DENMARK SMC Pneumatik A/S FINLAND SMC Pneumatics Finland OY FRANCE SMC Pneumatique SA GERMANY SMC Pneumatik GmbH HUNGARY SMC Hungary Ipari Automatizlsi Kft. IRELAND SMC Pneumatics (Ireland) Ltd. ITALY SMC Italia S.p.A. LATVIA SMC Pneumatics Latvia SIA NETHERLANDS SMC Pnuematics BV. NORWAY SMC Pneumatics Norway A/S POLAND SMC Industrial Automation Polska Sp.z.o.o.
Cylinder Bore Size MHS 2 D 50 32 to 63 Y59A Number of auto switches Nil S 2 pcs. 1 pc.
K) Insert the guide and element in order, and mount the O-ring for the bowl. If the support ring was removed, mount the support ring before the O-ring for bowl. Note: Ensure that the element inserting direction is correct. K) O-ring for the bowl Support ring Point Make sure that the O-ring for the element is facing upward. Guide -19- Doc. no.
.-230 Ryan Way, South San Francisco, CA 94080-6370-Main Office: (650) 588-9200-Outside Local Area: (800) 258-9200-www.stevenengineering.com VX3 L L VXA K VN K Conduit terminal: T LVC LVA F (Q) 34 LVH LVD LVQ 25 U LQ R C LVN G1/2 TI/ TIL E B A H D 2-P Port size (S) PA 2-M Thread depth N PAX L PB K (mm) Electrical entry P Port size B C A D E F H K L M N Grommet:G Conduit:C Conduit terminal:
Brown Brown Load Load Switch 1 Switch 1 Blue Blue Brown Brown Switch 2 Switch 2 Blue Blue Load voltage at ON = Power supply voltage Residual voltage x 2 pcs. = 24 V 4 V x 2 pcs. = 16 V Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V Example: Power supply is 24 VDC Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.
Brown Brown Load Load Switch 1 Switch 1 Blue Blue Brown Brown Switch 2 Switch 2 Blue Blue Load voltage at ON = Power supply voltage Residual voltage x 2 pcs. = 24 V 4 V x 2 pcs. = 16 V Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V Example: Power supply is 24 VDC Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.
Brown Brown Load Load Switch 1 Switch 1 Blue Blue Brown Brown Switch 2 Switch 2 Blue Blue Load voltage at ON = Power supply voltage Residual voltage x 2 pcs. = 24 V 4 V x 2 pcs. = 16 V Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V Example: Power supply is 24 VDC Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.
Brown Brown Load Load Switch 1 Switch 1 Blue Blue Brown Brown Switch 2 Switch 2 Blue Blue Load voltage at ON = Power supply voltage Residual voltage x 2 pcs. = 24 V 4 V x 2 pcs. = 16 V Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V Example: Power supply is 24 VDC Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.
:l 3/86 female thread Maintenance l--I space ~ Drain H I I Dimensions with mounting bracket Model Port size A B C D E F G I J 1 K 1 L 1 MI N I 01 P I 01 RI s B AMH850 11 /2 , 2 460.5 42 348 220 57.5 220 10 463.51 180 I 30 I 15 I 24 I 13 1120 1184 1 220 1110 1 18 I 6 6 SeriesAMH Made to Order Specifications ICD With Differential Pressure Gauge (GD40-2-01) I@ With IN-OUT Flange For ease of
[%F.S.] 2.5 2.0 1.5 1.0 0.5 0.0 0 1 2 3 4 5 Response time[s] [s] Operating pressure range Pressure [MPa] Operating range Fluid temperature [oC] When fluids with high temperature are used, the operating pressure range will be reduced. Operate within the range mentioned above.
screw M5 hexagon socket head cap screw M6 hexagon socket head cap screw M8 hexagon socket head cap screw Model AMP220 AMP320 AMP420 E OUT Maintenance space (mm) C B Bracket related dimensions Element service indicator related dimensions Port size (nominal size B) Model A D E I 66 80 90 J K L M N O P Q R S F 80 120 180 G 123 169 237 H AMP220 AMP320 AMP420 76 90 106 76 90 106 2 2.3 3.2 6 7
Dimensions/40 to 100 Basic/CA2YB 4-J 2-P (Rc, NPT, G) MM Width across flats KA G G D E H1 AL B1 K A N F N B C M S + Stroke H ZZ + Stroke (mm) Bore size (mm) 40 50 63 80 100 Stroke range Up to 500 Up to 600 Up to 600 Up to 700 Up to 700 B C A AL B1 D E F G H H1 J K KA M MM N P S ZZ 30 35 35 40 40 27 32 32 37 37 60 70 85 102 116 22 27 27 32 41 44 52 64 78 92 16 20 20 25 30 32 40 40 52 52 10
Dimensions/20 to 100 Basic/CG1YB GA GB 2-P (Port size) (Rc, NPT, G) 2-TC Section TA/TB Width across flats B1 H1 MM 0 -0.05 I 0 -0.05 TD E D E AL A TA TB 4-TC K F F H TE TF TG NA S + Stroke ZZ + Stroke KA C 0.1 8-J NA Section TA/TB (mm) Bore size (mm) 20 25 32 40 50 63 TC TDH9 TE TF TG M5 x 0.8 4 0.5 5.5 +0.08 0 8 +0.08 0 M6 x 0.75 5 1 6.5 10 +0.08 0 M8 x 1.0 5.5 1 7.5 12 +0.08 0 6 1.25 8.5
Dimensions/20 to 40 Basic/CM2YB Width across flats B2 2-P (Port size) (Rc, NPT, G) 2-NN H1 H2 G G Width across flats B1 I 2-Eh8 D MM AL A F F NA 1.5 1.5 N N K H S + Stroke ZZ + Stroke Boss-cut ZZ + Stroke (mm) A AL B1 B2 D E F G H H1 H2 I K MM N NA NN P S ZZ Bore size (mm) 20 25 32 40 0 0.033 0 0.033 0 0.033 0 0.039 18 15.5 13 26 8 13 8 41 5 8 28 5 M8 x 1.25 15 24 M20 x 1.5 1/8 62 116 20
h Nil L R S Operating direction Left Right a b h g d c f e Note) Lead wire length symbols 0.5m .
Mounting tap: M10 x 1.5 Pin diameter: 10H7 Nil S No shim With 2 mm shims C Port b side a side ( side) b side ( side) a side Port : Mounting tap : Pin diameter : Mounting tap : Pin diameter D Clamp arm b side S 66 b side Mounting tap E The positions of the mounting taps and pin holes are different depending on the body shape. Refer to Body shapes and positions of mounting sides.
These vortexes are stable under certain conditions, and their frequency is proportional to the flow velocity, resulting in the following formula. f = k x v f: Frequency of vortexes, v: Flow velocity, k: Proportional constant (determined by the vortex generator's dimensions, shape, etc.) Therefore, the flow rate can be measured by detecting this frequency.
Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V = 24 V 4 V x 2 pcs. = 16 V Example: Power supply is 24 VDC. Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.
If it does not locked easily, turn the handle s l i g h t l y c l o c k w i s e o r counterclockwise; then, push it until the orange colored line is no longer visible.
How to Find the Flow Rate (At air temperature of 20C) Choke flow: (P2 + 0.1)/(P1 + 0.1) ) 0.5 293 273 + t Q = 120 x S x (P1 + 0.1) x Subsonic flow: when (P2 + 0.1)/(P1 + 0.1) > 0.5 Q = 240 x S x (P1 P2)(P2 + 0.1) x 293 273 + t Q: Air flow rate [ /min (ANR)] S: Effective area (mm2) P1: Upstream pressure [MPa] P2: Downstream pressure [MPa] t: Air temperature [C] Note 1) Formulas above are