Piping Handbook - Hydrocarbon Processing -1968

5,PIP!NG

HANDBOOK

PIPING HANDBOOK

TABLE

OF

CONTENTS

STRESS ANALYSIS

A Simplified Computer Program For Pipe

Piping Design Method Beats Computer

Symmetrical Piping Arrangement Solves

Expansion Loop Design

Two-Phase

Page No.

Flow Distribution Problems

LAYOUT

Plant Layout And Piping Design For Minimum Cost Systems

What lnformation ls Essential For Good Piping Design

How To Design Yard

Piping

.

.

Locate Tower Nozzles Quickly

Piping Of Pressure Relieving Devices

MATERIALS

U.S. vs British And

Which Material For

.

.139 European Piping Specifications

Process Plant Piping?

4 5 10 20 24 25 34 39 49 59 66 67 73 80 81 88 94 96 EXPANSION

Find Best Pipe Expansion Loop Quickly

Expansion Joints: How To Select And Maintain Them

Spring Pipe Hanger Design Simplified

Piping Tierod Design Made Simple

THERMOWELLS

Procedures For The Piping Designer lnstallation And Soecifications

.,

. . lnstallation And Specifications

.,

. . .

Selection Of Thermowell lnsertion Lengths

PRESSURE DROP AND VIBRATION Simplified Utility Loop Balancing . .

Piping Design Stops Pulsation Flow

Find Line Pressure Drop By Nomograph

New Approach To Pipe Reactions

STEAM TRACING .

New Guide To Steam Tracing Design

100 101 105 111 120 121 125 131 133 138

A

Simplified

Compufer

Program

for

Pipe

Expansion

Loop

Design

Using

a

single

input

card

with

terms

and

measurements

common

to

any drafting

room,

computer

computation and

analysis

has

5:1

time

advantage over manual

methods

W. W. Shul!, and G.

N.

Bogel, Jr.

The Dow Chemical Co.. Ilouston

A

conputnn pRoGRAM _{has been written that}

_will

de-sign piping expansion loops with less cost and with less

elapsed time than ,existing rnanual methods.

It

requires a

single

input

card containing measurements common to any drafting room.

Mqnuol

l,lethod.

A

previous

articlel recornurends a

manual method for piping loop design except for critical lines (those with high or low operating temperatures and pressures andfor force sensitive connected equipment)

and except for piping which conveys hazardous or flam-nrable materials. These exclusions eliminate application of the author's semi-graphical methodl from most of the

piping in a petrochemical plant and require that the pip-ing designer both understand and have access to process

data. Even

if

process data were available, the hours

re-quired

to

segregate the hazardous senice and critical piping for solution by a stress consultant or specialist and

the queue hours for the attention of the stress consultant

or

specialist are

additional

factors which the authors omitted.

The authors' make the claim that,

for

non-excluded

piping, their graphical approximations are obtained

in

a

fraction of the time

it

takes to prepare the data for com-puter analysis. Readers will have to guess at the

engineer-ing and clock times used by the authors

in

arriving at

their conclusions.

In

our

trial

runs, assembling

of

data

took the same time and computation and analysis of re-sults from computer and hand calculations had a five to

one time advantage in favor of the computer, besides hav-ing the ability to better understand the effect of the vari-ous alternates.

It

is probable that the computer programs available to the authorst did have a complicated input for solution to simple problems; and considering only over-all

time lapse, their claim is often true for an answer to any

one single problem which happens to fit a manual method

which the designer uses with sufficient frequency to main-tain his speed and accuracy.

Effective use of computer programs also requires user

familiarity with their input requirements and their output capability. This learning eflort is less than the eflort of

learning any equivalent manual method. Moreover. one

rnan can examine pipe layouts for a plant, input the loops

on a computer program input form, and get answers for

one

to

several hundred problems

within

24 hours, and

in some offices within the hour. Well planned piping

de-sign jobs seldom need answers faster than 24 hours; but,

under planned conditions, the computer results can be

rnade available within minutes.

Resislonce ?o Computer Use.

It

has been our

experi-ence that people who resist using the computer as a tool

for the routine as well as the difficult tasks in design may

be described

in

two groups. The first group thinks that computer answers are either beyond question

or

are not subject to parametric study of input variables or

modifi-cation based

on

engineering judgment. This group has

A

SIMPLIFIED COMPUTER PROGRAM FOR

. .

.

S AND B ENGINEERTNG SEPVICES PIPE LOOP OESIGN PROCRAX {AINLINE PIPE LOOP DESIO -_sINGLE PLANE'2 TO 6 PIPE IEXAERS

rITH I TO 5 ELAOiS OR BENOS OF SAXE BENO RADIUS IRITTEN BY T.I.SHULL ANO G.N.AOGEL' HOUSTON TEIASI ilAY T967' PRL]GRA{ USFS FORNULAS ANO T BLES PPES€ilTEO IN TIPIPING OESIGN AND

ENGINEERINGI' THE GRINNELL COXPANYIINC't PFOVTO€NCEI RHOOE ISLANDT usA. SECOND EOtTION. 196f,.

ALL SUBS FILOI PROCEOUPE AS DETAIL€O ON PAGES 52t5f OF CRINNELL INPUT STAFTING IITH NY XEXBER AND €NDING IITH ANY f,EiAER' EACH XEXAEF HAS T'O JOINIS (TELD€O) IHICH ARE ANALYZED FOR IHE LO.EST ANO HIGHESI JOINI NUXBERS ARE CONSIDEREO IO tsE THE LOCATION OF IHE TNNER{OST PAIR OF GUIDES OR THE LOCAION OF

ANCHORS IF IHE lNPUT LENCIH TOUISIDE OF GUIOES' I 5 ZEKO. iHE EXCEPTTON IO THIS RULE TS FOR TIO XETBEF ISINGLE ELBOI' JOINIS FOR IBICH THE PROGRAI TILL INCREIENT ONE f,EMBERS LENCTH TO PROVIOE THE NECESSARY FLEXTBILTTY'

c.rf,xoN o(6) t6(6).vl 6),x( r2 );Y( r2l.xL( r2r,r( l2l,rxl l2ltrY(12,,R(12) r.s( r2t.Rxt t2),sT( t2), AAI 20 )

COTXON SIPTPI. TLA IXKTSAETAI XBARtYtsAR! XLP!YLPTPXYTP I XTPI Yt IMAX ICT' IFX! FY t CONSIX!CONSIYTOA'THX !FST{AXTDEMT ILEN!ALTsI

COHXON TLNCTXOFY.OLNG

coxf, oN tEx(5, t TRY( 5),BXl 5l.AY( 5),OXl 5l rAST (5' rBR'srtAXrKBD COIXON KPFI TKSTI(FNDIKFSI!KLOOPt J!K'N.KXAi

OOUBLE PRECISION PIPEIO

t.JI = OST TO.ENTROIO X OIRECTION Y(J) = DST to CENTROID Y DIRECTIoN II IJI: LFNGIB OF SEGXENI R(J) = X DIST FROY CENTFOIO 5(J) = Y OISI FROB CENTROIO PXY = PROO OF INIER I A

PIX = COU OF IdTERIA ABOUI X

PtY = xof, oF INIERIA ARoUT Y

CT = ExPANSION AT TEHP

XLGTYLG = t ANO Y DIST FND FOINT TO END POINT PI = P1PE IOXENI OF TNTERIA

Sf,P = PIPF SECT TON XODULUS

CENTROIO OF STFAIGHT LtNE IN PLANE OF PROJECTION - COLN A = LENGTH B= x OR Y DISI EoN r

COLNlarB, = A*A

( CENIRDTO OF BEND XK=FLEx.FACT'R=RADIUS OF aEND!X=DI5T TO AXISE0NS c coaNo(xKrRrxt= t.57axxrx aR

C PROD. OF INIEFIA LINE PARALLEL TO AN AXIS EON6 XL=LENGTH C PLNXY(xL!x.Yr= xL t r: Y

< PROO OF INTENIA AEND NEG IHEN AXIS I RADIAL OR ARC

( PENON(xKrR.x!Yl= xK +.lf7rFaRrR. xKI r.57 +R +x' Y

C PROD OF INTERIA EEND POS iHEN AXIS '

RADTAL AND AFC ( pBNOp(XK!R,r,y)= XXr.r37*R+R*R + iKal.57tR*XrY

C MOMENI OF INTERIA ST L'NE tN PLANE OF PROJ PARALLEL TO AXIS X

( PLMILNIxLTY) = XL aY:Y

C MOU OF INI SI LTNE PERPENDICULAR TO AXTS Y

( ppxtLN(iL,xl = iL*xllxL/I2.. XL+X1X C IDB OF tNT BEND FROM AItS Y

< p!taN (xxrRrxt _{= xK +to.r49aRrR+Rl.} l.s7! xK tR tx.x C PIPE SECIION NOOULUs DA = ACTUAL OTA. THK= THICXNESS IALL

s!( oA. rHKr sxx )=0. 25loarsMx*( DA+oa_2.4( DA.THK T+2.a(rHK.THKl ) . LAMBAOA XLAX F = RADIUS OF PIPE AEND

iL^{l THK. R. DA)=THK+R/l ( cA_rHK I *(DA-THxl +.2s) C FLFX. FA'TDA XLA = LAMNADA

FF ( xL A l= I .65/ xL A

C STRESS INIENSIFICAIION FACTORT AETAS' FOF IELDED ELBOIS OR BENDS' EEIAs( xLA l=0. 9/xLA*t O. 6667

C OIPE CROSS 5ECTION!L ICTAL AR-A PAREAI DATIH('=3. I4 I 6IIHK:(OA_TBK I c plpE ioH oF lNr

PMINTI OA'THK I=3. L 4I6*I DA-IHK)'+O3TIHK/4. C XBAR DISTANCE

C D l5 DISI v= vAFIAELE B=tsEND RAOIUS

C XPRI--A NON-ZEEN ENIPY IN CDL 7A PRINTS VALUES TN CO{{ON C LPRT--NON-ZERO ENIRY IN CI]L. 79 SHOYS LAST OF PIPE LODP5 USING C HEADER CAFD ANO PR'GRAil TILL EXPECT A NEt HEAOER CARD' C NPPI_-PFINIS STRESS A5 PtOF IEilBER LENGTH IS VARIED FOR OESIGN C PROALEqS. ENIER IN COL. AO DIGIT 1 FOP STNESS EACB.5 FEFIT

C DIGIT 2 FOR STRE5S EACH I. FEEII

C DIGTT 3 FOR SIREsS EACH I.5 FEEI! ETC'

C 6UIOE INOICAIES OI5IAN'E IN X OR Y DIRECTION OUISIDE THE SECIION A AO

C OEFINED FY Ol-D6 . o=x l=Y DIHECTIoNS x=Ol _'D3'D5

C T€5I DAIA FROM GFINNELL SIJCOlO EDITION PAGE 52. 1CT FROM PAGE 9 A 72 c A6 F5.O Fs.O Fs.O F5.O Fs.O Fs.O F5.O F5.O F5.O F5.O F5.O rtF4.IIF4-IlF4. JI1 C tD DIA TH( ALiST CT tsND D DT D2 03 D5 06 GUIOEVARYLVARYA PRT cP.52R1O.75.50 17675 996.50, {0. 24. 12. 14.

cP.5201o.75.50 17675 996.50.40. lO. 12. A. 2 lO.l4oOO CP.52RGRINNELL ANsiER5: ANCHoRS x=2795rY:lA67LB5. END HOMFNIS=24f64'-23970 FT-LB C FOR PROBLE{ P.52: MAi.HE'TD STRESS EET. JOINI 2

' 3 = 9O3OPS] t IOA'=29672 C {ANUAL-GRAPHICAL: 6REATESI STRAIcHT PtPE tS AI JOINI 2 = a7l4 Pst

c INEPIIA oF PIPE=212. !BAR=3a.63!YBAR=9.6oFT.

C sEcTloN MOOULUS=39.43 lxY=lO4s7. lx=941s. lY=21i69'

C

'OHPUTER RESULTS: XAX.BEND STRESS RET.JOINTS 2 A 3 = 9IA2 PSI C FOR PAOBLEM P.52: GREATEST JOINI SIREsS IS AT JOTNT 2 = A723 PSI

c INERTIA OF PIPE=2\?. XBAR=34.62'YBAR=9.59 Fr.

C sEaTloN HoDULUS=19.43 tXY=I0446. lX=9395. lY=2135'.

c 105 c I lo ttt at2 ttJ tr5 C I l6 c rlt I to too lor to! lo:l to2z IOal I O2a I 0?5 lo2, t 0t I O7 l06 ]L 32 a, t20 121 t22 c 23 2a 26 3t J2 ll fa 4a 5l 52 55 57 5A 65

little

knowledge

of

the difficulty

of

progran'ring

a

com-puter to satisfy every possible need or whim of the user.

Its members reject the computer as a tool when the pro-gram's logic requires flexibility

in

the

user's notion of

input and output content.

The second group has some experience rvith either

pro-graming

or

using the computer; yet, its members

over-6

tla

l4i

tro

estimate the

difficulty

of

communication

r.vith the computer. Both groups deter the economical emplo1'msn1

of

computers and thus restrain the development

of

the engineering sciences from the precomputer art of

apply-ing empirical relationships

to

approximate solutions for

-nvhich _{exact theoretical solutions exist. Fortunateln} mem-bership in either of these groups need not be perrnanent!

(ONT RL:2 rRITE (f,r r r5) lor RE-aD (rtl29, aA

5J

to2 icITE (r, r30l NPACETAA DO tOr J=rt 132 oo tor J=r53r2ll lOa AlJl=0.0

READ I tr lf I I PIPEIO,DA!IHKTALIST.CT tAFAOtD!kFSr!FSTMAx.<SNDtsNDilAx I TKDIRI OLNGT(PRIILPRTI NPRT

YFIIE (3! 1f2 I

IRITE (f, I IfJ} PIPEIDIOA'THXTALISTTCTTAFAOTDTKFSI'FSITAXtXSND.

Fig.

l-Pipe

expansion loop program. Note: Subroutines not included in the program listing solve step-by-step thE manual method illustrated in Lit. Cited 2, pages 52 and 53, using the functions duplicated as comment

T

Pf,OBLE X I D El,lTlF lC AT lOil .?a,aat+.cnt, ,tF€ ,1/tC, ,4 , t , , t LOOP lDEl'lTlFlCA-ilot I 6 ACTITAL 0wsl0E PIPE PIPE

rlU_ ALLOY-IELE sTncss (PSt, fr 2 c FACIOff AT OEB. E D ,t Itotu3 OF PIPE BEilD FIBST LOOP sE00ilD r.ooP TAIND LOP SEEIETT LEXGrll ITEEI) g, at FOUNil LOOP FIFTII LUn strTltLooP .= i= Er a-*E t ltax. L DE lBu I ax- IETL. IETGTII IFEET} ara LEUOTN IFEETI P al IErt6TE (FEETI t 5 LEXGTH (FEEN o tt LEI'IGIH IFEEN o A BF /0|'lD iIJIDE DITUETER lrilcHEsl 7 Ir[cr{Esl 2 1 ltnurEJ, llIERs-01 t7 _!i e

TYPICAL LOOP LAYOUT START AT ANY MEMBER,AND STOP

AT ANY MEMBER TO DEFINE LOOP

GEOMETRY - EETWEEN ANCHORS OR GUIOES

Fig. 2-lnput card with data for problem shown on Fig. 5.

The Progrom.

A

computer Program

in

FORTRAN

IV

was written (Fig. 1), just to see

if it

could be done, to

solve expansion Ioop design based on a published manual method.2 The program does

not

require manual

inter-'-'i1"

allow

solve

upon consultants.

If

the loop as input has too great an

anchor force or too great a stress, the program provides the required flexibitity by increasing the length of a

desig-nated member of the loop within user input limits. The Iocations

of

guides

normally

are

not

altered because

guides usually are mounted on supports whose locations are subject to controls other than piping flexibility.

Input Data. The entries for an individual piping loop

r"qri."

a single card (Fig.

2-80

letters

or

digits

maxi-mum) containing the following information:

.

PiPe loop identification name or number.

o

Outside diameter of the pipe and the wall thickness

l;:n"r)

. There

are no limitations on diameter or

thick-o

An allowable stress range at bend and terminal joints based upon code

or

connected

equipment limitation

(pounds/square inch).

o

A value of C obtained from the following equation:

^

Expansion, in.,/100

ft.

(E")

r-@

or from Literature Cited 2, page 9. This value combines

the lineal therrnal expansion

in

inches from 70o F to the

operating temperature with the tensile modrrlus

of

elas-ticity at 70o

F

(pounds/inch).

o

Radii

of

bends

(inches)-a single

bend radius is

applied to all bends.

o

Lengths between corners of the loop members within the innermost

pair

of

guides (feet).

A

"corner"

is

de-fined as the intersection

of

the center line extensions of straight pipe members (however short) which connect to

a 90" bend or square elbow. The input length must be

not less than the sum of bend

radii

(expressed

in

feet) incorporated

in

the particular member.

o

The number of a member whose length the program allel the

ex-eto

hold anchors and move guides.

o

The maximum permitted anchor force (pounds) and

the

X

or

Y

direction of this limitation.

o

The piping length (

most pair of guides and

axis

of

this

additional

contributes force because of thermal expansion.

An

additional header card precedes a grouP

of

pipe

loop entries. This card contains any desired alphabetic or numeric descriptive data desired

in

the output printout

(job title, user's name, accounting information, etc.).

Whol

The Progrom Solves. The program is capable

of expansion to handle many possibilities

of

piping loop design but implemented for PurPoses of this

documelta-tion

for

the following frequently used single-plane loop

types which have no external loads or restraints between anchors except guides:

o

U-shaped loop with unequal

or

equal legs plus up

to

trrzo tangent members

of

unequal or equal

length.

(Terms correspond

with

Literature Cited

2

terms for

U-shaped examples.)

o _U-bend expansion loops similar

to

above. IJse our term "legs" instead of the Literature Cited 2 term "tan-gents" for the expansion U-bend examples

in

Literature

Cited 2.

o

Simple two member loop with one elbow.

o

Z-shaped expansion loop.

o

Hooked Z-shaped with up

to

one tangent rnember.

Progrom Tesis. The program has been tested with book

problems' (Fig.

3)

and

with

the

examples

from

the

7

PIPE LDOP FlPE IALL ALLOI FACTOR C RADIUS LENGTHS OF ME{AERS (TO CENTER OF CORNERSI D- IHICH ANCHOR FORCE LIXIT , rnFNt- n D THtak- qTRFsS a nFG-F oF FFNn Dl n2 Df nd n< nA xaY vAeV .=Nn- I -Y ,=v nlo

rNcHEs _{l-lellEs -_e5!_}

--__--- !!!tsES- -EEEI- -EEEI- -EEqf- -EEEI- -EEEI- -EEE.M: EEE!

qlqE PouNDs FoRcE

75O-..._o <nn -E6b> SE qI_rnrNT . _)>

--E-6330. A020. 63A0, 962- 7426.

JOINT 2 HAS GREATEST STRESS ANcHoR F0RCE IN X-DIREcTION = 2a03. ANcHoR FoRCF IN Y-DIRFCTION= 147O. LENGTH OF sUM DF MEABERS

--

F' rrrn qtrH--Dl=D6 _LENG.IHS

-FEET FEFT FEET AX I S

COORDINATE OFFSET FEET MAXIMUM STRESS IS IN BEND LENGTH OUTSIOE AXIS DISTANCE .ll-uI_1,f.-_THRU .r I AErEF{ rpiHiB

' ^Hp r I O,|I X_AXIS Y-AXIS IN POUNDS/SO. INCH

MOMENT AT JOINT I = DIAGRAM OF PTPE LOOP MO{ENT AT JoINT a =

L4LOz.-_J5_{EXBEF "I_J6_ rl J7 ll - --L-Lq-ottrrrr -.+ | o-ooo ffEL MFMBER D2 MEMEER O4 | _J9__MEMRER D5__J r o_ | LOOP O. K. FROM STRESS STANDPOINT

Fig. 3-Test problem based on book2 input data.

PIPE LOOP PIPE TALL I|.OI FACTOK C PAOIUS LENGTHS OF MEMBERS (TO CFNTER OF CORNERS) D- IHICH ANCHOR FORCE LIf,IT

lrlclEs rNe,BES __PS!_ _{!NS!ES- -EEE!- -EEEI- -EEEI- -EEEI- -EEEI- -EEE! !S! EEEI} qgOE eQ!.Nqs FoRcE j.o9B lm lMo-a-@lo4Lp4

7t96. l6la7. l30aa. 2154f,. 24642. I tq6. o- o.

JOINT 5 HA5 GREAIEST STRESS ANCHOR FORCE IN X_DIRECTI ON - I2N6. ANCHOR FORCE IN Y-DIRECTION= LENGTH OF SUM OF MEMBERS COORDINATE OFFSET FEET MAXTMUM STRESS TS IN EEND LENGTH OUISIOE AXIS OISTANCE

FtLF PATH. _-D.!=s. LENGTHS

-JONI J+-THPU J I

FEET X_AXIS Y_AXT5

1NO 7 !@

IN POUNDS/SO.INCH FEET AX I S

MOMENT AT JOINI. DtaGRAM 0F PIPE LotP MOMENT AT JOINT IO =

+ leooo rEL l. ro.o@ Er MEMtsER D2 VEMEEi D4 / l2.ooo FEET J3 I - - - -- -A€G-INNIN5-:]Jl---::4EUBER-oJ:=:-:r+=l- -- . . l--cuIDE = | J' I o.o FEET EtBU N-E I I F#T- -- JrL=-I J9____MEMBER D5___JrO_l THIS I OIIP OVERSTRESSEO

Fig.

4-1"r,

problem based on Haque/Starczewskil input data.

PIPE LOOP PIPE TALL ALLOI FACTOR C RAOIUS LENGTHS OF MEMBERS (TO CENTER OF CoRNERSI D- THICH ANcHoR FoRcE LIHTT

J!-C!ES I-NCHE5 _{--eS.!.- ---} !!g!ES- _{-E-EE.L- -ECtrI- -EEEI- LEEEI- -EEEJ- -EEEL} lsi EEEI ggqE EgUNgS-FoRcE

t!Pu' OgL6.6fs O.'r0 -'5Oo. 35s.0 d.ooa r2.OO A.^0 a.OO _{a.OO 12.^O} O.O 2 4.5O t e2OO.e rtrH w^ovrNc _{xFMqFa I Fil.TH =} a-6 FFEY. eToFss I< r.:Ao- I aq/ea IU - aM.Hno FnorFc y-6ro ₌ ,^A9.. _{w-DIp - -O} IITH VARYING HE{BER LENGTH = a.s FEET' STRESS IS 30O79. LBS/SO.IN.r ANCHOR FORCES X-DIR.= la3l.. y-DlR.=

-O. YIT{ vAqvINA {EMBS _{I ENGrts =} o O FFFr- STREq( rq _'a^^1- LA<'sO.IN.- _^Ncpno FnR.FS y_DIor_

'5ra.,. v_DtF._ _o. IITH VARYING MEMBER LENGTB = 9.5 FEETr STRESS IS 26276. LAS/SO.IN.. ANCHOR FORCES X-DIR.= 1434.. Y-DIR.= _-O. !u+wYl!r: l.MHFo LE.Mg - rO- _{s| 3lS2-}

' pq"qr- : 12a6.€

IITH _{VARYING xEHtsER LENGTH =} lO.s FEET. STRESS IS 23251. LBS/so.IN.r ANCHoR FORCES X-DIR.= lr57.r y-DIR.= _-O.

-u Jl! JP 6473. 1456t. 12044. 19443. 21960, 21960. 1944a. 12044. 14561. 6473. o. o.

JOINT 6 HAS GREATEST STRESS ANCHOR FORCE IN X-OIRECTION = I045. ANCHOR FORCE IN Y-DIRECTIoN= LENGTH OF SUM OF MEMDERS COORDINAIE OFFSET FEET MAXTMUN STRESS IS IN BENO LENGTH OUTSIOE AXIS DISTANCE EL'rlo ptIB ot_DA til.-rs< Jot{ItL t THRU sl!'s 6 N0 , LOOp

FEET X-AXIS Y-AXTS IN POUNDS/SO. INCH FEET FEET AX T S

,rotu__

-^r.oo

UOVENT AT JOINT I - DIAGRAM OF PIPE LOOP MOMENT AT

aE. FOOr-POTJMIS Ar5- JOINT IO =

I I JI I ! l.ooo Ef MEMBER D4 ,l a tz.ooo FEEr J3 I ar u[Ytrac_:_I__---8ff8tre _^t )2_l CUTDE I I o. o FEET I

tt

l2.Qio FFEr Jll | _J9_____MEMBER D5___J I O_ | LOOP O. K. FRO! 5TRESS STANDPOTNT

Fig. S-Test problem based on Haque/Starczewskil input data.

About Ihe qa.rthors

'Wrr,lr,lrr _{\1r. Snur-r, is u, r,iril cnginecr} utith tlte Cor.poro,ta Enginer:.ring antl

C onstruc tion Sertices of The Dou

Cltenticol Co., HotLstott. He cletelops

qualitA tnctltods of ck:sign problem

sctlu-tions f or ci.-il, ntcclta.nicrLl, an.cl :--esscl

disciplines. Mr. Sltull holtls a, B.S.

cle-gree in nkc/l.anisa1 _{engineering front} Lottisionct Stote Uni.uersitu rtnd, Itcts

don.e gradu.ote toork at Tulane, tlLe Lini-uersity of Ho.ttston, and Georgle

lVaslt-ingtotr Uniuersity. He lm,s pS ,LJ(.al _s

c:t:-G. N. Bocrr,, Jn. is a seuiol" engineer

in the Corporate Engincering and

Con-struction Seruices of The Dous Chemicol Co., Houston. His utork inrolr-es

chenti-cal engineering, systems dnalAsis f or

effecti"-e use of coinpllters in design

engineering, liaison f or lrcat erclTan!.ler tlesign practices and metltods witlt Heo,t Transf er Researclt, Inc. Mr. Bogel ltolcls

a B.S. degree in chetnical enoineeri,ng

L

-J|I

from Teras A.&M. Uni"*ersity. He h,ns

wollced Lil.ptocess det:eloptnent, pilot ancl production plant supert:ision, pyocess and mccltanicctl design, and, project

en-gineering. He is a metnber of AIChE, ACS, ACM, TSPE, NSPE, Process Heat Exchu,nger Society, rntd the National

S ociety _f or O ceo,nogro,pltu.

Haquc/Starcze$.slii article (Figs.

,l

ancl

5). In

general,

the agreement rvith answers gir-en b,v our program asree

u.ith the problems and examples

\rithin +

2

perccnt.

Extrerle care \,vas necessary

to

make sure the problems

r'r'ere the same and that the results u,ere identifiable. For

the exarnple given

at

the botton'r

of

Page 204

in

the I{aqLrc,/Starczervski article,

our

stress using long radius

bencls and con-rputed intensification factor is in agreement,

but

onr anchor force is lorver than tlte square corner

solution bv

a

factor

of

tu o. The anchor force

for

our square corner solution

of

the sartre probleln agrees r.vith

the force from the computer program mentioned

in

the

Haque/Starczeu,ski article and is nine percent lo',ver than

their graphical method.

Square Corner Technique. The use by inclustry of the

square corner technique

with its

conselvative answers

(sometines by

a

factor

of

two

or

more) is difficult to

understand because

_it

tends to indicate problems rvhere

there are none and results in tvasting investment and

oper-ating capital for unnecessary expansion loops.

Note: The mainline program (Fig.

1)

with its sanple output (Fig.

3)

represents original u,.ork and is in no case

endorsed by our emplo),er or based on any program cllr-rently

in

use

_in

our ofTice.

1Ha.gu9,

_{)'-n",Jll?"r",t*n} r\,rethod Bcars

com-prrrer.'' FI; 16. No. -3. March l9ti7.

- 2 "Pipinq serond edirion, 1963. Grinncll Co., Providence,

Index!.ng Terms: Coefficients-6, Computers-10, Design-8, Diameter-6.

Expan-sions-6, L^engrh-6, Loops-9, _{Pipine-9. program,-i0, Radiui-6, Sr.*.F.-i,} ,:l

"(

perience _in tlrc Dotu Engirtcering Dr:pttttltt.emt in ci.-il ancl mcchanicctl engineering dr:si11tt. _{Hc is} q, _{nttutttr:r of} ASCE NSPZ', and ACI.

Piping

Design

Method

Beats

Computer

Symmetrical,

U-type

_PiPing

looPs

can

be

analyzed

for

flexibilitY

using

this

new graPhical

method.

lt

can be done

faster

than

the

time

required

to

prePare

the data

for

computer

analysis

M.

S. Hoque, Engineering Consultant, EMMCON,

London and

J.

Storczewski, Woodall Duckham, Ltd',

Crawley, England

ple easy

to

follow method is used and no extr'aordinary mathematical knowledge is required.

In

structural design, the imposed force on the system

is specified and the

deflection

piping

design, the deflection is

given

by the

imposed deflection is

determi

rigor-ous methods available

which

accu-racy, but their involved computational intricacies demand

the attention of an engineering mathematician or an

ex-perienced pipe stress analYst.

A

less time consuming, easy to use method is the only answer. One that would even win over the total time in-volved to solve a problem on a comPuter, and at the sarire

time produce acceptable results and also satisfy the re-quirements

of

the Code

for

Pressure Piping ASA 831.1

or British Standard Specification 3351-1961.

The authors present a method which can be used for

to

the solution'of

'lJ'

type symmetrical expansion loops.

It

procluces results reasonably accurate

in

almost a fraction

of the time

it

takes _{to Prepare} the

data

for computer

analysis.

The total deflection in a piping system is usually known' For example,

if

the pipe length of a symmetrical loop is

50 ft., i.e.

'U'

length betrveen anchor points, the thermal

expansion

4

inch per 100

ft.

of

lincar length, then the

total deflection

:

50/100(4)

:

2 inches. The height and

width of the loop are generally detennined by the space

available. Oncc the shape

of

the loop is decided by the

layout engineer, forces, moments, and stresses can easily

be found using the graphs presented in this article.

It

is suggested that the use

of

precise and analytical methods and also comPuter analysis shor.rld be limited to

only critical and hazardous lines. Critical lines are those

involved

with

high

or

low operating temperatures and

pressures and/or the type of equiprnent to which they are Connected. Hazardous lines are those concerned l'r'ith the

nature

of

fluid being conveyed,

highlf

inflammable etc.

Therefore, designers should segregate all critical and

haz-ardous lines rvhich demand

the

attention

of a

piping

stress

consultant or

specialist. For simpler, noncritical lines an approxirnate calculation method is permitted by

the Code for Pressure Piping ASA B31'1. The

Starczew-ski/Haque stored energl'method is an approrimate

solu-tion and the analvsis produces safe results.

Allowqble

Stress.

It

is

recommended

that the

stress

obtained bv this method should be compared by the code allor,r'able stress range S.1; rvhere

S,r

:

_I ( 1.25 ,Sc ₊ 0.25 Sr, )

S"

:

allorvable stress (S value ₎ in cold condition S,,

:

allorvable stress (S value) in the hot condition

S" and S,, are to be taken from tables in the applicable

sec-tions of the code

f :

stress-range reduction factor for cyclic conditions. IJse

a value

of

1.0 for one cycle pcr day or less. Consult

2.0

1.8

r.6

1.4

t.2 r.0 0.8

0.6

0.4

0.2 SCALE P

Weight and other sustained external loadings shall not

exceed Sa.

Pipe supports qnd restrqints are not considered in the

flexibility calculation.

It

is

assumed that the supports

which have not been considered

in

the analysis should

be located and designed so as not

to

interfere with the

flexibility of the system. The reactions computed by this method shall not exceed the limits which the attached equipment can safely sustain. Equipment such as. pumps, hrrbines and similar strain sensitive machines should re-ceive the manufacturer's approval; and the piping system

should be designed flexible enough to comply with their

recommendations.

SAMPLE PROBTE'VIS

The following data apply to all sample problems,

un-less otherwise stated.

PipeSize:3in.

Schedule

:

40

Operating Temperature

:

8600 F

E

:

Young's Modulus (cold)

:

27.9 (10") psi.

Thermal Expansion

:

7.37 in. per 100 ft. 1

:

Moment of Inertia

:

3.02 ina.

Z

:

Section Modulus

:

1.724 trf .

i :

Stress intensification factor

:

1.78

Material

:

Carbon Steel

Code

:

Power Piping

Se

:

Allowable Stress

:

16,800 psi.

Solution: Use Figures L and 2.

Fig. l-Force graph (above) and moment graph (right) for U-typ" pipe expansion loops.

Somple Problem

l.

For

a

simple

'IJ'

type expansion

loop as shown

in

Fig. 3, check maximum stress

in

the

loop, and

if

the calculated stress is much less than the allowable, suggest a loop which will produce a maximum

stress equal to or near Sa. Space does not permit the mod-ification of 11, but G can be modified.

Given: 3 inch carbon steel line, Sch. 40, thermal

ex-pansion 2 in. per 100 ft., temp. 325oF.

Data:,

I :

3.02 h.n;

Z

:

1.724 in.3;

i :

1.78. Allow-able Stress

Sr

:

18,000 psi (Power Piping).

Solution:

w10

B-

:_:1

'H10

?o 30 40 50 t00 200 300 400

500

a

5

rp00 2p00 3p00 4,000 5p00 rop00

ll

2.0 t.8

t.6

1.2

r.0

0.8 0.6 0.4

0.2 SCALE B

G10

o:7.-ro-1

z. T 50.u00

_,)

d = too,ooo E ut = o =

6o

_-

Total thermal

expansion:+

(100)

:2

inches.

Note:

_B

and a and other nomenclature have no con-nections with similar symbols used in piping design books.

These are used

in

this article just as symbols.

Step 1. Determine Force Fe using Fig. 1 (Force graph). Entei scale

p

with

_F

: t

then move vertically upwards

to the curye

a :

1, and now move horizontally

to

the

right to the line 8o

:

2 and then vertically move down

tJthe

line

FI

:

10

ft.

and horizontally from this point to scale Fal

I

and read the value Faf

I :

38-The Force Fa

:

38(I)

:

38(3.02)

:

114.76. SaY Fa

:

115 lb.

Step 2. Determine maximum moment in the bend using

Fig. 1 (Moment graph).

Enter B scale with 13

: l.

Move to curve

d

:

1 and

then horizontally on right to

H :

70

ft.

and vertically down to line F1

:

115 lb. From this point go

horizon-tally

to

the moment scale

to

rcad

BM,

(Bending

Mo-ment)

:

6,300 lb.-in.

Step 3. Calculate the maximum stress

in

the bend. .S,

:

Expansion stress

: U{ ,,

:

63oo

,r.rr)

_-

6.52opsi.

r.72+',

,S,

:

8,000 psi. S,

(

S.

500,000

The system is acceptable. The calculated Stress is less

than the allowable stress.

Step 4. Determine the flexible loop, rvhich

will

pro-duce a stress of 8,000 psi. Use Fig. 2.

.s.

18.000 Stress ratio

1_l_- _-2.i'6

sE

b,5zu

Find, y

:

0.475 from Fig. 2 when a

:

1 and F

:

1'

New y

:0.475

(stress ratio)

:0.475

(2.76)

:1.312.

Enter Fig.

2 with

new

y :

1.312, move to ₁₃

: l,

then travel vertically dorrnrvards to read ne\v

d :

0.25.

G Since a

:

PIPING DESIGN METHOD BEATS COMPUTER

l'-n

t23456789

a=G/H

Fig. 2-Variation of bending mometrt and stress with beta and alpha.

p:Y:

t,b/

_-

lo, tr

_-

to.

.H

The suggested loop will have the following dimensions: H

_-

l0 f.t.,W

_-

10ft., G

_-

2.5 ft.,and

U-

100 ft.

This new loop will produce a stress

:

18,000 psi.

Check stress:

Calculated

BM:

6300 lb.-in. Stress ratio

_-

2.76

BM (Stress ratio)

_-

6,300 (2.76)

_-

17,400 lb.-in.

New stress

_-

BM

,,

17'4oo

Z ,):

,12a

(1'78):lB,000Psi.

Somple Problem

2.

Calculation of forces at the nozzles, see Fig. 4.

,:(+)

_{-4/to_}

0.4

/c

\

a:l_

I

:8,/10-0.8

\H/

t-al U,

:

100 (55) :4.05 in. (Thermal expansion of 'U') From Fig.

I

find

FalI

:

l2O.

"'

Fe

-

120(I)

:

120(3.02)

:

362lbs.

Computer result

:

252 lbs.

Kellogg graphical

:

368 lbs.

Fa

:

Force

of

pipe

on

anchor

or

nozzle, caused by

thermal expansion.

Somple Problem

3.

See Fig. 5.

:0.4

:0

7.37 uo

:

100

(55)

:

4.05 in.

From Fig. 1, by interpolation,

FalI

:270.

:.

F.q.

:

270

(I) :

270 (3.02)

:

815 tbs.

By computer Fe:6421bs.

Somple Problem

4.

See Fig .6.

-

0.4

:0

l-it

oo:

100 (4) :0.295 in. Find force from Fig.

l,

FalI

:

20.

:'

Fe

_-

20

(I) :20

(3.02)

:

60.40lbs.

By computet

Ft

:

46.0 lbs.

From the above three cases,

it

is obvious that the result

by this method compared with computer analysis indicate

safe and reasonably accurate values; and for these shapes,

this method wins over computer analysis including data

preparation time.

Somple Problem

5.

See Fig. 7.

G10

a:-

_{- --:1}

H

10 7 .31

3:

-r55,-4.05in.

u

_-

_loo

Find from Fig. 1,

F^lI :

78. F.E

:

78

(I) :

7A (3.02)

:

236 ]bs.

M 1

_-

12,000 lbs, in.

Stress in the bend.

So-

M;!- qil

_{-- ':::?}

z ''

7.12+

(t.78)

:

12,400 psi.

S,i

:

16,800

psi.

Sr

(

S,r

The

reactions and stresses are

within the

allou,able

Iimits. The svstem is therefore adequately flexible.

Economic Loop Design.

If

the anchor forces are not the

factors which dictate the design, then the loop can be

quickly determined by the use of this method which will produce maximum moments and stresses equal to its al-lowable limits. No

trial

and error method is needed. By

using Fig. 2 a great deal of labor can be saved. This also

eliminates a large amount of mathematical computations

and reduces the chances of errors to a negligible degree.

The graph is self-explanatory and results are sufficiently accurate

for

most engineering purposes.

The

following example illustrates how Fig.

2

can be used efficiently.

oW4

'

_H

10

GO

A:-:

-H10

T

H

I

oW4 p:-:

-'_H10

GO

H10

O: G/H B=w/tt t.0 0.9

E

0.8

P

0.7

s06

u)

0.5 at)

H

0.4 F <n

6

0.3 F z. trJ

=

0.?

=

>-w10

6:_:

:

I

H

r0 0.09 0.08 0.07 0.06 0.05

t3

If

the calculation, based on either the Starczewski/Haque

method

or

an

analytical

method, indicates

that

the

moments and stresses exceed the allowable limits, then

Fig.

2

can be used

to

predict the guide distance "G" which would enable the shape to become flexible enough and to yield moments and stresses equal to the allowable limits.

Sompte Probtem

6.

A

loop which has the ratio a

:

GIH

:3

and

p

: IrylH:

5, and the solution indicates

the Bending Moment

:

50,000 lbs.

in. and

Expansion

Stress

:

30,000 psi. This exceeds allowable limits.

Allowable BM :25,000Ibs. in.

Allowable Sr

:

15,000 psi'

Determine, U-tyP" symmetrical expansion loop to yield

the maximum moment and stress equal to the allowable

limits.

Solution, Step 1. Determine from Fig. 2 when a

:

3

and

p

:

5. FolLw the arrows and read value _{of 1}

:

O.l7

which is the moment and stress factor on the left hand

vertical scale.

Since

the

allowable stress Sr

:

15,000 psi. and the calculated expansion stress SB

:

30,000 psi' then the

ratio

:

15,000/30,000

:

0.5.

Therefore, corrected y

:

0.17 (0.S;

:

g.g8t.

It

is assumed that F

:

W

lu :

5 remains unchanged'

Step 2. Determine new d.

Enter in Fig. 2 the corrected y

:

0.085 on the vertical scale from the

left

hand side

of

the graph, and move

horizontally to the right to the curve F

:

5

then move

vertically down to the e scale and read the new ot

:

5.7.

Now summarize the new values as follows:

a:

5.7

B:5

Expansion stress Sa

:

_{15,000 Psi.}

Bending Moment BM

:50,000

(0.5)

:

25,000, equals

the allowable moment.

Step 3. Determine the distance of the guide 'G'.

New a

:5.7 :

GIH

when 11 remains as before.

Therefore, the distance

of

the guide G

:

5.7 (H)

.

II

the

'G'

is increased to 5.7

(H)

then the new shape will yield a stress

of

15,000 psi. which is equal

to

allowable

stress.

Bosis For Method. This method assumes that the energy

is stored

in

a system because of bending, which is caused

by

the deflection due

to

the thermal condition

of

the

piping material.

Energy stored

: tY!

(1)

J2

Again:

It

is

accepted

that

when

a

piping system-is subjected

to

deflection

it

stores energy,

and

that

the stored energy must be equal to the work done upon the piPe'

The work done by a force upon a piping system

: y?

(2)

t4

Fig. 3-Simple U-type expansion loop for Example

l.

Fig. 4-Calculation of forces on nozzle, Sample Problem 2'

G:0

oo'

U

Fig.'A-Figure for Sample Problem 3.

Fig. 6-Figure for Sample Problem 4.

.rc'J

l*-u=,0'

U=55i

Fig. 7-Figure for Sample Problem 5.

where, F

:

Force and

X

:

Distance travelled

by

the

force in the direction of the force 'F'.

Also: Interndl energl' stored

by

the system is caused

by'

e

Compressive stresses

o

Tensile stresses

PIPING DESIGN METHOD BEATS COMPUTER .

Fig. &-Basis for the method starts with this piping layout.

'i-1

Fig. 9-The loop deforms from the cold to the hot position.

l.--G6---*j

_{l.-Gf ---i}

Fig. 10-The forces and moments acting on the Fig.

I

loop

are shown.

If

the energy stored by compressive and tensile stresses

are neglected then the total energy stored

will

be due to bending moment only.

See Figs. B, 9 and 10.

Energy due to Bending Moment

:

BM

:

I+

(3)

M

But d0

:

_EI

d*

(4)

Therefore, the equation will be transformed as follows:

ur,:l*!ru:l+W)

fMzdx

:Jzn

(s)

Deflection taken by the shape between

(a)

and

(Gr)

i.

equal to 8o/2

Fa6o/2

_f

U, a*

_I

G"

2 :l-Znr)n

(5A)

Thermal stresses

in

symmetrical expansion loops are

derived as follows: E,

f-i

ts-Lq lzl

---t

R1

Deflections 81, 82 and 8s are produced

by

elements

Lab, Rae and. cd respectively.

The system is confined between

2

anchor points.

Le:

Teft hand side anchor

Ra

:

right hand side anchor.

The following conditions apply

in

derivation:

o

That the whole system lies in one plane.

That the system is treated with square corners

inter-sections.

That

the system is composed

_of

straight elements of pipe of uniform size and thickness.

o

That the thermal expansion of a given element is

ab-sorbed by the elements orientated perpendicular to the direction of the deflection.

o

That the effect of dead weight, wind etc. are neglected.

o

That the clearance between guides and pipe is nil.

o

That the compressive stresses within the element are

neglected.

o

That

the flLxibility

of

the elbow caused by an oval

shape is neglected.

The three deflections

in

the horizontal plane are:

81

+

8,

*

8,

: 8o:

total deflection

The total deflection 66 must be absorbed by the system

contained between the points a and _fo which is tJle flexible portion of the system, see Fig. 9.

The loop

will

deform from a cold condition to a hot condition, when the system is subjected to the operating temperature, as shown

in

Fig. 9.

Fr

:

thrust acting on the loop

at

the point

,a,,

see

Fig. 10.

Mo:

bend'ng moment acting on the loop at the point 'a', see Figure 10.

From Fig. 9

it

is evident that the slope at the point Go

must be horizontal.

IIence the change

in

the slope between point

a

and

point Gs is equal

to

zero.

i.e

_ll

o'1"":o

F _o: Fo and _Go- G1 Fr

+

(6)

Mo:

M6

Bending moment, BM at point X

_-

X

_-

Me-

_{FEU) e)}

Slope and Bending Moment. The general relationship between slope and BM, is as follows:

dt: M,!:

(8)

EI

Therefore the total slope change between point a and

point Go is:

ll+l'"

X

:

Pipe Iength in general,

(s) Go I q. I -l I [,--1

f--0.---!

.:llr#):".

_llr-#,

*,),

*

ll*+*

*)*,"

But the change

in

slope between

a

and.

Go:

Q

*{*"*o)

+M,(H)

-L#L

*

Y* -

elnw)l:o

f

w1 fu,

r+twtl

*"1.

+H+

T):r"l;*-)

f H,+H(wt I

T

H2+H(w\

I

M,:Fo

_Lzrdffit

):'"tdc

+TiTT)

M-- F,'"

_"2(G+H)+w

H

*W

Now refer Equation 1 to 5a,

Further, (10) (II)

oruo:

4 EI 3o 2Fo EI 3o 2 FoH2 Er 60

w

:

r,

{,,

+ w),

\,

I'

I- W

'2

H 3

HW

-3-2

4(G+H)+2tr

I

(13) t

₊₂

p ₊ p,

+al4l2P

H2 1 -)-Hl-L2 _ tl/ -L

c

_{- rl}

H

(#)'

:"[-f,

*

Force, Moment, and Stress. The follou'ing units apply

in

the formula below:

Modulus of elasticity

:

E

(psi.) Resisting force

:

F,

:

F"

(lbs.) Moment of inertia of the pipe

:

1 (in.n)

Total thermal expansion between anchor points is in

inches.

Height of the loop

: Il

(ft.) Width of the loop

:

W

(ft.)

Distance of first grride

:

G (ft.)

Section modulus of pipe

: Z

(in.')

Stress intensification factor for the bend

: i :

0.9

_lh

where

h

:

tRlr'

See code for pressure piping ASA 831.1.

Then the force is

,

₊₂

+

'

(#)'

-(+)

W LLL,) 't-H

I

r+28+p'

--

3 , 2

I

4aa4l29

where _{F -} I,Y/H and, a : G/H.

I

,,-,

2B

+

+

+2p

4

1 4 B 2

'oouo

:

;*{ll

-;

o""f:'

+ll

w.-

rot)

*7'"*ll

,*"-'outo'f*""|

#

{rr

",

(r-ffi *r)'

(""

*

"

*[)

-

FoH

r#T,

(FoH, + FaHW)*

r,,

(F

*

L)\

I c+H+w/z

l

l, H,

zz\

ttr@+h+wF -

T(G+h +w-l

-

3

-

Tl

:

,o

*

rr'{

_212(G+H)+w)

_2(G+H)+W

(H

₊

14/)2 t1/ H +

+)

.

r*'(+

w H

t6

(12)

PIPING DESIGN METHOD BEATS COMPUTER The maimum stress can therefore be in a system either

at bend b or at the bend c.

It

is suggested that the moment at point b and point c be calculated, then take the largest moment of the two to calculate the expansion stress SE in the loop.

M

s,,

_-

_

(i)

Fig. l2-Loop designations for round corners.

Fig.

ll-For

round corners, tte

width, height, and guide distances must be modified.

Fig. l3-Symmetrical loop

with guide G distance from

bend.

Fig. l4-Symmetrical loop

with guide very near to the bend.

( 1e)

Loop Restroinls qnd Supporls. Design engineers should

make certain that the loop between the

two

guides is

made to function without any environmental obstructions

or restraints. Also that the system is fully supPorted and

no branch connections are made within the flexible

por-tion

of

the loop.

It

is not recommended

to

induce any

external loading upon the loop.

The

designers should

avoid locating

rigid

sections, such as large valves etc.,

within the loop.

A

good practice is to locate valves and other rigid sections near the guide or between the anchor

and the guide.

Line Size Limits. This method provides engineers with

reactions and stresses

that

are reasonably accurate for pipe

up to 6

inches

in

diameter. Lines above

6

inches

in diameter can be safely analyzed by this method. How-ever, the authqrs would like to point out that the results

thus obtained will be on the conservative side. Therefore, where the system is dictated by the space, reactions and

stresses or economic limitations, a precise analysis should be made.

Round Corners. Solutions to loops having round corners

can be solved as follows:

o

_If

the

square corner solution gives reactions and

stresses, which exceed acceptable limits,

o

_If

the radius of the bend is more than 1.5 pipe

diam-eters,

o

If

the line is above six inches

in

diameter,

.

Refer Figr,rres 11 and 12. Use Equations 20, 21,22, and 23, to modify the width (W)

_,

height

(I1),

and guide distance

(G)

respectively.

Use Figs.

I

and 2 to determine forces and moments.

Assumption:

R

:

radius of all the bends which

I{:

height of the loop which must

sides.

G

:

distance

of

the guides which both sides.

must be the same.

be equal on both

must be equal on

Fig.

l5-A

two-plane loop configuration.

Er

s^

l-n'-

" t_

^ 3456H3 I 1 I

Is

l

is

p

L+2tt+t2

, Z 4a*4*2F

the following forroula

(lbs. ₎ (15) derived. ( 16) (17) ( 1B) From equation (10) Mo: FoH

H+W

2(G+H) +W

W:u+1.57R(K/3)

H-h+t.s7R(K/3)

H-h+1.57R(K/6)

C:e*1.57R(K/6)

K:1.65/h -flexibility factor, K) 1, rf R)

rvhere h _- ' _{..' ; i:} thickness of pipe; r: mean

r2

pipe: R: radius of the bend,

1-rB ' Fo

It

T; )-,

+-O (lbs.-ft. t Note that Fo

:

F^ (20) (21) (22) I 2q\ (24) radius ol M": Mo- Fr(H) (lbs.-ft.)

Note: For force calculation use modified Height (H) of

loop. For moment calculation use original Height (II) of

ioop'. the the

ln

general: Stress

:

M _/Z

t7

For Figs. 13, t4, and 15,

u:

W

_-

2R;

h:

H

_-

2R;

g:G_R.

Symmetrical Loop. Refer to Figs. 12 and 13. Modify W,

H,

and G as follows:

For W, use Equation 20.

Eor

H,

use Equation 21.

For G, use Equation 23.

Symmetrical Loop. Refer

to

Fig.

14. Note

that

the

guide is very near the bend.

For W, use Equation 20.

For H, use Equation 21.

Since the guide is at or near the bend,

G:0.

Two-Plane Loop. Refer to Fig. 15.

For W, use Equation 20.

For H, use Equation 21.

For G, use Equation 23.

U-Loop With Equal Legs. Refer to Fig' 16.

u:W-2R.

h:H-R.

For W, use Equation 20.

For 11, use Equation 22.

G

:0.

Two-Pliane Loo,p

With

Y-Leg Longer Than Fitting to Fitting.

w:W-2R

hr:

Ht

_-

2R

hr:

H,

_-

2R

g:G_R

For W, use Equation 20.

H:.:

Ht

*

Hz. For G, use Equation 23.

Two-Plane Loop With Equal Legs.

u:

IU

_-

2R.

hr:

H,

_-

R.

Hr:

H,

_-

2R.

For W, use Equation 20.

For

IIr,

use Equation 22.

For H2, use Equation 21.

H:

Hr]-

Hr.

G

:0.

U-Loop

with

Equal Legs and Single Tangent. Refer

to Fig. 19 and solve the same as Fig. 16.

Grophicol vs. Computer Solufions. The following

ex-amples are given to illustrate the results obtained by this method and by the computer.

Example. Given:

4

in.

pipe, schedule 40; radius of

bend

:

0.5

ft.;

operating temperature

:

3500 F.;

mate-rial

:

ASTM

106 GR. B;

thermal

expansion

:

2.26

in./100 ft.; allowable stress

:

22,500 psi.; code

:

Power

Piping; moment

of

inertia

: I :

7.23 (in.a); section

modulus

:

Z'

:

3.22

(n.3);

stress intensification

tac-tor:i:1.95.

t8

Fig. 17-A twb-plane loop configuration with the Y leg longer than fitting to fitting.

Fig.

lLA

U.loop configuration

with equal legs.

Fig. 18-A two-plane loop with equal legs.

Fig.

19-A

UJoop corfiguration with equal legs and single tangent.

a:GfH :

12110

:

1.2;B

:

WIH

:

B/10

:

0.8; 8o

:

(2.261100) (180)

:

4.06 in.

From Fig. 1, Force Graph,

FII : r :

86

Force

: r(I)

:86(7.23)

:

620

From Fig. 1, Moment Graph,

BM

:

27,100 lb.lin.

stress

: +#

(1.95)

:

16,450 psi.

Fwi

PIPING DESIGN METHOD BEATS COMPUTER

.

. ized staff members to spend their valuable time only on

the analysis

of

critical and hazardous lines. Alsq

it

will help site engineers design and incorporate a loop in a rack

piping system, or in a long transmission line, without

seek-ing help from the design office.

Furthe4 the method

will

help project engineers

esti-mate piping flexibility in the proposal stage of the project. Symbols used:

Allowable Stress (psi.) Modulus of Elasticity (psi.)

Moment of Inertia (in.u) Section Modulus (in.3)

Force (lbs.)

Force component in the direction of axis.

Moment (lb./ft.)

Stress intensification factor. ,s/ E

I

Z F Fx

M

i

Fig. 20-An example configuration calculated by the graphical and computer methods.

Fig.

21-A

square corner configuration calculated by graphical and computer methods.

Method

Force, lbs.

Moment Max. stress

at guide at bi:nd, Remarks

Ib./in. psi.

About the quthors

M. S. HIQUB is an engineering

consul-tant as s o ciated.with E M M C O N, L ondon.

He specializes

in

pipe stress analysi,s,

piping lo,gout, and _fleribilitg analgsi,s i,n th,e lal1out stage. Mr. Haque receiued

a diploma

in

mechanical and electrical

engineet"ing _{from Dehri} Technical

In-stitute

of

India.

He

is

an

o,ssociate

nzember of ASME, Institute of Engi-neering Designers, Institute

of

Plant Engineers, Associate Fellou; of the In-stitute of Petroleum, and Associate of the Institute of Fuel.

He has had 16 gears enperience as a senior designer and

piping analyst usith such, _firms as Wellman Smith Ou;en

Eng. Co., Mattheu Hall & Co., Ltd., McKee Head

Wright-son, Ltd., and Constructors Joh.n Brou.tn, Ltd., all in London.

He also has fi,ae years _{fi,eld eupet'ience} on construction jobs

in Ind:ia.

J. St-q.nczowsKr 'is an engineet'wi,th Woodall-Duckham, Ltd,, Crawley, Eng-land. He o,ttended the Polislt Techruical

College

in

England and completed a B.Sc, mecltanical course at London

Urvi-oersity. He has done graduate work in

fluid dgnamics, mathematics, and

nuclear energA.

Mr.

Starczeu.tski has had, enperience

in

heat enchanger

de-sign, pressure aessel design, u-telding equipment dea elopment, special pu,t'pose

machine design and, other process equi,pment design. He h,as

uorked, u.titlt, such _firms as Constructors John Brown, Ltd.,

and, Caird & Raynet", Ltd,. both in London, Graphical 620 27,100 16,450 square corner

solution Computer 2 7,000

t6,47s

_{.t;?.?;,:to"*}

Example. Given: 6 in. pipe, schedule 40.

All other data as in example above.

8o

:

4.06 in.,

I :

28.1,

Z

:

8.50,

i

:

2.27

From Fig. 1, Force Graph,

r

:

86,

r(I) :

86(28.1)

:

2,410 lbs.

From Fig. 1, Moment Graph

BM

:

90,200 lb. in.

Stress

:

90,200 _18.5(2.27)

:

24,550

Force, Moment Stress,

lbs. lb./in. psi.

Graphical 2,4tO 90,200 21,550 sq. corner soln.

Computer 2,179 90,500 20,122 sq. corner soln

In

conclusion, the authors feel that the introduction of this time saving piping flexibility analysis method will help

piping design engineers solve most of the simpler config-urations. This

will

allow consultants and highly

special-6t7

Symmetrical Piping

Arrangement

_Solves

Two-Phase

Flow

Distribution

Problems

The

secret

to two-phase

distribution

in branched

piping

systems is

strict

adherence to symmetrical

piping

and

an

evenly dispersed

liquid

flow pattern

John L. Greene

The Fluor Corp., Ltd., Houston

A

rnequrNT ENcrNEEnrNc

problem

is

designing

branched piping systems

for

flow distribution, mist or

dispersed flow, and an over-all low pressure drop.

Con-trolling flow patterns, liquid distribution, flow

distribu-tion, and optimizing pressure drop need to be considered.

Recognizing

flow

polterns

in

two-phase flow

is

the

first part of the problem.1,3

In

a two-phase system, when gas flows

at

various rates, demonstrative types

of

flow patterns are developed.

In

general, these flows are

de-scribed _{as bubble, plug, stratified, wavy, slug,} annular, and spray, mist, or dispersed. Slug formation, plug flow, wavy flow, and stratified flow are shown in Figures

l,

2,

3. and

4,

respectively. For this discussion, slug flow is

defined as a mixture of liquid and gas that has a varying density with respect _{to time. Therefore,} the term slug flow

will

also include plug flow and

will

border on annular and bubble flow.

fn

engineering design the flow pattern must be deter-mined in every two-phase application.

fn

a service where pressure fluctuations cannot be tolerated, there can be no slug formation. For example, slug flow downstream of a

distillation column

will

cause pressure fluctuations and unstable operation,

or

downstream

of

catalytic reactors

it

can cause catalyst attrition.

Liquid Distribution. When two phases flow through the

same pipe, the gas flows faster than the liquid.

In

a

20

Fig. l-Shows slug flow; G:0.0085 lblsec;

L:

0.38 lb,/sec.*

Fig. 2-Shows plug flow; G:0.00421 lb,/sec; L:0.38 lblsec.*

Fig. 4-Shows stratffied flow; G

:

0.0081 lblsec; L

:0.38

lb/

sec.'

* _{Water and air at atmospheric conditions and in a 2-inch} OD horizontal pipe.'

ll

,'=T,'

(I) ELBOW PERPENDICULAR (2) ELBOW PARALLEL (3) TEE 8 CAP

TO HEADER TO HEADER TO HEADER

(GOOD) (POOR) (ACoEPTABLE)

Fig. S-Shows liquid distribution into a header:

(1)

elbow

perpendicular to header (good); (2) elbow parallel to header

(poor); (3) tee and cap to header (acceptable).

The problem of two-phase flow distribution

in

manifold

piping arrangements is frequently encountered

in

large plants, particularly around air coolers, parallel exchang-ers, etc.

The simplest solution to flow distribution is to provide a block valve

in

each branch line. From the standpoints

of valve and of pressure drop costs this is often

unattrac-tive. Therefore, the pressure drops through the system

must be depended upon to distribute the flow. As is shown

in Figure 6, if valves are not provided in each branch line of two-phase flow, then the layout should be symmetrical.

For comparison the preferred layouts

for

single-phase

flow are shown

in

Figure 7. The selection depends upon

the importance and duty of each service. Friction loss in the fittings was determined by using Bernoulli's Theorem

(Velocity head method) and velocity head coefficients

from the literature.5,6 To determine the pressure drop in two-phase flow when there

is

less gas

by

weight than Iiquid, two-phase flow correlations should be used.2

Applicotions. Savings can be realized in optimum

over-design of heaters, heat exchangers, etc.

In

large systems

horsepower usage from pumps and compressors can be

reduced.

With this knowledge of how to minimize pressure losses

in manifolds, plot plans can be laid out more efficiently.

The

preferred piping layout creates fewer

plot

plan

changes and shorter pipe runs.

Slug flow causes pressure fluctuations

in

the system.

Elimination of slug flow helps stabilize the unit. Slug flow can also cause major problems such as catalyst attrition.

Therefore, catalyst life can be increased. Minimizing

pres-sure drops and equalizing

liquid

and flow distribution

will increase yields and decrease capital and operating cost.

Exomple Problem. Distribution into and out

of

a

16-section air cooler with 5.3 pounds per square inch pres-sure drop (16-6' nozzles) and

with

the following flow:

LIQUID

Flow lb/hr.

Specific Gravity _@ Tem-perature & Pressure

Temperature oF

Viscositl, Cp.

VAPOR

Fis.

pass

&-Shows symmetrical piping and (2) two pass.

(2) _{TWO PASS}

in two-phase flow: (1) three

(I) GOOD DISTRIBUTION (2) FAIR DISTRIEUTION (3) POOR DISTRIBUTION

FiS.

Iayouts for single-pass flow in

two

d distribution, (2) fair

distri-buti

smooth turn the iiquid has a tendency to follow the out-side wall. The elbow

or

turn should be perpendicular to the manifold, as is shown in Figure 5.

If

thii

is not

possi-ble, a tee and cap or a mixing length after the turn may be used.

The liquid must be distributed into heat exchangers,

air

coolers and other types

of

equiprnent. Often

it

is necessary to rotate an elbow to the shell of an exchanger

in order to distribute the liquid on the baffling ,.r.a.rg"-ment. Tees and caps or mixing'lengths are also used on

the inlets to heat exchangers. On the outlet, liquid

dis-tribution is not usually important.

The severity

of

the operation and the

duty

(size of heat release)

of

the service as selected to provide liquid

distribution

are

determined

by an

economic balance.

Therefore, each case must be looked at individually.

Flow

distribution

in

monifold piping

systems

is

a

function

of

pressure losses through each lateral system.

IN

484,000 0.670 277 0.310

IN

OUT 652,000 0.678 150 0.412 OUT Flow lb/hr. Molecular weight Viscosity Cp. 533,000 8.49 0.01132 365.000 5.99 0.0099

Three cases r,vill be considered

to

determine

the

best

piping layout. Economics prevents putting valves in each

of the 16 sections.

Cqse.

l.

One header

with

16 ]aterals on the inlet and outlet is shorvn in Figure B.

(I) THBEE PASS

ST RATI FI ED

Flow Pattern. From Baker's1,2 two-phase flow

corre-lations the type of flow is determined aIter each lateral take-ofl and is shown

in

Figure 8. The flow patterns go

from mist to annular to slug to stratified flow. This flow pattern change is not acceptable.

Pressure Drop. The pressure drop calculations are a

trial and error procedure to determine the exact

distribu-tion. Using Bernoulli's Theorem and velocity head coeffi-cients from the literature5,6,7 the initial pressure and flow distributions (assuming equal distribution) are as follows for the first and sixteenth pass:

o

First Branch System: AP1

:

10.46 psi

including

cooler loss

o

Sixteenth Branch System. APro

:

8.24 psi including cooler loss

.'. Percent flow not distributed

:

4.89 percent

On a service with a large duty this 4.89 percent of mal-distribution

of

flow

is not

acceptable. Case

I

is

not

a

good system.

Cose

ll.

One tapered header with 16 laterals on the inlet

and outlet is shown in Figure 9.

Flow Pattern. From Baker's1'2 two-phase flow correla-tions the type of flow is determined to be in mist flow the

total length of the header on both the inlet and the outlet. Pressure Drop. The pressure losses were calculated the same way as Case

I

only the expansion and contraction

losses were considered.

'

First

Branch System:

aP,:1i.86

psi

including

cooler loss

o

Sixteenth Branch System: APre

:

9.55 psi including cooler loss

.'. Percent flow not distributed

:

4.64 percent

On a service with a large duty this 4.64 percent of mal-distribution of flow is not acceptable. There is 8.73 per-cent rnore pressure drop than

in

Case

I

and the tapered header is expensive. Case

II

is not a good system.

16-6.. LATERALS SPACED AT

8

SECTIONS

Fig. 8-Shows single 24-irch headers with 6-inch laterals (example problem-Case

I).

Fig. 9-Shows tapered (24

x

8-inch) headers with 6-inch

laterals (example problem-Case

II).

Fig.

l0-Shows semi-symmetrical manifold piping layout (example problem-Case

_III).

Cose

lil.

A

semi-symmetrical manifold piping system is

shown in Figure 10.

Flow Patterns:

AII

of

the piping

is

designed so that only mist flow is encountered. All turns into headers have

to be rotated corectly so that the liquid is evenly

dis-tributed.

Pressure Drop. The pressure losses were calculated as

in Case I.

o First

Branch System: AP,

:

9.34

psi

including cooler loss

o

Fourth Branch System:

APr:9.97 psi

including

cooler loss

Percent flow not distributed

:

1.96 percent

In Case

III

the florv is in the dispersed legion throughout the system. The flou, distribution is the best that can be

economically justified. This is good piping la.vout.

LITERATURE CITED

Indqirrg Tere: Ctrmputations-l0, Design-4,8, Distribution-7, Fluid Flow-4,7,

Layout-4,6. Liquid Phase-5, Piping-9, Vapor Phase-5.

NOTES

Plont

Loyout

qnd

Piping

Design

for

i,tinimum

Cost

Systems

Afier

process

ond

equipment ronditions

ore

sel, plont loyout

con

be

the

lorgest

single

cost

sover

in

HPI

plonts.

Line sizes

ond

pressure drops

depend

on

pipe

lengrh

ond configurotion.

Use

these guides

io

moximum

plping

system

economy

Robert Kern, The M. W. Kellogg Co., New York

PrprNc EcoNoMy _{is closely}

_related

_{to three areas of} plant design:

o

Equipment layout

o

Piping design

a. Line sizing and flow slntems

b. Piping layout, and

o

Piping details

These areas are interdependent; without an economical

Plonl

Loyout

qnd

Piping Economy. Plant layout can

be the biggest single cost saver in refinery and

petrochem-ical plant

design, after process and equipment design

posibilities have been exhausted. Savings can be rcalized

not only

in

piping but also

in

the cost of pumping com-pression and

utility

cost. Often

a

layout can eliminate

equipment

(for

example, pumps

with well

arranged

standbys).

The most important document issued

to

the layout

engineer

is

the process flow diagram (PFD). This has

to

be evaluated

for

an economical plant arrangement.

From

a

layout standpoint, three types

of

lines can be

distinguished.

Main Process Flow Lines. First, lines which represent

the main process flow. Such streams pass

through

furnaces,

reactors

and dryers, then they continue as

tower bottom and feed

inlet

to

the next tower, often

with exchangers and pumps between them. These lines

will

be

the

shortest

if

towers are arranged

in

process

flow sequence as close to each other as equipment sizes

and access space permits.

With

smaller interconnecting lines, towers can be located further apart without much increase

in

piping cost

il

other economies can thus be

realized.

For

example: the grouping

of

condensers

be-tween two towers can result

in

a

shortening

of

cooling water lines;

a

conrmon steam line can be designed for grouped reboilers. Grouped condensers and reflux drums

will

permit

a

common supporting structure. Figure 1

shows

an

example

of

alternative tower arrangements.

Many configurations are possible and justified

_if

shorten-ing

of

these process lines is the ultimate result.

Process flow is not always

a

simple straight through flow but can split into two or three streams, as is

oftL

done

with a

number

of

distillation columns. Subsidiary

circuits to process flow must also be considered such as

the refrigeration circuits

in

ammonia

or

ethylene units.

are generally large diameter lines and should have pref-erence over the

first

group which are

wually

smaller

process lines.