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The tubesheet of a firetube boiler is usually a flat plate. The tubesheet of awatertube boiler is part of the boiler drum. Single openings in circular vesselshave 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 foundin Section I Paragraph PG-52, Section IV Paragraph HG-350 and Section VIIIParagraph UG-53. The ligament rules only consider the material between theholes and do not consider the tube material wall thickness. The value of theligament efficiency found by these rules is used in the determination of theminimum required thickness and/ or the maximum allowable working pressurefor cylindrical components under internal pressure found in Paragraph PG-27and Paragraph UG-27

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

Longitudinal: located between the front and lengthwise holes along thedrum.

Circumferential: located between the holes and encircle the drum.

Diagonal: a special case because they are located between the holes and areoffset 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 notspaced to exceed two diameters centre to centre.

O O B B JECTIVE 3 3

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

OBJECTIVE 2

INTRODUCTION

LIGAMENTS

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The following symbols are used in the formulae for calculating ligamentefficiency:

P = longitudinal pitch of adjacent openings (mm)

p /

= diagonal pitch of adjacent openings (mm)p 1 = pitch between corresponding openings in a series of symmetricalgroups of openings (mm)

d = diameter of openings (mm)n = number of openings in length p 1 E = ligament efficiency

Use the formula:- p d

E p

= (2.1)

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

Use the formula:1

1

- p nd E

p= (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 inFig. 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 thefollowing formula:

/

/

- p d E

p 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 SectionVIII-1.

140 140 140 140 140 140 140

Longitudinal Line

FIGURE 1

Example of tubespacing with hole pitchequal in every row

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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 tubesheetUsing 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 patternshown in Fig.4 above and in Fig. 5 below. The openings are 63.5 mm diameteron a staggered pattern of three longitudinal rows on 76 mm circumferentialspacing and 116 mm longitudinal spacing. The maximum allowable workingpressure is 4100 kPa at a temperature of 250 C. Drum material is SA-516-55 andthe tube material is SA-209-T1. The openings are not located in or near anybutt-welded joint.

FIGURE 2

Example of tub espacing with hole pitchunequal in every secondrow

FIGURE 3

Example of tub espacing with hole pitchvarying in every secondand third row

FIGURE 4

Example of tub espacing with tub e holeson diagonal lines

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Solution

2 2Diagonal pitch 58 76

3364 5776

9140

95.6 mm

X = +

= +

=

=

Hole diameter (d) = 63.5 mmLongitudinal pitch (p) = 116 mm

Use equation 2.1-

116 - 63.5

1160.4526

p d E

p=

=

=

/ 95.6

116

0.824

p p

=

=

The point corresponding to these values on the diagram in Fig. PG-52.1, readfrom the y-axis, is 38%. As the point falls below the line of equal efficiency forthe 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

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Section I , paragraph PG-27.2.2. (2001)

( )

- 1 -PR

t C SE y P

= + (2.4)

P = 4100 kPa or 4.1 MPaR = 460 mmS = 108 MPa at 250 C for SA-516-55E = 0.38 as determined aboveC = 0y = 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 C

SE y P= +

=

=

=

= 48.885 mm

The minimum thickness of the drum shell would be 48.885 mm without anyallowance for manufacture or corrosion.

Note : The minimum thickness of this drum, plain, without being drilled fortubes would be 17.836 mm. Therefore, the drum could be manufacturedfrom two half shells; the tube sheet half being 48.885 mm thick, and thedrum half being 17.836 mm thick as shown in Fig. 6. Each half wouldmeet the conditions of rule PG-27.2.2.

DRUM

TUBESHEET

FIGURE 6

Example: Thickness of drum tu besheetDrum manufacturedfrom two half shells