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Modeling of rotary compensators

2017-11-14View Original

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I. Principle of compensators: To simulate any device or fitting in CII, it is necessary to have a thorough understanding of its working principle. As for rotary compensators, their appearance and internal structure are as follows: The thrust-free rotary compensator is a new type of compensator used for compensating for thermal expansion in thermal pipelines. The structure of the rotary compensator is shown in Figure (1). Its components mainly include an integral sealing seat, a sealing gland, large and small ends, friction-reducing centering bearings, sealing materials, and a rotating cylinder. When installed on thermal pipelines, two or more such compensators are used together to enable relative rotation that absorbs the thermal displacement of the pipelines, thereby reducing stress in those pipelines. The principle of its operation is shown in Figure (2). It can be seen from this that the rotary compensator has two characteristics: 1. It can only rotate around the axial direction of the pipeline ; 2. The rotation of the compensator mainly requires overcoming the frictional torque between the sealing surfaces ; 3. The two sealing surfaces are separate. Based on the first characteristic, during modeling, displacements in all directions except axial rotation should be constrained ; According to the second characteristic, what needs to be overcome primarily is the frictional load. The nature of friction is such that once static friction reaches its extreme value, it transforms into kinetic friction, and the value of kinetic friction remains relatively constant. The rotational friction torque Mk of a pair of rotating drums: Nominal diameter DN (mm): 100, 125, 150, 200, 250, 300, 350, 400; Ncm for Π-type configuration: 13735, 22103, 12303, 31258, 42329, 5605, 6171, 4056, 23373, 8030, 2623, 02623. Nominal diameter DN (mm): 450, 500, 600, 700, 800; Ncm for Π-type configuration: 38903, 00488, 30808, 22770, 41382, 25862, 32219, 08. H dimension: Nominal diameter DN (mm): 80, 100, 125, 150, 200, 250, 300, 350, 400, 450, 500, 600, 700, 800. Length of rotating drum (mm): 250, 250, 280, 300, 300, 300, 350, 350, 350, 380, 400, 400, 400, 400. H (mm) for Π-type configuration: 490, 550, 655, 730, 900, 1050, 1250, 1400, 1550, 1730, 1880, 2200, 2500, 2820. II. Key points of the model: (1) The compensator is simulated using rigid components with mass; it can be divided into upper and lower parts, each having half the length and mass. Due to the presence of sealing surfaces, these two rigid components are connected via CNODEs. Enter the following: (2) Limit constraints – since the two rigid components function as a single compensator unit, they certainly cannot move relative to each other; therefore, three limit constraints for X, Y, and Z are added. Enter the following: (3) The rotational compensator can only rotate around the axial direction of the pipe; it cannot rotate in the other two directions, so two rotational constraints are added. Enter the following: (4) The most important feature is that the RY2 constraint type available in CAESAR II can be used; RY2 is a bilinear constraint, and its stiffness curve along with the corresponding input fields are shown in the figure below. Given the structural characteristics of the rotary compensator, it can be seen that the stiffness of K1 should be infinite; once the external load reaches its critical value, the compensator will rotate axially. At this point, the load that needs to be overcome remains unchanged, while the stiffness value becomes K2=0. When K2 is set to 0 or left blank, the software assumes it to be rigid; a value of 1 indicates a very low stiffness. Therefore, for the rotation compensator, the input is as follows:
Reply #22017-11-24
The table tool isn’t good; it’s not very clear to read
Reply #32017-11-29
Could the original poster share other methods for installing expansion joints, such as those with large tie rods, corrective types, and universal types? There’s another question I’d like to ask: how is the stiffness of the expansion joint calculated or determined? During the training, the instructor said that we decide on the stiffness ourselves and then provide that value to the manufacturer; it’s sufficient as long as the manufacturer’s chosen stiffness value is lower than what we specified. I’m not quite sure about this
Reply #42017-12-08
I’ve learned that the more experienced, capable engineers who are willing to share their knowledge are truly deserving of respect. Thank you!

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