-
SME Code Calculations: Stayed Surfaces, Safety Valves,
Furnaces
Here is what you w i l l be able to do when you complete each
objective:
1. Calculate the required thickness and design pressure for
braced and stayed surfaces in pressure vessels and the minimum
required cross-sectional area of a stay.
2. Calculate the ligament efficiency method for two or more
openings in the pressure boundary of a pressure vessel.
3. Calculate the required size and capacity of safety valves and
safety relief valves.
4. Calculate required wall thicknesses of plain circular
furnaces, circular flues, and corrugated furnaces.
CHAPTER 2 Part A1 Revised 03/06 to conform with the 2004 ASME
Extract
Revised 03/06 to conform with the 2004 ASME Extract
-
Conforms with the 2004 ASME Extract Revised 03/06
51 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
This chapter uses ASME Sections I, IV, and VIII-1. Each of these
sections contains rules for braced and stayed surfaces, safety
valves, and furnaces or cylinders under external loads. The
objective of this chapter is not to produce a design engineer but a
Power Engineer with knowledge of the basic rationale of the
Codes.
Stress on a vessel with internal pressure. In cylindrical vessel
shells the stress set up by internal pressure longitudinally is
equal to twice the stress set up circumferentially.
vessel diameter internal pressureLongitudinal Stress = vessel
wall thickness 2
vessel end area x internal pressureCircumferential Stress =
vessel circumference x wall thickness
Example 1: Longitudinal and circumferential stress Determine the
stress longitudinally and circumferentially on the shell of a
vessel 4.5 m diameter, 35 mm thick and an internal pressure of 1350
kPa.
vessel diameter internal pressureLongitudinal Stress = vessel
wall thickness 2
4500 1.350 2 35
= 86.79 MPavessel end area internal pressureCircumferential
Stress =
vessel circumference
=
2
wall thickness
4500 1.354=
4500 354500 1.35=
4 35= 43.39 MPa
Note: The longitudinal pressure exerts a stress on the metal
circumferentially, and the radial pressure exerts a stress on the
metal longitudinally.
The strength of a vessel shell depends, therefore, on the
diameter and thickness and is independent of the length. The
candidate should consult the latest 2004 ASME Academic Codes
Extract: Section I; Section II, Part D; Section IV; and Section
VIII, Division 1, while studying this chapter.
INTRODUCTION
-
Conforms with the 2004 ASME Extract Revised 03/06
52 Revised Second Class Course Section A1 SI Units
Section I: The rules for stayed surfaces and staybolts can be
found in Paragraph PG-46 to Paragraph PG-49, Paragraph PW-19 and
Paragraph PFT-22 to Paragraph PFT-32. Section IV: The rules for
stayed surfaces and staybolts can be found in Paragraph HG-340 and
Paragraph HW-710 to HW-713. Section VIII-1: The rules for stayed
surfaces and staybolts can be found in Paragraph UG-47 to Paragraph
UG-50 and Paragraph UW-19. Stays are used in pressure vessels to
carry part or all of the pressure loading when it is desirable or
possible to reduce the span and/or the thickness of a tube sheet or
other pressure component. Opposite surfaces are tied together by
staybolts, tubes, or baffles that carry the pressure loading in
tension. Because bending moments, bending strength, and the tensile
strength of the stays now resist the pressure loading, the required
thickness of stayed surfaces may be less than that of surfaces
which are not stayed. The equation for flat-stayed surfaces is an
adaptation of the flat head equation, with the diameter replaced by
the distance stays. In this case, the C factor represents the
degree of restraint to rotation that the stay attachment provides.
The design pressure and thickness for stayed plates are calculated
by the following formulae:
2
2 t SCPp
= (1.1)
Pt pSC
= (1.2)
OOBBJJEECCTTIIVVEE 33 Calculate the required thickness and
design pressure for braced and stayed surfaces in pressure vessels
and the minimum required cross-sectional area of a stay.
OBJECTIVE 1
RELEVANT ASME CODE SECTIONS
FLAT STAYED SURFACES
-
Conforms with the 2004 ASME Extract Revised 03/06
53 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
t = minimum thickness of plate (mm) p = maximum pitch measured
between straight lines passing through the
centres of the staybolts in different rows. The lines may be
horizontal and vertical, or radial and circumferential (mm).
P = maximum allowable working pressure or internal design
pressure
(kPa). S = maximum allowable stress (MPa)given in Table 1A of
Section II,
Part D, or in Tables HF-300.1 and HF-300.2 in Section IV. C = a
constantthe value depends on details of the staybolt end design
as follows:
C = 2.1 for welded stays or stays screwed through plates not
over 11 mm in thickness with ends riveted over.
C = 2.2 for welded stays or stays screwed through plates over
11
mm in thickness with ends riveted over. C = 2.5 for stays
screwed through plates and fitted with single
nuts outside the plate, or with inside and outside nuts,
omitting washers, and for stays screwed into plates not less than
1.5 times the diameter of the staybolt (measured on the outside of
the staybolt diameter). If washers are used, they shall be at least
half as thick as the plate being stayed.
C = 2.8 for stays with heads not less than 1.3 times the
diameter
of the stays, screwed through plates or made with a taper fit
and having the heads formed on the stay before installing them, and
with the threaded ends not riveted over; the heads are made so as
to have a true bearing on the plate.
C = 3.2 for stays fitted with inside and outside nuts and
outside
washers where the diameter of the washers is not less than 0.4p
and the thickness is not less than the thickness (t) of the surface
being stayed.
Paragraph PG-46.2 states that the minimum thickness of plates to
which stays may be applied, in other than cylindrical or spherical
outer shell plates, is 8 mm except for welded construction covered
by PW-19.
-
Conforms with the 2004 ASME Extract Revised 03/06
54 Revised Second Class Course Section A1 SI Units
Example 2: Stayed flat plate - maximum pressure Calculate the
maximum allowable working pressure on stayed flat plates 12.5 mm
thick, with staybolts attached by fusion welding and pitched 154 mm
horizontally and vertically. The plate material is SA-516-55. The
average temperature is 200 0 C Solution: Use equation 1.1, see
Section I, PG-46.1
2
2 = t SCP p P = maximum allowable working pressure (MPa)
t = 12.5 mm p = 154 mm S = 108 MPa C = 2.2 for welded stays or
stays screwed through plates over 11 mm in
thickness with ends riveted over.
2
2
2
2
=
12.5 108 2.2 = 154
37193.75 = 23716
= 1.565 MPa = 1565 kPa
t SCPp
P
P
P
Example 3: Stayed flat plate - thickness Calculate the minimum
thickness for stayed flat plates, with staybolts screwed through
the plates and pitched 185 mm horizontally and vertically. The
plate material is SA-204-A, maximum allowable working pressure 6205
kPa and operating temperature of 3000 C. Solution Section I
PG-46.1
= Pt pSC
where P = 6.205 MPa S = 128 MPa p = 185 mm C = 2.2 (welded stays
or stays screwed through plates over 11 mm in
thickness with ends riveted over - assume that the plate will be
greater than 11 mm thick.)
-
Conforms with the 2004 ASME Extract Revised 03/06
55 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
=
6.205 = 185 128 2.2
= 185 0.022 = 185 0.148 = 27.46 mm
Pt pSC
t
ttt
The minimum required thickness of 27.46 mm is greater than 11 mm
and C = 2.2 is the correct factor to use.
The requirements are the same for Section I, Section IV and
Section VIII-1, only Section I references will be listed. Paragraph
PG-47 states specific requirements for staybolts or stays. A solid
stay of 200 mm or less in length shall be drilled with telltale
holes at least 5 mm diameter to a depth of at least 13 mm beyond
the inside of the plate. Hollow stays may also be used. This type
of stay can be found in the waterlegs of loco-type boilers.
Corrosion is likely in this area, and if a stay corrodes then a
'telltale' leak can be seen. Telltale holes are not required if the
staybolt is attached by fusion welding (PW-19.8). Paragraph PFT-26
states that the area supported by a stay is based on the full pitch
dimensions with the cross-sectional area of the stay subtracted.
The load carried by that stay is the product of the area supported
by the stay times the internal design pressure or MAWP (maximum
allowable working pressure). Therefore :
2
2
Stay load pressure ( ) (pitch area ( ) cross-sectional area of
stay ( ))( - )
P p aP p a
= = (1.3)
Paragraph PG-49 points at PFT-26 for computing the load on a
staybolt. This load is then divided by the maximum allowable stress
value from Table 1A of Section II, Part D. The result is multiplied
by 10%. Therefore: Minimum area of stay ( ) 1.10 (stay load/maximum
allowable stress ( ))
1.10 (stay load/ )a S
S= = (1.4)
STAYS AND STAYBOLTS
-
Conforms with the 2004 ASME Extract Revised 03/06
56 Revised Second Class Course Section A1 SI Units
Example 4: Minimum required area of stay Welded stays 30 mm
diameter will be used to support a flat plate 16 mm thick. The
pressure is 1350 kPa. The stays are spaced 200 mm horizontally and
vertically. The steel used for the stays and plate is SA-192 at a
maximum temperature of 300 C. Does the stay diameter meet the Code
requirements? Solution
Stay diameter of 30 mm. (a) = 0.7854 0.0302 = 0.707 10 -3 m2 P =
1350 kPa p = 200 mm = 0.2 m S = 91.9 MPa = 91900 kPa
Use equation 1.3 to determine the stay load.
2
2 -3
Stay Load ( - ) 1350 (0.2 - 0.707 10 ) 53.046
P p ax
= = =
Use equation 1.4 to determine the minimum required area of
stay.
-3 2
Minimum required area of stay 1.1 Stay load/ 1.1 53.046 / 91900
0.6 10 m
= = =
S
Use the equation below to determine the minimum diameter of the
stay.
-3
-3
Minimum diameter of stay 0.6 10 / 0.7854
0.764 10 (Ans.)
= = = 0.0276 m or 27.6 mm
The stated diameter of the stay is 30 mm; this is larger than
the minimum required diameter of 27.7 mm; therefore, the stay
diameter meets the Code requirements. Section IV, Paragraph HG-346
states that the firetubes in a firetube boiler may be used as
stays. The required thickness, maximum pitch and design pressure
for tubesheets with firetubes used as stays may be calculated by
the following formulae:
-
Conforms with the 2004 ASME Extract Revised 03/06
57 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
22
2 2
2
22
= + - 4
= + 4
= -
4
P Dt pCS
CSt DpP
CStPDp
t = the required plate thickness mm. p = the maximum pitch
measured between the centers of tubes in
different rows, mm. C = 2.7 for firetubes welded to plates not
over 11 mm thick C = 2.8 for firetubes welded to plates over 11 mm
thick S = the maximum allowable stress values given in Section IV,
Tables HF-
300.1 and HF-300.2 kPa P = the design pressure, kPa D = the
outside diameter of the tubes, mm.
The pitch of firetubes used as stays shall not exceed 15 times
the diameter of the tubes. Firetubes welded to tubesheets and used
as stays must meet the requirements of HW-713.
-
Conforms with the 2004 ASME Extract Revised 03/06
58 Revised Second Class Course Section A1 SI Units
The tubesheet of a firetube boiler is usually a flat plate. The
tubesheet of a watertube boiler is part of the boiler drum. Single
openings in circular vessels have been covered in Module 1.
Multiple openings, such as to be found in a tubesheet, present a
different case and are covered by ligament rules to be found in
Section I Paragraph PG-52, Section IV Paragraph HG-350 and Section
VIII Paragraph UG-53. The ligament rules only consider the material
between the holes and do not consider the tube material wall
thickness. The value of the ligament efficiency found by these
rules is used in the determination of the minimum required
thickness and/or the maximum allowable working pressure for
cylindrical components under internal pressure found in Paragraph
PG-27 and Paragraph UG-27 A ligament is the area of metal between
the holes in a tubesheet. The three types of ligaments are:
Longitudinal: located between the front and lengthwise holes
along the drum.
Circumferential: located between the holes and encircle the
drum.
Diagonal: a special case because they are located between the
holes and are
offset at an angle to each other. The rules of ligaments are
applicable to groups of openings in cylindrical-pressure parts that
form a definite pattern. These rules also apply to openings not
spaced to exceed two diameters centre to centre.
OOBBJJEECCTTIIVVEE 33 Calculate the ligament efficiency method
for two or more openings in the pressure boundary of a pressure
vessel.
OBJECTIVE 2
INTRODUCTION
LIGAMENTS
-
Conforms with the 2004 ASME Extract Revised 03/06
59 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
The following symbols are used in the formulae for calculating
ligament efficiency:
P = longitudinal pitch of adjacent openings (mm) p/ = diagonal
pitch of adjacent openings (mm) p1 = pitch between corresponding
openings in a series of symmetrical
groups of openings (mm) d = diameter of openings (mm) n = number
of openings in length p1 E = ligament efficiency
Use the formula:
- p dEp
= (2.1) when the pitch of the tubes on every row is equal (Fig.
1).
Use the formula:
1
1
- p ndEp
= (2.2) when the pitch of the tubes on any one row is unequal
(Figs. 2 and 3). For tube holes drilled along a diagonal, as shown
in Fig. 4, use the diagram in Fig. PG-52-1 to obtain the ligament
efficiency. (Fig. UG-53.5, Section VIII-1) Note: For holes along a
diagonal, Section IV, paragraph HG-350.4 provides the
following formula:
/
/
- p dEp F
= (2.3) where F is obtained from the chart in Fig. HG-321. This
method gives a higher efficiency than that obtained in Section I or
Section VIII-1.
140 140 140 140 140 140 140
Longitudinal Line
FIGURE 1
Example of tube spacing with hole pitch equal in every row
-
Conforms with the 2004 ASME Extract Revised 03/06
60 Revised Second Class Course Section A1 SI Units
130 140 130 140 130 140 130 140
Longitudinal Line270 mm
130 130 140 130 140 130 130 140 130
Longitudinal Line670 mm
140 mm160 mm
Longitudinal Line
Example 5: Thickness of drum tubesheet Using the rule in Section
I, determine the minimum thickness of a 920 mm I.D. (internal
diameter) cylindrical drum that has a series of openings in the
pattern shown in Fig.4 above and in Fig. 5 below. The openings are
63.5 mm diameter on a staggered pattern of three longitudinal rows
on 76 mm circumferential spacing and 116 mm longitudinal spacing.
The maximum allowable working pressure is 4100 kPa at a temperature
of 250 C. Drum material is SA-516-55 and the tube material is
SA-209-T1. The openings are not located in or near any butt-welded
joint.
FIGURE 2
Example of tube spacing with hole pitch unequal in every second
row
FIGURE 3
Example of tube spacing with hole pitch varying in every second
and third row
FIGURE 4
Example of tube spacing with tube holes on diagonal lines
-
Conforms with the 2004 ASME Extract Revised 03/06
61 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
Solution
2 2Diagonal pitch 58 76
3364 5776
9140 95.6 mm
X = += +==
Hole diameter (d) = 63.5 mm Longitudinal pitch (p) = 116 mm
Use equation 2.1 -
116 - 63.5 116
0.4526
p dEp
=
==
/ 95.6
116 0.824
pp
==
The point corresponding to these values on the diagram in Fig.
PG-52.1, read from the y-axis, is 38%. As the point falls below the
line of equal efficiency for the diagonal and longitudinal
ligaments, the diagonal ligament is the weaker.
Longitudinal Line
116 mm 116 mm
76
76
58 58
X
FIGURE 5
Solution - Thickness of a drum tubesheet
-
Conforms with the 2004 ASME Extract Revised 03/06
62 Revised Second Class Course Section A1 SI Units
Section I, paragraph PG-27.2.2. (2001)
( ) - 1 - PRt C
SE y P= + (2.4)
P = 4100 kPa or 4.1 MPa R = 460 mm S = 108 MPa at 250 C for
SA-516-55 E = 0.38 as determined above C = 0 y = 0.4 for ferritic
steel below 480 C
( )( )
- 1 -
4.1 460 108 0.38 - 1 - 0.4 4.1
1886 41.04 - 2.461886 38.58
(Ans.)
PRt CSE y P
= +=
=
== 48.885 mm
The minimum thickness of the drum shell would be 48.885 mm
without any allowance for manufacture or corrosion. Note: The
minimum thickness of this drum, plain, without being drilled
for
tubes would be 17.836 mm. Therefore, the drum could be
manufactured from two half shells; the tube sheet half being 48.885
mm thick, and the drum half being 17.836 mm thick as shown in Fig.
6. Each half would meet the conditions of rule PG-27.2.2.
DRUM
TUBESHEET
FIGURE 6
Example: Thickness of drum tubesheet Drum manufactured from two
half shells
-
Conforms with the 2004 ASME Extract Revised 03/06
63 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
The most important and also the most critical valve on a boiler
is the safety valve. Its only purpose is to protect the boiler by
automatically limiting the internal boiler pressure to a point
below its Maximum Allowable Working Pressure. To accomplish this,
one or more safety valves must be installed. When the valves open,
they must be capable of releasing all of the steam that the boiler
is capable of generating at maximum firing rates without exceeding
the specified maximum allowable pressure rise. By definition:
Safety valve is used for gas or vapour service. Relief valve is
used primarily for liquid service. Safety relief valve is suitable
for use as either a safety valve or a relief
valve. The rules for safety valves are found in Section I,
paragraph PG-67. Paragraph PG-67.1 states that each boiler shall
have at least one safety valve and if it has more than 47 m2 of
bare tube heating surface, it shall have two or more Paragraph
PG-70 states that the safety valve manufacturer determines the
maximum design capacity of the safety valve and the boiler
manufacturer determines the number of safety valves required by
Paragraph PG-67.1. Paragraph PG-67.2.2 states that for a waste heat
boiler the boiler manufacturer determines the minimum required
relieving capacity based on the heat produced by the auxiliary
firing or the waste heat recovery whichever is greater. This also
applies to boilers that are designed for duel fuel firing. Section
I. Paragraph PG-69 contains the rules and capacity tests that must
be met by a safety valve manufacturer to obtain the ASME Code
Symbol. Section I, Appendix A, paragraphs A-12 to A-17 show
examples illustrating the method of checking safety valve capacity
by measuring the maximum amount of fuel that can be burned per
hour.
Calculate the required size and capacity of safety valves and
safety relief valves.
OBJECTIVE 3
-
Conforms with the 2004 ASME Extract Revised 03/06
64 Revised Second Class Course Section A1 SI Units
Using the formula
0.75 2558
C HW = (3.1) W = mass of steam generated (kg/hr) C = total mass
or volume of fuel burned/hr (kg or m3) H = heat of combustion of
fuel (kJ/kg) from A-17
The sum of the safety valve capacities marked on the valves
shall be equal to or greater than W. Example 6: Mass of steam
generated A boiler at the time of maximum firing uses 730 m3 of
natural gas per hour. The boiler pressure is 1550 kPa gauge. What
is the mass of steam generated? Solution From A-17, natural gas has
an H value of 35 700 kJ/m3.
0.75 2558
730 35 700 0.75 2558
of steam per hour (Ans.)
C HW = =
= 7641 kg
Paragraph A-17 lists some specific heating values for various
types of fuels. The heating value of the fuel must be known to
solve equations determining the mass of steam that can generated in
a boiler. Section I. Appendix A. A-44 states that the minimum
safety valve relieving capacity may be estimated on the basis of
the kilograms of steam generated per hour per square metre of
boiler heating surface and waterwall heating surface, as given in
Table A-44. Section I. Appendix A-46 lists three methods that can
be used to check the safety valve capacity if the capacity cannot
be determined.
An accumulation test with all valves shut. By measuring the
maximum amount of fuel that can be burned in the
boiler and using this to compute the maximum amount of steam
that can be generated.
By measuring the maximum amount of feedwater that can be
supplied to
the boiler under maximum firing rates calculating the maximum
volume of steam that can be generated.
-
Conforms with the 2004 ASME Extract Revised 03/06
65 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
The strength of a plain furnace depends on the length, the
diameter, and the square of the thickness. The strength of a
corrugated furnace depends on the diameter and thickness.
Corrugated furnaces have the following advantages over plain
furnaces:
1. Stronger than a plain furnace of the same dimensions. 2.
Better expansion allowance using corrugations or ribs. 3. More
surface area for the same length, therefore better heat
transfer.
Examples of the form of furnace tubes in use are shown below in
Fig. 7 with common dimensions.
Manufacturers use different methods to produce corrugations in
the furnace tube. These patented designs have advantages over a
plain tube furnace. Plain furnaces: are often found in heating
boilers because of the simplicity of construction and low cost. The
design temperature of the furnace is specified as
Calculate required wall thicknesses of plain circular furnaces,
circular flues, and corrugated furnaces.
OBJECTIVE 4
FURNACES
PLAIN with RINGS
ADAMSON
152.4 mm
38 mm
FOX
203 mm
32 mm
MORISON
R
(r < 1/2R)
FIGURE 7
Circular Furnace Designs
-
Conforms with the 2004 ASME Extract Revised 03/06
66 Revised Second Class Course Section A1 SI Units
2600 C, but no design temperatures are specified for other
components of the heating boiler. This is left to the
designer/manufacturer. The rules for power boilers are to be found
in Section I, paragraph PFT-15 and 51. Ring-reinforced furnaces:
are found in Section I and Section VIII-1 vessels. The rules are
found in Section I, paragraphs PFT-16 and 17. Corrugated furnaces:
The rules for corrugated furnaces were originally developed in
England in the late 1800s. Since that time, riveted seams and
joints have been replaced by fusion welding which has improved the
use of this type of furnace. Small plain circular portions have
been added to the corrugated furnace tube for ease of construction.
The rules are found in Section I, paragraph PFT-18. Combined plain
circular and corrugated furnaces: have been produced but must
conform to the rules set out in Section I, paragraph PFT-19.
Section I states that the thickness of a plain circular furnace may
not be less than 8 mm. Section IV states that the thickness may not
be less than 6 mm. The difference in thickness is due to heating
boilers being constructed for low pressures. Furnaces are subjected
to external pressure. Section II, Part D, Subpart 3, Appendix 3
(Basis for Establishing External Pressure Charts) explains how
these rules were developed. The external pressure is equal to the
compressive stresses and buckling can occur below the elastic limit
if the wrong material or wrong wall thickness is chosen for a
specific service. The equations developed for this are similar to
those developed for column theory, where different relationships
exist for critical load depending on the length of the column. ASME
has used two equations to evaluate critical buckling pressures to
produce graphical charts that simplify the calculations needed for
a safe design. The charts are found in Section II, Part D, Subpart
3, Figure G, representing the geometric properties of the cylinder,
and Section II, Part D, Subpart 3, Figures CS-1 to CS-6
representing the material properties for carbon steels Section I,
paragraph PFT-51 outlines the procedure to determine the maximum
allowable working pressure of tubes, flues, plain circular,
Adamson, and ring-reinforced furnaces of firetube boilers. The
symbols defined below are used in the formulas for plain
furnaces:
PLAIN FURNACES
-
Conforms with the 2004 ASME Extract Revised 03/06
67 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
A = factor determined from Section II, Part D. Fig. Gused to
enter the applicable material chart in Section II, Part D. For
cylinders having (Do/t) values less than 10, see PFT-51.1.2
(b).
B = factor determined from the applicable material chart in
Section II, Part
D, Subpart 3 for maximum design metal temperature (kPa). Do =
outside diameter of cylindrical furnace or tube (mm). L = total
length of the plain furnace taken as the distance from centre
to
centre of weld attachments (mm). P = external design pressure
(kPa). Pa = maximum allowable design pressure (kPa). t = minimum
required furnace wall thickness (mm).
Procedure Step 1: Assume a value of t and determine the ratios
L/Do and Do / t Step 2: Enter Fig. G (Section II, Part D, Subpart
3) at the value of L/Do (y-axis)
For values of L/Do greater than 50, enter the chart at a value
of L/Do = 50
For values of L/Do less than 0.05, enter the chart at a value of
L/Do = 0.05
Step 3: Move horizontally to the line for the value of Do/t
determined in Step 1
(interpolation may be made for intermediate values of Do/t).
From this point of intersection, move vertically downward to
determine the value of Factor A (x-axis).
Step 4: Using the value of A determined in Step 3, enter the
applicable material
chart in Section II, Part D for the material under
consideration. Move vertically to an intersection with the
material-temperature line for the design temperature (interpolation
may be made between lines for intermediate temperatures).
Step 5: From the intersection obtained in Step 4, move
horizontally to the right
and read the value of Factor B. Step 6: Using the value of B
determined in Step 5, calculate the value of the
maximum allowable external pressure Pa using the following
formula:
( )4
3 /a o
BPD t
= (4.1)
-
Conforms with the 2004 ASME Extract Revised 03/06
68 Revised Second Class Course Section A1 SI Units
Step 7: For values of A falling to the left of the applicable
material/ temperature line, the value of Pa shall be calculated
using the following formula:
( )2
3 /a o
AEPD t
= (4.2) Step 8: Compare the calculated value of Pa obtained in
Step 6 or 7 with P. If Pa is
smaller than P, select a larger value for t and repeat the
design procedure until a value of Pa is obtained that is equal to
or just greater than P.
Section IV, paragraph HG-312 allows the use of the following
modified formula:
( ) /a oBP
D t= (4.3)
Section VIII-1, paragraph UCS-28 provides examples in the
Non-Mandatory Appendix L-3 using the same procedure and charts as
above. Paragraph UCS-28 (c) requires the use of Section I, PFT-19
rules for corrugated shells subjected to external pressure. Example
7: Plain furnace - wall thickness A plain circular furnace 2.0 m
long and 750 mm outside diameter is designed for an external
pressure of 103 kPa at 260 C. The furnace is constructed of
SA-285-C carbon steel. What is the required thickness of the
furnace wall? Solution Step 1: Assume wall thickness t = 10 mm, Do
= 750 mm, and L = 2000 mm. Calculate the ratios.
2000 750
2.667
750 10
75
o
o
LD
Dt
==
==
Step 2: Use Section II, Part D, Chart Fig. G. Step 3: The value
of A = 0.0008. Step 4: Use the value of A in Section II, Part D,
Chart Fig. CS-2. Step 5: SA-285-C has an E value of 186 x 103 kPa.
Value of B = 9500
-
Conforms with the 2004 ASME Extract Revised 03/06
69 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
Step 6: Use equation (4.1)
( )4
3 /4 9500 3 (75)
168.89 kPa
ao
BPD t
==
=
Step 7: As this value is greater than 103 kPa, assume a new
thickness of 7.5 mm
and repeat the procedure. Step 1: Calculate the ratios.
2000 750
2.667
750 7.5
100
o
o
LD
Dt
==
==
Step 2: Use Section II, Part D, Chart Fig. G. Step 3: The value
of A = 0.0005. Step 4: Use the value of A in Section II, Part D,
Chart Fig. CS-2. Step 5: SA-285-C has an E value of 186 x 103 kPa.
Value of B = 6800 Step 6: Use equation 4.1
( )4
3 /4 6800 3 (100)
90.667 kPa
ao
BPD t
==
=
Step 7: As this value is less than 103 kPa, the thickness is
unacceptable. Assume
a new thickness of 8 mm and repeat the procedure. Step 1:
Calculate the ratios.
-
Conforms with the 2004 ASME Extract Revised 03/06
70 Revised Second Class Course Section A1 SI Units
2000 750
2.667
750 8
93.75
o
o
LD
Dt
==
==
Step 2: Use Section II, Part D, Chart Fig. G. Step 3: The value
of A = 0.00058. Step 4: Use the value of A in Section II, Part D,
Chart Fig. CS-2. Step 5: SA-285-C has an E value of 186 x 103 kPa.
Value of B = 7800 Step 6: Use equation 4.1
( )4
3 /4 7800
3 (93.75) 110.9 kPa
ao
BPD t
==
=
Step 7: The value of Pa is slightly greater than 103 kPa;
therefore, a thickness of
8 mm is required. (Ans.)
Section I, paragraph PFT-18 contains the rule for determining
the maximum allowable working pressure for the most common types of
corrugated furnaces, such as the Leeds suspension bulb, Morison,
Fox, Purves, and Brown, having plain portions at each end not
exceeding 230 mm in length. Use the following formula:
CtPD
= (4.4)
CORRUGATED FURNACES
-
Conforms with the 2004 ASME Extract Revised 03/06
71 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
P = maximum allowable working pressure (kPa). t = thickness
(mm)not less than 8 mm for Leeds, Morison, Fox, and
Brown and not less than 11 mm for a Purves furnace. D = mean
diameter (mm). C = a constantthe value depends on the type of
furnace. C = 119 for Leeds furnaces, when corrugations are not more
than 200
mm from centre to centre and not less than 57 mm deep.
C = 108 for Morison furnaces, when corrugations are not more
than 200 mm from centre to centre and not less than 32 mm deep, and
the radius of the outer corrugation r is not more than one-half of
the radius of the suspension curve R (See Fig. 7 and Fig.
PFT-18.1).
Note: The mean diameter of the Morison furnace may be taken as
the
least inside diameter plus 50 mm. C = 97 for Fox furnaces, when
corrugations are not more than 200 mm
from centre to centre and not less than 38 mm deep. C = 97 for
Purves furnaces, when rib projections are not more than
230 mm from centre to centre and not less than 35 mm deep. C =
97 for Brown furnaces, when corrugations are not more than 230
mm from centre to centre and not less than 41 mm deep. Example
8: Corrugated furnace - wall thickness A Brown corrugated furnace
of 1065 mm mean diameter, fitted with plain end, 216 mm in length
is required to operate at a pressure of 860 kPa. The corrugations
are 222 mm from centre to centre and 41 mm deep. What is the
required thickness of the furnace wall? Solution Use equation
4.4
P = 860 kPa (0.86 MPa) D = 1065 mm C = 97 (Brown furnace with
corrugations not more than 230 mm from
centre to centre and not less than 41 mm deep).
CtPD
=
-
Conforms with the 2004 ASME Extract Revised 03/06
72 Revised Second Class Course Section A1 SI Units
So
0.86 1065 97
(Ans.)
PDtC
==
= 9.44 mm
The thickness calculated above is greater than the minimum
allowed thickness of 8 mm and is therefore acceptable. By carefully
following the procedures provided in Section I for calculating the
wall thickness of various furnace types, these calculations can be
a simple process.
-
Conforms with the 2004 ASME Extract Revised 03/06
73 Chapter 2 SME Code Calculations: Stayed Surfaces, Safety
Valves, Furnaces
The following questions provide the candidate with experience
using the ASME Codes. 1. A flat plate is stayed with welded
staybolts equally pitched both horizontally
and vertically. The plate is 12.2 mm thick and is made of
SA-285-B material. The maximum allowable pressure is 865 kPa, and
the operating temperature is 250 C. Calculate the pitch of the
stays.
2. Determine the maximum allowable working pressure in kPa for a
watertube
boiler drum. The drum plate thickness is 50.8 mm with an inside
radius of 500 mm. The longitudinal joint efficiency is 100%. The
material is SA-516-55 and the operating temperature is not to
exceed 300 C. The pitch of the boiler tube holes in the drum is 140
mm as shown in Fig 1. The diameter of the tube holes is 82.5
mm.
3. A boiler is to be converted from burning pulverized
semi-bituminous coal to
natural gas. At maximum load the boiler burns coal at a rate of
5.5 tonnes per hour. What is the maximum amount of natural gas that
can be burned per hour if the safety valves are re-rated to 3%
above their present setting?
4. A furnace is produced using the Fox corrugation system. The
furnace has a
mean diameter of 1118 mm and a maximum allowable working
pressure of 1375 kPa. The corrugations are 152.4 mm centre to
centre and have a suspension curve depth of 38 mm. The length of
the furnace is 2.5 m. The furnace material is carbon steel with a
minimum yield strength of 205 MPa.
(a) What is the minimum thickness of the furnace tube?
(b) What is the maximum allowable working pressure of a plain
furnace tube
2.5 m in length and 1118 mm outside diameter with the same
thickness?
CCHHAAPPTTEERR QQUUEESSTTIIOONNSS