Water Hydrocarbon Behaviour Cambel

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    ection5WATER-HYDROCARBON BEHAVIOR

    TABLE OF CONTENTSP GE

    WATER CONTENT 5.1Liquids 5.1Gases 5.5Hydrates 5.5Hydrate Inhibition 5.6

    HYDRATE INHIBITION IN LOW TEMPERATURE PROCESSING PLANT 5.8Valve Expansion Plant (LTS or fT Plant) 5.9Glycol Inhibition 5.10

    LIST OF FIGURESFIGURE P GE

    5.1(a) Solubility of Water in Hydrocarbons - SI Units 5.25.1(b) Solubility of Water in Hydrocarbons - English Units 5.25.2(a) Water Content of Gas - SI Units 5.35.2(b) Water Content of Gas - English Units 5.45.3(a) Hydrate Formation Conditions - SI Units 5.75.3(b) Hydrate Formation Conditions - English Units 5.75.4 Typical Mechanical Refrigeration Plant 5.95.5 Typical LTS Plant 5.95.6(a) Glycol-Water Freezing Point Curves - SI Units : 5.115.6(b) Glycol-Water Freezing Point Curves - English Units 5.11

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    W TERHYDROC RBON BEH VIOR

    NOTES:

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    ection 5WATER-HYDROCARBON BEHAVIOR

    WATER CONTENT

    iquidsAlthough we normally think water and oil do not mix, a small amount of water will dissolve inoil and other petroleum products. The solubility of water in various hydrocarbons is shown in Figures5.1(a) and 5.1(b). As you can see from the figures, more water will dissolve as the temperature rises.

    This is an important factor in making g a ~ o l i : 9 ~ o r ~ ~ e f i e d _ p e . 1 r o k u I a . : g ~ s . They normally aremade in a refinery or gasoline plant af'il"remperature around 38

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    WATER-HYDROCARBON BEHAVIOR

    - c:c: 00.0.0 - I'llI'll UU 0o '

    0.5

    0.4

    0.5

    0.4

    ( / ) (90.1

    OX -,-v Oe 0.1

    ' - 0 0.3 0.30 > 0>OI _.S 0- CllCl l _ca J:s: Q;

    - 0 0.2 0.2o ' oCll=1:g:s: r/'C5E

    .... J : g.-J : 0.S asC)

    2.0 c::0

    CI) 0 1.0-a 00 .....- CI)>- a.S CI):c 1U 0.5 g ~ n n

    0.0 0.040 60 80 100 120 140 160

    Temperature, ofFigure 5.1(b) Solubility of Water in Hydrocarbons - English Units

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    WATER CONTENT

    )

    / .

    .__. 1- -f". f f - +

    _. t-- j

    10000080000 ~ ~ . ~ - c l = o = J : i : L:: . ' - cc;Qi ~ c i = + I = c c i i ~ ~ c l _ ~ : : . c , . . ( ~ 600005000040000

    === ./.=;.:cl=L __ -+-V- - ---i . ~ : ; : -

    - . /

    === i

    ~ = I = Z .- = = 7 = = ~ 7-+. --t---....--f ==...::/ - -::/ ::t+++

    ./1

    = -

    / -1-/

    - -=. . . . .

    :=1

    .-/-1

    =

    500400

    3000

    - 300~

    20000UoL )

    : :: 6000s 5000_ 4000

    30000

    cryE~ 2000

    coo

    - 200 - - j --::Q)-oo 10080

    L I, I, - - , 1-+-+_ 0

    I... f-. r I r

    _ .. _ o =t._ -1. ._.

    3020 f---. ----t-.

    - ._ i- I 1 ; _ ... L,-i ' ,.) I JMC,1983 '10-40 -20 o 20 40

    Water Dew Point, C60 80 100

    Figure 5.2 a) Water Content of Gas - 81 Units

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    3000

    WATERHYDROCARBON BEHAVIOR

    I:::,0-')

    . .'

    2000

    600. 500400

    _.-. -- ----+---+ ---l -

    -------4 -----t----r-t-+---I

    100008000600050004000

    - 300t5n

    :2: 200:2:E.0

    LLoo1000 800'iii

    t-.,;,....

    100:::J 0-'- 50Q) 40Q)3: 30-

    r... :1-+ .,-

    - --- 20.cooo=,---.e.:

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    1

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    I ,

    -40 o 160 200 240

    Figure 5.2 b) Water Content of Gas - English Units --, ; - ~ . ,

    , \ .c , - ~ ' : : : '\. -, ...,

    J

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    WATER CONTENT

    asesNatural gas or other hydrocarbon vapors contain water in a vapor state (steam) just as air holds

    water in the form of humidity. The amount of water that gas will hold depends upon the temperatureand pressure as shown in Figures 5.2(a) and 5.2(b). Water must be removed from gas to preventcondensation in a pipeline or fuel consuming equipment.

    The most common use of the gas-water content curves are in designing and/or checking operation of gas dehydration facilities.

    Example 5.2: Gas enters a dehydration plant at a pressure of 4000 kPa [580 psi] and a temperature of 38C [lOOF]. The flow rate is three million m3 d [106 MMscf/d]. Howmuch water can the gas contain?

    Gas TemperatureGas PressureFrom Figures 5.2(a) and 5.2(b), water content at temperature and pressureGas flow rateTotal water in gas

    5 Units38C

    4000 cPa1480 kg/million m3

    3 million m3/d1480 x 3= 4440 kg/d

    English Units100F

    580 psi921b MMcf

    1 6 MMcf d92 x 106= 9752 Ibid

    A common form used to express water content of gas is the dewpoint. 1:J:le__ e . t . . e 5 L ~ U . L t h e temperature at which water will condense from gas. The relative humidity of gas at its dewpoint. ,. / v / / / / __ .,r ... . . ..... .. / ..... . .. . . .. . temperature is 100%. f gas containing water vapor is cooled, the temperature at which liquid firststarts to form is the dewpoint.

    The quantity of water vapor contained in gas at its dewpoint will depend upon the gas pressure.Figures 5.2(a) and 5.2(b). indicate the amount of moisture gas can contain at various temperatures andpressures. When gas is at its dewpoint, its water content will be that shown on the curves.

    Example 5.3: Gas leaving a dehydrator has a water content of 64 kg/million m 3 [4 IbINIMcf].Pressure is 5500 kPa [800 psi]. What is its dewpoint?

    Refer to Figures 5.2(a) and 5.2(b). Move up the left scale until you reach the water content ... 64 kg[4 lb]. At this point, move horizontally to the right until you intersect the pressure ... 5500 kPa [800psi]. Move down vertically to the temperature scale and read -lODe [14F].

    Figures 5.2(a) and 5.2(b). are to be used for pipeline quality gases. This means the gascontains small amounts of H2S CO and heavy hydrocarbons. For sour gases, different correlations arerequired. Gases with H2S and CO2 can hold more water than sweet gases.

    HydratesHydrocarbon fluids exhibit a peculiar characteristic under certain conditions of temperature and

    pressure when free water is present. A substance known as a hydrate will form. A hydrate is a frozenmixture of water and hydrocarbons that forms at a temperature well above the freezing point of water.Its physical form will vary from a gelatinous mush to solid ice. It can completely block a line orvessel, causing severe pressure drop, flow interruption, or even rupture.

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    WATER-HYDROCARBON BEHAVIOR

    A hydrate forms only in the presence of light hydrocarbon molecules (n-butane and smaller),CO2 and H2S. Water must be present in a liquid form. The approximate temperature and pressure atwhich hydrates form is shown in Figures 5.3(a) and 5.3(b). Hydrate formation is composition dependent and these curves should be used as a guideline only for other compositions. More detailed calculations can be found in other references.

    Free water must be present for hydrates to form. f the gas is properly dehydrated, no hydratewin form. Also, if a sufficient amount of alcohol or glycol is added to gas, hydrate formation can beprevented. This is called hydrate inhibition.

    Hydrates are of particular concern to gas processing plant operators because most plants haverefrigeration facilities to chill the gas to a temperature well below the hydrate formation point. f thegas is not dehydrated or the hydrates are not inhibited, hydrates can form in the heat exchangers andcan completely block the flow.

    The best way to avoid hydrate problems is to:1. Dehydrate the gas.2. Use chemical inhibition (alcohol or glycol).3. Keep the system temperature above the hydrate temperature.f hydrates have started to form in the system and the gas flow is not completely blocked, they

    can usually be removed by injecting methanol into the gas flow. Methanol injection ports are ofteninstalled upstream of heat exchangers, valves and turboexpanders in refrigeration systems for this reason. Hydrate formation is usually detected by measuring the differential pressure across the equipment.A rise in L ,p indicates hydrate formation.

    f the gas flow has been completely stopped by the hydrate plug, the system must be depressured and allowed to warm to ambient temperature so that the hydrate will melt. The depressurizationstep must be done very carefully so that differential pressure is not allowed to build up across thehydrate plug. This L ,p may cause the hydrate plug to break free and move through the piping at very./high speeds. Severe damage and injury may result from these hydrate bullets.ydrate Inhibition

    As discussed earlier, there are three primary methods of preventing hydrate formation in a gasproduction/processing system.1. Dehydrate the gas.2. Keep the temperature of the gas above the hydrate point.3. Inject a hydrate inhibitor.

    Method 1) is the most positive method since it removes the water from the gas. f free watercannot condense out in a gas system, hydrates cannot form. Method 1) is not always the most economical method of preventing hydrates however. Sometimes, due to the high cost of installing a dehydration unit, methods 2 or 3 are preferred.

    Method 2 is often used in gas production facilities to keep the gas temperature above thehydrate point downstream of the production choke. Water-bath heaters are frequently used to heat thegas. These are usually applied on a well-by-well basis. This method of preventing hydrates is obviously not effective in gas processing, especially low temperature processing.

    Method 3 is a very common method of preventing hydrates in both gas production and processing facilities. An inhibitor is a chemical that lowers the freezing point of water. (Lowering the freez-

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    WATER-HYDROCARBON BEHAVIOR

    ing point is the same as lowering the hydrate point.) Several chemicals can be used. The most common inhibitors are: . .,

    . >y V Ethylene glycol/ /2 Diethylene glycol

    3. MethanolEthylene glycol is by far and away the most common inhibitor in gas processing. Diethylene

    glycol is used occasionally, but its high viscosity is a problem in low temperature gas processing.Methanol is very popular for inhibiting hydrates on an intermittent basis in production facilities. It has.two main disadvantages, however. First, it cannot be recovered and all the methanol injected is losteither in the gas stream or the produced water. Secondly, most of the methanol injected vaporizes intothe gas and does nothing to prevent hydrates. As a result the methanol injection rates can be quitehigh.

    Methanol is cheaper than glycol and mixes more easily with the gas and water. This makes itvery popular in production facilities for inhibiting hydrates on a temporary basis usually in the wintertime) and for breaking hydrates which have already formed.

    We will discuss hydrate inhibition in this manual assuming that the inhibitor is ethylene glycol.Any time the word glycol is used in this section it will refer to ethylene glycol EG).

    HYDRATE INHIBITION N LOW TEMPERATUREPROCESSING PLANT

    When gas is processed in certain types of low temperature processing plants, glycol injection isfrequently used to prevent hydrate formation in the heat exchangers and the cold separator. Mechanicalrefrigeration plants and valve expansion plants often called Jff or LTX units) typically use glycolinjection systems. Glycol injection has been used to inhibit hydrates to temperatures as low as -40C[-40F].

    Ethylene glycol is injected into the stream as it enters heat exchangers in which the gas temperature is lowered. It mixes with water, which condenses from the gas as it is cooled, and therebyprevents the hydrate from forming. The glycol is recovered in the low temperature separator. It isdiluted with water that condenses as the gas is chilled. The diluted solution rich glycol) flows to areconcentrator, where the condensed water is boiled out. The reconcentrated glycol lean glycol) is thenpumped back to the injection points.

    Figure 5.4 shows a typical mechanical refrigeration gas processing plant.Inlet gas flows through a gas-to-gas exchanger where it is partially cooled with chilled gas

    leaving the plant. Glycol is injected into the inlet gas stream entering the gas-to-gas exchanger. Thepartially cooled inlet gas and glycol flowing out of the exchanger then enter a chiller where its temperature is lowered to the desired point with refrigerant - usually propane or Freon. Glycol is alsoinjected into the gas stream entering the chiller.

    The stream leaving the chiller is a mixture of gas, glycol, and liquid hydrocarbons. It flows toa low temperature separator where the liquids fall to the bottom. Cold gas leaves the separator andflows through the gas-to-gas exchanger where it partially cools the inlet gas. The outlet gas from theexchanger usually enters a pipeline.

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    ".t\1 t.' 1\.', r i i

    W TER CONTENTI. ' . : : ' ."

    20000

    18000

    16000

    14000", .....; ,,- ,} 'ydrates [1>':

    ,...- i? I -

    k ':/ "-" No Hydrates\ .... .> .*

    .1, [ / ',\........ ~ , : ; ..." vy , \ e( C,,' :-, . :.,. . I ' \; I"""'"

    Figure 5.3 a) Hydrate Formation Condit ions - 81 Units\3000

    2500

    2000ro'en0-eD 1500::;lfJlfJOJ0::

    1000

    500

    - - ._.

    "II

    IHydrates

    No HydratesJ

    vio'

    io'o o 10 20 30 40 50 60 70 80 90 100

    Temperature, ofFigure 5.3 b) Hydrate Formation Conditions - English Units

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    HYDRATE INHIBITION N LOW TEMPERATURE PROCESSING PLANT

    Refrig.Outlet VaporGas

    Gas GasExchanger

    30C[86 OF]InletGas

    LeanGlycol.l

    24C[76 OF]

    18C[0 OF] LiquidRefrig.

    Hydrocarbon

    Rich Glycol to Reconcentrator

    Figure 5.4 Typical Mechanical Refrigeration PlantGlycol, being heavier than liquid hydrocarbon, collects in the bottom of the low temperatureseparator and flows to a reconcentrator to remove the dilution water that condensed when the gas was

    chilled.Liquid hydrocarbons that collect in the low temperature separator flow to other processingfacilities to stabilize or segregate the various hydrocarbons in the stream.

    Valve Expansion Plant LTS or J TPlantIn a valve expansion plant the glycol injection system is identical to that in a refrigerated plant

    except that there is only one heat exchanger gas-to-gas exchanger) and the expansion across the valve,rather than a chiller, provides the secondary cooling for the gas. Figure 5.5 presents a typical LTS plantshowing the glycol injection points.

    \

    OutletGas

    30C 7 C[86 OF] [20 OF]lnlet Gas GasGas Exchanger

    18Ci 10 O ] HydrocarbonRich Glycol to Reconcentrator

    Figure 5.5 Typical LTS Plant

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    WATER-HYDROCARBON BEHAVIOR

    lycol InhibitionAlthough ethylene glycol is commonly used as an anti-freeze to prevent water from freezing in

    automobile engines, some mixtures of glycol and water will freeze at temperatures which occur in lowtemperature processing plants. In fact, pure ethylene glycol freezes at a temperature of 6C [22F]. JConsequently, the concentration of glycol that is injected in the gas stream must be carefully controlled'so that it will not freeze when it passes through the low temperature exchangers.

    Figures 5.6(a) and 5.6(b) indicate the freezing points of various glycol-water mixtures.As you can see from the glycol freezing point curves, a glycol concentration of 90 (10water) will freeze at a temperature of 25C [-13F]. A solution containing 47 glycol and the balance water will freeze at the same temperature. If a low temperature plant chills gas to 25C [_13P],the concentration of glycol used for injecting in the exchangers to prevent hydrate from forming mustbe above 47 and less than 90 , or the glycol will freeze.

    The concentration of the lean EG solution injected into the exchanger is typically around 75 wt . Theconcentration of the rich EG solution taken out of the cold separator can be detennined from theHammerschmidt equation below.

    (lOO)(d)(M)x ~ : : : : .(d)(M) + K

    Where: d = hydrate depression, C [OF] (hydrate temperature - system temperature)M = mol wt of inhibitor EG = 62DEG = 106Methanol = 32K = constant 1300 if d is in D2340 if d is in OFThe concentration calculated in the above equation is the minimum concentration of rich inhibi

    tor. I f the actual concentration is less than this, freezing will occur. t is usually best to inject enoughinhibitor so that the rich inhibitor concentration is greater than the number calculated above.Once the rich concentration has been determined, the inhibitor injection rate may be calculatedfrom the equation below.

    \ y.J\ ) : :;( ( \ 1 9; .- -ro I\ - /Where: mr = mass of inhibitor, kg [Ib] \ \... , < '1'\'\ >.f\mass of liquid water, kg [Ib]XL = lean inhibitor concentration, wt% r A ~ ; ) \ j J IXR = rich inhibitor concentration, wt \6, /C:;V

    . ,, . In the above equation the term XL - XR is called the glycol dilution ,; \ ::: \\ '? .\,\ . .. _ , C : ; ~ - ; C ~ -, ; ; ~ ;j/ The actual glycol injection rate in the previous example would be higher than the calculatedamount. The reason for this is the injection rate is the lowest possible rate needed to prevent hydrates.1/ To be on the safe side most people inject a higher glycol rate to be sure all of the glycol mixes with thegas. One way to estimate this higher rate is to assume a lower glycol dilution. For example, theminimum required dilution is 20 , but the actual injection rate would be calculated on a 10-15dilution.

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    HYDRATE INHIBITION N LOW TEMPERATURE PROCESSING PLANT

    o

    10

    U0a)~ ::J1ij 20 D~ ~ ~ ~ a.EI : : ~ ; ~

    30

    Shaded Areas lndicate Freezing Zone

    20 40 60 80 100Percent Glycol in Glycol Water Solution

    igure 5.6(a) Glycol-Water Freezing Point Curves - 81 Units

    oPercent Glycol in Glycol Water o l u t i o ~ ..

    ,

    Shaded Areas Indicate Freezing Zone

    .- - . I \0 20 40 60 80 10

    20

    20

    40

    LLo~

    ~ D 0a.E

    I

    igure 5.6(b) Glycol-Water Freezing Point Curves - English Units

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    WATER-HYDROCARBON BEHAVIOR

    Example 5 4: A gas flow rate of 2 x 106 std m3/d [70 MMsef/d] is to be cooled from 40C[104F] to -10C [14F] in a refrigeration system. How much 75 wt% EG mustbe injected to prevent hydrate formation in the heat exchanger if the exchangerpressure is 5800 kPa [840 psia]?

    Step 1 Determine hydrate temperature of gas.SI from Figure 5.3(a)English from Figure 5.3(b) TH = I5.5CTH =60F

    Step 2 Calculate the amount of free liquid water which condenses in system.

    From Figure 5.2(a)Water Content In 40C [104F]

    From Figure 5.2(b)Water Content Out -10C [14F]

    Water Condensedper day

    5 Units

    1137 kgl10 6std m3(1137)(2) =2274 kg/d

    English Units74 IblMMscf4 IblMMscf

    701bIMMscf(70)(70) =4900 IbidStep 3 Calculate the minimum rich EG solution concentration.

    5 UnitsXR = (100)(25.5)(62) = 55%(25.5)(62) + 1300

    English UnitsXR = (100)(46)(62) = 55%(47)(62) + 2340

    Step 4 Calculate the minimum amount of inhibitor to be injected per day.5 Units

    (2274)(55) = 6250 kg/d(75 - 55)

    English Units(4900)(55) = 13475 Ibid(75 - 55)

    NOTES:'

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