If the air pressure in the sealed tank of Figure were 30 kPa, what would the pressure gag...

If the air pressure in the sealed tank of Figure were 30 kPa, what would the pressure gage read at the tank bottom and how high would the water rise in the vertical tube?

FIGURE

Illustration for Example. When the valve is open, the water will rise in the vertical pipe to a certain height h that depends on the pressure in the tank.

Example

A sealed tank, shown in Figure, has a pocket of air trapped above the water, which is 1 m deep. A pressure gage at the bottom of the tank reads 30 kPa. Determine (a) the pressure head of water at the tank bottom, (b) the height that water will rise to in the vertical tube if the valve is opened, and (c) the pressure in the trapped air.

Solution

(a) Using Equation, we can compute the pressure head as follows:

Notice that the pressure head is greater than the depth of water in the tank. This means that the air in the tank must be exerting additional pressure, pushing downward on the water.

(b) The water would rise 3 m in the vertical tube, a height equal to the pressure head at the bottom of the tank.

(c) If the tank were open to the atmosphere, then 1 m of water depth would cause a pressure of only 9.8 kPa to register on the gage. The difference, or 30 - 9.8 = 20 kPa (rounded to two significant figures), must be exerted by the pressurized air in the sealed tank. Therefore, the air pressure in the tank is 20 kPa. (The pressure in a small volume of gas is considered to be uniform and does not depend on the height or depth of gas.)