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Sec~7 7 Inspection Data Evaluation, Analysis, and Recording 检验数据评估, 分析和记录


7 Inspection Data Evaluation, Analysis, and Recording 7.1 Corrosion Rate Determination 7.2 MAWP Determination 7.3 Required Thickness Determination 7.4 Assessment of Inspection Findings 7.5 Piping Stress Analysis 7.6 Reporting and Records for Piping System Inspection 7.7 Inspection Recommendations for Repair or Replacement 7.8 Inspection Records for External Inspections 7.9 Piping Failure and Leak Reports 7.10 Inspection Deferral or Interval Revision

Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang


7.1 Corrosion Rate Determination 腐蚀速率测定


7.1.1 Remaining Life Calculations 剩余寿命的计算 The remaining life of the piping system shall be calculated from the following formula:


7.1.2 Newly Installed Piping Systems or Changes in Service 新安装的管路系统或更改服务 以下为新安装的管路系统或更改服务时用来重新计算腐蚀率, 优先考虑秩序; 1. 相同的工艺流程服务,操作条件,类似的材料的数据做为参考, 2. 用户的体验或从相同管道系统公布的数据做为参考, 3. 三个月的服务后,通过使用厚度测量测定无损厚度测量做为参考.

Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang


7.1.3 Existing Piping Systems Corrosion rates shall be calculated on either a short-term or a LT basis. 腐蚀 速率应在短期或长期基础上的计算

计算 MAWP时,运用长期或短期腐蚀速率较大值 If calculations indicate that an inaccurate rate of corrosion has been assumed, the rate to be used for the next period shall be adjusted to agree with the actual rate found. 如果在计算腐蚀率是发现,前先的假设/计算是错误时,应当更改至真确的腐蚀率. (考试题)


7.2 MAWP Determination The MAWP for the continued use of piping systems shall be established using the applicable code. MAWP 计算应当按照所选的设计规范. 在这些计算中所用的残余厚度; 是实 际通过检查确定厚度管壁厚度减(-)去两倍的下一个检验日期估计腐蚀损失. TMAWP = T实际 - 2 x 估计腐蚀损失 估计腐蚀损失 = (LT或ST) x 检验间隔(法定或规范允许间隔). 小编: 两倍的估计腐蚀损失,可视为设定MAWP的安全系数.


For unknown materials, computations may be made assuming the lowest grade material and joint efficiency in the applicable code. 计算时用设计规范里材料 最低档次的材料屈服/强度和焊缝效率.


7.3 Required Thickness Determination所需壁厚厚度确定 The required thickness of a pipe shall be the greater of the pressure design thickness or the structural minimum thickness. 管线设计厚度是取两者的较大值: (1) 压力设计厚度或 (2) 结构的最小厚 度.


PMAWP = 计算 MAWP 时, 运用长期或短期腐蚀速率较大值

小编: 腐蚀量-C  最终腐蚀遗留量(c)  作业腐蚀量 (CR x Yrs) 腐蚀率 x 检验间隔(使用时间)  计算用作业腐蚀量 2x(CR x Yrs), 对腐蚀率估算不确定性的两倍的安 全系数


上述的C腐蚀余留量, 也许是管壁的内壁(工艺腐蚀),外壁(外部环境腐蚀)双腐 蚀. 计算时应当减去 C1 与 C2 (2013 June 考卷)


Question: A seamless NPS 10 pipe, ASTM A106 Grade B material operates at 750 psi and 700 degrees F (maximum). The thickness of the pipe as determined by the last inspection is 0.30". The pipe has been in service for 10 years. The original thickness (measured when installed) was 0.365". Two years previous to the 0.30" measurement the thickness of the pipe was measured to be 0.365". Two years previous to the 0.30" measurement the thickness of the pipe was measured to be 0.31". The next planned inspection is scheduled for 7 years. Using the worst corrosion rate (short or long term) determine what pressure the pipe will withstand at the end of its next inspection period?


Question: A blank is required between two NPS 10, 300 lb. class flanges. The maximum pressure in the system is 385 psi at 200 degrees F. A corrosion allowance of 0.175" is required. The inside diameter of the gasket surface is 9.25". The blank is ASTM A516 Grade 70 material with no weld joint. Calculate the pressure design thickness required for the blank.


For services with high risk, the piping engineer should consider increasing the required thickness to provide for unanticipated or unknown loadings, or undiscovered metal loss. 对于高风险的服务, 管道工程师应考虑增加所 需的厚度


7.4 Assessment of Inspection Findings 评估检查结果 Pressure containing components found to have degradation that could affect their load carrying capability [pressure loads and other applicable loads (e.g. weight, wind, etc., per API 579-1/ASME FFS-1)] shall be evaluated for continued service. 当压部件发现有可能影响部件承载能力的退化时, 必须评估其持续服务达标 的合适性. 评估方法有; 合适性评估

API 579-1/ASME FFS-1


API 579-1/ASME FFS-1


Assessment requires the use of a future corrosion allowance  Assessment of General Metal Loss—API 579-1/ASME FFS-1, Section 4.  Assessment of Local Metal Loss—API 579-1/ASME FFS-1, Section 5.  Assessment of Pitting Corrosion—API 579-1/ASME FFS-1, Section 6. In some cases will require the use of a future corrosion Allowance  Assessment of blisters and laminations-API 579-1/ASME FFS-1, Section 7 Assessment not requires the use of a future corrosion allowance  Assessment of weld misalignment and shell distortions- API 579-1/ASME FFS-1, Section 8.  Assessment of crack-like flaws- API 579-1/ASME FFS-1, Section 9.  Assessment of effects of fire damage-API 579-1/ASME FFS-1, Section 11.


7.5 Piping Stress Analysis 管道应力分析 Piping shall be supported and guided so that: 管道支撑和引导 1. its weight is carried safely, 安全足够的承载管道重量, 2. it has sufficient flexibility for thermal expansion or contraction 热膨胀收缩的灵活性, 3. it does not vibrate excessively.防范过度振动. 管道应力分析是早管系新建时的工程分析项, 在生产阶段多作为管系检验项,当检 验员观察到过度的振动与承载问题时,应当及时的咨询管线工程师意见与考虑 重新对受影响的管道做应力分析.


7.6 Reporting and Records for Piping System Inspection 管道系统检查的报告和记录 7.6.1 Permanent and Progressive Records 永久和逐行记录. 管道系统的所有者和使用者应当保持长期和逐行管路系统和泄压装置的记 录. 7.6.2 Types of Piping Records 管道纪录类型    

新制造,施工和设计信息 检验历史,记录 维修, 改造和重新评级信息 FFS 评估文件要求

7.6.3 7.6.4 7.6.5 7.6.6

操作及保养记录-生产运作和维护记录 计算机记录-便于检索,计算 MAWP,最低厚度等 管道单元(circuit)记录-详细记录材料,工艺参数, 维修记录等等 检查ISO图-确保检验重复性,等级分类,维修变动部位等等.


7.7 Inspection Recommendations for Repair or Replacement 检查建议维修或更换 维修或更换建议列表应当及时更新,建议列表跟踪系统应包括: 1. Recommended corrective action or repair and date, 建议纠正措施或维修和日期 2. Priority or target date for recommended action, 优先级或建议的行动目标日期 3. Piping system identifier (e.g. piping system or circuit number) that the recommendation affects. 管道系统标识符 A management system is required for tracking and reviewing outstanding recommendations on a periodic basis.需要一个管理制度, 用于跟踪和定期审查尚未落实的建议.


7.8 Inspection Records for External Inspections 外部检查检测记录 Results of external piping system inspections shall be documented. 7.9 Piping Failure and Leak Reports 管道故障和泄漏报告 Leaks and failures in piping that occur as a result of corrosion, cracking or mechanical damage shall be reported and recorded to the owner-user


7.10 Inspection Deferral or Interval Revision 检查延迟或间隔修订 没有既定的时间间隔内检查被认为是逾期查阅,除非 (1) 正式启动“延期 检讨” 建立可接受的替代检验计划或 (2) 经过适当的分析修改检验间隔 检查延迟 延长的检查日期是根据成文的风险分析过程成立,是一次性, 临时管道检验到期日期延长,不应被视为检查间隔调整. 间隔修订当管道条件和历史审查表明过于保守或不足, 检查间隔可以适 当修订, 延长或缩短.


API


Practice Problems # 1 1. Find the Maximum Allowable Working Pressure of a section of inservice piping. P = MAWP ? S = Maximum Allowable Stress 11,200psi t = Minimum thickness of pipe .245 inch E = Efficiency of longitudinal joint 80% D = Outside Diameter 6 inch P   2SEt/D


2. The thickness measured during the inspection of a section of pipe is .3875”. The pipe specification indicates that the pipe was originally 8” schedule 80 and the original nominal thickness was 0.5”. What is the long term corrosion rate if the pipe has been in service for 14 years?


3. The Owner/User orders a piping system that must carry 650 psi at 300ºF (temperature coefficient “Y”=.4), and the pipe must be 14” nominal pipe size. The stress of the ferritic material is 13,800 psi. The piping system will be operated in a slightly corrosive atmosphere and requires a 1/16” corrosion allowance. What is the minimum required thickness of this piping system with an efficiency of 100%?


4. If the diameter of a flange is 15.75�, what is the circumference? 5. If the base of a right triangle is 5 feet, and height is 7 feet, what is the length of the hypotenuse?

6. If P = 1500, E = .85, S = 13,500, Y = .4, and D = 28: find (t) in the following formula.


7. If the design pressure (P) of a piping system is 480psi and a pressure test of 1-½ times design pressure is to be applied, what would the test pressure (Pt) be if the stress at test temperature (St) is 13,800 psi and the stress at design temperature (S) is 8,400 psi?

9. A 4 inch test specimen is measured after a tensile test, and it is found that the length is now 4.575 inches. What is the percentage of elongation?


1. A section of piping, 38 feet long between the blank flanges, has been repaired and is ready for hydro, what is the volume, in cubic inches, of this piping section if the outside diameter is 8.625 inches and the thickness is .322 inches? 2. The piping engineer has determined that a fillet welded patch can be temporarily applied to the outside of an NPS 18 section of piping. The patch will be .5” thick and will be attached using a ½” leg length fillet weld. What would the theoretical throat of this fillet weld be? Throat .707Wleg 3. What is the total outside surface of a section of piping, requiring insulation, if the pipe is 32” outside diameter and 65 feet long?


4. A piping system was installed in 1954. The actual thickness at the time of installation was .7385�. The system had no thickness measurements taken until 1995, at which time the measured thickness was .603�. What is the resulting long term corrosion rate for this condition?


5. A newly constructed piping system is to be hydrostatically tested prior to being placed into service. The nominal thickness of the piping used in the system is .844�. The material stress value is 8500 psi at design temperature with quality factor of 1. The temperature coefficient (Y) is .4 and the outside diameter of the seamless piping is 10.75�. What is the required hydrostatic test pressure if the stress at test temperature equals 17,000psi? Use:


95) An eight-inch diameter piping system is installed in December 1979. The installed thickness if measured as 0.34". The minimum thickness of the pipe is 0.20". It is inspected 12/83 and the thickness is found to be 0.32". An inspection 12/87 reveals a loss of 0.01" from the 12/85 inspection. During 12/89 the thickness was found to be 0.29". The last inspection was during 12/95 and the thickness was found to be 0.26". What is the long-term corrosion rate of this system? a) b) c) d)

0.005”/year 0.0075”/year 0.00375”/year 0.0025”/year

96) Using the information in question 95, calculate the short-term corrosion rate: a) b) c) d)

0.005”/year 0.0075”/year 0.00375”/year 0.0025”/year


97)Using the information in questions 95 and 96, determine the remaining life of the system: a) b) c) d)

18 years 15 years 12 years 6 years

98)You have a new piping system that has just been installed. It is completely new and no information exists to establish a corrosion rate. Also, information is no available on a similar system. You decide to put the system in service and NDT it later to determine the corrosion rate. How long do you allow the system to stay in service before you take your first thickness readings? (2013 June) a) b) c) d)

1 month 3 months 6 months 12 months


99)After an inspection interval is completed and if calculations indicate that an inaccurate rate of corrosion has been assumed in a piping system, how do you determine the corrosion rate for the next inspection period? (2013 June) a) Check the original calculations to find out what the error is in the original assumption. b) Unless the corrosion rate is higher, the initial rates shall be used. c) The corrosion rate shall be adjusted to agree with the actual rate found. d) If the corrosion rate is higher than originally assumed, call in a corrosion specialist.


100)If a piping system is made up of unknown materials and computations must be made to determine the minimum thickness of the pipe, what can the inspector or the piping engineer do t establish the minimum thickness? (2013 June) a) The lowest grade material and joint efficiency in the applicable code may be assumed for calculations. b) Samples must be taken from the piping and testing for maximum tensile stress and yield strength will determine the allowable stress to be used. c) The piping made of the unknown material must be removed from service and current piping of known material must be installed. d) The piping of unknown material may be subjected to a hydrostatic stress tests while having strain gages on it to determine its yield strength and thus allowable stress.


101)A piping engineer is designing a piping service with high potential consequences if a failure occurs, i.e., a 350 psi natural gas line adjacent to a high density population area. What should he consider doing for unanticipated situations? a) b) c) d)

Have all his calculations checked twice. Increase the required minimum thickness. Notify the owner-user and the jurisdiction. Set up an emergency evacuation procedure.

102)When evaluating locally thinned areas, the surface of the weld includes __________ on either side of the weld or __________ times the minimum measured thickness on either side of the weld, whichever is greater. a) b) c) d)

0.5”, 3 1”, 2 2”, 1 1.5”, 1.5


102) When evaluating locally thinned areas, the surface of the weld includes __________ on either side of the weld or __________ times the minimum measured thickness on either side of the weld, whichever is greater. a) b) c) d)

0.5”, 3 1”, 2 2”, 1 1.5”, 1.5

这题个根据?


103)An inspector finds a thin area in a fabricated 24" diameter pipe. The thin area includes a longitudinal weld in the pipe and is 10 feet long and 2 foot circumferentially. Calculations show that with 0.85 joint factor, the pipe must be repaired, renewed, etc. or the pressure in the pipe must be lowered. The owner does not want to do any hot work on the pipe and he does not wish to lower the pressure. What other course could you follow? a) Write the results of the inspection up and leave it with the owner. b) Radiograph the weld 100 % and increase the joint factor to one. c) Insist that the weld be repaired or renewed or that the pressure be lowered. d) Call in a regulator agency to force the owner to repair, renew, etc. the line.


104)Piping stress analysis is done during the system's original design. How can the inspector make use of stress analysis information? a) An inspector cannot use this information. It is only meaningful to a piping engineer. b) It can be used to make sure the piping system was originally evaluated and designed correctly. c) It can be used to concentrate inspection efforts at locations most prone to fatigue or creep damage, and to solve vibration problems. d) The inspector should use this information to evaluate the need for conducting additional piping stress analysis.


105) You are inspecting a piping system. You find a significant loss of material (a major increase of corrosion rate) in gas oil piping (used as reboiler oil, temperature 500째F) on a Fluid Catalytic Cracking Unit. What is the best course of action for you to take? a) The losses may be reported to your supervisor for corrective response b) The losses should be recorded and reported in your final report after the unit has started. c) It shall be reported to the owner-user for appropriate action. d) Replace excessively thin piping and note replacement in the final report after unit start-up.


106)The __________ shall maintain appropriate permanent and progressive records of each piping system covered by API 570. a) b) c) d)

Inspector Owner-user Jurisdiction Examiner


8) A NPS 6 piping system is installed in December 1989. The installed thickness is measured at 0.719”. The minimum thickness of the pipe is 0.456”. It is inspected in December 1994 and the measured thickness is 0.608”. An inspection in December 1995 reveals a 0.025" loss from the December 1994 inspection. During December 1996, the thickness was measured to be 0.571". What is the long-term corrosion rate of this system? a) b) c) d)

0.01996”/year 0.02567”/year 0.02114”/year 0.03546”/year

9) Using the data in Question No. 8, calculate the short term corrosion rate in mils per year (M/P year) a) b) c) d)

0.0012 M/P year 0.012 M/P year 0.12 M/P year 12 M/P year


10)Using the information in Questions No. 8 and No. 9, determine the remaining life of the system a) b) c) d)

18 years 5.44 years 1.2 years 6 years

11)Using the information in Questions No. 10 and assuming an injection point in a Class 2 system with 7 years estimated until the next inspection what would the next interval be: a) b) c) d)

10 years 5 years 3 years 2.72 years


7.1.1 Remaining Life Calculations Corrosion Rate 腐蚀率

LT and ST corrosion rates should be compared to see which results in the shortest remaining life as part of the data assessment. The authorized inspector, in consultation with a corrosion specialist, shall select the corrosion rate that best reflects the current process (1) 长期/短期腐蚀率以最大值作为剩余使用寿命计算 或 (2) 授权检验员和管道工程师质询后,能反映现况最合适的腐蚀率


CR 腐蚀率用来干啥啊! a) 计算剩余使用寿命 b) 在职管道最高允许工作压力 ( c) 厚度检测最大计算时间间隔 (a/2)

在计算(b)最高允许压力, 或 (c) 厚度检验最大时间间隔 时, API570 运用 半衰期计算法作为安全系数:


API570, 6..3.3 Thickness measurements should be scheduled at intervals that do not exceed the lesser of one half the remaining life determined from corrosion rates indicated in 7.1.1.1 or the maximum intervals recommended in Table 2. Shorter intervals may be appropriate under certain circumstances. Prior to using Table 2, corrosion rates shall be calculated in accordance with 7.1.1.1. 7.1.1 Remaining Life Calculations The remaining life of the piping system shall be calculated from the following formula:

Table 2


在计算”剩余使用寿命”时 半衰期检验法不需要考虑 腐蚀率是,短期或长期腐蚀率较高值


服役压力管道厚度测量应在 预定的时间间隔执行:(以下较小值)  Table 2 厚度检查间隔  一半的剩余寿命确定腐蚀速率


计算MAWP

CR 为 长期/短期腐蚀率较大值 YRS 为 下次检验时间间隔 计算厚度为: 实际厚度(t) – 2xCRxYRS 下次检验时间间隔为剩余使用寿命的一半或Table2(较小值)


Table 4—Two Examples of the Calculation of MAWP Illustrating the Use of the Corrosion Half-life Concept


13)A seamless NPS 16 pipe, ASTM A135 Grade A material operates at 550 psi and 600 degrees F maximum. The thickness of the pipe as determined by the last inspection is 0.40". The pipe has been in service for 8 years. The original thickness at installation was measured to be 0.844". Two years previous to the 0.40" measurement the thickness of the pipe was found to be 0.54". Determine the greatest corrosion rate, i.e., short or long term in mils per year (M/P year). a) b) c) d)

55 M/P year 70 M/P year 0.70 M/P year 700 M/P year

0 years t= 0.844 6 years t= 0.54 8 years t= 0.40 Cr long= (0.844-0.40)/8 = 0.0555�/yr Cr Short= (0.54-0.40)/2 = 0.07�/yr


14)A seamless NPS 12 pipe, ASTM A106 Grade B material operates at 750 psi and 700 degrees F maximum. The thickness of the pipe as determined by the last inspection is 0.305”. The pipe has been in service for 13 years. The original thickness at installation was measured to be 0.405". Two years previous to the 0.305” measurement the thickness of the pipe found to be 0.316”. The next planned inspection is scheduled for 8 years. Using the appropriate corrosion rate determine what MAWP the pipe will withstand at the end of the next inspection period a) b) c) d)

720 psi 476 psi 611 psi 550 psi

Cr long = (0.405-0.305)/13 = 0.0077”/yr Cr short = (0.316-0.305)/2 = 0.0055”/yr Use Cr long = 0.0077”/yr Next inspection= 8 years P = 2x16700x(0.305-2x0.0077x8) / 12.75 = 476# B31.3-2010


15)A seamless NPS 6, ASTM A106 Grade A pipe operates at 300 degrees F and 765 psi. The allowable stress is 16,000 psi. Using the Barlow equation, determine the required thickness for these conditions a) b) c) d)

0.446” 0.332” 0.231” 0.158”

t = PD/2SE = 765x6.625 / (2x16000) 16)A seamless NPS 6, ASTM A106 Grade A pipe operates at 300 degrees F and 741 psi. The allowable stress is 16,000 psi. The owner-user specified that the pipe must have 0.125" for corrosion allowance. Using the Barlow equation, determine the required thickness for these conditions a) b) c) d)

0.278” 0.195” 0.325” 0.392”


17)A NPS 4 Schedule 80 (0.337" wall) branch connection is welded into a NPS 6 Schedule 40 (0.280" wall). A 0.375" reinforcing pad is used around the branch connection. The fillet weld sizes are as required by the Code. The branch connection is inserted into the header. The material of the branch and header is ASTM A672 Grade B70. What thickness would be used to determine whether heat treatment of the connection is required? (Express answer to nearest hundredth.) a) b) c) d)

0.768” 0.891” 0.998” 0.567”


331.1.3 Governing Thickness. ‌In the case of branch connections, metal (other than weld metal) added as reinforcement, whether an integral part of a branch fitting or attached as a reinforcing pad or saddle, shall not be considered in determining heat treatment requirements. “ Heat treatment is required, however, when the thickness through the weld in any plane through the branch is greater than twice the minimum material thickness requiring heat treatmentâ€? even though the thickness of the components at the joint is less than the minimum thickness. Thickness through the weld for the details shown in Fig. 328.5.4D shall be computed using the following formulas:


17)A NPS 4 Schedule 80 (0.337" wall) branch connection is welded into a NPS 6 Schedule 40 (0.280" wall). A 0.375" reinforcing pad is used around the branch connection. The fillet weld sizes are as required by the Code. The branch connection is inserted into the header. The material of the branch and header is ASTM A672 Grade B70. What thickness would be used to determine whether heat treatment of the connection is required? (Express answer to nearest hundredth.) a) b) c) d)

0.768” 0.891” 0.998” 0.567”

Used sketch(2) = Th + Tc Answer ?


19)A seamless NPS 10 pipe, ASTM A106 Grade B material, operates at 750 psi and 700 degrees F (maximum). The thickness of the pipe as determined by the last inspection is 0.30". The pipe has been in service for 10 years. The original thickness (measured when installed) was 0.365". Two years previous to the 0.30" measurement the thickness of the pipe was measured to be 0.31". Determine the greatest corrosion rate i.e. short or long term a) b) c) d)

0.0050 inches per year 0.0065 inches per year 0.0100 inches per year 0.0130 inches per year

Cr long = (0.365-0.3) /10 = 0.0065�/yr Cr short = (0.31-0.3)/2 = 0.005�/yr


20)A seamless NPS 10 pipe, ASTM A106 Grade B material operates at 750 psi and 700 degrees F (maximum). The thickness of the pipe as determined by the last inspection is 0.30". The pipe has been in service for 10 years. The original thickness (measured when installed) was 0.365". Two years previous to he 0.30" measurement the thickness of the pipe was measured to be 0.31". The next planned inspection is scheduled for 7 years. Using the worst corrosion rate (short or long term) determine what pressure the pipe will withstand at the end of its next inspection period ? a) b) c) d)

920 psi 540 psi 811 psi 750 psi

CR long = (0.365-0.3)/10 = 0.0065�/yr CR short = (0.31-0.3)/2 = 0.005�/yr MAWP = 2x 13900(0.30-2x 0.0065 x7-c) / 10.75 = 540psi



Api570 chapter 7