Chapter 3 N/\TURAL GAS TRANSMISSION ..ln lhi, cllapter, the general flow equaton for cornpressible flow in a pipeline will be derived mm basic principIes. Having obtained the general flow equation, the way in which flow efficiency is affected by varying different gas and pipeline pararneters will be examined. ni m:rent flow regirnes in gas transrnission systerns (i.e., partially turbulent and fully turbulent flow) will be presented. Sorne of the widely used transrnission equations and their applications, advantages, and lirnitations will be outlined. This will be followed by a dscussion ofpipes in series, pipeline looping, gas velocity, line packing, pipeline maximurn operating pressure, and sorne pipeline codeso The impact 01' gas temperature on the flow efficiency and gas temperature profile (Le., heat transfer fro"1 a buried pipeline and Joule-Thornpson eflects) will be discussed. Final/y, sorne rnajor econornic aspects and considerations in the design oi' gas pipeline will be presented. GENERAL IlOW EQUATION - STEADY STATE .. ;cction, the general flow equation for cornpressible fluids in a pipeline at steady-state cOlldltion is derived. Firs. the general Bernoulli equation will be obtained using a force balance on a segment of the pipeline. The Bemoulli equation is then used to derive the gcrH'ral Ilow equation for cornpressible fluids (natural gas) in a pipeline. , ;o!lsider a pipeline that transports a cornpressible fluid (natural gas) betwecn points 1 alld 1. al steady-state condition, as shown in Figure 3-1 wherc P = gas density P gas pressure A = pipeline cross-sectional area u gas velocity al steady-state condition dm o (3 - 1)dI where m is the rnass of gas flowing in the pipeline and I is time. The rnass flow rate ofgas at point 1 can be defined as in PI A I !JI (3 2) 57 58 Pipel ine Design and Construction: A Practical Approach CD U U2 Figure 1-1. Steady state flow of a compressble fluid in a pipeline Likewise, the mass flow rate at point 2 is m= P2 . A2 U2 It fuen follows that PI AI 'UI If the pipe has a constant diameter, then PI . U P2 . U2 or, in general m=pAu or m = p u A p'U e where e is a constant. It s also known that P v where ti the gas specific volume so u e (3 3) v From Newton's Law of Motion for a particle of gas moving in a pipeline [see Figure 3-2(a)]: dF= a dm where a = duldt is the acceleration: du du dydF = . dm = . P . A . dy pA du dt dt dt Natural Gas Transmission 59 .... 1 1'"dy dm =p.A.dy (g.,lI ] 2a. Movement of a gas partcJe in a pipeline and dv-=- = u d! hercfare dF = P . A . u . du In l, S. units, using the proportionality constant gc, the above equaton c0111d be written A A /ldF = - . u . p. du . du (3 - 4) ge gc l' The mpact of all exsting forces (i.e., pressurc, weight, friction, ctc.) cxcrted on n partic\e of gas in a nonhorizontal pipeline [Figure 3-2(b)] can be considered as follows: ;hc torces F and F2acting on the gas partic\e due lo he gas prcssure p and P2 can be J d t n e , ~ LIS: dF AdP1 and The force F3 exerted on the gas due to the weight Wof the gas particle s F3 = W sinO' in differential form dF dW sinO'3 wh('fr the weght of the gas is (3 - 5) 60 Pipeline Design and Construction: A Practical Approach dH w Figure 3-2b. Demonstration of all forces acting on a gas particle moving in a nonhorizontal pipeline (;, is local acceleration of gravity) fu rthcnn ore dll sin Cl' = ~ -dy whcre dI! is the change in elevation, Upon substitution for both dWand sin df\ = gL ,A . p . dH (3 ~ 6a) gc or gL AdF] = -- . - . dH (3 6b) gc v Fnally, the metion force is defined as: dF4 = ?T . D . dy . T (3 - 7) where ?TDdy is the surface area and T is the shear stress. The summation of all the forees acting on the elernent of the gas shouId be equal to zero, therefore: A U gL A. du +AdP + - . - dH +?TDdy . T O (3 - 8) gc V gc V - - - - - _ . ~ . ~ - Natural Gas Transmission 61 Thi; is the general fonn of the Bemoulli equation. In most cases, t is assumed that the nll values of gL and gc are egual. Then A u A ~ . . du + Ad? + - dH + 7rDdv . T gc V V ' O (3 9) Muaiply both sides by v/A: 1 7rDdyv - . udu + vd? + dH + -. T gc A = O (3 10) ,V!lli . /u = kinetic energy; vdP= pressure energy; dH = potential energy; (7rDdyx vIA)T = h;, " lf Iosses ]'1lr friction tenn or losses created by moving a fluid in a pipeline is defined by the i equation as follows: 2 2fu . dLdFFannng (3 - 11) gc D w/,;n; u average gas velocity friction factor n pipeline diameter L pipeline length Substituting the Fanning equation for Iosses in the general energy equation will n':.'>l.dl :n 2{u2 . udu + vd? + dH + -'-- . dL O (3 - 12) ge gc . D iding both sides of the eguation by v2: (3 13) 'e final fonn of the equaton can be obtaned by integrating each tenn, assuming u/v'" nllA = C == constant. Kinetic Energy Term j2 ~ . u 2 . du I gc V 2j ~ , ~ . du 1 gc V V since l!.=C v 62 Pipeline Design and Construction: A Practical Approach smce u v= e then Kinetic energy Pressure Energy T erm J2 dP 2= pdP IVI From the real gas law PV nZRT whcrc Z is the comprcssibility factor of the gas and R is the gas constant for m n and mp=-V lhc equation for the density of a gas is: PM p ZRT wheTe M is the average mole!?ular weight of the gas. AfteT substitution ioto IL pdP 2 Pl..l .dP ,ftvf J2 Zave . R . T I PdP ZR T ave M ~ ~ ~ Zave . R . Tave 2 wheTe Tave, is defined as follows: (3 14) (3 - 15) (3 16) (3 - 17) Natural Gas Transmission 63 '" lo. ~ O . 6 .E ,... J5 ';..., l5.0.5 '" ,...... :c; ';; ~ Q.E Eo o Figure 3-3. Compressibility factor for natural gases [Katz et al., 1959 Handbook of Gas Engneenng, reproduced with perrnssion from McGraw-Hill Co.} (Ti and T2 are the upstream and downstream gas temperatures) and Pave is obtained based on the relation l PdP: u u ._1.0 ___---0.98 12 14 15 Pseudo reduced pressure Pave 64 Pipelne Desgn and Construction: A Practical Approach or P avc = ~ [PI + P2 - ; ! ~ ~ J (3 - J8) (Pj and P2 are the upstream and downstream gas pressures). Having obtained Taye and P ave for the gas, the average compressibility factor, or Zave, can be obtained for lean sweet natural gases with an excellent accuracy using Kay's rule and compressibility factor charts. The Z factor can also be calculated with one of the widely used equations of state, such as AGA-8. BWRS, RK, SRK, or any other such formula. ro calculate Zave for a natural gas using Kay's rule, Taye and P ave ofthe gas are needed, and also pseudocritical pressure and temperature oftlte natural gas. Pseudocritical values can be obtained with Kay's rule as folIows: T ~ TCA ' YA + Tes' Ys + Tcc ' Yc + ... (3 19) P ~ = PCA . YA + PCB .YB + Pce . Yc + ... (3 - 20) where Tel = average pseudocritical temperature of the gas Pe' := average pseudocritical pressure of the g'!s T(:A, TeB, Tee,. critical temperature of each component PeA, PeB, Pec,. = critica! pressure of each component YA, YB, Ye, mole fraction of each component Finally, pseudoreduced pressure and temperature can be obtained as follows: P ayePr I -_ (3 - 21) and Tave (3 22)T ~ The values of Pr' and T/ can be used in compressibility factor charts to calculate Zave (as shown in Figure 3-3). I Example 3.1 IWhat is the compressibility factor for a natural gas with the following analysis at 1,000 psia and 100F? l Gas COMPONENT Mole % 85 10 5 Using Kay's rule to calculate pseudocritical properties (see Table 3-1): P ~ PCA ' r4 + PeB . YB +Pee . Yc+ T ~ = TC4 . YA + TCB ' fB + Tcc . fc+ P ~ = 666 x 0.85 +707 x 0.10 + 617.4 x 0.05 = 667.67 pSIa, pseudocritical pressure Natural Gas Transmission 65 = 343.3 x 0.85 + 549.8 x 0.10 + 666.0 x 0.05 = 380.085R pseudocritical temperature , Pa\'(' 1 , 000 Pr = F{ = 667.67 = 1.498 , T\\e 460 + 100 1.474Te = = -380.085 using the appropriate chart (see Figure 3.3), which covers both the p/ and range valuc, Zave 0= 0.85 is obtained. The model mentioned aboye for the calculation of the compressibility factor is a quick and accurate model for dry and sweet natural gases, and is most suitable for hand calculations. In arge gas transmssion networks with hundreds of pipe segments, while dividing each pipe into smalIer segments to consider temperature changes in the pipeline, this modeI becomes a cumbersome and time-consuming procedure to calculate the compressibility factor. For these networks, where all simulations are automated, apply accurate equations of state to calculate the compressibility factor. There are a arge number of equations of state that are suitable for a limited range (depending on pressure, temperature, and gas composition), but could deviate to produce inaccurate resuIts if used without these considerations. One such equation, which is commonly used in gas transmission systems and has a proven accuracy of better than 0.3% for compressbility lactor, is an AGA equation givcn by Starling and Savage (1994). This equation covers most of the existing pressure and tcmperaturc ranges used in gas transmission Iines. For further information on different equations of state (e.g., for gases containing H2S ami COl) refer to (Campbell el al., 1994). (,I\BlE 3-1. Critical propertes of constituents of natural gas (Courtesy Campbell Petroleum Series) Critical TE'mperaturE' Critical PrE'ssurE' ..('ompound Molecular Weight R K psia MPa .._ .. __.... CI 16.043 343 191 666 4.60 C2 30.070 550 305 707 4.8& C, 44.0n 666 370 617 4.25 iC4 58.124 734 408 528 3.65 ne. 58.124 765 425 551 3.80 iCs 72.151 829 460 491 3.39 nCs 72.151 845 470 489 3.37 nC6 86.178 913 507 437 3.01 nC7 100.205 972 540 397 2.74 114.232 1,024 569 361 2.49 nCq 128.259 1.070 595 332 2.29 nCjO 142.286 1,112 618 305 2.10 nCI 156.302 1,150 639 285 1.97 nC2 170.338 1,185 658 264 1.82 N2 28.016 227 126 493 3.40 CO2 44.010 548 304 1,071 7.38 H2S 34.076 672 373 1,300 8.96 O2 32.000 278 155 731 5.04 Hz 2.016 60 33 188 1.30 H20 18.015 1,165 647 3,199 22.06 Air 28.960 238 132 547 3.77 He 4.000 9 5 33 0.23 .----.. __ .. 66 Pipeline Design and Construction: A Practical Approach Potential Energy T erm lntegration of the potential energy tenn of Equation (3-13) will result in: 2dH 2 2 /2 (P' /vl) 2 . = p dH = -_.. d1/ (3 - 23) 1 1'2 I . 1 ZRT where 8.H 1/2 HI ' There is no simp.le mathematical relationship between e1evation change, gas pressure and gas temperature, so the relationship (P2,M2/Z2.R2,T2) can be taken out of the integral in the fonn of average values whik maintaining reasonable accuracy. Friction Loss Term The integral of the energy losses can be evaluated as follows: (3 - 24) where L is the pipeline length. The general tOrm of the flow equation is obtained by adding all the terms together and setting them equal to zero o (3 25) The aboye equation ean be further simplit1ed ifthe kinetie energy tcon is neglectcd (tor almost atl high-pressure gas transmission lines, the contribution ofthe kinetic energy term compared to the other tenns is insignificant). Therefore Al (2P (3 26)2RZ;vp Ta,." 2 The aboye equation ean be even further simplified upon the following substitutions: m 2 1fD2 A=- a pipe e A e (;)2 4 ' for Moreover, the gas relationship at a base or standard eondition is Pb Qb where Qb is the volumetrie gas flow. Ir and Natural Gas Transmission 67 .2 m hen Uas gravity s defined as G whclC /l4."ir'::::; 29. t fp(111 substitution and rearrangement to solve for C, Equation (3-26) would be 2 7 {2 ...2 5 2 . R . g z r PI 1"2 R. T .Z DQb = ....::._c b b 1 &.. ave (3 27)32 . T - 58Z ' T G . L . 1ave ave By fa king the square root 'of Qb, the general flow equation ofnatural gas in a pipeline is " - P; - VI 2.5 _____=--.....c.:.:.,,---. - . D (3 28)Zave . Tu"" . G . L 1 1; 1bove equation can be used in Imperial or S.L units; for any size or length of pipe; 1, 1; ar, partially turbulent or fully turbulent flow; and for low, medium, or high. ,\/stems. ution 01Parameters (Imperial Units): Q" gas flow rate at base conditions, MMSCFD or MCF/HR g( proportionality constant, 32.2 (lbm x ftIIbrx sec 2) /. -; compressibility factor at base condition Zb ,::::; I r temperature at base condton, 5200 R F" pressure at base condition, 14.7 psia P, = gas inlet pressure to the pipeline, psi a P2 gas exit pressure, psia G gas gravity, dimensionless = elevation change, ft Pave average pressure, psi a R= gas constant, 10.73, (psiaxft3llb molesxOR) average temperature, eR Zave compressibility factor at Pave, Tave, dimensionless L = pipeline length, ft or miles f = fuction coefficient, dimensionless r; \;,1 transmission factor, dimensionless . V J D = inside diameter of the pipeline, inch j I i O" ce -o -o :2. O ro U'\ 0;5' ::J llJ ::J o... d ::J U'\ ..........e (".o' ::J >TABLE 3-2. Formulas and transmission actors or commonly used flow equations (Courtesy JGT) '"O llJ "'" Equation Formulaa Transmissjon Factor ("

::;. Fritzscheb Q, 1720(k) llJ 5.145 -o .!og( 37 t) . -o a llJ ("Turbulent Q, "" 0.4696 4 log (3. 7Dlke) :;-Panhandle B Q, 2.431 16.49(R,,)oOI%1 Colebrook-White Q, = 0.4696 D" -410g(&+ I 10T Distribution Q, 0.6643 4.6! 9 (R,,)o 100 Mueller Q, = 0.4937 3.35 (Re)OIlU .._-_.---=-.. l Panhandlc A b 'T)Q, 6.872 Pipe Diameter, (in.) ce Pipe Diameter, (in.) Pole Q, 3i4 to I 1.732 3/4 lo 1 956 1 1 '4 to 1 1'2 1.905 1 1/4 to I 1/2 10.51 2 2.078 2 11.47 3 2.252 3 1243 4 2.338 4 12.90 Q )1)0 SpJtzglass (Hgh Pressure)d Qh 3415 Sptzglass (Low Pressure)d Q/, 3.550 Weymouth Q, = 1.3124 11.19 D!" z , The units 01' the quantities in all ofthese equations are: D in, h, in. WC; L tl; P" P2, Pb pSla; Q MCF:nr; J = Ibmift sec; and Tr, h R, c:b The constants 1.720 and 2450 include: ji 7.0 x 10'" lbm/f sec . ..., e lncludcs: Pb = 14,73 psia; Th 492'R; and Tr= 500'R. d The constants 3.415 and 3.550 indudc: Ph 14.7 psia; T 520 R; and TI 520'R. [' !l) Vl -! -< !l) :::l Vl :::l -' Vl Vl (5' :::l C' \.t:) \,.. l 70 Pipeline Design and Construction: A Practica I Approach Equation (3-28) can be written in the following form, taking all constants as e', therefore (3 - 29) In transmission lines, ifthe pipeline is horizontal or MI i8 insignificant compared to the value of - or 22 2)) SBG . 6.H . P aveP -- P 1 2 R T ave . Z ave ' then the elevation term can be omitted and Equation (3-29) becomes: Q c. __. (3 30). b Zave . rave . G . L YJ The aboye equation shows the efTect of jj and D on the flow _of gas in a pipeline. The expression IJ is the transmission factor and is an important para meter that represents the transmissivity of gas in a pipeline. Diameter is another major factor in pipeline design; it can be seen that ifthe diameter ofthe pipeline is doubled, the gas flow rate will be increased by (2)25 5.66 times. This demonstrates the importance of considering possible future expansions when selecting pipeline diameter. For example, if a 20 inches gas pipeline is changed to a 30 inches pipeline, gas flow rate is increased by almost 2.756 times, assuming the remaining parameters are constant. Unlike liquids, Equation (3-30) shows that a system operating at a lower temperature results in higher flows or lower pressure drops. I!1 contrast, higher temperatures will increase the gas viscosity, which will reduce fue flow capacity of the pipeline. The impact of other parameters, such as G and Z, will be discussed Jater. Considering the previous equation (3-30), Ql; can be calculated as Q2 C,,2. 15 b ---'---=--- . . D . TaV flow rate at base condition and it should be in MCF/hr. fIJen: (re _4 ') (0.65)855K = RL = 2.)52 x 10 . 2005280 (19.5)4856 K, = 0.052745 Qb 200.000 35.31 7.062 MCFI hr 2 ') n PI P; K, Qb (l, 000)2 -p; = 0.052745(7062(855 P2 = 522 psia Now if the system is looped: 'dal resistance of the second pipe is: 0.855 ( )K2 = R2 L 2.55 x 10-4 520 0.65 200 5280 (J 5.5)4.856 K2 0.160728 FOl' looped system, the total resistance is: Kr wl '1 1.855. Kr (0.052745)(0.160728) ( +0.1607281/1855 jl.855 Kr = 0.023430 f2 f2 K ()1855 I 2 r rtf (l, 000)2 = 0.023430(7062)1855 P2 = 823 psia ( downstream pressure after looping) 16", 200 Mile 20", 200 Mile L 10. Pipeline system after ooping 88 Pipeline Design and Construction: A Practical Approach Pipeline Segmental Looping In many cases, it may nol be necessary lo loop the entire pipeline to obtain Ihe desired flow or downstream pressure. Therefore, only a segment of the pipeline is looped lo meel requrements. Assume tha! the existing line has length L, diameler D), and a total resistance of K K', where inle! and exit pressures are p and P2, respectively. It is intended lo increase the existing gas Ilow rale from Q. to Q2 (i.e., Qb to Qb without any changes in downstream pressure. A pipeline loop with diameter D2 and length X will be added lo the existing pipeline in order to increase QI to Q2 without any changes lo the downstream pressure. The value of X, the length of the pipeline lo be looped lo the existing syslem, mus! be determined. Note that usrng, larger diameler pipes will reduce the required length of the segment to be looped. To obtain the total pipeline resistance, slart with one of the major transmission equations and conlinue to develop the equation to calculate the length ofthe loop. For this example, the Weymouth equation will be used. The Weymouth equation in Imperial Units, is written as: 2 2 0.000466 G . . L 2 PI - P2 = ~ - - - - ~ ~ . : - - : - : - ~ - _ . Q) or 2 2 2 PI K QI P2 and K Kt + Ki (total resistan ce of the single line) In the total resistance formula ofthe Weymouth equation, the value ofO.000466. GTris a constant, which could be assumcd as: e = 0.000466 . G Tr ex , ex e(L X) CL - ~ " . _ - ~ - .._K' --- K and K ::-::-;-- I ) 16'3' 2 -16/3' K 16jJ d 613 !DI DI D1 2 I J The equivalent resistancc for the looped segment is: \ (3 63) after simplification K" would be: ex ., (3 - 64) d/3 D8/3)( + 2 and KE Total = Ke + K ( i.e., pipes in series) Natural Gas Transmission 89 L-X p ""--------. .... QI K' I x Fi!{: \ 11. Pipeline segmental looping then __C__X_. + C(LX) (3 - 65) D 168/3 8/3) 2 3 ( DI +D2 I ami D \fiding the flow equations tor the existing pipeline and afier segmental looping: ur eL 1 = C(L.X)' (ciJ 2 .. ..__., (Di /J DI . : i !Inlly, the equation eould be written as follows: XL, .. (3 66) x", length of the pipeline to be looped, miles L length of the existing pipeline, miles Q = initial gas flow rate, MMSCFD Q2 = final gas flow rate, MMSCFD D existing pipeline inside diameter, inehes D2 == looped segment inside diameter, iDehes 'Por AGA fully turbulent equation, a value of 2.5 instead of 8/3 is used as the exponent i" ')c denominator of EquatioD (3-66). 90 Pipeline Design and Construction: A Practical Approach Example 3.6 A gas transmsson line of OD 16 inehes (ID 15.5 inches), Ql = 70,000 m3!hr, and L = 300 km is to be used without any pressure changes in the delvery of 120,000 mJ/hr of natural gas. What would be the length of pipeline with identical diameter that ShOllld be looped to the existing pipeline to satisfy lhe increased eapaeity (aceording to lhe Weymouth equalion). Solution: Using Equation (3-66). Q 70,000 mJ!hr Q2 = 120,000 m3br DI 15.5 inehes D2 = 15.5 inehes L 300 km X=?km 2 X = 300 ( 70.000) 120.000 1 ( 1 .)2l+m:D813 1 X =300(7/12)2 1=263.88 km, segment lo be looped ( ~ ) Sorne Important Considerations Regarding Pipeline Looping Equation (3-66) demonstrates that looping wi 11 inerease pipeline flow eapaeity without any ehanges to the upstream and downstream pressures. Likewise, if lhe flow is kept constant, adding a loop results in less of a pressure drop along the pipeline. However, Eqllation (3-66) also implies that the impaet of Jooping on the flow capaeity (or prCSSllre drop along the pipeline) is independenl of the location of the loop. In praetical pipeline operations this is nol lhe case, and the placement of the looping can have a :;ignificant impact on lhe response of lhe system. The behavior of the system is greatly affected by changes in temperature and the compressibility factor of lhe gas along the pipeline. There are two important parameters lo consider when ehoosing lhe location for a pipeline loop: temperature and pressure. Considering pipeline pressure, the magnitude of lhe pressure drop is higher al the downstream section of the pipeline because the gas has expanded. Hence, considering pressure alone, looping at the downstream portion of the pipeline is more effieient. However, temperature must also be considered. Al the upstream part of the pipeline, particularly downstream of a compressor station, the gas temperature is typically much higher than in other places along the line. Adding a loop in areas where the gas temperature is hotter increases the heat transfer from the pipeline to the immediate environment. This is especially true if the ground temperature is significantly less than the gas temperature. When higher rates of cooling occur, the pressure drop along the pipeline is considerably less. Therefore, it is most ofien recommended that under steady-state conditions, pipeline looping be in an upstream 1:) l' \ Natural Gas Transmission 91 region, suco as immediately downstream of a compressor station, especially if the gas is hot. It should be noled that a comprehensive simulation involving a gas temperature profile giving consideration lo elevation changes is always necessary to determine the exact location of looping in steady-state operations. TypicaJly, in situations wherc the gas temperature is very close to the ground tcmperaturc, or the differcnce is Jess than 5 .- 10C, temperature is no onger an important consideration when choosing a location for a loop. In !he following example. the results of hydraulic simulations for three different cases are presented to further clarify the effects of different loop locations for a pipeline at steady state. In al! simulations, Hydraulic Analysis and Resources Tool (HART) simulation software was used. It is TransCanada Pipeline's steady-state ,;imulation software developed in-house for use in designing the company's pipeline network. Parameters Jor Case Study: Pipeline tength = 100 km (62.15 miles) Gas ftow rate is constant = 289.542 MMSCFD Gas inlet pressure 1,200 psia Gas inlet temperature 45 oC or 113 "F Pipeline OD 20 inches Pipeline ID = 19.44 inches Pipeline roughness =o 750 IL inches Soil temperature 10 ('C (50 F) Cdse :: No loop P =1130Apsia p::: 1059psia P ::: 984.6 psia P = 905.5 psia ;/5 km 25 km 25 km 25 km T = 90.7"F (32.5'C) T 66.4"F (19.1C) T = 59.S"F (15.3"C) eIS':: 11: looping First 25 Km of the Pipeline (Upstream) p = p::: 1117.4psia p::: 1048.9psia p::: 976.5psia 25km 25 km 25 km 25 km T =113'F (45C) T = B024'F (26.8"C) T =69.e'F (20.9'C) T =62.4'F (16.9'C) T::: 57.4'F (14.1C) Case 111: looping Last 25 Km of the Pipeline (Downstream) p = 1200psia P = 1130.4psia P = 1059psia P 25 km 25 km 25 km 25 km T = 113"F (45"C) T ::: 90.7"F (32.6'C) T =76. fF (245"C) T = 66.4'F (19.1C) T = 60.3'F (15.7"C) For these three cases, it can be concluded that for the same gas flow rate, looping upstrearn of the pipeline gives the highest delivery pressure or the least pressure drop. ... 92 Pipeline Design and Construction: A Practical Approach 1250 1200 -. 1150 ... 1100 a I:! ji! 1050 1000 950 900 o 25 50 75 100 Di_{\uo) __case I - No Loop .... Case 11 Figure 3-12. Comparison of pressure drops for cases 1, 11, and III The collective results of the abovc cases are compared in Figure 3-12. PIr'ELlNE CAS VElOCITY _..,.._.- ---------------------------------IThe equation to detenllnc the gas veJocity in a pipeline is obtained as follows: 1 US Q,