Gas Hydraulics(Chapter 11)

8/13/2019 Gas Hydraulics(Chapter 11)

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Chapter 11 - Gas - Hydraulics• Gas pipe line hydraulics calculations• Equivalent lengths for multiple lines based on Panhandle A• Determine pressure loss for a low-pressure gas system• Nomograph for determining pipe-equivalent factors• How much gas is contained in a given line section• How to estimate equivalent length factors for gas lines• Estimating comparative capacities of gas pipe lines• Determination of lea!age from a gas line using pressure drop method• A quic! way to determine the si"e of gas gathering lines• Energy conversion data for estimating• How to estimate time required to get a shut-in test on gas transmission lines and

appro#imate a ma#imum acceptable pressure loss for new lines • How to determine the relationship of capacity increase to investment increase• Estimate pipe si"e requirements for increasing throughput volumes of natural gas• $alculate line loss using cross-sectional areas table when testing mains with air or

gas • %low of fuel gases in pipe lines• $alculate the velocity of gas in a pipe line• Determine throat pressure in a blow-down system• Estimate the amount of gas blown off through a line puncture• A practical way to calculate gas flow for pipe lines• How to calculate the weight of gas in a pipe line• Estimate average pressure in a gas pipe line using up and downstream pressures• $hart for determining viscosity of natural gas• %low of gas• &ultiphase flow• Nomograph for calculating 'eynolds number for compressible flow friction factor

for clean steel and wrought iron pipe

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Gas Pipe Line Hydraulics Calculations

Equations commonly used for calculating hydraulic data for gas pipe lines

Panhandle A.

Panhandle B.

Weymouth.

omenclature for Panhandle equations

( b ) flow rate* +$%DP b ) base pressure* psia



, b ) base temperature* ' , avg ) average gas temperature* ' P . ) inlet pressure* psiaP/ ) outlet pressure* psiaG ) gas specific gravity 0air ) .123

4 ) line length* miles5 ) average gas compressibilityD ) pipe inside diameter* in1h/ ) elevation at terminus of line* fth. ) elevation at origin of line* ftPavg ) average line pressure* psiaE ) efficiency factor

E ) . for new pipe with no bends* fittings* or pipe diameter changesE ) 2167 for very good operating conditions* typically through first ./-.8 monthsE ) 216/ for average operating conditions

E ) 2187 for unfavorable operating conditionsomenclature for Weymouth Equation

( b ) flow rate* +$%D, b ) base temperature* 'P b ) base pressure* psiaG ) gas specific gravity 0air ) .1234 ) line length* miles, ) gas temperature* ' 5 ) gas compressibility factor

D ) pipe inside diameter* in1E ) efficiency factor 0+ee Panhandle nomenclature for suggested efficiency factors3

!ample Calculations

( ) G ) 219, ) .22 %4 ) /2 milesP . ) /*222 psiaP/ ) .*722 psiaElev diff1 ) .22 ftD ) :12/9-in1, b ) 92 %P b ) .:1; psiaE ) .12

Pavg ) / < =0/*222 > .*722 - 0/*222 # .*722 < /*222 > .*72233 ) .*;9/ psia5 at .*;9/ psia and .22 % ) 218=71



Panhandle A.

Panhandle B.

Weymouth.

( ) 21:== # 07/2 < .:1;3 # ?0/2223 / - 0.*7223/ < 0219 # /2 # 792 # 218=73@.</ # 0:12/93 /199; )..*.2. &$%D

!ource

Pipecalc /12* Gulf Publishing $ompany* Houston* ,e#as1 Note Pipecalc /12 willcalculate the compressibility factor* minimum pipe BD* upstream pressure* downstream

pressure* and flow rate for Panhandle A* Panhandle C* eymouth* AGA* and $olebroo!-hite equations1 ,he flow rates calculated in the above sample calculations will differslightly from those calculated with Pipecalc /121 Pipecalc uses the Dranchu! et1 al1method for calculating gas compressibility1



Equivalent Lengths for Multiple Lines Based on Panhandle A

Condition "

A single pipe line which consists of two or more different diameters lines1

4et4 E ) equivalent length4 . * 4/ * 111 4n ) length of each diameter D . * D/ * 111 Dn ) internal diameter of each separate line corresponding to 4 . * 4/ * 111 4nE ) equivalent internal diameter

E#ample. A single pipe line* .22 miles in length consists of .2 mile .2-=<: in1 DF :2miles ./-=<: in1 D and 72 miles of //-in1 D lines1

%ind equivalent length 04 E3 in terms of //-in1 D pipe

4 E ) 72 > :20/.17 < ./1/73 :187=6 > .20/.17 < .21/73 :187=6

Condition ""

A multiple pipe line system consisting of two or more parallel lines of different diametersand different length1

4et4 E ) equivalent length4 . * 4/ * 4=* 111 4n ) length of various looped sectionsD . * D/ * D=* 111 Dn ) internal diameter of the individual line corresponding to length 4 . * 4/ *4 =* 111 4n1

4et4 E ) equivalent length4 . * 4/ * 4=* 111 4n ) length of various looped sectionsD . * D/ * D=* 111 Dn ) internal diameter of the individual line corresponding to length 4 . * 4/ *4 =* 111 4n1



when4 . ) length of unlooped section4 / ) length of single looped section4 = ) length of double looped sectionDE ) d . ) d /

then

4 E ) 4 . > 21/;99:04 / 3 > 4=0d./19.8/ < 0/d.

/19.8/ > d=/19.8/ 33.187=6

when d E ) d . ) d / ) d =

then 4 E ) 4 . > 21/;9::04 / 3 > 21.=2704=3

E#ample.

A multiple system consisting of a .7 miles section of =-8 7<8-in1 D lines and .-.2 =<:-

in1 D line* and a =2 mile section of /-8 7<8-in1 lines and .-.2 =<:-in1 D line1

%ind the equivalent length in terms of single ./-in1 BD line1

) 716 > .81. ) /:12 miles equivalent of ./-in1 BD pipe1

E#ample.

A multiple system consisting of a single ./-in1 BD line 7 miles in length and a =2 milesection of =-./ in1 BD lines1



%ind equivalent length in terms of a single ./-in1 BD line1

4 E ) 7 > 21.=27 # =2 ) 816/ miles equivalent of single ./-in1 BD line1

Determine Pressure Loss for a Lo !pressure Gas "ystem

se the +pit"glass equation for systems operating at less than . psig1

where( h ) rate of flow* in cubic feet per hour at standard conditions 0.:1; psia and 92o%3* scfhhw ) static pressure head* in inches of water +g ) specific gravity of gas relative to air ) the ratio of the molecular weight of the gas tothat of air d ) internal diameter of the pipe* inches4 ) length of the pipe* feet

E#ample$

Given the following conditions* find the flow in the system

hw ) 72 inches of water +g ) 21974 ) 722 feetd ) 91.8; inches

( h ) 66*:27 scfh



#omograph for Determining Pipe!equivalent $actors



%or turbulent flow* this handy nomograph will save a great deal of time in flowcalculations where different si"es of pipe are connected1 ,he advantage over a table ofvalues is in having all si"es of pipe e#pressed in relation to any other si"e1

E#ample 1.



.9-in1 # 7<.9-in1 in terms of /:-in1 # 7<.9-in1

0/=1=;7 < .71=;73:1;=7 ) .17/2= :1;=7 ) ;1/9

E#ample %. 'eciprocal values can be found when chart limits are e#ceeded1 /8-in1 BD in

terms of /:-in1 BD falls beyond the lower limits of the chartF however* /:-in1 BD in termsof /8-in1 BD gives a conversion factor of /1.1

Ho Much Gas is Contained in a Given Line "ection%

&ultiply the square of the inside diameter* in in1* by the gauge pressure* in lb<in1 / Fmultiply this by 21=;/F the answer is the appro#imate number of cubic ft of gas 0standardconditions3 in .*222 ft of line1

E#ample. How much gas in .*222 ft of .9-in1 schedule =2 pipe at =72 lb pressure

.71/7 # .71/7 # =72 # 21=;/ ) =2*/82 cubic ft of gas

E#ample. Appro#imately how much gas in eight miles of 1/72 wall /:-in1 pipe if the pressure is :22 psi

$ubic ft ) 0/=1730/=1730:223083071/83021=;/3 ) =*:;2*222 cubic ft

hen a section of line is blown down from one pressure to another* the total gas lost may be computed by the difference in the contents at the two pressures* using the above rule1

Ho to Estimate Equivalent Length $actors for Gas Lines



,his table shows the equivalent length factors for pipe lines of different diameters1 Bt is based on the Panhandle %ormula and the factors were obtained by varying the lengths of pipe and !eeping the other variables constant1

E#ample. %ind the equivalent length of a /:-in1 pipe line as compared with a ./ =<:-in1line1

Enter the chart at the top - /:-in1- 0/=122 BD3 and proceed downward to the ./1/72 parameter1 ,he equivalent length is /.1/1

,his means that under the same conditions of temperatures* pressures* specific gravities*etc1* one mile of ./ =<:-in1 pipe will flow the same amount of gas that /.1/ miles of /:-in1 will flow1 Another way of putting it* the pressure drop in one mile of /:-in1 pipe will

be the same as that in 12:8 miles of ./ =<:-in1 if the pressures* volumes* etc1* are thesame1

here a line is composed of several si"es of pipe diameters or where the lines are looped*this chart should prove useful1 Bn solving problems of this sort* the line is reduced to anequivalent length of some arbitrarily selected si"e pipe and created as a single line ofuniform si"e1

Estimating Comparative Capacities of Gas Pipe Lines



,o estimate capacity of one diameter line in terms of another read downward in theappropriate column to desired diameter line1

E#ample. A /:-in1 0/=1=;7-in1 BD3 pipe line has a capacity .92 that of a /2-in1 0.61722-in BD3 line1

,his chart is based on diameter factors using the Panhandle equation1 Equivalent inlet andoutlet pressures* lengths of line section* and uniform flow are assumed for both lines1

0%igures give capacity of one diameter pipe as percent of another13

ominal&iameter '() '*) '+) %() %*) %*) %+) 1,) 1() 1%-' *)

"nternal&iameter ' .% +) ''.% +) %/.'1% ) % .'0 ) %'.0 +) %'.+++) 1/. ++) 10. ++) 1 . ++) 1%.% +

=71/72I =9I .22 871; 9/1; :=19 =71/ ==1; //12 .917 ./12 917==1/72I =:I ..918 .22 ;=1. 7218 :.12 =61= /71; .61/ .:12 ;19

/61=./7I =2I .761; .=; .22 9619 791. 7=18 =71; /91= .61/ .21:

/71=;7I /9I //6 .6; .:6 .22 821; ;;17 721/ =;18 /;19 .:18

/=1=;7I /:I /87 /:: .;8 ./: .22 6912 9/1= :916 =:1/ .817

/=1222I /:I /68 /77 .89 ./6 .2: .22 971. :612 =71; .61/

.61722I /2I :78 =6/ /8; .66 .92 .7: .22 ;71= =712 /619

.;1722I .8I 928 7.6 =;6 /9: /.= /2: .=/ .22 ;=1/ =61=

.71722I .9I 8=7 ;.7 7// =9= /6= /8. .8/ ./; .22 7=18

./1/72I ./-=<:I .7:7 .=// 698 9;= 7:= 7/. /=8 /7: .87 .22

Determination of Lea&age $rom Gas Line 'sing Pressure DropMethod



,o determine the lea!age in &mcf per year at a base pressure of .:1: psia and atemperature of 92 %* ta!e as a basis a one mile section of line under one hour test1 sethe formula

4y ) 6 # D / 00P. < 0:92 > t. 33 - 0P/ < 0:92 > t/ 333

whereD ) inside diameter of the pipe* in1P . ) pressure at the beginning of the test in psiat. ) temperature at the beginning of the test in %P/ ) pressure at end of the testt/ ) temperature at end of the test

,he error in using this formula can be about : 1

E#ample.

A section of a .2-in1 pipe line one mile in length is tested using the pressure drop method1BD is equal to .21.=9 in1 Pressure and temperature at the start of the test were .6217 psigand 7/ %1 After one hour* the readings were .6212 psig and 7/ %1 $alculate the lea!age in&mcf<year1

,he atmospheric pressure for the area is .:1: psia1 ,hus

P. ) .6217 > .:1: ) /2:16 psia

P/ ) .6212 > .:1: ) /2:1: psia

4y ) 6 # 0.21.=93 / 00/2:16 < 0:92 > 7/33 - 0/2:1: < 0:92 > 7/333

) 06 # .2/1; # 2173 < 7./ ) 2162= &mcf per mile per year



A (uic& )ay to Determine "i*e of Gas Gathering Lines

Here is a short-cut way to estimate gas flow in gathering lines1 %or small gathering lines* theanswer will be within .2 of that obtained by more difficult and more accurate formulas1

where( ) cubic ft of gas per /: hoursd ) pipe BD in in1P. ) psi 0abs3 at starting pointP/ ) psi 0abs3 at ending point4 ) length of line in miles

E#ample. How much gas would flow through four miles of 9-in1 BD pipe if the pressure at thestarting point is :87 psi and if the pressure at the downstream terminus is /87 psi

Ans er.

( ) 072230/.930:223 < /

( ) /.*922*222 cubic ft per /: hours

sing a more accurate formula 0a version of eymouthJs3

( ) 08;.30./230:223 < /

( ) /2*622*222 cubic ft per /: hours

,he error in this case is ;22*222 cubic ft per /: hours* or about = 1



Energy Conversion Data for Estimating

,he accompanying data gives the high heating values of practically all commonly used fuels alongwith their relation one to the other1

E#ample. Bf light fuel oil will cost a potential industrial customer .8 cents per gallon and naturalgas will cost him ;7 cents per &cf* which fuel will be more economical

!olution. Assume the operator would use .2*222 gallons of light fuel oil per year at .8 cents pergallon1 ,he cost would be K.*8221 %rom the table* .2*222 gallons of fuel oil is equivalent to.2*222 # .=: ) .*=:2*222 cubic ft of natural gas1 At ;7 cents per &cf this would cost .*=:2 # 1;7) K.*2271

,he burning efficiency of both fuels is the same1 ,he savings by using natural gas would come toK;67 per year1

Ho to Estimate +ime ,equired to Get a "hut!in +est on Gas

+ransmission Lines and Appro-imate a Ma-imum Accepta.lePressure Loss for #e Lines

,hese two rules may prove helpful for air or gas testing of gas transmission lines1 ,hey are notapplicable for hydrostatic tests1 Lalues of the constants used by different transmission companiesmay vary with economic considerations* line conditions and throughputs1

,o determine the minimum time required to achieve a good shut-in test after the line has beencharged and becomes stabili"ed* use the following formula

Hm ) 0= # D / # 43 < P.

whereHm ) minimum time necessary to achieve an accurate test in hoursD ) Bnternal diameter of pipe in in14 ) length of section under test in milesP. ) initial test pressure in lb<in1 / gage

E#ample. How long should a .:-mile section of /9-in1 0/71=;7-in1 BD3 pipe line be shut in under a.*272 psi test pressure to get a good test



Hm ) 0= # 0/71=;73 / # .:3 < .*272 ) /71;7 hours or /7 hours and :7 minutes

hen a new line is being shut in for a test of this duration or longer* the following formula may beused to evaluate whether or not the line is ItightI

Pad ) 0H t # P . 3 < 0D # 6:63

wherePad ) acceptable pressure drop in lb<in1 / gaugeHt ) shut-in test time in hoursD ) internal diameter of pipe in in1P. ) initial test pressure in lb<in1 / gauge

E#ample. hat would be the ma#imum acceptable pressure drop for a new gas transmission lineunder air or gas test when a ::-mile section of =2-in1 0/61/72-in BD3 pipe line has been shut in for.22 hours at an initial pressure of .*.22 psig

Pad ) 0.22 # .*.223 < 0/61/72 # 6:63 ) : lb<in1/

Bf the observed pressure drop after stabili"ation is less than : psig* the section of line would beconsidered Itight1I

Note $orrections must be made for the effect of temperature variations upon pressure during thetest1

Ho to Estimate +ime ,equired to Get a "hut!in +est on Gas+ransmission Lines and Appro-imate a Ma-imum Accepta.le

Pressure Loss for #e Lines

,hese two rules may prove helpful for air or gas testing of gas transmission lines1 ,hey are notapplicable for hydrostatic tests1 Lalues of the constants used by different transmission companiesmay vary with economic considerations* line conditions and throughputs1

,o determine the minimum time required to achieve a good shut-in test after the line has beencharged and becomes stabili"ed* use the following formula

Hm ) 0= # D / # 43 < P.

whereHm ) minimum time necessary to achieve an accurate test in hoursD ) Bnternal diameter of pipe in in14 ) length of section under test in milesP. ) initial test pressure in lb<in1 / gage

E#ample. How long should a .:-mile section of /9-in1 0/71=;7-in1 BD3 pipe line be shut in under a.*272 psi test pressure to get a good test



Hm ) 0= # 0/71=;73 / # .:3 < .*272 ) /71;7 hours or /7 hours and :7 minutes

hen a new line is being shut in for a test of this duration or longer* the following formula may beused to evaluate whether or not the line is ItightI

Pad ) 0H t # P . 3 < 0D # 6:63

wherePad ) acceptable pressure drop in lb<in1 / gaugeHt ) shut-in test time in hoursD ) internal diameter of pipe in in1P. ) initial test pressure in lb<in1 / gauge

E#ample. hat would be the ma#imum acceptable pressure drop for a new gas transmission lineunder air or gas test when a ::-mile section of =2-in1 0/61/72-in BD3 pipe line has been shut in for.22 hours at an initial pressure of .*.22 psig

Pad ) 0.22 # .*.223 < 0/61/72 # 6:63 ) : lb<in1/

Bf the observed pressure drop after stabili"ation is less than : psig* the section of line would beconsidered Itight1I

Note $orrections must be made for the effect of temperature variations upon pressure during thetest1

Ho to Determine the ,elationship of Capacity /ncrease to /nvestment

/ncrease

E#ample. Determine the relationship of capacity increase to investment increase when increasing pipe diameter* while !eeping the factors P0in3* P0out3 and 4 constant1

!olution. ,he top number in each square represents the percent increase in capacity whenincreasing the pipe si"e from that si"e shown in the column at the left to the si"e shown by thecolumn at the bottom of the graph1 ,he lower numbers represent the percent increase in



investment of the bottom si"e over the original pipe si"e1 All numbers are based on pipe capable of.*222 psi wor!ing pressure1

Estimate Pipe "i*e ,equirements for /ncreasing +hroughput 0olumesof #atural Gas

'eflects the capacity to pipe si"e relationship when 4 0length3* P . 0inlet pressure3 and P / 0outlet pressure3 are constant1



Calculate Line Loss 'sing Cross!sectional Areas +a.le )hen +estingMains )ith Air or Gas

Bnitial volume ) 0P i < PC3 0A43

%inal volume ) 0P f < PC3 0A43

wherePi ) initial pressure* or pressure at start of test* psia



Pf ) final pressure* or pressure at end of test* psiaPC ) base pressure* psiaA ) cross-sectional area of inside of pipe in ft /

4 ) length of line in ft

,he loss then would be the difference of the volumes* found by the above formula* which could be

written

4oss during test ) A4 00P i - P f 3 < PC3

,o facilitate calculations* ,able . contains the cross-sectional areas in ft/ for some of the more popular wall thic!ness pipe from .-in1 through /:-in1 nominal si"e1

E#ample. Assume =*79/ ft of 81./7-in1 inside diameter pipe is tested with air for a period of /9hours1 ,he initial pressure was ..9 psig* the final pressure ../ psig* and the barometer was .:1:

psia in both instances1 Bt is desired to calculate the volume of loss during the test using a pressure base of .:197 psia1

Bf the area IAI is obtained from ,able .* and information as given above is substituted in theformula* we would have

4oss during test ) 21=92 # =*79/ # ?0..9 > .:1:3 - 0../ > .:1:3@ < .:197 ) =721. cubic ft

2a3le 1Characteristics of Pipe





$lo of $uel Gases in Pipe Lines



&.!. &a4is * +pecial ProMects* Bnc1* Cailey Bsland* &aine

,he rate of flow of fuel gases at low pressures* ordinary temperatures* and under turbulentconditions in pipe lines is reflected by an equation . based on sound theoretical considerations

q ) =1.87 # Delta0h3 217:= # d /19= < s21:98

whereq ) rate of flow of gas* cubic ft per minuteDelta0h3 ) pressure drop* in1 of water per .22 ft of piped ) actual inner diameter of +chedule :2 steel pipe* in1s ) specific gravity of gas at the prevailing temperature and pressure relative to air at 98 % and =2in1 of mercury1

,he equation can be solved readily and accurately by means of the accompanying nomographconstructed in accordance with methods described previously1 /

,he use of the chart is illustrated as follows1 At what rate will fuel gas with a specific gravity of219:2 flow in a .2-in1 +chedule :2 pipe line if the pressure is 21.92 in1 of water per .22 ft of pipe%ollowing the !ey* connect 21.92 on the Delta0h3 scale and 219:2 on the scale with a straight lineand mar! the intersection with the alpha a#is1 $onnect this point with .2 on the dJ scale fornominal diameters and read the rate of flow at 9/: cubic ft per minutes on the q scale1

+ome authorities give the e#ponent of s as 21:28 when for the data given the rate of flow as readfrom the chart would be decreased by /1; 0.; cubic ft per minute3 to attain 92; cubic ft perminute1



5eferences

.1 Perry* 1 H1* ed1*Chemical Engineers' Handbook * =rd ed1* p1 .786* &cGraw-Hill Coo! $o1*Bnc1* New Oor! 0.67231

/1 Davis* D1+1* Nomography and Empirical Equations * /nd ed1* $hapter 9* 'einholdPublishing $orp1* New Oor!* .69/1



Calculate the 0elocity of Gas in a Pipe Line

L ) ;:81;0(,3 < d / # P # 7/2

whereL ) velocity* feet per second( ) &cf<hour at standard conditionsd ) inside diameter of the pipe* inchesP ) pressure* psia, ) average gas temperature* '

E#ample. Given a pipeline having an inside diameter of :12/9 inches* a gas flow rate of .*222&cf<hour at a pressure of .22 psig* and an average gas temperature of 92o%* find the velocity ofthe gas1

!olution$

L ) 0;:81; # .*222 # 092 > :9233 < 0:12/9/ # .22 # 7/23

L ) :9.16 ft<sec

Determining +hroat Pressure in a Blo !do n "ystem

Bn designing a gas blow-down system* it is necessary to find the throat pressure in a blow-off stac!of a given si"e with a given flow at critical velocity1 ,he accompanying nomograph is handy touse for preliminary calculations1 Bt was derived from the equations

wherePt ) pressure in the throat* psia ) weight of flowing gas* lb<hr d ) actual BD of stac!* in1

' ) gas constant* ft lb< % ) $ p<$v

$ p ) specific heat at constant pressure* Ctu<lb< %$ v ) specific heat at constant volume* Ctu<lb< %, . ) absolute temperature of gas flowing* ' ) 0 % > :923



sing the following values for natural gas* ' ) 881888 and ) .1=.* the above equation reducesto

E#ample. ,o use the accompanying nomograph* assume that the rate of flow is !nown to be /122# .2 9 lb<hr* the BD of the stac! is ./ in1 and the flowing temperature is :82 '1 +tart from theabscissa at the 'ate of %low* * and proceed upward to the BD of the pipe1 %rom this intersection

proceed to the right to the flowing temperature of :82 '1 ,urn upward from this intersection to thePt scale and read the answer - =67 psia1

Estimate the Amount of Gas Blo n 1ff +hrough a Line Puncture



,o calculate the volume of gas lost from a puncture or blowoff* use the equation

( ) D / P .

where ( ) volume of gas in &cf<hr at a pressure of .:16 psi* 92 % and with a specific gravity of2192

D ) diameter of the nipple or orifice in in1P ) absolute pressure in lb<in1 / at some nearby point upstream from the opening

E#ample1 How much gas will be lost during a five minute blowoff through a /-in1 nipple if theupstream pressure is .*222 psi absolute

( ) D / P .

( ) 0/3 / # .*222

( ) :*222 &cf<hr

( 7 min ) :*222 # 7 < 92 ) === &cf

A Practical )ay to Calculate Gas $lo for Pipe Lines

Here is a short cut way to calculate gas flow in pipe lines1 Bt is based on the eymouth formula1At 92 % and specific gravity of 2192* the answer will be accurate1 %or every .2 variation intemperature* the answer will be . error1 %or every 212. variation in specific gravity* the answerwill be three-fourths percent in error1

%ormula

where( ) cubic ft of gas per /: hoursF 8;. is a constantd ) pipe BD in in1P. ) psi 0abs3 at starting point

P/ ) psi 0abs3 at ending point4 ) length of line in miles

E#ample. How much gas would flow through one mile of 9-in1 BD pipe if the pressure at thestarting point is :87 psi and if the pressure at the downstream terminus is /87 psi

!olution.



) 08;.30./230:223

) :.*822*222 cubic ft per /: hours

E#ample. hat would be the flow through four miles of .2-=<:-in1 quarter-in1 wall pipe using thesame pressures

( ) 89*=72*222 cubic ft per day

Ho to Calculate the )eight of Gas in a Pipe Line

5ule. %ind the volume of the pipe in cubic ft and multiply by the weight of the gas per cubic ft1 ,ofind the latter* multiply the absolute pressure of the gas time = and divide by .*2221

,he basis for the latter is that gas with a specific gravity of 192 at ;2 % weighs =129 lb<cubic ft at.*222 psia1 And everything else remaining equal* weight of the gas is proportional to absolute

pressure1 ,hus to find the weight of gas at* say* 722 psia* multiply = # 0722 < .*2223 ) .17 lb1 0.17=is more accurate31

E#ample. %ind the weight of gas in a .*/72-ft aerial river crossing where the average pressurereads 9/71= on the gaugesF the temperature of the gas is ;2 % and specific gravity of gas is 1921

!olution. $hange psig to psia

9/71= > .:1; ) 9:2 psia

eight of gas per cubic ft

=129 # 09:2 < .*2223 ) .1678 lb<cubic ft



Lolume of .*/72 ft of /=-.<:-in1 BD pipe ) =*987 cubic ft

eight of gas ) ;*/.; lb1

,his method is fairly accurateF here is the same problem calculated with the formula

) 0L30.::30P abs3 < ',

where ) weight of gas in lbL ) volume of pipePabs ) absolute pressure of gas' ) universal constant .*7:: Q molecular weight of gas, ) temperature of gas in '

0,o find the molecular weight of gas* multiply specific gravity # molecular weight of air* or in thiscase* 19 # /8167 ) .;1=;13

) 0=*98730.::309:23 < 0.*7:: Q .;1=;30;2 > :923

) ;*/26 lb

Estimate Average Pressure in Gas Pipe Line 'sing 'p and Do n"tream Pressures

,o find the average gas pressure in a line* first divide downstream pressure by upstream pressureand loo! up the value of the factor R in the table shown1 ,hen multiply R times the upstream

pressure to get the average pressure1

E#ample. %ind the average pressure in a pipe line if the upstream pressure is ;;7 psia and thedownstream pressure is :/2 psia1

Pd < Pu ) :/2 < ;;7 ) 217:

Bnterpolating from the table*

R ) 21;8 > : < 7 0212/3 ) 21;69

Pa4erage ) 21;69 # ;;7 ) 9.; psia

,his method is accurate to within / or = psi1 Here is the same problem calculated with the formula



Paverage ) 9.7 psia

Chart for Determining 0iscosity of #atural Gas



$lo of Gas



%or gas flow problems as encountered in oil field production operations* the %anning equation for pressure drop may be used1 A modified form of this equation employing units commonly used inoil field practice is .*/*=*:*7*9

P ) %40&$%D3 / +,5 < /2*222 d 7Pav

'e ) /21.:0&$%D3+ < dS

whereP ) pressure drop* psia% ) friction factor* dimensionless4 ) length of pipe* ft&$%D ) gas flow at standard condition+ ) specific gravity of gas, ) absolute temperature 0 % > :923Pav ) average flowing pressure* psiad ) internal pipe diameter* in1

'e ) 'eynolds number* dimensionlessS ) viscosity* centipoises

Bn actual practice* empirical flow formulas are used by many to solve the gas flow problems offield production operations1 ,he eymouth formula is the one most frequently used since resultsobtained by its use agree quite closely with actual values1 'ecent modification of the formula* byincluding the compressibility factor* 5* made the formula applicable for calculation of high

pressure flow problems1

,he modified formula is as follows

( s ) :==1:7 # 0, s < Ps3 # d/199;

# 00P.

/

- P /

/

3 < 4+,53.</

where( s ) rate of flow of gas in cubic ft per /: hours measured at standard conditionsd ) internal diameter of pipe in in1P. ) initial pressure* psi absoluteP/ ) terminal pressure* psi absolute4 ) length of line in miles+ ) specific gravity of flowing gas 0air ) .123, ) absolute temperature of flowing gas

2a3le 16alues of d %.((0 for use in Weymouth7s gas-flo formula



+tandard conditions for measurements

, s ) standard absolute temperaturePs ) standard pressure* psi absolute5 ) compressibility factor

Lalues of e#pression d /199; for different inside diameter of pipe are given in ,able .1



Another formula quite commonly used is the one determined by %1H1 liphant

where( ) discharge in cubic ft per hour :/ ) a constantP. ) initial pressure in lb<in1/* absoluteP/ ) final pressure in lb<in1/* absolute4 ) length of line in milesa ) diameter coefficient

,his formula assumes a specific gravity of gas of 2191 %or any other specific gravity multiply thefinal result by

,he values of diameter coefficients for different si"es of pipes are given below

%or pipes greater than ./ in1 in diameter the measure is ta!en from the outside and for pipes ofordinary thic!ness the corresponding inside diameters and multipliers are as follows



%or determining gas flow rates for specific pressure drops for pipe si"es used in productionoperations* ,ables =* :* and 7 may be used1 ,he tables are calculated by the eymouth formulafor listed inside diameters of standard weight threaded line pipe for the si"es shown1

,ables =* :* and 7 give the hourly rates of flow of 21;2 specific gravity gas* flowing at 92 % andmeasured at a standard temperature of 92 % and at a standard pressure base of .:1: psi > : o"1

gauge pressure of .:197 psi1 %or specific gravity of gas other than 21;2 and for flowingtemperature other than 92 % correction can be made by use of the factors shown in ,able /1 %orinstance* to determine the flow of gas of specific gravity 2129 flowing at the temperature of /2 %the value obtained from the gas flow ,ables =* :* and 7 would be multiplied by .1./* the factorobtained from ,able /1

2a3le %2emperature - specific gra4ity multipliers









,o find from ,ables =* : and 7 the volume of gas delivered through any length 4* the volumefound from the table should be multiplied by . < sqrt0431 4 must be e#pressed as a multiple of thelength heading the table in use1 %or instance if the delivery is to be determined from a :*722 ft line*

the volume found for a .*222 ft line in ,able = should be multiplied by .sqrt0:1731 At the bottomof each of the two tables in question multipliers are given to be used as correction factors1

se of ,ables =* :* and 7 may be illustrated by the following e#ample1

E#ample.

hat is the delivery of a /-in1 gas line* =*922 ft long with an upstream pressure of /22 psi and adownstream pressure of 72 psi ,he specific gravity of gas is 212;F the flowing temperature* ;2 %1

%rom ,able = the volume for .*222 ft of line* for 21;2 specific gravity gas and 92 % flowing

temperature for /22 and 72 psi up and downstream pressure* respectively* is .2617 &$% per hour1$orrection factor for the pipe length* from the same table is 17=1 $orrection factor* from ,able /for specific gravity and flowing temperature is 166.1 ,herefore

Delivery of gas ) .2617 # 17= # 166. ) 7;*7./ cubic ft<hr

Multi!phase $lo

As stated at the beginning of this chapter* the problems of the simultaneous flow of oil and gas orof oil* gas* and water through one pipe have become increasingly important in oil field productionoperations1 ,he problems are comple# because the properties of two or more fluids must beconsidered and because of the different patterns of fluid flows* depending upon the flowconditions1 ,hese patterns* in any given line* may change as the flow conditions change* and theymay be coe#istent at different points of the line1 =*7*;

Different investigators recogni"e different flow patterns and use different nomenclature1 ,hosegenerally recogni"ed are

.1 Cubble flow - bubbles of gas flow along the upper part of the pipe with about the samevelocity as the liquid1

/1 Plug flow - the bubbles of gas coalesce into large bubbles and fill out the large part of thecross-sectional area of the pipe1

=1 4aminar flow - the gas-liquid interface is relatively smooth with gas flowing in the upper part of the pipe1

:1 avy flow - same as above e#cept that waves are formed on the surface of the liquid1



71 +lug flow - the tops of some waves reach the top of the pipe1 ,hese slugs move with highvelocity1

91 Annular flow - the liquid flows along the walls of the pipe and the gas moves through theenter with high velocity1

;1 +pray flow - the liquid is dispersed in the gas1

,he above description of flow patterns has been given to emphasi"e the first reason why the pressure drop must be higher for the multi-phase than the single-phase flow1 Bn the latter the pressure drop is primarily the result of friction1 Bn the multi-phase flow* in addition to friction* theenergy losses are due to gas accelerating of waves and slugs1 ,hese losses are different fordifferent patterns* with patterns changing as a result of changes in flow conditions1 ,he secondreason is the fact that with two phases present in the pipe* less cross-sectional area is available foreach of the phases1 As previously discussed* the pressure drop is inversely proportional to the fifth

power of pipe diameter1

,he literature on multi-phase flow is quite e#tensive1 'eference ; gives ;/ references on thesubMect1



2 o-Phase Hori8ontal 9lo

ver twenty correlations have been developed by different investigators for predicting the pressure drop in the two-phase flow1 nder the proMect sponsored Mointly by the AmericanPetroleum Bnstitute and American Gas Association at the niversity of Houston* five of thesecorrelations were tested by comparing them with /*9/2 e#perimental measurements culled frommore than .7*222 available1

,he statistical part of the report on the proMect concludes that of the five correlations tested* the4oc!hart-&artinelli correlation shows the best agreement with the e#perimental data* particularly



for pipe si"es commonly used in oil field production operations18 ,he method may be summari"edas follows =*:

.1 Determine the single-phase pressure drops for the liquid and the gas phase as if each one ofthese were flowing alone through the pipe1 ,his is done by use of previously givenequations1

/1 Determine the dimensionless parameter

where DP4 and DPG are single-phase pressure drops for liquid and gas respectively1

=1 ,he method recogni"es four regimes of the two-phase flow as follows%low 'egimes

Gas 4iquid,urbulent 4aminar,urbulent ,urbulent4aminar 4aminar4aminar ,urbulent

:1 ,he type of flow of each of the phases is determined by its 'eynolds number 0%igure /31

71 Determine factor % from a chart as a function of R for appropriate flow regime191 ,he two-phase pressure drop is then

,he single-phase pressure drop of either gas or liquid phase can be used1 ,he results are the same1

%or the two-phase flow problems in oil field production operations* certain simplifications can beused if only appro#imate estimates are desired

.1 Bnformation which would permit determining the volume of oil and gas under flowconditions from equilibrium flash calculation is usually unavailable1 ,he following

procedure may be used

Gas volume is determined by use of the chart in %igure . which shows gas in solution foroil of different gravities and different saturation pressures 0in this case the flowline pressures31 Bf the gas-oil ratio is !nown for one pressure the ratio for another pressure can be determined from the chart as follows

Assume gas oil ratio of 922 cubic ft<bbl at 622 psi for a =2o APB oil1 Determine the gas oilratio at .*=22 psi1 %rom the chart* at 622 psi* the gas in solution is /27 cubic ft<bbl and at.*=22 psi is =22 cubic ft<bbl* an increase of 67 cubic ft<bbl1 ,herefore* at .*=22 psi the gasoil ratio will be 922 - 67 ) 727 cubic ft<bbl1



Bncrease in volume of oil because of the increase of the dissolved gas can be obtained fromcharts for calculation of formation volume of bubble point liquids1 However* forappro#imate estimates in problems involving field flowlines* this step frequently can beomitted without seriously affecting the validity of results1

/1 Bn a maMority of cases here under consideration* the flow regime is turbulent-turbulent1

,herefore the chart in %igure shows the factor for only this regime1

,he procedure is illustrated by the following e#amples1

E#ample 1. Determine two-phase pressure drops for different flow ratios and different flow pressures under the following conditions

il - =. APBLiscosity - 7 cp at .22 %+eparator pressure - 622 psigGas oil ratio - 9;2 cubic ft<bbl%low line - / in1

ellhead pressure of .*=22 psig was assumed and calculations were made for flow rates of .22and 72 bbl per day1 ,he single phase pressure drops were calculated for the gas and liquid phases*each of them flowing alone in the line* according to the %anning equations1



$alculation of pressure drop for liquid phase

Hf ) f 4 0bbl<day3 / 0lb<gal3 < .8.*6.9 D 7

wheref ) friction factor

4 ) length of pipe* ftD ) inside diameter of pipe* in1

$alculation of pressure drop for gas phase see pp1 /82-/8.1

%or the two flow rates and the two pressures* these single-phase pressure drops were calculated to be as follows

ith two pressure drops determined for two flow line pressure for each of the flow rates* lines can be drawn showing pressure drop for different flow line pressures for these two rates 0%igure =31%rom these tow lines relationships can be established for pressure drops for other rates and flowline pressures1 ,he procedure is as follows

+tart with* for instance* 822 psia line1 %or the rate of 72 bpd at 822 psia flow line pressure* the pressure drop from the chart is 12;: psia<.22 ft1 %rom the 72 bpd rate on the scale in the upper partof the chart* go downward to the line representing 12;: psia pressure drop1 ,his gives one point ofthe 822 psia line1

%or the rate of .22 bpd the chart indicates that at 822 psia flowline pressure* the pressure drop is 1.;7 psia<.22 ft1 %rom the .22 bpd rate in the upper part of the chart* go downward to the 1.;7

psia<.22 ft line1 ,his gives the second point of the 822 psia line1 ,he line can now be drawnthrough these two points1 4ines for other pressures can be constructed in a similar way1

,he chart issued as follows %or conditions represented by this particular chart* what is the two- phase pressure drop for the rate of 92 bpd and average flow line pressure of .*222 psia



+tart at the rate of 92 bpd on the upper scale of the chart1 Go downward to the intersection with the.*222 psia pressure* then to the left* and read the two-phase pressure drop as 21;7 psia<.22 ft1

E#ample %. Bf pipe diameter is to be determined for the desired pressure drop and the !nownlength of line* flow rate* and the properties of fluids* then the problem cannot be solved directly1,he procedure is as follows

An estimate is made of the diameter of the two-phase line1 ,he best approach is to determine* forthe conditions* the diameter of the line for each of the phases flowing separately1 Bn a maMority ofcases* the sum of these two diameters will be a good estimate of the diameter for the two-phase

line1 $alculations are then made for the two-phase pressure drop using the assumed diameter1 Bfthis drop is reasonably close to the desired one* the estimate is correct1 Bf not* a new estimate must be made and the calculations repeated1

"nclined t o-phase flo



Bn hilly terrain additional pressure drop can be e#pected in case of the two-phase flow1 4ittle published information is available on the subMect1 vid Ca!er suggested the following empiricalformula .2

Hori8ontal three-phase flo

Lery little is !nown about the effect on the pressure drop of addition to the gas-oil system of athird phase* an immisicible liquid* water1 %ormation of an emulsion results in increased viscosities1%ormulas are available for appro#imate determination of viscosity of emulsion* if viscosity of oiland percent of water content of oil is !nown1 ,he point is* however* that it is not !nown what

portion of the water is flowing in emulsified form1

%rom one published reference .. and some unpublished data* the following conclusions appear to be well founded

.1 ithin the range of water content of less than .2 or more than 62 * the flow mechanismappears to approach that of the two-phase flow1

/1 ithin the range of water content of from ;2 to 62 * the three-phase pressure drop isconsiderably higher than for the two-phase flow1

%or solutions of three-phase flow problems of oil field flow lines* the following approach has beenused1 ,he oil and water are considered as one phase and the gas as the other phase1 Previouslygiven calculations of a two-phase drop is used1 ,he viscosity of the oil-water mi#ture isdetermined as follows

S4 ) 0L o # S o > L # S 3 < 0Lo > L 3

where S 4 * So * and S are viscosities of mi#ture of oil and of water respectively* and L o and L w represent the corresponding volumes of oil and water1

No test data are available to determine the accuracy of this approach1 ntil more information isdeveloped on the subMect the method may be considered for appro#imate estimates1



5eferences

.1 Cameron Hydraulic Data * .=th ed1* Bngersoll-'and $ompany* New Oor!* .69/1/1 Flow of Fluids Through al!es" Fittings and #ipe * ,echnical Paper No1 :.2* Engineering

Division* $rane $ompany* $hicago* .67;1=1 +treeter* Lictor 4* Handbook of Fluid Dynamics * &cGraw-Hill Coo! $ompany* Bnc1* New

Oor!* .69.1:1 Prof 1E1 DurandJs Discussions in Handbook of the #etroleum $ndustry * David ,1 DayFohn iley T +ons* Bnc1

71 Ca!er* vid I+imultaneous %low of il and Gas*I The %il and &as ournal * uly /9* .67:191 $ampbell* Dr1 ohn &1* IElements of %ield Processing*I ,he il and Gas ournal*

December 6* .67;1;1 $ampbell* Dr1 ohn &1* IProblems of &ulti-phase Pipe 4ine %low*I in I%low $alculations

in Pipelining*I The %il and &as ournal * November :* .676181 Duc!ler* A1 E1* ic!s* BBB* &oye* and $leveland* '1 G1* I%rictional Pressure Drop in ,wo-

Phase %low*I ($ChE ournal * anuary* .69:161 4oc!hart* '1 1* and &artinelli* '1 $1* IProposed $orrelation of Data for Bsothermal ,wo-

Phase* ,wo-$omponent %low in Pipes*I Chemical Engineering #rogress :7 =6-:8*0.6:63*.21 vid Ca!erJs discussion of article IHow phill and Downhill %low Affect Pressure Drop

in ,wo-Phase Pipe 4ines*I by Crigham* 1 E1* Holstein* E1 D1* and Huntington* '1 41F The%il and &as ournal * November ..* .67;1

..1 +obocins!i* D1 P1* and Huntington* '1 D1* I$oncurrent %low of Air* Gas* il and ater ina Hori"ontal 4ine*I Trans ()*E 82 .*/7/* .6781

./1 Ceal* $arlton* I,he Liscosity of Air* ater* Natural Gas* $rude il and Bts AssociatedGases at il %ield ,emperatures and Pressures*I #etroleum De!elopment and Technology *AB&E* .6:91

.=1 +wit"er* %1 G1 #ump Fa+ * Gould Pumps* Bnc1

.:1 #ump Talk * ,he Engineering Department* Dean Crothers Pumps* Bnc1* Bndianapolis* Bnd1

.71 +wit"er* %1 G1*The Centrifugal #ump * Gould Pumps* Bnc1

.91 5alis* A1 A1* IDonJt Ce $onfused by 'otary Pump Performance $urves*I Hydrocarbon #rocessing and #etroleum ,efiner * Gulf Publishing $ompany* +eptember* .69.1

#omograph for calculating ,eynolds num.er for compressi.le flofriction factor for clean steel and rought iron pipe



Bac:ground

,he nomograph 0%igure .3 permits calculation of the 'eynolds number for compressible flow andthe corresponding friction factor1 Bt is based on the equation



where

) flow rate* lb<hr +g ) specific gravity of gas relative to air qs ) volumetric flow rate* cubic ft<sec 0at .:1; psia and 92 %3m ) fluid viscosity* centipoise

Bf the flow rate is given in lb<hr* the nomograph can be used directly without resorting to %igure /to obtain qs1 %igure / converts volume flow to weight flow rates if the specific gravity of the fluidis !nown1





E#ample. Natural gas as /72 psig and 92 % with a specific gravity of 21;7* flows through an 8-in1schedule :2 clean steel pipe at a rate of .*/22*222 cubic ft<hr at standard conditions1

%ind the 'eynolds number* and the friction factor1

At +g ) 21;7 obtain q ) 96*222 from %igure .* S ) 212..1

Connect With ;ar: or 5ead

) 96*222 S ) 212.. Bnde#

Bnde# d ) 8 in1 'e ) 7*222*222

'e ) 7*222*222 hori"ontally to 8 in1 f ) 212.:

!ource

Flow of Fluids Through al!es" Fittings" and #ipes * ,echnical Paper No1 :.2* = - .6* $rane$ompany* $hicago* Bllinois 0.67;31

Documents

Gas Hydraulics(Chapter 11)