Presentation on theme: "? What Do I use? Numasizing. Compressed air is an energy medium that must be conserved, Numasizing has been developed to use this resource more effectively."— Presentation transcript:
Compressed air is an energy medium that must be conserved, Numasizing has been developed to use this resource more effectively. 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.
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.
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
The Numasizing Process is... A.Establishing Customer Objectives 1.Increase Productivity 2.Optimum Energy Utilization 3.Minimum Component Size
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.
Generating Numasizing Computer Bank INTO…. The Numasizing Process is... 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.
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
Knowledge of a pneumatic circuit design begins with an understanding of the term C v 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. The larger the C v --- the greater the flow. Typically, the greater the C v of the entire circuit, the faster the devices in the circuit will respond. CvCvCvCv
C VV =Valve C V C VFC = Flow Control C V C VO = Orifice C V C VP = Pipe C V C VC =Cylinder Port C V C VF = Fitting C V C VV = 4.0C VFC = 4.5 2.99 C VO = 1.12 1.05 C VP = 4.8C VC = 6.0C VF = 7.0 1.03 1.01 2.99 2.47 2.19 2.06 2.000 C VV = 4.0C VFC = 4.5 C VO = 4.37 C VP = 4.8C VC = 6.0C VF = 7.0 1.000 C VV C VFC C VS 2 2 2 1 1 1 + = 4 4.5 2.99 2 2 2 1 1 1 + =
Relation Between C vs & Actuator Stroke Time (T) n Stroke time is inversely proportional to C vs n Stroke time is directly proportional to system exhaust volume 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) CvsT (sec) 1 4.0 2 2.0 4 1.0 8 0.5
C v of Fittings are Established by Calculating an Equivalent Conductor Length Utilizing a Factor C v 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 C v of the CONDUCTOR. Formula: L e ’ = n K d” Equivalent Length = number of fittings x K (feet) factor x ID (inches) Formula: L e + L c = L t Equivalent Length + Conductor Length = Total Length
C v of Fittings are Established by Calculating an Equivalent Conductor Length Utilizing a Factor C v 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
ELBOWS BRANCH TEE K = 5 BENDS 10 x I.D. minimum K = 0 K = 1.5 RUN TEE x K = 0 K = 1 Y Fitting 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 90Â° K = 5.0 The more significant the change in flow direction, the greater the restriction and therefore the greater the K factor.
Formula: C v (port)= 23 x D s D s = smallest I.D. in a port 2 To calculate C v we need to determine the smallest I.D. (D S ) 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 C v. What is the cylinder’s C v ? C 3.0 3/8 NPT 1/4 NPT ID.364 - 1/4 NPT Pipe, SCH.40 v C 3.0 C 5.6v ID.493 - 3/8 NPT Pipe SCH. 40
Controlling Dimension.406” ID - Controlling Dimension Ftg. 1/2 Tube to 3/8 NPT ID =.406 1/2 Steel Tube -.43” ID 3/8 NPT How shall we decide which ID conductor to use? The C v of a cylinder port is based on the smallest I.D. entering that port
.406” ID 3/8 NPT Controlling Dimension 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. The C v of a cylinder port is based on the smallest I.D. entering that port Ftg. 1/2 Tube to 3/8 NPT ID =.406
Flow C d = 0.98C v = 28.8d 2 C d = 0.92C v = 27.0d 2 C d = 0.82C v = 24.1d 2 C d = 0.80C v = 23.5d 2 C d = 0.72C v = 21.2d 2 C d = 0.65C v = 19.1d 2 C d = 0.61C v = 18.0d 2 C d = 0.53C v = 15.6d 2 Various orifice discharge coefficients C d and their related C v All openings = d” (diameter)
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. Conductor I.D. Time 0 0 3/16” I.D. Tubing 1/16” I.D. Tubing 1/8” I.D. Tubing
The Bends - a practical example: Given: 3/8 NPT Sch.40 pipe has an I.D. of 0.493” or essentially 1/2” Conductor LengthC v 10’ (3/8 NPT)3.0 20’ (3/8 NPT)2.1 30’ (3/8 NPT)1.75 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.
The most common way of adjusting a cylinder’s speed is with a FLOW CONTROL. 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 MODEL SELECTION AND FLOW CAPACITY (C V ) CHART ** For optional bottom port, add B to model number (e.g..., 4FC3B A flow control cannot conserve compressed air, reduce force, or speed up a circuit. If utilize bottom port option: 2.44 2.86 6.36 7.95
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. EP
L e L x F L f P e = 75 P r = 75 Loads on a Cylinder 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 L f. Work that the cylinder performs is expressed as L e for extend and L r for retract directions of movement. The crucial load, however, is L x --- the EXHAUST BACK PRESSURE LOAD. F L f L xL r P e = 75 P r = 75
Return Stroke LOAD 10 # in 0.45 SEC 75 PSIG Extend Stroke Supply Exhaust LOAD 60 # in.4 SEC 75 PSIG SAMPLE PROBLEM: Move a load in the specific times, given 75 psig supply.
Cylinder Response Time in Seconds Exhaust Supply 75psig P dm =36psi Pressure differential at inception of motion P d =Running pressure differential varies during cycle (see curve) T d =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 (T y ). T m =0.26 Time cylinder piston is in motion T e or T r =0.40 Time to extend or Time to retract(T d +T m ) T y =0.008 Solenoid time delay P=10psi Pressure drop from Supply Pressure P d T d T y P P dm m T End of Stroke Pressure PSIG Time - Seconds 0 0 TT re or
L e L x F L f P e = 49 P r = 21 F L f L xL r P e = 49 P 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.
Extend Stroke LOAD 60 # in 0.40 SEC 21 PSIG 49 PSIG Supply Exhaust LOAD 10 # in 0.45 SEC Return Stroke 21 PSIG 49 PSIG Now that the ideal pressures have been selected, what results can be observed?
Cylinder Response Time in Seconds Exhaust Supply 21psig 49psig P dm =36psi Pressure differential at inception of motion P d =Running pressure differential varies during cycle (see curve) T d =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 (T y ). T m =0.26 Time cylinder piston is in motion T e or T r =0.40 Time to extend or Time to retract (T d +T m ) T y =0.008 Solenoid time delay P=2 psi Pressure drop from Supply Pressure P dm P d End of Stroke T m T d Pressure PSIG Time - Seconds 0 0 P T y TT re or
Dual Pressure Response Time in Seconds P dm =36psi Pressure differential at inception of motion P d =Running pressure differential varies during cycle (see curve) T d =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 (T y ). T m =0.19 Time cylinder piston is in motion T e or T r =0.28 Time to extend or Time to retract (T d +T m ) T y =0.008 Solenoid time delay P=0.4 psi Pressure drop from initial Supply Pressure Exhaust Supply 20psig 75psig Exhaust Supply 75psig Initial: PdPd End of Stroke TdTd TyTy Pressure PSIG Time - Seconds 0 0 00 End of Stroke P dm Time saved from initial pressure conditions (0.11) TmTm T e or T r P
Survey # Pe/Pr Bore Te/Tr CPM FC’s $ Cost/M $ Capital Expenses Expenses $ Operating Cost HP Required 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 C v ‘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
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 L e and L r loads. L x --- the exhaust back pressure load --- is determined by the ability of the cylinder’s ports to allow exhaust to escape; the C v of the cylinder.
Effective Orifice --- Valves: VALVE C v As with cylinders, VALVE C v is determined not by the port of the valve, but by the smaller orifice of the fitting in the port. Port Size (NPT) 10/321/81/43/81/23/411 1/4 Valve Size MK3.18.35* MK7.2.4* MK 22.214.171.124* MK 126.96.36.199.5* MK 55.62.02.854.05.05.55* *there is no improvement in the valve’s C v beyond this point Consider: System C v equals the combination of Cylinder C v ~ Conductor and Fittings C v ~ Valve C v and any additional devices in the circuit
Critical Pressure Ratio: P 80 “ “ “ “ “ “ “ “ = 1 P + P 2 PP 2 P Q 0 10 20 30 40 45 50 60 70 80 70 60 50 40 35 30 20 10 0 28 38 43 45.5 46 “ 1 At start, there is 80 psig available and the needle valve is closed. As the needle valve is slowly opened, in P increments of 5 and 10 psi, the flow is noted. OBSERVE Even with a larger P --- FLOW DOES NOT CONTINUE TO INCREASE. Air has reached CRITICAL FLOW and FLOW cannot increase. Terminal velocity occurs at the CRITICAL PRESSURE RATIO =.53 P 2 P 1 P 1 =.47 P Q (SCFM) P 1 P 2 P Needle Valve Flow meter
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 P 1 =.47 P
Terminal Velocity for a Cylinder: 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 C v of the system and maintain the critical pressure ratio. Therefore: HOW FAST CAN AN AIR CYLINDER CYCLE? 8 0 1 2 3 4 5 6 7 x1000r/min H C F E
If Energy Waste Was This Obvious, You’d Put A Stop To It
Air Leaks: Annual Cost of Compressed Air Leakage Through Orifices of Various Diameters 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 PSIG1/323/641/163/321/8 24 hrs/day - 365 days/yr. 30¢/1000 SCFM (6.75¢/kwh) Compressed air is ENERGY that must be CONSERVED.
Resource Cost: Average Compressor H.P. In Use based upon 4 s.c.f.m./H.P. - 30¢/1000 s.c.f. - 6.75¢/kwh Compressed Air Cost 4000 3750 3500 3250 3000 2750 2500 2250 2000 1750 1500 1250 1000 750 500 250 100020003000400050006000 1 shift/day - 250 days/year 2 shifts/day - 250 days/year 3 shifts/day - 250 days/year 24 hours/day - 365 days/year Cost to Generate Compressed Air ($1,000’s) Horsepower
Total Energy Production: 1990 FUEL SOURCES -- Production of ENERGY 38,470,000 BPDOE (Barrels Per Day Oil Equivalent) (81.44 Quadrillion BTU's or Quads) Electrical Segment of Industry 4,860,000 BPDOE (10.26 Quads) 34% 37%20%16% CONSUMPTION
Electricity Production: 10% Savings 107,000 BPDOE (.22 Quads) 20% Savings 214,000 BPDOE (.45 Quads) compared to Prudhoe Bay Oil Fields 1,800,000 BPD 1990 FUEL SOURCES -- Production of ELECTRICITY 14,230,000 BPDOE (Barrels Per Day Oil Equivalent) (30.1 Quadrillion BTU's or Quads - 37% of US. Energy System of 81.26 Quads) If Numasized, a minimum savings of 15% would be realized. Compressed Air Segment 1,070,000 BPDOE (2.26 Quads) 22%
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).
Formulae: Flow of electrons is very similar to flow of air molecules Think of CONDUCTANCE as the reciprocal of RESISTANCE Whose formula came first? 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), P 1 and P 2 are pressure expressed in PSIA. Same General Formula E = I R E R I = E C d ELECTRICITY OHM’S LAW I = Q = P C V PNEUMATICS BERNOULLI’S LAW Q = 22.48 C V P x P 2 T 1 G Q = 22.48 C V.47 P 1 x.53P 1 528 x 1 (AT CRITICAL FLOW) Q =0.488 P 1 C V Q =(K) P C V