Pneumatic Valves. Valve Actuators

Transcription

1 Air Preparation Units Symbol Description Pneumatic Valves Symbol Description Symbol Cylinders Description Filter / Separator with manual drain 3-Position, 4-Way, APB ports closed, center pos. Standard double acting Filter / Separator with automatic drain Oil Removal Filter Automatic Drain Lubricator less drain Lubricator with manual drain Lubricator with automatic filling 3-Position, 4-Way, CE 5-Ported cylinder ports open to exhaust in center position 3-Position, 4-Way, PC 5-Ported pressure ports open to exhaust in center position Quick Exhaust Single Acting Double Rod Spring Return Ram Type Telescope Tandum Air Line Pressure Regulator adjustable, relieving Air Line Pressure Regulator pilot controlled, relieving Filter / Regulator (piggyback) manual drain relieving (with gauge) Filter / Regulator (piggyback) auto drain relieving Air Line Combo F-R-L simplified Pneumatic Valves Symbol Description Check Flow Control Relief Valve -Position, -Way -Position, 3-Way Shuttle Valve Actuators Symbol Description complete simplified Manual general symbol Push Button Lever Pedal or Treadle Mechanical cam, toggle, etc. Spring Detent line indicates which detent is in use Piezo Solenoid Internal Pilot Supply Remote Pilot Supply Duplex Lines and Functions Symbol Description connected disconnected Solid Line Main Line Dashed Line Pilot Line Dotted Line Exhaust or Drain Line Center Line Enclosure Outline Lines Crossing (90 intersection not necessary) Lines Joining (90 intersection not necessary) Lines Joining Flow Direction hydraulic medium Flow Direction gaseous medium Energy Source Line with Fixed Restriction Line with Adjustable Restriction Flexible Line Plugged Port, Test Station, Power Take-off Quick Disconnect Without Checks -Position, 4-Way And / Or Composite solenoid and pilot or manual override connected disconnected Quick Disconnect With Checks -Position, 4-Way 5-Ported And / Or Composite solenoid and pilot or manual override and pilot connected disconnected Quick Disconnect With One Check 1 Parker annifin Corporation

2 Design Torque Design torque represents the maximum torque that an actuator must supply in an application. This maximum is the greater of the Demand Torque or the Cushion Torque. If the demand torque exceeds what the actuator can supply, the actuator will either move too slowly or stall. If the cushion torque is too high, the actuator may be damaged by excessive pressure. Demand torque and cushion torque are defined below in terms of load, friction, and acceleration torque. Equations for calculating demand torque and cushion torque for some general applications are provided on the following pages. T - Torque The amount of turning effort exerted by a rotary actuator. TD - Demand Torque This is the torque required from the actuator to do the job and is the sum of the load torque, friction torque, and acceleration torque, multiplied by an appropriate design factor. Design factors vary with the applications and the designers knowledge. Equation 4-3) TD = Ta + Tf + TL TL - Load torque This is the torque required to equal the weight or force of the load. For example, in Fig. 4-8a, the load torque is 563 Nm (5000 lb-in.); in Fig. 4-8b the load torque is zero; in Fig. 4-8c the load torque is 563 Nm (5000 lb-in.). The load torque term is intended to encompass all torque components that aren t included in the friction or acceleration terms. Tf - Friction torque This is the torque required to overcome friction between any moving parts, especially bearing surfaces. In Fig. 4-8a, the friction torque is zero for the hanging load; in Fig. 4-8b the friction torque is 775 Nm (6880 lb-in) for the sliding load; in Fig. 4-8c the friction torque is zero for the clamp. Equation 4-4) Tf = mwr Ta - Acceleration Torque This is the torque required to overcome the inertia of the load in order to provide a required acceleration or deceleration. In Fig. 4-8a the load is suspended motionless so there is no acceleration. In Fig. 4-8b, the load is accelerated from 0 to some specified angular velocity. If the mass moment of inertia about the axis of rotation is I and the angular acceleration is a, the acceleration torque is equal to Ia. In Fig. 4-8c there is no acceleration. Some values for mass moment of inertia are given in Table 4. Some useful equations for determining a are listed in Table 5. Equation 5 below shows the general equation for acceleration torque. Equation 4-5) Ta = Ia TC - Cushion Torque This is the torque that the actuator must apply to provide a required deceleration. This torque is generated by restricting the flow out of the actuator (meter-out) so as to create a back pressure which decelerates the load. This back pressure (deceleration) often must overcome both the inertia of the load and the driving pressure (system pressure) from the pump. See applications. Equation 4-6) Tc = Ta + PrV Tf ± TL θ The friction torque Tf reduces the torque the actuator must apply to stop the load. The load torque TL may add to, or subtract from the torque required from the actuator, depending upon the orientation of the load torque. For example, a weight being swung upward would result in a load torque that is subtracted. Warning: Rapid deceleration can cause high pressure intensification at the outlet of the actuator. Always insure that cushion pressure does not exceed the manufacturer s pressure rating for the actuator. KE Kinetic Energy (1/ Jmω ) This is the amount of energy that a rotating load has. The rotator must be able to stop the load. All products have kinetic energy rating tables. Choose the appropriate deceleration option (i.e., bumper, cushions, shock absorbers, etc.) that meets or exceeds the kinetic energy of the load. Parker annifin Corporation

3 Rotary Actuator Torque Requirements Demand Torque Examples A) Example of load torque The load is held motionless as shown. 10 in. TD = Tα + Tf + TL Tα = 0 Tf = 0 TL = (500 lb)(10 in) = 5,000 lb-in TD = 5,000 lb-in Figure 4-8a 500 lbs. B) Due to friction and acceleration The 500 lb rotating index table is supported by bearings with a coefficient of friction of 0.5. The table s acceleration a is rad/ sec. The table s mass moment of inertia I is,330 lb-in-sec. TD = Tα + Tf + TL Tα = Iα = (,330 lb-in-sec)(/sec) = 4,660 lb-in Tf = µwrb = 0.5 (500 lb)(55 in) = 6,880 lb-in TL = 0 TD = 4,660 lb-in + 6,880 lb-in = 11,540 lb-in 60 in. 55 in. 500 lbs. Figure 4-8b Bearings C) Load torque example 10 in. 500 lb. Force TD = Tα + Tf + TL Tα = 0 Tf = 0 TL = (500 lb)(10 in) = 5,000 lb-in TD = 5,000 lb-in Figure 4-8c 3 Parker annifin Corporation

4 Rotary Actuator Moments of Inertia Table 4: Mass Moments of Inertia Rectangular prism Circular cylinder = 1 m( b + c ) = 1 ma I y = 1 m( c + a ) I y = I z = 1 m( 3a + L ) I z = 1 m( a + b ) Thin rectangular plate = 1 m( b + c ) I y = 1 mc I z = 1 mb Circular cone = 3 ma 10 I y = I z = 3 m( 1 a + h ) 5 4 Sphere = l y = l z = ma 5 Thin disk = 1 mr Parallel Axis Theorem: I p = l + md When acceleration is constant: P d G I y = I z = 1 mr 4 I p = Mass moment of inertia about an axis parallel to a centroidal axis I = Mass moment of inertia about a centroidal axis m = Mass d = Distance between axes When velocity is constant: θ = ω 0 t + 1 αt θ = ω 0 t + 1 ω t t ω = ω 0 + αt ω = (ω 0 + αθ) 1/ α = θ t α = (ω t - ω 0 ) θ α = (ω t - ω 0 ) t θ = ωt t θ ω t = time = angular position = angular velocity at time = t ω 0 = angular velocity at time = 0 α = angular acceleration 4 Parker annifin Corporation

5 Rotary Actuator Basic Equations Basic Velocity, Acceleration, Kinetic Energy And Torque Equations Equations below are based on triangular velocity profile. ωmax =.035 α = α = ωmax Θ 57.3 ωmax (t ) K.E. = 1/ Jm ω Ta = α Jm Θ t Tf = W Us Distance from pivot point to center of external bearings TL = Torque arm length WL cos (φ) to C.G. of load Where φ = Angle between torque arm and horizontal plane Where: Θ = Angle of Rotation (Degrees) t = Time to rotate through Θ (sec) ω = Angular velocity, radians/sec α = Angular accelerations (radians/sec ) WL = Weight of load (lbf) Ta = Torque to accelerate load (lb-in) Us = Coefficient of static friction Jm* = Rotational mass moment of inertia (lb-in-sec) Tf = Torque to overcome friction (lb-in) TL = Torque to overcome effects of gravity *Use I values from Table 4. Coefficients of Friction Material* µs µk Steel on Steel Steel on Steel (lubricated) Aluminum on Steel Copper on Steel Brass on Steel Teflon on Steel *dry contact unless noted 5 Parker annifin Corporation

6 Selecting the proper vacuum cup CAUTION:! Selecting the type of vacuum cup, material, and size suitable for an application is important to the overall vacuum system. Calculating the forces involved for each application is recommended to determine the vacuum cup size. It should be noted that these calculations are basic theoretical guidelines and each application must be tested for actual results. With all vacuum applications, certain practical assumptions concerning cup materials, environmental conditions, and product characteristics to name a few, may not be consistent with the performance. Again, the user should determine the efficiency, performance, and safety factor of the cup selection. Calculating pad diameter and forces Mass The term mass is a quantity of matter and its ability to resist motion when acted on by an external force. The magnitude of an object is represented as a certain number of kilograms (kg) and is symbolized as m. The easiest way to determine the mass of an object is to measure the weight with a scale within the earth s gravitational field (ag = 9.81m/sec ). Likewise, outside of any gravitational field, a mass could potentially be weightless. Forces For vacuum applications, force is a vector quantity in a defined direction either horizontal or vertical. The standard international unit of force is measured in Newtons (N) which is the equivalent of (kgm/sec ). The force can be calculated by measuring the effect of a change in acceleration on a mass. Newtons Law: F(N) = mass(kg) x ag(m/sec ) Consider an object with a mass of 10kg. The gravitational force on this object would be: F(N) = 10kg x 9.81m/sec = 98.1 N Acceleration Acceleration is the change in velocity of a moving object. Acceleration is a vector, a directional quantity expressed in units of meters per second squared (m/ sec ) and symbolized as a. To explain the magnitude of acceleration consider an object with a change in velocity of meters per second (m/sec) over a 4 second time frame. The acceleration can be calculated with: a = velocity a = 6m/sec a = 3m/sec time sec This is considered an average acceleration. Coefficient of friction Certain values for coefficient of friction should be taken into consideration when calculating the combined forces in motion. Actual values between suction cups and surfaces are difficult to determine. Therefore, coefficient of friction values from published charts, should be used as a reference to adjust the safety factors accordingly. Lifting forces When calculating lifting forces, safety factors of for horizontal lifts and 4 for vertical lifts are minimum values. Applications with irregular shapes, difficult surfaces, and backward motions will require increased safety factors. F: orizontal Lift FV: Vertical Lift orizontal lifting force Apply Newtons Law to calculate the force on a 10kg mass with a change in acceleration of 3m/sec and a safety factor of. F(N) = mass(kg) x (a g+ a) x S F(N) = 10kg x (9.81m/sec +3m/sec ) x F = 56. N 3m/sec 10kg 10kg F Vertical lifting force Apply Newtons Law to calculate the force on a 10kg mass with a dry surface, a change in acceleration of 3m/sec and a safety factor of 4. FV(N) = mass(kg) x (a g+ a) x Sv FV(N) = 10kg x (9.81m/sec +3m/sec ) x 4 FV = 5.4 N 3m/sec FV Combined vertical lift and horizontal motion Calculate the force on a 10kg mass with a dry surface, a change in acceleration of 3m/sec, and a change in travel acceleration of m/sec. FM(N) = FV + F FM(N) = [(10kg x m/sec ) x 4] + [10kg x (9.81m/sec + 3m/sec ) x ] FM(N) = (80kgm/sec ) + [56kgm/sec ] FM(N) = 6400kgm/sec + 65,536kgm/sec FM = 68. N 10kg F 3m/sec m/sec 6 Parker annifin Corporation

7 Analyze the forces Using the previous examples, consider an application where 4 cups have been selected to transfer the product. Take the orizontal Lifting Force (F) of 56. N and divide by the number of cups (4) to obtain the individual force for each cup. 56. (N) = N / Cup 4 Referring to the chart below, at 60% vacuum, select a force greater than N. The appropriate selection is a 40mm diameter cup which has a theoretical lifting force of 76.9 N. The same calculation can be applied to the Vertical Lifting Force and the Forces in Motion examples to determine the cup diameter. To convert Pounds (lbf) to Newton (N), multiply lbf x 4.4. Theoretical lifting force per cup lbf (N) Cup Diameter mm Calculate the diameter of the cup Calculate the cup diameter for horizontal lift at 60% of full vacuum using the information from the previous page. D = 35.7 m (a g + a) x S Pv x n D = ( ) x 61 x 4 D = mm Referring to the chart below, at 60% vacuum, select a cup diameter equal to or greater than 37mm. The appropriate selection is a 40mm diameter cup which has a theoretical lifting force of 76.9 N. Vacuum level 3 ing 6 ing 9 ing ing 15 ing 18 ing 1 ing 4 ing 7 ing -1.5 PSIG -3 PSIG -4.5 PSIG -6 PSIG -7.5 PSIG -9 PSIG PSIG - PSIG PSIG Area 10. kpa 0.3 kpa 30.5 kpa 40.6 kpa 50.8 kpa 61 kpa 71.1 kpa 81.3 kpa 91.4 kpa cm 10% 0% 30% 40% 50% 60% 70% 80% 90% (0.0) (0.04) (0.04) (0.06) (0.08) (0.10) (0.) (0.14) (0.14) (0.03) (0.06) (0.10) (0.13) (0.16) (0.19) (0.) (0.5) (0.8) (0.10) (0.0) (0.9) (0.39) (0.49) (0.59) (0.69) (0.78) (0.88) (0.0) (0.40) (0.60) (0.80) (1.00) (1.0) (1.40) (1.60) (1.80) (0.9) (0.58) (0.87) (1.0) (1.40) (1.70) (.00) (.30) (.60) (0.39) (0.78) (1.18) (1.60) (.00) (.40) (.70) (3.10) (3.50) (0.5) (1.0) (1.54) (.00) (.60) (3.10) (3.60) (4.10) (4.60) (0.80) (1.60) (.40) (3.0) (4.00) (4.80) (5.60) (6.40) (7.0) (1.80) (3.60) (5.41) (7.0) (9.00) (10.8) (.6) (14.4) (16.) (.60) (5.0) (7.79) (10.4) (13.0) (15.6) (18.1) (0.8) (3.3) (3.0) (6.40) (9.60) (.8) (16.0) (19.) (.4) (5.6) (8.8) (5.00) (10.0) (15.0) (0.0) (5.0) (30.0) (35.0) (40.0) (45.0) (7.0) (14.4) (1.6) (8.8) (36.0) (43.) (50.4) (57.6) (64.8) (9.80) (19.6) (9.4) (39.) (49.0) (58.9) (68.6) (78.5) (88.) (.9) (5.6) (38.5) (51.) (64.0) (76.9) (89.6) (103) (115) (0.1) (40.0) (60.1) (80.0) (100) (0) (140) (160) (180) (8.9) (57.6) (86.5) (115) (144) (173) (0) (31) (59) (45.) (90.0) (135) (180) (5) (70) (315) (360) (405) (51.4) (10) (154) (05) (56) (308) (359) (410) (461) (65.1) (130) (195) (59) (34) (389) (454) (519) (583) (7.5) (144) (17) (89) (361) (434) (506) (578) (650) (97.) (194) (91) (387) (484) (581) (678) (775) (871) (116) (30) (346) (461) (576) (69) (807) (9) (1037) (181) (360) (541) (70) (900) (1081) (60) (1441) (160) (31) (640) (961) (79) (1601) (19) D (mm) = Diameter of Cup m (kg) = Mass ag = 9.81m/sec a = Motion Acceleration S = Safety Factor Pv (kpa) = Operating Vacuum Pressure n = number of Cups (41) (56) (880 7 Parker annifin Corporation

8 Selecting a generator size The choice of Generator Series depends on the system requirements for components and overall performance for the application. Inline Generators offer the basic function for creating vacuum flow. Adding integrated components such as automatic blow off Controls, Vacuum and Blow-off Solenoids, Pressure Sensors, Check Valves and Filters are options that can reduce overall mounting space, reduce cycle time and can offer air conservation functions as well as emergency stop modes. For guidelines on selecting Vacuum Generators by features, consult the highlight features in the Generator Index section for each Series Section. A vacuum source can only achieve and hold a degree of vacuum that sustains the amount of leakage into the vacuum system. In most cases, it is the leakage through the product and by of the cup seal that limits the system degree of vacuum. Products with high product leakage are Porous Applications. The degree of vacuum that can be obtained with this type of product can vary and tends to be below 10 ing. Products with low or no leakage are called Non- Porous Applications. It can be assumed that the maximum degree of vacuum of the system is the maximum degree of the vacuum generator. Due to design cycle time and safety requirements, a lower degree of vacuum is generally chosen other than the maximum obtainable degree of vacuum. Chart 1 lists different units of measure for vacuum with typical application levels The system degree of vacuum must be determined by product testing. Evacuation time The size of the generator generally refers to either the Evacuation Time or the Vacuum Flow Rates of the generator and varies by the size of the nozzle / diffuser. Evacuation Time is the time required to evacuate the air out of a vacuum system to specific degree of vacuum. Typically, this degree of vacuum is a value where it is safe to move a product in a pick and place application and is determined by the design engineer. Evacuation Time can also be considered response time of the system. A typical Evacuation Time chart for a generator series is shown in Chart. The time to achieve a given degree of vacuum in a 1 cubic foot volume is listed in seconds for each Generator. Chart : evacuation time Series / nozzle diameter Air supply pressure Air consumption Chart 1: Basic vacuum pressure measurements units Negative gauge pressure PSIG Absolute pressure PSIA Atmospheric pressure at sea level Typical porous vacuum level Typical non-porous vacuum level Perfect vacuum (zero reference pressure) Inches of mercury ing Example: A pick and place application requires a 0.5 secs for creation of 18 ing of vacuum in the vacuum system. The vacuum system volume, which includes tubing and cups, is 0.00 ft 3. The evacuation time charts are given for a 1 cubic foot (ft 3 ) volume. To use these charts, convert the time requirement of the system to an equivalent time for a 1 cubic foot (ft 3 ) volume. In this example, 1 cubic foot (ft 3 ) is 500 times the system volume of 0.00 ft 3. Multiply the system time requirement by 500 (500 x 0.5 secs = 5 seconds). Any generator with a evacuation time of less than 5 seconds to attain 18 ing can be chosen for this application. A CV- 15-S will meet the requirements for this application. A (-) listed means the generator will not obtain a higher degree of vacuum than the level of the first (-). Evacuation time in sec / ft3 * to reach different vacuum levels (ing) PSI SCFM CV-05S CV-05LS CV-10S CV-10LS CV-15S CV-15LS CV-0S CV-0LS * 1 ft 3 = 8.31 liters 8 Parker annifin Corporation

9 Vacuum flow A typical Vacuum Flow chart for a generator series is listed in Chart 3. The vacuum flow rate at given degree of vacuum is listed in SCFM for each Generator. This chart is generally used to determine the change of degree of vacuum given a change in vacuum flow rate of a generator. Example. A CV-15S can only obtain 9 ing. The vacuum flow rate at 9 ing is 1.50 SCFM. This means that the cup Chart 3: Vacuum flow (SCFM) Nozzle diameter Tubing reference SAE x 5.4 = mm ing CV-05S CV-05LS CV-10S CV-10LS CV-15S CV-15LS CV-0S CV-0LS Tubing ID SAE mm 5/64 3/3.38 1/ /3 4 3/ / /16 8 3/ / Tubing length (L) In. M In. x 54 = M Pad volume reference (pv) PFG Ø Cu in. L Nozzle diameter to tubing diameter to cup diameter reference For each application, the size of the nozzle diameter, vacuum tubing I.D., and maximum cup diameter must be practical in relationship to each other. The chart to the right is a quick reference to aid in selecting the vacuum tubing I.D. and nozzle diameter given the maximum cup diameter. As an example, one 60mm cup with meters in tubing length would require a minimum 6mm I.D. vacuum tube and a 1.5mm nozzle. The same 60mm cup with 3.5 meters in tubing length would require a minimum 8mm I.D. vacuum tube and a.0mm nozzle to achieve an equivalent performance. seal and product leaks 1.50 SCFM of air. This generator can maintain the leak rate of 1.50 SCFM. Choosing a generator with more flow at 9 ing will increase the degree of vacuum in the system because the generator can overcome more leakage. In this case,the vacuum flow rates are linear since this CV generator is a single stage venturi generator. Replacing a CV-15S with CV-0S will increase the degree of vacuum in the system to approximately 16. ing. The CV-0S now maintains 16. ing at a flow rate of 1.50 SCFM. PBG Ø Cu in. L ft 3 = 8.31 liters 1 ft 3 = 178 in 3 Tubing I.D. (mm) Nozzle Dia. 14 (mm) Tubing Length (m) PCG Ø Cu in. L Maximum Cup Diameter (mm) Parker annifin Corporation

10 Valve Technical Saving Money and Space by Sizing Your Valves Properly This catalog gives you a flow rating (Cv) for each valve in the Parker annifin line. You can plug your requirements into the following simple formula, and determine the Cv needed to do the job. By not oversizing, you ll save space and money, and you ll ensure the valve you select will do the job. Converting the Job Requirements Into Cv (Capacity Co-efficient). Cylinder Area Cylinder Compression A (Sq. In.) X Stroke X Factor X (Table ) Cv = (See Table 1) (In.) (Table ) Stroke Time (sec.) x 8.8 Let s work through an example: We want to extend a 3-1/4" bore cylinder which has a " stroke in one second, and we have a supply pressure of 80 PSI to do the work. ere s what we know: Cylinder Area for a 3-1/4" Bore, from Table sq. in. Cylinder Stroke... in. Stroke Time Required in Seconds...1 sec. Compression Factor at 80 PSI, from Table A Constant for 80 PSI, from Table Substituting in the formula, we have: 8.30 x x 6.4 x.048 Cv = = x 8.8 Any valve, therefore, which has a Cv of at least 1.06, will extend our cylinder the specified distance in the required time. Choosing the Valve Series Your next step is to choose a basic valve design to do the job. For a quick guide to valve designs, see Table 3. aving selected the basic valve design, consult the Capacity Co-efficient (Cv) tables which describe the individual valve capacities. Selecting the Valve Model, Options and Accessories aving determined Cv, series, port size, flow-path configuration (pre-determined by circuit design), and actuation method, you re ready to choose the exact valve model number. Read the pertinent catalog pages; note the exact model numbers, options and accessories you want. Then phone or write your Parker annifin air valve distributor. They will give you prompt, accurate service. Note: Need circuit design help? Contact your local Parker annifin distributor. They are backed up by our regional Sales Engineers and offices. Between them, you ll find answers to all of your questions. Table 1 Effective Square-Inch Areas for Standard-Bore-Size Cylinders Bore Size Cylinder Area (Sq. In.) Bore Size Cylinder Area (Sq. In.) 3/4".44 4".57 1" /" /8".99 5" /4" 1.3 6" /" " /4".41 8" 50.7 " " /" 4.91 " /4" " /8" 10.3 Table Compression Factors and A Constants Inlet Pressure (PSIG) Note: Use A constant at 5 PSI rp for most applications. On very critical applications, use A at PSI rp. You will find in many cases, a 10 PSI rp is not detrimental, and can save money and mounting space. 1 GT * Tabulated values are the solution of where T is for.48 (P1 P) P 68 F and G =1 for Air. Table 3 Characteristics of the Major Valve Designs A. Poppet 1. igh flow capacities 3-Way and 4-Way. Minimum lubrication requirements 3. Fast response 4. Self-cleaning poppet seats 5. Pressures of 15 to 150 PSIG (modifications for vacuum to 50 PSIG) B. Spool Valves (WCS) 1. Low friction 3-Way and 4-Way. Lower operating pressures 3. Fast response 4. Less wear 5. Long Cycle Life - Under pressure, radial expansion of the seal occurs to maintain sealing contact with the valve bore 6. Non-Lube Service - No lubrication required for continuous valve shifting 7. Bi-Directional Spool Seals - Common spool used for any pressure, including vacuum C. Packed Bore 1. Wide range of flow capacities 4-Way. Wide range of flow-path configurations 3. Pilot-operated models available 4. Pressures of vacuum to 150 PSIG D. Rotary Or 1. Inexpensive Reciprocating Disc. Versatility in manual actuation 4-Way, manually operated Cv Capacity Co-efficients (sometimes called Flow Factors). Each flow path through the valve has its own Cv value. All Cv ratings for each valve cataloged on this page are listed on the front side of this sheet. Cv = Q GT.48 (P1 P) P Cv = Q x A (Table ) Compression Factor A Constants for Various Pressure Drop* PSI rp 5 PSI rp 10 PSI rp Q = Flow in Standard Cubic Feet per minute (14.7 PSIA at 60 F) P1= Inlet Absolute Pressure (gauge pressure ) P = Outlet Absolute Pressure (gauge pressure ) Note: P must be greater than.53 x P1 G = Specific Gravity of flowing medium (Air, G =1) T = Absolute Temperature of Air (460 + F.) 10 Parker annifin Corporation