Pneumatics Technology

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Pneumatics Technology
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Why Pneumatics (Advantages)

• Generally less expensive than hydraulic
  applications
• Cleaner (leaks arent as much of a problem)
• Lighter fluid less power involved in overcoming
  inertial forces
• Fluid readily available
• Fluid is not flamable or toxic

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Pneumatics (Drawbacks)

• Power levels are much lower than hydraulics
  because pressure levels are restricted (explosive
  hazard of a tank)
• Air is compressible. This is the BIG difference
  between pneumatica and hydraulcics, and makes it
  impossible to do precise position control of
  cylinder extension or motor rotation (tends to
  be all or nothing)

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How we will approach Pneumatics

• Compressible fluid basics
• Units and conventions
• Perfect gas law
• Sizing Compressors
• Sizing Receivers
• Determining Compressor Power
• Conditioning the air

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Our working fluid. AIR

• 21 Oxygen
• 78 Nitrogen
• About 1 CO2 (were changing this)
• Plus a bit of a few other gasses

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Air Pressure and UNITS!

• The stack of air from the top of the atmosphere
  to the earth surface exerts 14.7 lbs on each
  square inch of surface at sea level, when the
  temp is 68 degrees F
• That is called absolute pressure or pressure
  absolute, also noted as PSIA
• That is equivalent to 101,000 Pa or 101 kPa

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Pressure Examples

• Gage pressure (indicated on most of our common
  pressure gages) is pressure above atmospheric
  pressure
• If given psi, assume that means psig
• Then the absolute pressure the gage pressure
  14.7 ( or gage pressure in kPa 101 kPa)
• 120 psi or 120 psig 134.7 psia
• 800 kPa (gage) 901 kPa absolute

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Negative Pressure (Vacuum)

• 0 psig (pounds per square inch gage) 0 psi
  vacuum and 14.7 psia (absolute)
• -4.7 psig 4.7 psi vacuum or 10 psia
• -8.0 psig 8 psi vacuum and 6.7 psia
• -14.7 psig 14.7 psi vacuum and 0 psia
• 0 psia is as low as we can go. You cant have
  more than 14.7 psi vacuum (and that much is
  rather hard to get)

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Absolute Temperature

• We will also find it important to be able to
  determine absolute temperatures
• Absolute temp (English units) is in the Rankine
  Scale
• ºR ºF 460
• Absolute temp in SI units uses the Kelvin scale
• ºK ºC 273
• Remember these or keep them handy!!!

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Temp Examples

• What temperature does water boil at in the
  Rankine scale?
• 212 ºF 460 672 ºR
• What about the Kelvin scale?
• 100 ºC 273 373 ºK
• What about freezing?
• 32 ºF 460 492 ºR
• 0 ºC 273 273 ºK

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Pressure, Temperature and Volume of Air are Linked

• We can calculate a lot of things if we know the
  volume occupied by a gas (air), the absolute
  pressure, and the absolute temperature
• The General Gas Law states

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Perfect Gas Law Example

• Air in the 65 in3 piston chamber of a diesel
  engine starts at 32 psi gage (turbod)
• The compression stroke reduces the volume from 65
  cu in to 3.5 in3
• The air before compression was 180 ºF
• At the end compression the pressure is 1638 psia
• What is the temp of the air at the top of the
  compression stroke assuming we dont lose heat to
  the cylinder, piston or head?

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Solution

• Convert ALL units to absolute
• 180 ºF 180460 ºR 640 ºR
• 32 psi gage 46.7 psia
• Now use the perfect gas law and solve for T2
• T2 (P2V2)T1/(P1V1)
• T2 (1638)(3.5)(640)/(46.7)(65)
• T2 1208 ºR 749 ºF

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Free Air and Standard Air

• In most of our applications we will be talking
  about compressed air. How do we quantify
  compressed air
• Even though our fluid is compressed, in our
  actual use we will measure the air volume before
  compression and express volumes in either Free
  Air, or Standard Air

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Free Air and Standard Air

• Free Air means that we are speaking in terms of
  what volume our air would occupy at the
  temperature and pressure near our system inlet
• Standard Air means that we are speaking in
  terms of the volume our air would occupy at
  standard conditions of 14.7 psi and 68 degrees F.

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Free Air and Standard Air

• Because actual inlet conditions (temp and ambient
  pressure) can vary by location and from hour to
  hour, ratings of things like compressors are
  usually given in Standard conditions and called
  SCFM (standard cubic feet per minute)
• If your inlet conditions are known and
  substantially different than standard (Denver)
  you may do calculations in CFM of Free Air rather
  than standard air

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Compressor Example

• 30 cfm (compressed volume) of air at 125 psig and
  90º F are supplied to a receiver (tank) by a
  compressor
• If inlet conditions are 14.7 psia and 70º F, how
  many cfm of free air must the compressor take in?
• We use the perfect gas law solved for V1
• V1 V2 (P2/P1)(T1/T2)

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Compressor Example

• If we divide both sides of that equation by time
  we get one that involves flow rate, which is what
  we are interested in
• The if Q is V/t in cubic feet per minute
• Q1 Q2 (P2/P1)(T1/T2)
• Note that this expresses free air flowrate in
  terms of compressed air flow and the inlet and
  compressed pressures and temperatures

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Compressor Example

• We must use absolute temps and pressures
• 90º F 550º R, and 70º F 530º R
• 125 psig 139.4 psia
• Then
• Q1 30 cfm x (139.4 / 14.7)(530/550)
• Q1 275 cfm of free air

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Sizing of Receivers(How big a tank do I need?)

• Plant requirements
• What are your average needs of compressed air?
• What are your peak requirements?
• Compressor must supply at LEAST the average
  requirements. Peak need may come from the
  receiver, which acts as a reservoir
• Note that pressure in the tank will fall during
  peak usage, so time of peaks is a factor

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Tank Formula (guideline)

• Vr 14.7 (t)(Qr Qc) / ( Pmax Pmin)
• Vr tank Volume in ft3
• t time time that the receiver must supply a
  given flow rate (minutes)
• Qr flow rate from the tank
• Qc flow rate from the compressor
• Pmax Maximum pressure in the receiver
• Pmin Minimum pressure in receiver allowed

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Tank Sizing Example

• We need 20 scfm from a receiver and want 6
  minutes between the time the compressure shuts
  down at 100 psi and when it resumes operation at
  80 psi
• Use the formula from the previous slide with Qc
  0, t 6, Pmax 100 psi, and Pmin80 psi
• Vr 14.7(6)(20)/(100-80) 88.2 ft3 660 gal
• Note that we didnt use absolute pressures
  because the difference is the same (20 psi)

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More of Same Example

• What if our compressor is supplying 5 scfm during
  this time?
• Vr 14.7 (6)(20 5)/(100-80) 66.2 ft3
• Vr 495 gal
• SI version of the equation
• Vr 101 (t)(Qr-Qc)/(Pmax Pmin)
• Vr m3, Q m3/min, P kPa, t minutes

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How much Power to drive the compressor?

• English Units Formula
• Pin absolute pressure at the compressor inlet
  (psia)
• Pout absolute pressure at the compressor outlet
  (psia)
• Q flow rate (SCFM) taken in

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How much Power to drive the compressor?

• S.I. Units Formula
• Pin absolute pressure at the compressor inlet
  (kPa abs)
• Pout absolute pressure at the compressor outlet
  (kPa abs)
• Q flow rate (m3/min) taken in

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Compressor Power Example

• Determine actual power required to run a
  compressor that delivers 100 scfm of air at 100
  psig. Overall compressor efficiency is 75
• First use the HP equation for power
• We need absolute pressures
• Standard pressure is 14.7 absolute
• 100 psig 10014.7 psia

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Compressor Power Example

• HP (14.7 x 100)/65.4 x (114.7/14.7)0.286-1)
• HP 18.0 HP (this is pneumatic power)
• Actual HP will need to be adjusted for the
  compressor efficiency (75 or 0.75)
• HPactual HPpneumatic / 0.75 24.0 HP
• Electric motor would need to be at least 24 HP to
  supply the pneumatic power

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Conditioning the Compressed Air

• Filters
• Regulators
• Lubicators
• FRL

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Conditioning the Compressed Air

• Silencers or Mufflers
• Reduces the high noise levels associated with
  exhausting air at high pressures or flow rates

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Conditioning the Compressed Air

• Aftercoolers
• Cools air between the compressor and the receiver
• Can remove much of the moisture

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Conditioning the Compressed Air

• Air Dryers
• Absorbtion dryers
• Use calcium chloride or other pellets
• Material converts to a liquid with the water
  which is removed
• Adsorbtion
• Moisture adsorbed onto alumina or silica gel
• Must be regenerated
• Refrigeration dryers
• Chill the air to condense additional moisture

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Commercial Chiller Air Dryer
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Moisture Content of Saturated Air
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Moisture Example

• Compressor delivers 100 scfm at 100 psig to a
  system
• Entering air is saturated (100 R.H.) and is 80º
  F
• If the compressor runs 8 hrs /day how much water
  is delivered to the system (if not removed) each
  day?

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Moisture Example

• Look at the moisture chart. Our conditions are
  80 degrees and 14.7 psia (standard)
• Follow the 80º line to the top left
• The air contains 1.58 lbs H2O per 1000 ft3
• Moisture flow rate when running is
• Air flow rate x moisture content
• 1.58 lb/1000 ft3 x 100 ft3/min 0.158 lb H2O/min

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Moisture Example

• 0.158 lb/min x 60 min/hr x 8 hr/day 75.8
  lb/day
• One gallon of water weighs 8.34 lb so.
• 75.8 lb/day / 8.34 lb/gallon 9.09 gal/day
• This will be baaaaad for cylinders and air motors
  in our tooling. Best get rid of most of this

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More Moisture Example

• How much moisture delivered to the system if we
  aftercool the air to 80º F ?
• Back to the moisture chart
• Now we have 80º F again, but we are now at 100
  psig. Go up at that pressure to the 80º line and
  then to the left to find that this cooled air can
  hold 0.20 lb H2O / 1000 ft3
• Now we do the same calculation as before

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More Moisture Example

• 0.20 lb/1000 ft3 x 100 ft3/min 0.02 lb H2O/min
• 0.02 lb/min x 60 min/hr x 8 hr/day 9.6
  lb/day, or 1.15 gallons per day
• Better
• Now how about if we use a chiller dryer which
  cools the air to 40º F before the receiver?

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More Moisture Example

• Back to the chart
• At 100 psig and 40º has about 0.05 lb moisture /
  1000 ft3
• Since we have 100 scfm/ minute again we have
  0.005 lb/min
• 0.005 lb/min x 60 min/hr x 8 hr/day is
  2.4 lb/day or 0.29 gal per day
• Since our lines will not go below 40º F we should
  never condense water in our system