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Ripple compensator
- Categories:Industry Knowledge
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- Time of issue:2016-02-17 15:13
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(Summary description)Applications Acarbonsteelpipeline,nominaldiameter500mm,workingpressure0.6MPa,mediumtemperature 300°C,thelowestambienttemperatureis-10°C,andthecompensatorinstallationtemperatureis20°C.Accordingtothepip
Ripple compensator
(Summary description)Applications Acarbonsteelpipeline,nominaldiameter500mm,workingpressure0.6MPa,mediumtemperature 300°C,thelowestambienttemperatureis-10°C,andthecompensatorinstallationtemperatureis20°C.Accordingtothepip
- Categories:Industry Knowledge
- Author:
- Origin:
- Time of issue:2016-02-17 15:13
- Views:
Information
Applications
A carbon steel pipeline, nominal diameter 500mm, working pressure 0.6MPa, medium temperature
300°C, the lowest ambient temperature is -10°C, and the compensator installation temperature is 20°C. According to the pipeline layout (as shown in the figure), an internal pressure type corrugated compensator should be installed to compensate the axial displacement X=32mm, lateral displacement Y=2.8mm, angular displacement θ=1.8 degrees, known L=4m, the number of fatigue failures of the compensator is considered 15,000 times, try to calculate the bearing A force.
Solution: (1) According to the axial displacement of the pipe X=32mm.
Y=2.8mm.
θ=1.8 degrees.
The axial displacement X0=84mm of 0.6TNY500×6F is found from the sample,
Lateral displacement: Y0=14.4mm. Angular displacement: θ0=±8 degrees.
Axial stiffness: Kx=282N/mm. Lateral stiffness: Ky=1528N/mm.
Angular stiffness: Kθ=197N·m/degree. Use the following relationship to determine whether this compensator meets the requirements of the title:
Substitute the above parameters into the above formula:
(2) The amount of pre-deformation △X for the compensator is:
Because △X is positive, it must be "pre-stretched" by 13mm before leaving the factory.
(3) Calculation of bearing A force:
Internal pressure thrust: F=100·P·A=100×0.6×2445=14600(N)
Axial elasticity: Fx=Kx·(f·X)=282×(1/2×32)=4512(N)
Lateral elasticity: Fy=Ky·Y=1528×2.8=4278.4(N)
Bending moment: My=Fy·L=4278.4×4=17113.6(N·m)
Mθ=Kθ·θ =197×1.8=354.6(N·m)
Resultant bending moment: M=My+Mθ=17113.6+354.6=17468.2(N·m)
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