fall. $\endgroup$
    – Bernhard
    Aug 23, 2015 at 9:08

If you remove the bottom you no longer have a container. The pressure is atmospheric and you just have gravity at work.

  • $\begingroup$ So pressure ceases to exist when the bottom is removed? And the pressure at the top equals the pressure at the bottom? But $\frac{\partial v}{\partial t} = g$ is not correct either because unless more water flows in, after a few seconds the container (which is now a tube of some length) is empty and $v = 0$ everywhere..$\frac{\partial v}{\partial t} = g$ just requires an initial condition which is $v(z, 0) = 0$ which at the top of container hold true for all time due to the water flowing out. So how can that be reconciled with $\frac{\partial v}{\partial t} = g$? $\endgroup$
    – sonicboom
    Aug 22, 2015 at 13:50
  • $\begingroup$ I don't follow. Yes when all the water exists the tube there will be no more water in the tube. The water will accelerate in the tube and continue to accelerate once it exists the tube. $\endgroup$
    – paparazzo
    Aug 22, 2015 at 16:23
  • $\begingroup$ @sonicboom If you consider the container only, you will have to account for the mass flowing out as well. Think of what your control volume for the balance equation is. $\endgroup$
    – Bernhard
    Aug 23, 2015 at 9:10

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