heat transfer through pipe support.pdf

302 Escoe: Piping and Pipeline Assessment Guide

Residual Heat Transfer Through Pipe Shoes

Heat transfer through plate surfaces is simpler than more complex surfacesbecause they can be handled with one-dimensional equations that are simpleto apply. Based on Figure 5-26, we consider the heat balance down throughthe shoe as follows:

Writing in equation form, we have for one-dimensional steady state flow:

Eq. 5-57

For the conduction process, Dt 5 ti 2 tpFor the convection process, Dt 5 tp 2 to

Substituting these into Eq. 5-57, we have

kmAm1 ti 2 tp

L 2 5 hoAp(tp 2 to)

kmAm 1 Dt

L 2 5 hoAp(Dt)

1Heat conducted through

shoe to base plate 2 5 1Heat loss by convection from

shoe to outside air 2

t1

L

P P

tp

Width

Figure 5-26. Heat transfer through pipe shoe.

Piping Support Systems for Process Plants 303

90°F

15″

113°F

132°F

508°F

803°F

803°F

888°F

to = 90°F

858°F

ti = 900°F

Figure 5-27a. Thermal gradients through a pipe clamp, clevis, and supporting rod.

Solving for tp, we have

Eq. 5-58

where Am 5 (P 3 length of shoe) 3 2, in.2

Ap 5 base width 3 length of shoe, in.2

ho 5 free convection coefficient for shoe to air, Btu/hr-ft-8FK 5 thermal conductivity of shoe material, Btu/hr-ft-8FL 5 shoe height, in.

Like the analysis for cylinders, the free convection coefficient (ho)can be substituted with a forced convection coefficient. However, mostpipe shoes are normally protected by enough equipment and structuresto prevent direct wind from blowing continuously on the shoe for anylength of time. Of course, this depends on each individual case.Figures 5-27a and 5-27b show thermal gradients for various simplepipe support configurations.

tp 5kmAmti 1 hoApLto

(kmAm 1 hoApL), 8F


12′′ φ SCH 40 5′′ Calcium silicateinsulation

to = 90°F

P = 0.35 in.

Base width

= 8 in.

L = 10 in.

ti = 750°F

Figure 5-28. Pipe supported on a shoe with temperatures shown.

Example 5-3: Heat Transfer Through a Pipe Shoe

A 12 in. process header shown in Figure 5-26 is supported by a shoe 14 in.long. The process fluid is at 7508F, and it is desired to determine the temper-ature of the bottom of the shoe base plate where Teflon is mounted toaccommodate pipe movement. The Teflon cannot withstand temperature ator greater than 4008F. Referring to Figure 5-28 and using Eq. 5-58, we have

900ti = 900°F = Process fluid temperature

800

1

2

3

4

5

6 7

700

600

500

400

300

200

5

L [inches]L

t°F

10 15 7

2′′

Typ6543

L

2 1

3′′

Insulation

ti

1

2

3

45

67

Figure 5-27b. Thermal gradient through pipe clamp support.


where ho 5 3.0 Btu/hr-ft2-8F for carbon steel in still airkm 5 26.0 Btu/hr-ft-8F, thermal conductivity of pipe materialL 5 10.0 in.

Am 5 (0.375)(14) 5 5.25 in.2

Ap 5 (8.0)(14) 5 112 in.2

to 5 908F

tp 5 306.3038F

Thus, the Teflon under the base plate is adequately protected. The amountof heat loss through the shoe base plate is

q 5 hoAp (tp 2 to)

Example 5-4: Emergency Constant Spring Replacement

The author was called out in the middle of the night when it was dis-covered that a constant spring hanger had bottomed out and become arigid hanger on a hot superheated steam line. The engineering contractorrepresentative discovered that the existing spring that bottomed out wasdesigned only for 1.5 in. of travel, whereas 3.5 in. of travel is necessary,according to the pipe stress computer runs. The design temperature was9008C, and the operating temperature was 7808C. The contractor placeda chain to support the pipe temporarily with operational monitoring.There were few options, and a chain support acting like a come-alongcould only be tolerated for a short time. Constant springs have a long

q 5 504.706 Btu

hr

q 5 (3.0) Btu

hr-ft2-8F 1 112 in.2

144 in.2 2 ft2 (306.303 2 90) 8F

tp 5

(26.0) Btu

hr-ft-8F 1 5.25 in.2

144 in.2 2

ft2 (750)8F 1 3.0 Btu

hr-ft2-8F 1 112 in.2

144 in.2 2

ft2 1 10

12 2

ft (90)8F

3(26.0) Btu

hr-ft2-8F 1 5.22 in.2

144 in.2 2

ft2 1 (3.0) Btu

hr-ft2-8F 1 112 in.2

114 in.2 2 ft2 1 10.0

12 2

ft4

tp 5kmAmti 1 hoApLto

(kmAm 1 hoApL), 8F


delivery time half way around the world, so the option taken was to visitthe junk yard and try to find a constant spring close to the requireddesign.

The required spring was for a 1254 kg load on the super steam header.A constant spring was found in the junk yard rated for 1889 kg loadwith the same travel, 3.5 in. This spring and the one being replaced werethe classical offset-slider crank mechanism design described previously,designed and fabricated by the same spring vendor. The spring was asize 28 shown in the spring table in Figure 5-29, rated for 4445 lbf

(1889 kg).When the spring was installed, its supporting force should have been

in balance with the portion of the piping weight at which it was rated.The spring was preset to the cold position at the specified load. Thespring had a turnbuckle, thus allowing for normal piping elevationadjustments. In our case, the actual piping load differed from thecalculated load. The spring manufacturer claimed that an adjustment of15 to 20% of the specified load could be made by turning the load adjust-ment bolt. Each division on the adjustment scale equals 5% of the ratedload, according to the spring manufacturer’s technical bulletin (catalog).Referring to Figure 5-30, we see the adjustment bolt setting has eightgraduations.

Turning the bolt clockwise decreases the load 5% for each gradua-tion; hence, four graduations represent a 20% decrease in the load.Similarly for a counterclockwise turn of the adjustment bolt, the loadsetting is increased. Thus, the load of 1889 kg reduced by 20%becomes

1889 kg(0.8) 5 1511.2 kg

The acquired spring is adjusted by locking the spring assembly in thecold position and adjusting the load nut to match the required loading.The required loading is 20% over the operating load of the former spring.Since this load is 1254 kg, the adjusted load is

1254 3 1.2 5 1504.8 kg or 1505 kg

The 1505 kg is close enough to the 1511 kg reduction.The load adjustment scale on the spring housing had ten graduations

for the total adjustment range, as shown in Figure 5-31.This is not to be confused with the scale on the load adjustment bolt.

This scale indicates the amount of spring travel. Normally the actual isless than the total available travel to account for excursions. To translatethe process temperatures and load requirements on the spring scale, we


Constant Support Size Selection Table

Hgr. Total Travel (Inches)

Size Maximum Rated Load in Pounds For Each Size

4000

4100 2 21⁄2 3 31⁄2 4 41⁄2 5 51⁄2 6 61⁄2 7 71⁄2 8 81⁄2 9 91⁄24200

1 90 75 60 55 45 40 35 35 30 25 20 20

130 100 85 70 65 55 50 45 40 35 30 30

2 175 140 115 100 85 60 70 65 60 50 45 40

3 235 190 160 135 120 105 95 85 80 70 60 55

4 325 260 215 185 165 145 130 120 110 95 80 70

5 445 360 295 255 225 200 180 160 150 130 110 100

6 600 485 405 345 305 270 245 220 205 175 150 135

7 755 605 505 435 380 335 305 275 255 235 215 205 190 180 170 160

8 905 725 605 520 455 405 365 330 305 280 260 245 230 215 205 190

9 1090 875 725 625 545 485 435 395 365 335 310 290 275 255 245 230

10 1310 1045 875 750 655 585 525 475 440 405 375 350 330 310 290 275

11 1465 1175 980 840 735 650 585 535 490 450 420 390 365 345 325 310

12 1620 1295 1080 925 810 720 650 590 540 500 465 435 405 385 350 345

13 1810 1450 1210 1035 905 805 725 660 605 560 520 485 455 430 405 385

14 2000 1605 1335 1145 1005 890 805 730 670 620 575 535 500 475 445 425

15 2240 1790 1495 1280 1120 995 900 815 750 690 640 600 560 530 500 475

16 2475 1980 1655 1415 1240 1100 990 900 825 765 710 660 620 585 550 525

17 2770 2215 1850 1585 1385 1230 1110 1010 925 855 790 740 695 655 615 585

18 3060 2445 2045 1750 1530 1365 1225 1115 1020 945 875 815 765 720 685 650

19 3400 2720 2265 1940 1670 1510 1360 1240 1135 1050 975 905 850 800 755 720

20 3735 2990 2490 2130 1870 1660 1495 1360 1245 1150 1070 995 935 880 830 790

21 4120 3295 2745 2350 2060 1830 1650 1500 1375 1270 1180 1100 1030 970 915 870

22 4500 3500 3000 2575 2250 2000 1800 1640 1500 1385 1285 1200 1125 1050 1000 950

23 4950 3650 3300 2830 2475 2200 1980 1600 1650 1525 1415 1320 1240 1165 1100 1045

24 5400 4320 3600 3085 2700 2400 2160 1965 1800 1665 1545 1440 1350 1270 1200 1140

25 5940 4755 3960 3395 2970 2640 2375 2160 1980 1830 1700 1585 1485 1400 1320 1255

26 6480 5185 4320 3710 3240 2880 2595 2355 2160 1995 1855 1730 1620 1525 1440 1370

27 5710 4755 4075 3565 3170 2855 2595 2380 2195 2040 1905 1785 1680 1585 1505

28 6230 5190 4445 3890 3460 3115 2830 2595 2395 2225 2075 1945 1830 1730 1645

29 6845 5705 4890 4280 3805 3425 3110 2855 2635 2445 2285 2140 2015 1905 1805

30 7645 6380 5465 4780 4250 3825 3475 3190 2945 2735 2555 2395 2250 2125 2020

31 8660 7215 6180 5405 4810 4330 3935 3600 3330 3090 2885 2705 2545 2405 2285

32 9480 7895 6760 5915 5265 4740 4305 3945 3640 3380 3150 2950 2780 2630 2500

33 10515 8770 7510 6575 5845 5260 4780 4380 4045 3750 3500 3285 3095 2920 2775

34 11700 9640 8265 7225 6425 5775 5250 4815 4450 4130 3845 3610 3405 3210 3050

35 10700 9160 8020 7130 6415 5830 5345 4940 4585 4220 4010 3775 3565 3385

36 11760 10060 8820 7840 7055 6415 5880 5430 5040 4600 4410 4150 3920 3720

37 13055 11180 9790 8700 7835 7120 6530 6030 5595 5170 4895 4605 4350 4130

38 14350 12305 10760 9565 8610 7830 7175 6625 6150 5745 5380 5060 4785 4545

39 15985 13700 11985 10650 9585 8725 7995 7375 6850 6390 5990 5640 5330 5060

40 17615 15095 13205 11740 10565 9610 8810 8125 7545 7040 6605 6215 5870 5575

41 19310 16545 14480 12875 11580 10530 9655 8910 8275 7720 7240 6815 6435 6115

42 21000 18000 15750 14005 12600 11455 10500 9690 9000 8400 7875 7415 7005 6650

43 22180 19140 16740 14880 13390 12170 11160 10300 9565 8925 8370 7875 7445 7070

44 24000 20280 17725 15750 14175 12885 11825 10910 10135 9450 8860 8335 7885 7485

45 21700 19050 16925 15235 13850 12705 11725 10890 10160 9525 8960 8470 8045

46 23250 20370 18100 16295 14815 13585 12535 11640 10870 10185 9585 9050 8600

47 22325 19845 17860 16240 14890 13740 12755 11910 11160 10510 9925 9520

48 24290 21590 19425 17660 16190 14945 13870 12960 12140 11435 10795 10260

49 26200 23290 20960 19055 17465 16125 14970 13980 13100 12330 11645 11065

50 28115 24990 22490 20450 18740 17300 16065 15005 14060 13230 12495 11870

51 30590 27195 24475 22250 20395 18830 17490 16320 15300 14390 13600 12935

52 33065 29400 26460 24045 22050 20360 18910 17640 16540 15550 14700 14000

53 35900 31920 28745 26120 23940 22110 20530 19145 17955 16885 15960 15180

2 21⁄2 3 31⁄2 4 41⁄2 5 51⁄2 6 61⁄2 7 71⁄2 8 81⁄2 9 91⁄2

Figure 5-29. Spring support manufacturer’s load table.

NOTE: The minimum

load is the load for the

preceding size


Figure 5-31. Elevation view—looking into the face of the spring.

Load adjustment scaleEach mark from zero isa 5% adjustment

Side walls ofspring (Typ)

Load adjustment nut

∆

∆

∆ = Amount of extra travel to prevent spring bottoming out

HotCold

Travel pointer

Operating range

3.5″ Total travel

12 3 5 6 7 8

910

4

S

Figure 5-30. Spring travel scale on side of spring.

made the following calculations to see if the spring had enough travel tooperate safely.

The 9008C represents 3.5 in. of total travel on the spring. This temper-ature is the very maximum that could be expected in an excursion. Thusfor the operating case, which had a process temperature of 7808C, thetravel incurred for this case is

9008C

3.5 in.5

7808C

x Q x 5 3.033 in.


On each side of the scale, there are 3.033/2 5 1.5165 in. Thus, the numberof graduations on the spring scale on the housing is

number of graduations 5 (1.0 2 0.857)(10 graduations)5 1.429 graduations on scale

D 5 the amount of extra travel to prevent the spring from bottoming out

Thus,

This represents the number of graduations on the spring scale betweeneach side of the operating range and the design, or overall travel. Nowthe process temperature at the time of installation is 7508C. We calculatethe difference between the installed position and the operating condition(d). Thus, we have

Now,

The number of graduations on the scale is

(1.0 2 0.857)(10 graduations) 5 1.429 graduations – differencebetween 3.0 in. of travel and3.5 in. of travel

Thus, the total adjustment for the installed temperature of 7808C is

d 1 D 5 1.51650 1 0.7580 5 2.27 in. (Actual adjustment)

1 3.0

3.5 2 5 0.857 Q D 51.5165 in.

25 0.758 in.

d 53.033 in.

25 1.5165 in.

9008C

3.5 in.5

7508C

x Q x 5 3.033 in.

D 51.429

25 0.71 graduation

1 3.0

3.5 2 5 0.857


Now we know the spring has enough travel at the desired load tooperate safely. The spring was set and placed in use where it operatedfor ten months before the new spring arrived to replace it. The authormonitored the spring during the entire period, and it performed verywell. If a replacement spring had not been found in surplus, then theother alternatives would have been to shut down for months, which theplant management found unacceptable, or to fabricate a spring hangerfrom scratch.

Example 5-5: Pipe Header Simple Support

Problem: A 30 in. f crude oil header is simply supported with the piperesting on the steel. A company procedure calls for saddle-type supportswith pads for piping 30 in. NPS (nominal pipe size) and larger. In casesaddle pipe support cannot be installed, the line shall be analyzed forlocalized stresses. The question was: is the pipe resting on the steeloverstressed? If it is overstressed, why has it not failed, as it has been inoperation for thirteen years?

Solution: If the company procedure specifies a saddle support for lines30 in. and larger, then this standard must be adhered to. Regarding itsstress state, the pipe resting on the steel at a point is not a contact stress.Contact stresses are called Hertz stresses and are to applied solid bodies(e.g., meshing gears or ball bearings). They are not used for hollowsurfaces (e.g., pipes or vessels). One classic example of this is the Zickanalysis of horizontal drums supported on two saddles—contact stressesare not considered in the analysis.

In the case of the pipe resting on a steel support, the situation is like aring. The cross section of the pipe resting on a solid surface represents abulkhead or supporting ring in a pipe, supported at the bottom andcarrying a total load (W) transferred by tangential shear (v) distributedas is shown in Roark Table 9.2 Case 20 [Reference 6]. The Roark modelused in Case 20 is called a tangential shear, where the vertical reactionat the pipe support is applied to counter the pipe’s weight, whichincludes the metal weight and content. The support reaction causesshear to flow around the pipe wall to replicate the pipe behaving like abeam. Rings are very common in structural design. This case is shownin Figure 5-32.

The Roark Table 9.2 Case 20 gives a bending moment due to the shearstress caused by the weight of the pipe and the internal fluid. This bending


moment acts through the wall of the pipe. Being a bending moment, it hascompression and tensile components through the cross section of the pipe.The internal pressure stress acts only in tension. Thus, when the pressurestress is combined with the shear bending moment stress, one side of thewall is added and the other side of the wall is subtracted from the tensilepressure stress. The sign convention on the shear bending moment is notrelative, as the stress will be one value on each respective surface (we addtensile stresses and add and subtract tensile and compressive stresses).This is shown in Figure 5-33.

When solving for the shear bending moment (I), the Roark Table 9.2Case 20, the resisting cross section acts in the plane of the paper. The effec-tive width of this cross section is determined as shown in Figure 5-34.

Shear acts at a 458 angle; thus, the resisting cross section would be 458

diagonal lines drawn from the point of reaction to the axis of the pipe. Asseen in Figure 5-34, the effective length is 2Ro. Typically, the effective

Pipe wall

Pressure tensile load

Shear bending stress

Figure 5-33. Hoop stress and shear moment bending stress in the pipe wall.

W

A

V

x

Figure 5-32. Roark Table 9.2 Case 20.


length of the cross section would be minimum of 2Ro or 40tnom, wheretnom is the nominal pipe wall thickness. The bending stress induced bythe shear bending moment is calculated as follows:

be 5 MAX[40tnom, 2Ro], see Figure 5-34 Eq. 5-59

Eq. 5-60

where S 5 section modulus of the pipe, in.3

tnom 5 pipe nominal wall thickness, in.

This formula is derived from the relationship

where I 5 moment of inertia, in.4

In our case be 5 30 in., as 40tnom 5 40(1.358) in. 5 54.32 in. Thus, theshear moment bending stress is

Eq. 5-61

The bending moments vary around the circumference of the pipe. Shownin Figure 5-35 is the bending moment distribution for the pipe loadedwith water.

s 5

Mshear

s

S 5

I

C5

bd3

12

d

2

5

bd2

6, for a rectangular cross-sectional area

S 5

(be)t2nom

6

Figure 5-34. The effective length of pipe resisting the shear moment bending stress.

Reaction force

Ro

2Ro

Pipe

45″

C C


The maximum bending moment occurs at 105.328 from the top point ofthe pipe. We loaded the pipe with crude oil using a specific gravity of 0.825for the operational case. It was assumed that the pipe would not be operatingat full pressure loaded with water. Combining the internal pressure stressand the shear moment bending stress for values of the angle from 08 to 1808,we have the following combined stresses for pipe loaded with crude oil:

Combined stresses for pipe loaded with crude oil

Angle x Total Stress on Inner Wall, psi Total Stress on Outer Wall, psi

0 16,924.6 14,452.1

15 17,194.2 14,182.4

30 17,960.3 13,416.3

45 19,908.7 12,278.0

60 20,417.2 10,959.5

75 21,676.8 9,699.9

90 22,617.8 8,758.8

105 22,989.8 8,366.8

120 22,580.6 8,796.0

135 21,244.7 10,132.0

150 18,923.6 12,453.0

165 15,661.2 15,715.5

180 11,607.7 19,769.0

The Section Modulus Conjecture: The criteria mentioned earlier concern-ing the effective area resisting the bending moment induced by shear insupporting the pipe has been a source of conjecture for many years. In thesolution of a ring loaded in shear and inducting a moment into the plane ofthe pipe, the three-dimensional section resisting the bending moment is notconsidered because the ring is a two-dimensional problem. In Figure 5-34the effective length of the cross section resisting the bending moment

Figure 5-35. The shear bending moment across a pipe resting on a simple support.

0

0

0

0


induced by the shear is shown as 2Ro. The effective cross section is themaximum of 40tnom or 2Ro; however, in structural applications, the mini-mum value of 40tnom or 2Ro is often applied. In mechanical application,where we have internal pressure combined with the weight and thermalloads, the stresses can be significantly high—sometimes several timesthe ultimate strength of the pipe material. However, in reality, these pipesresting on simple supports do not fail, meaning that there is a fallacy in theequations—the theory does not match reality.

One author that addresses this phenomenon is Bednar [Reference 7],where in his brilliant work he discusses this problem on page 170. Bednarstates that, to utilize the derived equation for the bending moment andbring the resulting stresses in the shell in agreement with actual measuredstresses, we use what he describes as a “fictitious resisting width” of shellplate that is the smaller of 4Ro or L/2, where Ro is the shell radius and Lis the length between supports. Of course, Bednar is discussing horizontalvessels supported on saddles, whereas here we are discussing continuouspiping, like a continuous beam. Bednar’s Figure 6.5 on page 168 is thesame ring solution we use from Roark, but they are derived from differentsources.

Another author, Troitsky [Reference 8] uses the same solution for thering where the shear induces a bending moment from the load at thepoint of contact with the pipe support, with one exception. He is workingwith conveyor tube structures supported by ring girders. Troitsky uses theresisting section on pages 12–19 with the following expression:

Eq. 5-62

Where r is the cylinder radius, t is the cylinder thickness, b is the widthof the ring girder, and c is the effective width resisting the shear-inducedbending moment.

Based on these criteria, we use the following to compute the effectivewidth to obtain the section modulus, as follows:

Eq. 5-63

where R 5 mean radius, CA 5 corrosion allowance, and be is defined inEq. 5-59. Thus, Eq. 5-60 becomes

Eq. 5-64

Now the stress criterion must be considered. In this case the pipe issubjected to primary and secondary stress. Thus, the maximum stress

S 5beff1t2nom2

6

beff 5 be 1 1.56√R(tnom 2 CA)

c 5 1.56√rt 1 b


intensity smax is based on the primary or local membrane stresses plusprimary bending stress plus the secondary stress (Pm or PL 1 Pb 1 Q),using the nomenclature of the ASME Section VIII Division 2Appendix 4. The value of smax cannot exceed 3Sm, where all stressesmay be computed under operating conditions. The following criterionmust be met:

3Sm # 2(SMYS)

The value of 3Sm is defined as three times the average of the tabulatedSm values for the highest and lowest temperatures during the operationcycle. In the computation of the maximum primary-plus-secondarystress intensity range, it may be required to consider the superposition ofcycles of various conditions that produce a total range greater than therange of any individual cycle. The value of 3Sm may vary with eachcycle, or combination of cycles, being considered since the maximumtemperatures may be different in each case. Thus, care must be used toensure the applicable value of 3Sm for each cycle, and combination ofcycles, is not exceeded except as permitted by ASME Section VIIIDivision 2 Appendix 4 Paragraph 4-136.4. This requirement is seldomrequired in piping systems, unless different services are used in the pipewith differing temperatures and pressures.

Since the pipe considered is operational, we can use the API 579Fitness-for-Service recommended practice for the allowable stress, whichis 20,000(1.2) 5 24,000 psi from Figure B.1 of the API 579. For this case,the pipe wall thickness is large enough to accommodate the stress levels.The output of the computer run for the angular position of 105.238, thepoint of maximum bending moment, is as follows.

Looking at Figure 5-36, we find the stresses in the 30 in. pipe areacceptable because, as mentioned previously, this is the angular position ofmaximum bending moment induced by the shear. However, if we takeanother pipe with a thinner wall, the stress magnitudes will differ. Considera 24 in. line load with the same crude oil at the same facility. The pipe hasa 3/80 wall and is seamless, making the tnom, less mill tolerance, 0.328 in.The results are as follows for the maximum bending moment at the sameangular position of 105.238.

We will now consider a 240 f pipe with a nominal wall thickness of0.328 in. resting on a solid surface like we did for the 300 f pipe. The samemethodology is used and the results are shown in Figures 5-37a and b.

The reader will notice that the outside wall is in compression and theinside wall is in tension because of the bending moment. Thus, only part ofthe wall is of significant stress, not the entire wall of the pipe. This being


if or(x , 0, x 2 p . 1E-10) then caution 5 ‘Ang_Err else caution 5 ‘_call get_tab(matnum, matl, G, E, nu)

if ring 5 ‘thin then call get_tn_con(I, A, R, F, E, G; alpha, beta)

if ring 5 ‘thick then call get_tk_con(h, R, F, nu; alpha, beta)

k2 5 1 2 alpha

k1 5 k2 1 beta

call case(W, R, I, x, k1, k2; v, LTM, LTN, LTV, M, N, V, MA, MC, NA, VA, DH, DV,delL, case)

plot 5 given(‘plot, plot, ‘y)

if and(solved(), plot ,. ‘n) then call genplot(W)

Di 5 Do 2 2 · tnom

be 5 MAX(40 · tnom, 2 · Ro)

sinner 5 sp 2 s

souter 5 sp 1 s

sallow 5 3.0 · Sm

If 3 · Sm , 2 · SMYS then OK 5 ‘acceptable

If 3 · Sm . 2 · SMYS then OK 5 ‘unaccpetable

sp 5 3 P

Ej 4?33 Do

2 ? tnom 4 2 0.44

s 5M

S

S 5beff? tnom2

6

beff 5 be 1 1.56 ?√R ?(tnom 2 CA)

Wreaction 5 Am?L ?12 ?0.283 1 3 p

4 4?3Di24?r ?L ?12

Am 5 3 p

4 4?3Do2 2 Di24

Ri 5Di

2

Ro 5Do

2

Figure 5-36a. Equation sheet for 30 in. f pipe.

Rules Sheet

Rules


Input Name Output Unit Comment

Table 9.2: Roark’s Formulas

Formulas for Circular Rings

case ‘CASE_20 Reference Numberplot ‘y Generate plots? ‘n 5 no

(Default 5 yes)caution ‘_ Caution Message

17 matnum Material Number (SeeMaterial Table)

matl “Steel - A.S.T.M. Material nameA7-61T”

E 2.9E7 psi Young’s Modulusnu 0.27 Poisson’s Ratio

‘thin ring ‘thick or ‘thin: thick/thin ring7107.502 W lbf Applied total load1943 I in.4 Area moment of inertia of

X-section11.813 R in. Mean radius of centroid of

X-section2 F Shape Factor of X-section

THIN RINGS:27.83 A in.2 Area of X-section1.087E7 G psi Shear Modulus of elasticity

THICK RINGS:h in. Distance from Centroidal Axis

to Neutral Axis

DH 1.2938739E-4 in. Change in horizontal diameterDV 22.4236034E-4 in. Change in vertical diameterv 184.791 lbf/in. Tangential shearalpha 0.500310 Hoop-stress Deformation

Factorbeta 2.670776 Radial Shear Deformation

FactorAT SECTION:

105.23 x deg Angular PositionN 22450.334 lbf Internal ForceV 3.129E-2 lbf Radial ShearM 215247.877 lbf-in. Internal moment

AT A:NA 1696.791 lbf Internal ForceVA 0 lbf Radial ShearMA 24.143 lbf-in. MomentMC 13358.654 lbf-in. Moment at CdelL 22.0957714E-4 in. Increase in LOWER radius

LOAD TERMS:

Variables Sheet

Figure 5-36b. Variable sheet showing results and answers for 30 in. f pipe.



LTM 10065.9581776 lbf-in.LTN 22004.5965509 lbfLTV 1637.2296434 lbfk1 3.170466 Simplifying constantsk2 0.499690Di 27.284 in. Inside diameter of pipe

30 Do in. Outside diameter of pipe1.358 tnom in. Nominal pipe wall thickness

Ro 15 in. Outside radius of pipeRi 13.642 in. Inside radius of pipe

3.1416 p Metal area of pipe x-sectionWreaction 31272.705042 lbs Combined of metal weight

plus water50 L ft Length between pipe supports.03 r Density of fluid, lbs/in.3

be 54.32 in. Effective length of pipe wallresisting bending due toshear

S 18.616281 in.3 Section modulus of resistingpipe section to shearmoment

s 2819.061411 psi Bending stress due to shearmoment

Am 122.195158 in.2 Cross sectional area of pipe1400 P psi Internal pressure1 Ej Weld joint efficiency

sinner 15722.978937 psi Total stress in inner shell2 pressure 1 shearbending

sp 14903.917526 psi Internal pressure stresssouter 14084.856114 psi Total stress in outer shell

2 pressure 1 shearbending

beff 60.5681950 CA

sallow 60000 psi Allowable stress of pipematerial at temperature

20000 Sm Allowable stress for primaryand secondary

35000 SMYS psi Specified Minimum YieldStrength of pipe material

OK ‘acceptable

Figure 5-36b. cont’d.

the case, the high stress does not exist through the wall of the pipe, thusavoiding plastic deformation.

As mentioned previously, there are hundreds of pipes resting on flat solidsurfaces that do not fail or exhibit plastic deformation. The 30 in. f pipementioned was checked for cracks and deformation, and neither existed. The


if or(x , 0, x 2 p . 1E-10) then caution 5 ‘Ang_Err else caution 5 ‘2call get_tab(matnum, matl, G, E, nu)

if ring 5 ‘thin then call get_tn_con(I, A, R, F, E, G; alpha, beta)

if ring 5 ‘thick then call get_tk_con(h, R, F, nu; alpha, beta)

k2 5 1 2 alpha

k1 5 k2 1 beta

call case(W, R, I, x, k1, k2; v, LTM, LTN, LTV, M, N, V, MA, MC, NA, VA, DH, DV,delL, case)

plot 5 given(‘plot, plot, ‘y)

if and(solved(), plot ,. ‘n) then call genplot(W)

Di 5 Do 2 2 · tnom

be 5 MAX(40 · tnom, 2 · Ro)

sinner 5 sp 2 s

souter 5 sp 1 s

sallow 5 3.0 · Sm

If 3 · Sm , 2 · SMYS then OK 5 ‘acceptable

If 3 · Sm . 2 · SMYS then OK 5 ‘unaccpetable

sp 5 3 P

Ej 4?33 Do

2 ? tnom 4 2 0.44

s 5M

S

S 5beff? tnom2

6

beff 5 be 1 1.56 ?√R ?(tnom 2 CA)

Wreaction 5 Am?L ?12 ?0.283 1 3 p

4 4?3Di24?r ?L ?12

Am 5 3 p

4 4?3Do2 2 Di24

Ri 5Di

2

Ro 5Do

2

Figure 5-37a. Equation sheet for 24 in. f pipe.

Rules



Table 9.2: Roark’s Formulas

Formulas for Circular Rings

case ‘CASE_20 Reference Numberplot ‘y Generate plots? ‘n 5 no

(Default 5 yes)caution ‘_ Caution Message

17 matnum Material Number (SeeMaterial Table)

matl “Steel-A.S.T.M. Material nameA7-61T”

E 2.9E7 psi Young’s Modulusnu 0.27 Poisson’s Ratio

‘thin ring ‘thick or ‘thin: thick/thin ring7107.502 W lbf Applied total load1943 I in.4 Area moment of inertia of

X-section11.813 R in. Mean radius of centroid of

X-section2 F Shape Factor of X-section

THIN RINGS:27.83 A in.2 Area of X-section1.087E7 G psi Shear Modulus of elasticity

THICK RINGS:h in. Distance from Centroidal Axis

to Neutral Axis

DH 1.2938739E-4 in. Change in horizontal diameterDV 22.4236034E-4 in. Change in vertical diameterv 184.791 lbf/in. Tangential shearalpha 0.500310 Hoop-stress Deformation

Factorbeta 2.670776 Radial Shear Deformation

FactorAT SECTION:

105.23 x deg Angular PositionN 22450.334 lbf Internal ForceV 3.129E-2 lbf Radial ShearM 215247.877 lbf-in. Internal moment

AT A:NA 1696.791 lbf Internal ForceVA 0 lbf Radial ShearMA 24.143 lbf-in. MomentMC 13358.654 lbf-in. Moment at CdelL 22.0957714E-4 in. Increase in LOWER radius

LOAD TERMS:

Figure 5-37b. Variable sheet for 24 in. f pipe.

pipe has been in service for 13 years and has seen all possible operationcycles. The same is true for the 24 in. f pipe and many others. It is con-cluded that this methodology adequately predicts the behavior of a pipe rest-ing on a simple support, or flat surface. However, many companies consider

Variables sheet



LTM 10065.9581776 lbf-in.LTN 22004.5965509 lbfLTV 1637.2296434 lbfk1 3.170466 Simplifying constantsk2 0.499690Di 23.344 in. Inside diameter of pipe

24 Do in. Outside diameter of pipe.328 tnom in. Nominal pipe wall thickness

Ro 12 in. Outside radius of pipeRi 11.672 in. Inside radius of pipe

3.1416 p Metal area of pipe x-sectionWreaction 7107.502462 lbs Combined of metal weight

plus water30 L ft Length between pipe supports.03 r Density of fluid, lbs/in.3

be 24 in. Effective length of pipe wallresisting bending due toshear

S 0.485396 in.3 Section modulus of resistingpipe section to shearmoment

s 231413.258210 psi Bending stress due to shearmoment

Am 24.392689 in.2 Cross sectional area of pipe550 P psi Internal pressure1 Ej Weld joint efficiency

sinner 51315.209430 psi Total stress in inner shell2 pressure 1 shearbending

sp 19901.951220 psi Internal pressure stresssouter 211511.306991 psi Total stress in outer shell

2 pressure 1 shearbending

beff 27.0707300 CA

sallow 60000 psi Allowable stress of pipematerial at temperature

20000 Sm Allowable stress for primaryand secondary

35000 SMYS psi Specified Minimum YieldStrength of pipe material

OK ‘acceptable

Figure 5-37b. cont’d.

it good engineering practice to place pipes 30 in. f and larger on saddle sup-ports. It is also noted that, for large diameter thin cylinders resting on theground in stock yards and construction sites, angle beams 908 to one anotherare tack welded to the inside surface to prevent excessive ovaling.

In conclusion, a saddle support is required for the pipe if constructed new.


Note: The internal pressure stress was computed from the following for-mula, which is developed from Eq. 2-13(a), where Y 5 0.4 and D 5 Ro/2:

where W 5 1 for temperature less than 9508F, as in this case.

Solving for the stress we have the following:

where Do 5 outside diameterRo 5 outside radius

References

1. A. Keith Escoe, Mechanical Design of Process Systems, Vol. 1,2nd edition, 1994, Gulf Publishing Company, Houston, pp. 81–83.

2. A. Keith Escoe, Mechanical Design of Process Systems, Vol. 1,2nd edition, 1994, Gulf Publishing Company, Houston, pp. 132–135.

3. J. P. Holman, Heat Transfer, 7th edition, 1990, McGraw PublishingCompany, New York, pp. 335, 338.

4. Nicholas P. Cheremisinoff, Heat Transfer Pocket Handbook, 1984,Gulf Publishing Company, Houston, p. 106.

5. Alan E. Chapman, Heat Transfer, 3rd edition, 1974, MacmillanPublishing Company, New York.

6. Warren C. Young and Richard G. Budynas, Roark’s Formulasfor Stress and Strain, 7th edition, 2002, McGraw-Hill PublishingCompany, New York.

7. Henry H. Bednar, Pressure Vessel Design Handbook, 2nd edition,1986, Van Nostrand Reinhold Company, New York.

8. M. S. Troitsky, Tubular Steel Structures, Theory and Design, 2nd edition,1990, The James Lincoln Arc Welding Foundation.

s 5P

E 1 Do

2t2 0.42

t 5PRo

sEW 1 0.4P

heat transfer through pipe support.pdf

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