Uses: Axial double bellows (FS) for axial compensation, the compensation capacity, more economical is its own characteristics Model: The company DN32-DN1000, pressure level: 0.1Mpa-2.5Mpa Connection: 1, flange connection 2, take over the connection ?? goods axial compensation: 30mm-450mm First, the model example: two, for example: 0.6FS100 × 20F said: working pressure of 0.6MPa, diameter DN = 100mm, wave number 20, flanged The double bellows. Second, use: bellows compensator, to meet pipeline axial compensation, the compensation capacity, more economical is its own characteristics. Flange press machine standard JB81-59 supply, but also according to the national standard, standard or other standards-based supplier according to user requirements. Third, the force of the fixed bearing: Pressure thrust: Fp = 100 · P · A (N) Axial stretch: Fx = Kx · X (N) Where: P: maximum working pressure or the maximum test pressure MPa A: effective area (check samples) cm2 Kx: compensator axial stiffness N / mm X:? compensator using axial compensation mm Fourth, the application, for example: a steel pipe, nominal diameter DN = 500mm, working pressure P = 0.4MPa, medium maximum temperature Tmax = 250 ℃, the minimum temperature Tmin = 0 ℃, the number of fatigue N = 1500 times. Compensation required distance between the fixed bearing pipe L = 85m, take over the connection, test selection and calculation of bearing loads. 1, the selection, the amount of compensation required pipe: X = a · L · △ T = 0.0133 × 85 (250-0) = 283mm where: a is the linear expansion coefficient, taking 0.0133mm / m · ℃ (check sample choose 0.6FS500 × J) 2, bearing pressure:? check sample 0.6FS500 × 12J compensator A = 2445cm2 Kx = 141N / mm pressure thrust: Fp = 100 · P · A = 100 × 0.6 × 2445 = 146700N axial stretch : Fx = Kx · X = 141 × 283 = 39903N bearing force: F = Fp + Fx = 146700 + 39903 = 186603N?