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Retiring thickness of piping system

Retiring thickness of piping system

Retiring thickness of piping system

The required thickness is described in ASME B31.3. This can be calculated from internal pressure of piping system. However, the result (especially in Class 150# piping system) is very far from the thickness of comercial pipe. Do we have other factors to be considered? Please comment about the method to specify the practical retiring thickness of piping system.

RE: Retiring thickness of piping system

Two other major factors additive to pressure design (but others exist):
-sustained loads and support span
-corrosion allowance
Most codes specify also minimum thicknesses to insure minimum general robustness of system.

Online tools for structural design

RE: Retiring thickness of piping system

Pan, hi again. Prex have the reason, when you calculate the required thickness you need to estimate a STRUCTURAL THICKNESS also, is refer to the minimun thickness neccesary resist the weight of the pipe distance between supports, the weight of valves and the weight of the fluid.
Genereally, in 150# systems, the pressure it's too low, my recomendation is began with 0.125" (normal corrosion allowance) plus the required thickness per B31.3 (involve Material, pressure and temperature) equation,  e.g .085" (is't looked like to the value that made you writte the question? )and get a round value 0.125 + .085" = 0.235.
Now you have a value similar to commercial pipe.

                T-Min Calculation  

       Design: B31.3                       PIPE
Material Code: A106B                                   Design Temp: 150.0 °F
MSpec: A106       GR: B       P No: 1    Cl/Type:        Notes:

    T-Min has been calculated using the variables listed below.
  t  = Pressure design wall thickness, in
  P  = Internal design gauge pressure (150 lb/in²)
  S  = Applicable allowable stress (20000 lb/in²)
  E  = Quality Factor, long weld joint factor (1.00)
  y  = Temperature Factor (0.40)
  D  = Outside Diameter (6.625 in)

          P * D                    150 * 6.625
  t = ------------- =  ----------------------------= 0.025 in
     2 * (SE + Py)     2 * [(20000 * 1.00) + (150 *0.40)]

As you can see 0.025 it's too low, but the corrosion allowance, the weight of the pipe etc. where is?

Are here: (this is an example)

Structural T-Min has been calculated using the variables listed below.

 Ro = Outer Radius of Pipe (3.313 in)
 Ri = Inner Radius of Pipe (3.000 in)
 SG = Specific Gravity of Fluid in Pipe (1.000)
  D = Pipe Support Spacing (25.00 ft = 300.0 in)
  L = Max Center Load on Pipe (250.00 Lb)
  S = Material Stress of Pipe (20000.00 lb/in²)
  A = Additional Corrosion Allowance for Pipe (0.125 in)

 WP = Pipe Weight / in = ((Ro*Ro)-(Ri*Ri))*Pi*0.2833 = (1.7557 lb/in)
 WF = Fluid Weight / in = Pi * Ri * Ri * SG * 0.03613 = (1.0215 lb/in)
 WL = Load Bending Moment = L * D / 4 = (18750.00 in-lb)
  X = Max Bending Moment = WL + (WP + WF)*D*D/8 = (49993 in-lb)

                          X * Ro * 4
  t = A + Ro - ((Ro**4) - ----------- )**(0.25) = 0.200 in
                            S * Pi

and now .200" looks comercial? I hope this help you.

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