CALCULATE WITH AIRCYLINDERS

PIPING

VALVES

EXPOSED LENGTH OF PISTON ROD, IN.Tons 10 20 40 60 70 80 100 120

12 34 134 1316 1116

1 58 78 118 114 138

112 1116 1516 1316 138 112

2 34 1 134 1716 1916 1316

3 1316 78 118 138 1916 158 178

4 1516 1 1316 112 158 134 2 214

5 1 118 1516 1916 134 178 218 238

712 1316 114 1716 134 178 2 214 212

10 138 1716 158 178 2 218 2716 234

15 1116 134 178 218 214 238 21116 3

20 2 2 218 238 212 258 278 314

30 238 2716 212 234 234 278 314 312

40 234 234 278 3 3 314 312 334

50 318 318 314 338 312 312 334 4

75 334 334 378 4 4 418 438 412

100 438 438 438 412 434 434 4-7/8 5

150 538 538 538 512 512 512 534 6

Table shows minimum suggested rod diameters for various rod diameters and unsupported lengths. Any cylinder mounting other than vertical can cause bending stress on the rod when extended, just from the weight of the rod and cylinder itself. Therefore, trunnion mounting should be used in these applications to help balance the cylinder weight when the rod is extended.

PISTON ROD STRENGTH

A cylinders extended piston rod can buckle if subjected to a heavy load. The accompanying chart suggests minimum rod di-ameters for various loads when the rod is extended and unsup-ported. The values are based on no side load or bending stress at any point along the rod.

The exposed length of rod is shown at the top of the table. This length is typically lon-ger than the stroke length of the cylinder. The vertical scale shows the load on the cylin-der and is in English tons (1 ton = 2,000 lb). If the rod and front end of the cylinder bar-rel are rigidly supported, then a smaller rod will be sufficient; use the column that is 1/2 the length of the actual piston rod. If pivot-to-pivot mounting is used, double the actual length of the exposed rod and use the suggested rod diameter.

CYLINDER FORCEPiston

diameter,in.

Roddiameter,

in.

Effectiveareain.2

Pressure, PSI

60 70 80 90 100 110 120

112 None 1.77 106 124 142 159 177 195 23058 1.46 88 102 117 132 146 161 190

1 0.99 59 69 79 89 98 108 128

2 None 3.14 188 220 251 283 314 345 37758 2.83 170 198 227 255 283 312 340

1 2.35 141 165 188 212 235 259 283

212 None 4.91 295 344 393 442 491 540 58958 4.60 276 322 368 414 460 506 552

1 4.12 247 289 330 371 412 454 495

3 None 7.07 424 495 565 636 707 778 84858 6.76 406 431 540 608 676 744 814

314 None 8.30 498 581 664 747 830 913 996

1 7.51 451 526 601 676 751 826 902

138 6.82 409 477 545 613 681 818 818

4 None 12.57 754 880 1006 1131 1257 1283 1508

1 11.78 707 825 943 1061 1178 1296 1415

138 11.09 665 776 887 998 1109 1219 1330

5 None 19.64 1178 1375 1571 1768 1964 2160 2357

1 18.85 1131 1320 1508 1697 1885 2074 2263

138 18.16 1089 1271 1452 1634 1816 1997 2179

6 None 28.27 1696 1979 2262 2544 2827 3110 3392

138 26.79 1607 1875 2143 2411 2679 2946 3214

134 25.90 1552 1811 2069 2328 2586 2845 3104

8 None 50.27 3016 3519 4022 4524 5027 5530 6032

138 48.79 2927 3415 3903 4391 4879 5366 5854

134 47.90 2872 3351 3829 4308 4786 5265 5744

10 None 78.54 4712 5498 6283 7069 7854 8639 9425

134 76.14 4568 5329 6091 6852 7614 8375 9136

2 75.40 4524 5278 6032 6786 7540 8294 9048

12 None 113.10 6786 7917 9048 10179 11310 12441 13572

2 110.00 6598 7697 8797 9896 10996 12095 13195

212 108.20 6491 7573 8655 9737 10819 11901 12983

DETERMINING CYLINDER FORCE

The table below shows cylinder forces in pounds for both extension and retraction. Numbers in color text represent extension forces, using the full piston area. Numbers in black show retraction forces with various rod sizes. These values are theoretical, derived by calculation.

Pressures in the top row of the chart represent differential pressures across the two cylinder ports. In practice, the air-supply line must supply another 5% of pressure to make up for cylinder loss, and must supply about 25% to 50% additional pressure to compensate for flow losses in lines, fittings, and valves so the cylinder will attain suf-ficient travel speed. Most manufacturers recommend designing a system 25% greater than theoretical calculations.

Cylinder forces are shown for extension and retraction. Values in blue represent extension, where pressure is applied to the cap end of the cylinder (no rod area). Values in black text represent retraction forces, with pressure applied to the cylinders cap end.

ESTIMATING CYLINDER SPEED

Cylinder speed is difficult to estimate because of flow losses within the system from piping, fittings, and flow paths and restriction in the particular valves used. Flow losses cause a drop in pressure, which directly affects the force output. The sum of all flow losses, pressure required for the force output, and the available inlet pressure must all be known to accurately determine maximum speed of the cylinder. Because these losses cannot be accurately determined by calcula-tion, detailed simulation and physical testing are the only ways to accurately determine cylinder speeds. However, general rules can be applied to approximate air cylinder speed.

The first general rule is to choose a cylinder that will allow at least 25% more force than what is required. For extremely fast operations, choose a cylinder that will provide 50% more force than required. This will leave 25% or 50% of inlet pressure to satisfy system losses.The second rule is to select a directional control valve that has the same port size as the cylinder it will be operating. Larger valves of-ten have internal flow capacity that is the same as the connection size, whereas the internal flow capacity of smaller valves is typically much less than the connection size. However, always compare man-ufacturers flow rating and other specifications to help ensure that the system will deliver the best combination of speed, output force, and efficiency.

APPROXIMATE CYLINDER SPEEDS

Bore, in.Valve orifice diameter, in.

132 116 18 14 38 12 34 11 6 15 37

118 5 12 28 85112 3 7 16 502 4 9 28 70

212 6 18 45 723 4 12 30 48

314 3 10 24 37 794 7 17 28 605 4 11 18 40 826 3 7 12 26 558 4 7 15 32

10 4 9 2012 3 6 14

Values in the table are in in./sec and represent approximate speeds under average conditions where the force required is 50% of available from a constant 80- to 100-psi inlet pressure and the directional valve internal flow area is equal to its port area. Acceleration distance is assumed to be relatively short compared to a sufficiently long stroke.

ESTIMATED TRAVEL SPEED OF A LOADED AIR CYLINDER

The chart at right gives theo-retical SCFM air flow in SCFM through sharp-edged orifices. In actual practice, only about 23 of this flow is achieved. Assume 75% of line pressure is actually working on the load; the remain-ing 25% is consumed by flow losses, valves, connecting lines, etc. Use the accompanying ta-ble to approximate flow through a sharp-edged orifice.

Calculate 75% of line pres-sure, in psi, and find it in the left-most column in the accompany-ing table. Scan across the table to the column corresponding to the port size of the control valve. Because valves do not contain sharp-edged orifices, divide this flow value in half. After de-termining the flow, convert it to cfm at the pressure required to move the load. Travel speed can be estimated from this value.

APPROXIMATE AIR FLOW THROUGH AN ORIFICEOrifice diameter, in.

Pressure across

orifice, psi164 132 116 18 14 38 12 5/8 34 78 1

5 0.062 0.249 0.993 3.97 15.9 35.7 63.5 99.3 143 195 2546 0.068 0.272 1.09 4.34 17.4 39.1 69.5 109 156 213 2787 0.073 0.293 1.17 4.68 18.7 42.2 75.0 117 168 230 3009 0.083 0.331 1.32 5.30 21.2 47.7 84.7 132 191 260 339

12 0.095 0.379 1.52 6.07 24.3 54.6 97.0 152 218 297 38815 0.105 0.420 1.68 6.72 26.9 60.5 108 168 242 329 43020 0.123 0.491 1.96 7.86 31.4 70.7 126 196 283 385 50325 0.140 0.562 2.25 8.98 35.9 80.9 144 225 323 440 57530 0.158 0.633 2.53 10.1 40.5 91.1 162 253 365 496 64835 0.176 0.703 2.81 11.3 45.0 101 180 281 405 551 72040 0.194 0.774 3.10 12.4 49.6 112 198 310 446 607 79345 0.211 0.845 3.38 13.5 54.1 122 216 338 487 662 86550 0.229 0.916 3.66 14.7 58.6 132 235 366 528 718 93860 0.264 1.06 4.23 16.9 67.6 152 271 423 609 828 108270 0.300 1.20 4.79 19.2 76.7 173 307 479 690 939 122780 0.335 1.34 5.36 21.4 85.7 193 343 536 771 1050 137190 0.370 1.48 5.92 23.7 94.8 213 379 592 853 1161 1516

100 0.406 1.62 6.49 26.0 104 234 415 649 934 1272 1661110 0.441 1.76 7.05 28.2 113 254 452 705 1016 1383 1806120 0.476 1.91 7.62 30.5 122 274 488 762 1097 1494 1951130 0.494 1.98 7.90 31.6 126 284 506 790 1138 1549 2023

This table helps determine approximate flow through a sharp-edged orifice using line pressure, orifice size of the valve, and applying correction factors.

Air consumption by a cylinder is calculated using geometry and using compres-sion ratio, which is simply the applied pressure expressed in absolute units di-vided by atmospheric pressure. The concept is best illustrated with an example.Determine the air consumption of a 2-in. bore cylinder with a 4-in. stroke operat-ing 30 complete cycles (extend and retract) per minute at 80-psig inlet pressure.1. First, find the area of the piston by converting the bore into the area of a

circle: (2 in. / 2)2 = 3.14 in.2

2. Determine air consumption in a single stroke: 3. 14 in.2 4 in. = 12.56 in.3

3. Determine consumption per complete cycle (in most cases, the volume dis-placed by the piston rod can be ignored): 12.56 in.3 2 = 25.12 in.3/cycle

4. Determine volume of 80 psig air consumed in one minute: 25.12 in.3 30 cycles/min = 753.6 in.3/min of 80 psig air

5. Convert in.3 to ft.3: (753.6 in.3/min) (1728 in.3/ft3) = 0.436 ft.3/min

6. Convert air compressed to 80 psi to uncompressed (free) air: (80 psig + 14.7 psia) 14.7 psia = 6.44, which is the compression ratio.

7. Determine volume (ft3) of free air used per minute: 0.436 ft3 6.44 = 2.81 ft3 of free air used per minute.

8. Therefore, the consumption rate of a 2-in. bore cylinder with a 4-in. stroke operating 30 cpm from 80-psi air is 2.81 scfm of free air. (Standard conditions are 70 F at sea level.)

CYLINDER AIR CONSUMPTION Cylinder

borePressure, psig

60 70 80 90 100 110 120 130 140 1501.50 0.009 0.010 0.012 0.013 0.015 0.016 0.017 0.018 0.020 0.0212.00 0.018 0.020 0.022 0.025 0.027 0.029 0.032 0.034 0.036 0.0392.50 0.028 0.032 0.035 0.039 0.043 0.047 0.050 0.054 0.058 0.0623.00 0.039 0.044 0.050 0.055 0.060 0.066 0.070 0.076 0.081 0.0873.25 0.046 0.053 0.059 0.065 0.071 0.078 0.084 0.090 0.096 0.1024.00 0.072 0.081 0.091 0.100 0.110 0.119 0.129 0.139 0.148 0.1585.00 0.113 0.128 0.143 0.159 0.174 0.189 0.204 0.219 0.234 0.2496.00 0.162 0.184 0.205 0.227 0.249 0.270 0.292 0.314 0.335 0.3578.00 0.291 0.330 0.369 0.408 0.447 0.486 0.525 0.564 0.602 0.642

10.00 0.455 0.516 0.576 0.637 0.698 0.759 0.820 0.881 0.940 1.00012.00 0.656 0.744 0.831 0.919 1.010 1.090 1.180 1.270 1.360 1.450

DETERMINE AIR VOLUME REQUIRED

The figures in the table are for cylinders using standard piston-rod diameters. Air consumption was calculated assuming the cylinder would dwell momentarily at the end of each stroke, allowing air to fill up the cylinder to set system pressure. If the cylinder strokes be-fore compressed air fills it, air consumption will be less than what is shown in the table.

Assuming pressure losses through piping, fittings, and other com-ponents will be about 20% to 25%, ensure that the cylinder bore selected will overcome the load at 75% of available system pressure. Without this surplus pressure the cylinder may not travel at its de-sired speed. As an example, calculate the flow, in scfm, of a cylinder in a punching operation that moves a 2,250-lb weight 60 times per minute through a 6-in. stroke.

By selecting a 6-in. bore cylinder, the 2,250-lb force is produced from a pressure of 80 psi. As a general rule, you should provide about 20% higher pressure (20 psig), to account for system losses and set the regulator at 100 psig. Then, using the table at right, make the following calculation:0.249 x 6 (stroke) x 60 (cycles per minute) = 89.6 scfm

Knowing the supply pressure, go to the corresponding column. Each value represents a 1-in. extend-and-retract cycle. Multiply that value by the actual stroke and by the number of the actual cycles per minute. The result will be the flow, in scfm, for the application.

CALCULATING COMPRESSED AIR CONSUMPTION RATE

PNEUMATIC PIPE SIZEFlow,scfm

Length of run, ft Compressor hp25 50 75 100 150 200 300 500 100

6 12 12 12 12 12 12 12 34 34 118 12 12 12 34 34 34 34 1 1 330 34 34 34 34 1 1 1 114 114 545 34 34 1 1 1 1 114 114 114 7 1260 34 1 1 1 114 114 114 1 12 1 12 1090 1 1 114 114 114 114 1 12 1 12 2 15

120 1 114 114 114 1 12 1 12 1 12 2 2 20150 114 114 114 1 12 1 12 2 2 2 2 12 25180 114 1 12 1 12 1 12 2 2 2 2 12 2 12 30240 114 1 12 1 12 2 2 2 2 12 2 12 3 40300 1 12 2 2 2 2 2 12 2 12 3 3 50360 1 12 2 2 2 2 12 2 12 2 12 3 3 60450 2 2 2 2 12 2 12 3 3 3 3 12 75600 2 2 12 2 12 2 12 3 3 3 3 12 4 100750 2 2 12 2 12 3 3 3 3 12 3 12 4 125

Pipe sizes listed in this table assume a 100-psi pneumatic system to carry air at a 1-psi loss per 100 ft. A conservative approximation considers each pipe fitting equivalent to 5 ft. of pipe. Flow capaci-ties at pressures other than 100 psi will be inversely proportionate to pressure, as calculated using Boyles Law using absolute pressure values.

PNEUMATIC PRESSURE LOSS

Free air, cfm

in. in 1 in. 114 in. 112 in.

80 psi 125 psi 80 psi 125 psi 80 psi 125 psi 80 psi 125 psi 80 psi 125 psi

10 0.45 0.30 0.11 0.08 0.04 0.0220 1.75 1.15 0.40 0.28 0.15 0.0830 3.85 2.55 0.90 0.60 0.30 0.2040 6.95 4.55 1.55 1.05 0.45 0.3050 10.50 7.00 2.40 1.60 0.75 0.50 0.18 0.1260 3.45 2.35 1.00 0.70 0.25 0.1770 4.75 3.15 1.35 0.90 0.35 0.23 0.16 0.1080 6.15 4.10 1.75 1.20 0.45 0.30 0.20 0.1490 7.75 5.15 2.25 1.50 0.56 0.40 0.25 0.17

100 9.60 6.35 2.70 1.80 0.65 0.45 0.30 0.20125 15.50 9.80 4.20 2.80 1.05 0.70 0.45 0.32150 23.00 14.50 5.75 4.00 1.45 1.00 0.65 0.45175 23.00 8.10 5.45 2.00 1.30 0.90 0.60200 10.90 7.10 2.60 1.75 1.15 0.80250 4.05 2.65 1.80 1.20300 5.80 3.85 2.55 1.70350 7.90 5.15 3.55 2.35400 10.30 6.75 4.55 3.05450 5.80 3.80500 7.10 4.70

Values in the table are approximate pressure losses at specified flows for every 100 ft. of clean Schedule 40 steel pipe. Losses will increase as residue builds up on the inner surface of pipe over time.

AIR-FLOW LOSS THROUGH PIPEFlow,scfm

Pipe size, in. NPT12 34 1 114 112 134 2 2 12

5 12.7 1.2 0.510 50.7 7.8 2.2 0.515 114 17.6 4.9 1.120 202 30.4 8.7 2.025 316 50.0 13.6 3.2 1.4 0.730 456 70.4 19.6 4.5 2.0 1.135 621 95.9 26.6 6.2 2.7 1.440 811 125 34.8 8.1 3.6 1.945 159 44.0 10.2 4.5 2.4 1.250 196 54.4 12.6 5.6 2.9 1.560 282 78.3 18.2 8.0 4.2 2.270 385 106 24.7 10.9 5.7 2.9 1.180 503 139 32.3 14.3 7.5 3.8 1.590 646 176 40.9 18.1 9.5 4.8 1.9

100 785 217 50.5 22.3 11.7 6.0 2.3110 950 263 61.1 27.0 14.1 7.2 2.8120 318 72.7 32.2 16.8 8.6 3.3130 369 85.3 37.8 19.7 10.1 3.9140 426 98.9 43.8 22.9 11.7 1.4150 490 113 50.3 26.3 13.4 5.2160 570 129 57.2 29.9 15.3 5.9170 628 146 64.6 33.7 17.6 6.7180 705 163 72.6 37.9 19.4 7.5190 785 177 80.7 42.2 21.5 8.4200 870 202 89.4 46.7 23.9 9.3220 244 108 56.5 28.9 11.3240 291 128 67.3 34.4 13.4260 341 151 79.0 40.3 15.7280 395 175 91.6 46.8 18.2300 454 201 105 53.7 20.9

Find the value from the table according to the pipe size and flow in scfm, then divide this value by the compression ratio. Compression ratio is calculated by adding atmospheric pressure (nominally 14.7 psia) to the gauge pressure and dividing the sum by atmospheric pressure.

Multiply the quotient by the actual length of pipe, in feet, then divide by 1,000. The result is the pressure loss, in psi.

FRICTION IN AIR HOSEHose ID,

in.

12

Flow, scfm

Pressure, psi

50 60 70 80 90 100 11020 1.8 1.3 1 0.9 0.8 0.7 0.630 5 4 3.4 2.8 2.4 2.3 240 10.1 8.4 7 6 5.4 4.8 4.350 18.1 14.8 12.4 10.8 9.5 8.4 7.660 23.4 20 17.4 14.8 13.3 1270 28.4 25.2 22 19.3 17.680 34.6 30.5 27.2 24.6

34 20 0.4 0.3 0.2 0.2 0.2 0.2 0.130 0.8 0.6 0.5 0.5 0.4 0.4 0.340 1.5 1.2 0.9 0.8 0.7 0.6 0.550 2.4 1.9 1.5 1.3 1.1 1 0.960 3.5 2.8 2.3 1.9 1.6 1.4 1.370 4.4 3.8 3.2 2.8 2.3 2 1.880 6.5 5.2 4.2 3.6 3.1 2.7 2.490 8.5 6.8 5.5 4.7 4 3.5 3.1

100 11.4 8.6 7 5.8 5 4.4 3.9110 14.2 11.2 8.8 7.2 6.2 5.4 4.9

1 30 0.2 0.2 0.1 0.1 0.1 0.1 0.140 0.3 0.3 0.2 0.2 0.2 0.2 0.250 0.5 0.4 0.4 0.3 0.2 0.2 0.260 0.8 0.6 0.5 0.5 0.4 0.4 0.370 1.1 0.8 0.7 0.7 0.6 0.5 0.480 1.5 1.2 1 0.8 0.7 0.6 0.590 2 1 1.3 1.1 0.9 0.8 0.7

100 2.6 2 1.6 1.4 1.2 1 0.9110 3.5 2.6 2 1.7 1.4 1.2 1.1

Pressure drop per 25 ft. of hose. Factors are proportionate for longer or shorter lengths.

PRESSURE LOSS THROUGH PIPE FITTINGSPipe size, NPT

Gate valve

Long radius elbow

Medium- radius elbow

Standard elbow

Angle valve

Close-return bend

Tee (through

side)

Globe valve

12 0.31 0.41 0.52 0.84 1.1 1.3 1.7 2.534 0.44 0.57 0.73 1.2 1.6 1.8 2.3 3.51 0.57 0.77 0.98 1.6 2.1 2.3 3.1 4.7

114 0.82 1.1 1.4 2.2 2.9 3.3 4.4 6.5112 0.98 1.3 1.6 2.6 3.5 3.9 5.2 7.82 1.3 1.7 2.2 3.6 4.8 5.3 7.1 10.6

212 1.6 2.2 2.8 4.4 5.9 6.6 8.7 13.13 2.1 3.0 3.6 5.7 7.7 8.5 11.4 17.14 3.0 3.9 5.0 7.9 10.7 11.8 15.8 23.75 3.9 5.1 6.5 10.4 13.9 15.5 20.7 31

This table gives values for air-pressure flow losses through screw fittings expressed in the equivalent lengths of straight pipe of the same diameter. For example, flow resistance of a 2-in. gate-valve flow resistance is the same as that for 1.3 ft. of straight pipe.

VALVE SIZING AND CV

Coefficient of velocity, CV, is widely used to compare flow rating of valves. The greater the coefficient, the higher the flow rating. Often it is useful to convert CV into scfm (standard cubic feet per minute) or scfm to CV. Although CV represents flow capacity at all pressures, scfm represents flow at a specific air pressure. The accompanying table relates CV to scfm for pressures commonly encountered with industrial pneumatic systems.

To obtain flow in scfm at a particular pressure, divide the CV value by the appropriate factor from the table. For example, to determine the output in scfm of a valve with a CV of 0.52 when operated at 80 psi, simply divide 0.52 (the valves CV) by 0.212 (the factor from the table corresponding to 80 psi):

To determine CV from scfm, simply multiply the scfm value by the factor corresponding to the appropriate pressure.

ONE SIZING METHOD

Different methods can be used for sizing air valves, and one popular method is explained here. You may consider oversizing the valve by 20% to 25% to compensate for the inevi-table losses in a pneumatic system.

A formula and table will yield the valve velocity coefficient (CV) for operating a cylinder within a specific cycle period:

Where:A = bore area, in.2 S = stroke, in.CP = pressure-drop constant (from table), psiFC = compression factor (from table)t = time, sec

COMPRESSION FACTOR AND P CONSTANTSInlet Pressure,

psiCompression

factor, FCConstant for pressure drop (CP)

P @ 2 psi P @ 5 psi P @ 10 psi

10 1.6 0.102

20 2.3 0.129 0.083 0.066

30 3.0 0.113 0.072 0.055

40 3.7 0.097 0.064 0.048

50 4.4 0.091 0.059 0.043

60 5.1 0.084 0.054 0.040

70 5.7 0.079 0.050 0.037

80 6.4 0.075 0.048 0.035

90 7.1 0.071 0.045 0.033

100 7.8 0.068 0.043 0.031

110 8.5 0.065 0.041 0.030

120 9.2 0.062 0.039 0.029

CV TO SCFM CONVERSIONAir pressure,

psig 40 50 60 70 80 90 100

Factor 0.0370 0.0312 0.0270 0.0238 0.0212 0.0192 0.0177

Use P at 5 psi to determine pressure-drop constant for most industrial applications.Use 2 psi for critical applications or 10 psi to conserve cost and mounting space.

CV =A X S X CPXFC

T X 29

Copyright 2015 by Penton Media Inc.

0.520.212

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