Fluid Dynamics: Answering Pressure in a Full Tank

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SUMMARY

The discussion centers on calculating the lowest pressure in a full tank of water when a valve at the bottom is opened while the top valve remains closed. The height from valve to valve is 30 feet. The pressure is determined using the formula: 14.7 psi - (ρ)gh, where ρ is the density of water, g is the gravitational acceleration (32.2 ft/s²), and h is the height (30 ft). It is concluded that at a height of 40 feet, the minimum pressure drops below the vapor pressure of water, making the vessel unfeasible at that height.

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  • Familiarity with the concept of vapor pressure
  • Basic grasp of gravitational acceleration (32.2 ft/s²)
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shannajo03
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I understand the concept of this question, I think, but I don't know how to go about getting an answer.

A tank is completely full of water with no air. A valve at the bottom is opened, while one on top remains closed. What is the lowest pressure that the tank will reach. From valve to valve the height is 30ft.

I understand that as the water drains, it will pull a vacuum inside the tank, which will eventually collapse at a low enough pressure. I also know that the water will vaporize at some point due to the lack of pressure in the tank, but I don't know if they want the vapor pressure (no temp was given) or if there is some calculation to do.

Any guidance is appreciated!
 
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For anyone who is wondering, I found out the answer. You assume atmospheric pressure pushes up into valve B, so the pressure in the tank is 14.7psi-(rho)gh where rho is the density of water, g is gravitational accel. (32.2ft/s^2), and h is 30ft. The point was that on the second part (h=40ft), the minimum pressure is below the vapor pressure of water, so the vessel is not possible at that height.
 

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