# Code Calculation-ASME Section I

July 9, 2017 | Author: Devendra Bangar | Category: Pipe (Fluid Conveyance), Boiler, Sphere, Pressure, Structural Steel

#### Short Description

Descripción: Code Calculation-ASME Section I...

#### Description

Code Calculations – ASME Section I Section 2 – Module 2 Feb2004

Learning Outcome When you complete this learning material you will be able to: Using the 1983 ASME Section I Code Extract (SI), and related tables, calculate the design thickness and pressure of boiler tubes, drums, and piping, and calculate the capacities of safety valves.

Learning Objectives You will specifically be able to complete the following tasks: 1. Given the tube material specification numbers, and other necessary parameters, use the formulae in PG-27.2.1 to calculate either the minimum required wall thickness or the maximum allowable working pressure for a boiler tube. 2. Given the material specification, construction method, and other necessary parameters, use the formulae in PG-27.2.2 to determine the required thickness and or maximum working pressure for boiler drums, headers, or piping. 3. Given the required specifications and operating conditions, use formula PG-29.1 to calculate the required thickness of a seamless, unstayed dished head. 4. Given the required specifications and operating conditions, use the formulae in paragraphs PG-29.11 and PG-29.12 to calculate the required thickness of an unstayed, full-hemispherical head. 5. Given the design and steam generation capacity of a boiler, use information in paragraphs PG-67 to PG-71 to calculate the minimum relieving capacity of the boiler safety or relief valve.

Note: All information and questions posed in this module are in SI units, per ASME 1983 edition and will be referenced to the SI Extract, as supplied with the course.

PE3-2-2-29

OBJECTIVE 1 Given the tube material specification numbers, and other necessary parameters, use the formulae in PG-27.2.1 to calculate either the minimum required wall thickness or the maximum allowable working pressure for a boiler tube.

SYMBOLS USED IN THE FORMULAE OF PG-27 The symbols in the formulae to be used in this module are found in Paragraph PG-27.3 and are defined as follows. It is extremely important that the correct units be applied when performing the calculations: t

=

minimum required thickness (millimetres, mm). (Also see PG-27.4, Note 7)

P

=

maximum allowable working pressure (megapascals, MPa). (Note - this refers to gauge pressure)

D

=

outside diameter of cylinder (millimetres, mm)

R

=

inside radius of cylinder (millimetres, mm)

E

=

efficiency of longitudinal welded joints or of ligaments between openings, whichever is lower. The values allowed for ‘E’ are listed in PG-27.4, Note 1. This is a factor that has no units, (for example, the value of ‘E’ for seamless cylinders is 1.00)

S

=

maximum allowable stress value, at the operating temperature of the metal, as listed in the Table PG-23.1, (megapascals, MPa). See PG-27.4, Note 2. The tables are located in an Appendix near the back of the 1983 Code Extract. (For example, the max. allow. working stress for SA-192, at 400°C, is 73 MPa)

C

=

minimum allowance for threading and structural stability, (millimetres, mm). See PG-27.4, Note 3

e

=

thickness factor for expanded tube ends (millimetres, mm). See PG-27.4, Note 4

y

=

a temperature coefficient: This factor has no units and has a value between 0.4 and 0.7. The values allowed for y are listed in PG-27.4, Note 6, (for example, for ferritic steel at 550°C, the value of ‘y’ is 0.7)

PE3-2-2-30

BOILER TUBE CALCULATIONS To calculate the required minimum wall thickness or the maximum allowable working pressure of ferrous boiler tubing, up to and including 127 mm O.D., the following formulae, as given in PG-27.2.1, are used: Minimum thickness: t

=

PD + 0.005 D + e 2S+P

⎛ 2 t - 0.01 D - 2 e ⎞ Maximum pressure: P = S ⎜⎜ ⎟⎟ ⎝ D - ( t - 0.005 D - e ) ⎠ Example 1 (to find tube wall thickness):

Calculate the minimum required wall thickness of a superheater tube. The tube is 76 mm O.D. and is connected to a header by strength welding. The maximum allowable working pressure is 4150 kPa gauge and the average tube temperature is 400°C. The tube material is alloy steel with specification SA-213-T11. Solution:

From PG-27.2.1, the formula to use is:

t =

PD + 0.005 D + e 2S+P

First, find all the required factors for the formula: given that: P = 4150 kPa = 4.15 MPa D = 76.0 mm from Codes:

e = 0 (from PG-27.4, Note 4, strength welded) S = 103 MPa = stress value at 400°C for SA-213, T-11 *

*Note: This 103 MPa value for S is found in Table PG-23.1. First locate the specification number, SA-213 T11, in the left column under the headings “Spec. Number” and “Grade or Class” (page 96 of extract). Then scan across the table to the “400” column under “For Metal Temperatures Not Exceeding °C” The corresponding value is 103 MPa.

PE3-2-2-31

Now, complete the calculation by substituting all factors into the formula:

t =

= =

4.15 × 76 + 0.005 × 76 + 0 2 × 103 + 4.15 315.4 + 0.38 210.15 1.88 mm (Ans.)

Information concerning the type of material used and the construction of the tube can be found in PG-9. The student should check PG-6 and PG-9 before starting calculations. The information in these sections will direct the student to the correct section of Table PG-23.1 by indicating if the metal is carbon steel, low alloy steel, or high alloy steel. PG-6 deals with steel plate, PG-9.1 deals with boiler tubes or pressure containing parts, PG-9.2 deals with all superheater parts. These sections will also help to correctly select the values for E and e (as per PG-27, Note 1 and Note 6). Note: This value for the thickness of the tube is exclusive of manufacturer’s tolerances. (See PG-16.5) Example 2 (to find maximum allowable working pressure):

Calculate the maximum allowable working pressure, in kPa, for a watertube boiler tube, which is 73.5 mm O.D. and has a minimum wall thickness of 4.71 mm. The tube is strength-welded into place in the boiler and is located in the furnace area of the boiler. Tube material is carbon steel, SA-192, with a mean wall temperature of 280°C. Solution:

From PG-27.2.1, the formula to use is: ⎛ 2 t - 0.01 D - 2 e ⎞ P = S ⎜⎜ ⎟⎟ ⎝ D - ( t - 0.005 D - e ) ⎠ First, find all the required factors for the formula: given that:

t = 4.71 mm D = 73.5 mm

PE3-2-2-32

from Codes:

e = 0 (from PG-27.4, Note 4) S = 79 MPa = stress value at 375 °C for SA-192, carbon steel *

*Note: This 79 MPa value for ‘S’ is found in Table PG-23.1. First, note that PG-27.4 states “tube temperature will not be taken as less than 370°C when absorbing heat”. Since this tube is in the furnace, it is absorbing heat. Now, find SA-192 in the table and scan across to find the temperature. You’ll notice that there is no column for 370°C, so take the next higher temperature, which is 375°C. Use the value of 79 MPa from this column. Note: In general, when a temperature given in a problem does not appear in Table PG-23.1, select the next higher temperature from the table DO NOT INTERPOLATE BETWEEN VALUES.

Now, substitute the values of all factors into the formula: P = 79 × (2 × 4.71) – (0.01 × 73.5) – (2 × 0) 73.5 – (4.71 – 0.005 × 73.5 – 0)

= 79 ×

9.42 – 0.735 – 0 73.5 – (4.71 – 0.3675)

= 9.921 MPa (Ans.) In both Example 1 and Example 2, the tubes were strength-welded into place. In this case the value of ‘e’ is zero. In calculations involving tubes expanded into place, the appropriate value of ‘e’ would be inserted into the formula. (See PG-27.4, Note 4)

Self-Test Problems 1. Calculate the minimum required wall thickness of a boiler tube, which is strength-welded to a header. The maximum allowable working pressure is 4450 kPa, and the mean wall temperature is 370°C. The tube material is SA-192 and the outside diameter is 50 mm. (Ans. 1.62 mm) 2. Calculate the maximum allowable working pressure for a watertube boiler tube 76 mm O.D. and 3.25 mm minimum thickness, which is strength-welded to the drum. Tube material is SA-192 and the tube temperature does not exceed 370°C. (Ans. 6.2 MPa)

PE3-2-2-33

OBJECTIVE 2 Given the material specification, construction method, and other necessary parameters, use the formulae in PG-27.2.2 to determine the required thickness and or maximum working pressure for boiler drums, headers, or piping.

PIPING, DRUM and HEADER CALCULATIONS

PG-27.2.2 (see page 3 of the 1983 Section I Extract) gives the formulae that are used to calculate the required minimum thickness or the maximum allowable working pressure of ferrous piping, drums, and headers. The size of each component may be stated as the outside diameter or as the inside radius. The formulae that are applied differ in each case, and are as follows: To find the minimum thickness

• If the outside diameter is given, use:

t =

PD + C 2SE + 2yP

• If the inside radius is given, use:

t =

PR + C S E - (1 - y) P

To find the maximum working pressure

• If the outside diameter is given, use:

P =

2 S E (t - C) D - 2 y (t - C)

• If the inside radius is given, use:

P =

S E (t - C) R + (1 - y) (t - C)

PE3-2-2-34

Example 3 (to find the required thickness of a boiler drum):

Calculate the minimum required thickness, in mm, of a welded boiler drum having an inside diameter of 1.5 m. The drum welds are finished flush with the surface of the plate. The drum plate is carbon steel, SA-516-65, and the metal temperature will not exceed 250°C. The maximum allowable working pressure is 4500 kPa gauge. The efficiency of the ligaments between the tube holes is 0.5. Solution:

The inside diameter is given and therefore the formula from PG-27.2.2, for inside radius can be used can be used:

t = given that: P = R =

PR + C S E - (1 - y) P 4500 kPa gauge = 4.5 MPa D/2 = 0.75 m = 750 mm

E = 0.5 from Codes: C = 0 (from PG-27.4, Note 3: drum is larger than NPS=4) S = 112 MPa (Table PG-23.1 for SA-516-65 at 250°C) y = 0.4 (PG-27.4, Note 6, ferritic at temperature below 475°C) substituting these values into the equation:

t =

PR + C S E - (1 - y) P

=

4.5 MPa × 750 mm +0 112 MPa × 0.5 – (1 – 0.4) × 4.5 MPa

=

3375 MPa mm 56 MPa - 0.6 x 4.5 MPa

=

3375 MPa mm 56 MPa - 2.7 MPa

=

3375 MPa mm 53.3 MPa

= 63.32 mm (Ans.)

PE3-2-2-35

Example 4 (to find the maximum working pressure of a boiler drum):

Calculate the maximum allowable working pressure for a welded drum if the plates are 25 mm thick and of material SA-299. The inside diameter of the drum is 988 mm and the joint efficiency is 100%. Assume the steam temperature will not exceed 400°C. Solution:

The inside diameter is given and therefore the formula from PG-27.2.2, for inside radius, can be used;

P = given that:

S E (t - C) R + (1 - y) (t - C)

t = 25 mm R = D/2 = 494.0 mm E = 1.0 (PG-27.4, Note 3)

from Codes: C = 0 (from PG-27.4, Note 3) S = 108 MPa (Table PG-23.1 for SA-299 at 400°C) y = 0.4 (PG-27.4, Note 6, temperature less than 400°C) substituting these values into the equation:

P =

S E (t - C) R + (1 - y) (t - C)

=

108 MPa × 1(25 mm – 0) 494 mm + (1 – 0.4)(25 mm – 0)

=

2700 MPa mm 509 mm

= 5.305 MPa (Ans.)

PE3-2-2-36

Example 5 (to find the required thickness of a header):

Calculate the required thickness, in mm, of a superheater outlet header, operating at 500°C and having a maximum allowable working pressure of 17 MPa. The header material is SA-335-P7 and the outside diameter is 457.2 mm. Solution:

The outside diameter is given and therefore the formula from PG-27.2.2, for outside diameter should be used: PD t = + C 2SE + 2yP given that: P = 17.0 MPa D = 457.2 mm from Codes: C = 0 (from PG-27.4, Note 3) S = 63 MPa (Table PG-23.1 for SA-335-P7 at 500°C) y = 0.5 (PG-27.4, Note 6) E = 1.0 (PG-27.4 Note 1) substituting these values into the equation:

t =

PD + C 2SE + 2yP

=

17 MPa × 457.2 mm +0 (2 × 63 MPa × 1) + (2 × 0.5 × 17 MPa

=

7772.4 MPa mm 143 MPa

= 54.35 mm (Ans.)

PE3-2-2-37

Example 6 (to find the required thickness of a high-pressure boiler pipe):

Calculate the minimum thickness required for a seamless steel feedwater pipe of material SA-209, grade T1. The outside diameter of the pipe is 323.85 mm and the operating pressure and temperature are 5200 kPa and 500°C respectively. The pipe is plain-ended. Assume that the material is an austenitic steel. Note: Plain-end pipe is that which does not have its wall thickness reduced when joined to another pipe. For example, pipe lengths welded together rather than joined by threading are classed as plain-end pipes. Solution:

The outside diameter is given and therefore the formula from PG-27.2.2, for outside diameter should be used: PD t = + C 2SE + 2yP given that: P = 5.2 MPa D = 323.85 mm from Codes: C = 0 (from PG-27.4, Note 3; 4 inch nominal and larger) S = 69 MPa (Table PG-23.1 for SA-209-T1, at 500°C) y = 0.4 (PG-27.4, Note 6; austenitic steel at 500°C) E = 1.0 (PG-27.4 Note 1; seamless pipe as per PG-9.1) substituting these values into the equation:

t =

PD + C 2SE + 2yP

=

5.2 MPa × 323.85 mm +0 (2 × 69 MPa × 1) + (2 × 0.4 × 5.2 MPa)

=

1684.02 MPa mm 138 MPa + 4.16 MPa

= 1684.02 MPa mm 142.16 MPa = 11.85 mm (Ans.)

PE3-2-2-38

Note on Manufacturer’s Tolerance: The calculated thickness in Example 6 does not include the manufacturer’s tolerance. Since the manufacturing process does not produce absolutely uniform wall thickness, an allowance is added, which is called the manufacturing tolerance. This is usually done by increasing the minimum required thickness, as calculated in the formula, by 12.5%. Example 7 (for minimum thickness of steam piping):

Calculate the required minimum thickness (in mm) of steam piping which will carry steam at a pressure of 4300 kPa gauge and a temperature of 370°C. The piping is plain-end, 273.05 mm O.D.; the material is low alloy steel, SA-335 P11. A manufacturer’s tolerance of 12.5% must be added to the pipe. Solution:

The outside diameter is given and therefore the formula from PG-27.2.2, for outside diameter should be used: PD t = + C 2SE + 2yP given that: P = 4.3 MPa D = 273.05 mm from Codes: C = 0 (from PG-27.4, Note 3) S = 103 MPa (Table PG-23.1 for SA-335-P11, at 370°C) y = 0.4 (PG-27.4, Note 6; ferritic steel at 475°C) E = 1.0 (PG-27.4 Note 1) substituting these values into the equation:

t =

PD + C 2SE + 2yP

=

4.3 MPa × 273.05 mm +0 (2 × 103 MPa × 1) + (2 × 0.4 × 4.3 MPa)

=

1174.115 MPa mm = 5.61 mm 206 MPa + 3.44 MPa

multiply by 1.125 to add the manufacturers tolerance of 12.5%: t = 5.61 x 1.125 = 6.31 mm (Ans.)

PE3-2-2-39

Self-Test Problems 3. Calculate the minimum required plate thickness of a welded boiler drum having an inside radius of 935 mm and a maximum design working pressure of 9020 kPa. The plate material is SA-516 grade 70 and metal temperature does not exceed 320°C. Weld reinforcement on the longitudinal joints has been removed flush with the surface of the plate. (Ans. 72.96 mm) 4. Calculate the minimum thickness required for a welded steel pipe of material SA-209 grade T1b, plain end. The outside diameter of the pipe is 273.05 mm and the operating pressure and temperature are 2000 kPa and 400°C, respectively. (Ans. 3.04 mm) 5. A steam header between the boiler and first stop valve is to be fabricated of 6 inches NPS pipe. The material specification is SA-369 FPA seamless pipe. The operating pressure will be 8440 kPa at 420°C. The pipe will be joined by full penetration welds and will be fully radiographed. Calculate the minimum thickness of the pipe wall if the manufacturer’s tolerance is 12.5%. (Ans. 11.98 mm) 6. A 137 mm thick boiler drum is made of SA-515-70 steel and has a ligament efficiency of 0.66. If the steam temperature is 280°C and the inside diameter of the drum is 1.6 m, what will the maximum operating pressure be, in kPa? (Ans. 6500 kPa)

PE3-2-2-40

OBJECTIVE 3 Given the required specifications and operating conditions, use formula PG-29.1 to calculate the required thickness of a seamless, unstayed dished head.

The following Paragraphs from PG-29 must be considered when performing calculations on dished heads. •

Paragraph PG-29.1 states that the thickness of a blank, unstayed dished head with the pressure on the concave side, when it is a segment of a sphere, shall be calculated by the following formula:

t =

5PL 4.8 S E

The symbols in this formula are defined as follows: t

= minimum thickness of plates (mm)

P = maximum allowable working pressure (MPa) L = radius (mm) to which the head is dished, measured on the concave side mm S = maximum allowable working stress (MPa), using values Table PG-23.1 E = efficiency of the weakest joint used in forming the head (not including the joint that joins the head to the shell) to the shell) PW-12, Joint Efficiency Factors, states that for welded joints an efficiency of 1.0 (that is, 100%) may be used, provided all weld reinforcement on the joint is removed substantially flush with the surface of the plate. Otherwise a joint efficiency not to exceed 90% shall be used. Seamless heads have an efficiency of 100%. •

Paragraph PG-29.2 states: “The radius to which the head is dished shall be not greater than the outside diameter of the flanged portion of the head.” If two different portions of the head are dished to different radii, then the longer radius shall be used as the value of ‘L’ in the formula.

PE3-2-2-41

Paragraph PG-29.3 states that when a head, dished to a segment of a sphere, has a flanged-in manhole or access opening that exceeds 152 mm in any dimension, then its thickness must be 15 %, or 3.2 mm, whichever is greater, more than the thickness of a blank unstayed head as calculated by the formula in PG-29.1. Note: This would apply to a manhole such as is found on the end of a boiler drum.

Paragraph PG-29.6 states that no head, except a full-hemispherical head, shall be of lesser thickness than required for a seamless shell of the same diameter.

Paragraph PG-29.5 states that in the case of a dished head with a flanged-in manhole, if the dish radius ‘L’ is less than 80% of the diameter of the shell to which the head is attached, then, when calculating the thickness by:

t =

5PL 4.8 S E

the value of ‘L’ must be made equal to 80% of the shell diameter. In addition, the thickness thus calculated must be increased by the greater of 15% or 3.2 mm (PG-29.3) to compensate for the flanged-in manhole. This method of calculation will give the minimum thickness for any form of head having a flanged-in manhole. Example 8:

Calculate the thickness of a seamless, unstayed dished head with pressure on the concave side, having a flanged-in manhole 280 mm by 380 mm. The head has a diameter of 1235 mm and is a segment of a sphere with a dish radius of 1016 mm. The maximum allowable working pressure is 1380 kPa, the material is SA-285 C and the metal temperature does not exceed 204°C. Solution:

First thing to check: Is the radius of the dish at least 80% of the radius of the shell, per PG-29.5? 1016/1235 = 0.823 = 82.3 % This is greater than 80%, so the value of L in the formula will be 1016 mm. Use the formula from PG-29.1:

t =

5PL 4.8 S E

given that: P = 1.380 MPa L = 1016 mm (radius of the curvature of the sphere)

PE3-2-2-42

from Codes: S = 95 MPa (Table PG-23.1 for SA-285-C, at 250 °C) substituting these values into the equation:

t =

=

5PL 4.8 S E 5 × 1.380 × 1016 4.8 × 95 × 1

= 15.37 mm This would be the thickness of a blank head, that is a head with no manhole. In this case there is a manhole and it exceeds the 152 mm allowed by PG-29.3. Therefore, the thickness must be increased by 15% or by 3.2 mm whichever is greater. 15% of 15.37 mm = 0.15 x 15.37 = 2.306 mm But this is less than 3.2 mm, so the thickness must be increased by 3.2 mm. Therefore, the required thickness is: 15.37 mm + 3.2 mm = 18.57 mm (Ans.) Example 9:

Calculate the thickness of a seamless, blank unstayed dished head having pressure on the concave side. The head has a diameter of 1067 mm and is a segment of a sphere with a dish radius of 915 mm. The maximum allowable working pressure is 2068 kPa and the material is SA-285 A. The metal temperature does not exceed 250°C. Solution:

First thing to check: Is the radius of the dish at least 80% of the radius of the shell, per PG-29.5? 915/1067 = .857 = 85.7% This is greater than 80%, so the value of L in the formula will be 915 mm. Use the formula from PG-29.1:

t =

5PL 4.8 S E

PE3-2-2-43

given that: P = 2.068 MPa L = 915 mm (radius of the curvature of the sphere) from Codes: S = 78 MPa (Table PG-23.1 for SA-285, at 250°C) E = 1.0 (for seamless heads) substituting these values into the equation: t = 5 × 2.068 × 915 4.8 × 78 × 1

= 25.27 mm From PG-29.6, the head in this example must be as thick as, or thicker than, a seamless shell of the same diameter. Therefore, before we can confirm that the calculated thickness of 25.57 mm is adequate, we must determine the shell thickness. Calculate the shell thickness using the appropriate formula from PG-27.2.2 where:

C = 0 y = 0.4

then:

t = =

PD + C 2SE + 2yP 2.068 × 1067 +0 2 × 78 × 1.0 + 2 × 0.4 × 2.068

= 14.00 mm Since 25.27 is greater than 14.00 mm, the head thickness of 25.27 mm, as calculated before, is adequate.

PE3-2-2-44

Example 10:

Calculate the thickness of the head in Example 9 if it has a flanged-in manhole. Solution:

The thickness of the blank head in Example 2 is equal to 25.27 mm. According to PG-29.3, this thickness must be increased by the greater of 3.2 mm or 15%. 25.27 mm x 0.15

= 3.79 mm

Since this is greater than 3.2 mm, increase the thickness by 3.79 mm: Head thickness = 25.27 mm + 3.79 mm = 29.06 mm (Ans.)

Self–Test Problems 7. Calculate the thickness required for a dished seamless head, which is attached to a boiler having a shell diameter of 1200 mm. The head has a flanged-in manhole with one dimension equal to 160 mm. The head is a segment of a sphere with a dished radius of 1120 mm. The head material is SA-285 Grade C, the maximum allowable working pressure is 1930 kPa and the steam temperature does not exceed 260°C. (Ans. 27.26 mm) 8. Calculate the thickness of a seamless blank unstayed dished head having pressure on the concave side. The head has a diameter of 830 mm and is a segment of a sphere with a dish radius of 615 mm. The maximum allowable working pressure is 1650 kPa, the material is SA-299 and the metal temperature does not exceed 200°C. (Ans. 8.78 mm)

PE3-2-2-45

OBJECTIVE 4 Given the required specifications and operating conditions, use formulae in paragraphs PG-29.11 and PG-29.12 to calculate the required thickness of an unstayed, fullhemispherical head.

When a boiler head is in the form of a complete hemi-sphere, termed “full-hemispherical”, the requirements of Paragraph PG-29.11 apply. This paragraph states that the minimum required thickness for a blank, unstayed, full-hemispherical head with the pressure on the concave side shall be calculated by one the following two formulae:

t =

PL 1.6 S E

(1)

t =

PL 2 S E - 0.2 P

(2)

Formula 1 is normally used. However, formula 2 may be used if the head exceeds 13 mm thickness and is used for shells or headers that are designed according to PG-27.2.2, and if the head is attached by fusion welding or is integrally formed on a seamless shell. •

Paragraph PG-29.12 states if a flanged-in manhole, meeting code requirements, is placed in a full-hemispherical head, then the thickness of the head is calculated using the same formula as for a head dished to the segment of a sphere (per PG-29.1), with a dish radius equal to 80% of the shell diameter and with the added thickness for the manhole. That is, the following formula is used, where the value of ‘L’ in the formula is 80% of the diameter of the shell.

t =

5PL 4.8 S E

Example 11:

Calculate the minimum required thickness, in mm, for a blank, unstayed, full-hemispherical head, with the pressure on the concave side. The head is fabricated from seamless material and is double butt welded to the shell. All reinforcement is removed and fully radiographed. The radius to which the head is dished is 700 mm, maximum allowable working pressure is 4000 kPa, and the head material (SA-285 C) will not reach a temperature greater than 340°C

PE3-2-2-46

Solution:

Use the formula from PG-29.11:

t =

PL 1.6 S E

given that: P = 4.0 MPa L = 700 mm (radius of the curvature of the head) from Codes: S = 95 MPa (Table PG-23.1 for SA-285, at 350 C) E = 1.0 substituting these values into the equation: t =

4.0 × 700 1.6 × 95 × 1

= 18.42 mm (Ans) Example 12:

A seamless, welded, full-hemispherical head is welded to a boiler shell that has an inside diameter of 1100 mm. Maximum working pressure is 3500 kPa, the material is SA-226, and operating temperature is 300°C. The head has a flanged in manhole that meets code requirements. Calculate the minimum required thickness for the head. Solution:

With this question PG-29.12 applies. Therefore use the formula from PG-29.1: (per PG-29.12)

t =

5PL 4.8 S E

given that: P = 3.5 MPa from codes: S = 81 MPa (Table PG-23.1 for SA-226, at 300°C) E = 1.0 L = 880 mm (80% of 1100 mm per PG-29.1)

PE3-2-2-47

substituting these values into the equation:

t =

5 × 3.5 × 880 4.8 × 54 × 1

= 39.61 mm From PG-29.3, due to the manhole, this thickness calculation must be increased by the greater of 15% or 3.2 mm. 39.61 mm x 0.15 = 5.94 mm Since this is greater than 3.2 mm, the thickness must be increased by this amount: t = 39.61 mm + 5.94 mm = 45.55 mm (Ans.)

Self–Test Problems 9. Calculate the minimum required thickness for an unstayed full-hemispherical head with the pressure on the concave side if the head has the following specifications: Inside diameter = 1.0 m Pressure = 1500 kPa Temperature = 285°C Plate specification is SA-285 C The head is fabricated from seamless material and is double butt welded to the shell. All weld reinforcement is removed and has a flanged-in manhole that complies with the code. (Ans. 45.39 mm) 10. What is the minimum required thickness for a blank, full-hemispherical head if the material of construction is SA-515-65, operating temperature is 425°C, pressure is 1800 kPa, and the head is dished to a radius of 870 mm? (Ans. 12.23 mm)

PE3-2-2-48

OBJECTIVE 5 Given the design and the steam generation capacity of a boiler, use information in paragraphs PG-67 to PG-71 to calculate the minimum relieving capacity of the boiler safety or relief valve.

PG-67 – PG-72: SAFETY VALVES (and Safety Relief Valves)

Paragraphs PG-67 to PG-72 of ASME Code, Section 1, deal with safety valves and safety relief valves. In particular, these sections cover the following topics: •

PG-67 Boiler Safety Valve Requirements: the types and numbers of safety valves required on the various types of boiler (that is, the boiler proper)

PG-68 Superheater Safety Valve Requirements: locations and capacities of superheater and reheater safety valves

PG-69 Testing: rules for the testing of safety valve capacities by manufacturers

PG-70 Capacity: methods and requirements for relieving capacity of safety valves

PG-71 Mounting: required methods for attaching safety valves to boilers

PG-72 Operation: guidelines for the operating ranges of safety valve popping and blowdown pressures

SAMPLE EXCERPTS RE SAFETY VALVES

The code requirements for safety and safety relief valves contain some very specific and technical data The requirements differ significantly between different types of boilers, with special references being made to specific types of boilers, such as electric, waste heat, oncethrough, high-temperature water boilers, and organic fluid vaporizer generators. All rules cannot be covered here, and the student must review the sections to understand where the special mentions are made. However, the following are a sample of some of the rules, with respect to power boilers. Each sample is only a partial quote of its respective paragraph and the student is advised to read the entire paragraph in the 1983 Section I Extract or in the Code itself. •

PG-67.1: “Each boiler shall have at least one safety valve or safety relief valve and if it has more than 46.4 m2 of bare tube water-heating surface …….it shall have two or more safety valves or safety relief valves…”

PE3-2-2-49

PG-67.2: “The safety valve capacity for each boiler shall be such that the safety valve, or valves, will discharge all the steam that can be generated by the boiler without allowing the pressure to rise more than 6% above the highest pressure at which any valve is set and in no case more than 6% above the maximum allowable working pressure. The safety valve capacity shall in compliance with PG-70 but shall not be less than the maximum designed steaming capacity as determined by the manufacturer..”

PG-68.2: “The discharge capacity of the safety valve, or valves, on an attached superheater may be included in determining the number and size of safety valves for the boiler ………. provided the discharge capacity of the safety valve, or valves, on the boiler, as distinct from the superheater, is at least 75% of the aggregate valve capacity required.”

PG-68.4: “Every reheater shall have one or more safety valves ……. The capacity of reheater safety valves shall not be included in the required relieving capacity for the boiler and superheater.”

PG-69.1: “Capacity test data reports for the initial certification of each valve model, type, and size, signed by the manufacturer and authorized observer witnessing tests, shall be submitted to the National Board of Boiler and Pressure vessel Inspectors for certification.”

PG-69.2: [paraphrased] (for a particular safety valve design). Tests shall be made (by the manufacturer) to determine the lift, popping, and blowdown pressures and capacities …….. A coefficient (of discharge) shall be established as follows:

K =

actual steam flow theoretical steam flow

The average K from the tests will be taken as the coefficient of design and shall be used for determining the relieving capacity of all sizes and pressures of the design, in the following formula: For flat seat valves: W = (0.00525 x π D L P x K) x 0.90 where:

W D L P K

= = = = =

mass of steam/h (kg) seat diameter (mm) lift at 103% of set pressure (mm) (1.03 x set gauge pressure) + 100 (kPa abs) average coefficient of discharge

Note: There are other formulae for 45 deg seats and for nozzle-type safety valves.

PE3-2-2-50

PG-70: CAPACITY

PG-70.1 states that “the minimum safety valve or safety relief valve relieving capacity (for other than electric boilers, waste heat boilers, organic fluid vaporizer generators, and forcedflow steam generators with no fixed steam and water line) shall be determined 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 the Table PG-70.” Table PG-70 is as follows: TABLE PG-70 MINIMUM KILOGRAMS OF STEAM PER HOUR PER SQUARE METRE OF SURFACE Firetube Watertube Boilers Boilers Boiler heating surface: Hand fired

25

30

Stoker Fired

35

40

Oil, gas, or pulverized fuel fired

40

49

Hand fired

40

40

Stoker fired

49

59

Oil, gas, or pulverized fuel fired

69

79

Waterwall heating surface:

PG-70.1 also states that “the minimum safety valve or safety relief valve relieving capacity for electric boilers shall be 1.6 kg/h/kW input. Example 13:

A stoker-fired firetube boiler has 62 m2 of heating surface. How much steam must the safety valve on this boiler be capable of discharging per hour? Solution:

From Table PG-70, a stoker-fired firetube boiler must have a safety valve that capacity of 35 kg/h per metre of heating surface. therefore:

Capacity (kg/h) = heating surface (m2) x 35 kg/h/m2 = 62 m2 x 35 kg/h/m2 = 2170 kg/h (Ans.)

PE3-2-2-51

Example 14:

A watertube boiler is gas-fired and has 65 m2 of boiler heating surface, plus 85 m2 of waterwall surface. What is the minimum required relieving capacity for all the safety valves? Solution:

From PG-70, a gas fired watertube boiler must have safety valve capacity of 49 kg/h for each m2 of boiler surface, plus 79 kg/h for each m2 of waterwall surface. Therefore:

Total capacity

= capacity for boiler + capacity for waterwalls = (65m2 x 49 kg/h/m2) + (85m2 x 79 kg/h/m2) = 3185 kg/h + 6715 kg/h = 9900 kg/h (Ans.)

Example 15:

A watertube boiler, equipped with a superheater, has two safety valves on the steam drum and one safety valve on the superheater. The boiler is fired on pulverized coal and has 70 m2 of boiler surface, 95 m2 of waterwall surface, and 20 m2 of superheater surface. What is the minimum combined relieving capacity permitted for the steam drum safety valves? Solution:

From PG-70, a pulverized-fired watertube boiler must have a safety valve capacity of 49 kg/h for each m2 of boiler surface, plus 79 kg/h for each m2 of waterwall surface. Therefore:

Total capacity = capacity for boiler + capacity for waterwalls = (70 m2 x 49 kg/h/m2) + (95 m2 x 79 kg/h/m2) = 3430 kg/h + 7505 kg/h = 10935 kg/h

But, from PG-68.2, the boiler safety valves must have a minimum of 75% of the total capacity. So: Min. capacity of boiler valves = 10935 kg/h x .75 = 8201 kg/h (Ans.) Please note: Superheater area is NOT included in heating surface for capacity calculations

PE3-2-2-52