🛢️Unlocking the Secrets of ASME SEC VIII Div 1, UG-37 & UG-40: Reinforcement Requirements & Limits 📐
10/17/23, 11:31 PM
🛢️Unlocking the Secrets of ASME SEC VIII
Div 1, UG-37 & UG-40: Reinforcement
Requirements & Limits 📐
Abhishek Singh
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Founder SCOOTOID eLearning Platform for Mechanical Engineers
Published Sep 13, 2023
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.
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🛢️Unlocking the Secrets of ASME SEC VIII Div 1, UG-37 & UG-40: Reinforcement Requirements & Limits 📐
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.
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🛢️Unlocking the Secrets of ASME SEC VIII Div 1, UG-37 & UG-40: Reinforcement Requirements & Limits 📐
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."
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🛢️Unlocking the Secrets of ASME SEC VIII Div 1, UG-37 & UG-40: Reinforcement Requirements & Limits 📐
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.
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🛢️Unlocking the Secrets of ASME SEC VIII Div 1, UG-37 & UG-40: Reinforcement Requirements & Limits 📐
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
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🛢️Unlocking the Secrets of ASME SEC VIII Div 1, UG-37 & UG-40: Reinforcement Requirements & Limits 📐
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
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🛢️Unlocking the Secrets of ASME SEC VIII Div 1, UG-37 & UG-40: Reinforcement Requirements & Limits 📐
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.
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🛢️Unlocking the Secrets of ASME SEC VIII Div 1, UG-37 & UG-40: Reinforcement Requirements & Limits 📐
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.
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🛢️Unlocking the Secrets of ASME SEC VIII Div 1, UG-37 & UG-40: Reinforcement Requirements & Limits 📐
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.
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To find the Value of F, we use the below graph. The value of F depends on the value
of θ.
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Areas providing Reinforcement in Set-on Nozzle
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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.
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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
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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
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fr1
🛢️Unlocking the Secrets of ASME SEC VIII Div 1, UG-37 & UG-40: Reinforcement Requirements & Limits 📐
= 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
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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
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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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Static Equipment Design
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Matias Nicolas Sainz
1w
Jefe de ingeniería SP Filtraciones
Thanks for sharing. The explanation was really clear.
I would like to consult you for the value "E1" in the calculation of excess area of the shell. According
to understand if the nozzle is not located on any welding should be considered equal to 1, this is
so? On the other hand, in the calculations of surplus areas in both the shell and the nozzle consider
E=1, so I understand according to the code you have to consider them as seamless, is this correct?
I really hope I’m not a bother. I think you have pretty clear concepts and I think you could help me
understand better.
Thank you!
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1w
RAKSHIT BHATT
API 510/570, FFS Engineer at Asset Integrity Engineering
So as per UG-40, even if there is material available outside the limits of reinforcement, we can not
consider them in the area compensated for the opening?
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1mo
Girish Dhameja
Sr. Designer-Static at Special Technical Services (Oman)
Thanks for sharing
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·
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1mo
ZUBER PATEL
Sr. Manager at Godrej & Boyce Mfg. Co. Ltd I Ex - L&T Heavy Engineering Ltd
It's really easy to understand the reinforcement concept along with code requirements !
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·
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kedar shelar
1 Reaction
1mo
Static design engineer
Very useful
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