PTE 526 NATURAL GAS ENGINEERING - PowerPoint Presentation

1. PTE 526 NATURAL GAS ENGINEERINGBYENGR. J. O. OWOLABICHEMICAL AND PETROLEUM ENGINEERING DEPARTMENT AFE BABALOLA UNIVERSITY ADO-EKITI,EKITI STATE, NIGERIA

2. LEARNING MATERIALS1. Natural Gas Production Engineering, C.U.Ikoku.2. Gas Production Operations, H. Dale Beggs, Gulf Pub. Company, Houston, Texas, USA. 1984.3. Gas Processing and Conditioning, Vols 1 & 2, J.M. Cambell.

3. PHYSICAL CONSTANTS OF HYDROCARBON GASES

4. COMPRESSIBILITY FACTOR CHART

5. PSEUDO-CRITICAL PROPERTIES OF GASES AND CONDENSATES

6. WORKED EXAMPLE 1Calculate the specific gravity and the gas deviation factor of Salford Field gas from its composition. Reservoir pressure and temperature are 3250 psia and 213oF respectively.ComponentMol. FractionMol. WtPcTcMethane0.861216.04673343Ethane0.059130.07708550Propane0.035844.09617666Butane0.017258.12550766Pentanes0.005072.15400846CO20.001044.011070548N20.020728.02492227TOTAL 1.0000NOTE: The Tc is in Rankin = oF + 460

7. SOLUTIONComponentMol Fract (xi)Mol. Wt(Mi)MixiPcxiPcTcxiTcMethane0.861216.0413.81673579.59343295.39Ethane0.059130.071.7870841.8455032.51Propane0.035844.091.5861722.0866623.84Butane0.017258.121.005509.4676613.18Pentanes0.005072.150.364002.458464.23CO20.001044.010.0410701.075480.55N20.020728.020.5849210.182274.70Total1.000019.15666.67374.40

8. SOLUTION The specific gravity of the gas: γg = 19.15 /28.997 = 0.661 The pseudocritical pressure and temperature are 666.67 psia and 374.40 R Pr = 3250 / 666.67 = 4.87 Tr = 673 / 374.40 = 1.80 The gas deviation factor (z) from chart = 0.91

9. DETERMINATION OF VISCOSITY By CorrelationsCompute AMW of the gas yiMiDetermine the µgi of the gas mixture at 1 atmosphere using chart. If the pressure of the mixture is above 1 atm thenCalculate pseudo critical temperature/pressureCalculate reduced temperature/pressureDetermine the viscosity ratio (g/ gi) from chartCalculate gIf the gas contains non-HC components, then they should be corrected using chart.

10. VISCOSITY CHART FOR 1 ATM

11. VISCOSITY CHART FOR 1 ATM

12. EXERCISE 1Calculate the value of the viscosity for the dry gas with a specific gravity of 0.818 at reservoir temperature of 220oF and reservoir pressure of 2100 psig.

13. ESTIMATION OF GAS RESERVESThree methods are availableVolumetric Method Material Balance Method Pressure Decline MethodVOLUMETRIC METHODThis is applied in a new field for rough estimates. No production history is required. We need only geologic data like porosity, water saturation etc.

14. Volumetric Method ContdWhere: A = Area of the reservoir (Acres) h = Formation thickness (ft)  = Porosity Swi = Initial water saturation Bgi = Initial gas formation volume factor G = Initial Gas in place

15. Volumetric Method ContdGas produced at any given condition (Gp) isRecovery Factor

16. This is used for reservoir that has produced long enoughThere are two cases:Without water production/influx With water production/influxMATERIAL BALANCE METHOD

17. Material Balance in Gas ReservoirsThe conservation of mass may be applied to gas reservoirs to give the following MB:The MB may be made in terms of moles of the gas as:If: Vi = initial gas pore volume in cu.ft, pi = final pressure after producing Gp SCF of gas We = water encroached into the reservoir (cu.ft)

18. We = water encroached into the reservoir (cu.ft)Wp = water produced (cu.ft)Vf = final volume after producing Gp SCF of gasThen,Replacing eqns (1) and (2) by their equivalents using the gas law, we have Or

19. But,Hence,Replacing Vi by its equivalent G/Big the equation becomes:Dividing through by Bgi and expanding

20. If Bgf and Bgi are expressed in cu.ft/SCF instead of SCF/cu.ft, they will be in the numerator and the equation reduces to:GpBgf is the volume of the produced gas at the pressure pf, G(Bgf – Bgi) is the change in volume of the initial gas when expanded from pi to pf and We and BwWp are the volumes of water influx and production respectively.

21. The effect of connate water expansion and formation compaction can quite easily be incorporated into the gas reservoir MB. It has already been shown that :Where HCPV = GBgiThe MB for gas reservoir becomes:

22. For volumetric reservoir the production volume equals the expansion volume and neglecting the effects of water expansion and pore volume reduction, eq (7) becomes:

23. PRESSURE DECLINE METHODA plot of P/Z versus Gp will produce a straight line of slope (TPb/TbVi) and intercept at Gp = 0. Note that the reservoir gas volume can be put in units of scf by the use of Bg.Thus Vi = GBgi Thus both G and Pi can be obtained graphically.

24. WORKED EXAMPLE 2Calculate the water influx and residual gas saturation in water-drive reservoirs. Given that:Initial Bulk reservoir volume = 415.3 cu.ftAverage porosity = 0.172, Average connate water saturation= 0.25Initial pressure = 3200 psia, Bgi = 0.005262 cu.ft/SCFFinal pressure = 2925 psia Bgf = 0.005700 cu.ft/SCFCumulative water production = 15200 STB, Bw = 1.03 bbl/STB Gp = 935.4 MM SCF, Bulk volume invaded by water at 2925 psia = 13.04 MM cu.ft

25. SOLUTIONFrom Eq (5), This much water has invaded 13.04 MM cu.ft of bulk rock, which initially contained 25% connate water. Then the final water saturation of the flooded portion of the reservoir is:

26. SOLUTION CONTDAnd the residual gas saturation Sgr = 32%.

27. COMPRESSION OF NATURAL GASAdd energy to the gas. The type of compression station will depend on the purpose for which the gas will be used. For field/gathering stations (0 – 750 psig). Between field and transmission lines (1300 psig). EOR (6000 psig).Storage field station > 4000 psig. Distribution units (20–100psig)Types of compressorsPositive Displacement Compressors - Reciprocating Compressors - Rotary BlowersContinuous flow compressors - Centrifugal Compressors - EjectorsReciprocating compressors are the most commonly used in gas industry

28. COMPONENTS OF COMPRESSORPistonCylinderSuction/Discharge ValvesConnecting RodImpellerShaftDiffuserVoluteCompression CycleIdeal CycleActual Cycle

29. MULTISTAGE COMPRESSOR ARRANGEMENT Compressor Ratio r = 123PsPdFor design purposes r < 6For practical purposes r  4- Optimum Number of stages is given byWhere: Pd = discharge pressure (psia) Ps = suction pressure (psia) n = No of stages required

30. COMPRESSOR DESIGNThis involvesThe determination of compressor capacityThe determination of power requirementsDetermination of compressor capacityWhere q = Flow capacity (ft3/min, scfd) d = Piston diameter (ft) L = Stroke length (ft) S = Compressor speed (rpm) Ev = Volumetric efficiency

31. The volumetric efficiency is calculated fromZ1 = gas derivation factor at suction conditionZ2 = gas derivation factor at discharge conditionK = Cp/Cv = Isentropic exponentr = compression ratio = p1/p2A = factor to allow for leakage, friction usually between 0.03 and 0.06.C = clearance, which varies from 0.04 to 0.16

32. DETERMINATION OF COMPRESSOR HORSEPOWERTheoretical HP is determined by 3 ways Analytical method Mollier diagram Quickie estimate

33. WORKED EXAMPLE 3 A single acting reciprocating compressor having a piston diameter of 4 inches and a stroke length of 6 inches is to be used to compress gas from 100 psig and 100oF to 400 psig. Calculate the flow capacity in ft3/min and scfd if the compressor runs at a speed of 500 rpm. Given that : k= 1.3, C = 6% , A = 5%, T2 = 753oR, Z1 = 0.98, Z2 = 0.95

34. SOLUTIONr = p1/p2 = 414.7/114.7 = 3.62Z1/Z2 = 0.98/0.95 = 1.03 Ev = 1 – 0 .05 – 0.06 [1.03 (3.62)0.77 – 1] = 0.85 q = [π(4/12)2(6/12)(500)(0.85)]/4 = 18.54 ft3/minTo convert to ft3/day = 18.54 ft3/min x 1440 min/day = 26704 ft3/dayqsc = q/BgBg = 0.0283 ZT/P = 0.0283(0.98)(560)/114.7 = 0.135 ft3/scfqsc = 26704/0.135 = 1.97 x 105 scfd

35. Where: W = Power required HP/MMscf Pb = Pressure at standard condition (psia) Tb = Temperature standard condition (oR) T1 = Suction Temperature (oR) Zs = Gas deviation factor at suction condition Analytical Method

36. Mollier Diagram Method W = 0.0432 Hwhere W = Power required (HP/MMSCFD) H = Enthalpy change (BTU/Ib-mol)Note: Enthalpy-Entropy diagrams apply only to the gaseous state. If a gas is cooled below its dew point, condensation occurs and the heat removed can not be determined directly from the diagram.

37. Mollier diagrams method Contd

38. WORKED EXAMPLE 4Calculate the horsepower required to compress 1 MMscf of a 0.6 gravity gas from 100 psia and 60oF to 1600 psia using the analytical method. Assume that a two stage compressor is used and that the gas is cooled back to 60oF between stages. Also find the final discharge temperature. Other data are: K = 1.28, Pb = 14.65, Tb = 60oFUse the Mollier diagram method to solve the above example.

39. SOLUTION (Analytical Method)The total compression ratio is 1600/100 = 16. Hence, r = 4 for each compressor stage. First Stage:P1 = 100, P2 = 400, Z1 = 0.985, Z1(k – 1)/k =0.22 w1 = 75.8 Hp/MMscfd

40. Solution (Analytical Method) ContdThe gas temperature after compression from 100 to 400 psia is:The gas is cooled to 80oF before entering the second stage.Second StageP1 = 400, P2 = 1600, Z1 = 0.94, Z1(k – 1)/k =0.21

41. Solution (Analytical Method) Contdw2 = 71.6 hp/MMscfd T2 = Tfinal = 273oFThe total power requirement is:Hp = q(w1 + w2) = 1 MMscfd (75.8 + 71.6) = 147.4

42. SOLUTION (Mollier Diagram Method)First Stage:The first stage compresses the gas from 100 to 400 psia at constant entropy. Enter p1=100 and T1= 80oF and read H1= 380 Btu/Ib-mole. Proceed vertically (ΔS = 0) to p2 = 400 psia and read H2= 1990, T2 = 260oF. Follow the 400 psia line to T3 = 80oF and read H3 = 220. The power required for first stage is:w = 0.0432 ΔH1-2 = 0.0432(H2 – H1)w = 0.0432 (1990 – 380) = 70 Hp/MMscfd

43. Solution (Mollier Diagram Method) ContdSecond Stage:From point 3 where p = 400, T = 80 and H = 220 proceed vertically to the final pressure of 1600 psia and read T4 = 274oF, H4 = 1920 Btu/Ib-mole.The power required for the second stage is:w2 = 0.0432 (H4 – H3) = 0.0432 (1920 – 220) = 73.4 Hp

44. The total horsepower required is then:Hp = q(w1 + w2) = 1 MMscfd (70 + 73.4) = 143.4

45. WORKED EXAMPLE 5What is the amount of heat that must be removed in cooling 1 Ib-mole of 0.6 gravity gas at 200 psia from 600oF to 100oF ?If this gas is then heated from 100oF to 300oF, how much heat is required per Ib-mole of gas?

46. SOLUTIONAt 200 psia/600oF, H = 6100 Btu/Ib-moleFollow constant pressure to 100oF, H = 500 Btu/Ib-mole.Heat removed = 6100 – 500 = 5600 Btu/Ib-mole.At 200 psia/300oF, H = 2600 Btu/Ib-moleHence Heat Added = 2600 – 500 = 2100 Btu/Ib-mole.

47. QUICKIE CHARTSUsed for making quick estimates. They give higher values than other methods. Should not replace accurate methods. bHp =ORbHp =Where V = Inlet capacity of compressor (mmcfd) q = Inlet capacity of compressor (cfm) Pb = Standard pressure (Psia) Tb = Standard temperature (oR) T1 = Inlet temperature of compressor (oR) (bHp/MMcfd) = Factor determined from chart bHp = Power requirement (HP)

48. NATURAL GAS METERINGTo determine volumetric flow rateTo determine pressure loss for a particular flow rate Equipment used Orifice meter Turbine meterPilot tubeCritical flow prover

49. CHOICE OF MEASURING EQUIPMENTFactors to consider in the choice of measuring equipment are:VolumeAccuracyUseful lifeRange of flowTemperatureMaintenance requirementAvailability of powerNature of fluidCost of operationSafety of the device

50. ORIFICE METERINGMeans of measuring the PD caused by a change in velocity of the gas as it passes through a restriction placed in the pipe.Gas flow rate in scf/hr is given as: Where: qsc = gas flowrate scf/hr C = Orifice constants hw = differential press across the orifice (inches H2O) pf = flowing pressure (Psia) and C = Fb *Fpb *Ftb *Fg* Ftf*Fr*Fpv*Fm*YThe orifice constant depends primarily on the basic orifice factor Fb. Many of the other terms are negligible or essentially equal to one.

51. Basic Orifice Factor (Fb):The value of Fb is found from tables. Two standard are provided in gas measurement – flange taps and pipe taps.Pressure-Base Factor (Fpb): The Fpb factor corrects the value of Fb for cases where the pressure base used is not 14.73 psia. It may be determined by the equation Fpb = 14.73/Pb.Temperature-Base Factor (Ftb):The Ftb factor corrects for any contract wherein the basic temperature is not 520oR (60oF). This factor may be computed by the formula Ftb = Tb/520.

52. Specific-Gravity Factor (Fg):The Fg is to correct the basic orifice equation for those cases where the specific gravity of the gas is other than 1.00. The equation is Fg = γg-0.5 . Flowing Temperature Factor (Ftf): For flowing temperature correction if the flowing temperature of gas is other than 600FReynolds Number Factor (Fr): It accounts for variation of the discharge coefficient with Reynolds number. Value obtained from chart

53. Supercompressibility factor (Fpv): It corrects the variation from the ideal gas law: Expansion Factor (Y): It accounts for change in gas density as the pressure changes across the orifice. The change is usually small and therefore ignored Manometer Factor (Fm): For correction when mercury-type manometer is used. It accounts for different heads of gas above the two legs of the manometer. Usually small value and therefore insignificant.

54. WORKED EXAMPLE 6A meter run that is equipped with flange taps and a 3000 inch orifice has an inside diameter of 6.065 inches. The static pressure obtained from the downstream tap reads 80 psia and the average differential pressure is 49.5 inches of water. If the pressure and the temperature bases are 14.9 psia and 60oF respectively, calculate the flow rate in standard cubic feet per hour. The gas gravity is 0.60 and the flowing temperature is 65oF.

55. SOLUTIONCalculate C’ = Fb *Fpb *Ftb *Fg* Ftf*Fr*Fpv*Fm*YFrom tables, for d= 3.000 and D= 6.065,Fb= 1891.9Fpb= 14.73/Pb = 14.73/14.9 = 0.989Ftb= 1.000Fg= (γg)-0.5= (0.6)-0.5 = 1.291Ftf= (520/Tf)0.5 = (520/525)0.5 = 0.995

56. From the tables, b= 0.0332Fr= 1+ b/(hw*pf)1/2= 1+ 0.0332/(49.5*80)1/2= 1.001For γg = 0.6, T = 65oF, P= 80 psia, the value calculated for Z is 0.990.Fpv= (Z)-0.5 = (0.990)-0.5 = 1.005Fm= 1.000To determine Y, β= d/D = 3.000/6.065 = 0.495 hw/Pf = 49.5/80 = 0.619

57. SOLUTIONFrom the tables, Y = 1.0037 (requires interpolation)C’=1891.9(0.989)(1)(1.291)(0.995)(1.001)(1.005)(1.000)(1.0037) = 2426.9qsc= C’√(hwPf) = 2426.9√(49.5)(80) = 152,721 scf/hr = 3.665 MMscfd

58. WORKED EXAMPLE 7A metering system is required to measure approximately 8.5 MMscfd of 0.62 gravity gas at a line pressure of 250 psig. The meter run is to be made of 8 inches pipe (7.981 in ID). Determine the size of the orifice plate to give a differential of about 50 inches. Flowing temperature averages about 80oF.

59. SOLUTIONFor hw = 50,For an approximation, all the terms that make up the value of C’ will be ignored in this case, except Fg and Ftf.From the Fb table for flange taps, for D = 7.981, we have: d=3.500 for Fb of 2537.7 and d= 3.375 for Fb of 2352.0, .

60. To determine Y, β= d/D = 3.000/6.065 = 0.495 hw/Pf = 49.5/80 = 0.619hence: a 3.500 inch orifice plate would be selected to obtain an hw reading of approximately 50 inches at the design flow rate

61. TRANSPORTATION OF NATURAL GASMainly by (1) Pipeline (2) LiquefactionPipeline:Trans-Sahara pipeline (Algeria to Europe)West African Gas Pipeline (WAGP)Soviet Union to EuropeAlaska to Southern USALiquefactionLNG Plants

62. PIPELINE DESIGNFactors to consider:Terrain – land, swamp, seaLocationDistance of transportationNature and volume of gasUse of gasElevation of routeScope of expansionReserves available Design software packages are available

63. GAS FLOW IN PIPELINESThe Value of C depends on the units used T d L q CPsia 0R in Mi scfd 77.54Psia 0R in ft scfd 56.38Psia 0R in ft MMscfd 5.638 x 10-3Kpa 0K M m M3/d 1.149 x 106Where f = friction factor d = Diameter of pipeline (inches) L = Length of Pipeline (miles)

64. EVALUATION OF FRICTION FACTORFor Weymouth Equation =For Panhandle A Eequation =For Panhandle B Equation = Pipeline flow equation without ‘f’ becomes:Where the values of a1- a5 are

65. a1 a2 a3 a4 a5Weymouth: 433.50 1.000 0.5000 0.5000 2.667Panhandle A: 435.87 1.0788 0.5394 0.4604 2.618Panhandle B: 737.00 1.0200 0.5100 0.4900 2.530 q = ft3/day T = 0R  = psia L = miles d = inchesWeymouth equation is used for pipeline diameter of d 16 inches

66. WORKED EXAMPLE 8Using the following data, calculate the flow capacity of the pipeline using the Weymouth equation and Panhandle B equation. P1= 847 psia, P2 = 600 psia, d = 25.375 in, L = 100 miles, Gas gravity = 0.67, T = 505oR, Z = 0.846, Tb= 520oR, Pb = 14.7 psia, E = 1.0

67. SOLUTION Tb/Pb = 520/14.7 = 35.374Weymouthq = 433.5(35.374)1.0(8.366)0.5(1/0.67)0.5(25.375)2.667 = 301,610,000 ft3/dayPanhandle Bq = 737.0935.374)1.02(8.366)0.51(1/0.67)0.49(25.375)2.53 = 359,732,857 ft3/day

68. PIPELINES IN SERIESIf a pipeline or gathering line consists of sections of different diameter pipe, the flow capacity of the entire pipeline can be calculated by first determining an equivalent length of a line of some arbitrary diameter that would have the same flow capacity as the system. For a series pipeline: T = 1 + 2 + 3 If the friction factor is not required, the Pressure Drop becomes:Where a3 and a5 are obtained from the previous table. For example, if Weymouth equation is used, since a3 = 0.5 and a5 = 2.667, the equation becomes:ΔP1ΔP2ΔP3

69. PIPELINES IN PARALLELD1L1q1qtqtD3L3q3D2L2q2For pipes in parallel, the total flow rate is the sum of the rates in the individual pipes or qT = q1 + q2 + q3If the length of the individual pipes are the same, the total flow capacity is calculated from:

70. PIPELINES IN PARALLEL

71. PIPELINES IN PARALLELFor pipes in parallel, the total flow rate is the sum of the rates in the individual pipes or qT = q1 + q2 + q3If the length of the individual pipes are the same, the total flow capacity is calculated from:When applied without friction factor, the above equation becomes:

72. FIELD HANDLING OF NATURAL GASA typical well stream is a high velocity, turbulent and constantly expanding mixture of gases, hydrocarbon liquids, water vapour, free water, solids and other contaminants. Field processing of NG consists of four basic processes.Separation of the gas from free liquids and entrained solids.Processing the gas to remove condensable and recoverable hydrocabon vapours.Processing the gas to remove condensable water vapour to avoid hydrate formation.Processing the gas to remove other undesirable components such as hydrogen sulphide and carbon dioxide.

73. DEHYDRATION OF NATURAL GASRemoval of water vapour from the natural gas. Maximum allowable is 7 Ib of H20/MMscf of gasTwo processes are used:Absorption (gas-liquid) Process Adsorption (gas-solid) Process

74. Dehydration of Natural Gas(Absorption Process)

75. Dehydration of Natural GasTEG is used as the absorbing agent. Process same as acid gas removal

76. Dehydration of Natural Gas

77. Dehydration of Natural Gas(Adsorption Process)Drying of gas using molecular sieve beds to adsorb water to prevent ice & hydrate formation at low temperatures. This is a gas-solid process.Adsorption ProcessAdsorption describes any process in which molecules from the gas are held on the surface of a solid by surface forces.AdsorbentsBauxiteAluminaMolecular SievesSilica gelCarbon (Charcoal)

78. DESIGN CAPACITY FOR ECONOMIC LIFEBauxite: 4 – 6 kg water per 100 kg desiccantAlumina: 4 – 7 kg water per 100 kg desiccantMolecular Sieves : 9 – 12 kg water per 100 kg desiccantSilica gel: 7 – 9 kg water per 100 kg desiccantIn normal operation, outlet dewpoint will be below -40oC until water breakthrough.

79. 79 of 52Regen.GassupplyRegen Gasto U-1100NG fromU-1100DEHYDRATION UNITU 1200NG toU-1300Two adsorbers on line anytimeand one on regeneration/stand-by in rotating sequence.C-1201 BC-1201 CC-1201 A

80. DESIGN INPUT CRITERIA FOR ABSORBER1 Gas flow rate (MMscfd)2 Specific gravity of gas3 Gas inlet pressure (psig)4 Maximum working pressure of absorber (psig)5 Gas inlet temperature6 Outlet gas water content required (Ibm/MMscfd)7 Vap/Liquid equilibrium correlation for TEG-Water system

81. What you needGas capacity/Absorber size chartWater content chartTEG mol % vs mol fraction of water (x)Activity coefficient (γ)chartAbsorber efficiency (Ea)Absorption factor (A)

82. Absorber Gas capacity (Trayed Column)

83. Absorber Gas Capacity (Packed Column)

84. Absorber Material Balance

85. Absorber Material Balance

86. USING THE RELATIONSHIP:Ea = Absorption efficiencyK = 1.33 x 10-6 (W) (γ)

87. Water Content of Natural Gas (oilfield unit)

88. Water Content of Natural Gas (SI unit)

89. Rewriting the above equation in terms of water content we have:Where: Wo = (W) (γ) (xo)γ = activity coefficient for water in TEG system (from chart)Xo = mole fraction water in TEG-Water system (from chart)Wo = water content at saturationWN+1 = Inlet gas water contentW1= exit gas water contentEa = absorption efficiency (ie fraction of component absorbed)K = 1.33 x 10-6 (W) (γ)

90.

91.

92.

93. Water Dew point vs temperature

94. Concentration of Rich TEG leaving the bottom of the Contactor is: Lw = glycol to water circulation rate (gal TEG/Ib water) ρi = specific gravity x (8.34) ρi = density of lean TEG solution (Ibm/gal)Rich TEG = conc of TEG in rich solution from contactor, %/100Lean TEG = conc of TEG in lean solution to contactor, %/100Number of traysActual No of trays = No of theoretical trays / tray efficiencyWhere tray efficiency = 25% for bubble cap & 33.3% for valve tray

95. Dew points of TEG solution

96. Stripper sizing L = Lw x Wi x q / 24 Where L= glycol circulation rate (gal/hr) Lw = glycol to water circulation rate (gal TEG/Ib water) Wi = water content of inlet gas (Ib water/MMscfd) q = gas flow rate (MMscfd)Reboiler duty Ht = 2000 x L Ht = total heat load on reboiler (Btu/hr) Overall size of reboiler A = Ht / 7000 (based on heat flux of 7000 Btuh/ft2) (Max ht flux across fire box is 8000 Btu/h/ft2) A = total area of firebox

97. Chart for Number of Trays

98. HEAT LOSS FROM REBOILER/STRIPPER Hl = 0.24 x As x (Tv – Ta) where Hl = overall ht loss from reboiler/stripper (Btu)As = total exposed surface area of reboiler/stripper (ft2)Tv = temp. of fluid in the vessel (oF)Ta = min. ambient air temperature (oF) 0.24 = ht loss from large insulated surfaces (Btuh/ft2)

99. Worked Example 9Calculate the circulation rate of 98.7 wt% lean TEG needed to dry 106 std m3/d (1762 kg mol/hr) of natural gas at 7.0 MPa and 40oC in a six tray absorber (1.5 theoretical tray) to achieve an exit gas water content of 117 kg /106 std m3. The inlet water content is 1100 kg/ 106 std m3.

100. Solution:Given: Wt % of TEG-Water system: Determine mol fraction of water in TEG system using chart. (xo = 0.099).Determine the activity coefficient (γ) using chart. (γ = 0.66)Determine water content of the gas if not given. Calculate water content at saturation Wo = (1100)(0.66)(0.099) = 71.9kg/106 std m3. Calculate absorption efficiency Ea.Determine the absorption factor A, knowing the number of theoretical trays using chart. For N = 1.5, and Ea = 0.956, A = 7.

101.

102.

103.

104. Calculate the TEG circulation rate (L)L = (A)(K)(VN+1) but K = 1.33 x 10-6 (W) (γ) = 1.33 x 10-6 (1100) (0.66) = 0.000967 and VN+1 = 1739 kmol/h Hence L = (7.3)(0.000967)(1739) = 12.3 kmol/h.8) Mol wt of the lean TEG solution (Mol fract. of water x Mol wt of water + Mol fract. of TEG x Mol wt of TEG) (0.099 x 18) + (0.901 x 150) = 137 kg/kmol

105. L = 12.3 x 137 = 1685 kg TEG/hDensity of TEG = 1.12 g/cm3 = 1.12 kg/ litre TEG circulation rate = 1685 / 1.12 = 1500 litre/hAmount of water absorbed in 1 hr = (1100 – 117) / 24 = 41 .0 kg Hence TEG to water circulation rate = 1500 / 41.0 = 36.6 litre / kg H2O

106. Worked Example 10Size a trayed-type glycol dehydrator for a field installation to meet the following requirements:Gas flow rate = 10.0 MMscfdGas specific gravity = 0.7Operating line pressure = 1000 psigMaximum working pressure of contactor = 1440 psigGas inlet temperature = 100oFOutlet gas water content = 7 Ib H2O/MMscfGlycol to water circulation rate = 3.0 gal TEG/Ib H2OLean glycol concentration = 99.5% TEGNo of theoretical trays = 1.5

107. SOLUTIONDetermine the size of the contactor using chart: For P= 1000 psig and Qg = 10.0 MMscfd; the contactor diameter = 24-in OD.2) Determine inlet water content from chart: For P = 1000 psig = 1014.7 psia and T = 100oF Inlet water content = 61 Ib H2O/MMscf3) Determine the density of lean glycol: ρ = (S.G)( density of water) = (1.111)(8.34) = 9.266 Ib/gal

108. 4) Calculate the concentration of TEG out of contactor: = 96.0%No of trays = No. of theo. Trays / tray efficiency = 1.5 / 0.333 = 4.50 = 5.0 (next whole number of tray should be used).

109. Calculate glycol circulation rate: L = Lw x Wi x q / 24 = (3)(61)(10) / 24 = 76.25 gal/hr7) Calculate Reboiler heat load: Ht = (L)(ρ)(CP)(ΔT) Ht = 2000 x L = (2000) ( 76.25) = 152,500 Btu/hr

110. ADSORBER DESIGN CONSIDERATIONS 1 Cycle time (< 1hr to 8hr) 2 Gas flow rate 3 Desiccant capacity 4 Required dewpoint 5 Total amount of water to be removed 6 Regeneration requirements 7 Pressure drop limitations 8 Adsorption performance of the adsorber

111. Adsorption Related Equations

112. PROCEDURE FOR ADSORBER SIZING CALCULATIONSThere are a series of dependent variables. Some must be fixed for design purposes. The general procedure used are:1. Estimate an allowable superficial velocity2. From the inlet water content, calculate the load water to be adsorbed per cycle, assuming all inlet water is removed.3. Determine vessel diameter. Choose OD of vessel to be used. Subtract vessel wall thickness x 2 to give ID. Then subtract for internal insulation (if used) to find desiccant bed diameter4. Find actual “Vg” from equation. Check to see if it falls within proper velocity limits. If not, change diameter so that it does and find new “Vg”

113. 5 Determine “x” from equation and charts for xs6 Determine minimum bed length hT7 Check tB from equation to see if cycle length chosen is satisfactory. If not, adjust and repeat steps 1 – 6. Note water loading “q” is independent of cycle length.8 Use the results of the above for regen. CalculationsGiven: Gas flow rate Inlet water content Operating Temp and Pressure Gas composition or other gas properties

114. We can determine the followings:(1) Minimum Bed DiameterWhere Vg = 540 Dp / ρg (Dp = avg particle diameter)(2) Determination of Bed Length (ht) ht = [(127.4) (wt of water adsorbed/cycle) /(ρB) (D2)(x)] Where D = bed diameter ( ft or m) x = useful capacity wt.of water (Ib or kg) ρB = bulk density of desiccant (Ib/ft3 or kg/m3) If wt of water adsorbed is not known, use the relationship:

115. (3) Desiccant capacity X = desiccant useful capacity kg water per 100 kg desiccant xs = capacity at saturation kg water per 100 kg desiccant ht = bed length hz = zone length (depends on the characteristics of the adsorbent)For Silica gel : Molecular Sieve: hz = 0.6 hz (Si gel)Alumina hz = 0.8 hz (Si gel) A = constant (141 or 375) hz = zone length ( cm or in) q = water loading (kg/h/m2 or Ib/h/ft2) Vg = superficial velocity (m/min or ft/min) R.S.= % relative saturation of inlet gas(assume 100% if saturated)

116. Relative Water saturation charts

117. (4) Total Water Load (q): q = 0.054[(Flowrate)(W) / D2] Where q = water loading (Ib/h/ft2 or other units) Flowrate = gas rate (MMscfd) D = bed diameter ( ft) W = water content (Ib/MMcsfd)(5) Breakthrough Time (tB) tB = [(0.01) (x) (ρB ) ( ht )] / q where tB = breakthrough time ( hr) ρB = bulk density of desiccant (Ib/ft3) ht = minimum bed length (ft or m) q = water loading (Ib/h/ft2 or other units)

118. REGENERATION SYSTEMThe total heat load is the sum of 1. heat to desiccant 2. heat to hydrocarbon and water 3. heat to vessel 4. heat to inert ball NB: Take Sp Ht of steel = 0.5 kJ/kg/oC Sp Ht of balls = 1.0 kJ/kg/oCIf vessel is insulated, omit (3) and add 10% to the calculated 1,2 and 4.

119. (1) Heat load If “m” is the regeneration gas flow rate (kg/h) the available energy to heat the vessel and its contents to the required temp is: Q = m x Sp Ht x TH – T1 T1 = temp of inlet wet gas TH = (19 – 38oC) > Tm (Values above 315oC should not be used) Tm = maximum regen temp ( 175 – 260oC)Fuel gas required V = Q / (0.7 x Gas Heating Value)

120. Worked example 110.27 x 106 std m3 (10 MMscf/d) of a 0.6 relative density natural gas is to be dehydrated. The wet gas enters saturated at 6.9 Mpa (1000 psia) and 38oC (100oF). The vendor proposes a unit composed of 2.76cm (30 in) O.D. towers containing silica gel beds 4.57 m (15 ft) in length. After allowing for shell and internal insulation thickness, the bed diameter is 64.8 cm (25.5 in). Does this meet company criteria including a gas superficial velocity not exceeding 9.15 m/min (30 ft/min)? The water content of the inlet gas is 1021 kg/ 106 std m3 (61 Ib/MMscf). Z = 0.88 and bulk density of gel is 721 kg/m3.

121. SolutionCalculate the total water load per cycle. Water absorbed = (0.27)(1021) / 3 = 91.9 kg/cycle Calculate the useful capacity (x) Calculate hz4) Calculate the gas velocity (vg)

122. 5) Calculate the water loading (q)Hence hz

123. xs = (16)(0.9) = 14.46) Calculate the Capacity at saturation (from chart) Calculate the Desiccant Capacity 8) Determine the minimum bed length (hB)

124. 9) Determine the breakthrough time (θB)