2 edited body

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 $$\frac{\partial p}{\partial z} = 0$$$$\frac{\partial v}{\partial z} = 0$$.

So

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

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?

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 $$\frac{\partial p}{\partial z} = 0$$.

So

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

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?

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 $$\frac{\partial v}{\partial z} = 0$$.

So

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

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?

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# The pressure in a container of water is based on depth. So what happens if I remove the bottom of the container?

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 $$\frac{\partial p}{\partial z} = 0$$.

So

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

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?