Inspection of subsea production systems · of subsea prbduction systems for . exploitation of these deep water petroleum ' re~erves ' poses new inspection problems for the U.S. Geological

bull bullbull -

INSPECTION OF SUBSEA PRODUCTION SYSTEMS

William Walters and Stacy Gehman 1

ACKNOWLEDGEMENTS

This report describes work done under technology transfer authorization for the Conservation Division of the US Geological Survey

The authors also thank Shell Oil Co and Exxon Company USA for their cooperation in providing detailed information

on their approach~s to subsea production without which this program would not middothave been possible

1 Introduction

The tremendous demand for oil and gas in the United States has

dep 1 eted our oil and gas reserves to an extent which has caused concern

for many years Efforts to maintain American independence from foreign

oil suppliers albng with the rising costs of imported oil have

made possible the consideration for development of marginal or

previously unprofitable oil and gas fields in the deeper waters

of the outer cont1i nenta 1 she 1 f (OCS)

The current ~evelopment of subsea prbduction systems for

exploitation of these deep water petroleum re~erves poses new

inspection problems for the US Geological Survey (USGS)

This report presents the results of a study f~nded by USGS to

explore techniques for inspection of sea floor completion and

production equip~ent A technique requirjng reither divers nor

submersibles has been identified and its deta~lcd application for

the Exxon subsea production system (SPS) is p~esented in Appendix A

to this report The technique uses production tubing within the I

sys tern and could bullsubstantially reduce i nspect i on expense and 1

danger to i nspectors which would otherwise be encountered during

first-hand inspecticns This technique is explained in detail

in Sections 4 and 5 Section 6 contains a review of OCS orders

applicable to subs~a systems with suggest~d g~nera1 inspection

procedures Appendix B inc 1 udes the test procedure suggested to I

USGS by ShellLockheed for their subsea well control system

along with recormiendations by MDL

2 Types of Subsea Svstems

Various approaches are being taken to produce oi 1 from deeper

water using subsea sxstems There are three basically different

approaches to deep water production

1 The complet~ly submerged remote controlled system

Production equipment specifically designed for operation in the deep

water environment ismanifolded on a submerged template mounted on

the ocean floor Maintenance is by pump dmm tools and manned or

remote manipulatorsbull The Exxon Subsea Production System SPS)

discussed in Appendix A is of this type

2 The atmosphtiric wellhead Subsea production equipment is

located in a chamber bullmaintained at at111osphe1 ic pressure The

chamber is accessibl~ to workers from transfer capsules or

submersibles This approach alloKs the use of conventional I

completion equipmentand maintenance of thi1t equipment by personnel

in a dry one-atmosphere environment The ShellALockheed system

discussed in the App~ndix B of this type I

3 The surface-platformdeep water wellheaJ system Innovative

designs for surface iroducti on p 1 a tforms whi th aie being developed for I

use in deep water present unique structural evaluation problems

These systems are ou~side the scope of this repo~t

All these systems are being designed for de~th capability of

1000 ft and beyond but most are presently condned to 300 ft or I

less for ease of experimental testing Altlough 1all these systems I

present cha 11 eng i ng problems for inspection the Exxon SPS was chosen

for detailed investigation because 1t provided a range of inspection

problems representat1ve of subsea p1middotoduction middotsystems Sufficient I

design data 1vere available on this system to permit a thorough

evaluation of these problems

3 Inspection Techn igues and lI9~~-di t_E_~

As part of ongotng OCS lease m~nagement proQram USGS personnel

are responsible for ensuring that all offshore e~ploratory and production

operations are performed in a safe and poll11tionpreventing manner

according to pub 1 is hed OCS orders This re~pons tbil i ty is carried out

by both sch0dulrd c11q 1msched11lelt1 inspccticlaquos of all offshore oprrating

facilities In some cdses these i11specticir1s can be nade by visual

observation or by re~ding instruments while prescribed test procedures

are carried out In other cases the inspcctor must question the

lessee concerning installation of critical equipment or testing without actually observing the item in question For obvious reasons it is

preferable to make first-hand ~bservations whenever possible In I

subsea operations~ this brings up the question of the techniques and

procedures required to perform these inspections One very direct ~ - j

technique would be to employ submersible vchiCles having windows for

viewing of underwater equipment by trained inspectors Howevcr this -

method will evide~tly be very costly requiri19 not only the submersible

but also a specia1ly ~quipped ~urfoce suppo~tlaunching vessel Esti shy

mates for employi~g submersible vehicles are as high as $18 millionlf

for development of a complete system including ship submersible and I

special tools The submersible cost alone 1ould be in the area of

$1 to $2 mill ion each Thus complete independence from reliance on I I middot-middot bull - - -- 1 __ middot~

the p~oducers comes deuroarly nor() v11lid d9ampmtnf- Several of the more sophisticated subsea production systems feature

personnel transfer capsules At the risk of increasing dependence --~-middot---

on producers these company-operated submersible enclosures could be

used IStill another alternative is the possibility of leasjo_g StJ~- J 7 mersi~li_~ and supp_g_rlships from comrnerc~a_l or~an_izatio~_s as heli- _ 1 _

copter transport is currently obtained by USGS to discharge its present Jltj

inspection responsibilities This wo~ld tend to defray the high purshy

chase cost and al so minimize rel i ancc on producers middot- l -middot

However the safety of the inspectors performing the underw(lter

_inspecion is of pri~ary concern-W~l-1 knmm po~~ntial dan-ers g~n~r-1111e(~JJry ally related to life support structura1 integrity and surface comshy

munications exist when r~anned submersibles arp used in such operations

Accidents and human error occurring while submerged can be vastly more

serious than if they occurred in the notTial e~vironment The use of

manned submersibles would also require a regular inspection of the

submersible ard its ancillary equipment to assure safety The added

burden of such a program if undertaken is self-evident

lfunpublished correspondence Dr J D Stochiw NUC to J Meek HDL 23 Jan 1976

-Another alternative is the use of an unmanned submersible The J ~ 1

i 1t~ use of unmanned reniote controlled underwater vehicles 1~ith television -middot

I -- - IL

and manipulators would eliminate personnel danger Hov1ever extra costs and possible technical limitations make a third alternative deshy

sirable if feasi 1

ble This alternative is tomiddot provide remote read-out 71 1shy1 bullbull -- ____ - ___~ lt l

1so that inspectors may perform the inspections from aboard the pro- -~r l middot

duction-supp_ort vess9ls or deep-water platforms already in use by the middot i

lessees The alter~ative would require that t_echniques be used to -~

sense the item i1

n question at the subsea v1el1head or production equ1 shyI J I

ment and transmi~t this critical information to the surface This I

sensing may req~ire fnstallation of pgsition pressure and thermal I ) t

transducers in ~lv_gs pipes pumps etc located in the subsea enshy1 I -~middot-middotmiddotmiddotmiddot--- ---shy

vi ronment Tele111etering the information to the surface could involve

ultrasonic hydrltaulic electric-wire or other transmission techniquesI

The information 1di sp 1ays could in lcude vi sual gages or meters and comshy

Iputer controlled cathode-ray tubes (CRTs)

The consid~rations discussed above place inspection techniques

and procedures 1ln three categories ie those performed by manned

submersibles by unmanned submersible and by remote controldisplay lrf Ultimately it liould appear that emphasis should be placed on sensors

1 _on the subsea s11ste111 with data display or readout at the surface-- ---

vessel or platfdrm Increased use of sensors however will require 10

the development of methods for their calibration A method for cali shy

brating pressure transducers and determining existence and magnitude of

leakage through closed valves is presented in the next secion of this

report This approach v1ould eliminate the need for very costly submersible

vehicles and th~ support ships required for I

them as well as avoid a -fJ possible so4rce

of risk to USGS inspectors --- I I

4 Calibration and Jn~fiectiq~hni_~

Production or service tubing that is accessible at the surface

can be used formiddot calibration of pressure sensors and determining valve

leakage if a tu~ing path to the component tq be monitored can be isoshy

lated by remotely actuated va1ves If a closed volume of fluid can

be obtained in this way net leakage into or out of the volume can

be determined by a pressure rise or drop respectively which is meashy

sured at the surface If leakage occurs the magnitude of the leak

can be determined by measuring the flow rate required to hold the I

surface pressure constant while bleeding fluid from or pumping fluid

into the tubing(assuming the fluid in the tubing is in thermal equilishy brium with the surroundings) If there is no leakage or if the leakage

is very small the process may be used to calibrate subsurface presshy

sure transducers Surface readings of pressure in the closed volume

can be used to calibrate the remote read out of subsurface transducers I

monitoring the pressure in the volume if th~ surface readings are comshy

pensated for the hydraulic gradient in the enclosed fluid This com-middot

pensation and a~sociated errors are discuss~d in the following section

This technique for calibration and inspection of subsea transducers I

and valving requires only that the inspector have access to the pipe- middot

lines that connect the subsea system to its middotsurface production and I

distribution systems Closed volumes are oqtained in these pipelines

by appropriately setting (turning on or off) various valves in the

pipeline and subsea systems Appendix A of this report applies this

technique to the Exxon SPS and shows that the necessary closed volumes

can be obtained for this system to permit inspection of all necessary

points

- - ~e~ Df1 J 11 ~ 5 Pressure Transducer Ca1i brati on l)t a() fh IS

If a pressure transducer at the bottom o~ a long column of fluid

(liquid or gas) is to be calibrated by applying pressure at the topI

then the pressure 1read at the top must be cor~ected for the fluid

pressure gradient in the column to obtain a correct value for presshy sure at the transducer If themiddot transducer depth bene~th the surface

is le_ss than about I

1000 ft and the calibrating fluid is a liquid

then a constant liquid density may be assumed with Jess than one pershy cent errQJ The Bressmicrore correction is then given by Pgh where P is lt--middot -- the fluid densityi g t~e acce)eration of gravity and h is the height

of the fluid colu~n ~n terms of specific gravity (relative to water) I

SpGr the pressure c~rrection (in psi) is 0~434 ~ SpGr x h

For greater accurhcy or for greater transducer depths the cornshy

pressibility of the fluid must be considered I

Figure 1 is taken from

API STD 1101 (Man4al of Petroleum Measurement bullStandards Chapter 5

Section 2) and presents the compressibility o~ hydrocarbons as percent

change in volume 9r density per 1000 psi pressure change for a range

of API gravities and temperatures Dividing ~he compressibility pershy

centage obtained from figure 1 by 100000 results in a factor F such

d P = FdP p

where dp is the c~ange in density associated vlith a change in pressure

dP In psi

middot l middot11 _I ] ii J l ff - middot1middot -middot

lmiddotshyltl

ltfl lbullJ JJ n 0 w 0

Dr~1 r-----~d ir f ~cmiddotmiddotmiddot t- E 1 1 0~-~0n f r AC-lgt Ilei nd C L C_fJJ middotcmiddotr -cmiddotibili~v of Liquid Ii-C-xmiddotirlmiddot~-1~ lbullo= AJ 5 I1middotJ ~7middot~i 1~iJ

~t-tbulln Con1r1middot~)11a11~ of Jiqr1d fl1r bullr11lon_~0middot1H1almmmiddotb~~____

~ -t c~ Ibull_~- 4-r r sngt llI

Integrating both sides results in

= F(P-Ps1

p = eF(P-P~)or PS ( 1 )

where the subscript s indicates surface cdnditions Equation (1)

relates the density 2t any depth in the fluid column to the pressure at that point

For a small change in fluid depth dP =pgdh (2)

where dh is the change in depth Substituting eq~ation (1) into (2)

dP =p geF(P-Ps)dh s

-F(P-P ) or p gdh= e s dP s

Integrating from the surface to a total depth D

-F(P-P ) -F(P -P )1 s e D s + 1= - e = shyf fI

-Rearranging we have

It is interesting to note that gp 5D is the pressure head expected for an incompressible fluid If FgrsDlaquo l (that is the fluid is nearly incompressible) then equation 3 simplifies to

Po-Ps = gPsD

(note lnl+x) ~ x for x laquo 1)

For an example of the type of error that might occur if compressibility

is not considered assume gp 5D = 4000 psi (as it would for an ~gproxi mately 10000 ft deep subsea completion) and that F = 2 x 10 psi Thus

4Po-Ps = -5 x 10 ln(l-008) = 4170 psi

If compressibility had been ignored the pre~sure error would have been 170 psi This analysis has assumed that temperature is constant with depth and that the compressibility of hydrocarbon liquids does not depend on pressure The APl STD 1101 values for compressibility were based on data from zero to 1000 psig

so more data may be reshyquired to assure pressure independence

For cases in which the subsea conditions have been specified (for example setting a downhole pressure for cal~bration of a transducer) the surface presure P required to establish the condition is unknown s Thus the surface density re qui red to solve equation 3 is also unknown A derivation similar to the above shows that the hydrostatic pressure

--difference in terms of subsea conditions is

P0-Ps = ~ ln 0 + fp 0gD) (4)

The density of the oil is most usually known at atmospheric pressure

so the density at pressure P0is found from the atm~spheric density p 0

from equation 1

P0 0 FP = r e D (5)

The use of the above equation assumes a constant temperature so that density varfotions in the tubing are caused only by pressure The

density at atmosp~eric pressure and the desired temperature are found by using table 23 API STD 2540 (ASTM-IP Petroleum Measurement

Tables) to correct the standard density measured at 60degF relative to I

water at 60degF (Sp Gr 6060degF) Specific gravity may be found from

API gravity by usi 1

ng Tible 3 from API STD 2540 or from the equationI I

Sp Gr 6060~F = 1415(APl Gravity at 60degF + 1315)

In the follo~ing paragraphs several examples will be worked to I

demonstrate the uSJe of the equations and to show t~e effect of a

temperature gradi~nt in the oil from the ocean floor to the surface I

The example will i 1nclude two different oils 20deg and 60deg API 6060degF)

at two temperatur~s (70degF and 120degF) and two completion depths

(l000 ft and 101 000 ft) The equations wil1 be solved for surface

pressure assuming a sea bottom transducer is being calibrated at l000 psi

for the 1000 ft 1deep completion and at 5000

psi for the 10000 ft

deep completion The 1000 ft- depth was chosen because H represents I

present capa bi 1 i ties for subsea completion l Q 000 ft las chosen as a maximum depth fqr completions in the foreseeable future (in

areas such as the ~leutian Basin)

i I __ _l__

-Example 1

1000 ft completion -- = 1000 psiP0 20deg API 6060~F -- Sp Gr = 09340

Temp From Fig 1 From Table 23 (API STD 2540) I

70degF F = 039 x 10-5 ps1middot-T Sp Gr = 09305 -5 -1 J

120degF F = 050 x 10 psi Sp Gr bull 09130 At 1000 psi gpl~OO = gp x lOOO = 0434 x (=~middot Gr )0 x lOOO

At 70F gplOOO =0434 (09305) e039 x cO x 1000 1 bull

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

o0398 poundft 1Po - PS =-F- ln (1 + F~D gD)

At 70degFmiddot P P - l ln (1 + 039 x 10-5 x 0405 x 1000) D - S - 039 x 10-S

= 405 psi Ps = lopo - 405 = 595 psi

At 120degF Po - p = l ln (l + ~50 x 10-5 x 0398 x 1000) s 050 x 10-5

= 398 psi P~ = 1000 - 398 = 60 psi

Thus if the $urface pressure P5 were setat 599 psi the pressure

at the sea bottom transducer 1~ould be 1000 4 psi even if the temperature

of the 20deg API gravity oil varied between 70 and 120degF

Example 2 I

1000 ft completion -- P~ = 1000 psi I

60deg API 6060deg F -- Sp Gri = 0 7389 From Pi_g__j_ From Table 23 (API STD 25401

70degF F = 0~80 x 10-5 psi-1 Sp Gr bull 07344 -5 -1

120degF F = l10 x 10 psi Sp Gr bull 07117 At 1000 psi pg middot = gp eF x 1000 = 0434middotx (Sp Gr) eF x 1000

1000 o middot o i 080 ~ 10~5 x 1000

At 70degF 9P]OOO bull 0434 (07344) e bull 0321 psift I

1 i o x i o~s x 1 oooAt 120degF gP1000 bull 0434 (07117)e

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

At 120degF P~ --1----~5 ln (1 + 1 1lO x 10-5 x 0312 x 1000) 110 x 10

= 311 psi Ps = 689 psi

Thus if the surtace pressure P5 were set atmiddot684psi the pressure at the sea bottom transducer would be 1000 plusmn_ 5 psi ~ven if the temperature of the 60deg API gravity oil varied between 70deg and 120degF

From the foregoing examples it is apparrnt that for subsea compleshytions up to 1000 ft there is little effect on the calculated

pressure at a suhsea transducer due to compressibility and temperature bull

The pressure ca~ be accurately calculated from ~ I bullr

PD - PS = 0434 x Sp Gr x D where the specific gravity of the oil at sta~dard conditions has been corrected to the observed temperature (using an average between the subsea and surface oil temperatures) and 1 i ne pressure Example 3

10000 ft c~mpletion -- P0 = 5000 psi 20deg API 6060degF Sp Gr = 09340

From Fi9 l From Teble 23 (API STD 2540)

70deg F = 039 x l0-5psi-l Sp Gr = 09305 120deg F = 050 x l0-5psi-l Sp Gr = 09130

F x 5000 F x 5000At 5000 psi gp5poo = gp e = 0434 x

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

50 x 5000At 120F 9P~ooo = 0434 (09130)e0middot x 10 -5

= 0406 psift 1

PD - PS= -F- ln (l + Fp gD)0 __L__-=s 1nAt 70degF P ~ (l + 0~39 x 10-5 x 0412 x 10000

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

At l20F P - P =--1-- ln (l + 050 x 10-5 x 0406 x 10000) D S 050 x l0- 5

- -p~ = 4019 psi P = 981 psi ~

Thus if the 1surface pressure r5 were set at 947 the pressure at the

sea bottom trans~ucer would be 5000 +- 34 psi even if the temperature of

the 20degAPI gravity oil varied between 70deg and l20degF

Example 4

I 110000 ft ~omp etion P f 5000 psi0

60deg API 6060degF Sp Gr = 07389 Temp

70degF

l 20degF

From Fig l I ~ -5 -1

F = Oi80 x 10 psi I middot -5 -1F = 110 x 10 psmiddotiI

From Table 23 (API----middotmiddot Sp Gr = 07344

Sp Gr= 07117

STD 2540)

middot F x 5000 F x 5000At 5000 ps1 1 gp 5000 = gp = 0~34 x (Sp Gr)e e0 0

80At 70degF gp middot = 0434 (o 7344)e0middot x 10-5 x 5000

~ooo = o332 psift

_ llxl0_ 5 x5oooAt 120degF gp5000 - 0434 (07117)e

= 0326 psift P ~ P ~ t ln (1 + fp gD)D S D

At 70degF Po_ p = ~__]-____ ln (l + Q80 x lo- 5 x 0332 x 10000) s 080 x 10-~

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

At l20degF PD - PS = l 5 ln (l + l l x 10-5 x 0326 x 10000) llOxlOshy

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

Thus if the surface pressure P5 were set at 1760 psi the presshy

sure at the sea ~ottom transducer would be 5000 + 37 psi even if the I shy

temperature cf the 60deg API gravity oil varied between 70deg and 120degF

From these pxample calculations for a 10000 ft completion it is

apparent that temperature variations cause r~latively small errors in I

sea floor ca 1 i bration pressures even for sue~ a great depth If the

pressures are needed more acet1rately than this then the temperature

profile in the tubing would be needed

It has thus been shown that relatively simple calculations of

liquid hydrostatic head result in accurate pressure values for calishy

bration of subsea transducers I

The calibration of gas pressure transdutersmiddot is somewhat different

For low pressures and shallow depth little correction need be made to

the pressure measured at the surface For a gas specific gravity SG

(relative to ai~ at 20degC and 147 psia) pressure P(psia) and depth I

D(ft) the pressure correction AP is given by

-5AP = 356 x 10 x SG x P x D

This equation assumes that the ideal gas law holds and that tPP

is small The equation nay be modified to include the super-compresshy

sibility factor z as follows 0

AP= 356 x 10-Sx SG x D x Pz

For a 1000 ft deep subsea completion line pressure of 1000 psia

(assuming SG ~ 1 and z = 1) AF= 36 psi It is thus apparent that even with large errors in specific gravity and z the pressure correction

for hydrostatic head of the gqs is small for depths up to 1000 ft

Again for a line pressure of 1000 psi but a 10000 ft deep comshy

pletion llP = 3~0 psi so the assumption of PP being small no longer

holds An integration would thus be needed to determine the transducer

pressure and t~e vaHable z factor should be included in thi~ integral Thus a closed form solution similar to tha~ found for liquids cannot

be obtained an~ numerical integration is necessary However for a wide range of conditions the equations pre~ented for both liquids and

I II gases a11 ow acc4ra te cal i bratfon of sea bottom pressure transducers

through the production tubing~ I

6 J_nspection_B$9YJrcment_~and Te~t Proc~dures j

The OCS Ord~rs ~ere reviewed for those item~ pertinent to a subsea system The OCS inspection requirements and the suggested general

inspection procedures for these items are discussed in the following

paragraphs Tubing PLtg_

Inspection Requirements (OCS Order No 5)I

1 T~e s~~tained liquid leakage flow should not exceed I

400 ccmin 2 The ga~ leakage flo~1 should nGt e)ceed 15 cu ftmin

General Pro~middotedure

The first step i~ this procedure is to isolate a column of fluid

between the tubing plug and the inspector with the valves opened as

for production except that the tubing plug is in pl ace Any fl ow reaching the inspector at the surface is a direct indication of leakage past the tubing plug Thermal equilibrium may b~ required if the

maximum allowabl~ leakage is too small Theflow volume can be measured at the surface in the same manner as for conventionlt1 platform systems

middotShut-in Tubi9_ls_e_sur~

Inspection Requirements (OCS Order No I

1 Wells with a shut-in tubing ptessure of 4000 psig or

greater sha11 be equj pped with a subsurface-~ontro11 ed subsurface safety device

2 When the shut-in tubing pressure declines below 4000 I

psig a remotely sontrolled subsurface safety device shall be installed

when the tubing is first removed and reinstalled

General Procedure

To determine if the shut-i~ tubing pressure i~ greater than 4000 psig I

a pressure sensor should be located ~1here the shut-in pressure can be sensed

A convenient arrangement is to locate a pressure s~nsor between the master

valve and wing valve on each tubing With the downhole safety valve and

master valve opened and the wing valve close~ the shmicrot-in tubing pressure

as indicated by the sensor is noted

Calibration of the pressure sensor can be accomplished by applying

a known pressure ~t the surface to a closed cdlumn of fluid connecting I

the surface with the sensor with corrections calculated as explained

in Section 5

Subsurface Safety Device

Inspection Requirements (OCS Order No 5) 1 A surface-controlled subsurface safety device shall be

test operated every six months

2 A subsurface-controlled subsurface safety device shall

be removed inspected and repaired every 12 mdnths 1

General Procedure To test a surface-controlled subsurface safety device the well

should be opened for production and the command then given to close

the subsurface Sifety device The leakage flqw volume if any is

measured at the surface in the same manner as for conventional plat shy

form sys terns bull

Pressure Relief Valves Inspection Requtrements (0CS Order No 8)

All pressure reli-f valve~ shall be either bench tested for opershy

ation or tested with an external pressure source (if the valve is somiddot

equipped)annually Jtbt Wf SJJS~+ General Procedure

To test a pressure relief valve for operation the vessel or line

should be pressunizeq to the pressure necessary to open the valve by

using the method suggested in Section 5 Uriless remote position inshy

dicators are provided a means to determine ~hether or not the valve

has actually ope1ed would have to be devised bullfor each subsea system

by observing the effect on the pressure sensors Pressure Sensors

Inspection Requirements (OCS Order No 8)

All pressure sensors shall be tested at least once each month

General Procedure

To remotely test and cal i~rate a subsea bullpressure sensor it is

necessary to

(1) isolate a closed column of fluid irom the surface to

the subsea sensor by properly arranging the ~ys terns va1ves

(2) determine the specific gravity compressibility and the

column height of the fluid

(3) select the pressure required at the subsea sensor

(4) apply the appropriate pressure at the surface end of

the column and

(5) compare the telemetered press4re signal or indication

with the pressur~ applied to the sensor

Automatic Hellhead Safety Devices -~~__--r---

Inspection 8equiremcnts (OCS Order No 8) All automatic wellhead safety devices shall be tested for operation

and holding pres~ureonce each1month

bull )

General Prooedure

To test the operation of the automatic well head safety devices

the out-of-the-tolerance condition (usually ~igh or low pressure) which actuates the devices should be simulated Urless position 1ndicators

are provided a means to deterrl1ine whether 011 not the valves have

actually closed ~ust be devised for each indi1vidu~l system because of

the peculiarities

in each syste m J I

L~ 6 - Jlaquo4 sfSf~ rCheck Valves ~J0141d 11tJt NC Uf~ l(I -v-

Inspection Requirements (OCS Order No ~)

A11 check valves on a11 fl ow 1i nes sha11 be tested for opera ti on

and holding pressure once each month

General Procedures

To test the check valve ft will have to be back pressured The

check valve fails to operate properly if any back flo~1 through the I

valve occurs

~iguid Level Shut-in Controls I

Inspection Requirements (9cs Order No 8)

All liquid-l 1

evel shut-in controls shall be tested once each month

by raising or lowering liquid level across the leyel-control detector

General Pro~edure I

The preferable way to test the liquid-level sh~t-in controls is to

simulate the out-of-tolerance conditions (1 iquid Jevel either low or I

high) at the sensor This sim~lation should cau~e either the inlet

shutoff va1ve to close or the discharge shutoff valve to close In

this manner both the control YStem and the automatif valves are tested I

A means will have to be devised for each subsea system to determine I

if the out-of-tolerance condition actually occurred at the sensor if

it was properly Jetected and i~ it caused thd proper actuation I

Vessel Automatic~Inlet-Shutoff Valves I

Inspection ~equirement (OCS Order No 8)I

bull ) ~ J) All automatic~~~~~-shutoff valves actuated by vessel~~

~fsensorj shalil be tested for operation onle each month General Procedure

Described above under gene~al procedure for Liquid-Level Shut-

In Controls Vessel AutomaJi c-Di sc_b~_g_e-Sh_ll__Dff__Y_ii_l ves

Inspection RequireiT1ent (OCS Order No 8) All automatic-discharge-shutoff valves actuated by vessel lowshy

level sensors shall be tested for operation qnce each month

General Proc~dure

Described above under Liquid-Level Shut-tn Controls

High-tempera_ure Coniprc2_()r-Sh_t1tdow_ri_Co~_o1s Inspection Requirement (OCS Order No 8)i

High temperature controls which protect the compressor against

abnormal pressure~ soltiY by such temperature

safety devices shall be tested annually

General Procedure Because of the danger involved in elevating the temperature suffi shy

ciently to cause + temperature device to actuate it is recommended that each temperature device be pre-tested and replaced each year rather than be tested remotely

Oil Spil 1 Detection E_g_t_pmen~

Inspection Requirements (OCS Order 7) All platforms and structures are require~ to have curbs gutters

and spill pans connected to a tank or sump fhis equipment is required to collect all hydrocarbon spillage on the platform It is recommended that this requirement be extencjed to require all unenclosed subsea

equipment to be covered by inverted spill pans with hydrocarbon sensors

and sumps to rem~ve spillage

General Procedure To test the hydrocarbon spi 11 age detection and removal system

a controlled spill should be made in each ar~a or section of spill

- bull

pans while th~ SUljlPI

pumps are monitored for proper operation This I I

metbod will require a signal to the surface to indicate directly when the pumps are running or indirectly by indicating a pressure rise in

the affected pipeline due to the pun1ping action The indirect method is preferablebec~use the pressure rise not only indicates the pumps are running b~t indicates they also are pumpi~g fluid One additional safeguard should ~e employed A TV camera should be used to observe

the spill pans in the vicinity of the spill to assure that they are not leaking due to damage or inadequacy of de~ign Hydrocarbon S~nsors

IInspection Requirements

Hydrocarbon sensors are not covered in tbe OCS Qrders presently

in effect General Procedure The general test procedure for the oil-spill-detection equipment

above will test the hydrocarbon sensors for operation In addition

I bull some means of knowing when each separate hydrocarbon sensor in the

spill pan assemblies hus sensed hydrocarbon will provide additional I

protection by warning of system degradation b~fore the system is comshypletely inoperable

7 Summary and RecQ)enda tions This report deal~ with the inspection problems that have arisen

as a result of the present trend of the offshore oil industry towards

subsea completion and production of deep water petroleum reservoirs I

Inspection techniques for completion and prod~ction equipment are inshyvestigated but inspection of fiser systems submersibles and strucshy

tures are not inc~uded in the scope of this wprk

A technique ~as been detailed for calibr1ttion of subsea sensors I

and for verifying the proper OReration of va1ves The technique reshy

quires only that the inspector 1 have access to production tubing at

- bull

the surface By actuating subsurface control valves from the surface

the inspector can isolate a closed fluid path1 to the point of interest I

Pressures read at the surface (when corrected for hydraulic gradients)

may then be used to ca 1 i brate subsurface tran1sducers Flow rates

measured at the surface can be used to I

indicate leakage rates bull

subsurface valves I

but fluid in the tubing must be allow

ed to reach

sufficient therma1l equilibriumwith its surroundings so that a false leakage rate is riot indicated due to thermal expansion of the fluid

I bull

This technique has been used to work ou~ an inspection procedure

for the Exxon Sulisea Production System The 1procedure demonstrates I I

that the necessany closed fluid paths can be obtained to permit inshyspection of all necessary points Thus from 1a theoretical point of et

j9aYYew 1 t has ~en established that accurate euroVa 1ua tion of subsea con- J

ditions can be made from the surface (io implement the procedures I

it would first be necessary to conduct an experimental program with field

tests to determi~e operational problems) tAtti e

Inspection of subsea production systems

ACKNOWLEDGEMENTS

This report describes work done under technology transfer authorization for the Conservation Division of the US Geological Survey

The authors also thank Shell Oil Co and Exxon Company USA for their cooperation in providing detailed information

on their approach~s to subsea production without which this program would not middothave been possible

1 Introduction

I -- - IL

sirable if feasi 1

sense the item i1

points

d P = FdP p

dP In psi

lmiddotshyltl

= F(P-Ps1

p = eF(P-P~)or PS ( 1 )

dP =p geF(P-Ps)dh s

Po-Ps = gPsD

4Po-Ps = -5 x 10 ln(l-008) = 4170 psi

P0-Ps = ~ ln 0 + fp 0gD) (4)

from equation 1

P0 0 FP = r e D (5)

i I __ _l__

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

1 Introduction

I -- - IL

sirable if feasi 1

sense the item i1

points

d P = FdP p

dP In psi

lmiddotshyltl

= F(P-Ps1

p = eF(P-P~)or PS ( 1 )

dP =p geF(P-Ps)dh s

Po-Ps = gPsD

4Po-Ps = -5 x 10 ln(l-008) = 4170 psi

P0-Ps = ~ ln 0 + fp 0gD) (4)

from equation 1

P0 0 FP = r e D (5)

i I __ _l__

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

I -- - IL

sirable if feasi 1

sense the item i1

points

d P = FdP p

dP In psi

lmiddotshyltl

= F(P-Ps1

p = eF(P-P~)or PS ( 1 )

dP =p geF(P-Ps)dh s

Po-Ps = gPsD

4Po-Ps = -5 x 10 ln(l-008) = 4170 psi

P0-Ps = ~ ln 0 + fp 0gD) (4)

from equation 1

P0 0 FP = r e D (5)

i I __ _l__

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

I -- - IL

sirable if feasi 1

sense the item i1

points

d P = FdP p

dP In psi

lmiddotshyltl

= F(P-Ps1

p = eF(P-P~)or PS ( 1 )

dP =p geF(P-Ps)dh s

Po-Ps = gPsD

4Po-Ps = -5 x 10 ln(l-008) = 4170 psi

P0-Ps = ~ ln 0 + fp 0gD) (4)

from equation 1

P0 0 FP = r e D (5)

i I __ _l__

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

I -- - IL

sirable if feasi 1

sense the item i1

points

d P = FdP p

dP In psi

lmiddotshyltl

= F(P-Ps1

p = eF(P-P~)or PS ( 1 )

dP =p geF(P-Ps)dh s

Po-Ps = gPsD

4Po-Ps = -5 x 10 ln(l-008) = 4170 psi

P0-Ps = ~ ln 0 + fp 0gD) (4)

from equation 1

P0 0 FP = r e D (5)

i I __ _l__

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

points

d P = FdP p

dP In psi

lmiddotshyltl

= F(P-Ps1

p = eF(P-P~)or PS ( 1 )

dP =p geF(P-Ps)dh s

Po-Ps = gPsD

4Po-Ps = -5 x 10 ln(l-008) = 4170 psi

P0-Ps = ~ ln 0 + fp 0gD) (4)

from equation 1

P0 0 FP = r e D (5)

i I __ _l__

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

d P = FdP p

dP In psi

lmiddotshyltl

= F(P-Ps1

p = eF(P-P~)or PS ( 1 )

dP =p geF(P-Ps)dh s

Po-Ps = gPsD

4Po-Ps = -5 x 10 ln(l-008) = 4170 psi

P0-Ps = ~ ln 0 + fp 0gD) (4)

from equation 1

P0 0 FP = r e D (5)

i I __ _l__

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

lmiddotshyltl

= F(P-Ps1

p = eF(P-P~)or PS ( 1 )

dP =p geF(P-Ps)dh s

Po-Ps = gPsD

4Po-Ps = -5 x 10 ln(l-008) = 4170 psi

P0-Ps = ~ ln 0 + fp 0gD) (4)

from equation 1

P0 0 FP = r e D (5)

i I __ _l__

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

= F(P-Ps1

p = eF(P-P~)or PS ( 1 )

dP =p geF(P-Ps)dh s

Po-Ps = gPsD

4Po-Ps = -5 x 10 ln(l-008) = 4170 psi

P0-Ps = ~ ln 0 + fp 0gD) (4)

from equation 1

P0 0 FP = r e D (5)

i I __ _l__

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

Po-Ps = gPsD

4Po-Ps = -5 x 10 ln(l-008) = 4170 psi

P0-Ps = ~ ln 0 + fp 0gD) (4)

from equation 1

P0 0 FP = r e D (5)

i I __ _l__

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

P0-Ps = ~ ln 0 + fp 0gD) (4)

from equation 1

P0 0 FP = r e D (5)

i I __ _l__

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

-Example 1

=0405 p~~

-5 0 ( ) eo 50 x 10 x 1000

At 120 F gplOOO ~ 0434 09130

= 405 psi Ps = lopo - 405 = 595 psi

= 398 psi P~ = 1000 - 398 = 60 psi

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

Example 2 I

1000 o middot o i 080 ~ 10~5 x 1000

I = 03]2 psift

PD - Ps = F1 1 n (1 + F PD9 D)

1At 70degF PD

- Ps = _ ln (1 +1080 x io-5 x 0321 x 1000)5

080 x 10 = 321 psi

PS bull 679 psi

= 311 psi Ps = 689 psi

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

(=~middotGr ) e0 0

At 70F 9P5poo = 0434 (09305)e039 x 10 x 5000

= 0412 psift I

= 0406 psift 1

P5 039 x 10 = 4087 psi ~ 5000 - 4087 = 913 psir5

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

- -p~ = 4019 psi P = 981 psi ~

Example 4

70degF

l 20degF

Sp Gr= 07117

STD 2540)

~ooo = o332 psift

= 3277 psi I P5 ~ 5000 - 3277 = 1723 psi

= 3203 psi

PS = 5000 - 3202 = 1797 psi

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

shybull

-5AP = 356 x 10 x SG x P x D

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

General Procedure

in Section 5

form sys terns bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

General Procedure

necessary to

the column and

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

bull )

General Prooedure

the peculiarities

in each syste m J I

General Procedures

valve occurs

All liquid-l 1

General Pro~edure I

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

General Proc~dure

- bull

subsurface valves I

ed to reach

I bull

- bull

subsurface valves I

ed to reach

I bull

- bull

subsurface valves I

ed to reach

I bull

Documents

Inspection of subsea production systems · of subsea prbduction systems for . exploitation of these deep water petroleum ' re~erves ' poses new inspection problems for the U.S. Geological