SMC Corporation of America
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Lock unit model MWB32-UT MWB40-UT MWB50-UT MWB63-UT MWB80-UT MWB100-UT Applicable rod size [mm]*1 12 f8 16 f8 20 f8 20 f8 25 f8 30 f8 Bore size of combinable cylinder [mm] 32 40 50 63 80 100 Lock holding force*2 (Max. static load) [N] 630 980 1,570 2,450 3,920 6,080 Made to order common specifications With coil scraper (-XC35), Made of stainless steel (-XC68) *1 The applicable rod size affects

Mounting: B (Basic), Model: 63 (20mm), Thread Type: TN (NPT), Rod Length: None, Made to Order: XC35 (w/Coil Scraper)

Lead wire length: 1000 mm ZM-K1LO . 1 pc. VJ10-36-1A-10 . 2 pcs.

Assuming that the tube I.D. is 2 mm, the piping capacity is as follows: V = /4 x D2 x L x 1/1000 = /4 x 22 x 1 x 1/1000 = 0.0031 L (2) Assuming that leakage (QL) during adsorption is 0, find the average suction flow rate to meet the adsorption response time using the formula on the front matter 17.

Host SOH UT ENQ 32H Check sum CR (H) (L) Chemical Thermo-con SOH UT STX 32H Internal sensor temp.

Assuming that the tube I.D. is 2 mm, the piping capacity is as follows: V = /4 x D2 x L x 1/1000 = /4 x 22 x 1 x 1/1000 = 0.0031 L (2) Assuming that leakage (QL) during adsorption is 0, find the average suction flow rate to meet the adsorption response time using the formula on the front matter 17.

Assuming that the tube I.D. is 2 mm, the piping capacity is as follows: V = /4 x D2 x L x 1/1000 = /4 x 22 x 1 x 1/1000 = 0.0031 L (2) Assuming that leakage (QL) during adsorption is 0, find the average suction flow rate to meet the adsorption response time using the formula on the front matter 17.

Assuming that the tube I.D. is 2 mm, the piping capacity is as follows: V = P/4 x D2 x L x 1/1000 = P/4 x 22 x 1 x 1/1000 = 0.0031 L (2) Assuming that leakage (QL) during adsorption is 0, find the average suction flow rate to meet the adsorption response time using the formula on the front matter 8.

COM. 13 (+) 13 () Voltage limit V, 1 minute, AC 1000 Positive common Negative common Insulation resistance M/km, 20C 5 or more Note) When using a valve with no polarity, either positive common or negative common can be used. Note) The minimum bending radius of the D-sub connector cable is 20 mm.

Up to 600 Up to 1000 Up to 600 Up to 1000 Up to 750 Up to 1000 Up to 750 Note) End boss is machined on the flange for E.

Assuming that the tube I.D. is 2 mm, the piping capacity is as follows: V = /4 x D2 x L x 1/1000 = /4 x 22 x 1 x 1/1000 = 0.0031 L (2) Assuming that leakage (QL) during adsorption is 0, find the average suction flow rate to meet the adsorption response time using the formula on the front matter 17.

Assuming that the tube I.D. is 2 mm, the piping capacity is as follows: V = /4 x D2 x L x 1/1000 = /4 x 22 x 1 x 1/1000 = 0.0031 L (2) Assuming that leakage (QL) during adsorption is 0, find the average suction flow rate to meet the adsorption response time using the formula on the front matter 17.

Assuming that the tube I.D. is 2 mm, the piping capacity is as follows: V = /4 x D2 x L x 1/1000 = /4 x 22 x 1 x 1/1000 = 0.0031 L (2) Assuming that leakage (QL) during adsorption is 0, find the average suction flow rate to meet the adsorption response time using the formula on the front matter 17.

Assuming that the tube I.D. is 2 mm, the piping capacity is as follows: V = /4 x D2 x L x 1/1000 = /4 x 22 x 1 x 1/1000 = 0.0031 L (2) Assuming that leakage (QL) during adsorption is 0, find the average suction flow rate to meet the adsorption response time using the formula on the front matter 17.

R O D B O R E S I Z E 7/16-20 7/16-20 7/16-20 3/4-16 3/4-16 3/4-16 1-14 1-14 1 1/4-12 1 1/2-12 1 7/8-12 5/8 5/8 5/8 1 1 1 1-3/8 1-3/8 1-3/4 2 2-1/2 S TA N D A R D S T R O K E 150 1.5 200 2 250 2.5 325 3.25 400 4 500 5 600 6 800 8 1000 10 1200 12 1400 14 E 1 G 1 3/8 H 1 3/4 J 2 K 2 1/2 L 3 M 3 1/2 Z Please consult P/A Customer Service for larger sizes.