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1. Numasizing ? What Do I use?

2. What Is Numasizing? • Compressed air is an energy medium that must be conserved, Numasizing has been developed to use this resource more effectively.

3. What Is Numasizing? • Numasizing is a technique which takes into account all of the physical specifications of a circuit and results in a tailor made circuit designed to the customer’s targets.

4. What Is Numasizing? • Numasizing is not based on theoretical approach or mathematical model. A data base of over 250,000 test firings of cylinders allows Numatics to predict components and pressures with confidence.

5. The Numasizing Process is... A + B + C = D A. Establishing Customer Objectives B. Delineating All Circuit Specifications C. Selecting All Circuit Components D. Results in Obtaining Customer Goals

6. The Numasizing Process is... A. Establishing Customer Objectives 1. Increase Productivity 2. Optimum Energy Utilization 3. Minimum Component Size

7. The Numasizing Process is... B. Delineating All Circuit Specifications 1. Extend Load & Retract Load 2. Available Pressure 3. Desired Extend & Retract Times 4. Conductor Length 5. Required Cylinder Stroke, etc.

8. INTO…. The Numasizing Process is... Numasizing Computer Bank Generating C. Selecting All Circuit Components All components in the pneumatic system starting at the valve through the exhaust port such as the valve, fittings, conductor, regulator and actuator.

9. The Numasizing Process... D. Results in Obtaining Customer Goals 1. Actuator Response Time 2. Compressed Air Cost 3. Air Consumption 4. Pressure Optimization 5. Compatible Sizes

10. Effecting Pneumatics

11. Knowledge of a pneumatic circuit design begins with an understanding of the term Cv is essentially a dimensionless number used to express the CONDUCTANCE VALUE of a pneumatic device. All fixed orifice devices in a pneumatic system have a conductance value and therefore a certain capability to flow. Cv The larger the Cv --- the greater the flow. Typically, the greater the Cv of the entire circuit, the faster the devices in the circuit will respond.

12. 1 1 1 + = 2 2 2 CVV CVFC CVS 1 1 1 + = 2 2 2 4 4.5 2.99 CVO = 1.12 CVV = 4.0 CVFC = 4.5 CVP = 4.8 CVC = 6.0 CVF = 7.0 2.99 1.05 1.03 1.01 1.000 CVV = Valve CV CVFC = Flow Control CV CVO = Orifice CV CVP = Pipe CV CVC = Cylinder Port CV CVF = Fitting CV CVO = 4.37 CVV = 4.0 CVFC = 4.5 CVP = 4.8 CVC = 6.0 CVF = 7.0 2.99 2.47 2.19 2.06 2.000

13. Stroke time is inversely proportional to Cvs Stroke time is directly proportional to system exhaust volume Relation Between Cvs & Actuator Stroke Time (T) Cvs T (sec) 1 4.0 2 2.0 4 1.0 8 0.5 1,092 ft/sec (vel of air - unconfined) 400 ft/sec (vel of air - in a straight conductor) 80 in/sec (vel of actuator piston & rod)

14. Cv of Fittings are Established by Calculating an Equivalent Conductor Length Utilizing a K Factor Once the ID and the line length has been determined --- plus all the fittings equivalent length has been added, we can calculate the Cv of the CONDUCTOR. Formula: Le’ = n K d” Equivalent Length = number of fittings x K (feet) factor x ID (inches) Formula: Le + Lc = Lt Equivalent Length + Conductor Length = Total Length

15. Cv of Fittings are Established by Calculating an Equivalent Conductor Length Utilizing a K Factor Fitting K Values: Device K Factor Swing Check Valves, fully open 11 90 Degree Standard Elbow 2.5 45 Degree Standard Elbow 1.3 90 Degree Long Radius Elbow 1.5 Reducer (1 Size) 1.5 Enlarger (1 Size) 2 “Y” Fitting 1 Standard Tee Flow Through Run 1.5 Flow Through Branch 5

16. K = 1 Y Fitting K = 1 R=3 x I.D. Long Radius K =1.5 R=1.5 x I.D. Std. Elbow K = 2.5 R=0 x I.D. SHARP 90° K = 5.0 BRANCH TEE K = 5 ELBOWS 10 x I.D. minimumK = 0 RUN TEE K = 1.5 x BENDS The more significant the change in flow direction, the greater the restriction and therefore the greater the K factor. K = 0

17. 3/8 NPT 1/4 NPT C ID .493 - 3/8 NPT Pipe SCH. 40 ID .364 - 1/4 NPT Pipe, SCH .40 C 3.0 v 5.6 v Formula: Cv (port)= 23 x Ds Ds = smallest I.D. in a port C 3.0 2 What is the cylinder’s Cv ? To calculate Cv we need to determine the smallest I.D. (DS) of whatever component is in the port of the cylinder. Reducing the port size with a bushing reduces the flow capability of the port. The smallest orifice will determine the effective Cv .

18. Ftg. 1/2 Tube to 3/8 NPT ID = .406 3/8 NPT .406” ID - Controlling Dimension 1/2 Steel Tube - .43” ID The Cv of a cylinder port is based on the smallest I.D. entering that port How shall we decide which ID conductor to use?

19. The Cv of a cylinder port is based on the smallest I.D. entering that port Ftg. 1/2 Tube to 3/8 NPT ID = .406 3/8 NPT .406” ID 1/2 Nylon Tube = .38” ID -- NOW becomes the Controlling Dimension for the CYLINDER PORT. A fitting that accepts steel tubing will have a larger effective orifice than a fitting for plastic or poly tubing because the wall thickness for plastic / poly tubing is so much greater.

20. 2 Flow Cd = 0.98 Cv = 28.8d 2 Flow Cd = 0.92 Cv = 27.0d 2 Flow Cd = 0.82 Cv = 24.1d 2 Flow Cd = 0.80 Cv = 23.5d 2 Flow Cd = 0.72 Cv = 21.2d 2 Flow Cd = 0.65 Cv = 19.1d 2 Flow Cd = 0.61 Cv = 18.0d 2 Flow Cd = 0.53 Cv = 15.6d All openings = d” (diameter) Various orifice discharge coefficients Cd and their related Cv

21. Optimized Drilling for Intersecting Holes

22. 3/16” I.D. Tubing 1/8” I.D. Tubing ConductorI.D. 1/16” I.D. Tubing 0 0 Time There is an optimum conductor ID for each application. Choose the best commercially available size. Too small a conductor ID and there is RESTRICTION; too large and it becomes a VOLUME CHAMBER. The LENGTH of the conductor now must be considered.

23. Given: 3/8 NPT Sch.40 pipe has an I.D. of 0.493” or essentially 1/2” Conductor Length Cv 10’ (3/8 NPT) 3.0 20’ (3/8 NPT) 2.1 30’ (3/8 NPT) 1.75 The Bends - a practical example: The direct distance between the valve and the cylinder is 10’. We choose to pipe the circuit with an additional 10’ (because it looks better). If we used (8) 90 degree standard elbows (K=2.5) between the valve and cylinder, it would have an equivalent length of 10’: Le’ = nKd” Le’ = 8 x 2.5 x 1/2” = 10’ We have penalized the cylinders’ speed by reducing the efficiency of the circuit. By identifying the weak link, we can improve our circuit.

24. TO CYLINDER FREE FLOW C L BOTTOM PORT MODEL SELECTION AND FLOW CAPACITY (CV) CHART If utilize bottom port option: 2.44 2.86 6.36 7.95 Size Model Number * Free Flow Controlled Flow 1/4 NPTF 3/8 NPTF 1/2 NPTF 3/4 NPTF 2FC2 3FC2 4FC3 5FC3 2.3 2.7 6.0 7.5 2.0 2.4 5.5 6.0 ** For optional bottom port, add B to model number (e.g..., 4FC3B The most common way of adjusting a cylinder’s speed is with a FLOW CONTROL. A flow control cannot conserve compressed air, reduce force, or speed up a circuit.

25. E P ACTUATOR ACTUATOR PRESSURE PRESSURE EXHAUST EXHAUST APPLIED BLOCKED Quick Exhaust Valve The most common device selected for improving cylinder speed is a Quick Exhaust Valve. This device has a limited life -- the disc is slammed (full line pressure) twice every cycle. No breaking or control of the cylinder will occur either. Quick exhaust allows the cylinder to exhaust at the cylinder port, not back through the valve or through a flow control.

26. F L L x e L f P P e = 75 r = 75 L F x L r L f P P e = 75 r = 75 On any cylinder, there are three LOADS resisting movement. The cylinder must overcome each load before it can extend or retract. Let’s examine each load separately. Cylinders, typically, must have some friction in sealing the piston against the cylinder wall and also the rod --- Load Lf. Work that the cylinder performs is expressed as Le for extend and Lr for retract directions of movement. The crucial load, however, is Lx --- the EXHAUST BACK PRESSURE LOAD. Loads on a Cylinder

27. LOAD 60 # in .4 SEC Extend Stroke Return Stroke Supply Exhaust LOAD 10 # in 0.45 SEC 75 PSIG 75 PSIG SAMPLE PROBLEM: Move a load in the specific times, given 75 psig supply.

28. T T or e r T T d m Exhaust T 75psig y 75psig Supply P D P dm P d Pressure PSIG End of Stroke 0 0 Time - Seconds Cylinder Response Time in Seconds Pdm = 36psi Pressure differential at inception of motion Pd = Running pressure differential varies during cycle (see curve) Td = 0.14 Time delay primarily due to exhaust preload, a minimum delay due to static friction of cylinder seals and a negligible solenoid time delay (Ty). Tm = 0.26 Time cylinder piston is in motion Te or Tr = 0.40 Time to extend or Time to retract (Td+Tm) Ty = 0.008 Solenoid time delay DP = 10psi Pressure drop from Supply Pressure

29. F L L x e L f P P e = 49 r = 21 L F x L r L f P P e = 49 r = 21 Pressures Fit The Load Select the cylinder pressures based on required force. Consider what the cylinder must do in each direction and select the required pressure for each individual action.

30. LOAD 60 # in 0.40 SEC Extend Stroke Supply Exhaust Return Stroke 49 PSIG 21 PSIG 49 PSIG 21 PSIG Now that the ideal pressures have been selected, what results can be observed? LOAD 10 # in 0.45 SEC

31. T T or e r T T d m Exhaust T 21psig y 49psig Supply P D Pressure PSIG P d P dm End of Stroke 0 0 Time - Seconds Cylinder Response Time in Seconds Pdm = 36psi Pressure differential at inception of motion Pd = Running pressure differential varies during cycle (see curve) Td = 0.14 Time delay primarily due to exhaust preload, a minimum delay due to static friction of cylinder seals and a negligible solenoid time delay (Ty). Tm = 0.26 Time cylinder piston is in motion Te or Tr = 0.40 Time to extend or Time to retract (Td+Tm) Ty = 0.008 Solenoid time delay DP = 2 psi Pressure drop from Supply Pressure

32. Te orTr Exhaust 20psig Td Time saved from initial pressure conditions (0.11) Tm Ty 75psig Supply P D Initial: Exhaust 75psig 75psig Supply Pd Pressure PSIG End of Stroke End of Stroke Pdm 0 00 0 Time - Seconds Dual Pressure Response Time in Seconds Pdm = 36psi Pressure differential at inception of motion Pd = Running pressure differential varies during cycle (see curve) Td = 0.09 Time delay primarily due to exhaust preload, a minimum delay due to static friction of cylinder seals and a negligible solenoid time delay (Ty). Tm = 0.19 Time cylinder piston is in motion Te or Tr = 0.28 Time to extend or Time to retract (Td+Tm) Ty = 0.008 Solenoid time delay DP = 0.4 psi Pressure drop from initial Supply Pressure

33. $ Capital Expenses HP Required $ Operating Cost $ Cost/M Survey # FC’s Pe/Pr Bore CPM Te/Tr 2 75/75 1.50 .40/.45 70 yes 476 257 4.75 1.34 2B 75/75 2.00 .23/.14 163 no 1733 287 2.08 4.89 3 49/21 1.50 .40/.45 70 no 229 230 4.75 0.65 4 61/35 1.50 .40/.45 70 no 289 217 4.75 0.81 5 70/31 1.25 .40/.45 70 no 219 190 4.75 0.62 Compressor pressure for all surveys are kept at 100 PSIG Valve Cv‘s and costs for survey #’s 2, 2B, 3, 4 and 5 are respectively, 2.05 - $85, 2.05 - $85, 0.41 - $58, .25 - $45, 0.25 - $45 3/8” NPT Pipe conductor cost $25 and 1/8” NPT costs $20 Cylinder costs for 2.0”, 1.5” & 1.25” bores are respectively $152, $102 and $75 A pair of flow controls cost $20 and a pair of regulators cost $50 Survey #2 ($85+$25+$102+$20+$25 = $257) O.E.M. would select #5 because of minimum capital investment Survey #2B ($85+$25+$152+$102+$50 = $287) End user would select #2B because of low cost/pc Survey #3 ($58+$20+$102+$50 = $230) Facilities would select #3 because of lowest pressure Survey #4 ($45+$20+$102+$50 = $217) demands or #5 because of minimum HP requirement Survey #5 ($45+$20+$75+$50 = $190) depending on the major objective of facilities engineer

34. 2-1/2 Seconds 40 Minutes 300 Hours Dynamic Seals With Lubrication Flow of O-Ring into Metallic Surfaces “The theory has been proposed and generally accepted that the increase of friction on standing is caused by the rubber O-ring flowing into the microfine grooves or surface irregularities of the mating part. As a general rule for a 70º durometer rubber against an 8 micro-inch surface, the maximum break-out friction which will develop in a system is three times the running friction.” Friction is always a factor with dynamic seals. Obviously, the more dynamic seals --- the greater the friction. We can assume the cylinder has been sized properly to overcome the Le and Lr loads. Lx --- the exhaust back pressure load --- is determined by the ability of the cylinder’s ports to allow exhaust to escape; the Cv of the cylinder.

35. Effective Orifice --- Valves: As with cylinders, VALVECvis determined not by the port of the valve, but by the smaller orifice of the fitting in the port. Port Size (NPT) 10/32 1/8 1/4 3/8 1/2 3/4 1 1 1/4 Valve Size MK 3 .18 .35* MK 7 .2 .4* MK 8 .3 .8 1.0* MK 15 .4 1.1 1.4 1.5* MK 55 .6 2.0 2.85 4.0 5.0 5.55* *there is no improvement in the valve’s Cv beyond this point • Consider: System Cv equals the combination of • Cylinder Cv ~ Conductor and Fittings Cv ~ Valve Cv • and any additional devices in the circuit

36. Q (SCFM) P P 1 2 Flow meter P Needle Valve P P P = + 1 2 P P P Q 2 1 80 “ “ “““““““ 0 10 20 30 40 45 50 60 70 80 80 70 60 50 40 35 30 20 10 0 0 28 38 43 45.5 46 “ “ “ “ Terminal velocity occurs at the CRITICAL PRESSURE RATIO P P P 2 = .47 = .53 P 1 1 Critical Pressure Ratio: At start, there is 80 psig available and the needle valve is closed. As the needle valve is slowly opened, in DP increments of 5 and 10 psi, the flow is noted. OBSERVEEven with a larger DP --- FLOW DOES NOT CONTINUE TO INCREASE. Air has reached CRITICAL FLOW and FLOW cannot increase.

37. P P = .47 1 Terminal Velocity: A jet engine takes air in, compresses the air with fuel, and then ignites the mixture. At some point, air cannot enter the engine any faster. No matter how much more fuel is added --- the plane cannot go any faster. The limiting factor is AIR FLOW, supply to the engine. In a pneumatic circuit, the cylinder reaches TERMINAL VELOCITY when the air cannot enter and/or exit the circuit any faster. This occurs at

38. H 4 5 3 6 2 C 1 7 F 8 0 x1000r/min E Terminal Velocity for a Cylinder: Therefore: HOW FAST CAN AN AIR CYLINDER CYCLE? • IF a properly sized cylinder had • air supplying and • exhausting at the • CRITICAL PRESSURE DROP RATIO THE PNEUMATIC CYLINDER WOULD BE AT • TERMINAL VELOCITY for that particular circuit. • To improve cycle time we must improve the smallest Cv of the system and maintain the critical pressure ratio.

39. If Energy Waste Was This Obvious, You’d Put A Stop To It

40. Retract Pressure (PSIG) 90 80 70 60 50 40 30 0% 5.0 9.2 14.8 19.1 24.0 28.9 10.0 14.1 19.8 24.0 28.9 33.9 18.4 24.0 28.2 33.9 38.8 29.6 33.9 38.8 43.7 38.1 43.0 47.9 47.8 52.9 57.8 90 80 70 60 50 40 30 Extend Pressure (PSIG) Benefits of Dual Pressure: Why NUMASIZE? Percent Cost Savings Attainable Utilizing Numasizing inIndustrial Pneumatic Systems

41. PSIG1/32 3/64 1/16 3/32 1/8 30 $ 78 $172 $310 $ 689 $1239 50 $112 $252 $448 $1009 $1784 70 $146 $330 $566 $1316 $2337 90 $181 $403 $722 $1611 $2891 110 $216 $486 $861 $1931 $3444 24 hrs/day - 365 days/yr. 30¢/1000 SCFM (6.75¢/kwh) Air Leaks: Annual Cost of Compressed Air LeakageThrough Orifices of Various Diameters Compressed air is ENERGY that must be CONSERVED.

42. 4000 3750 24 hours/day - 365 days/year 3500 3250 3000 Cost to Generate Compressed Air ($1,000’s) 2750 3 shifts/day - 250 days/year 2500 2250 2000 2 shifts/day - 250 days/year 1750 1500 1250 1 shift/day - 250 days/year 1000 750 Horsepower 1000 2000 3000 4000 5000 6000 500 250 Resource Cost: Compressed Air Cost Average Compressor H.P. In Use based upon 4 s.c.f.m./H.P. - 30¢/1000 s.c.f. - 6.75¢/kwh

43. 1990 FUEL SOURCES -- Production of ENERGY OIL COAL NAT. GAS HYDRO NUCLEAR 41.9% 23.3% 24.0% 3.5% 7.0% 16,120,000 BPD 8,960,000 BPDOE 9,230,000 BPDOE 1,350,000 BPDOE 2,690,000 PBDOE OTHER 0.3% 38,470,000 BPDOE (Barrels Per Day Oil Equivalent) 120,000 PBDOE (81.44 Quadrillion BTU's or Quads) Transportation 10,400,000 PBDOE 27% INDUSTRIAL COMMERCIAL RESIDENTIAL 14,200,000 BPDOE 7,700,000 BPDOE 6,170,000 BPDOE Electrical Segment of Industry 4,860,000 BPDOE (10.26 Quads) 34% Total Energy Production: 37% 16% 20% CONSUMPTION

44. 1990 FUEL SOURCES -- Production of ELECTRICITY OIL COAL NAT. GAS HYDRO NUCLEAR 4.2% 56.8% 9.8% 9.5% 18.9% 597,000 BPD 8,081,000 BPDOE 1,392,000 BPDOE 1,350,000 BPDOE 2,690,000 PBDOE 14,230,000 BPDOE (Barrels Per Day Oil Equivalent) OTHER 0.8% 120,000 PBDOE (30.1 Quadrillion BTU's or Quads - 37% of US. Energy System of 81.26 Quads) ELECTRICITY 14,230,000 BPDOE (30.1 Quads) DISTRIBUTION OTHER If Numasized, a minimum savings of 15% would be realized. 390,000 PBDOE (0.88 Quads) 3% Compressed Air Segment INDUSTRIAL COMMERCIAL RESIDENTIAL 4,860,000 BPDOE 4,290,000 BPDOE 4,690,000 BPDOE 1,070,000 BPDOE (2.26 Quads) 22% (10.26 Quads) (9.0 Quads) (9.96 Quads) 34% 30% 33% 10% Savings 107,000 BPDOE (.22 Quads) 20% Savings 214,000 BPDOE (.45 Quads) compared to Prudhoe Bay Oil Fields 1,800,000 BPD Electricity Production:

45. Conceptual View Point: The preceding two slides/charts are based on Energy Information Administration (EIA) surveys. There were some slight differences between the 1990 and the updated figures for 1990 as published in the October 1991 report. Some variations also showed up when comparing the figures of National Business, Electric World, Pipe and Gas Journal, etc. with those of EIA. Since the deviations were minor (mostly due to independent rounding), we have reconciled all of them as to render uniform results. From a conceptual viewpoint and from the ultimate potential savings and increased productivity aspect, they have no bearing as the differences are miniscule. Also, we have incorporated in these figures the fact that 1,033,100 BTU’s (EIA from utilities) is required to generate 100 Kwhrs of electricity (due to losses in conversion and transmission), while 300 Kwhrs are needed to produce 1,024,000 BTU’s of heat (1 Kwhrs = 3413 BTU’s).

46. PNEUMATICS BERNOULLI’S LAW Q = 22.48 CVP x P2 T1 G Q = 22.48 CV .47 P1 x .53P1 528 x 1 (AT CRITICAL FLOW) Q = 0.488 P1CV Q = (K) P CV ELECTRICITY OHM’S LAW E = I R E R I = E Cd I = Q = P CV Same General Formula Formulae: Those familiar with electricity will recognize Ohm’s Law. In pneumatics, FLOW is represented by Q, T is temperature in Rankin, G is the specific gravity (assume 1 for air), P1 and P2 are pressure expressed in PSIA. • Flow of electrons is very similar to flow of air molecules • Think of CONDUCTANCE as the reciprocal of RESISTANCE • Whose formula came first?

47. Your Most Important Design Tool