Hydraulics

Hydraulics

FIRE SCIENCE

Harnessing Our Laws of Nature

A tmospheric pressure and hydrostatic liquid pressure are used for thousands of applications in our industrial world which is full of hydraulic pumps and systems. Pressure applied through liquids has a nature that is characterized as being smooth and even in the operation of valves, switches, and levers. Pressure applied to a confined liquid from without, and/or the creation of a negative pressure from within, permit what appears to be a miraculous creation or multiplication of energy in the form of a force that can be used to hold back or deliver liquids, draft, lift hundreds of pounds of steel, etc.

It is hard to believe that much of what we daily encounter in fire service hydraulics is based upon principles explained hundreds of years ago. These natural laws were revelations of the wonderful ways that pressure can be transmitted and utilized—and they still are today.

DRAFTING OF WATER

Have you ever wondered how the pump on your apparatus can lift water from a pond or a river? Perhaps you haven’t. But if you were told that negative pressure played a part, would that mean anything to you?

You may answer, who cares? What’s the difference? All you have to do is to push a group of buttons in a certain way and the pump does the work.

Yet, for those firefighters who do see the value of studying hydraulics, the principle of drafting water is a wonderfully simple application of common sense—and one of nature’s forces known as atmospheric pressure.

Atmospheric pressure

A long time ago, the weight of the earth’s atmosphere (the air’s weight) and the earth’s surface area was calculated. When the weight of the atmosphere was divided by the earth’s surface area, it was found that approximately 14.7 pounds of the atmosphere rested on each square inch of the earth’s surface. The figure 14.7 pounds per square inch (psi) is when the sea and the land is at the same level.

If you are on top of a mountain, there would be less atmosphere above you, consequently, less weight and, therefore, less than 14.7 psi. If you are at the bottom of a deep valley, there will be more atmosphere above you and more than 14.7psi impacting on every square inch of surface. Atmospheric pressure, therefore, is a source of natural energy ready to be used.

In Figure 1, with 14.7 psi being put on the suction connection, the water will only raise to Point A because at this level the atmospheric pressure within the suction equals the atmospheric pressure forcing the water up into the suction. If the water is to go higher, the air pressure within the suction connection must be reduced.

Negative pressure

Figure 2 shows the suction connection of a pumper in which the air pressure has been reduced to 5 psi. Here, the water has risen to over 22 feet because 9.7 psi of atmospheric pressure has been removed from inside the suction connection above the water (14.7 —9.7 = 5 psi). With less pressure inside the hose, atmospheric pressure acts on the water, pushing it upward like a drinking straw effect.

The 9.7 pounds that has been removed is referred to as negative pressure. Negative pressure is equal to atmospheric pressure minus actual pressure (9.7 psi = 14.7 — 5). Therefore, we can say that negative pressure is the difference between atmospheric pressure (14.7 psi) and the actual pressure that is in a contained space (5 psi), providing that the difference (9.7 psi) is the result of an actual pressure of 14.7 psi or less. The following will help illustrate what we have been discussing:

By referring to Figure 2 and to the chart on negative pressure, it can be seen that the negative pressure created in the suction connection really is responsible for pressure of an equal amount to push the water up into the pump. When the 9.7 psi was removed from the suction connection, it allowed 9.7 psi of atmospheric pressure in the water to move the water upward.

By experimentation, it is known that one pound of pressure is created by a column of water 2.304 feet in height. Therefore, we can say that for every pound of pressure there is the ability to lift water 2.304 feet. Applying this knowledge to our 9.7 psi of upward energy, we find that the water will move up into the suction to a height of 22.35 feet (9.7 X 2.304). This relationship can be expressed as:

Lift = negative pressure X 2.304

Vacuum

Negative pressure is a difficult concept for many people to cope with. It really is an indication of how much nothingness there is in our suction. Another term that is more frequently used to describe the absence of air in a contained space is vacuum. If we remove all the air from the suction, all of the 14.7 psi of atmospheric pressure will also be removed. There will be 14.7 psi of negative pressure and a perfect vacuum.

Figure 3 shows how high a perfect vacuum will allow atmospheric pressure to push a column of water upward. The glass tube has its entrance below the water’s surface in a container that is open to the atmosphere. Prior to the water moving upward, the tube was under a perfect vacuum. The water has risen to approximately 34 feet.

Lift = negative pressure x 2.304

— 33.9 = 14.7 x 2.304

Inches of mercury

If instead of using water in the container in Figure 3, mercury was used, the atmospheric pressure would have pushed the mercury up to a height of approximately 30 inches. Using the ratio of the height of water as compared to the height of mercury, one inch of mercury represents a lift of 1.13 feet of water.

Figure 1Figure 2Figure 3

By looking at a pump’s compound suction gage, a pump operator can read inches of mercury when a vacuum is created or pounds of pressure under normal operations. When beginning to lift water (draft) he should take careful note of the inches of mercury that the gage shows. For each inch, the pump can lift water approximately 1.13 feet. If it is necessary to lift the water 15 feet, the operator must get a reading of a little more than 13 inches of mercury. This relationship can be expressed as:

L = Hg” x 1.13

where L is lift of water in feet; Hg” is inches of mercury; and 1.13 is feet of water lift per inch of mercury. Another variation of the basic formula to find lift is:

Hg” = 15/1.13 or Hg” = 13.27

By experimentation, it is known that one inch of mercury will indicate a pressure of approximately .492 psi. Therefore, there will be approximately a half pound of negative pressure within a pump for each inch of mercury on the compound gage. This relationship can be shown as:

Negative pressure = Hg” x .492

As discussed previously, negative pressure can be used to determine lift. However, since inches of mercury are indicated on a compound gage, inches of mercury are the most common and efficient way to determine lift.

Figure 4

The beauty and simplicity associated with the use of atmospheric pressure, one of nature’s sources of energy, in the act of drafting water has been used in one form or another since ancient times. It is still used in our modern day of technology, and it will always work as long as our atmosphere remains unchanged, a proper operating system is in place, and a negative pressure can be created.

PASCAL’S LAW AND HYDRAULIC PUMPS

Another natural, yet fantastic phenomenon that has been used for hundreds of years to lift, push, etc., has more than proved its worth in fire service applications by enabling a small hydraulic pump no bigger than a large cooking pot to raise an aerial ladder that weighs hundreds of pounds. Defined today as our modern discipline of physics, it also allows a dry pipe sprinkler valve to hold back six pounds of water pressure with just one pound of air pressure.

The use of this natural phenomenon stems from a discovery credited to Blaise Pascal during the 1630s. Pascal’s law states that pressure set up in a liquid exerts a force equally in all directions, and that this pressure vector is perpendicular to all containing surfaces, e.g., the sides and bottom of an open glass. His law has become two of the six basic principles of pressure used with fluids (liquids and gases):

  1. Liquid pressure is exerted in a perpendicular direction to any surface on which it acts.
  2. Pressure applied to a confined liquid from without is transmitted in all directions without reduction in intensity.

Force, pressure, and area

Figure 4 helps explain Pascal’s law and indicates why the aerial will go up and the air pressure will keep the water back. To completely understand and appreciate the principles illustrated in Figure 4, it is necessary to review the meaning of force, pressure, and area.

Force is an energy that is acting on something, for example, the 400 pounds of weight in Figure 4 that is pushing down on Piston A.

Pressure is an amount of energy acting on one square inch of surface area.

Area is a measurement concept that allows us to visualize the size of a two-dimensional surface. For example, if we want to order carpet for a floor, we would want to know how much carpet is needed to cover the floor. Area can be expressed in square miles, yards, feet, or inches. Pressure is most commonly expressed in square inches, so we usually express area in square inches to keep the units of measurement in agreement. The relationship of force, pressure, and area can be stated as:

F = P x A

where F equals force in pounds; P equals pressure in psi; and A equals area in square inches.

Hydrostatic force

Piston A in Figure 4 has a downward force of 400 pounds and an area of 4 square inches. Pressure can be determined by a variation of the force equation as follows:

Continued on page 60

Continued from page 58

Piston A is applying 100 psi of pressure to a contained liquid. The piston is, therefore, the instrument that employs Pascal’s law as stated in the two principles of pressure. The container actually fills up with a pressure of 100 pounds on each square inch of surface, and, as can be noticed in Figure 4, the pressure is perpendicular to the surface that it acts on.

The movement of Piston A has a distinct influence on Piston B and is the meaning of Pascal’s law. Since there are 20 square inches on the surface of Piston B and Piston A is creating a pressure of 100 psi within the liquid, there will be 2,000 pounds of upward force at Piston B. This is found simply by multiplying 100 psi by 20 square inches of surface area. Piston B will lift 2,000 pounds with an investment of 400 pounds of force on Piston A.

The influence Piston A has on Piston B can be expressed as:

Therefore: Pressure that will react on Piston B’s surface equals 100 psi.

Hydrostatic force used to induce pressure

Pressure can be placed in a hydraulic system through the application of a force created by the weight of water (see Figure 5). There are 20 cubic feet of water in the pipe. Since a cubic foot of water weighs 62.5 pounds, there are 1,250 pounds of weight (force) at the base of the pipe where it enters the tank that already contains water. The pipe has a cross sectional area of 25 square inches. The pressure created by the 1,250 pounds of force is:

There are 50 psi on every square inch of the tank’s surface. This is in addition to the pressure created by the water already stored in the tank. For example, if there are 1,000 cubic feet of water in the tank, there will be 62,500 pounds of force resting on the tank’s base (62.5 X 1,000 = 62,500). The tank’s base has an area of 14,400 square inches. Pressure can be calculated as:

Figure 5

By adding the pressure entering the tank from the pipe (50 psi) and the pressure already in the tank (4.34 psi), the total pressure at the base of the tank, on every square inch, will be 54.34 psi. This is another application of Pascal’s law that can and is frequently used in contained hydraulic systems such as sprinkler pressure tanks.

Dry pipe valve

Sprinkler systems in buildings that are either unheated or inadequately heated and are subject to freezing temperatures frequently make use of dry pipe valves. There is water in the sprinkler system from the source (a water main) to the dry pipe valve. From the valve on, there is no water in the sprinkler system. Instead, there is air under a low pressure. Air will not freeze, and the sprinkler pipes will not burst as they could if they were full of water, which expands during freezing.

Figure 6 is a crude representation of a dry pipe valve presented to focus on yet another use of Pascal’s law in fire protection. Remembering that air is a fluid and that the principles of Pascal’s law apply to gases as well as liquids, we can see why one pound of air pressure can keep back six pounds of water pressure. One pound of air pressure is in contact with twenty square inches of area while six pounds of water pressure is in contact with only two square inches of area. The relationship in terms of offsetting forces is as follows:

The one pound of air pressure induces a force of 20 pounds, which is greater than the force of 12 pounds presented by the water under higher pressure. The lower air pressure will keep the dry pipe clapper valve shut. Pascal’s law permits this sprinkler system to employ low air pressures and thus operate at a more cost effective level.

Figure 6

Six principles of fluid pressure

  1. Liquid pressure is exerted in a perpendicular direction to any surface on which it acts.
  2. At any given point beneath the surface of a liquid, the pressure is the same in all directions—downward, upward and sideways.
  3. Pressure applied to a confined liquid from without is transmitted in all directions without diminution (reduction in intensity).
  4. The pressure of a liquid in an open vessel is proportional to the depth of the liquid.
  5. The pressure of a liquid in an open vessel is proportional to the density of the liquid.
  6. Liquid pressure on the bottom of a vessel is unaffected by the size and shape of the vessel.

Trip pressure

The air pressure or water pressure needed to open a dry pipe clapper valve (trip pressure) can be determined through the use of what we have discussed. For example, consider the following problem: There are 60 psi of pressure on the 4-square-inch surface on the wet side of a dry pipe valve. There are 24 square inches of surface area on the air side of the dry pipe clapper valve. What is the trip air pressure needed to keep the clapper valve closed?

In order for the clapper valve to stay shut, the force on the air side must at least equal the force on the water side. Instead of force, the equivalent of force was used or P x A. In effect, we are saying that force air side equals force water side. Ten psi of air pressure will keep the valve closed. However, in practice, excessive air pressure is carried to provide a margin of safety against the clapper valve being unseated due to unusual pressure surges in the water supplying the system. Yet, if there is too much air pressure, there may be a delay in getting water into the sprinkler system during a fire.

WINNING THE FIRE SERVICE LEADERSHIP GAME, an unusual new book on fire service management by Hugh J. Caulfield, is being published this month by Fire Engineering Books. Principles of management essential to a successful career in the fire service appear in the book as vivid scenes acted out in fire stations. The book can be ordered for $19.95 from Fire Engineering Books, Box C757, Brooklyn, NY 11205.

No posts to display