API 510 Chapter 9 ASME VIII Pressure Design
9.1 The role of ASME VIII in the API 510 syllabus
ASME VIII-I Rules for the Construction of Pressure Vessels is a long-established part of the API 510 syllabus (or ‘body of knowledge’ as they call it). Strictly, it is concerned only with the new construction of unfired vessels (i.e. not boilers or fired heat exchangers) and so contains no reference in it at all to what happens to vessels after they are put into service.
Now you see the problem; the API 510 syllabus is almost the antidote to ASME VIII in that its role is to give guidance of what to do with vessels that have been in service for some time. In doing this it has to deal with vessels that may be corroded, have been re-rated or in some way are not the same as when they were new.
Fortunately, the answer is fairly straightforward and centres around the situation when a vessel is being repaired or altered; the title of API 510 is Inspection, Repair and Alteration remember. API 510 sees vessel repairs and alterations as a straightforward remanufacturing exercise that should follow the requirements of ASME VIII-I, just as when it was built. The added difficulty of doing this on-site rather than in the manufacturing shop brings two implications:
- An API 510 qualified inspector has to know some relevant points of ASME VIII.
- There may be some areas in which ASME VIII either physically cannot be followed (because the vessel is already built) or in which an alternative, perhaps less conservative, solution is adequate.
Figure 9.1 shows the solution. Note how the blanket coverage of repairs by ASME VIII is overridden in a few well-chosen areas written into API 510.
9.2 How much of ASME VIII is in the API 510 syllabus?
ASME VIII is a complex multilayered network of intrigue. It is all held together by a rather confusing system of clause numbering, littered with a nested interlinking network of cross-references. It is no doubt a proven, competent code, but it can look confusing if you don’t deal with it regularly or, even worse, have never seen it before.
The good news is that not much of the content of ASME VIII is included in the API 510 syllabus (formally called the ‘body of knowledge’ remember). You can think of it, simplistically, as involving only a few minor abstracts plucked out of the code and used as the subject of a small family of examination questions. Look at it carefully and you will conclude that the scope of these questions is less, and the content is more straightforward, than the API body of knowledge infers.
Practically, the subjects chosen from ASME VIII relate in some way to the repair or replacement of vessel components. This makes sense, as this is what API 510 is all about. While the individual topics are not difficult, the way in which they fit together is important to understand. Figure 9.2 shows the situation – note how all the individual topics are not linear, as such, but act together to specify the ‘design’ of a repaired
or replaced pressure vessel component. For simplicity, we will look at each of these in turn.
9.2.1 Material choice
The API 510 syllabus takes a fairly one-dimensional view of material choice. Unlike the API 570 syllabus, which utilizes ASME B31.3, ASME VIII contains no material data at all. Instead, it references the full data set of ASME II (d), which thankfully is not in the API 510 syllabus. The main material property that API 510/ASME VIII is concerned with is that of resistance to brittle fracture. The fundamental issue is therefore whether a material is suitable for the minimum design metal temperature (MDMT) for which a vessel is designed. This topic is covered by clause UCS-66 of ASME VIII.
9.2.2 Vessel design features
The main ASME VIII design topics required included in the API 510 syllabus are:
- Internal pressure in shells and heads (clauses UG-27 and UG-32)
- External pressure on shells (clause UG-28)
- Nozzle compensation (mainly figure UG-35.1)
- Nozzle weld sizing (mainly figure UW-16)
All four of these topics are heavily and artificially simplified for the purpose of producing API 510 exam questions. Without this simplification, calculation questions would be just too complex for a 4-minute exam answer.
9.2.3 RT grades
ASME VIII has a fairly unique approach to vessel design in that for every specified vessel ‘design requirement’ there is not just one but up to four possible design solutions. Put another way, if you specify an ASME VIII-I vessel for a specific set of pressure–temperature criteria you could legitimately receive up to four different ‘designs’, all of which are fully compliant with the specific requirements of the code.
The difference comes from the concept of ‘RT grade’. This works off three simple principles:
- ASME VIII allows a degree of freedom in the amount (scope) of NDE done on a vessel.
- The amount of NDE that is actually chosen is denoted by the RT grade: RT-1, RT-2, RT-3 or RT-4, as set out in clause UG-116 of ASME VIII-I.
- The RT grade that is chosen defines the joint efficiency E, which is used for the shell and head design calculations. A high joint efficiency of 1 (when grade RT-1 is chosen) gives the smallest wall thickness.
22.214.171.124 RT grade terminology
This is not the easiest concept of ASME VIII to understand on first reading. Figure 9.3 shows the situation. Note how both RT-1 and RT-2 can be referred to as ‘full’ radiography grades, even though strictly they are not. It is easier to just accept this, rather than look too deeply into the logic behind it. Grade RT-3 is referred to as the ‘spot’ radiography grade whereas RT-4 means ‘less than spot RT’, which also includes no RT at all.
As you can see from Fig. 9.3, the most awkward grade to understand is RT-2. We will explain this later in Chapter 10; for the moment just think of it as a ‘full’ RT category.
9.3 ASME VIII clause numbering
ASME VIII uses a system of clause numbering that can look confusing when you first encounter it. The full code is divided into multiple sections designated by letters (UG, UW, UCS, UHT, UF, UB, UNF, UCI and similar). Most of these are not required for the API 510 syllabus; the only ones you actually need parts of are as follows.
- UG: the G denotes general requirements.
- UW: the W denotes welding.
- UCS: the CS denotes carbon steel (so you can expect this
section to be concerned with materials of construction and their heat treatment).
- UHT: the HT denotes heat treatment (this part contains specific requirements for heat-treated ferritic steels).
- Appendix 1: supplementary design formulae
- Appendix 3: definitions
Remember that not all of the ASME VIII code pages provided in your API 510 document package are actually needed. Only specific paragraphs are in the examination scope. Of the above, the majority of the technical details relating to welding are contained in section UW. A few others appear in section UCS.
We will now look at one of the major topics of the UG section of ASME VIII. This section deals predominantly with design. The first topic to be covered (internal pressure loadings) acts as an introduction to the design concepts of ASME VIII. Subsequent sections look at the related subjects of external pressure loadings, nozzle compensation, pressure testing and materials issues such as impact test and heat treatment.
9.4 Shell calculations: internal pressure
Shell calculations are fairly straightforward and are set out in UG-27. Figure 9.4 shows the two main stresses existing in a thin-walled vessel shell.
Hoop (circumferential) stress
This is the stress trying to split the vessel open along its length. Confusingly, this acts on the longitudinal weld seam (if there is one). For the purpose of the API 510 exam this is the governing stress in a shell cylinder. The relevant UG-27 equations are:
P = maximum design pressure (or MAWP).
t = minimum required thickness to resist the stress.
For API 510 Exam purposes, the important stress for calculation purposes is the circumferentail(hoop) stress as this is the one that governs the required shell thickness
These only apply to thin-walled shells. That is all the API 510 exam questions normally cover
Figure 9.4 Vessel stresses
S = allowable stress of the material. This is read from ASME IID tables or, more commonly, given in the exam question (it has to be as ASME IID is not in the syllabus).
E = joint efficiency. This is a factor (between 0.65 and 1) used to allow for the fact that a welded joint may be weaker than the parent material. It is either read off tables (see UW-11 and UW-12 later) or given in the exam question. You can think of E as a safety factor if you wish.
Ri = the internal radius of the vessel. Unlike some other design codes ASME VIII-I prefers to use the internal radius as its reference dimension, perhaps because it is easier to measure.
A key feature of Ri is that it is the radius in the corroded conditions (i.e. that anticipated at the next scheduled inspection). Don’t get confused by this – it is just worked out in this way. If a vessel has a current Ri of 10 in and has a corrosion rate (internal) of 0.1 in/years, with the next scheduled inspection in five years, then:
Current Ri = 10 in
Ri in 5 years = 10 in + (5 x 0.1 in) = 10.5 in corroded condition
Hence 10.5 in is the Ri dimension to use in the UG-27 equation.
Axial (longitudinal) stress
This is the stress trying to split the vessel in a circumferential plane; i.e. trying to pop the head off the shell. It is approximately half the magnitude of the hoop stress and so not a ‘governing’ design parameter (at least for the purpose of the API 510 exam). You can see that the equation for it is in UG-27 – but probably they rarely, if ever, appear in exam questions. Now note the following specific points.
Looking at the formula for the cylindrical shell in UG-27 (c) (1) for circumferential stress, there are two limitations applied to it:
- The thickness must not exceed one-half of the inside radius, i.e. it is not a thick cylinder.
- The pressure must not exceed 0.385SE, i.e. not be high pressure. In practice this is more than about 4000 psi for most carbon steel vessels.
If the first requirement applies, i.e. you are dealing with a thick cylinder, then a completely different set of equations is needed. Don’t worry about them – they won’t be in the exam.
9.4.1 Shell calculation example
The following information is given in the question.
R = inside radius of 30 in (Ri )
P = pressure of 250 psi (MAWP)
E = 0.85 (type 1 butt weld with spot examination as per UW-12)
S = 15 800 psi
What minimum shell thickness (which may be called tmin or trequired) is necessary to resist the internal MAWP?
Using thickness (t) = PR/(SE–0.6P) from UG-27
Thickness = 250 x 30 / [15800 x 0.85 – (0.6 x 250)]
t = 0.565 in ANSWER
Remember that this is exclusive of any corrosion allowance that you decide to add.
Now, as an exercise, look at section UG-16 (c) covering mill undertolerance. Note how, strictly, you are allowed an undertolerance of 0.25 mm or 6 % of design thickness while still using the design pressure. Don’t worry about this because if exam questions require you to use this undertolerance allowance, they will mention it in the questions.
Now work the equation from the other perspective, where you are given a shell material thickness and you have to determine the maximum acceptable pressure (MAWP):
Given t = 0.625 in
Using pressure (P) = SEt/(R + 0.6t) from UG-27
Pressure (P) = 15 800 x 0.85 x 0.625 /
[30 + (0.6 x 0.625)]
MAWP = 276 psi ANSWER
9.5 Head calculations: internal pressure
- 9.5.1 Head thickness Pressure vessels have the pressure enclosed by a head that has the pressure acting on the inside of the head. There are a number of different types of heads that can be used:
These are all in the API 510 syllabus except for the toriconical design. Figure 9.5 shows the shapes. The formulae for the required minimum thickness are given in ASME VIII UG-32 (d) to (g).
9.5.2 Ellipsoidal heads UG-32 (d)
The ratio of major to minor axis is set at 2:1 for a standard ellipsoidal head. This is the type assumed by the UG-32 (d) formula (see Fig. 9.6). When the ratio of the major to minor axis is not 2, then a factor K is incorporated in the formula. Calculations involving this case are not common in the API 510 exam.
The formulae used for the required minimum thickness of ellipsoidal heads in UG-32 (d) are:
Minimum thickness (t) = PD / (2SE – 0.2P) Maximum pressure = 2SEt / (D + 0.2t) where
P = internal pressure
D = inside diameter (notice how the diameter replaces the Ri used for shells)
E = joint efficiency
S = allowable stress of the material
Ellipsoidal head calculation example
Here is an example for a 2:1 ellipsoidal head, using similar figures from the previous example. Guides:
D = inside diameter of 60 in
P = pressure of 250 psi (MAWP)
E = 0.85 (double-sided butt weld with spot examination (UW-12))
S = 15800 psi
What thickness is required to resist the internal pressure?
Using t = PD/(2SE—0.2P)
Thickness (t) = 250 x 60 / [(2 x 15 800 x 0.85)
– (0.2 x 250)]
t = 0.56 in ANSWER
Again, we can work the equation from the other perspective, where you are given a material thickness and you have to determine the MAWP.
Assuming a given head thickness of 0.625 in
Using maximum pressure (MAWP) = 2SEt / (D + 0.2t)
Pressure = P = 2 x 15 800 x 0.85 x 0.625/
[60 + (0.2 x 0.625)]
P = 279 psi ANSWER
9.5.3 Torispherical heads UG-32 (e)
Note the two restrictions on physical dimension of the head, given in UG-32(e) (see Fig. 9.7): .
- Knuckle radius = 6 % of the inside crown radius
- Crown radius = outside diameter of skirt
These are long-standing restrictions based on well-established design principles for torispherical heads. The formulae for the required minimum thickness and MAWP given in UG-32 (e) are: Thickness t = 0.885PL/(SE – 0.1P)
Pressure P = SEt/(0.885L + 0.1t)
L = inside spherical radius of the flatter part of the head (called the crown radius).
Torispherical head example
L = inside spherical (crown) radius of 30 in
P = pressure of 250 psi (MAWP)
E = 0.85
S = 15 800 psi
Thickness required (t) = 0:885 x 250 x 30
15 800 x 0:85) – (0:1 x 250)
t = 0.495 in ANSWER
Alternatively, to find P using a given head thickness of 0.625 in:
Pressure (P) = 15 800 x 0:85 X 0:625
(0:885 x 300) + 0:1 x 0:625)
P = 315 psi ANSWER
9.5.4 Hemispherical heads UG-32 (f)
Limitations apply of thickness and pressure as before. The formulae for the required minimum thickness are given in UG-32 (f) (see Fig. 9.8):
t = PL/2SE – 0.2P
P = 2SEt/(L + 0.2t)
This time, L is the spherical inside radius (note that there is no crown or knuckle radius as the head is hemispherical; i.e. a half circle). Hemispherical head example Given:
Internal pressure (P) = 200 psi
Allowable stress (S) = 15 000 psi
Spherical radius (L) = 60 in
Joint efficiency (E) = 1.0
Required thickness (t) = 200 x 60 / [(2 x 15000 x 1) – (0.2 x 200)]
t = 0.4 in ANSWER
Alternatively, calculating the maximum allowable pressure for a given thickness of, say, 0.5 in: Pressure (P)=2 x 15 000 x 1 x 0.5/60 + (0.2 x 0.5)
P = 250 psi ANSWER
9.5.5 Conical heads (without knuckle) UG-32 (g)
Only conical heads without a moulded knuckle are covered in the API 510 syllabus. A cone with a knuckle is known as a toriconical head. They are covered in UG-32 (g) but are not in the API 510 syllabus. In practice, conical heads of any type rarely appear as exam questions, but here they are for reference.
The formulae for the required minimum thickness given in UG-32 (g) are:
t = PD/2 cos α (SE – 0.6P)
P = 2SEt cos α/[(D + 1.2 t cos α)]
where alpha (α) is the half-cone angle of the cone.
Example of conical head calculation Given:
Internal pressure (P) = 300 psi
Inside diameter of cone (D) = 40 in
llowable stress (S) = 12 000 psi
Joint efficiency (E) = 0.85
Cone half angle (α) = 30°
Cosine of 30° = 0.866
Calculating required thickness (t):
t = PD/2 cos α (SE – 0.6P)
Thickness (t) = 300 x 40/[2 x 0.866 x (12 000 x 0.85 – (0.6 x 300)]
t = 0.69 in ANSWER
Alternatively calculating the maximum allowable pressure for a given head thickness of, say, 0.75 in:
P = 2SEt cos α/[(D + 1.2 t cos α)]
Pressure (P)=2 x12000 x 0.85 x 0.75 x 0.866/ [40 + (1.2 x 0.75 x 0.866)]
P = 325 psi ANSWER
In recent years, conical head questions have appeared rarely, if ever, in the API 510 exam.
9.5.6 Corrosion allowance
Remember that none of the thicknesses calculated in the above calculations have included any allowance for the corrosion of the material. This must be added to the required thickness obtained from the calculation. Normally it is left to the particular owner/user to set his or her own requirements. This can vary greatly, depending on the type of material, from 0.25 in for carbon and low alloy steels to no corrosion allowance at all for stainless steel type materials.
The corrosion allowance is important when calculating the remaining life of a vessel under API 510 guidelines (as we will see in later calculations).
Now try these familiarization questions on shell and head calculations.
9.6 Set 1: shells/heads under internal pressure familiarization questions
9.7 ASME VIII: MAWP and pressure testing
We will now look at another of the major topics of the UG section of ASME VIII, the determination of MAWP (maximum allowable working pressure). This fits together with internal and external pressure calculations and influences the related subject of pressure testing. MAWP calculations are mathematically straightforward but cover a few different interconnected areas such as:
- Weld joint efficiency
- Static pressure head
- Calculating the MAWP for a corroded vessel and/or . Calculating the required minimum thickness of a vessel subject to a known internal pressure
The last two in the above list are simply different ways of looking at the same thing. Both use the concept of a weld joint efficiency and a method to take into account the effect of static head pressure in the vessel, as well as pressure imposed from an external source or process.
MAWP is an acronym used only in ASME and API codes. It is comparable, but not identical, to the term ‘design pressure’ preferred by most other non-US codes (see Fig. 9.9, and Fig. 4.7 in Chapter 4). Think of this way of understanding the ASME view of it:
- Design pressure is a nominal value of pressure provided by (for example) a process engineer or contractor to a vessel designer.
- This pressure is the minimum required in order for the vessel to fulfil its process function. . The vessel designers then respond by designing a vessel based around MAWP because this is the parameter referred by ASME VIII when calculating the required thickness (tmin) of the pressure envelope components.
- From the above you can see that MAWP must be equal to or greater than ‘design pressure’. It cannot be lower or the vessel will not meet its design requirement. You can see this relationship between design pressure and MAWP set out in ASME VIII UG-98 and API 576
In US code, Design pressure is an almost nominal value chosen at the pre design stage to describe the process function requirments of a vessel before code calculations are carried out
9.7.2 Where is MAWP measured?
By convention (and only by convention), MAWP is always measured at the top of a vessel. You can think of it as shown in Fig. 9.9 with the upper point of a vessel (vertical or horizontal) being fitted with a ‘MAWP gauge’.
If the vessel is full of water the weight of the water (called the static head) causes the pressure at the bottom of the vessel to be greater than that at the top. Hence when designing the thickness of the bottom of the shell and the lower head the additional pressure due to the static head must be taken into account, over and above that pressure from the MAWP.
9.7.3 Pressure testing
The most common pressure test is the standard hydraulic test covered in ASME VIII UG-99 (see Fig. 9.10). Note how ASME use the term hydrostatic test – this term is strictly more applicable to atmospheric storage tanks, but that’s what they use. Whereas, in earlier editions, ASME had used 1.5 x MAWP as the standard multiplier for test pressure this has now been amended to the following:
Test pressure (hydraulic) = 1.3 x design pressure x ratio of material stress values
Ratio of material = stress at test temperature (normally ambient)
stress values stress at design temperature
Remember that this test pressure is measured at the highest point of the vessel.
These stress values are taken from the tables of material stress values in ASME II(D) but are given in exam questions. Note that where a vessel is constructed of different materials that have different stress values, the lowest ratio of stress values is used.
This can be seen in an example:
Design pressure (MAWP) = 250 psi
Design temperature = 750°F
Material: carbon steel SA516-60
Allowable stress value at room test temperature = 15 000 psi
Allowable stress value at 750°F = 13 000 psi
Ratio of stress values = 15 000/13 000 = 1.154
Test pressure = 1.3 x 250 x 1.154
= 375 psi ANSWER
The test medium for a hydrostatic test is normally water but other non-hazardous fluids may be used provided the test temperature is below its boiling point. Where combustible
fluids such as petroleum distillates are used with flash points of less than 110 °F (43 °C), these should only be used for tests carried out near atmospheric temperatures.
9.7.4 The hydrostatic test procedure
ASME VIII section UG-99 (g) gives requirements for the test procedure itself. This is a fertile area for closed-book examination questions. An important safety point is the requirement to fit vents at all high points to remove any air pockets. This avoids turning a hydrostatic test into a pneumatic test, with its dangers of stored energy.
Figure 9.11 shows the hydrostatic test procedure. A key point is that the visual inspection of the vessel under pressure is not carried out at the test pressure. It must be reduced back to MAWP (actually defined in UG-99 (g) as test pressure/1.3) before approaching the vessel for inspection. If it was a hightemperature test (> 120 °F), the temperature must also be allowed to reduce to this, before approaching the vessel.
Once the pressure has been reduced, all joints and connections should be visually inspected. Note how this may be waived provided:
- A leak test is carried out using a suitable gas.
- Agreement is reached between the inspector and manufacturer to carry out some other form of leak test.
- Welds that cannot be visually inspected on completion of the vessel were given visual examination prior to assembly (this may be the case with some kinds of internal welds).
- The contents of the vessel are not lethal.
In practice, use of these ‘inspection waiver points’ is not very common. Most vessels are tested and visually inspected fully as per the first sentences of UG-99 (g).
A footnote to UG-99 (h) suggests that a PRV set to 133 % test pressure is used to limit any unintentional overpressure due to temperature increases. Surprisingly, no PRV set to test pressure is required by the ASME code; you just have to be careful not to exceed the calculated test pressure.
9.7.5 Pneumatic test
UG-100 covers the requirements for pneumatic testing. Look at the main points in Fig. 9.12. As a basic principle, ASME VIII does not recommend using pneumatic testing as an arbitrary alternative to a hydraulic test. The idea is that pneumatic testing is only used when it is absolutely necessary, when any of the following occurs: .
- The vessel is not designed or supported in a way that it can be filled with water.
- A vessel cannot tolerate the presence of water
A previous hydraulic test has already been carried out (so the pneumatic test is only required as a type of leak test).
With pneumatic testing, the major consideration is safety. As air is compressible, a lot of energy will be stored in the vessel that can catastrophically release on failure of the vessel. Great care and hazard assessment is therefore needed before carrying out any pneumatic test.
To minimize any risk of brittle fracture, the test temperature should be at least 30 °F above the minimum design metal temperature (MDMT) and UG-100 gives a specific test procedure as follows.
The test pressure is lower than that for the hydro test:
Test pressure (pneumatic) = 1.1 x design pressure (MAWP) ratio of material stresses
Ratio of material = allowable stress at test temperature (usually ambient)
stress values allowable stress at design temperature
As before, the stress values come from ASME II(D) so should be given in the exam question.
Section UG-100 (d) gives requirements for the pneumatic test procedure itself. This is a fertile area for closed-book examination questions. The steps are:
- Pressurize to 50 % test pressure.
- Increase in steps of approximately 10 % of test pressure until the required pressure is reached.
- Reduce to test pressure/1.1 and perform the visual inspection.
The only requirement for the test time is that the pressure must be held ‘long enough to allow leakage to be detected’. In a similar way to hydraulic testing there are some ‘unusual conditions’ waivers allowed for the final visual inspection stage. It may be waived provided that: .
- A leak test is carried out using a suitable gas.
- Agreement is reached between the inspector and manufacturer to carry out some other form of leak test.
- All those welds that cannot be visually inspected on completion of the vessel are given visual examination prior to assembly.
- The vessel contents are not lethal.
YOU CAN IGNORE SECTION UG-101 ‘PROOFTESTING’ AS IT IS NOT IN THE API 510 EXAMINATION SCOPE.
9.7.6 Test gauges UG-102
This is fairly straightforward and commonsense. Note that the requirements are based on a single pressure gauge to measure the test pressure. The requirements are:
- The gauge must be connected directly to the vessel (not via a network of pipes).
- It should be visible from where the pressure is applied (if it isn’t, then an additional gauge must be provided).
- For large vessels, gauges that record the pressure measurements are recommended. .
- The gauge should have an indicating range of not less than 1 times the test pressure and not more than 4 times the test pressure. This is to make sure that it gives an accurate reading.
- All gauges should be calibrated against either a dead weight tester or master gauge.
Now try these familiarization questions on MAWP and pressure testing.
9.8 Set 2: MAWP and pressure testing familiarization questions
9.9 External pressure shell calculations
9.9.1 External pressure So far, we have only looked at the design for vessels under internal pressure. We will now look at the topics of ASME VIII (section UG-28) covering vessels under external pressure. Many vessels are subject to vacuum conditions or have jackets, which can apply an external pressure to the shell (see Fig. 9.13). Details are given in UG-28 of ASME VIII for cylindrical vessels that may or may not have stiffening rings. Typical forms of cylindrical shells are shown in ASME VIII figure UG-28.1
External pressure calculations are completely different to those in UG-27 and UG-32 for internal pressure on heads and shells. This is because the mode of failure is completely different. Vessels under external pressure fail by buckling, a catastrophic (and fairly unpredictable) mechanism that is
much more complex than the tensile stress failures that result from internal pressure.
Consequently, the concepts of allowable stress and joint efficiency do not play the same part that they do in internal pressure calculations on shells. Thankfully, the API 510 exam syllabus is limited to calculations relating to external pressure on cylindrical shells only. Heads are not covered (they are far too complicated).
Instead of simple stress equations formulae, external pressure calculations involve reading off charts to obtain two special factors A and B, and then using these factors in simple formulae to calculate either:
- The wall thickness (t) required to resist a given external pressure (Pa) or
- The maximum external pressure (Pa) that a vessel of given wall thickness (t) can resist.
9.9.2 External pressure exam questions
The API exam questions only require you to do a shortened method of assessment of external pressure conditions. This is because the two charts necessary to find the A and B calculation factors are not in ASME VIII (they are hidden away in ASME II, which is not part of the examination document package). In previous years the API 510 exam question paper occasionally contained extracts from these charts but more recently these have simply been replaced by given values of parametersAandB, making things much easier.
The easiest way to understand the UG-28 calculations themselves is to look at this worked example. Figure 9.14 shows the parameters for a vessel under external pressure operating at 300 °F:
- t = thickness of the shell = 0.25 in .
- L = distance between stiffeners = 90 in
- Do = shell outside diameter = 180 in
The first step is to calculate the values of the dimensional ratios (L/Do) and (Do/t):
L/Do = 90/180 = 1/2///2, Do/t = 180 / 0.25 = 720
In a real design situation, these ratios would then be plotted on charts to give values of A and B. In this example, the charts would give values of A = 0.000 15 and B = 2250 (remember that you will generally be given these in an exam question).
From UG-28, the safe external pressure (Pa) is then calculated from the equation below:
Pa = 4B/3 (Do/t)=4 x 2250/3 x 720 = 4.2 psi Conclusion – the vessel is not suitable for full vacuum duty (-14.5 psi ).
Now try these familiarization questions.
9.10 Set 3: vessels under external pressure familiarization questions
9.11 Nozzle design
Two aspects of vessel nozzle design are included in the API 510 syllabus: nozzle compensation and weld sizing (see Fig. 9.15). By necessity the subjects covered are highly simplified in order to fit the required format of the exam questions. Over recent years the exam questions in this area have become both simpler and fewer in number, which is good news.
From a practical perspective, these subjects are in the API 510 syllabus owing to their relevance to vessel repairs. It is reasonable to expect that an inspector should be able to check design dimensions relating to a new or altered nozzle.
9.11.1 Nozzle compensation
Exam questions on nozzle compensation centre around ASME VIII figure UG-37.1. In true ASME code style this figure incorporates a formidable amount of information on a single page. Note the figure itself. What it actually shows is a ‘half-section diagram’ – two halves separated by a vertical centreline (see Fig. 9.16). The left-hand side shows the configuration in which the nozzle is ‘set through’ the shell. You can ignore this as the set-through configuration is specifically excluded from the API syllabus. The right-hand side is in the syllabus – here the nozzle is set on to the shells, i.e. does not project through. The ASME VII term for this is abuts.
Now look at the equations and text below the diagram. This is (perhaps not so obviously) divided into two scenarios, this time separated by an imaginary horizontal line approximately halfway down the page. The top half covers the situation where the nozzle does not have a reinforcing element (compensation pad) and the lower half applies to when a reinforcing pad has been installed. Note some other points about this diagram.
The area cut out to accommodate the nozzle is given the symbol A.
The code uses the principle of the area replacement method (see Fig. 9.17). This is simple enough; the area cut out (A) must be replaced by metal available or added in other areas to restore the strength of the component. This replacement can be taken from a combination of four sources:
- Excess material (above trequired) available in the shell (called A1).
- Excess material (above trequired) available in the nozzle (called A2).
- Material available in the welds. The number of welds obviously depends on whether or not there is a reinforcing pad fitted. These are called A41 and A42 (note that A43 as shown in ASME VIII figure UW-37.1
The Area Replacement Method Requires that:
Are Removed(A) must be compensated for by the sum of
i.e You must have more spare area available than you have removed. if not , you add a reinforcing pad to make up the difference
is not relevant as set-through nozzles are not in the syllabus).
Material available (called A5) in the reinforcing pad – if one is fitted.
If the amount of compensation area available is more than that area removed (A), then the nozzle is adequately compensated, so no further compensation is required. This can be expressed in equation form, as shown near the bottom of figure UG-37.1.
An opening is adequately reinforced if: A1 + A2 + A3 + A41 + A42 + A5 > A
While not inherently difficult, a full compensation calculation involving all these variables is much too long for a 4-minute exam question, so actual exam questions tend to be heavily simplified, involving: .
- Calculation of ‘area required’ A only.
- Calculations where most of the parameters are given, so only simple maths is required.
- Question about reinforcing limits.
9.11.2 Reinforcing limits
Reinforcing limits are the linear distances from a nozzle, beyond which adding reinforcement becomes ineffective. In practice, the size (diameter) of nozzle compensation pads is frequently chosen to coincide with the reinforcing limit, thereby achieving the most ‘efficient’ design.
Figure 9.18 shows the reinforcement limits, extracted from the code figure UG-37.1. Note how:
- There are two linear limits: one axially ‘along’ the vessel and the other extending radially outwards ‘up the nozzle’.
- Both limits have two options for their calculated value. The axial limit uses the larger of its two options while the radial limit uses the smaller of its two options. In practice, it is usually the first term of each option that ‘governs’, but in an exam situation it is best to check both just in case.