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Search Results "MWBB100TN-UT-1000-XC68"

Mounting: B (Basic), Model: 100 (30mm), 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.

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.

or less 1000 or less XC22 Fluororubber seal 140 1600 or less 1000 or less 1000 or less XC26 Clevis pins with flat washer 160 Double clevis pin and double knuckle pin made of stainless steel 1600 or less 1200 or less 1200 or less XC27 180 2000 or less 1200 or less Rod side trunnion XC30 200 2000 or less 1200 or less With coil scraper XC35 250 2400 or less 1200 or less Hard chrome plated

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.

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 SMC Customer Service for larger sizes.