The MWB-UT is a stand alone lock unit compatible with tie-rod cylinders and is ideal for applications requiring intermediate stops, emergency stops, and drop prevention. The MWB series is an improved design over the MNB series with its improved ease of maintenance with separate lock and cylinder construction. The exhaust locking design allows for the lock release to be manually operated with
Operations Manual: Lock unit Lock unit specification Lock unit model MWB32-UT MWB40-UT MWB50-UT MWB63-UT MWB80-UT MWB100-UT Applicable rod size mm *2 12 f8 16 f8 20 f8 20 f8 25 f8 30 f8 Locking operation type Exhaust locking Fluid Air Proof pressure 1.5MPa Max. operating pressure 1.0MPa Min. operating pressure 0.3MPa Ambient and fluid temperature -1070 (No freezing) Operating fluid temperature Lubrication Not required
Mounting: B (Basic), Model: 100 (30mm), Thread Type: TF (G)
Mounting: Basic, Model: 30mm, Thread Type: G, Rod Length: -, Made to Order: -
Unit model MWB32-UT MWB40-UT MWB50-UT MWB63-UT MWB80-UT MWB100-UT Applicable rod size 12 f8 16 f8 20 f8 20 f8 25 f8 30 f8 MWB-UT CP96N C96N Lock Unit Accessories Auto Switch Made to Order Double Acting, Single Rod Double Acting, Double Rod Double Acting, Single Rod Double Acting, Double Rod Material Carbon steel/Stainless steel Surface treatment Hard chrome plating: 10 m Chamfer the rod
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
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.
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.