ASME B31.3 PRESSURE DESIGN OF EXTRUDED OUTLET HEADER

 PRESSURE DESIGN OF EXTRUDED OUTLET HEADER

 

In this blog, we will learn to do pressure design of extruded outlet header as per ASME B31.3

First thing to know how we calculate pipe thickness under internal pressure.

Below formula is applicable when pipe thickness is less than outside diameter of pipe / 6 (t<D/6). For instance, pipe outside diameter is 323.8 and pipe thickness is 9.53 then the D/6 value will be 33.93 which is more than the ‘t’ value (9.53). so the condition is met (t<D/6).

t = calculated thickness of pipe under the internal pressure

P = inter design gage pressure

D = outside diameter of pipe

S = allowable stress value which we get from Table A-1 or Table A-1M, Table A-1 is used when the design temperature is given in ‘F’ and design pressure is given in ksi meanwhile Table A-1M is used for SI Unit system, when the design temperature is given in ‘C’ and design pressure is given in MPa.

 

 

E = weld joint quality factor

Weld joint quality factor for seamless pipe will be 1. it depends on weld joint type and NDT we do after the weld. In this code Table A-1A and Table A-1B fully deal with weld joint quality factor. In above picture the values are given with respect to pipe weld joint configuration and NDT examination.

W = weld joint strength reduction factor.

It deals with welded pipe strength reduction due to elevated temperature.

 

In above picture, it is shown the weld strength reduction factor for carbon steels will remain 1 till 593 degree Celsius temperature.

Y = coefficient value for ferrite and austenitic material when the pipe thickness meet the criteria t<D/6. Values in below table mentioned for ferrite and austenitic materials.

 

 

And if the pipe thickness does not meet the criteria t<D/6, in other words if the pipe wall thickness is meet the the criteria t>D/6 or t=D/6, then Y coefficient value shall be calculated with the formula as given below.

d = inside pipe diameter

c = corrosion and erosion allowance.

Corrosion allowance depends on service life of piping. Usually in this code 1.5mm to 3mm is considered. Normally per year corrosion is considered 0.1mm to 0.3mm and ideally 0.15mm and then calculate corrosion allowance as mentioned below.

Total corrosion allowance = corrosion per year x service life

                                          = 0.15 x 20 = 3mm

Therefore, it is calculated the 3mm corrosion allowance is considered for 20 years of piping service life.

 

 

EXTRUDED OUTLET CALCULATION

We use extruded type hot tap fitting for new tie-in connection, benefit for using the extruded fitting is to avoid branch to run weld (groove weld).

WHAT IS REINFORCEMENT

Reinforcement means to use the techniques and procedures to strengthen pipes to withstand with internal pressure, mechanical stresses to enhance durability and lifespan of piping.

Extruded tee is a kind of special tee, which we form by extrusion and header project outlet

Below diagram, illustrate nomenclature of the extruded tee reinforcement.

b = Stand for branch

T_b = Stands for branch thickness

T_h = Stands for header thickness including corrosion allowance but not mill tolerance

D_h = Outside diameter of header

d_x = inside diameter of extruded outlet

d₂ = half width of reinforcement zone, which is equal to d_x

h_x = height of extruded outlet, that must be equal to or greater than r_x (external radius of extruded outlet)

r_x = external radius of extruded outlet

L₅ = height of reinforcement zone

    = 0.7 x SQRT(D_b x T_x)

 

 

 

Tx = Corroded finish thickness of extruded outlet, measured at a height equal to r_x above the outside surface of the header.

 

Limitation on Radius r_x:

1.    Minimum r_x – the lesser of 0.05D_b or 38mm (1.5”)

2.    Maximum r_x shall not exceed as mentioned below

a.    For Db < DN200 (NPS8), 32mm (1.25”)

b.   For Db ≥ DN200 (NPS8), O.1Db + 13mm (0.5”)

Note: Above-mentioned conditions shall not be met through machining the extruded out.

Now, we will identify the area that is required to be replaced with the reinforcement is A_1, which is need to be calculated. It is thickness of header pipe multiply inner diameter of branch.

Its general understanding we have to have a reinforcement more than or equal to the area that is cut from the header pipe.

Therefore, the area to be required for the reinforcement will be

A₁ = Kt_hd_x

 

K is obtained as per below criteria

 

1.   1. For D_b/D_h > 0.6, K = 1

2.    2. For 0.6 ≥ D_b/D_h > 0.15, K = 0.6 + 2/3(D_b/D_h)

3.    3. For D_b/D_h ≤ 0.15, K = 0.7

 

 

t_h is the value we calculate same as we do calculate pipe thickness. Corrosion allowance and mill tolerance will not be added.

Let’s assume the header nominal thickness is 13.49mm

T̅_h = Header nominal thickness, which is 13.49mm, it is 12.5% of T_h

Mill tolerance = 12.5%, that value is 1.68mm

Therefore,

T_h = 13.49mm – 1.68mm = 11.8mm

t_h  value let’s assume we calculated 2.5mm

Corrosion allowance is 0 mm

So the portion of header thickness between (T_h – c) - t_h is known as excess thickness which is later we add in available reinforcement area.

So, the excess thickness in header will be (11.8 – 0) – 2.5 = 9.3mm

In the same way we calculate the excess thickness available in branch, for reference screenshot is given below 

 

Let’s assume the branch pipe nominal thickness is 9.53mm

T̅_b = branch pipe nominal thickness, which is 9.53mm, it is 12.5% of T_b

Mill tolerance = 12.5%, that value is 1.19mm

Therefore,

T_h = 9.53mm – 1.19mm = 8.34mm

t_h  value let’s assume we calculated 2.1mm

Corrosion allowance is 0 mm

So the portion of header thickness between (T_h – c) - t_h is known as excess thickness which is later we add in available reinforcement area.

So, the excess thickness in header will be (8.34 – 0) – 2.1 = 6.24mm

 

We will look into the available areas in reinforcement, and sum of all available areas shall be greater or equal to required reinforcement area A₁

Available area = A₂, A₃ & A₄ are shown in reinforcement diagram

A₂ + A₃ + A₄ ≥ A₁

Now we will go into detail of available areas for reinforcement.

A₂ = this area will be the excess area which is available in header. We will see how we formulate the excess area available in header.

A₂ = (2d₂ – d_x)(T_h – t_h – c)

A3 = this area will be the excess area which is available in branch. We will see how we formulate the excess area available in branch.

A₃ = 2L₅(T_b – t_b – c)

A₄ = this area will be available as excess thickness in the extruded outlet lip

A₄ = 2r_x[T_x – (t_b – c)]

Hence all available areas have been drawn in detail, so we will make sure the below condition is satisfied and met the criteria, if not we will increase the thicknesses to meet the condition.

A₂ + A₃ + A₄ ≥ A₁

 

For More About Branch reinforcement