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## Fluid Mechanics - Pascal's Principle and Hydrostatic Pressure

The key assumption we make here is that the only force we need to consider on the fluid is the force exerted on the small piston (150 pounds). If this is truly the only force acting on the fluid, then it will likewise be the only source of fluid pressure, and pressure will simply be equal to force divided by area (150 pounds ÷ 3 square inches = 50 PSI).

However, when we are dealing with tall columns of fluid, and/or dense fluids, there is another force we must consider: the weight of the fluid itself. Suppose we took a cubic foot of water which weighs approximately 62.4 pounds, and poured it into a tall, vertical tube with a cross-sectional area of 1 square inch:

Naturally, we would expect the pressure measured at the bottom of this tall tube to be 62.4 pounds per square inch, since the entire column of water (weighing 62.4 pounds) has its weight supported by one square inch of area.

If we placed another pressure gauge mid-way up the tube, though, how much pressure would it register? At first you might be inclined to say 62.4 PSI as well, because you learned earlier in this lesson that fluids naturally distribute force throughout their bulk. However, in this case the pressure is not the same mid-way up the column as it is at the bottom:

The reason for this apparent discrepancy is that the source of pressure in this fluid system comes from the weight of the water column itself. Half-way up the column, the water only experiences half the total weight (31.2 pounds), and so the pressure is half of what it is at the very bottom. We never dealt with this effect before, because we assumed the force exerted by the piston in the hydraulic lift was so large it “swamped” the weight of the fluid itself. Here, with our very tall column of water (144 feet tall!), the effect of gravity upon the water’s mass is quite substantial. Indeed, without a piston to exert an external force on the water, weight is the only source of force we have to consider when calculating pressure.

An interesting fact about pressure generated by a column of fluid is that the width or shape of the containing vessel is irrelevant: the height of the fluid column is the only dimension we need to consider. Examine the following tube shapes, all connected at the bottom:

Since the force of fluid weight is generated only along the axis of gravitational attraction (straight down), that is the only axis of measurement important in determining “hydrostatic” fluid pressure. The fixed relationship between the vertical height of a water column and pressure is such that sometimes water column height is used as a unit of measurement for pressure. That is, instead of saying “30 PSI,” we could just as correctly quantify that same pressure as 830.4 inches of water (”W.C. or ”H2O), the conversion factor being approximately 27.68 inches of vertical water column per PSI.

As one might guess, the density of the fluid in a vertical column has a significant impact on the hydrostatic pressure that column generates. A liquid twice as dense as water, for example, will produce twice the pressure for a given column height. For example, a column of this liquid (twice as dense as water) 14 inches high will produce a pressure at the bottom equal to 28 inches of water (28 ”W.C.), or just over 1 PSI. An extreme example is liquid mercury, which is over 13.5 times as dense as water. Due to its exceptional density and ready availability, the height of a mercury column is also used as a standard unit of pressure measurement. For instance, 25 PSI could be expressed as 50.9 inches of mercury (”Hg), the conversion factor being approximately 2.036 inches of vertical mercury column per PSI.

The mathematical relationship between vertical liquid height and hydrostatic pressure is quite simple, and may be expressed by either of the following formulae:

Where,

P = Hydrostatic pressure in units of weight per square area unit: Pascals (N/m2) or lb/ft2

ρ = Mass density of liquid in kilograms per cubic meter (metric) or slugs per cubic foot (British)

g = Acceleration of gravity (9.8 meters per second squared or 32 feet per second squared)

γ = Weight density of liquid in newtons per cubic meter (metric) or pounds per cubic foot (British)

h = Vertical height of liquid column

Dimensional analysis vindicates these formulae in their calculation of hydrostatic pressure. Taking the second formula as an example:

As you can see, the unit of “feet” in the height term cancels out one of the “feet” units in the denominator of the density term, leaving an answer for pressure in units of pounds per square foot. If one wished to set up the problem so the answer presented in a more common pressure unit such as pounds per square inch, both the liquid density and height would have to be expressed in appropriate units (pounds per cubic inch and inches, respectively).

Applying this to a realistic problem, consider the case of a tank filled with 8 feet (vertical) of castor oil, having a weight density of 60.5 pounds per cubic foot. This is how we would set up the formula to calculate for hydrostatic pressure at the bottom of the tank:

If we wished to convert this result into a more common unit such as PSI (pounds per square inch), we could do so using an appropriate fraction of conversion units: