Large pull rod transverse corrugated compensator(DLB) Purpose:
The large pull rod transverse corrugated compensator can compensate for the lateral and angular displacement of curved pipelines

Model:
Our factory produces DN100-DN5000 with pressure levels ranging from 0.1Mpa to 2.5Mpa

Connection method: 1. Flange connection 2. Pipe connection

Product horizontal compensation amount: according to user requirements

1Large pull rod transverse corrugated compensatorModel example:

Example: 0.6DLB500-FB-1500
Indicating: The nominal diameter is 500mm, the working pressure is 0.6MPa, and the length is 1500mm. It is a large pull rod transverse corrugated compensator connected by stainless steel flange.

IILarge pull rod transverse corrugated compensatorinstructions:
The large pull rod transverse corrugated compensator can compensate for the lateral and angular displacement of bent pipe sections.

3、 Calculation of the force exerted on the support by the large pull rod transverse corrugated compensator:
Lateral elasticity: Fy=Ky· Y bending moment: M=Fy· L
In the formula: Ky lateral stiffness N/mm Y actual lateral displacement (deformation) mm
The distance m from the L-pipe rack to the midpoint of the compensator.

4Large pull rod transverse corrugated compensatorApplication example (without considering the correction of compensation amount and stiffness by temperature)
A certain carbon steel pipeline with a nominal diameter of 500mm, a working pressure of 0.6MPa, a medium temperature of 350 ℃, an ambient low temperature of -10 ℃, and a compensator installation temperature of 20 ℃. According to the pipeline layout (as shown in the figure), a large pull rod transverse compensator needs to be installed to compensate for the 160mm lateral displacement of the AB pipe section. The number of fatigue failures of the compensator is considered to be 15000 times, and the force on the support AC is calculated.

Solution: (1) According to the lateral displacement of AB pipe section
Y=160mm， 0.6DLB500-FB-2000 can be selected for sample inspection, with a lateral compensation of Y0=193mm and a lateral stiffness of Ky=22N/mm
(2) Calculation of stress on supports A and C
Consider when calculating; Cold Tight” And the cold tightening amount can be taken as half of the actual compensation amount, which is 1/2Y.
FA=Ky· 160/2=22 × 80=1760 (N) MA=FA × 2=1760 × 2=3520 (N· m)
FB=Ky· 160/2=22 × 80=1760 (N) MB=FB × 2=1760 × 2=3520 (N· m)
Namely: A and C supports respectively bear a thrust of 1760 Newtons and 3520 Newtons; The force distance of meters.
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