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# U-Tube

A glass U-tube is shown in its "Initial Condition," 1). The tube is filled with mercury to the level "0 - - 0", its legs are open at their tops to the atmosphere. The cross-sectional areas, left and right, are AL and AR. Beside the U-tube is a cup containing mw grams of water.

The event is that the water is poured into the right leg whereupon the fluids, under gravity, attain a second state of equilibrium. We expect the level of mercury in the left leg to rise above "0 - - 0." Represent the densities of water and mercury as: ρw and ρm. Obtain an algebraic expression for the final height of mercury in the left leg, hm.

Solution:  A needed first step is to sketch an "anticipated" final condition of the system and to apply some dimensions to it. It is not necessary that you draw the "proper" sketch but simply that you get something to start with. For instance see Figure 2). To proceed we look at the system/event in its final condition in terms of its properties: geometry, mass and momentum.

1. Water Column:  Water is incompressible; its mass is constant and its density is known. Consequently the water volume is the same, initially and finally - the change of water is change in shape. In the final configuration, we a first equation, (i), that has one unknown quantity, L:

2. Fluid Volume Above "0 - - 0":  Initially in the U-tube, the mercury volume is unknown. But once the water is added some fluid will reside above the "0 - - 0" datum. Mercury on the left and water on the right.
3. Momentum of the static fluids:  Newton's 2nd Law applied to static fluids is called the hydrostatic equation:

Hence there are three equations with three unknown quantities: L, hw and hm. We leave it here.

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