Pneumatic Technology

1
Pneumatic Technology

• Valves and Valve Issues

2
Orifices and Valve Sizing

• Valves act as orifices
• Flow is more complicated than with liquid
• Basic flow equation for a pneumatic valve looks a
  lot like the one for hydraulic flow

3
Valve Flow Equations (English and SI)

• Q is SCFM or std m3/min
• Cv is a flow capacity constant
• P1 is pressure upstream
• psia or kPa absolute
• P2 is pressure downstream
• T1 is upstream temperature
• ºR, or ºK

4
Choked Orifice Flow

• Orifice flow equations work to a point
• When the downstream pressure, P2, is 53 of the
  upstream pressure, P1, then the downstream
  airflow reaches the speed of sound and the
  orifice is choked
• Increases in upstream pressure, or decreases in
  the downstream pressure will no longer increase
  the valve flow

5
Valve Sizing Example

• Air at 80ºF passes through a ½ inch diameter
  orifice having a flow capacity constant, Cv, of
  7.4. If the upstream pressure is 80 psi what is
  the maximum possible flow-rate in scfm of air?
• T1 80 460 540ºR
• P1 80 psi 14.7 psi 94.7 psi
• Max flow occurs when the orifice is choked

6
Valve Sizing Example

• Choked orifice implies downstream pressure is 53
  of upstream pressure
• P2 0.53 P1 0.53 (94.7) 50.2 psia
• The use the flow equation

7
Another ExampleSize a Valve

• Your clamping cylinder requires 43 psi to hold
  the part
• You need a flowrate of 21 scfm to clamp quickly
• First, what is the maximum system pressure
  (beyond which flow will not increase)?
• 0.53 x P1 P2 Note P2 43 14.7 (psia)
• P1 57.7 / 0.53 108.9 psia 94.2 psig

8
Pressure Regulators
9
Directional Control Valves

• Example 3-Position double acting solenoid
  actuated valve
• Note that unlike hydraulic valves these vent
  return air out two ports rather than one T
  port. Lets label these as E for Exhaust

A
B
P
E
E
10
Cylinder Air ConsumptionExample

• Double acting cylinder with 1.75 in dia piston
  and 0.75 in dia rod and 6 in stroke
• Cycle the cylinder 30 times per minute
• Air supply is 100 psig and 80º F
• Determine air consumption (SCFM)
• Implies air at standard temp (68º F and 14.7 psia)

11
Solution

• First determine volume of compressed air used
  (Q2) of 100 psig, 80º F air
• Each cycle we extend and retract 6 inches
• Extension volume p D2/4 x stroke
• Retraction vol p /4(D2 d2rod) x stroke
• Vol p 1.752/4 x 6 p/4(1.752 0.752)x6
• Vol of compressed air is 26.21 in3/ stroke
• 26.21 in3/stroke x 30 strokes/min 786 in3/min
  (1728 in3/ ft3) so 0.455 ft3/min

12
Solution

• Q1 Q2(P2/P1)(T1/T2)
• P2 100 14.7 psia 114.7 psia
• P1 14.7 psia,
• T2 80 460 540 ºR
• T1 68 460 528 ºR
• Q1 0.455 CFM (114.7/14.7)(528/540)
• Q1 3.47 SCFM

13
Lets save some and get a raise Example

• You extend your cylinder as in the previous
  example to do the task
• Do you need 100 psig to retract the cylinder?
  Probably not. Lets use a 2nd regulator set at
  10 psig and connect our DCV to use this lower
  pressure to retract
• How many scfm do we need now?

14
Lets save some and get a raise Example

• Extension still takes the same amount of air
  (14.43 in3/stroke x 30 /min 433 in3/min or
  433/1728 ft3/min) 0.25 ft3/min
• Convert to scfm Q1 0.25 CFM (114.7/14.7)(528/540
  )
• Q1 on extension 1.91 scfm
• Now look at retraction compressed volume
• Q2 p/4(1.752 - .752) x 6 in x 30/min 353
  in3/min or 0.204 ft3/min

15
Lets save some and get a raise Example

• But now we want Q1 figured with 10 psig to
  retract. 10 psig 24.7psia
• Q1 0.204 cfm (24.7/14.7)(528/540)
• Q1 0.335 scfm
• Total air for this cylinder is now
• 1.91 scfm 0.335 scfm 2.24 scfm
• Compared to 3.47 scfm in the initial problem
• Save 1.22 scfm x 60min/hr x 16 hrs/day x 250
  days/ year 292,800 scfm per year