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 surfaces because they can be handled with one-dimensional equations that are simple to apply. Based on Figure 5-26, we consider the heat balance down through the shoe as follows:

1

2 1

Heat conducted through Heat loss by convection from 5 shoe to base plate shoe to outside air

2

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

1 2 5 h A (Dt) Dt L

Eq. 5-57

o p

For the conduction process, Dt 5 ti 2 tp For the convection process, Dt 5 tp 2 to Substituting these into Eq. 5-57, we have

1 t 2L t 2 5 h A (t

kmAm

i

p

o p p

2 to)

t1

L P

P

tp Width Figure 5-26. Heat transfer through pipe shoe.

Piping Support Systems for Process Plants

303

Solving for tp, we have tp 5

kmAmti 1 hoApLto , 8F (kmAm 1 hoApL)

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-8F K 5 thermal conductivity of shoe material, Btu/hr-ft-8F L 5 shoe height, in. Like the analysis for cylinders, the free convection coefficient (ho) can be substituted with a forced convection coefficient. However, most pipe shoes are normally protected by enough equipment and structures to prevent direct wind from blowing continuously on the shoe for any length of time. Of course, this depends on each individual case. Figures 5-27a and 5-27b show thermal gradients for various simple pipe support configurations.

90°F 113°F

to = 90°F

15″ 132°F 508°F 803°F 803°F ti = 900°F

888°F 858°F

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

304


900

Insulation

ti = 900°F = Process fluid temperature 1

800 ti

700

t °F

600

22

500

3

400

44 5

300

66

7

200

2′′ Typ 5

10 L [inches]

15

3′′ 7 6543 2 L

1

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

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 temperature of the bottom of the shoe base plate where Teflon is mounted to accommodate pipe movement. The Teflon cannot withstand temperature at or greater than 4008F. Referring to Figure 5-28 and using Eq. 5-58, we have 5′′ Calcium silicate insulation

12′′ φ SCH 40

to = 90°F ti = 750°F

L = 10 in.

P = 0.35 in.

Base width = 8 in. Figure 5-28. Pipe supported on a shoe with temperatures shown.


tp 5

305

kmAmti 1 hoApLto , 8F (kmAm 1 hoApL)

where ho 5 3.0 Btu/hr-ft2-8F for carbon steel in still air km 5 26.0 Btu/hr-ft-8F, thermal conductivity of pipe material L 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 in. Btu 112 in. 10 ft (750)8F 1 3.0 ft 1 2 ft (90)8F 1 5.25 2 1 2 144 in. hr-ft -8F 144 in. 12 Btu 5.22 in. Btu 112 in. 10.0 3(26.0) hr-ft -8F 1 144 in. 2 ft 1 (3.0) hr-ft -8F 1 114 in. 2 ft 1 12 2 ft4

(26.0) tp 5

2

Btu hr-ft-8F

2

2

2

2

2

2

2

2

2

2

2

2

2

2

tp 5 306.3038F Thus, the Teflon under the base plate is adequately protected. The amount of heat loss through the shoe base plate is q 5 hoAp (tp 2 to) Btu q 5 (3.0) hr-ft2-8F Btu q 5 504.706 hr

1

112 in.2 144 in.2

2 ft (306.303 2 90) 8F 2

Example 5-4: Emergency Constant Spring Replacement The author was called out in the middle of the night when it was discovered that a constant spring hanger had bottomed out and become a rigid hanger on a hot superheated steam line. The engineering contractor representative discovered that the existing spring that bottomed out was designed 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 was 9008C, and the operating temperature was 7808C. The contractor placed a chain to support the pipe temporarily with operational monitoring. There were few options, and a chain support acting like a come-along could only be tolerated for a short time. Constant springs have a long

306


delivery time half way around the world, so the option taken was to visit the junk yard and try to find a constant spring close to the required design. 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 load with the same travel, 3.5 in. This spring and the one being replaced were the classical offset-slider crank mechanism design described previously, designed and fabricated by the same spring vendor. The spring was a size 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. The spring had a turnbuckle, thus allowing for normal piping elevation adjustments. In our case, the actual piping load differed from the calculated load. The spring manufacturer claimed that an adjustment of 15 to 20% of the specified load could be made by turning the load adjustment bolt. Each division on the adjustment scale equals 5% of the rated load, according to the spring manufacturer’s technical bulletin (catalog). Referring to Figure 5-30, we see the adjustment bolt setting has eight graduations. Turning the bolt clockwise decreases the load 5% for each graduation; hence, four graduations represent a 20% decrease in the load. Similarly for a counterclockwise turn of the adjustment bolt, the load setting 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 the cold 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 is less than the total available travel to account for excursions. To translate the process temperatures and load requirements on the spring scale, we

307


NOTE: The minimum load is the load for the preceding size

Constant Support Size Selection Table Hgr. Size

4000 4100 4200 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

Total Travel (Inches) Maximum Rated Load in Pounds For Each Size

2

21⁄2

3

31⁄2

4

41⁄2

5

51⁄2

6

90 130 175 235 325 445 600 755 905 1090 1310 1465 1620 1810 2000 2240 2475 2770 3060 3400 3735 4120 4500 4950 5400 5940 6480

75 100 140 190 260 360 485 605 725 875 1045 1175 1295 1450 1605 1790 1980 2215 2445 2720 2990 3295 3500 3650 4320 4755 5185 5710 6230 6845 7645 8660 9480 10515 11700

60 85 115 160 215 295 405 505 605 725 875 980 1080 1210 1335 1495 1655 1850 2045 2265 2490 2745 3000 3300 3600 3960 4320 4755 5190 5705 6380 7215 7895 8770 9640 10700 11760 13055 14350 15985 17615 19310 21000 22180 24000

55 70 100 135 185 255 345 435 520 625 750 840 925 1035 1145 1280 1415 1585 1750 1940 2130 2350 2575 2830 3085 3395 3710 4075 4445 4890 5465 6180 6760 7510 8265 9160 10060 11180 12305 13700 15095 16545 18000 19140 20280 21700 23250

2

21 ⁄2

3

31 ⁄2

45 65 85 120 165 225 305 380 455 545 655 735 810 905 1005 1120 1240 1385 1530 1670 1870 2060 2250 2475 2700 2970 3240 3565 3890 4280 4780 5405 5915 6575 7225 8020 8820 9790 10760 11985 13205 14480 15750 16740 17725 19050 20370 22325 24290 26200 28115 30590 33065 35900 4

40 55 60 105 145 200 270 335 405 485 585 650 720 805 890 995 1100 1230 1365 1510 1660 1830 2000 2200 2400 2640 2880 3170 3460 3805 4250 4810 5265 5845 6425 7130 7840 8700 9565 10650 11740 12875 14005 14880 15750 16925 18100 19845 21590 23290 24990 27195 29400 31920 4 1 ⁄2

35 50 70 95 130 180 245 305 365 435 525 585 650 725 805 900 990 1110 1225 1360 1495 1650 1800 1980 2160 2375 2595 2855 3115 3425 3825 4330 4740 5260 5775 6415 7055 7835 8610 9585 10565 11580 12600 13390 14175 15235 16295 17860 19425 20960 22490 24475 26460 28745 5

35 45 65 85 120 160 220 275 330 395 475 535 590 660 730 815 900 1010 1115 1240 1360 1500 1640 1600 1965 2160 2355 2595 2830 3110 3475 3935 4305 4780 5250 5830 6415 7120 7830 8725 9610 10530 11455 12170 12885 13850 14815 16240 17660 19055 20450 22250 24045 26120 51 ⁄2

30 40 60 80 110 150 205 255 305 365 440 490 540 605 670 750 825 925 1020 1135 1245 1375 1500 1650 1800 1980 2160 2380 2595 2855 3190 3600 3945 4380 4815 5345 5880 6530 7175 7995 8810 9655 10500 11160 11825 12705 13585 14890 16190 17465 18740 20395 22050 23940 6

61⁄2

7

235 280 335 405 450 500 560 620 690 765 855 945 1050 1150 1270 1385 1525 1665 1830 1995 2195 2395 2635 2945 3330 3640 4045 4450 4940 5430 6030 6625 7375 8125 8910 9690 10300 10910 11725 12535 13740 14945 16125 17300 18830 20360 22110 61⁄2

25 35 50 70 95 130 175 215 260 310 375 420 465 520 575 640 710 790 875 975 1070 1180 1285 1415 1545 1700 1855 2040 2225 2445 2735 3090 3380 3750 4130 4585 5040 5595 6150 6850 7545 8275 9000 9565 10135 10890 11640 12755 13870 14970 16065 17490 18910 20530 7

71⁄2

8

205 245 290 350 390 435 485 535 600 660 740 815 905 995 1100 1200 1320 1440 1585 1730 1905 2075 2285 2555 2885 3150 3500 3845 4220 4600 5170 5745 6390 7040 7720 8400 8925 9450 10160 10870 11910 12960 13980 15005 16320 17640 19145 71⁄2

20 30 45 60 80 110 150 190 230 275 330 365 405 455 500 560 620 695 765 850 935 1030 1125 1240 1350 1485 1620 1785 1945 2140 2395 2705 2950 3285 3610 4010 4410 4895 5380 5990 6605 7240 7875 8370 8860 9525 10185 11160 12140 13100 14060 15300 16540 17955 8

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

81⁄2

9

91⁄2

180 215 255 310 345 385 430 475 530 585 655 720 800 880 970 1050 1165 1270 1400 1525 1680 1830 2015 2250 2545 2780 3095 3405 3775 4150 4605 5060 5640 6215 6815 7415 7875 8335 8960 9585 10510 11435 12330 13230 14390 15550 16885 81⁄2

20 30 40 55 70 100 135 170 205 245 290 325 350 405 445 500 550 615 685 755 830 915 1000 1100 1200 1320 1440 1585 1730 1905 2125 2405 2630 2920 3210 3565 3920 4350 4785 5330 5870 6435 7005 7445 7885 8470 9050 9925 10795 11645 12495 13600 14700 15960 9

160 190 230 275 310 345 385 425 475 525 585 650 720 790 870 950 1045 1140 1255 1370 1505 1645 1805 2020 2285 2500 2775 3050 3385 3720 4130 4545 5060 5575 6115 6650 7070 7485 8045 8600 9520 10260 11065 11870 12935 14000 15180 91⁄2

308

Escoe: Piping and Pipeline Assessment Guide 3.5″ Total travel Operating range

∆

S 1

2

3

4

5

6

Hot

7

8

9

∆ 10

Cold Travel pointer

∆ = Amount of extra travel to prevent spring bottoming out Figure 5-30. Spring travel scale on side of spring.

made the following calculations to see if the spring had enough travel to operate safely. The 9008C represents 3.5 in. of total travel on the spring. This temperature is the very maximum that could be expected in an excursion. Thus for the operating case, which had a process temperature of 7808C, the travel incurred for this case is 9008C 7808C Q x 5 3.033 in. 5 x 3.5 in. Load adjustment nut Side walls of spring (Typ)

Load adjustment scale Each mark from zero is a 5% adjustment Figure 5-31. Elevation view—looking into the face of the spring.


309

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

1 2 5 0.857 3.0 3.5

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, D5

1.429 5 0.71 graduation 2

This represents the number of graduations on the spring scale between each side of the operating range and the design, or overall travel. Now the process temperature at the time of installation is 7508C. We calculate the difference between the installed position and the operating condition (d). Thus, we have 9008C 7508C 5 Q x 5 3.033 in. x 3.5 in.

d5

3.033 in. 5 1.5165 in. 2

Now,

1 2 3.0 3.5

5 0.857 Q D 5

1.5165 in. 5 0.758 in. 2

The number of graduations on the scale is (1.0 2 0.857)(10 graduations) 5 1.429 graduations – difference between 3.0 in. of travel and 3.5 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)

310


Now we know the spring has enough travel at the desired load to operate safely. The spring was set and placed in use where it operated for ten months before the new spring arrived to replace it. The author monitored the spring during the entire period, and it performed very well. If a replacement spring had not been found in surplus, then the other alternatives would have been to shut down for months, which the plant management found unacceptable, or to fabricate a spring hanger from scratch.

Example 5-5: Pipe Header Simple Support Problem: A 30 in. f crude oil header is simply supported with the pipe resting on the steel. A company procedure calls for saddle-type supports with pads for piping 30 in. NPS (nominal pipe size) and larger. In case saddle pipe support cannot be installed, the line shall be analyzed for localized stresses. The question was: is the pipe resting on the steel overstressed? If it is overstressed, why has it not failed, as it has been in operation for thirteen years? Solution: If the company procedure specifies a saddle support for lines 30 in. and larger, then this standard must be adhered to. Regarding its stress 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 hollow surfaces (e.g., pipes or vessels). One classic example of this is the Zick analysis of horizontal drums supported on two saddles—contact stresses are not considered in the analysis. In the case of the pipe resting on a steel support, the situation is like a ring. The cross section of the pipe resting on a solid surface represents a bulkhead or supporting ring in a pipe, supported at the bottom and carrying a total load (W) transferred by tangential shear (v) distributed as is shown in Roark Table 9.2 Case 20 [Reference 6]. The Roark model used in Case 20 is called a tangential shear, where the vertical reaction at the pipe support is applied to counter the pipe’s weight, which includes the metal weight and content. The support reaction causes shear to flow around the pipe wall to replicate the pipe behaving like a beam. Rings are very common in structural design. This case is shown in Figure 5-32. The Roark Table 9.2 Case 20 gives a bending moment due to the shear stress caused by the weight of the pipe and the internal fluid. This bending


311

A V x

W Figure 5-32. Roark Table 9.2 Case 20.

moment acts through the wall of the pipe. Being a bending moment, it has compression and tensile components through the cross section of the pipe. The internal pressure stress acts only in tension. Thus, when the pressure stress is combined with the shear bending moment stress, one side of the wall is added and the other side of the wall is subtracted from the tensile pressure stress. The sign convention on the shear bending moment is not relative, as the stress will be one value on each respective surface (we add tensile 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 9.2 Case 20, the resisting cross section acts in the plane of the paper. The effective 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. As seen in Figure 5-34, the effective length is 2Ro. Typically, the effective

Pressure tensile load Pipe wall Shear bending stress

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

312

Escoe: Piping and Pipeline Assessment Guide Pipe

2Ro C

Ro

C

45″

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

length of the cross section would be minimum of 2Ro or 40tnom, where tnom is the nominal pipe wall thickness. The bending stress induced by the shear bending moment is calculated as follows: be 5 MAX[40tnom, 2Ro], see Figure 5-34 S5

(be)t2nom 6

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 bd 3 12 I bd 2 5 , for a rectangular cross-sectional area S5 5 C d 6 2 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, the shear moment bending stress is

s5

Mshear s

Eq. 5-61

The bending moments vary around the circumference of the pipe. Shown in Figure 5-35 is the bending moment distribution for the pipe loaded with water.

Piping Support Systems for Process Plants 0

0

0

0

313

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

The maximum bending moment occurs at 105.328 from the top point of the pipe. We loaded the pipe with crude oil using a specific gravity of 0.825 for the operational case. It was assumed that the pipe would not be operating at full pressure loaded with water. Combining the internal pressure stress and 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

0 15 30 45 60 75 90 105 120 135 150 165 180

Total Stress on Inner Wall, psi

Total Stress on Outer Wall, psi

16,924.6 17,194.2 17,960.3 19,908.7 20,417.2 21,676.8 22,617.8 22,989.8 22,580.6 21,244.7 18,923.6 15,661.2 11,607.7

14,452.1 14,182.4 13,416.3 12,278.0 10,959.5 9,699.9 8,758.8 8,366.8 8,796.0 10,132.0 12,453.0 15,715.5 19,769.0

The Section Modulus Conjecture: The criteria mentioned earlier concerning the effective area resisting the bending moment induced by shear in supporting the pipe has been a source of conjecture for many years. In the solution of a ring loaded in shear and inducting a moment into the plane of the pipe, the three-dimensional section resisting the bending moment is not considered because the ring is a two-dimensional problem. In Figure 5-34 the effective length of the cross section resisting the bending moment

314


induced by the shear is shown as 2Ro. The effective cross section is the maximum of 40tnom or 2Ro; however, in structural applications, the minimum value of 40tnom or 2Ro is often applied. In mechanical application, where we have internal pressure combined with the weight and thermal loads, the stresses can be significantly high—sometimes several times the ultimate strength of the pipe material. However, in reality, these pipes resting on simple supports do not fail, meaning that there is a fallacy in the equations—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. Bednar states that, to utilize the derived equation for the bending moment and bring the resulting stresses in the shell in agreement with actual measured stresses, we use what he describes as a “fictitious resisting width” of shell plate that is the smaller of 4Ro or L/2, where Ro is the shell radius and L is the length between supports. Of course, Bednar is discussing horizontal vessels supported on saddles, whereas here we are discussing continuous piping, like a continuous beam. Bednar’s Figure 6.5 on page 168 is the same ring solution we use from Roark, but they are derived from different sources. Another author, Troitsky [Reference 8] uses the same solution for the ring where the shear induces a bending moment from the load at the point of contact with the pipe support, with one exception. He is working with conveyor tube structures supported by ring girders. Troitsky uses the resisting section on pages 12–19 with the following expression: c 5 1.56√rt 1 b

Eq. 5-62

Where r is the cylinder radius, t is the cylinder thickness, b is the width of the ring girder, and c is the effective width resisting the shear-induced bending moment. Based on these criteria, we use the following to compute the effective width to obtain the section modulus, as follows: beff 5 be 1 1.56√R(tnom 2 CA)

Eq. 5-63

where R 5 mean radius, CA 5 corrosion allowance, and be is defined in Eq. 5-59. Thus, Eq. 5-60 becomes beff 1t2nom2 S5 6

Eq. 5-64

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


315

intensity smax is based on the primary or local membrane stresses plus primary bending stress plus the secondary stress (Pm or PL 1 Pb 1 Q), using the nomenclature of the ASME Section VIII Division 2 Appendix 4. The value of smax cannot exceed 3Sm, where all stresses may be computed under operating conditions. The following criterion must be met: 3Sm # 2(SMYS) The value of 3Sm is defined as three times the average of the tabulated Sm values for the highest and lowest temperatures during the operation cycle. In the computation of the maximum primary-plus-secondary stress intensity range, it may be required to consider the superposition of cycles of various conditions that produce a total range greater than the range of any individual cycle. The value of 3Sm may vary with each cycle, or combination of cycles, being considered since the maximum temperatures may be different in each case. Thus, care must be used to ensure the applicable value of 3Sm for each cycle, and combination of cycles, is not exceeded except as permitted by ASME Section VIII Division 2 Appendix 4 Paragraph 4-136.4. This requirement is seldom required in piping systems, unless different services are used in the pipe with differing temperatures and pressures. Since the pipe considered is operational, we can use the API 579 Fitness-for-Service recommended practice for the allowable stress, which is 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, the point of maximum bending moment, is as follows. Looking at Figure 5-36, we find the stresses in the 30 in. pipe are acceptable because, as mentioned previously, this is the angular position of maximum bending moment induced by the shear. However, if we take another pipe with a thinner wall, the stress magnitudes will differ. Consider a 24 in. line load with the same crude oil at the same facility. The pipe has a 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 same angular position of 105.238. We will now consider a 240 f pipe with a nominal wall thickness of 0.328 in. resting on a solid surface like we did for the 300 f pipe. The same methodology 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 the inside wall is in tension because of the bending moment. Thus, only part of the wall is of significant stress, not the entire wall of the pipe. This being

316


Rules Sheet Rules 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 Do Ro 5 2 Di Ri 5 2 p Am 5 ? Do2 2 Di2 4

3 43

4

Wreaction 5 Am?L?12?0.283 1

3 4 4 ? 3Di 4 ? r ?L?12 p

2

be 5 MAX(40 · tnom, 2 · Ro) beff 5 be 1 1.56? √R?(tnom 2 CA) beff?tnom2 6 M s5 S P Do sp 5 ? Ej 2 ?tnom S5

3 4 33

4 2 0.44

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


317

Variables Sheet Input

Name

17

Output

case plot

‘CASE_20 ‘y

caution matnum

‘_

matl

“Steel - A.S.T.M. A7-61T” 2.9E7 0.27

Unit

Table 9.2: Roark’s Formulas Formulas for Circular Rings Reference Number Generate plots? ‘n 5 no (Default 5 yes) Caution Message Material Number (See Material Table) Material name

‘thin 7107.502 1943

E nu ring W I

11.813

R

2

F

27.83 1.087E7

A G

in.2 psi

h

in.

105.23

Comment

psi lbf in.4 in.

DH DV v alpha

1.2938739E-4 22.4236034E-4 184.791 0.500310

in. in. lbf/in.

beta

2.670776

x N V M

22450.334 3.129E-2 215247.877

deg lbf lbf lbf-in.

NA VA MA MC delL

1696.791 0 24.143 13358.654 22.0957714E-4

lbf lbf lbf-in. lbf-in. in.

Young’s Modulus Poisson’s Ratio ‘thick or ‘thin: thick/thin ring Applied total load Area moment of inertia of X-section Mean radius of centroid of X-section Shape Factor of X-section THIN RINGS: Area of X-section Shear Modulus of elasticity THICK RINGS: Distance from Centroidal Axis to Neutral Axis Change in horizontal diameter Change in vertical diameter Tangential shear Hoop-stress Deformation Factor Radial Shear Deformation Factor AT SECTION: Angular Position Internal Force Radial Shear Internal moment AT A: Internal Force Radial Shear Moment Moment at C Increase in LOWER radius LOAD TERMS:

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

318 Input

30 1.358 3.1416 50 .03

1400 1

0

Escoe: Piping and Pipeline Assessment Guide Name

Output

Unit

LTM LTN LTV k1 k2 Di Do tnom Ro Ri p Wreaction

10065.9581776 22004.5965509 1637.2296434 3.170466 0.499690 27.284

lbf-in. lbf lbf Simplifying constants

15 13.642

in. in. in. in. in.

31272.705042

lbs

L r be

54.32

in.

S

18.616281

in.3

s

2819.061411

psi

Am P Ej s inner

122.195158

in.2 psi

15722.978937

psi

sp s outer

14903.917526 14084.856114

psi psi

beff CA s allow

60.568195

20000

Sm

35000

SMYS OK

ft

60000

Comment

psi

psi

Inside diameter of pipe Outside diameter of pipe Nominal pipe wall thickness Outside radius of pipe Inside radius of pipe Metal area of pipe x-section Combined of metal weight plus water Length between pipe supports Density of fluid, lbs/in.3 Effective length of pipe wall resisting bending due to shear Section modulus of resisting pipe section to shear moment Bending stress due to shear moment Cross sectional area of pipe Internal pressure Weld joint efficiency Total stress in inner shell 2 pressure 1 shear bending Internal pressure stress Total stress in outer shell 2 pressure 1 shear bending Allowable stress of pipe material at temperature Allowable stress for primary and secondary Specified Minimum Yield Strength of pipe material

‘acceptable

Figure 5-36b. cont’d.

the case, the high stress does not exist through the wall of the pipe, thus avoiding plastic deformation. As mentioned previously, there are hundreds of pipes resting on flat solid surfaces that do not fail or exhibit plastic deformation. The 30 in. f pipe mentioned was checked for cracks and deformation, and neither existed. The

Piping Support Systems for Process Plants Rules if or(x , 0, x 2 p . 1E-10) then caution 5 ‘Ang_Err else caution 5 ‘2 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 D Ro 5 o 2 Di Ri 5 2 p Am 5 ? Do2 2 Di2 4

3 43

4

Wreaction 5 Am?L?12?0.283 1

3 43 4

p ? Di2 ? r ?L?12 4

be 5 MAX(40 · tnom, 2 · Ro) beff 5 be 1 1.56? √R?(tnom 2 CA) beff?tnom2 6 M s5 S P Do sp 5 ? Ej 2?tnom

S5

3 4 33

4 2 0.44

s inner 5 sp 2 s s outer 5 sp 1 s s allow 5 3.0 · Sm If 3 · Sm , 2 · SMYS then OK 5 ‘acceptable If 3 · Sm . 2 · SMYS then OK 5 ‘unaccpetable Figure 5-37a. Equation sheet for 24 in. f pipe.

319

320


Variables sheet Input

17

Name

Output

case plot

‘CASE_20 ‘y

caution matnum

‘_

matl

“Steel-A.S.T.M. A7-61T” 2.9E7 0.27

Unit

Table 9.2: Roark’s Formulas Formulas for Circular Rings Reference Number Generate plots? ‘n 5 no (Default 5 yes) Caution Message Material Number (See Material Table) Material name

‘thin 7107.502 1943

E nu ring W I

11.813

R

2

F

27.83 1.087E7

A G

in.2 psi

h

in.

105.23

Comment

psi lbf in.4 in.

DH DV v alpha

1.2938739E-4 22.4236034E-4 184.791 0.500310

in. in. lbf/in.

beta

2.670776

x N V M

22450.334 3.129E-2 215247.877

deg lbf lbf lbf-in.

NA VA MA MC delL

1696.791 0 24.143 13358.654 22.0957714E-4

lbf lbf lbf-in. lbf-in. in.

Young’s Modulus Poisson’s Ratio ‘thick or ‘thin: thick/thin ring Applied total load Area moment of inertia of X-section Mean radius of centroid of X-section Shape Factor of X-section THIN RINGS: Area of X-section Shear Modulus of elasticity THICK RINGS: Distance from Centroidal Axis to Neutral Axis Change in horizontal diameter Change in vertical diameter Tangential shear Hoop-stress Deformation Factor Radial Shear Deformation Factor AT SECTION: Angular Position Internal Force Radial Shear Internal moment AT A: Internal Force Radial Shear Moment Moment at C 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 operation cycles. The same is true for the 24 in. f pipe and many others. It is concluded that this methodology adequately predicts the behavior of a pipe resting on a simple support, or flat surface. However, many companies consider

Piping Support Systems for Process Plants Input

24 .328 3.1416 30 .03

550 1

0

Name

Output

Unit

LTM LTN LTV k1 k2 Di Do tnom Ro Ri p Wreaction

10065.9581776 22004.5965509 1637.2296434 3.170466 0.499690 23.344

lbf-in. lbf lbf

12 11.672 7107.502462

lbs

L r be

24

in.

S

0.485396

in.3

s

231413.258210

psi

Am P Ej s inner

24.392689

in.2 psi

51315.209430

psi

sp s outer

19901.951220 211511.306991

psi psi

beff CA sallow

27.070730

20000

Sm

35000

SMYS OK

Comment

Simplifying constants in. in. in. in. in.

ft

60000

321

psi

psi

Inside diameter of pipe Outside diameter of pipe Nominal pipe wall thickness Outside radius of pipe Inside radius of pipe Metal area of pipe x-section Combined of metal weight plus water Length between pipe supports Density of fluid, lbs/in.3 Effective length of pipe wall resisting bending due to shear Section modulus of resisting pipe section to shear moment Bending stress due to shear moment Cross sectional area of pipe Internal pressure Weld joint efficiency Total stress in inner shell 2 pressure 1 shear bending Internal pressure stress Total stress in outer shell 2 pressure 1 shear bending Allowable stress of pipe material at temperature Allowable stress for primary and secondary Specified Minimum Yield Strength of pipe material

‘acceptable

Figure 5-37b. cont’d.

it good engineering practice to place pipes 30 in. f and larger on saddle supports. It is also noted that, for large diameter thin cylinders resting on the ground in stock yards and construction sites, angle beams 908 to one another are tack welded to the inside surface to prevent excessive ovaling. In conclusion, a saddle support is required for the pipe if constructed new.

322


Note: The internal pressure stress was computed from the following formula, which is developed from Eq. 2-13(a), where Y 5 0.4 and D 5 Ro /2: t5

PRo sEW 1 0.4P

where W 5 1 for temperature less than 9508F, as in this case. Solving for the stress we have the following:

s5

P E

1

Do 2 0.4 2t

2

where Do 5 outside diameter Ro 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 Publishing Company, 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, Macmillan Publishing Company, New York. 6. Warren C. Young and Richard G. Budynas, Roark’s Formulas for Stress and Strain, 7th edition, 2002, McGraw-Hill Publishing Company, 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.

Heat Transfer Through Pipe Support.pdf

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