In the first module (Basic Structure) we examined fluid dynamics and the relationship between fluid velocity and pressure. Although this will play a role in our study of plumbing, in this section we will concentrate on the relationship between height and pressure. In the following section we will review fluid dynamics and examine the role that viscosity, adhesion and turbulence play in water flow through pipes.
Pressure and Height
Let's look at a simple cylindrical glass filled to a height h with a liquid. The glass must support the weight of the liquid. (The glass must also be supported by something, such as the table shown.) But what part of the glass does this? Believe it or not, if the sides of the glass are vertical, it is only the bottom of the glass that supports the weight!
The sides do exert a force on the liquid. But that force can only be horizontal. Here's why. An important property of a liquid is that it cannot support shear forces within itself. It cannot hold a shape. The molecules of a liquid can slide freely by one another. You can hold a solid, such as piece of paper, at the edge and the rest of the solid will stay in place because each molecule in the solid holds onto the one next to it. But you cannot hold a "piece of liquid" from the edge. The molecules will simply slide by each other. You must hold liquids in a container, where the bottom, top and side surfaces can act in a way to support the liquid.
For a static fluid (and slow moving fluids as well), the surface of the fluid's container can only exert a force perpendicular to the surface of the container. The diagram shows the force(s) that the glass exerts on the liquid and the equal but opposite force(s) that the liquid exerts on the container. Since the pressure is defined as the force per unit area, P = F/A, the diagram also represents the pressure exerted. Notice that (as you learned earlier in SCI 100) the pressure increases with depth. The reason is really quite straightforward. To see why, let's calculate the force and pressure exerted on the bottom of the glass by the liquid.
The total force exerted on the bottom of the glass by the liquid is exactly equal to the weight of the liquid. Weight is just mass times gravity (W = mg) and the mass is the mass density (r = mass/volume) times the volume (m = r V). The liquid has the shape of a cylinder, so its volume is just the area of the top (or bottom), which we will label A, times the height h. So, V = Ah. And finally, the pressure on the bottom of the glass is the liquid's weight divided by the area of the bottom. Put it all together and we have:
mass (m) = r V = rAh
Pressure (P) = Force / Area = (rAh) g / A
P = rgh
Although the calculation above specifically found the pressure at the bottom of the glass, it gives the pressure at any depth. We can visualize an imaginary plane surface (think of it as composed of a layer of molecules) at any depth. The liquid below the plane always supports the weight of the liquid above the plane. So the pressure on that plane is rgh . And on what is this pressure exerted? This pressure is exerted on every element of the fluid by all other elements around it. It has no direction, or it has every direction.
Problem:
We will deal with the Cistern in more detail in the next module, but a simple calculation will help understand just how important pressure is to designing the cistern. In many house designs, the cistern is built into the hillside and becomes the back wall of the bottom floor. You know that every few years the cistern should be cleaned and cracks repaired. Rather than place an access hole in the top, you've decided it would much more convenient to place the access door at the bottom of the back wall. You want to be able to easily get in and out, so you choose to make the door a modest 80 cm tall by 50 cm wide. What force would be acting on this door if the cistern was full?
Solution: Let's estimate that for a full cistern, the water height would be 2 meters above the floor. The depth from the top of the water to the center of the door would be 1.75 m. and the pressure at the center of the door would be
P = rgh = 1000 kg/m3 x 9.8 m/s2 x 1.75 m = 1.72 x 104 Pa (Recall that Pa = Pascal = N/m2).
If we take this as the average pressure over the door, the total force would be
F = PA = 1.7 x 104 N/m2 x (.8 m x .5 m) = 6,900 N.
Which is about 1500 lb! Maybe it's not such a good idea after all.
Gauge Pressure
The equation for pressure we have been using is missing a potentially important term. It is the pressure at the surface of the liquid. For the glass of liquid, you probably assumed that it was open to the atmosphere ... a not unreasonable assumption. There is atmospheric pressure all around us. We are at the bottom of a sea of air. At sea level, the atmospheric pressure averages about 105 Pa or about 15 psi (pounds per square inch).
NOTE: Standard Atmosphere Pressure is defined to be 1.01 x 105 Pa = 14.7 psi.
It varies as cold and warm fronts move in and out of the area, but the variation is typically only a few percent and changes over periods of hours or days. (The lowest pressures you are likely to ever encounter at sea level would be in the eye of a major hurricane. Even for Lenny, Marilyn and Hugo, the eye pressure was less than 10% below this average value.)
So did we make a mistake not including it in the derivation above? Not really, because atmospheric pressure also acts at the bottom of the glass. Let's denote the pressure at the surface from which h is measured as Po. The pressure at a depth h is actually the sum of the surface pressure and the rgh term.
Problem: 
Let's consider an example of gauge pressure. It is common knowledge among snorklers that you must "equalize" your mask as you dive. Equalizing is a technique in which you exhale into the mask to make the pressure inside the mask equal to the water pressure outside the mask. What happens if you dive to a depth of 2.0 m without equalizing?
Solution: At the surface, the air trapped inside your mask is at atmospheric pressure, Po, and the pressure outside is the same. But at a depth h, the outside pressure is Po + rgh. The net pressure on the mask is just the difference. See if you can show that there will be a net pressure difference of 2.0 x 104 Pa. (The density of sea water is 1030 kg/m3, a bit higher than freshwater.)
A typical mask has dimensions of about 15 cm by 10 cm. So you should be able to show that creates a net force of 300 N or about 67 lb. That might be a bit uncomfortable.
Now consider the cistern problem from above. Should we have considered atmospheric pressure in that problem? Why or why not?
The Water Pump System
The most common water delivery system in the Caribbean involves a cistern and water pump. The details of how a water pump works is left as an option for a special project. For now, simply think of it as device that moves water. The typical setup is quite interesting. The pump "pulls" water from the cistern and pushes it into a holding tank that has an air pocket trapped inside. There is a pressure sensor that turns the pump on when the pressure in the tank falls below a minimum set level and turns it off when a maximum set pressure level is achieved. Most systems allow the user to set the levels. For discussion, we'll use a typical setting of 20 psi and 40 psi for the lower and upper levels, respectively.
The diagram above shows a 2-story house. The fixtures represent showers, sinks, toilets, etc. The pressure at the surface of the water in the tank is the Po term in our equation. Note that in this case, it is not atmospheric! Pump pressures are essentially always given as gauge pressures. That is, the minimum pressure level of 20 psi is really 20 psi relative to the atmosphere. The absolute minimum pressure in the tank would be about 20 psi + 15 psi = 35 psi. But the pressure just outside the fixtures is also one atmosphere, so it is the 20 psi difference that we will use to determine how high the water can be pushed.
The 20 psi gauge pressure will be able to push water from the top surface in the holding tank to a maximum height hmax, given by P = 20 psi = rghmax. The fixtures will need to be below this level if water is to reach them. Assuming all fixtures are below this height, the minimum gauge pressure at the fixtures will be
Pfixture = 20 psi - rgh
Note the minus sign! That is because the fixtures are above the tank. The gauge pressure at the fixtures will be less than that of the holding tank. (You could think of the "depth" h as being negative.) For the first floor h = h1 and for the second floor h = h2. Both h1 and h2 must be below hmax, or the gauge pressure at the fixture would be negative. If a fixture is below the holding tank, then the pressure would be 20 psi + rgh. As we will see in the next section, the pressure at the fixture determines the speed at which water will flow out of the fixture.
Air in the Tank
What is the role of the air pocket trapped in the holding tank? Why doesn't the outlet from the pump connect directly to the pipe carrying water to the fixtures? The answer lies with the compressibility of gases and the incompressibility of liquids.
You have may noticed that the pump usually does not turn on immediately after you open a faucet or start the shower. Similarly, it does not turn off as soon as you stop the faucet or shower. The diagram to the right will help you see why this is so. The first picture in the diagram represents the point at which the pressure in the tank has dropped to 20 psi. The pump turns on and begins pumping water into the tank. As the tank fills, the air is compressed. When the volume of the trapped air is compressed to about half the volume in the first picture, the pressure will be 40 psi and the pump will turn off. As water is used, the air pocket expands, the volume increases and the pressure decreases. The pump starts again once the pressure drops to 20 psi when the first air volume is reached. The difference in the volume of the air pocket between these two pressure levels is the volume of water than can be used before the pump turns on again. (For a small home system, this difference in volume typically corresponds to a few gallons of water.) You have probably noticed how the speed of the water coming from a fixture slowly drops until the pump comes on again. But the variation is usually not an inconvenience.
However, if the pump was connected directly to the faucets, the behavior would be very different. Assume there is 40 psi pressure at the outlet of the pump. Since liquids are essentially incompressible, the pressure would drop almost instantly when a faucet was opened. It would drop to zero gauge pressure after only a fraction of a second. The water velocity would drop to nothing during this time, until the pump comes on in response to the pressure drop. The water would begin flowing again and the pump would remain on until the faucet was closed. Those who have experienced this situation when a faulty holding tank loses the air pocket know why this is undesireable. The rapid change in pressure creates a few problems. First, the initial drop in pressure followed by the pump starting often creates a burst of water that splatters. Second, the rapid change in pressure creates a strain in the pipes and its joints. The "clanging" response of the pipes is called "water hammer". (It is a problem that must be dealt with in larger apartment buildings by using special pressure relief devices.) Even over a short period of time, water hammer can cause pipes to fail, often in the areas of joints or bends.
Problem: You are building a two story home with a bathroom on the second floor. The cistern is just below the first floor. Will a water pump that maintains a minimum gauge pressure of 20 psi be adequate for your home? (You will need to estimate the heights.)
Solution: The pressure in the holding tank "pushes" water from the holding tank to the fixtures in the bathroom. Let's estimate how far this might be. If the holidng tank is at the same level as the floor of the cistern, it could be about 2 m below the first floor. The second floor will be about 3 meters above the first and the faucet about a meter above that. So the total distance between the holding tank and the fixtures could be about 6 meters. The pump can supply a pressure difference of 20 psi. In order to raise water 6 meters, the gauge pressure at the outlet pipe needs to be P-Po = 20 psi = rgh . Let's convert this pressure to Pa. Since 1 atm = 14.7 psi = 1.01 x 105 Pa, you should be able to show that 20 psi = 1.4 x 105 Pa. Solving for h we get h = 14 m. Even if our estimate is off by a meter or two, the pump should have no problem pushing water to the second floor fixtures.
Here's a good rule of thumb. One atmosphere of pressure corresponds to a depth of about 10 m or 34 ft of fresh water.
Using this rule of thumb, we can set up a simple ratio to find the height (h) to which 20 psi can lift the water.
We could also have used 34 ft to find that 20 psi could raise water 46 ft.
Buoyancy
From common sense (or perhaps because your did well in SCI 100), you probably know that objects will float in a liquid when they are less dense than the liquid. (This also applies to gases as well. That is why hot air balloons can float in the surrounding cooler and denser air.) The upward force that the liquid exerts on these objects is called the buoyancy force. But even objects that don't float also have a buoyancy force. Try lifting a rock or piece of loose coral from the sea floor. It will feel lighter while submerged and heavier when lifted out of the water. The reason is easy to understand and was first noted by Archimedes. (There is evidence that earlier cultures certainly understood buoyancy, but Archimedes described it scientifically.) Archimedes' Principle states:
Since the weight of the fluid is w = mg and the mass is related to the displacement volume and density through m = rVdisp, we have
When any part of an object is in a liquid, the pressure that the surrounding liquid exerts at any point on the surface of the object is exactly the same as it was before the object was there. The liquid that was displaced by the object was being supported by the rest of the liquid around it. Since the pressure is strictly a function of depth, the surrounding liquid pushes on the object in exactly the same way it did the displaced liquid. Hence, the net force exerted on the object is equal to the weight of the displaced liquid. If the object is more dense than the liquid, then its weight will be greater than that of the displaced liquid even after it is completely submerged. It will sink to the bottom. However, if the object is less dense, then as the object is lowered into the liquid, it will displace its own weight before it becomes completely submerged. In this case:
This is true even for the massive cruise ships that visit our islands. The ships displace a volume of water whose weight is exactly the same as that of the ship. But the steel from which they are made is much more dense than water. (Steel is about 8 times more dense than water.) So how do you think the ships manage to float?
The Water Closet
The water closet is a old, nicer name for a toilet. Buoyancy plays an important role in the tank of the most common type of toilet. Most tanks will have a flapper and a float ball. Both rely on buoyancy to operate.
When you flush a toilet, the lever you turn (or button you push) pulls up on a hinged flapper. The flapper is a rubber flap with an air pocket inside that covers the outlet hole at the bottom of the tank. When lifted, water flows out of the tank and into the toilet bowl. The flapper stays up away from hole because it is "slightly" buoyant. But once the water level drops to the level of the flapper, it begins to fall. Usually, it falls back over the hole before the water level drops completely to the hole.
The float ball floats at the top of the water level and is connected to the water inlet valve. As soon as the water level in the tank begins to drop, the float ball drops and turns on the inlet valve. (The water drains into the bowl much faster than it comes in through the inlet.) Once the water level drops to the point that the flapper falls back over the outlet hole, the tank begins to fill. Once the float ball reaches a certain level, the inlet is turned off and the toilet is ready for the next flush. Next time you flush a toilet, take the cover lid off and watch how it works. The mechanical workings of most toilets are easy to see.
Are you ready for a few QUESTIONS about fluid statics?
The next topic is Fluid Dynamics.
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