Expansion & Piping Flexibility

Chapter 9

Piping flexibility Fundamentals Flexibility analysis Identifying lines with adequate flexibility Simplified flexibility analysis methods Computerized piping flexibility analysis Special considerations for specific piping systems

Piping Flexibility definition Piping flexibility is one of the most important, least understood functions of piping design. Today flexibility analysis, or stress analysis as it is often called, is delegated to a computer. Consequently, the piping designer’s responsibility is normally limited to a quick check to determine if the piping layout is within reasonable tolerances. If the quick check shows it is outside these limits, he turns the problem over to stress specialists. A stress specialist translates data to an input sheet, turns this over to the computer group and later receives a computer output sheet. Then if piping system is too rigid, the stress specialist may suggest corrective modifications; but the ultimate responsibility rests with the designer. A computer run, including formulation of input and diagnosis of output, is expensive but certainly warranted if quick check method suggests a need for it. However, thousands of dollars are wasted on computer analysis of lines of visibly adequate flexibility or of lines which would be accepted if quick check method were used. Piping designers who know nothing abut flexibility analysis are quick to request a full stress analysis rather than take a chance. Although several books could be written on piping stress analysis, this chapter must be limited to presenting a quick check method and explaining it so that all readers can understand and apply it to their problems. The competent piping designer will make every effort to provide adequate flexibility in this piping using a minimum number of fittings. When a quick check determines that the system is not flexible enough, he reviews the system to determine whether or not he can redesign, may be adding an elbow or two to increase flexibility. Then he uses the quick check method again. If his system proves adequate, he has saved the expense of full analysis. If the system is still slightly rigid and the designer believes the computer analysis may prove his system to be within flexibility allowances then he should ask for the stress analysis. The quick check method has a built-in safety factor; many arrangements which prove slightly stiff by the quick check method actually come out OK when full analysis is made.

Purpose of Analysis A hot piping system will expand or elongate. A cold piping system will contract or shrink. Both of these actions create stress problems. A stress analysis determines the forces at anchor points, stresses in the piping system, and bending moment at any point. For any of these factors an allowable is known. For any force generated at an anchor point, often an equipment nozzle, there must be at least an equal resisting force. If a system throws 20 000 pounds of force into an anchor designed to withstand 15 000 pounds of force, it will give. If the anchor happens to be an equipment nozzle this give means a rupture and possibly an explosion and fire. Before designing piping systems for adequate flexibility, the designer must know what forces are allowable.

Page 2 of 26

1. Piping Flexibility Design It is preferable to provide adequate flexibility in piping systems by using loops or other confiigurations constructed of pipes and fittings. Sometimes space or cost is prohibitive and movement must be absorbed by expansion joints such as the bellows, Barco ball or sliding type. Piping flexibility should always be achieved with the minimum number of anchors and guides feasible. Axial expansion joints must be guided on each side and anchored at the end of pipe runs to withstand hydrostatic testing thrust. U-type loops be anchored on both sides of the pipe run to work. Lines which are to be purged by steam or hot gas must be checked to make sure that they will be adequately flexible during the purging operation. Close relief systems and hot blowdown or pumpout system must also be carefully though out. Temperatures in start-up lines often surpass operating temperatures. Exchanger by-pass may still be cold while the inlet and outlet lines are already hot, resulting in excessive stresses. Always review systems at their worst operating conditions such as during start-up when a hot line will be feeding into a cold tower or vice versa. Unit systems, such as a steam system having a larger diameter header in the unit pipeway and smaller size branches feeding into equipment, are often provided with unnecessary and expensive expansion loops and anchors. Before loops are designed, every effort should be made to make the branches flexible enough to withstand header expansion. An anchor placed near the center of the run can be used to direct header expansion by forcing 50% of the expansion to either side of and away from the anchor, distributing expansion along the header and thereby simplifying branch flexibility. Flexibility must be considered from the begining in every system designed. During the original layout - which may be freehand on a scratch pad - flexibility must be uppermost in the piping designer’s mind. Aquick check should be made of all lines 8” or larger during this intial stage. If flexibility appears to be a problem, smaller lines should be reviewed and quick checked, too. If a line is too stiff and must be rerouted, the time to find out is early in the design. Many dollars and manhours can be wasted making finished drawings of piping systems which are too stiff. When this happens and is finally discovered, the designer must start over completely. One of the many differences between a draftsman and a designer is that while the draftsman can only draw the pipe, the designer knows flexibility, process, instrumentation, flow of fluids and many other specifics which must be considered during the piping layout and design stage.

Page 3 of 26

2. The Quick Check Method The purpose of the quick check method is to determine whether or not a piping system is adequately flexible without the formal calculations required for a full stress analysis. Ordinarily if a system is within the quick check guidelines, no further flexibility analysis is required. To make use of this method the piping designer must establish some basic facts. Anchor points must be known or assumed. Design pressures and temperatures, expansion coefficients for the metal involved and any branch or equipment restraints must be included. Special design conditions such as start-up, cyclic operation, steam tracing, etc. must be known. The lengths of pipe solved for are based on standard weight carbon steel pipe. Aloy pipeing, other than austenitic stainless steel and aluminum, may be laid out on the same basis. When pipe wall thickness differs from standard weight, a correction factor of rationing moment of inertias must be applied. In this case :

Minimum h(adjusted )!

moment of inertia,pipespecified xmin imum h in theformula moment of inertia,std.wt.pipe

When anchor expansion adds to the thrust from the L leg, a correction ratio of linear expansion must be applied. In this case :

Minimumh(adjusted) !

Anchor movement "legLmovement xminimum hinthe formula legLmovement

The minimum length of h leg required to provide an adequately flexible member must be tested by applying a factor A which corrects for design temperature. (See Table 5-1 for definitions) The formula is : Minimum h = A x Do For the A coefficient of carbon steel pipe, see Table 5-2. For L-shaped configurations (see Figure 5-1) the following formulas apply :

Figure 5-1 L-shaped configuration

Minimum h2 = 0.0025 Do L T

Page 4 of 26

400 (h) 2 Do T = A x Do

Minimum L !

Test

h

Table 5-2 A Coefficients for Carbon Steel Design Temp. oF

A

Design Temp. oF

A

150 200 300 400 500

0.4 0.6 1.0 1.4 1.8

600 700 800 900 1000

2.20 2.50 2.80 2.95 3.15

Example No. 1 L-shaped configuration.

Pipe = 6” Sch. 40 Do = 6.625”, so use 7” L = 30’ - 0” T = 650o - 70 o = 580 oF

Applying the above formula the minimum h is calculated : h2 = 0.0025 x Do x L x T = 0.0025 x 7 x 30 x 580 = 304.5 h = 17’-5” which is the minimum short leg allowed Test h for minimum length using the A coefficient : h = A x Do A at 650 oF (interpolated) is 2.35 so h = 2.35 x 7 = 16’-5” 16’-5” < 17’-5” so use the larger figure as the minimum.

Page 5 of 26

Example No. 2 L-shaped configuration.

Pipe Do T

= 12” Sch. 60 = 12.75” so use 13” = 500o - 70 o = 430 oF

Solve for h and L Test minimum h :

Minimum L !

h = A x Do = 1.8 x 13 = 23.4’

400 ( h) 2 400 ( 23. 4 ) 2 ! ! 39 ' 13 x 430 Do T

Since this pipe is not standard weight, the minimum h must be adjusted by the ratio of moment of inertias.

Factor !

400 . 5 I (12" Sch. 60 ) ! ! 1. 435 279. 3 I (12" Std. Wt .)

then h (adjusted) = 23.4’ x 1.435 = 32.6’

Since the minimum h has grown to 32.6’, maximum L must be recalculated.

400 ( h) 2 400 ( 32. 6 ) 2 ! ! 76. 5 ' Maximum L ! 13 x 430 Do T Since L must be greater than h, L can be as short as 33’ or as long as 76.5’. The shorter lengths will cause smaller forces on connecting equipment.

Page 6 of 26

Adaptation of L-Shaped Method The L-shaped formulas are easily adaptable to more complicated shapes in one or three planes.

Example No. 3 Single plane problem. See Figure 5-2.

Figure 5-2

Step No. 1.

Single-plane configuration.

Determine L, the major length of line at right angles between anchors. a + c + e is greater than b - d + f , so L = a + c + e

Step No. 2.

Solve for minimum h using the formula described in example 2.

Step No. 3.

Compare minimum h required with

b 2 " d2 " f 2 .

Figure Error! No text of specified style in document.:1

If the square root of the sum of squares equals or exceeds the required minimum h, flexibility is sufficient.

Page 7 of 26

Example No. 4 Three plane problem. See Figure 5-3.

Figure 5-3

Three-plane configuration.

Step No. 1.

Determine distances between anchors in the horizontal plane at right angles,and the vertical distance. North-South distance = a + e East-West distance = c + g Vertical distance =b-d+f

Step No. 2.

Determine L, the longest distance, a + e + > c + g > b - d + f therefore L = a + e. Note : This example assumes L to be a + e , L must be the largest of the above three sums.

Step No. 3.

Determine h, the shortest distance. The sum of legs b + c + d + f + g must equal or exceed h. These are the legs at right angles to L.

Page 8 of 26

Formula for Z-Shapes The basic formulas for determining minimum h and maximum L apply to Z-shaped configurations : Minimum h2 = 0.0025 Do L T

Minimum h ! 0. 05 D o L T

400 (h) 2 Do T To test for minimum h a test for leg ratio must also be made. Referring to Figure 5-4 : Maximum L !

Figure 5-4

Z-shaped configuration.

B # 4 L !B" C C Test both minimum h and leg B as equal to A x Do.

Example No. 5 Z-shaped configuration, see Figure 5-4 :

Pipe Do T L

= 8” Sch. 40, = 8.625” , so use 9” = 300o - 70 o = 230 oF = 30’-0”

To solve for minimum h : h2 = 0.0025 Do L T = 0.0025 x 9 x 30 x 230 = 155.25’ $ h = 12’-5.5” Test for minimum h :

h = A Do = 1 x 9 = 9’-0” minimum for h and B

Test B and C length ratio; assume B = 25’-0” , C = 5’-0”

B 25 ! ! 5 # 4 , so it is satisfactory. 5 C Page 9 of 26

Formula for U-Shapes Formulas for U-shapes differ somewhat from the L and Z shapes. For U-shaped configurations with equal legs h, the formulas are noted below. Refer to Figure 5-5.

Figure 5-5 U-shaped configuration with equal legs.

Minimum h2 = 0.0016 Do L T Maximum L !

Minimum h ! 0. 04 D o L T

or

625 (h ) 2 Do T

To test for minimum h : h !

A Do 1. 25

Example No. 6 U-shape with equal legs, see Figure 5-5 :

Pipe Do T L

= 14” Sch. 30, = 14” = 470o - 70 o = 400 oF = 30’-0”

To solve for minimum h : h2 = 0.0016 Do L T = 0.0016 x 14 x 30 x 400 = 16.4’ say 16’-6” Test for minimum h : h !

A Do 1. 25

!

1. 68 x 14 1. 25

! 18. 8 ' use 19 ' %0 "

Since 19’-0” is larger than 16’-6”, minimum h becomes 19’-0”.

Page 10 of 26

For U-shapes with unequal legs, see Figure 5-6, the following formulas apply :

Figure 5-6 U-shaped configuration with unequal legs.

Minimum h2 = 0.0021 Do ( L1 - L2 ) T

or

Minimum h ! 0. 045 D o ( L1 % L 2 ) T

2

500 ( h) Do T To test for minimum h + L2 = A x Do Maximum L1 % L 2 !

Example No. 7 U-shape with unequal legs, see Figure 5-6 :

Pipe Do T L1 L2

= 14” Sch. 30, = 14” = 470o - 70 o = 400 oF = 25’-0” = 5’-0”

To solve for minimum h : Minimum h ! 0. 045 D o ( L1 % L 2 ) T ! 0. 045 14 ( 25 % 5 ) 400 Test for minimum h : h + L2 = A x Do = 1.68 x 14 = 23.5’ Minimum L1 also equals 23.5’, so minimum h + L2 = 15 + 5 = 20’.

! 15 '

Then 15’ is too short. So use 20’-0” as minimum h.

When L2 and h are known and L1 needs to be solved for, use the approach given in example 3, L-shaped configuration, changing h (above) to L and L1 and L2 (above) to components of minimum h. The formula then becomes : Minimum h ! 0. 05 D o L T

where h ! a 2 " b 2 Page 11 of 26

Formula for Expansion Loop Figure 5-7, Typical expansion loop, depicts a typical loop found in most hot piping systems which have too much expansion to be absorbed by a straight line. For this configuration :

Figure 5-7 Typical expansion loop.

Minimum h = 0. 02 Do L T

Minimum h2 = 0.0004 x Do x T x L

or

2500 (h) 2 Maximum L = Do T Minimum W = 0.5 h Preferred W = 1.5 h To test for minimum h : h =

A Do 1. 25

Example No. 8 Expansion loop, see Figure 5-7:

Pipe Do T L

= 12” Std. wt. = 12.75” so use 13” = 520o - 70 o = 450 oF = 200’

To solve for minimum h : h2 = 0.0004 x Do x T x L = 0.0004 x 13 x 200 x 450 = 468

$ h = 21.6’

Minimum W = 0.5 h = 0.5 x 21.6 = 10.8’

To test for minimum h =

A Do 1. 88 x 13 = 19.5’ = 1. 25 1. 25

(To find A, see Table 5-2)

Use the larger 21.6’. Page 12 of 26

Expansion Loop Stress and Anchor Force As an expansion loop example, forces are transmitted to the two anchor points. To calculate the anchor force and maximum pipe stress of the loop, refer to Figure 5-8 and the following example. This example assumes use of LR elbows and is restricted to line size of 4” through 14”. For force at each anchor : Maximum stress :

F = C E IP S = K E D

Where C = Constant to be obtained from Table 5-3. E = Expansion to be absorbed by the loop, in inches (should be limited to 10”). IP = Moment of inertia, inches4. D = Outside diameter of pipe in inches. K = Constant to be obtained from Table 5-4.

Figure 5-8 Guided expansion loop.

Length of H in feet 20 19 18 17 16 15 14 13 12 11 10

Table 5-3 Values for Constant C 1.64 1.92

2.20 2.47 2.75 3.03 3.41 4.02 4.83 5.86 7.10 10

1.54 1.46 1.39 1.80 1.70 1.61 2.06 1.95 1.84 2.31 2.19 2.08 2.57 2.44 2.31 2.83 2.68 2.55 3.18 3.02 2.88 3.76 3.54 3.36 4.51 4.25 4.04 5.45 5.51 4.89 6.60 6.23 5.91 15 20 25 Length of W in feet

1.33 1.56 1.78 2.00 2.22 2.44 2.74 3.20 3.84 4.66 5.65 30

Page 13 of 26

Length of H in feet 20 19 18 17 16 15 14 13 12 11 10

Table 5-4 Values for Constant K 337 361

385 409 433 457 487 528 582 725 7.10 10

295 270 248 318 290 267 341 310 285 364 330 304 387 350 322 410 372 340 437 396 362 475 428 391 521 471 429 577 520 474 642 578 526 15 20 25 Length of W in feet

230 246 263 279 296 313 330 361 395 436 484 30

Page 14 of 26

3. Need for Formal Stress Analysis The need for formal stress analysis should be determined by designers with extensive flexibility experience. Hot piping connecting to strain-sensitive equipment such as pumps, compressors and turbines shall be closely reviewed for possible full analysis. For other systems full stress analysis is required when the following criteria is not satisfied :

DE & 0.03 2 (L % U) Where

D E L U e

= Nominal pipe size, in inches. = Expansion to be absorbed, in inches (E = U e) = Developed length of line axis, in feet = Anchor distance, in feet = Length of straight line joining the anchors = Coefficient of expansion (See Table 5-5).

Example No. 9

For quick-check stress analysis

Example for quick-check stress analysis Figure (5-9) See Figure (5-9): Pipe = 6” Sch. 40 Carbon Steel, Design Temp. = 400 oF.

D = 6”

DE (L % U)

2

&

0. 03

Page 15 of 26

STEP 1 Establish the distance between anchors in plan and elevation, in feet and decimals of a foot. X = Total line length away from A2 = 8’ + 4’ = 12’ Y = Vertical elevation difference = 6’ - 2’ - 2’ = 2’ (Difference in elevation between A1 and A2)

Z = Total line length away from A1 = 10’ + 6’

= 16’

STEP 2 Determine length U, the straight length between points A1 and A2.

U!

x2 " y 2 " z2 !

122 " 22 " 162 ! 20.1'

(say 20’)

STEP 3 Determine the expansion to be absorbed, E = U e, where U = Anchor distance, in feet = Length of straight line joining the anchors = 20’ e = Coefficient of expansion (See Table 5-5) From Table 5-5 @ 400 oF e = 2.7 inch per 100 ft, so e = 0.027 inch per ft. Then E = U e = 20 x 0.027 = 0.54” STEP 4 Determine value of L, the total length of line. L = 6 + 10 + 2 + 8 + 6 + 2 + 4 = 38’ STEP 5 Solve the formula which must be equal to or less than 0.03 or full stress analysis is needed.

DE (L % U)

2

&

0. 03

6x0.54 3.24 ! ! 0.01 & 0.03 , 2 '38%20 ( 18 2

So the configuration is satisfactory. Page 16 of 26

When piping connects to equipment nozzles which expand and contract due to temperature, the nozzle movement must be considered and added to expansion (E) calculations in the direction they occur. Referring to Figure (5-9), should anchor point A1 become an equipment nozzle and expand downward 0.375” and in direction Z toward A2 by 2” , the calculations must be modified. Expansion must be figured for net lengths of X, Y, and Z and anchor movements applied. STEP I:

)X

Calculate expansion in direction X : Since there is no anchor movement in direction X :

! 12 x 0.027" ! 0.324"

) X ! 0.324" " 0 ! 0.324"

STEP II: Calculate expansion in direction Y:

) Y ! 2 x 0.027" ! 0.054"

) Y ! 0.054"%0.375"! Since A1 is moving downward 0.375” So use 0.321” as this becomes the net anchor movement.

% 0.321"

STEP III: Calculate expansion in direction Z : Since A1 is moving 2” in the direction Z toward A2:

) Z ! 16 x 0.027"! 0.432" ) Z ! 0.432" " 2" ! 2.432"

STEP IV: Calculate expansion in direction U:

E!

)X " )Y " )Z

STEP V:

2

2

2

!

0.324 2 " 0.3212 " 2.432 2 ! 2.45"

Solve the basic formula using the value for E:

DE (L % U) 2

&

0. 03

6 x 2.45 14.7 ! ! 0.045 > 0.03 (38%20) 2 324

which is larger than 0.03, so a stress analysis required. Page 17 of 26

Table 5-5 Coefficients of Expansion Linear Thermal Expansion Between 70 OF and Indicated Temperature, inches per 100 feet* Material Temp. O F -325 -300 -275 -250 -225 -200 -175 -150 -125 -100 -75 -50 -25 0 25 50 70 100 125 150 175 200 225 250 275 300 325 350 375 400 425 450 475 500 525 550 575 600 625 650 675 700 725 750 775 800 825 850 875 900 925 950 975 1000 1025 1050 1075 1100 1125 1150 1175 1200 1225 1250 1275 1300 1325 1350 1375 1400 1425 1450 1475 1500

Carbon Steel Carbon-Moly Low Chrome (thru 3Cr-Mo)

5Cr-Mo thru 9Cr-Mo

-2.37 -2.24 -2.11 -1.98 -1.85 -1.71 -1.58 -1.45 -1.30 -1.15 -1.00 -0.84 -0.68 -0.49 -0.32 -0.14 0 0.23 0.42 0.61 0.80 0.99 1.21 1.40 1.61 1.82 2.04 2.26 2.48 2.7 2.93 3.16 3.39 3.62 3.86 4.11 4.35 4.60 4.86 5.11 5.37 5.63 5.90 6.16 6.43 6.70 6.97 7.25 7.53 7.81 8.08 8.35 8.62 8.89 9.17 9.46 9.75 10.04 10.31 10.57 10.83 11.10 11.38 11.66 11.94 12.22 12.50 12.78 13.06 13.34

-2.22 -2.10 -1.98 -1.86 -1.74 -1.62 -1.50 -1.37 -1.23 -1.08 -0.94 -0.79 -0.63 -0.46 -0.30 -0.13 0 0.22 0.40 0.58 0.76 0.94 1.13 1.33 1.52 1.71 1.90 2.10 2.30 2.50 2.72 2.93 3.14 3.35 3.58 3.80 4.02 4.24 4.47 4.69 4.92 5.14 5.38 5.62 5.86 6.10 6.34 6.59 6.83 7.07 7.31 7.56 7.81 8.06 8.30 8.55 8.80 9.05 9.28 9.52 9.76 10.00 10.26 10.53 10.79 11.06 11.30 11.55 11.80 12.05

Austenitic Stainless Steels 18Cr-18Ni -3.85 -3.63 -3.41 -3.19 -2.96 -2.73 -2.50 -2.27 -2.01 -1.75 -1.50 -1.24 -0.98 -0.72 -0.46 -0.21 0 0.34 0.62 0.90 1.18 1.46 1.75 2.03 2.32 2.61 2.90 3.20 3.50 3.80 4.10 4.41 4.71 5.01 5.31 5.62 5.93 6.24 6.55 6.87 7.18 7.50 7.82 8.15 8.47 8.80 9.13 9.46 9.79 10.12 10.46 10.80 11.14 11.48 11.82 12.16 12.50 12.84 13.18 13.52 13.86 14.20 14.54 14.88 15.22 15.56 15.90 16.24 16.58 16.92 17.30 17.69 18.08 18.47

12 Cr 17 Cr 27 Cr

25Cr20Ni

Monel 67Ni – 30Cu

3½ Nickel

Aluminum

-2.04 -1.92 -1.80 -1.68 -1.57 -1.46 -1.35 -1.24 -1.11 -0.98 -0.85 -0.72 -0.57 -0.42 -0.27 -0.12 0 0.20 0.36 0.53 0.69 0.86 1.03 1.21 1.38 1.56 1.74 1.93 2.11 2.30 2.50 2.69 2.89 3.08 3.28 3.49 3.69 3.90 4.10 4.31 4.52 4.73 4.94 5.16 5.38 5.60 5.82 6.05 6.27 6.49 6.71 6.94 7.17 7.40 7.62 7.95 8.18 8.31 8.53 8.76 8.98 9.20 9.42 9.65 9.88 10.11 10.33 10.56 10.78 11.01

-3.00 -2.83 -2.66 -2.49 -2.32 -2.15 -1.98 -1.81 -1.60 -1.39 -1.18 -0.98 -0.78 -0.57 -0.37 -0.16 0 0.28 0.51 0.74 0.98 1.21 1.45 1.70 1.94 2.18 2.43 2.69 2.94 3.20 3.46 3.72 3.98 4.24 4.51 4.79 5.06 5.33 5.60 5.88 6.16 6.44 6.73 7.02 7.31 7.60 7.89 8.19 8.48 8.70 9.07 9.37 9.66 9.95 10.24 10.54 10.83 11.12 11.41 11.71 12.01 12.31 12.59 12.88 13.17 13.46 13.75 14.05 14.35 14.65

-2.62 -2.50 -2.38 -2.26 -2.14 -2.02 -1.90 -1.79 -1.59 -1.38 -1.18 -0.98 -0.77 -0.57 -0.37 -0.20 0 0.28 0.52 0.75 0.99 1.22 1.46 1.71 1.96 2.21 2.44 2.68 2.91 3.25 3.52 3.79 4.06 4.33 4.61 4.90 5.18 5.46 5.75 6.05 6.34 6.64 6.94 7.25 7.55 7.85 8.16 8.48 8.80 9.12 9.44 9.77 10.09 10.42 10.75 11.09 11.43 11.77 12.11 12.47 12.81 13.15 13.50 13.86 14.22 14.58 14.94 15.30 15.66 16.02

-2.22 -2.10 -1.98 -1.86 -1.74 -1.62 -1.50 -1.38 -1.23 -1.08 -0.93 -0.78 -0.62 -0.46 -0.30 -0.14 0 0.22 0.40 0.58 0.76 0.94 1.13 1.32 1.51 1.69 1.88 2.08 2.27 2.47 2.69 2.91 3.13 3.34 3.57 3.80 4.03 4.27 4.51 4.75 4.99 5.24 5.50 5.76 6.02 6.27 6.54 6.81 7.08 7.35 7.72 8.09 8.46 8.83 8.98 9.14 9.29 9.45 9.78 10.11 10.44 10.78

-4.68 -4.46 -4.21 -3.97 -3.71 -3.44 -3.16 -2.88 -2.57 -2.27 -1.97 -1.67 -1.32 -0.97 -0.63 -0.28 0 0.46 0.85 1.23 1.62 2.00 2.41 2.83 3.24 3.67 4.09 4.52 4.95 5.39 5.83 6.28 6.72 7.17 7.63 8.10 8.56 9.03

Gray Cast Iron

0 0.21 0.38 0.55 0.73 0.90 1.08 1.27 1.45 1.64 1.83 2.03 2.22 2.42 2.62 2.83 3.03 3.24 3.46 3.67 3.89 4.11 4.34 4.57 4.80 5.03 5.26 5.50 5.74 5.98 6.22 6.47 6.72 6.97 7.23 7.50 7.76 8.02

Bronze

Brass

Wrought Iron

70Cu30Ni

-3.98 -3.74 -3.50 -3.26 -3.02 -2.78 -2.54 -2.31 -2.06 -1.81 -1.56 -1.32 -1.25 -0.77 -0.49 -0.22 0 0.36 0.66 0.96 1.26 1.56 1.86 2.17 2.48 2.79 3.11 3.42 3.74 4.05 4.37 4.69 5.01 5.33 5.65 5.98 6.31 6.64 6.96 7.29 7.62 7.95 8.28 8.62 8.96 9.30 9.64 9.99 10.33 10.68 11.02 11.37 11.71 12.05 12.40 12.76 13.11 13.47

-3.88 -3.64 -3.40 -3.16 -2.93 -2.70 -2.47 -2.24 -2.00 -1.76 -1.52 -1.29 -1.02 -0.75 -0.48 -0.21 0 0.35 0.64 0.94 1.23 1.52 1.83 2.14 2.45 2.76 3.08 3.41 3.73 4.05 4.38 4.72 5.06 5.40 5.75 6.10 6.45 6.80 7.16 7.53 7.89 8.26 8.64 9.02 9.40 9.78 10.17 10.57 10.96 11.35 11.75 12.16 12.57 12.98 13.39 13.81 14.23 14.65

-2.70 -2.55 -2.40 -2.25 -2.10 -1.95 -1.81 -1.67 -1.49 -1.31 -1.13 -0.96 -0.76 -0.56 -0.36 -0.16 0 0.26 0.48 0.70 0.92 1.14 1.37 1.60 1.83 2.06 2.29 2.53 2.77 3.01 3.25 3.50 3.74 3.99 4.25 4.50 4.76 5.01 5.27 5.53 5.80 6.06 6.32 6.59 6.85 7.12 7.40 7.69 7.91 8.26 8.53 8.81 9.08 9.39

-3.15 -2.87 -2.70 -2.53 -2.36 -2.19 -2.12 -1.95 -1.74 -1.53 -1.33 -1.13 -0.89 -0.66 -0.42 -0.19 0 0.31 0.56 0.82 1.07 1.33 1.59 1.86 2.13 2.40 2.68 2.96 3.24 3.52

* These data are for information, and it is not to be implied that materials are suitable for all the temperatures shown.

Cold Spring in Piping Cold spring is the term used for springing in the cold position a per cent of the calculated expansion. Figure 5-10 shows an L-shaped configuration with solid lines indicating the neutral position. As the long leg gets hot the line would expand 3” away from the anchor, forcing the elbow out 3”. By using 50% cold spring, cutting 1; “ out of leg X, the line will be sprung in the cold position. This reduces forces considerably, applying reduction in both the cold and hot positions since the elbow now only moves 1; “ beyond the neutral position. Cold spring is often abused by inexperienced piping designers who specify small amounts of cold spring on their drawings. The designer must remember that pipe fabrication tolerances normally are between 1/16” and 1/8”. A cold spring of 1/8”, or even 1/4”, is ignored unless a special hold to no tolerance note is added to the drawing - and the designer must be aware that this will be a costly note. This type of cold spring is rarely justified. Cold spring is applied to piping systems for these four basic reasons : 1. When required by detailed stress analysis. 2. To improve resultant forces and moments although not required by stress analysis. 3. To maintain adequate pipe spacing. 4. Misalignment correction. Any small amount of cold spring may be detailed stress analysis. When necessary, drawings must show the specified amount to the closest 1/16”. Cold spring may be specified when not mandatory to improve resultant forces and moments such as hot lines connecting to rotating equipment. Less than 1/4” is not specified.

Figure 5-10. Cold spring in piping Page 19 of 26

To maintain adequate pipe spacing and clearances in pipe racks and pipe groups, cold spring of 1” or more is specified. Normal pipe spacing will allow for more than 2” expansion so clod spring of less than 1” is seldom justified. Misalignment correction cold spring is used only for physical appearance and should never be specified for less than 3/4”. Normally this is used when a perfectly flexible line would grow as much as 4” - 6” and might appear to be out of line with a parallel line. Small amounts of growth here would not be visually recognized.

Page 20 of 26

PIPING STRESS ANALYSIS AND SUPPORT LOADS Computerized Method

Types of Computer programs. The microcomputer has become the daily tool and workstation for the piping stress analyst. Files which contain data for piping stress analysis are created, edited, and saved at this work station. These files are later transferred to the mini- or mainframe computer for the calculation of piping stresses and support loads. Most of computer programs for piping stress analysis such as ADLPIPE, NUPIPE, and SUPERPIPE were developed for use on mainframe computers. With the introduction of many powerful microcomputers in the mid-1980s, microcomputer-based programs for piping stress analysis were also developed such as AUTOPIPE and CAESAR II. Some of these new programs are menu driven and user friendly. They help save engineering time and cost. In general, these computer programs may be divided into four classes : 1. Programs that can perform pressure, thermal expansion, dead weight, and external forces (e.g. wind) analysis for ASME Section III, Class 2, 3, ANSI/ASME B31.1, B31.3, B31.4, B31.5, B31.8, NEMA, API-610, and API-617-piping. (AUTOPIPE and CAESAR II have response spectrum and SAM analysis capability. However, there is a limit on the number of analysis which can be performed in the same computer run because of the memory capability of microcomputers.) 2. Programs that can perform seismic, independent support motion, thermal transient, and time-history analyses in addition to those mentioned in item 1 for ASME Section III, Class 1, 2, 3, ANSI/ASME B31.1, and B31.3 piping. Programs such as ADLPIPE, ME101, NUPIPE, PIPESD, and SUPERPIPE are in this class. 3. General-purpose programs such as ANSYS. ANSYS is a general-purpose finite element analysis program which can perform static and dynamic analysis; elastic and plastic analysis; steady-state and transient heat transfer; steady -state fluid flow analysis; and nonlinear-history analyses. There are 40 different finite elements available for static and dynamic analysis. Dynamic analyses can be performed either by model superposition or direct integration. 4. Specialized programs such as PIPERUP. PIPERUP performs nonlinear elastic-plastic analyses of piping systems subjected to concentrated static or dynamic time-history forcing functions. These forces result from fluid jet thrusts at the location of a postulated break in high-energy piping. PIPERUP is an adaptation of the finite element method to the specific requirements of pipe rupture analysis.

Method of Analysis. The piping system is modeled as a series of masses connected by massless springs having the properties of the piping. The mathematical model should include the effects of piping geometry changes, elbow flexibilities, concentrated weights, changes in piping cross sections, and any other parameters affecting the stiffness matrix of the model. Mass point spacing should follow the guidelines specified above. Valves should be modeled as lumped masses at valve body and operator, with appropriate section properties for valve body and valve topworks. Rigid supports, snubbers, springs, and equipment nozzles should be modeled with appropriate spring rates in particular degree of freedom. Stress-intensification factors should be input at the appropriate locations (elbows, tees, branch connections, welds, etc.). Piping distributed weight should include pipe weight, insulation weight, and entrained fluid weight. Once an accurate model is developed, the loading conditions are applied mathematically : 1. Statically applied loads (dead weight, wind loads, pressure thrust, etc.). 2. Thermal expansion. 3. Statically applied boundary condition displacements (seismic anchor movement, LOCA containment displacement, etc.). 4. Response spectrum analysis (seismic, etc.). 5. Dynamically applied boundary condition displacements (LOCA motion, etc.). 6. Dynamically applied forcing functions (steam hammer, etc.). The results of the analyses should be examined in order to determine if all allowables are met (i.e., piping stress, valve acceleration, nozzle loads, etc.). The loads must be combined using the appropriate load combinations and submitted to structural designers for their analysis. y condition displacements (LOCA motion, etc.).