Hydrostatic Pressure

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Contents

Introduction

In the previous section (Fluids and Solids Introduction) the concept of pressure was given. When dealing with fluids, it is important to realize that they exert pressure perpendicular to any contacting surface. In addition, when the fluid is at rest that pressure is isotropic, i.e. it acts with equal magnitude in all directions. If the forces were not balanced, the fluid would move in the direction of the resulting force to stabilize it. This concept was first formulated by Blaise Pascal in 1647 and would later be known as Pascal's law.

Hydrostatic pressure

Consider what happens as one goes under water. At the surface, we will just the ambient pressure around us, but as one goes down, we feel that pressure along with the pressure of the water. In fact, the further we go down, the higher the pressure gets because more and more water is pushing down above us. We can generalize this idea if we also consider what would happen if it weren't water, but say, mercury. Mercury is much heavier than water and so we would expect it to cause even a stronger pressure as we go under it. Now also consider what happens if we did the same scenario on the moon, the moon's gravity is far weaker than that of the earth and so we would expect gravity wouldn't pull the liquid down as hard and the pressure wouldn't go up as quickly as we went under. In general, all these concepts can be summarized in the following formula,

\ P_{hydrostatic} = \rho g h +P_{a}

where P is the hydrostatic pressure (Pa), ρ is the liquid density (kg/m3), g is gravity (m/s2), h is the height of liquid above (m), and Pa is the atmospheric pressure (Pa).


1. What is the hydrostatic pressure at a depth of 5m in a fresh water lake?

100 KPa
125 KPa
150 KPa
175 KPa
To answer this problem we will need to use the hydrostatic pressure equation along with a few constants. We will need to know what atmospheric pressure is (101 KPa) and what the density of fresh water is (1 g/cm cubed, or 1000 kg/m cubed). Keeping our units consist, we get total pressure = atmospheric + pgh = 101000 + (1000)(9.8)(5) = 150000Pa.

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Gauge Pressure

When we measure the pressure inside a tire, what are we actually measuring? Inside the tire the air is at a certain pressure (heavily compressed at say 4 atmospheres) and outside, while we have become accustomed not to notice it, there is an air pressure (say at ground level or 1 atmopshere).

Thus as the air inside the tire flows through the nozzle to the outside it is really going from 4atms to 1atm, with a difference of 3atms. This difference is what we call the Gauge Pressure- it is the pressure above the ambient level.

Similarly, submarines measure gauge pressure as they go deeper down, and thus do not include the pressure at sea level, only the changes in pressure.

 P_{total} = P_{atm} + P_{gauge}


As you can see from above then, when we are dealing with liquids, the gauge pressure is simply the ρgh term in the previous formula.


1. What will happen to the gauge pressure of a car's tires as it drives up a mountain?

The gauge pressure will decrease
The gauge pressure will increase
The gauge pressure will remain the same
the gauge pressure will increase but then decrease at a certain height
If we consider what gauge pressure is, it is the difference between the total pressure and the atmospheric pressure. As a car drives moves up, the total pressure in the tires doesn't change, but the surrounding atmospheric pressure does decrease the higher the car gets. Thus to compensate for this change, the gauge pressure must be increasing such that the total remains constant.

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Pascal's Law

In researching the properties of fluids, pascal formulated a theory that was later to become Pascal's Law. Simply stated, a change in the pressure of an enclosed incompressible fluid is conveyed undiminished to every part of the fluid and to the surfaces of its container.

It is important to recognize the effect of this law. Applying a pressure at one point in the fluid will result in that same amount of pressure being conveyed elsewhere in the container. And, because the fluid is incompressible, if we apply enough pressure to say push in or move 2L of volume, then the fluid will equally push back 2L of volume somewhere else.

Mathematically, we can look at the pressure we apply and the resulting pressure the liquid conveys as,

\begin{align}P_{1} &= P_{2}\\
\frac{F_1}{A_1} &= \frac{F_2}{A_2}\end{align}

But we can also consider what is happening to the volume of liquid we are displacing in say the pipe as,

\begin{align}V_{1} &= V_{2}\\
{A_1}{d_1} &= {A_2}{d_2}\end{align}

Thus, if we apply a pressure to one end of the pipe which just so happens to push the contents a distance d1, then the liquid at the other end will be pushed out a distance d2 (which may or may not be equal to d1 depending on if the cross sectional area of the pipe at both ends is the same or not). We can also take these formulas one step further now, and recognize that we can combine them to get,

{F_1}{d_1} = {F_2}{d_2}



Which is a lot like just saying the work we apply is equal to the work the liquid does, which is exactly what it does. This concept is taken advantage of in hydraulic pumps. Have you ever wondered how a car can be lifted so easily at the mechanics? Well consider what would happen if we applied a small force for a very long distance? The work done in exchange by the liquid could be a huge force (enough to lift a car) but over a small distance. And this is in fact how it works.

The hydraulic Pump

Using the concepts above, it is easy to see that just like levers and pulleys, we can use hydrostatic pressure to our advantage in performing difficult work. Consider the system above were there are two compressible pads connected through a liquid system. If the left has an area that is 5 times smaller than the right area (say 50 and 10) then pushing the pad on the left with 100N would result in a pressure of (100/10) = 10Pa. This pressure, according to pascal's principle is equal to the pressure on the second, larger pad. Thus, if the second pad's pressure is 10Pa and its area is 50, then the force it is exerting is 500N. (Notice what has happened, the ratio of area is 5, and now the ratio in force is 5!) What does this mean? A small force can be converted into a huge force, but what is the catch? Well, work can't just get easier and in fact it does not, it stays the same. To get the large pad (or piston) to move 10m, we have to push the small pad 50m. The reason for this is that energy must always be conserved. Work is force times distance, and since the force is increased on the larger piston, the distance the force is applied over must be decreased. The work of the small piston, 100 newtons multiplied by 50 meters is 5000 joules, which is the same as the work of the large piston, 500 newtons multiplied by 10 meters.


1. Force = __________ x ____________ ?

flow x distance
work x time
pressure x flow
pressure x area
The only answer choice which provides meaningful units is pressure (F/m2) x area (m2)

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