Asme Calculations Examples Efficiency in Holes

Revised 03/06 to conform with the 2004 ASME

CH A PT ER 2

SME Code Calculations: Stayed Surfaces, Safety Valves, Furnace

Here is what what you w i l l be able to do when when you complete complete each objecti objective: ve:

1. Calculate the required thickness and design pressure pressure for braced braced and s surfaces in pressure vessels and the minimum required cross-sectiona of a stay. 2. Calculate the ligament efficiency method for two or more openings in pressure boundary of a pressure vessel. 3. Calculate the required size and capacity of safety valves and safety valves.

4. Calculate required wall thicknesses thicknesses of plain circular furnaces, circular circular f and corrugated furnaces.

Chapter 2 • SME Code Calculations: Stayed Surfaces, Safety Valves, Furnaces

INTRODUCTION 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 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. Longitudinal Stress =

vessel diameter × internal pressure vessel wall thickness × 2

Circumferential Stress =

vessel end area x internal pressure 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.

Longitudinal Stress =

= =

vessel diameter × internal pressure

vessel wall thickness × 2 4500 ×1.350 2 × 35 86.79 MPa

Circumferential Stress = π

=

4

vess vessel el end end area area × inter interna nall press pressure ure vessel circumference × wall thickness

× 4500 2 ×1.35

× 4500 × 35 4500 ×1.35 = 4 × 35 π

= 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.

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Revised Second Class Course • Section A1 • SI Units

OBJECTIVE O B BJ J ECTIVE ECTIVE 3 1

Calculate the required thickness and design pressure for braced a stayed surfaces in pressure vessels and the minimum required cross-sectional area of a stay.

RELEVANT ASME CODE SECTIONS

Section I: The rules for stayed surfaces and staybolts can be found in Para PG-46 to Paragraph PG-49, Paragraph PW-19 and Paragraph PFTParagraph PFT-32.

Section IV: The rules for stayed surfaces and staybolts can be fou Paragraph HG-340 and Paragraph HW-710 to HW-713.

Section VIII-1: The rules for stayed surfaces and staybolts can be fou 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 it is desira desirable ble or possible possible to reduce the span and/ or the th e thickness thickness of a tube or other pressure component. Opposite surfaces are tied together by stay tubes, or baffles that carry the pressure loading in tension. Because be moments, bending strength, and the tensile strength of the stays now res pressure loading, the required thickness of stayed surfaces may be less tha of surfaces which are not stayed.

FLAT STAYED SURFACES

The equation for flat-stayed surfaces is an adaptation of the flat head equ with the diameter replaced by the distance stays. In this case, the represents the degree of restraint to rotation that the stay attachment provid

The design pressure and thickness for stayed plates are calculated b following following formulae: 2

t SC


t = minimum thickness of plate (mm) p = maximum maximum pitch measured between straight lines passing through thr ough the th e centres of the staybolts in different rows. The lines may be horizontal and vertical, or radial and circumferential (mm). P = maximum maximum allowable allowable working working pressure or o r internal inter nal design design pressure p ressure (kPa). S = maximum maximum allowable allowable stress (MPa)—given (MPa)—given in Table 1A of Section II, I I, Part D, or in Tables HF-300.1 and HF-300.2 in Section IV. C = a constant—the constant— the value depends on detail d etailss of the staybolt end design design as follows: C = 2.1 for welded welded stays stays or stays stays screwed screwed through thr ough plates not over o ver 11 mm in thickness with ends riveted over. C = 2.2 for welded stays stays or stays stays screwed screwed through thro ugh plates plates over 11 mm in thickness with ends riveted over. C = 2.5 for stays screwed screwed through plates and fitted with single 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 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 3.2 for stays fitted with inside inside and and outside nuts and outside washers where the diameter of the washers is not less than 0.4 p and the thickness is not less than the thickness (t ) of the surface being stayed.

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Example 2: Stayed flat plate - maximum pressure Calculate the maximum allowable working pressure on stayed flat plates 12. thick, with staybolts attached by fusion welding and pitched 154 mm horizo and vertically. vertically. The plate material is is SA-516-55. SA-516-55. The average temperat temp erat 0 200 200 C Solution: Use equation 1.1, see Section I, PG-46.1 2

P

=

t SC 2

p P = max m ax imum allowable work work ing pressure pressure (M Pa) t = 12.5 mm p = 154 mm S = 108 MPa C = 2.2 for welded welded stays stays or stays screwed through plates over 11 m thickness with ends riveted over. 2 t SC P = 2 p P = P =

12.52 ×108 × 2.2 1542 37193.75

23716 P = 1.565 MPa =

1565 kPa

Example 3: Stayed flat plate - thickness Calculate the minimum thickness for stayed flat plates, with staybolts sc through the plates and pitched 185 mm horizontally and vertically. The material is SA-204-A, maximum allowable working pressure 6205 kP operating temperature of 300 0 C. Solution Section I PG-46.1 t = p

where P = 6.205 MPa

P SC


t = p t = 185

P SC

6.205 128 × 2.2

t = 185 × 0.022 t = 185 × 0.148 t = 27.46 mm

The minimum required thickness of 27.46 mm is greater than 11 mm and C = 2.2 is the correct factor to use.

STAYS AND STAYBOLTS

The requirements are the same for Section I, Section IV and Section VIII-1, only Section Section I references will 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 stays may also also be used. This type of stay can be found foun d in the th e waterlegs waterlegs of locotype boilers. Corrosion Corro sion is likel likely y in this area, and if a stay stay corro corrodes des then a 'telltale' '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 :

Stay load = pressure ( P) ×(pi (pitch tch area ( p2) −cross -sec -section ional area of stay tay ( a))

= P × ( p - a) 2

(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:

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Example 4: Minimum required area of stay Welded stays 30 mm diameter will be used to support a flat plate 16 mm The pressure is 1350 kPa. The stays are spaced 200 mm horizontally vertically. The steel used for the stays and plate is SA-192 at a max temperature of 300° C.

Does the stay diameter meet the Code requirements? Solution Stay diameter of 30 mm. (a) = 0.7854 0.7854 ×0.030 × 0.0302 = 0.70 0.707 7 × 10 -3 m P = 1350 1350 kPa p = 200 mm = 0.2 m S = 91.9 MPa = 91900 91900 kPa

Use equation 1.3 to determine the stay load. Stay Load

= P × ( p 2 - a) = 1350 × (0.2 2 - 0.707 x 10-3 ) = 53.046

Use equation 1.4 to determine the minimum required area of stay. Minimum required area of stay = 1.1 × Stay load/ S 53.046 / 91900 91900 = 1.1 × 53.046

= 0.6 × 10-3 m2 Use the equation below to determine the minimum diameter of the stay. Mini Minimu mum m diam diamet eter er of stay stay

=

0.6 0.6 ×10-3 / 0.78 0.7854 54

= =

0.764 × 10 -3 0.0276 m or 27.6 mm (Ans.)

The stated diameter of the stay is 30 mm; this is larger than the min required diameter of 27.7 mm; therefore, the stay diameter meets the requirements.

Section IV, Paragraph HG-346 states that the firetubes in a firetube boile be used as stays. The required thickness, maximum pitch and design pressu tubesheets with firetubes used as stays may be calculated by the foll


t =

p =

⎛ P ⎞ ⎛ 2 ⎜ CS ⎟ + ⎜ p ⎝ ⎠ ⎝

2 D ⎞

π

⎟

4 ⎠

⎛ CSt 2 ⎞ ⎛ π D 2 ⎞ ⎜ ⎟ +⎜ ⎟ P ⎝ ⎠ ⎝ 4 ⎠ 2

P =

CSt

⎛ π D 2 ⎞ p - ⎜ ⎟ ⎝ 4 ⎠ 2

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 welded to plates p lates not over 11 mm thick C = 2.8 for firetubes welded welded to plates p lates over 11 11 mm thick S = the maximum maximum allowable allowable stress values values given given in Section IV, I V, Tables HF300.1 and HF-300.2 kPa P = the design design press pr essure, ure, 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.

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OBJECTIVE O B BJ J ECTIVE ECTIVE 3 2

Calculate the ligament efficiency method for two or more openings the pressure boundary of a pressure vessel.

INTRODUCTION

The tubesheet of a firetube boiler is usually a flat plate. The tubesheet watertube boiler is part of the boiler drum. Single openings in circular v have been covered in Module 1. Multiple openings, such as to be foun tubesheet, present a different case and are covered by ligament rules to be in Section I Paragraph PG-52, Section IV Paragraph HG-350 and Section Paragraph UG-53. The ligament rules only consider the material betwee holes and do not consider the th e tube material wall wall thickness. thickness. The value value ligament efficiency found by these rules is used in the determination o minimum required thickness and/ and/ or the t he maximum maximum allowable allowable working working pr for cylindrical components under internal pressure found in Paragraph P and Paragraph UG-27

LIGAMENTS

A ligament is the area of metal between the holes in a tubesheet. The three t of ligame ligaments nts are:

Longitudinal: located between the front and lengthwise holes along th drum. Circumferential: located between the holes and encircle the drum.

Diagonal: a special case because they are located between the holes and offset at an angle to each other.

The rules of ligaments are applicable to groups of openings in cylind pressure parts that form a definite pattern. These rules also apply to opening spaced to exceed two diameters centre to centre.


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The following symbols are used in the formulae for calculating ligament efficiency: P = longitudinal pitch of adjacent openings (mm) p / = diagonal diagonal pitch of adjacent adjacent openings op enings (mm) (mm) p1 = pitch between b etween corresponding correspon ding openings in a series series of symmetrical symmetrical groups of openings (mm) d = diameter diameter of opening op eningss (mm) n = number of openings in length p1 E = ligament ligament efficiency

Use the formula: E

=

p - d

p when the pitch of the tubes on every row is equal (Fig. 1).

(2.1)

Use the formula: E

=

p nd 1 -

p1 when the pitch of the tubes on any one row is unequal (Figs. 2 and 3).

(2.2)

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 h oles along along a diagonal, diagonal, Section Section IV, paragraph paragraph HG H G -350.4 -350.4 provides the following following formula: E

=

p / - d

(2.3)

/

p F

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

1 40

14 0

FIGURE 1

Example of tube spacing with hole p equal in every row

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FIGURE 2 Example of tube spacing with hole pitch unequal in every second row

130 140

130 140 130 140 130 140

270 mm Longitudinal Line

FIGURE 3

1 30 13 0 14 0

130 140 130 130 140 130

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


160 mm

Example 5: Thickness of o f drum drum tubesheet tubeshee t Using the rule in Section I, determine the minimum thickness of a 920 mm (internal diameter) cylindrical drum that has a series of openings in the p shown in Fig.4 above and in Fig. 5 below. The openings are 63.5 mm dia on a staggered pattern of three longitudinal rows on 76 mm circumfe spacing and 116 mm longitudinal spacing. The maximum allowable w pressure is 4100 kPa at a temperature of 250° C. Drum material is SA-516-5 the tube material is SA-209-T1. The openings are not located in or nea butt-welded joint.


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Solution

1 16 m m

116 mm

FIGURE 5

Solution - Thicknes a drum tubesheet

76 X 76

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Longitudinal Line Diago Diagona nall pitch pitch X =

582 + 76 2

=

3364 + 5776

= =

9140 95.6 mm

Hole diameter (d) = 63.5 mm Longitudinal pitch (p) = 116 mm Use equation 2.1 p - d E = p

= =

116 - 63.5 63.5 116 0.4526

/

p

p

=

95.6

=

0.824

116

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 diagonal and longitudinal liga ligaments, ments, the diagonal diagonal ligament ligament is the th e weaker. weaker.

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Section I, paragraph PG-27.2.2. (2001) t

=

PR SE - (1 - y ) P

+ C

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 t

= = = = =

PR SE - (1 - y ) P

+ C

4.1 × 460 108 × 0.38 - (1 - 0.4 ) × 4.1 1886 41.04 41.04 - 2.46 2.46 1886 38.58 48.885 mm (Ans.)

The minimum thickness of the drum shell would be 48.885 mm withou allowance for manufacture or corrosion.

Note: The minimum thickness of this th is drum, plain, plain, witho without ut being dril tubes would be 17.836 mm. Therefore, the drum could be manufac from two half shells; the tube sheet half being 48.885 mm thick, a drum half being 17.836 mm thick as shown in Fig. 6. Each half w meet the conditions of rule PG-27.2.2.

FIGURE 6 DRUM

Example: Thickness of drum tubesheet Drum manufactured from two half shells


OBJECTIVE 3 Calculate the required size and capacity of safety valves and safety relief valves.

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 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 m 2 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

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Using the formula W

=

C × H × 0.75

2558

(3.1)

W = mass mass of steam generate generated d (kg/ (kg/ hr) C = total mass mass or volume volume of fuel fuel burned/ hr (kg or m 3) H = heat of combustion of fuel (kJ (kJ// kg) kg) from A-17 A-17

The sum of the safety valve capacities marked on the valves shall be equa greater than W .

Example 6: Mass of steam generated A boiler at the time of maximum firing uses 730 m 3 of natural gas per hou boiler pressure is 1550 kPa gauge. What is the mass of steam generated? Solution From A-17, A-17, natural gas gas has an H value of 35 700 700 kJ/ kJ/ m 3. W

= = =

C × H × 0.75

2558 730 × 35 700 × 0.75 2558 7641 kg of steam per hour (Ans.)

Paragraph A-17 lists some specific heating values for various types of fuel heating value of the fuel must be known to solve equations determining the of steam that can generated in a boiler.

Section I. Appendix A. A-44 states that the minimum safety valve rel capacity may be estimated on the basis of the kilograms of steam generate hour per square metre of boiler heating surface and waterwall heating surfa given in Table A-44. Section I. Appendix A-46 lists three methods that can be used to check the 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 th boiler and using this to compute the maximum amount of steam tha be generated.


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OBJECTIVE 4 Calculate required wall thicknesses of plain circular furnaces, circular flues, and corrugated furnaces.

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.

FIGURE 7

152.4 mm

Circular Furnace Designs

38 mm FOX PLAIN with RINGS

203 mm 32 mm

R

(r < 1/2R) ADAMSON

MORISON

Manufacturers use different methods to produce corrugations in the furnace

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260 2600 C, but no design temperatures are specified for other components heating boiler. This is left to the th e designer/ designer/ manufacturer. The rules for power boilers a found in Section I, paragraph PFT-15 and 51.

Ring-reinforced furnaces : are found in Section I and Section VIII-1 v The rules are found in Section I, paragraphs PFT-16 and 17.

Corrugated furnaces : The rules for corrugated furnaces were developed in England in the late 1800s. Since that time, riveted seams and have been replaced by fusion welding which has improved the use of this ty furnace. Small plain circular portions have been added to the corrugated fu tube for ease of construction. The rules are found in Section I, para PFT-18.

Combined plain circular and corrugated furnaces : have been produc must conform to the rules set out in Section I, paragraph PFT-19.

PLAIN FURNACES

Section I states that the thickness of a plain circular furnace may not be les 8 mm. Section IV states that the thickness may not be less than 6 mm difference in thickness is due to heating boilers being constructed fo pressures.

Furnaces Furn aces are subjected subjected to external external pressure. Section Section II, Part D, D , Subp Subp Appendix 3 (Basis for Establishing External Pressure Charts) explains how rules were developed. The external pressure is equal to the compressive st and buckling can occur below the elastic limit if the wrong material or w wall wall thickness is is chosen for a specific specific service. service. The equations developed fo are similar to those developed for column theory, where different relation exist for critical load depending on the length of the column. ASME has two equations to evaluate critical buckling pressures to produce graphical that simplify the calculations calculations needed for a safe design. design. The charts are fou Section II, Part D, Subpart 3, Figure G, representing the geometric propert the cylinder, and Section II, Part D, Subpart 3, Figures CS-1 to CS-6 represe the material properties for carbon steels

Section I, paragraph PFT-51 outlines the procedure to determine the max


A = factor determined from Section II, Part D. Fig. G—used to enter the applicable material chart in Section II, Part D. For cylinders having (Do / t) values values less less than 10, see PFT-51.1.2 (b). (b). B = factor determined from the th e applicable applicable material chart in Section Section II, II , Part D, Subpart 3 for maximum design metal temperature (kPa). D o = outside diameter of cy cylindrica lindricall furnace or tube (mm). L = total tot al length of the plain plain furnace taken as the distance from centre to centre of weld attachments (mm). P = external design design pressure (kPa). (kPa). Pa = maximum maximum allowa allowable ble design design pressure (kPa). (kPa). t = minimum required required furnace wall wall thickness (mm). (mm).

Procedure Step 1: Assume a value of t and determine the ratios L / D o and D o / t y-axis) Step 2: Enter Fig. G (Section II, Part D, Subpart 3) at the value of L / D o ( y • For values of L / D o greater than 50, enter the chart at a value of L / D o = 50 • For values of L / D o less than 0.05, enter the chart at a value of L / D o = 0.05

Step 3: Move horizontally to the line for the value of D /o t determined in Step 1 Do// t ). (interpolation may be made for intermediate values of Do ). From this point of intersection, move vertically downward to determine the value x -axis). of Factor A ( x 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

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values of A falling falling to the th e left of the th e applicable applicable material/ material/ temper Step 7: For values line, the value of Pa shall be calculated using the following formula: Pa

=

2 AE 3 ( Do / t )

Step 8: Compare the t he calculated calculated value of Pa obtained in Step 6 or 7 with P smaller smaller than t han P, select a larger value for t and repeat the design proce until a value of Pa is obtained that is equal to or just greater than

Section IV, paragraph HG-312 allows the use of the following modified for Pa

=

B

( Do / t )

Section VIII-1, paragraph UCS-28 provides examples in the Non-Man Appendix L-3 using the same procedure and charts as above. Paragraph UC (c) requires the use of Section I, PFT-19 rules for corrugated shells subjec external pressure.

Example 7: Plain furnace - wall thickness A plain circular furnace 2.0 m long and 750 mm outside diameter is design an external pressure of 103 kPa at 260° C. The furnace is construct SA-285-C carbon steel. What is the required thickness of the furnace wall?

Solution Step 1: Assume wall thickness t = 10 mm, D o = 750 mm, and L = 2000 mm Calculate the ratios. L Do

=

2000 750

= 2.667 Do t

= =

750 10 75

Step 2: Use Section II, Part D, Chart Fig. G.


equation (4.1) Step 6: Use equation 4 B Pa = 3 ( Do / t )

= =

4 × 9500 3 × (75) 168.89 kPa

Step 7: As this value is greater than 103 kPa, assume a new thickness of 7.5 mm and repeat the procedure. p rocedure. Step 1: Calculate Calculate the th e ratios. ratio s. L 2000 Do

=

750

= 2.667 Do t

= =

750 7.5 100

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. B = 6800 6800 Step 5: SA-285-C has an E value of 186 x 10 3 kPa. Value of B

Step 6: Use equation 4.1 4 B Pa = 3 ( Do / t )

= =

4 × 6800 3 × (100) 90.667 kPa

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.

Calculate th ratio Step 1 Calculate

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L Do

=

2000 750

= 2.667 Do t

= =

750 8 93.75

Step 2: Use Section II, Part D, Chart Fig. G.

0.00058. Step 3: The value of A = 0.00058. A in Section II, Part D, Chart Fig. CS-2. Step 4: Use the value of A B = 7800 7800 Step 5: SA-285-C has an E value of 186 x 103 kPa. Value of B

Step 6: Use equation 4.1 4 B Pa = 3 ( Do / t )

= =

4 × 7800 3 × (93.75) 110.9 kPa

Step 7: The value of Pa is slightly greater than 103 kPa; therefore, a thickn 8 mm is required. (Ans.)

CORRUGATED FURNACES

Section I, paragraph PFT-18 contains the rule for determining the max allowable working pressure for the most common types of corrugated fur such as the Leeds suspension bulb, Morison, Fox, Purves, and Brown, h plain portions at each end not exceeding 230 mm in length.


P = maximum maximum allowa allowable ble working working pressure (kPa). (kPa). t = thickness (mm)—not (mm)—not less than 8 mm for Leeds, Leeds, Morison, Fox, and Brown and not less than 11 mm for a Purves furnace. D = mean diameter (mm). C = a constant—th constant—thee value value depends on the type of furnace. 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 Moriso Morison n fur furnaces naces , 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 corrugation r is not more than t han one-half one-half of the radius of the suspension curve R (See Fig. 7 and Fig. PFT-18.1).

diameter of the th e Morison Morison furnace furn ace may be taken as the Note : The mean diameter 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 Ex ample 8: Corr Corrugated ugate d furnace furnace - wall thickness thickne ss 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 (0.86 MPa) D = 1065 mm C = 97 (Brown furnace with corrugations not more than 230 mm from

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So t

= = =

PD C 0.86 × 1065

97 9.44 mm (Ans.)

The thickness calculated above is greater than the minimum allowed thickn 8 mm and is therefore acceptable.

By carefully following the procedures provided in Section I for calculatin wall thickness of various furnace types, these calculations can be a s process.


C H H A P T T E E R R Q U U E E S S T T I I O O N N S S 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) (a) What is the minimum thickness of the furnace tube? (b) What is the maximum allowa allowable ble working pressure of a plain furnace tube 2.5 m in length and 1118 mm outside diameter with the same thickness?

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Asme Calculations Examples Efficiency in Holes

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