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Information about 2 Construction

Published on December 23, 2008

Author: jacobrajeev

Source: slideshare.net

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PROJECT SCOPING DATA Project scoping data worksheet for major facilities Project Name 2.8 Describe type of soil—sandy, rocky, caliche, etc. 2.9 Does the site drain naturally? If fill material is Location required, what is the availability? Is sand-asphalt mix Latitude available? Longitude • Offshore D Onshore D Arctic 3.0 Drafting D Marsh Site elevation, ft Project description 3.1 Will the drawings need to be sealed by a professional engineer? 3.2 Determine drawing forms to be used and type of drawings required—single line or double line. 4.0 Structural and architectural 1.0 Type of contract required 4.1 Will all exposed steel need to be galvanized or have 1.1 Describe the type of contract that the project will be other protective coating? based on. 4.2 Is sand, rock, cement, and ready-mixed concrete 1.2 What is the completion date? available locally? Price? 1.3 What type permits will be required and who will 4.3 Requirements for building design: obtain them? Steel 1.4 Will an environmental impact statement be required? Concrete/brick 1.5 Describe currency requirements if in a foreign country. Cooling and heating requirements 1.6 If the project is located in a foreign country, does the Other requirements country have regulations/codes that will need to be complied with? 5.0 Piping 2.0 General engineering 5.1 How will storm water, waste water, and sewage be disposed of? Will an oil-water separator be required? 2.1 Does the client have engineering standards and 5.2 Will connections to any existing pipelines be specifications, or will it be necessary to develop required? these, or can the Engineering Contractor's specifica- 5.3 Are there any facilities outside the job site that must tions and standards be used? be given consideration in the development of a plot 2.2 Will in-country engineering concerns, such as design plan? institutes, be required to provide engineering services? 5.4 What is the pressure rating for any existing mani- 2.3 If in-country engineering facilities are required, how folds, suction lines, or outgoing pipelines? much assistance will be required? 5.5 Are there any special requirements for isolating the 2.4 Describe any existing above or underground obstruc- facility? tions at the job site. 5.6 Will scraper traps be required? Incoming, outgoing, 2.5 Obtain copies of any applicable local codes dealing or both? with air and/or water pollution, sanitary systems, 5.7 How many relief headers are required? electrical systems, and structures. Obtain copies of 5.8 If system relief valves are required, where will they any regulations dealing with waste disposal. relieve to? 2.6 Is soil bearing data available? 5.9 Will a fire water system be required? Is a source 2.7 If the project involves a pipeline that transports of water available? If a source is available, what is the heat-sensitive fluid, are any soil thermal conductivity capacity of the source? data available? 5.10 Is the water fresh, brackish, or seawater?

6.0 Electrical 9.5 Is the location near an established seaport? If not, determine the nearest seaport and logistics for moving 6.1 Is power available from the local power grid? If yes, material from the seaport to the job site. This should specify voltage, frequency, phase, and system capacity. include any weight limitations, load width restrictions, 6.2 If power generation will be required in connection and requirements for moving heavy equipment at with the project, will it be a requirement to connect to night during periods of light road traffic. a local grid? 9.6 Where will material be stored pending installation? 6.3 If local power is available, define the approximate 9.7 What procedure will be used for turning material over location where the power will enter the job site. to the contractor? 6.4 Will the local power entity furnish and install any substation equipment that will be required? 6.5 If power is to be supplied from a local power grid, 10.0 Special considerations attempt to determine the short-circuit capacity of the system. 10.1 Is heavy equipment available locally and does it have the capacity to lift the heaviest piece of 6.6 Will across-the-line starting of large electric motors be material? permitted? 6.7 If across-the-line starting is not permitted, what is 10.2 Is an adequate supply of sufficiently skilled local the maximum horsepower that will be permitted for labor available? across-the-line starting? 10.3 If the project is in a foreign location, is there a limit on the use of expatriates? 10.4 What is the procedure for obtaining a work permit? 7.0 Instrumentation equipment 10.5 What is the availability of contractors that may be working in the area? 7.1 What type of control panel will be required? 10.6 Will a construction camp be required? 7.2 Define electrical classification that will be required 10.7 If a construction camp is required, what is the for a control house if a control house is required. availability of a local catering service? 7.3 Are local communications facilities available? De- scribe. 11.0 Environmental information 8.0 Equipment 11.1 Ambient Temp. (0F) Max. Min. 11.2 Relative humidity 8.1 Have any long-delivery items been placed on order? Wet bulb (0F) Max. Min. List items and delivery schedule. Dry bulb (0F) Max. Min. 8.2 Who will supply general equipment specifications? 11.3 Water temp, surface (0F) Max. Min. 8.3 Will in-country purchasing be required? 11.4 Water temp, bottom (0F) Max. Min. 8.4 List design criteria for equipment sizing. 11.5 Potential for icing Y or N 8.5 Define spare equipment requirements. If equipment 11.6 Rainfall (inches/hour) Max. Min. is purchased outside the country where the project is 11.7 Prevailing wind direction located, consideration should be given to transit time 11.8 Velocity Max. Min. from country of origin to the job site. 11.9 Weather window (time of year) 8.6 Define the design life for the facility. 8.7 Will oil storage tanks be required? Will secondary seals be required? Will tank mixers be required? 12.0 Fluid characteristics 9.0 Material purchasing 12.1 Crude oil 12.2 Sediment and water, % 9.1 If a preferred vendor list is available, obtain a copy of 12.3 Free water the list. 12.4 Salt content (#/1,000 bbls) 0 9.2 If such list is not available, will it be a requirement to 12.5 API gravity @ F and 0 develop a list for the client's approval? @ F (2 required) 9.3 Determine the shipping address. 12.6 Viscosity cP @ °F and 9.4 If material is shipped to a foreign location, who will cP @ °F (2 required) 0 handle receipt of the material and clearance with local 12.7 Oil pour point F customs? 12.8 Reid vapor pressure

12.9 Produced water: oil & grease (ppm) 12.15 Water content #/MMSCF or dew point 0 12.10 Total suspended solids (mg/kg oil) F 12.11 Natural Gas 12.16 Max. heating value (BTU/SCF) 12.12 MOL % inerts 12.17 Min heating value (BTU/SCF) 12.13 MOL % CO 2 12.18 NGL content gal/MCF or dew point 0 12.14 ppm H2S or grains/100 SCF F 13.0 Offshore Environmental Information Parameter 1 Year 5 Year 25 Year 100 Year Waves Direction Significant wave height, ft Maximum wave height, ft Crest elevation, ft Significant max. wave period, sec Peak period, sec Tides Astronomical tide, ft Storm surge, ft Wind @ 30 ft elevation above MLW 1 -minute average, kt 1 -hour average, kt 2-second gust, kt Current Surface speed, kt 3 ft from bottom, kt Wave force coefficients C D with marine growth C D with no marine growth CM Marine growth to 150 ft, inches Sea temperature, 0 F Maximum surface Minimum surface Maximum bottom Minimum bottom Air temperature, 0 F Maximum Minimum Relative humidity Wet bulb, 0 F Dry bulb, 0 F Notes/Comments RIGHT-OF-WAY How to determine the crop acreage included in a right-of-way strip Multiply the width of the strip in feet by the length in Example. A right-of-way 35 ft wide crosses a cultivated rods; divide this by 2,640 to obtain the acreage. If the ends field for a length of 14 rods; how many acres of crop were of the strip are not parallel, use the length of the center line destroyed? of the right-of-way. 35 x 14/2, 640 = 0.18 acres, or almost V5 acre.

Example. A right-of-way 50 ft wide crosses a field for 330 50 x 330/2, 640 = 6.25 acres. rods. How many acres of crop were destroyed? The rule is exact, not an approximation. Clearing and grading right-of-way: labor/equipment considerations To estimate labor crew and equipment spread for clearing 7. Providing sufficient and proper lighting where and grading operations, the following items, as they may required. apply to a given project, should be given consideration: 8. Providing guards where required. 9. Preserving all trees, shrubs, hedges and lawns where 1. Removal of trees, brush, and stumps. required. 2. Grubbing and removal of stumps that are in the way 10. Grading irregularities where required. of the ditch. 11. Preserving topsoil for replacement, through all 3. Disposal of all debris, including method of disposal and cultivated or improved fields and pastures, to its length of haul. original position. 4. Clearing area spoil a sufficient distance from the ditch 12. Proper grading of the terrain so as to allow passage line so that the spoil-bank from the ditching operations of loaded trucks and equipment hauling materials and will not fall in any foreign material that might become so ditching operations can be properly performed. mixed with the excavated spoil. 13. Protecting and preserving existing drainage facilities. 5. Cutting of merchantable timber into standard lengths 14. Protecting any existing structures or pipelines. and stacked along the right-of-way for disposition by 15. Protecting any telephone or utility lines and keeping others if specifically required by the right-of-way them in service. agreement. 16. Cutting through fences and hedges where required 6. Providing temporary walks, passageways, fences, or and replacing these when necessary. other structures so as not to interfere with traffic. 17. Installing gates and fences where required. Estimating manhours for removing trees NET MANHOURS-EACH NET MANHOURS-EACH Softwood Trees Hardwood Trees Softwood Trees Hardwood Trees Average Tree Average Tree Diameter Open Congested Open Congested Diameter Open Congested Open Congested in Inches Area Area Area Area in Inches Area Area Area Area Cross-cut Saws Chain Saws

Manhours include ax trimming, cutting down with cross- Manhours do not include hauling, piling and burning of cut saws or chain saws, and cutting into 4-ft lengths for the trees or branches, or the removal of stumps. tree diameter sizes as listed previously. Manhours are average for various heights of trees. Estimating manhours for removing tree stumps NET MANHOURS-EACH NET MANHOURS-EACH Oper. Powder Oper. Powder Item Laborer Engr. Man Total Item Laborer Engr. Man Total Grub & Removal by Hand Blast & Pull with Tractor 8quot; to 12quot; diameter 6.00 6.00 8quot; to 12quot; diameter 0.83 0.11 1.50 2.44 14quot; to 18quot; diameter 7.50 7.50 14quot; to 18quot; diameter 1.05 0.23 2.33 3.61 20quot; to 24quot; diameter 9.00 9.00 20quot; to 24quot; diameter 1.50 0.30 3.38 5.18 26quot; to 36quot; diameter 11.20 11.20 26quot; to 36quot; diameter 2.11 0.42 4.76 7.29 Manhours include excavating and removing by hand or blasting and removing with cables and tractors. Manhours do not include burning or removal from premises. Clearing and grading right-of-way Equipment Spread Code description NUMBER OF UNITS IFOR L = Light—light brush and grass, no trees. 50 Linear Ft 80 Linear Ft 100 Linear Ft Equipment Width Width Width M = Medium—considerable brush of larger size. Description L M MH H L M MH H L M MH H MH = Medium Heavy—large brush and small trees. H = Heavy—much small brush, many small trees, and D8 Tractor 1 1 1 1 1 1 2 2 1 2 3 3 W/Dozer occasional large trees. 07 Tractor 0 0 1 1 1 1 1 1 1 1 2 2 W/Dozer Labor Crew Truck-2 1 / 2 2 2 2 2 2 2 2 2 2 2 3 3 NUMBER OF MEN FOR Ton Dump 50 Linear Ft 80 Linear Ft 100 Linear Ft Truck— 1 1 2 2 1 2 3 3 2 3 4 4 Personnel Width Width Width Pick-up Description L M MH H L M MH H L M MH H Ripper or 1 1 1 1 1 1 1 1 1 2 2 2 Brushrake Foreman 1 1 2 2 1 2 3 3 2 3 4 4 Operator 1 1 2 2 1 2 3 3 2 4 5 5 Mechanic 1 1 1 1 1 1 1 1 1 1 1 1 Swamper 1 1 2 2 1 2 3 3 2 4 5 5 Above equipment spread should be ample for clearing Truck Driver 2 2 2 2 2 2 2 2 2 2 3 3 and grading 1 mile of right-of-way per 10-hour day for Laborer 10 15 20 30 15 25 30 40 20 35 40 50 the width and conditions outlined. Haul trucks are based Total Crew 16 21 29 39 21 34 42 52 29 49 58 68 on round trip haul of 2 miles. If brush and trees are to be burned on site, omit above dump trucks. Small tools such as saws, axes, etc., must be added as required for Above total crew should be ample for clearing and grading the individual job. 1 mile of right-of-way per 10-hour day for the width and

conditions outlined. Crew spread includes cutting, stacking MH = Medium Heavy—large brush and small trees. or piling, loading, and hauling a round-trip distance of 2 H = Heavy—much small brush, many small trees, and miles. If burning is necessary or permitted, substitute fire occasional large trees. tenders for dump truck drivers. See Clearing and Grading equipment spread for number of dump trucks. Source Code description Page, J. S., Cost Estimating Man-Hour Manual for Pipelines L = Light—light brush and grass, no trees. and Marine Structures, Gulf Publishing Co., Houston, M = Medium—considerable brush of larger size. Texas, 1977. DITCHING How many cubic yards of excavation in a mile of ditch? Multiply the width in inches by the depth in inches by 12 x 30 x 1.36 = 490 cubic yards per mile. 1.36; the answer is cubic yards per mile. The rule is correct within about % of 1%, actually, the errors in depth and width are much greater than this. To get Example. How many cubic yards of excavation in a mile the cubic yards per 1,000 ft, as in computing rock ditching, of 12-in. ditch 30-in. deep? use 0.257 instead of 1.36. Shrinkage and expansion of excavated and compacted soil Ever notice how the spoil from a ditch occupies a greater Volume of undisturbed clay soil: volume than the ditch itself? There's a reason. Excavate sand and it expands about 10%. Ordinary soil expands about 25 3.5 x 5.0 = 17.5 cu ft per lineal foot of ditch. percent and clay expands about 40%. Volume of the spoil will be 143% of the undisturbed Here's a summary of the bulk you can expect from volume. excavated soil: Volume of spoil = 1.43 x 17.5 = 25 + cu ft per lineal ft. Type soil Undisturbed Excavated Compacted Through mechanical compaction this volume can be Sand 1.00 1.11 0.95 reduced to 15.75 cu ft. Ordinary earth 1.00 1.25 0.90 Clay 1.00 1.43 0.90 Example. Find the volume of loose spoil from a pipeline ditch 42 inches wide, 60 inches deep. The excavation is through clay. Ditching and trenching: labor/equipment considerations In determining the labor crew and equipment spread 1. Ditching or trenching for buried pipelines should be for ditching and trenching operations, the following should in accordance with the following table of minimum be given consideration should they apply to the particular width and coverage for all soil formations. project: 2. In rock, cut ditches at least 6 in. wider.

3. If dirt-filled benches are used, ditch should be Nominal Pipe Minimum Normal Minimum excavated deeper to obtain proper coverage. Size Inches Width Inches Coverage Inches 4 22 30 4. Trench should be excavated to greater depth when 6 26 30 required for proper installation of the pipe where the 8 26 30 topography of the country warrants same. 10 26 30 5. Repair any damage to and maintain existing natural or 12 30 30 other drainage facilities. 14 32 30 6. Do not open ditch too far in advance of pipelay crew. 16 36 30 7. Obtain permits for blasting. 18 38 30 8. When blasting, use extreme caution and protection. 20 40 30 9. Clean up blasted rock to prevent damage to coated 24 44 30 pipe. 30 50 30 36 52 36 42 58 36 CONCRETE WORK How to approximate sacks of cement needed to fill a form To obtain a close estimate of the number of sacks of Example. How many 94-lb. sacks of cement will be cement that will be required, first determine the volume required to fill a form for a concrete base 10 ft by 10 ft (cubic feet) to be filled in the form. Divide the volume by if it is to be 6 in. thick? 4.86 to approximate the number of sacks of cement needed. 10 x 10 x 0.5 — = 10.3, or 11 sacks needed. 4.86 What you should know about mixing and finishing concrete To determine proper mix, divide the constant 44 by the S = 6.28 x 2 x 0.035 sum of the parts of cement, the parts of sand, and parts of = 0.44 cubic yard of sand gravel to determine the number of bags of Portland cement required. Multiply the number of bags of cement as G = 6.28 x 4 x 0.035 determined above by parts of sand and the constant 0.035 = 0.88 cubic yard of gravel to calculate the number of cubic yards of sand needed. To determine cubic yards of gravel needed, multiply bags of cement by parts of gravel x 0.035. To increase or decrease slump of concrete, add or subtract 1 gallon of water per cubic yard of mix and subtract or add Example. Calculate the quantities of cement, sand, and 20 Ib. of aggregate to maintain yield. gravel required for 1 cubic yard of 1:2 :4 concrete. To adjust from no air to air-entrained concrete and maintain strength, reduce water V4 gallon per sack of cement C= 44 and reduce sand 10 Ib. per sack of cement for each 1% of entrained air. 1+ 2+ 4 = 6.28 bags of cement

PIPE LAYING How to determine the degrees of bend in a pipe that must fit a ditch calling for a bend in both horizontal and vertical planes Rule. To find the number of degrees in the combination X2 = 25 bend, square the side bend and the sag or overbend; X = 5° add them together and extract the square root. The answer will be the number of degrees necessary to make the pipe fit the ditch. Example. Determine the bend to make in a pipe whose ditch has a 9° sag and 12° sidebend. Let X = the unknown Example. Determine the bend to make in a pipe whose angle: ditch has a 3° overbend and a 4° side bend. Let X = the unknown angle: X2 = 92 + 122 X2 = 3 2 + 4 2 X2 = 225 X2 = 9 + 16 X = 15° How to bend pipe to fit ditch—sags, overbends, and combination bends UNLIKE SIGNS LIKE SIGNS ARE ARE ADDED SUBTRACTED SLOPE 0 0 OO' Note: 6 * Side Bend Left Must Be Considered In Making 'Combination' ELEVATION 8* SIDE BEND RIGHT 6 ° SIDE BEND LEFT PLAN To make straight sags or overbends fit the ditch (see draw- ing), add angles of unlike signs and subtract those of like signs. Example. (Sta. 1 + 40) Example. (Sta. 2 + 40) Slope + 100OO7 Slope - I0OO7 SIoPe-IS0OO7 Slope- I0OO7 (add) H0OO7 overbend (1) 60OO' side bend left (2) (sub.) 140OO' overbend Combination bend = 11° + ( x 6°) = 13°00' overbend left. In cases involving either a sag or overbend, in addition to a side bend, the rule becomes: Make the combination bend Note: This rule gives an error of approximately 1° for a equal to the largest angle plus lfo of the smallest. maximum bend of 18.5°.

Pipe bending computations made with hand-held calculator Maximum code radii for pipe cold bends requires printer for calculations Frank E. Hangs, Sovereign Engineering, Inc., Houston Table 1 Cold bending pipe is subject to provisions of Liquid Examples of computations using the cold bend program Petroleum Transportation Piping Systems, B31.4, and Gas Transmission and Distribution Piping Systems, B31.8. These codes stipulate a minimum bending radius for each size pipe EXflHPLE I EXflHPLE 1 COHT. in addition to requirements for thinning, flattening, etc. XEQ quot;BENDquot; LLT=? The following program (Table 1), written for the Hewlett 5.8888 RUN PIPE BENDS Packard 41C/CV calculator, addresses the geometry of VLT=S.8388 fabricating offset bends, sloping scraper traps, connections, i l TRIPL=? HLT=33.78S8 28.9888 R N U and direction changes common to all pipelines. Minimum D=? IH. EXflHPLE 2 code radii are calculated for each size of oil or gas line. The 12.7586 RUN GflS PL=>12? Y?i N? XEQ 17 vertical distance between below and above ground center H RUH DIRECTION lines is matched to a design distance with given tangents V=? CHHHGE 8.8886 RUN (5 ft is a convenient tangent for pipe bending machines). T=? L TURN=? 36.8888 RUH The bend angle, overall horizontal distance, and total length 5.8888 RUH R=? of pipe are calculated for each configuration. ii INCR=? 88.8888 RUH 8.5888 RUH T=? A printer is a must. The prompting feature of this 5.8888 RUN calculator asks for specific input data. The Results Recap RESULTS RECflP RESULTS routine prints out calculated data and inputs. Each item is il=25.5888 identified for permanent record. Additional printouts of R,'B=18.8886 ¥=13.2188 R=19.1258 H=49.3381 Results Recap may be made by XEQ quot;DATA.quot; H=35.4929 L=Sl.8879 L=37.&235 Fig. 1 and Example 1 are a typical offset bend. V CBLC=S.8312 V GIVEH=S.8886 Fig. 2 and Example 2 are a conventional direction change. L TURN=38.0888 R=88.8888 Fig. 3 and Example 1 show receiver and launcher scraper D IH=12.7588 trap connections, where the traps are not level. D FT=I.8625 EXflHPLE 3 T=5.8888 Fig. 4 and Example 3 are a crossing under a foreign line, *1. INCR=8.5880 XES -BEHD- where offset bends are used. XEQ 15 SCRfiPERTRHPS PIPE BENDS The oil and gas codes differ in bending requirements. These TRflP i = ? i TRIflL=? are defined by the radius of bend/diameter ratio. The oil code, 5.8888 RUN 8.8888 RUN B31.4, specifies the R/D ratios as follows: for 12 in. and less, R TRflP? Y? H? D=? IH. use 18; 14 in., 21; 16 in., 24; 18 in., 27; 20 in. and larger, use 30. V RUN 14.8888 RUN GflS PL=>12? Y?, H? The gas code, B31.8, specifies for all pipe 12 in. and H RUN larger, the ratio is a constant of 38.1987. (The code states VR=7.5227 V=? HR=37.1487 2.5138 RUN quot;longitudinal axis of the pipe shall not be deflected more LR=38.6925 T=^ than I1Z2 degrees in any length along the pipe axis equal to TRflP 4=5.8888 5.8888 RUN LRT=? i INCR=? the diameter of the pipe.quot;) 18.0888 RUH 8.5888 RUH The program prompts: Gas pipeline greater than 12 in. VRT=6-6511 RESULTS RECflP Yes? or No? (Gas PL > > 12Y?N?). Is line gas or oil? When this HRT=47.1827 information and other data are keyed in, the proper radius is XEQ 15 41=10.5888 SCRfiPERTRfiPS R/D=21.8888 determined and the results computed and printed out. R=24.5888 If the scraper trap routine is desired, it should be run TRflP *=? H=28.7621 5.8888 RUN L=28,9797 immediately after quot;Bendquot; while pertinent data are in V CflLC=2.6429 R TRRP? Y? W> storage—XEQ 15 for each receiver and each launcher. H RUN V GIVEH=2.5138 Here again the program asks if this is a receiver (R trap) Y? D IH=14.8808 VL=8.3942 or N? Note that trap angle is requested for R & L traps (they HL=33.8878 D FT=I.1667 could differ), and the length of the traps may not be equal. LL=35.3545 T=5.8888 TRflP 4=5.8888 L INCR=8.5888

Table 1 Examples of computations using the cold bend program (continued) 81HBL quot;BEHB- 64 38.1972 125 RCL 85 198 CLR 256 RCL 17 322 RCL 18 388 PROHPT 62 ODV 65 XEQ 13 126 COS 191 quot;Zl INCR=quot; 257 - 323 + 339 STO 32 63 SF 12 127 * 192 RRCL 11 258 RCL 18 324 STO 24 398 RCL 38 04 quot;PIPE quot; 66*LBL 13 128 + 193 RVIEH 259 + 325 RCL 13 391 COS 65 quot;HBENDSquot; 67 STO 03 129 STO 12 194 CLR 268 STO 21 326 RCL 19 392 CHS 06 PRfi 68 RCL 07 138 RCL 18 195 BEEP 261 RCL 13 327 + 393 1 07 RDV 69 * 131 4 196 RDV 262 RCL 19 328 STO 25 394 + 08 CF 12 70 STO 08 132 * 197 RDV 263 - 329 RDV 395 RCL 31 09 quot;Zl TRIRL=?quot; 71 GTO 85 133 RCL 88 198 STOP 264 STO 22 338 RDV 396 * 10 PROHPT 134 2 265 RDV 331 -VR=- 397 RCL 38 11 STO 05 72*LBL 05 135 * 199*LBL 15 266 quot;VL=- 332 RRCL 23 398 SIN 12 -D=? IN.quot; 73 quot;V=?quot; 136 RCL 05 288 SF 12 267 RRCL 28 333 RVIEH 399 RCL 32 13 PROHPT 74 PROHPT 137 57.2 958 281 -SCRRPERquot; 268 RVIEH 334 CLR 488 * 14 STO 06 75 STO 09 138 / 282 -!-TRRPS- 269 CLR 335 -HR=quot; 401 + 15 12 76 quot;T=?quot; 139 * 283 PRR 278 quot;HL=quot; 336 RRCL 24 402 STO 33 16 / 77 PROHPT 148 + 284 CF 12 271 RRCL 21 337 RVIEH 483 RCL 38 17 STO 07 78 STO 18 141 STO 13 285 RDV 272 RVIEH 338 CLR 494 SIN 18 quot;GfiS PL=>12?quot; 79 quot;Zl INCR=?quot; 286 -TRRP Z=?quot; 273 CLR 339 -LR=quot; 405 RCL 31 19 quot;I- Y?, N?quot; 88 PROHPT 287 PROHPT 274 -LL=quot; 348 RRCL 25 486 * 20 RON 81 STO 11 142+LBL quot;D fiTfiquot; 288 STO 84 275 RRCL 22 341 RVIEH 487 RCL 32 21 PROHPT 143 RDV 289 COS 276 RVIEH 342 CLR 488 + 22 ASTO Y 144 CLR 218 CHS 277 CLR 343 -TRRP Z=quot; 489 RCL 38 23 fiOFF 82»LBL 81 145 -RESUL TS quot; 211 1 278 quot;TRRP Z=quot; 344 RRCL 84 419 COS 24 -Y- 83 RCL 08 146 quot;I-RE CRP- 212 + 279 RRCL 84 345 RVIEH 411 RCL 32 25flSTOX 84 2 147 RVIE U 213 RCL 88 288 RVIEH 346 CLR 412 * 26 X=Y? 85 * 148 CLR 214 * 281 CLR 347 -LRT=?- 413 + 27 GTO 12 86 RCL 85 149 RDV 215 STO 15 282 quot;LLT=?quot; 348 PROHPT 414 STO 34 28 12 87 COS 158 quot;Zl= 216 RCL 18 283 PROHPT 349 STO 81 415 RCL 38 29 RCL 06 88 CHS 151 RRCL 85 217 RCL 84 284 STO 82 358 RCL 84 416 57.2953 38 X<Y? 89 1 152 RVIEN 218 SIN 285 RCL 84 351 SIN 417 / 31 GTO 86 98 + 153 CLR 219 * 286 SIN 352 * 418 RCL 31 32 14 91 * 154 quot;R/D 228 STO 16 237 * 353 CHS 419 * 33 RCL 06 92 RCL 18 155 RRCL 03 221 RCL 88 288 RCL 28 354 RCL 23 420 RCL 32 34 X=Y? 93 2 156 RVIE H 222 RCL 84 289 + 355 + 421 2 35 GTO 87 94 * 157 CLR 223 SIN 298 STO 26 356 STO 28 422 * 36 16 95 RCL 85 158 quot;R=quot; 224 * 291 RCL 82 357 RCL 24 423 + 37 RCL 96 96 SIN 159 RRCL 88 225 STO 17 292 RCL 04 358 RCL 81 424 STO 35 33 X=Y? 97 * 168 RVIE U 226 RCL 18 293 COS 359 RCL 84 425 RDV 39 GTO 08 98 + 161 CLR 227 RCL 84 294 * 368 COS 426 quot;RESULTSquot; 40 18 99 STO 14 162 quot;H=quot; 228 COS 295 RCL 21 361 * 427 XEQ -PRR- 41 RCL 06 188 FS? 81 163 RRCL 12 229 * 296 + 362 + 423 RDV 42 X=Y? 191 GTO 83 164 RVIEH 238 STO 18 297 STO 27 363 STO 29 429 -V=- 43 GTO 09 182 RCL 89 165 CLR 231 RCL 84 298 RDV 364 RDV 430 RRCL 33 44 19.99 183 X O Y 166 quot;L=quot; 232 57.2958 299 quot;VLT=quot; 365 -VRT=- 431 RVIEH 45 RCL 06 184 X<=Y? 167 RRCL 13 233 / 388 RRCL 26 366 RRCL 28 432 CLR 46 X)Y? 185 GTO 82 168 RVIE U 234 RCL 88 381 RVIEH 367 RVIEH 433 -H=- 47 GTO 18 186 GTO 83 169 CLR 235 * 382 CLR 368 CLR 434 RRCL 34 178 -V CR LC=quot; 236 STO 19 303 quot;HLT=- 369 quot;HRT=quot; 435 RVIEH 48*LBL 86 187*LBL 02 171 RRCL 14 237 -R TRRP?quot; 304 RRCL 27 378 RRCL 29 436 CLR 49 18 188 RCL 11 172 RVIE H 238 quot;H Y? N?quot; 385 RVIEH 371 RVIEH 437 -L=- 50 XEQ 13 189 ST+ 85 173 CLR 239 RON 386 CLR 372 CLR 438 RRCL 35 118 GTO 81 174 -V GIV EN=quot; 248 PROHPT 387 RDV 373 RDV 439 RVIEH 51*LBL 07 175 RRCL 89 241 RSTO Y 388 RDV 374 STOP 448 CLR 52 2! HItLBL 83 176 RVIE H 242 ROFF 389 STOP 441 RDV 53 XEQ 13 112 RCL 18 177 CLR 243 -Y- 375*LBL 17 442 RDV 54*LBL 08 113 2 178 RDV 244 RSTO X 318*LBL 16 376 SF 12 443 -Z TURN=quot; 55 24 114 * 179 -D I N=quot; 245 X=Y? 311 RCL 14 377 quot; DIRECTION quot; 444 RRCL 38 56 XEQ 13 115 RCL 88 188 RRCL 86 246 GTO 16 312 RCL 15 378 -h CHRNGEquot; 445 RVIEH 116 2 181 RVIEH 247 RCL 14 313 - 379 PRR 446 CLR 57*LBL 09 117 * 182 CLR 248 RCL 15 314 RCL 16 388 CF 12 447 -R=- 53 27 113 RCL 85 183 quot;D F T=quot; 249 - 315 - 381 -Z TURN=?quot; 448 RRCL 31 59 m 13 119 SIN 184 RRCL 87 258 RCL 16 316 STO 23 382 PROHPT 449 RVIEH 128 • 185 RVIEH 251 + 317 RCL 12 383 STO 38 458 CLR 69*LBL 18 121 + 186 CLR 252 STO 28 318 RCL 18 384 -R=?- 451 RDV 61 38 122 RCL 18 187 quot;T=quot; 253 RCL 12 319 - 385 PROHPT 452 RDV 62 XEQ 13 123 2 188 RRCL 18 254 RCL 18 328 RCL 17 386 STO 31 453 STOP 63*LBL 12 124 * 189 RVIE U 255 - 321 + 387 quot;T=?quot; 454 .END.

10^4-in. foreign line Bend 14-in. oil line -view Clearance 2 ft Minimum HT of Note: Offset bend bend = quot;Vquot; on each side. Origin Bend Note: This distance can be readily Figure 4. Crossing under foreign lines with offset bends. determined by inspection. Figure 1. Cold offset bends. 09 V Vertical distance desired (ft between center lines) 10 T Tangent (ft) 11 2$. lncr. Bend angle increment in degrees 12 H Horizontal distance, overall (ft) 13 L Length of pipe (ft) 14 VCaIc. Vertical distance calculated (ft) 15 R(1-cos Trap 16 T sin Trap 4. Turn 17 R sin Trap 4 18 T cos Trap 4 Origin Trap 19 R * 57.2958 Figure 2. Pipe direction change. Vertical distance to launcher trap 20 VL connection (ft) Horizontal distance to launcher trap Legend & registers 21 HL connection (ft) Reg. no. Length of pipe to launcher trap 00 Not used 22 LL connection (ft) 01 LRT Length of receiver trap (ft) Vertical distance to receiver trap 02 LLT Length of launcher trap (ft) 23 VR connection (ft) 03 R/D Radius of bend/pipe diameter (ft/ft) Horizontal distance to receiver trap 24 HR connection (ft) 04 Trap 4 Scraper trap angle in degrees Length of pipe to receiver trap 05 *1 Bend angle in degrees 25 LR connection (ft) 06 D—in. Pipe diameter (in.) Vertical distance to end of launcher 07 D-ft Pipe diameter (ft) 26 VLT trap (ft) 08 R Bend radius (ft) Horizontal distance to end of launcher 27 HLT trap (ft) Vertical distance to end of receiver Receiver Launcher 28 VRT trap (ft) Horizontal distance to end of receiver 29 HRT trap (ft) 30 4 Change direction desired in degrees 31 Radius Change direction (ft) Trap 32 Tangent Change direction (ft) 33 V Vertical distance calculated, change direction Pipe below ground 34 H Horizontal distance calculated, change direction Note: Aboveground pipe ends are modified as shown for traps not level. 35 L Length of pipe calculated, change Face to face of valve included in LRT and LLT. direction Figure 3. Scraper traps.

The direction change, Example 2 (XEQ 17), can be run at any time since it prompts for input data. If this is a code bend, then R must be determined for pipe size and service quot;Bendquot; program. Put in calculator, Size 35, XEQ quot;Bend.quot; of line by quot;bendquot; (use results of previous run for same Key in prompted data and R/S each time. Key in a trial diameter or key in data as shown on Example 1 to obtain R). angle. Use a larger angle for smaller pipe as 15° for 12-in., Use this minimum radius or any larger radius. Points on a 10° for 30-in. Results Recap prints out the calculated values circle of given radius may be calculated for any angle. and inputs. Each quantity is identified. Additional printouts Subdivide angle, and for each subdivision key in R and let may be made by XEQ quot;Data.quot; T = O, thus determining points on curve. If one inadvertently puts in a larger trial angle than needed for a solution (i.e., calculated V approximates given Formulas for reverse bends V), only one calculation is made and printed out for this angle. If calculated V is too large, store smaller trial 4 in 05, V = 2R(I - cos 4_1) + 2Tsin 4_1 XEQ 01, to get proper result. Remember to clear Flag 01 before resuming normal H = 2R sin £1 + 2Tcos 4_1 + 2T quot;Bendquot; calculations. L 2R = 57Sk + 4T Parts of quot;Bendquot; routine may be used for small pipe in noncode work to determine bend angle for offsets. Do not XEQ quot;Bendquot; for noncode work as this determines R specified by code. Formulas for scraper trap connections Procedure: XEQ clearing GTO quot;Bend.quot; Now select a radius (try: R = 1 8 D : 31A in. OD: R = Receiver. 18 3.5/12 = 5.25). This can be changed if necessary. The V VR = V - R(I - cos Trap 2^_)T sin Trap 4 distance, offset, is determined by design configuration. Store R in 08, V in 09. Assume trial angle, say 20° for small pipe, HR = H - T + Rsin Trap 4 + Tcos Trap 4_ store in 05. Let T = 0.5 ft, store in 10. SF 01. XEQ 01. (Only one calculation is performed.) Inspect results. Do V given and V calc appear reasonably close? If not, take another quot;Fixquot; by changing Launcher. R or angle (store new values, XEQ 01). This routine can be continued to a satisfactory solution. A point will be reached VL = V - R(I - cos Trap 4_)+T sin Trap 4_ where R seems reasonable; choose an angle less than HL = H - T + Rsin Trap 4_ + Tcos Trap 4 apparent solution, store in 05. CF 01, store 0.5 in 11, XEQ 01, and zero in. This returns to the iterative process for a more precise solution. T and R can be changed to suit. quot;Bendquot; routine may be used to calculate bends for crossing under foreign lines. (See Example 3.) Receiver trap. The direction change routine (XEQ 17) can be used for code and noncode work. For code: determine minimum VRT = VR - LRT sin Trap 4. bend radius from quot;Bendquot; for pipe size and for gas or oil line. HRT = HR + LRT cos Trap 4 Use R from previous example or XEQ quot;Bend,quot; and key in data as in Example 1 for desired diameter. Launcher trap. For noncode work: user may employ a radius that is suitable in his judgment. Do not use quot;Bendquot; program as this VLT = VL - LLT sin Trap 4

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