# Hi every one, My question is what should be a pressure in a open end

by anwaar
Tags: pressure
 P: 1 Hi every one, My question is what should be a pressure in a open end tube,in which one end is connected to a pump and the other end is dipped in a liquid container. the pressure of the end which is connected to the pump. thanx. if some one has some idea. working. My view is the pressure at this end must be equal to the atmospheric pressure minus the pressure of the liquid. i could not find ans to this question. my assumption is base on pressure in a manometer. thanx
 P: 419 I think the concept you're looking for is buoyancy. For example, why does a steel ship float even though steel is more dense than water? One nice example that might help you is the example of a U-shaped tube with water filling the bottom up to a height h, where one end of the tube is open to the atmosphere and the other is attached to a piston. (Remember the force on the piston is equal to the cross-sectional area of the piston times the pressure in the tube, so pressure and force are freely interchangeable.) Implicit in this experimental setup is that the water and air are in a gravitational field--without gravity g, the water would float around freely in the air. Gravity is crucial to this entire argument. When no force is applied to the piston, the water level in both sides of the tube is equal, because the piston end and the open end are in equilibrium with ambient pressure. When a force is applied (downward) to the piston, the water level in the piston side moves down while the water level in the open side moves up. Let's assume both sides have the same cross-sectional area, so the piston side water level moves down by δ while the open side water moves up by δ. How can we relate this to the pressure? If you stare at this long enough, you'll realize that the work required to cause this difference in water levels is exactly equal to the work required to lift the displaced chunk of water by δ. If the cross sectional area of the piston is α and the water has density ρ, this is equivalent to lifting the mass m=ραδ up a height δ, which we can calculate easily using W=mgh, the usual expression for gravitational potential. Instead of doing the calculus to find W = ∫ F dx = ∫ p(x)*α dx, we can simply equate the gravitational force of the lifted chunk with the piston force. mg = F => (ραδ)g = pα thus ρδg = p. If you understand this argument, you should be able to figure out the answer to your question.

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