PIPELINE ENGINEERING

Transcription

1 Mechanical Energy Balance PIPELINE ENGINEERING FLUID FLOW V g z + vdp + W o = F (1-1) potential energy expansion work Ketic energy Work added/ Sum of friction change change subtracted by losses compressors or pumps/expanders Note that the balance is per unit mass. In differential form gdz + vdp + VdV = δwo δf (1-) Rewrite as follows Divide by dl (L is the length of pipe) ( ) dp = ρ g dz V dv δf + δ W o (1-3) dp dl Tot g dz V dv F W = ρ + o dl ρ + dl ρ δ δ ρ δ L δl (1-4) or: ( δ W δl o dp dp dp dp = + + (1-5) dl dl dl dl Tot elev accel frict is usually ignored, as the equation applies to a section of pipe) The above equation is an alternative way of writg the mechanical energy balance. It is not a different equation. The differential form of the potential energy change is

2 dl φ dz g dz dl = g s φ (1-6) Friction losses: We use the Fanng or Darcy-Weisbach equation (Often called Darcy equation) δf = V f D dl (1-7) an equation that applies for sgle phase fluids, only (two phase fluids are treated separately). The friction factor, turn, is obtaed from the Moody Diagram below. Figure 1-1: Moody Diagram Friction factor equations. (Much needed the era of computers and excel) Lamar Flow f = 16 Re (1-8)

3 Turbulent Flow f = Ṙe (1-9) a smooth pipes: a=0. Iron or steel pipes a=0.16 Turbulent Flow 1 ε 51. = log10 + f 37. D Re f (Colebrook eqn) (1-10) Equivalent length of valves and fittgs: Pressure drop for valves and fittgs is accounted for as equivalent length of pipe. Typical values can be obtaed from the followg Table. Table 1-1: Equivalent lengths for various fittgs. Fittg Le D 45 O elbows O elbows, std radius 3 90 O elbows, medium radius 6 90 O elbows, long sweep 0 90 O square elbows O close return bends O medium radius return bends 50 Tee (used as elbow, enterg run) 60 Tee (used as elbow, enterg branch) 90 Gate Valve (open ) 7 Globe Valve (open ) 300 Angle Valve (open) 170 Pressure Drop Calculations Pipg is known. Need pressure drop. (Pump or compressor is not present.) Incompressible Flow a) Isothermal (ρ is constant) dp dz dv df =- g +V + dl ρ dl dl dl Tot (1-11) for a fixed φ V constant dv = 0 3

4 δ δ L = D F V f (1-1) L p = ρ g Z + V f + F D (1-13) b) Nonisothermal It will not have a big error if you use ρ(t average ), v(t average ) Exercise 1-1: Consider the flow of liquid water (@ 0 o C) through a 00 m, 3 pipe, with an elevation change of 5 m. What is the pressure drop? Can the Bernoulli equation assumg compressible flow be used for gases? The next figure illustrates it. 4

5 Figure 1-: Error Bernoulli equation In conclusion, if p p p usg the assumption of compressibility is OK. Compressible Flow (Gases) a) Relatively small change T (known) 5

6 For small pressure drop (somethg you can check after you are done) can use Bernoulli and fanng equation as flows Then V gdz +vdp+d = -df (1-14) but G V=v A, where g 1 V df dz + dp+ dv = - v v v v (1-15) V = Velocity (m/sec) v = Specific volume (m 3 /Kg) G = Mass flow (Kg/sec) A = Cross sectional area (m ) Then, g 1 G dv df G dl dz + dp+ = - = -f v v A v v A D (1-16) Now put tegral form g dz dp G dv G fdl v v = A V (1-17) A D Assume T +T T av = (1-18) pav = p + p 3 p p p + p which comes from p av = p pdp p dp (1-19) f = av f(t,p )+ f(t,p ) (1-0) The tegral form will now be 6

7 ρ av dp G V G g z+ + v = A V A L ln f av (1-1) D Now use p v = Z R T M pavm, where M: Molecular weight. Then ρ av =, which leads to: Z RT av av dp M M ρ v Z RT Z RT p av = p dp = ( p p ) = ( p p ) (1-) av av av av av Therefore; ρ ρav V avg z+ ( p p ) + ln = fav av G G L p A V A D (1-3) but, V V Z = Z T T p (1-4) p Then ρ ρav Z T p avg z+ ( p p ) + ln = fav av G G L p A Z T p A D (1-5) To calculate Z av Kay s rule is used. This rule states that the reduced pressure and temperature of the gas is obtaed usg the average pressure and temperature (as above calculated) and a pseudo critical pressure and temperature. pav pr = (1-6), pc Tav Tr = (1-7), T C In turn the critical pressure and temperatures are obtaed as molar averages of the respective components critical values., p = yp (1-8) C i C, i i, C yt i C, i i With these values the Z factor comes from the followg chart: T = (1-9) 7

8 Figure 1-3: Natural Gas Compressibility Chart Equation (1-5) can be further simplified. First neglect the acceleration term because it is usually small compared to the others, to obta: 8

9 ρ ρav G avg z ( p p ) fav av L + + = 0 p A D (1-30) Form this equation we can get G, as follows: G M p av 5 g z π D M ( p p ) ZavRT av = 3 favl ZavRTav ZavRTav (1-31) G But the volumetric flow at standard conditions is given by psq= ZsRTs where the M subscript s stands for standard conditions. Therefore: Q ( ) av p p g z 5 R Zs T s ZavRT av D Mps ZavTavL fav π = 3 M p (1-3) Now, if z =0, we get Q ( p p ) π R Z T D = 3 5 s s Mps ZavTav L fav (1-33) which can be rearranged as follows: p p = K Q (1-34) 64 MpsZavTav fav where K = L 5 π R Z T D and is known to be W L, a product of a resistant factor W s s 64 Mps ZavTav fav times the length L. With this, we have W =. 5 π R Z T D s s To calculate pressure drop we recognize that average pressures are a function of unknown. Then we propose the followg algorithm: p, which is a) Assume b) Calculate (1) p and calculate K 64 (1) p av Mp Z T f ( i) ( i) ( i) () i s av av av = 5 π R Zs Ts D (i + 1) ( i) p = p K Q b) Use formula to get a new value 9

10 d) Contue until p - p (i+1) (i) p (i) ε Dependg on the choice of friction term expression, several formulas have been reported for equation (1-34). They are summarized the followg table. Table 1-: Different forms of compressible flow equations 10

11 Table 1- (contued): Different forms of compressible flow equations Exercise 1-: Natural gas (84,000 std m 3 /hr at 49 atm and 38 o C) is sent from a gas refery to a city, through a 16 pipele. The distance is 170 Km. The gas reaches the other end at ground temperature, (5 o C). The gas to have the followg molar fractions: Methane: 98%, ethane: 1.%, propane: 0.75%, and water: 0.05%. We also assume Re~ and ε/d = As a first approximation, we recommend usg the Panhandle A equation: p p = K Q W= T s /D (Wilson G.G., R. T. Ellgton and J. Farwalther, 1991, Institute of Gas Technology Education Program, Gas distribution Home Study Course), where s is the gas gravity (=M gas /M air )=0.65 for natural gas), T is o R and D ches) What is the pressure drop? 11

12 Heat Transfer Effects To account for temperature changes due to heat transfer, we use total energy balance gdz +d( vp ) +VdV + du = δ q + δ wo (1-35) where the followg is identified: - Potential energy change: g dz - Rate of work done on the fluid element by pressure forces: d(vp) - Ketic energy change: VdV - Internal energy changes: du - Heat transfer: δ q. This is given per unit mass flowg (Kcal/h)/(m 3 /h) - Work added: δ wo. This term is due to pumps and compressors. Sce we will treat these separately, this term is usually set to zero for pipes. But the heat δ q is given by teractions with the ambient surroundgs: π D δ q= U( To T) dl (1-36) G where U is the heat transfer coefficient, T o is the side pipe temperature, π DdL = da (see next figure) and G is the flowrate. Figure 1-4: Area element Then, (ignorg δw o because there are no pumps) to get: ( ) V U To -T DdL gdz +dh+d = G (1-37) Integrate and solve for h (use T av the heat transfer equation) ( ) U To Tav π DL V V h = h + g z z G ( ) 1 (1-38) But G V v A Z RTav G = = av (1-39) p M A 1

13 Fally, to obta the let temperature, one would need to obta it form the enthalpy and pressure the let The procedure suggested is then: a) Assume T, p b) Use mechanical energy balance to obta (1) c) Use total energy balance to obta h (1) d) get temperature T e) Go to b) and contue until convergence T =T ( p,h ) (1-40) (1) p SCENARIO II One has a turbe or Compressor/pump and needs W o. We use total energy with δq = 0 and dz =0 Integratg, one obtas: w o dv dh = δ wo - (1-41) W V = = h+ G (1-4) In this expression, we have w o given Joules/Kg, W Joules/sec and h Joules/Kg. Thus, the work of the compressor/pump is given by: V W = G h+ (1-43) For compressors, W is positive, while for turbes, it is negative. However, h is known for liquids because enthalpy does not vary much with pressure. In addition, there isn t much temperature change pumps). However, for gases, h is much harder to obta. Therefore we go back to the Mechanical Energy equation for pumps/compressors. Indeed, the Bernoulli equation gives V W = G vdp+ G vdp (1-44) where the acceleration term has been neglected. For pumps, the density is constant, so one obtas: 13

14 W = G p (1-45) ρ For compressors, one needs to obta an expression of volume knowg that the evolution is isentropic (or nearly isentropic). Thus, pv n = constant (n=cp/cv for ideal gases n>cp/cv for n real gases). Substitutg v = ps n vs ρ tegrate to get n 1 n n p W = G pv 1 n+ 1 p (1-46) The above expression does not clude the compressibility factor. A better expression, which cludes the efficiency, is n 1 n n Z + Z 1 p W = G RT 1 n+ 1 η a p (1-47) The efficiency factor is usually between 60 to 80% and normally given by the manufacturer. One expression for such factor is: η = a T p p T n 1 n T 1 (1-48) Fally, the let temperature is obtaed from pv = p v (1-49) n n Usg the gas law to obta v / v terms of temperatures and pressures, substitutg and rearrangg, on e obtas: n 1 n n 1 p n T T = = p [ CR] (1-50) where CR is the compression ratio. Normally, manufacturers recommend not exceedg 300 o F at the let. 14

15 Exercise 1-3: The natural gas of exercise 1-, is available origally at atm. Calculate the compression work needed to reach delivery pressure (49 atm) usg one compressor. Calculate the let temperature and determe the duty needed to cool the gas down to the correspondg let conditions. Is it acceptable to use one compressor? We now discuss the compression ratio. This is limited compressors to the range 1. to 6. Extra compressors should be added if the CR >6, and after-coolers need to be added to control the temperature. If more than one compressor is to be used, the practice is to use the same CR for all. Exercise 1-4: Consider two compressors. - Write the power expression for each one assumg the gas is cooled down to its let temperature after compression. - Add both expressions to obta the total work as a function of the termediate pressure (the rest should not be a variable) - Take first derivative and obta the desired result that CR 1 =CR Exercise 1-5: Obta the set of compressors needed to compress the gas of exercise 1- properly, that is, limitg the temperature and usg the right CR. 15

16 Two Phase Flow Two phase flow has several regimes, which are depicted the next figure: Dispersed Annular Stratified Froth Wavy Slug Plug Bubble The two extreme cases are: Figure 1-5: Two phase flow regimes - Bubble: Vapor and Liquid Equilibrium (Benzene 40%, Toluene 60%) - Dispersed: Liquid and gas (air and benzene). The latter is common gas pipeles; the former is common crude pipeles, especially light crudes. 16

17 One important thg to recognize is that except for the extreme cases, the phases travel at different velocities. Typical velocities are shown the next table: Table 1-3: Typical velocities of two phase flows REGIME LIQUID VEL(ft/sec) VAPOR VEL.(ft/sec) Dispersed Close to vapor > 00 Annular <0.5 > 0 Stratified < Slug 15 (But less than vapor vel.) 3-50 Plug < 4 Bubble To predict the flow patterns, one needs to use the Baker Plot (next Figure) for horizontal pipes (there is a similar one for vertical pipes). Figure 1-6: Two phase flow regimes transitions Wg W L In this diagram, we have Gg =, Gl = A W λ, (W lb/h, A ) which are the g superficial velocity of the vapor and the liquid, respectively. In turn, the parameters are given by λ = ρρ L g (with densities given lb/ft 3 1/ µ ) ψ = L /3 σ ρ (with the surface tension L given dyn/cm and the viscosity cp) We note that: 1) λ and ψ depend on the fluid property only ) G l depends on the ratio of flows (Known beforehand. Not a design parameter) 17

18 3) G g depends on the vapor/gas superficial velocity. It can be modified changg the diameter 4) Transition boundaries are not at all that sharp. From this diagram, we notice that followg change of regimes a pipe. As the pressure drop is large, then the density of the vapor is lower. 1) λψ ρ g Glλψ ρg Abscissa decreases Gg 1 ) Ordate creases 1 1 λ ρ g λ ρ g Thus trajectories are always "up" and "to the left". Thus a bubbly flow may become, plug, slug or annular, an annular may become wavy or dispersed, dependg on the startg position the plot, and so on. PRESSURE DROP Lockart and Martelli (1949) developed one of the first correlations. It is based on multiplyg the pressure drop obtaed by considerg the vapor phase occupyg the whole pipe, by a factor ptwophase = φ pvaporphase (1-51) b In turn, the correction factor is given byφ = ax, where gives some typical values of the correspondg constants: X pliquidphase =. The followg table p VaporPhase Table 1-4: Constants for Lockart and Martelli s correlation a b Bubble Slug Stratified (horizontal) Plug Annular D() D() We notice that there are several more modern correlations, which will be explored later. In turn, the pressure drop due to gravity, is given by dp = ε gρg + (1 εg) ρ l g sθ dl gravity (1-5) where ε g is the (void) fraction of gas. We omit the pressure drop due to acceleration. 18

19 Hydrate Formation Hydrates are crystalle structures between water and hydrocarbons. One typical example is given the figure below: Cage of water molecules CH 4 molecule the center Figure 1-7: Methane Hydrate The next figure shows the Pressure-Temperature diagram of water-hydrocarbon systems. Curve 1-1 represents the curve for vapor pressure of the hydrocarbon. Figure 1-8: Generic Hydrate P-T diagram 19

20 The next figure shows some specific cases of hydrocarbons: Figure 1-9: Hydrate P-T diagram for various hydrocarbons Clearly, high pressure pipeles, favorable thermodynamic conditions for hydrate formation can be encountered. It is therefore important to keep md that these conditions need to be avoided. These hydrates can be prevented from formg through heatg, pressure change (not a choice pipeles) and the troduction of hibitors. These hibitors are salts, alcohols, glycols, ammonia and MEA. The most widely used is methanol. The next figure shows the depression of hydrate formation temperature observed for various hydrates. Figure 1-10: Hydrate temperature formation depression 0

21 Pipele Costs Historical pipele and compressors stalled cost data were obtaed from the Oil & Gas Journal special report on Pipele Economics, September 3, 001. Pipele per mile cost distribution for different pipe diameters and compressor stalled cost for different horsepower requirement are plotted the followg figure. All cost figures are updated to 005 dollars usg Marshal & Swift cost dexes y = 43.x Figure 1-11: Pipe average cost (k$/mile) vs. ID y = 1.65x Figure 1-1: Compressor cost (k$) vs. horsepower Fixed Capital Investment were calculated by addg the stalled cost of a pipe length (assumed 5000 miles) and the cost of all required recompression stations. The Fixed Capital Investments obtaed are then divided by the pipe length to obta a per mile cost profile for different flow rates. The curve the next figure shows that this cost profile takes a logarithmic shape. 1

22 B$ B$ B$ y = x B$ B$ B$ B$ 0.00 B$ 0.00 B$ B$ B$ BSCFD Figure 1-13: Pipele fixed cost (b$/mile) vs. capacity (BSCFD) A lear correlation gives the followg form: FCI ( B$ / mile) = * Capacity( BSCFD) Operatg costs for pipeles were estimated as follows; an average of 5 operators is assumed to be the requirement for each compression station, with an hourly wage of $1. Direct supervisory and clerical labor is assumed to be 0% of operatg labor. Compressor fuel requirement is estimated at 8,000 Btu / BHP-HR, and fuel cost at $.5 per million Btu. Matenance cost is assumed to be 7% of the FCI for compressors and 3% for pipes while surance is 1% for compressors and 0.4% for pipes. Operatg cost per pipele mile versus capacity is plotted the next figure: B$ B$ B$ y = 7E-05x + 4E-05 B$ B$ B$ Figure 1-14: Pipele per mile annual operatg cost vs. capacity This estimate should be reasonable with ab 40% accuracy. Similar lear approximation to that of the FCI is assumed. Lear regression was used to estimate the operatg cost

23 dependence of the capacity, ignorg capacities less than 100 MMSCFD. This gave a general correlation of the followg form: Oper. Cost( B$ / mile / year) = * Capacity( BSCFD) Exercise 1-9: Consider the pipele of Exercise 1-8: - Vary the pipele maximum pressure (100 psia) to some lower and higher value. Adjust the diameter accordgly and calculate the number of recompression stations. - Calculate the cost. Can you say that 100 psia is the right pressure? Pipele Loopg Pipele loopg is the practice of designg pipeles with segments run parallel. This practice creases the pipele flow capacity with alterg the fal pressure. If temperature is close to ambient temperature, the location of a loop does not change the fal delivery pressure. However, when temperature changes substantially, then the location of a loop has an fluence. Thus, these cases, for example, it is recommended to loop the upstream region, where the gas is hotter. This allows the gas to cool down faster and therefore crease the delivery pressure. Consider the followg example: A 100 Km length (0 OD) pipele is used to send 89 MMSFD at an let pressure of 1,00 psia and a temperature of 45 o C. The pipe roughness is 750 µ ches, and a soil temperature of 10 o C. Three alternatives were studied for this pipele. a) No loopg, b) Loopg the first 5 Km, and c) Loopg the last 5 Km. The results of a simulation are shown the next figure: Figure 1-15: Results from Loopg 3

24 Exercise 1-10: Verify the results of figure 1-15 usg the simulator. Retrograde condensation One very common phenomenon pipeles is retrograde condensation. Consider the P-T plot of the next figure. It corresponds to a gas with the followg composition: Methane: %, Ethane: 3.5%, Propane: 0.585%, n-butane: 0.16%, i-butane: 0.11%, pentane: 0.055%, i- pentane 0.05%, hexane: 0.09%, heptane: 0.04%, octane: 0.03%, nonane: 0.01%, CO : , N : 1.34%. Assume a 15, 00 Km pipele starts at 60 atm and 15 o C. If the external temperature is 5 o C (U=1 BTU/hr-ft o F), then it is clear that there will be liquid formation this pipele, even if the operation is isothermal. Figure 1-16: P-T diagram of example gas and retrograde condensation Interestgly, if the pressure at the other end is low enough, then the liquid might vaporize aga. This means that the pressure drop regime side the pipe might change and one has to be careful performg the simulations. Exercise 1-11: Generate the answers for the above example usg the simulator. Change the pipe diameter and the length to verify the statements. 4

25 Pipele Optimization Process J-Curve Analysis Conventional Pipele design methods, which rely mostly on hand calculations or at best on simple spreadsheet suggest that the compressor size and the pipe diameter be varied and the cost of service ($/(m 3 *Km) for the first year be plotted as a function of flowrate. Typical assumptions are that there is no volume buildup the pipe, the time value of money is neglected and that the facilities are designed to susta the flows. For example, Figure 1-17 shows one such exercise performed for three different diameters and parametric at different maximum operatg pressures (MOP) and compression ratios. Efficient operatg ranges that are flat are preferred Figure 1-17: J-Curves for various diameters Exercise 1-1: Expla why J-curves go through a mimum. 5

26 Optimization Parameters J-Curves are a simplistic first approach but one that can provide a first approximation to the right diameter and compressors. Thus one needs to establish - Re: In most cases this is defed by a variety of other factors and given to the designer. - Pipele Initial Capacity: Most pipeles are constructed takg to account the fact that demand at the receivg end(s) will crease through time. Thus, one is faced with the decision of designg for future capacity and underutilize the pipele for some time or design for current or more short term capacity and use loops to expand later. - Expansions: If capacity expansions are considered, then they need to take place through loopg. Not only the new loop has to be designed, but its timg and capacity be selected. - Maximum operatg pressure: This choice has already been considered constructg the J-Curves. However, more complex situations, one is faced with multiple delivery pots with different delivery pressures, etc. - Pipe Size: This choice has already been considered the J-Curve selection but needs to be revisited anyway view of the fluence of the other factors. - Load Factor: This factor is the ratio between the average daily volume delivered divided by the peak volume. If this ratio is too small, then storage facilities for ventory holdg (salt caverns, underground caverns, abandoned reservoirs, etc if available, or large LNG or high pressure, CNG, storage tanks) are more convenient than more powerful compressors and larger diameters. The issue to resolve is when these are substituted by ventory holdg sites. In addition, the question remas where these holdg sites should be located. - Compressor Station Spacg: While an earlier exercise suggests that when multiple compressors are used it is best to keep the compression ratio equal, this hypothesis needs verification. Reciprocatg compressors are chosen when the power requirement is smaller than 5,500 Kw. Compression ratios recommended for centrifugal compressors are generally the range of 1.5 to 1.35 and smaller than 1.5 for reciprocatg compressors. Exercise 1-13: Consider the pipele of Exercise 1-8: - Assume two compression stations will be used. - Determe (by spection and usg a simulator what is the best compression ratio for each compressor. Aim at mimizg total work only. - Heatg and Coolg: Clearly, coolg leads to significant savgs because pressure drop is reduced. However, money needs to be spent to stall and run the coolers. Thus the trade-off needs to be resolved. 6

27 Exercise 1-14: Consider the shown the followg figure Supply: 5,7,000 m3/d 638 Km Km 650 Supply: 10,407,840 m3/d,148,00 m3/d,83,000 m3/d 18,40 m3/d 6,595,00 m3/d 336,,000 m3/d 384,000 m3/d 134,400 m3/d 3,617,400 m3/d 64,600 m3/d - The pipg is the ground and is not sulated. Assume a ground temperature of 5 o C and a ground conductivity of 0.7 W/(m o C). The gas elevation profiles are provided the followg table: Km Elevation (m) The gas ((1.9% methane, 5% Ethane, % propane, 1% n-butane and 0.1% n- pentane) is supplied at the two pots dicated the diagram at 1,367 kpa and 35 o C the first station (Km 0), and 150 kpa and 30 o C the second (Km 143). - Determe usg simulations a) Pipg diameter, b) Compressors at the supply station, c) coolg required. Do not use a pressure above 5,600 Kpa. Use cost data provided above. - Will new compressors be needed/beneficial? 7

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