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SOLVED PROBLEMS

 

PROBLEM 1: Because of a leak in a lucried gasoline storage tank, water has seeped into the depth as shown in figure. The specific gravity of the gasoline is S.G=0.68.Determine the pressure at the gasoline-water interface and at the bottom of the tank. Express the pressure in units of , and as a pressure head in meters of water?

SOLUTION:

Since we are dealing with liquids at rest, the pressure distribution will be hydrostatic, and therefore the pressure variation can be found from the equation

With p₀ corresponding to the pressure at the free surface of the gasoline, then the pressure at the interface is

If we measure the pressure relative to atmospheric pressure (gauge pressure), then p₀ = 0, and hence

It is noted that a rectangular column of water 3.4m tall and 1m² in cross section weighs 33.32 kN.

A similar column with a 1mm² cross section weighs 0.033N.

We can now apply the same relationship to determine the pressure at the tank bottom, that is

And,

PROBLEM 2: A spherical balloon of diameter 1.75m and total mass 1.4 kg is released in the atmosphere. Assuming that the balloon does not expand and that the temperature lapse rate in the atmosphere is . Determine the height above sea-level to which the balloon will rise. Atmospheric temperature and pressure at sea-level are 15°C and 101 KPa respectively, for air

SOLUTION:

Density of balloon=

Therefore, Balloon rises until atmospheric density

Now, we know

 

…………………………………… (1)

Since,

l

…………………………………… (2)

Again,

From (2) ,

Therefore solving we get,

Z = 839.986m