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Fluid Power. Fluid Pressure. One of the most important measurements of fluid power is pressure. We will examine atmospheric pressure, vacuums, and other factors that effect fluid pressure. Atmospheric Pressure. Atmospheric pressure is the reference point from which we measure fluid pressure.
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Fluid Pressure • One of the most important measurements of fluid power is pressure. • We will examine atmospheric pressure, vacuums, and other factors that effect fluid pressure.
Atmospheric Pressure • Atmospheric pressure is the reference point from which we measure fluid pressure. • The air surrounding the earth applies a force of 14.7 psi on the surface of the earth. We will round this figure off to 15 psi.
GAGES • Most gages are set so that atmospheric pressure registers as 0 psi. • Some gages show pressure both above and below this point. • When we use a gage in which 0psi equals atmospheric pressure, we add the word gage to the reading.
GAGES • Therefore, Atmospheric pressure is 0 pounds per square inch gage (psig) • The g is often dropped for convenience. • When you see a reading in psi, it actually means psig.
We normally measure pressure when we are using it to create motion. • In most fluid power systems, no motion will occur until we overcome atmospheric pressure. • We are usually concerned with pressures above atmospheric pressure.
VACUUMS • Any pressure below atmospheric pressure. • Atmospheric pressure is about 15 psi. • Mercury(Hg) is a liquid metal that weighs ~.5 lbs. per cubic inch. • A 1-sq. inch column of Hg 30inches high weighs ~15 lbs. • This column weight equals Atmospheric pressure. • We can use this column to measure fluid pressure.
VACUUMS Figure 21-1 • Tube A is open at the top. Therefore, atmospheric pressure (15lbs.) is the same inside and out. • Tube B is sealed at the top and 1/3 of the air has been pumped out. As a result, the pressure on the inside drops to 10 lbs. • Outside pressure is still at 15 lbs.
VACUUMS • Atmospheric pressure will push the Hg into the tube, until the pressure inside equals 15 lbs. • To achieve balance of pressures, enough Hg to produce 5 lbs must enter the tube. • At 1/2psi per inch of Hg, 10”produces 5psi. • The weight of the Hg and pressure equal AP, the column becomes stationary.
VACUUM Tube C shows a further removal of air (5psi) The Hg now stands at 20”. This height is equal to 10psi Together the air inside the tube and the Hg equal atmospheric pressure.
VACUUM Tube D shows all of the air removed-0psi. This is an example of a perfect vacuum. It now takes ???? Hg to balance the pressure? 1”Hg=1/2 lb. x=15(AP)/.5 x=30”
Vacuum • Any pressure below atmospheric pressure will result in Hg being pushed up the tube in the amount equal to a equivalent weight. • We can use this fact to measure pressures below AP. • Any pressure below AP is called a vacuum, and is measured in inches of Hg.
GAGES Fig. 21-2 • Gage A is a regular gage. It has been calibrated to reflect the AP(15psi) as zero, since we are usually concerned with pressure above AP. We must remember a regular gage measures pressures increased above AP
GAGES Fig. 21-2 • Gage B is a compound gage. It reads pressure both above and below AP. • Notice that the needle is set at 0 psi. This reading has been calibrated to set the gage to reflect any rise or fall in AP. • Notice the readings of the gage below 0 are expressed in inches of Hg.
GAGES Fig. 21-2 • Gage C is an absolute gage. It registers AP at 15 psi. • WHEN READING THIS TYPE OF GAGE WE MUST ADD THE WORD GAGE! • AP is expressed as 15 pounds per sq. inch absolute,or 15 psia. • A perfect vacuum is 0 psia. • Figure 21-3 shows a comparison
Pressure of Confined Fluids • A force acting on a confined liquid exerts outward pressure equally in all directions. • This is the most basic principle of fluid power.
Pressure of Confined Fluids • Figure 21-4 shows a Bourdon Gage. Look closely at the bourdon tube. • The shape of the tube is such that the outer surface is greater than the inner surface. As we increase the pressure on the inside of the tube, the tube, by design, begins to straighten.
Static Head PressureFigure 21-5 • A tank of liquid produces a varying amount of pressure from top to bottom. • The weight of the liquid and height of tank determine the pressure • The pressure developed by the weight of the liquid is called “Static Head Pressure”. • The pressure a column of liquid produces is the same regardless of area covered. • Note that the pressure in the middle of the column is less than that at the bottom.
Pneumatics • Gases and liquids share many properties. • They are different in regard to volume. • Liquids have a definite volume • Gases completely fill any container they occupy.
Pneumatics • Volume is the amount of space displaced. • Since liquids and gases behave differently in this area, pressure and temperature effect liquids and gases differently. • As pressure increases, the volume of a gas decreases- the volume of a liquid remains the same. Figure 21-6
Compression of Gases • As force is applied to a gas confined in a container, the gas molecules are pushed closer together. This is called compression. • Molecules resist compression- the more the molecules are pushed together, the more they push back and try to move apart. • This effort to move apart produces pressure.
Compression of Gases • Scientists have found that the pressure of a gas is inversely proportional to its volume. • This means that any change in pressure produces an equal but opposite change in volume • Example: If the pressure is doubled, the volume will be half of the original pressure. Figure 21-7
BOYLES LAW This relationship between pressure and volume is known as Boyles Law- “The volume of a gas varies inversely with the pressure applied to it, provided that temperature remains constant.”
BOYLES LAW Stated Mathematecicaly: P1V1= P2V2 Where: P1= original pressure V1= original volume P2= new pressure V2= new volume Figure 21-7
BOYLES LAW • NOTE: To use the formula we must use ABSOLUTE PRESSURE rather than gage pressure. • To convert gage pressure (psig or psi) we must add 15 lbs. This is the correction for AP. • After solving the problem, convert the reading back by subtracting 15. Figure 21-8
Charles’ Law • Temperature also effects the volume of gases; we must know the temperature at which a volume was measured. • Jaques Charles explained the relationship between temperatur and volume of a gas.
Charles’ Law • Charles showed that the volume of a gas is directly proportional to the temperature- that is, any change in temperature produces an equal change in volume. However, this occurs only if the pressure remains the same. • Examle: if a gas’s temp. doubles, its volume will also double as long as pressure does not change.
Charles’ Law • This mat be mathematically stated: V1/T1= V2/T2 Where V1= original volume T1= original temp. V2 = new volume T2 = new temp. Figure 21-10
Charles’ Law • To calculate change in volume due to temp., WE MUST EXPRESS ALL TEMP. ACCORDING TO AN ABSOLUTE TEMPERATURE SCALE. • On this type of scale 0o is the temp. at which a material has no heat ( no molecular motion). • We call this temp. Absolute Zero. Figure 21-9
ABSOLUTE TEMPERATURE SCALE • To change regular Fahrenheit temps. To absolute Fahrenheit temps. Simply add 460o. Figure 21-9
Charles’ Law • Temperature affects the pressure of a gas in the same way that it affects its volume.- If the volume remains constant, the pressure will increase or decrease in proportion to a temperature-pressure relationship.
Charles’ Law • We can substitute pressure for volume in Charles’ law: P1/T1= P2/T2 Absolute pressure and temp. must be used in all calculations Where: P1= original pressure T1= original temp. P2 = new pressure T2 = new temp.
Hydraulics • Liquids differ from gases in that they have a definite volume- Under normal conditions, the cannot be compressed. This property of liquids permits a direct and efficient transfer of force. Liquids may be used to multiply force- Very high MA are possible in hydraulic power systems.
HydraulicsTransferring Force Figure 21-11 • Pressure applied to a liquid exerts that applied force equally in all directions. • As a result, the pressure applied to piston A is transferred to piston B. • Changes in the shape or size of the fluid’s container do not affect the transmission of force.
HydraulicsTransferring Force Figure 21-12 &13 Remember from our online manual that - pressure = force/area
Multiplying Force Figure 21-12 &13 We can use a fluid’s ability to transmit forces to produce a gain in MA. Figure 21-13 how this is done: A force of 50 lbs. is applied to piston A. Piston a has an area of 5sq. Inches. Therefore, the pressure throughout the liquid is 10 psi.- F=P x A
Multiplying Force Figure 21-13 Piston B has an area of 20 sq. inches. The force of 10 psi pushes up on this piston. We can calculate the output force with the formula: F= PxA or 10x20 or 200pounds output compared to 50 lbs. input. We have gained a MA of 200:50 or 4:1.
Multiplying Force Figure 21-13 The same rules apply for the movement as we observed in our simple machines- F1x D1 = F2x D2 To move piston B 1 inch, piston A will have to move 4 inches.
Fluid Flow • We can also see that the larger the piston, the more fluid it will take to move it. The movement of fluid is known as Flow. • We measure flow in gallons per minute (gpm). • The flow of gasses is measured in cubic feet per minute (cfm)