??? Unlocking the Secrets of ASME SEC VIII Div 1, UG-37 & UG-40: Reinforcement Requirements & Limits ??

UG-37: Reinforcement Required for Openings in Shell and Formed Heads

Pressure vessels require openings for various purposes such as inlet, outlet, instrumentation, etc. Adding an opening weakens the structure, which needs to be addressed. The weakening of the vessel is due to the removal of material from the shell. The removal of material causes a reduction in the cross-section area and volume of the shell. This removed material is to be added back to the vessel and it is generally added back in form of a Reinforcement Pad.

The area can be compensated by adding material at a different location but there is no point in adding material far from the opening as it will not provide any support to the lost material due to the opening. In the figure below we can see that the material added next to the opening is the correct way to compensate for the material lost due to openings.

Basic Methodology Behind Compensating Material lost due to Opening

We will consider the volume lost in the area by looking at the cross-section area of the opening. Now we can see in the figure below that the opening area is divided into two equal parts and placed right next to the opening. So, the area lost in the opening is now added back to the shell and if we revolve the area we added then we will get the RF Pad that is required to compensate for the opening.

Set-in & Set-on Nozzle

In pressure vessel design, set-in and set-on nozzles refer to different ways of connecting pipes to the vessel.

A set-in nozzle is a type of nozzle that is welded to the inside of the pressure vessel wall, with the pipe extending into the vessel. This type of nozzle is also sometimes referred to as a "welded-in nozzle."

A set-on nozzle, on the other hand, is a type of nozzle that is welded to the outside of the pressure vessel wall, with the pipe extending away from the vessel. This type of nozzle is also sometimes referred to as a "welded-on nozzle."

Both set-in and set-on nozzles can be reinforced with pads or other forms of reinforcement to ensure the structural integrity of the vessel. So based on the configuration of the opening the reinforcement requirement changes. We will look into different requirements of reinforcement later. The choice between set-in and set-on nozzles will depend on various factors, such as the intended use of the vessel, the operating conditions, and the required flow rates.

Required Area for Compensation

So now, the required thickness is calculated using the ASME formula for circumferential stress, and this gives the minimum thickness required to take care of the pressure without any corrosion allowance. Once the corrosion allowance is added, the resulting thickness is known as the design thickness, which is then compared to the available thickness in the market, known as nominal thickness.

Required thickness (tr) = 14.5 mm

Corrosion Allowance (c) = 3 mm

Design Thickness = 17.5 mm

Nominal Thickness (t) = 18 mm

Now, let us try to calculate the area lost due to the opening.

Area Lost = d x t

But what is the value of t?

a)????? Required Thickness

b)????? Design Thickness

c)????? Nominal Thickness

Let us try to understand which is the correct thickness to use in the calculation of the area lost. Nominal thickness is not considered for any calculation because it includes corrosion allowance and margin over that, and is not the exact area that is lost. Design thickness is also not considered because it contains corrosion allowance, which cannot be part of any calculation. The remaining thickness is the required thickness, which is the minimum thickness required to take care of the pressure and is the exact area that is lost. Therefore, the correct answer is Required Thickness.

So, Area Lost = d * tr which is the Required Area.

This calculation is based on engineering judgment and does not involve code.

Now, let us look at the code. The Required Area as per code is:

A = d tr F + 2 tn tr F (1 - fr1)

d – Finished Diameter of the opening

tr – Required Thickness of the Shell

tn – Nozzle Wall Thickness

fr1 – Ratio of Nozzle Allowable Stress to Shell Allowable Stress

F – Correction Factor that compensates for the variation in internal pressure stresses on different planes with respect to the axis of a vessel

Now, why the code formula looks completely different ? The answer lies in understanding importance of each term:

Understanding fr1

Set-in Nozzle

In the above figure, what will be the value of d. As per our engineering judgment calculation of area lost, we considered d as the diameter of the opening and not the nozzle inside diameter but as per code, d is the finished diameter of the opening which means the nozzle inside diameter. In the set-in nozzle, the material of the nozzle is inserted and hence we can see that some amount of area lost is compensated by the nozzle wall that is inserted which will also depend on the nozzle material which is used. So, to consider that we have the factor fr1.

Value of fr1 = Sn/Sv = Nozzle Allowable Stress/Shell Allowable Stress

If in Set-in Nozzle, the material of the Nozzle and Shell is the same then,

fr1 = Sn/Sv = 1

So,

A = d tr F + 2 tn tr F (1 - fr1)

A = d tr F + 2 tn tr F (1 - 1)

A = d tr F

Set-on Nozzle

In Set-on Nozzle there is no nozzle coming into the area lost a portion of the shell. Hence there is no point in considering the Nozzle Allowable Stress in the formula as the nozzle does not play any role in this type of configuration. Hence the Value of fr1 = 1 for the Set-in type of Nozzle.

Hence,

A = d tr F + 2 tn tr F (1 - fr1)

A = d tr F + 2 tn tr F (1 - 1)

A = d tr F

Understanding F

As we know that stress acts differently in different plans of the vessel. As we saw in UG-27 that the thickness due to circumferential stress and longitudinal stress are different as the stress in the longitudinal plane is more than compared to stress in the circumferential plane.

So, if we cut the Vessel through Longitudinal Plane then we can see how the nozzle look.

Now, if we cut the Vessel through Circumferential Plane then we can see how the nozzle look

So, if your design uses the Circumferential Plane the stress will be different than in the Longitudinal Plane. Hence to compensate for this difference in stress we use the value of F.

Let us see how to find the value of F.

If we see the Nozzle from the top view as shown in the figure above, we will see the nozzle opening as shown on below. Here, Plane 1 is Longitudinal Shell Axis and the blue line in the above figure is the plane from which we will cut and design. θ is the angle between the Longitudinal Shell Axis and the Cutting Plane.

To find the Value of F, we use the below graph. The value of F depends on the value of θ.

Areas providing Reinforcement in Set-on Nozzle

In Set-on nozzles, the diameter of the opening and the thickness of the nozzle are the two key parameters. The required thickness of the nozzle is represented by the blue color in the diagram. Any additional thickness beyond this required thickness will act as a reinforcement area. This means that the nominal thickness of the nozzle will also participate in the reinforcement of the area. The shell also has a nominal thickness which will act as a reinforcement area.

The reinforcement pad is provided in the design for further reinforcement. The welds that are present in the nozzle will also participate in the reinforcement area. Therefore, all of these areas will take part in the provided reinforcement. It is important to note that only the reinforcement pad is not the only provided reinforcement.

Areas providing Reinforcement in Set-in Nozzle

In a set-in nozzle, the ID of the nozzle will be the opening diameter. Reinforcement provided by areas in a Set-on type of nozzle will also act as reinforcement areas in a Set-in type of nozzle. Additionally to those other areas like the inside of the set-in nozzle, there is no area required since both side pressures are equal. Therefore, the complete projection inside the vessel will take part in the reinforcement. The reinforcement pad and welds will also take part in the provided reinforcement.

UG-40: Limits of Reinforcement

Understanding the limits of reinforcement is crucial for static engineers as it helps to determine the maximum dimensions where we can place the reinforcement. Reinforcement provided outside the limits of reinforcement will not be considered as reinforcement. In the code, there is a rectangle box which is the limit of the reinforcement. The dimensions of that rectangle box can be seen in the figure below.

Example

Given Data

MOS of Shell – SA-516 Gr.70

MOS of Nozzle – SA-106 B

P – 1 MPa

Pe – 0.1013 MPa

T – 250 degree C

CA – 3 mm

Shell ID – 2000 mm

Nominal Thickness of Shell (t) – 12 mm

Nozzle Schedule – 80

Nozzle Diameter (d) – 200 mm

Projection Inside (h) – 25 mm

Projection Outside – 150 mm

Solution

Calculate fr1

fr1??????? = Sn / Sv

??????????? = 117.9 / 137.9

??????????? = 0.855

Calculate F

θ = 0 degree C

Hence, F = 1

Calculate tr

tr = PR/SvE - 0.6P

tr = (1 x 1000)/(137.9 x 1) -(0.6 * 1)

tr = 7.305mm

Nozzle Thickness

tn = 13.7 – 3 = 9.7 mm

Area Required(A)

A = d tr F + 2 tn tr F (1 - fr1)

A = (200 x 7.305 x 1) + [2 x 9.7 x 7.305 x 1 x (1-0.855)]

A = 1481.54 mm2

Area A1

Case 1:

A1 = d (E1 t - F tr) - 2 tn (E1 t - F tr) x (1-fr1)

A1 = 200 x (1 x 9 – 1 x 7.305) – 2 x 9.7 x(1 x 9 – 1 x 7.3052) x (1 - 0.855)

A1 = 334.23 mm2

Case 2:

A1 = 2 (t + tn) ( E1 t - F tr ) - 2 tn (E1 t - F tr ) ( 1 - fr1 )

A1 = 2 x (9 + 9.7) x (1 x 9 - 1 x 7.305) - 2 x 9.7 x (1 x 9 - 1 x 7.305) x (1 - 0.855)

A1 = 58.62 mm2

Therefore,

A1 = max( 334.23 , 58.62 )

A1 = 334.23 mm2

Area A2

Case 1:

A2 = 5 (tn - trn) x fr2 t

A2 = 5 (9.7 – 0.85) x 0.855 x 9

A2 = 340.5 mm2

Case 2:

A2 = 5 (tn - trn) (2.5 tn + te) x fr2 tn

A2 = 5 (9.7 – 0.85) (2.5 x 9.7 + 12) x 0.855 x 9.7

A2 = 13303.29 mm2

Therefore,

A2 = min( 340.5 , 13303.29 )

A2 = 340.5 mm2

Area A3

A3 = ( 2 ti fr2 ) x min(2.5xt, 2.5 x ti, h)

A3 = ( 2 x 6.7 x 0.855 ) x min(22.5, 16.75, 25)

A3 = 191.9 mm2

Area A41

fr3 = (lesser of Sn or Sp)/Sv = 0.855

A41 = (leg)2 x fr3

A41 = (10)2 x 0.855

A41 = 85.5 mm2

Area A42

fr4 = Sp/Sv = 1

A42 = (leg)2 x fr4

A42 = (10)2 x 1

A42 = 100 mm2

Area A43

fr2 = Sn/Sv = 0.855

A43 = (leg)2 x fr2

A43 = (10)2 x 0.855

A43 = 85.5 mm2

Area A5

A5 = (Dp - d - 2tn) te fr4

A5 = (350 - 200 - 2 x 9.7) x12 x 1

A5 = 1567.2 mm2

Safe Design Check

For Safe Design,

Area Available ≥ Area Required

A1 + A2 + A3 + A41 + A42 + A43 + A5 ≥ A

334.23 + 340.5 + 191.9 + 85.5 + 100 + 85.5 + 1567.2 ≥ 1481.54

2704.83 ≥ 1481.54

Hence Design is Safe

Design for External Pressure

As per UG-37 (d)(1)

A = 0.5 {d tr F +2tn tr F (1-fr1)}

F = 1

tr = Thickness required for External Pressure

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