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IIHinge ripple compensatorApplication example: |
Installation temperature of a certain heat pipe: 20 ℃, diameter 500, working pressure 0.6MPa (6kgf/cm2), low temperature -10 ℃, expansion coefficient of carbon steel pipeline a=13.2 × 10 ˉ 6/℃. |
Design and calculate the installation form of the pipeline as follows: |
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1. Determine displacement:
△X=al△T=13.2×10ˉ6(80+80)×103×(120-(-10°))=275mm
2. Choose compensator: Compensator I selection: 0.6JY500 × 6J& theta; 0=± 8° (Rated displacement 2 × 8°) LⅠ=1.1m, Kθ Ⅰ=197N· m Compensator II and III selection: 0.6JY500 × 4& theta; 0=± 5° LⅡ=LⅢ=0.9m, Kθ Ⅱ=Kθ Ⅲ=295N· M/degree |
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LA (greater than or equal to) ≥ 997.69mm&hellip& hellip; (1) Generally, Lp ≥ 1.5DN × 4+L Ⅰ+L Ⅱ+L Ⅲ&hellip& hellip; (2) (Lp is the installation length of the compensation section) According to equation (1), take the margin and circle the LA value upwards: LA=1200(If the installation span requirement allows, it is better to have LA slightly larger.) Actual working angular displacement
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3Hinge ripple compensatorConsider the issue of cold tightness:
In order to ensure the overall stress condition of the pipeline, the compensator is usually cold tightened during design and installation. The cold tightening capacity is calculated according to the following formula: |
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So perform cold tightening installation in the opposite direction of the work displacement (cold tightening amountδ=74.04mm)
4. Calculate the load-bearing capacity of the support:
Next, analyze the stress situation of fixed supports G1 and G2 and guide brackets D1 and D2. a、 Expansion joint deformation torque. |
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b. G1 point: D1 point: Fy=Fz=0 Fx=Fy=Fz=0, Mx=My=0 Fx=- V=-1899N Mz=MII=973.5N· m Mx=My=Mz=0 G2: Fx=V=1899N D2: Fx=Fy=Fz=0,Mx=My=0 Fy=Fz=0, Mx=My=Mz=0, Mz=- MII=-973.5N· m Force diagram: |
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Universal corrugated compensator, provides free design and installation solutions Looking forward to your call: 0317-8294124 |
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