Fluid Mechanics in Mechanical Engineering, Part II, Fluid Statics Continued

     Last week we talked about basic concepts of fluid statics, using the example of a hydraulic jack.  This week we’ll continue talking about fluid statics and explore another example.

     Let’s say you have a 10 foot deep swimming pool and you want to know what the water pressure is at its bottom.  Your common sense may tell you that that pressure can be measured by examining the weight of the water itself, and you would be partially correct.  But there is another factor that you may not have considered to be significant to the calculation. 

     At this point let me pose a question.  How do we breathe?  Your answer is directly related to measuring that pressure at the bottom of the pool.  You see, contrary to popular belief, air is not weightless, and because it has weight it impacts objects, creating an external pressure.  This pressure is known as atmospheric, or barometric, pressure.  When you breathe, your diaphragm, which is moved by a muscle, allows the lung cavities to expand or contract accordingly.  When the diaphragm is lowered the lung cavities are increased in volume, allowing the atmospheric pressure of your surroundings to come rushing in to fill the space that has been created.

     Along those lines, it was discovered some time ago that in order to determine the pressure at a given point within a body of water, we must consider not only the depth of the water itself, but also the atmosphere that is above it, bearing down on it.  This is illustrated in Figure 1.



Figure 1 – A Swimming Pool Filled With Water


     In equation form, this relationship looks like this: 

            P = (Atmospheric Pressure) +

                           (The Specific Weight of Water) x (The Depth of the Water)

     Now, it’s known that atmospheric pressure at sea level is about 14.7 pounds per square inch.  So let’s say that this is the pressure being exerted on the surface of the water in our example by the weight of the air above it.

     As for the second component of our equation, the water, its specific weight is held to be a constant of approximately 0.036 pounds force per cubic inch.  Now there is only one thing left to do before solving our equation, and that is to convert the water’s depth from feet into inches. This must be done so that the units of depth (inches) match the units of specific weight (pounds force per cubic inch) in our calculation.  The depth of the water would therefore be 10 feet times 12 inches per foot, or 120 inches.  Now we can return to our equation, insert these values, and solve for the pressure at the bottom of the pool: 

P = 14.7 lbf/in2 + (0.036 lbf/in3) x (120 in) = 19.02 lbf/in2 = 19.02 psi,

where “psi” is an abbreviation for pounds per square inch.

     Notice how the pressure at the bottom of the pool doesn’t depend on how wide or long the pool is?  It only depends on its depth.  This means the deeper into a body of water that you go, the more water weight will be bearing down upon you, that is to say, the water pressure increases.  This is why submarine hulls have a propensity to collapse if they dive too deep.  The pressure from the water above gets to be too great compared to the air pressure inside the submarine, and the metal of its hull stresses to the breaking point.  It’s like squeezing an egg in your hand.

     This wraps things up for the topic of fluid statics.  Next week we’ll continue with our fluid mechanics series and talk about fluid dynamics.  This area of fluid mechanics involves fluids that move, like water moving through pipes and air flowing over airplane wings.  



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