Gas processing and compression

Quick refresher

Sales gas is defined by the water and hydrocarbon dew points, the content of hydrogen sulfide, mercaptan and total sulfur, carbon dioxide and oxygen, and the heating value. Each requirement guards against its own threat: hydrates and corrosion, liquid in the pipe, toxicity and corrosion, and a shortage of energy for the consumer.
Gas in the reservoir is saturated with water. The water content grows with temperature almost as fast as the vapour pressure of water, and falls as pressure rises. On cooling, the extra water drops out in gathering lines (pipes that carry gas from the wells), after chokes and in pipelines.
We prevent hydrates by heating, drying or inhibitors (chemicals that stop hydrates forming). The more methanol or glycol there is in the water, the more it lowers the hydrate formation temperature. The Hammerschmidt formula works for methanol up to about 20–25% by mass.
Gas is dried by absorption in glycol (triethylene glycol, TEG, or diethylene glycol, DEG), which takes up the water like a sponge, or by adsorption, in which water sticks to the surface of solid grains such as silica gel and zeolites. Adsorbents give the lowest dew points, which are needed before deep cooling.
Liquid hydrocarbons are recovered by cooling the gas: by throttling (letting the gas expand through a choke, which cools it) in low-temperature separation, with refrigeration units, or with turboexpanders (turbines in which the gas expands and cools). Throttling needs a free pressure drop, and as the field is depleted there is less of it.
Gas heats up when compressed, so we compress it in stages, with cooling between the stages and about equal compression ratios (the ratio of outlet pressure to inlet pressure). The number of stages is set first of all by the limit on the discharge temperature. Each further stage saves less power.

Key relationships

y_w = \frac{p_s(T)}{p} + \beta(T)

mole fraction of water in saturated gas (Bukacek)

\Delta T = \frac{K\,W}{M\,(100 - W)}

lowering of the hydrate formation temperature (Hammerschmidt), $K = 1297$ °C

p_i = y_i\,p

partial pressure of a component

r = \left(\frac{p_{\text{out}}}{p_{\text{in}}}\right)^{1/n}

compression ratio of each stage when there are $n$ equal stages

T_2 = T_1\left[1 + \frac{r^{(k-1)/k} - 1}{\eta}\right]

discharge temperature of a stage

Common pitfalls

Putting the gauge pressure (measured from atmospheric pressure) into the water content calculation instead of the absolute pressure (measured from zero).
Quoting a water dew point without the pressure: the same water in the gas gives different dew points at different pressures.
Using the Hammerschmidt formula for methanol concentrations above 25%: it quickly loses accuracy and gives too little inhibitor.
Counting the methanol only in the water: a noticeable part of it goes into the gas and the condensate.
Choosing the number of compression stages by power alone and forgetting the limit on the discharge temperature.

After this lesson you can

list the requirements for sales gas and explain what each of them protects against
calculate the water content of gas with the Bukacek correlation, and the amount of water that drops out on cooling
calculate the concentration of a hydrate inhibitor with the Hammerschmidt formula and the amount of inhibitor needed, and name the limits of the method
describe glycol and adsorption drying, the removal of acid components (hydrogen sulfide and carbon dioxide), and the recovery of liquid hydrocarbons by cooling, and estimate their main quantities
choose the number of compression stages from the discharge temperature and calculate the power of a compressor station
describe what gas processing capacity Kazakhstan is building, and why less than half of the gas produced becomes sales gas

What sales gas must be like

Gas comes out of the reservoir saturated with water vapour. It carries liquid droplets (water and condensate) and heavy hydrocarbons that condense on cooling. It often contains hydrogen sulfide, mercaptans (sulfur compounds with a bad smell) and carbon dioxide, and sometimes nitrogen and sand. The gas transport system and the consumers accept gas only on four conditions. It cannot form liquid or hydrates in the pipe, even in the coldest conditions of its operation. It does not corrode metal. It is safe. And it gives the required amount of heat when burned.

Water dew point: The temperature at which the first drop of water appears in the gas at a given pressure. The less water in the gas, the lower the dew point. As the pressure rises, the same water drops out at a higher temperature, so the dew point is always quoted together with the pressure.
Hydrocarbon dew point: The temperature at which hydrocarbon liquid starts to condense from the gas at a given pressure. It is set mainly by the heavy components, pentanes and heavier: even a fraction of a per cent of them raises the dew point by tens of degrees.
Water content: The mass of water vapour in one cubic metre of gas at standard conditions, g/m³ (in oilfield units, pounds of water per million standard cubic feet, lb/MMscf).

Requirements for gas for main gas pipelines in the Technical Regulation of the Eurasian Economic Union TR EAEU 046/2018

Indicator: Mole fraction of oxygen
Requirement: not more than 0.020%
Conditions and exceptions: —

Indicator: Mole fraction of carbon dioxide
Requirement: not more than 2.5%
Conditions and exceptions: —

Indicator: Hydrogen sulfide
Requirement: not more than 0.007 g/m³
Conditions and exceptions: in Kazakhstan, up to 0.020 g/m³ until 1 January 2030

Indicator: Mercaptan sulfur
Requirement: not more than 0.016 g/m³
Conditions and exceptions: in Kazakhstan, up to 0.036 g/m³ until 1 January 2030

Indicator: Total sulfur
Requirement: not more than 0.030 g/m³
Conditions and exceptions: in Kazakhstan, from 1 January 2030

Indicator: Lower heating value (the heat released when the gas burns)
Requirement: not less than 31.80 MJ/m³
Conditions and exceptions: 7600 kcal/m³

Indicator: Water dew point
Requirement: not higher than −10.0 °C
Conditions and exceptions: at 3.92 MPa, temperate climate; cold climate: −20.0 °C in winter, −14.0 °C in summer

Indicator: Hydrocarbon dew point
Requirement: not higher than −2.0 °C
Conditions and exceptions: at the working pressure, temperate climate; cold climate: −10.0 °C in winter, −5.0 °C in summer

Indicator: Solid impurities
Requirement: not more than 0.001 g/m³
Conditions and exceptions: —

Indicator	Requirement	Conditions and exceptions
Mole fraction of oxygen	not more than 0.020%	—
Mole fraction of carbon dioxide	not more than 2.5%	—
Hydrogen sulfide	not more than 0.007 g/m³	in Kazakhstan, up to 0.020 g/m³ until 1 January 2030
Mercaptan sulfur	not more than 0.016 g/m³	in Kazakhstan, up to 0.036 g/m³ until 1 January 2030
Total sulfur	not more than 0.030 g/m³	in Kazakhstan, from 1 January 2030
Lower heating value (the heat released when the gas burns)	not less than 31.80 MJ/m³	7600 kcal/m³
Water dew point	not higher than −10.0 °C	at 3.92 MPa, temperate climate; cold climate: −20.0 °C in winter, −14.0 °C in summer
Hydrocarbon dew point	not higher than −2.0 °C	at the working pressure, temperate climate; cold climate: −10.0 °C in winter, −5.0 °C in summer
Solid impurities	not more than 0.001 g/m³	—

The regulation “On the Safety of Combustible Natural Gas Prepared for Transportation and (or) Use” was adopted by Decision No. 74 of the Council of the Eurasian Economic Commission of 14 September 2018 and has been in force since 1 January 2022. The table is in the version of Council Decision No. 38 of 23 May 2025, which came into force 180 days after publication. The concentrations refer to 20 °C and 101.325 kPa. The Kazakhstan exceptions for hydrogen sulfide and mercaptan sulfur apply to gas that is used only inside the country. For fields put into operation in Kazakhstan before 2005 inclusive, less strict dew point limits are allowed by agreement between the supplier and the receiver. The hydrocarbon dew point is not measured if the gas has 1.0 g/m³ or less of C₅₊ hydrocarbons (pentanes and heavier). The regulation does not apply to gas exported beyond the customs territory of the Union.

SchematicProcessing of sour gas, which contains hydrogen sulfide (schematic): from the inlet separator through the removal of acid components, drying and the recovery of liquid hydrocarbons to the compressors and the gas pipeline. The set and the order of the blocks depend on the gas composition and pressure. At a late stage of development a booster compressor (a compressor for low-pressure gas) is also placed at the inlet. In the labels, NGL means natural gas liquids and LPG means liquefied petroleum gas.

Where this is done depends on the gas and on the distances. At gas and gas condensate fields, we bring the gas to the requirements at integrated gas treatment units. If the wells are far away, we do part of the operations (separation and the injection of inhibitor) at preliminary gas treatment units. Sour gas, and gas rich in ethane, propane and butanes, goes to gas processing plants (GPPs). They turn it into sales gas, liquefied petroleum gases (LPG), the wide fraction of light hydrocarbons (natural gas liquids, NGL), stable condensate (condensate that is safe to store) and sulfur. When the reservoir pressure falls, we place booster compressor stations at the inlet. We collect the associated gas of oil fields from the separation stages (Oil treating: separators, dehydration, desalting and stabilisation) and compress it for delivery, processing, gas lift or injection (from the wellhead to sales products).

Water in gas: water content and dew point

Gas in the reservoir is in contact with water and is saturated with its vapour. If the gas and the water behaved ideally, the mole fraction of water in the gas (the share of water molecules among all the molecules) would equal the ratio of the vapour pressure of water $p_s$ to the gas pressure $p$ (Raoult's and Dalton's laws, the laws of ideal mixtures). This gives the main rules. $p_s$ grows very fast with temperature: it doubles about every 12–14 °C between 20 and 60 °C. And a rise of the gas pressure “squeezes” the water out. At high pressures the gas holds more water than the ideal value, because water molecules are attracted to the surrounding gas molecules. The empirical Bukacek correlation (a formula fitted to measurements) for natural gas gives this correction.

Water content of natural gas saturated with water (Bukacek correlation)

y_w = \frac{p_s(T)}{p} + \beta(T),\qquad \beta = 10^{\,2.0179 - \frac{3083.87}{1.8\,T - 0.07}},\qquad W = 748.9\,y_w

where:

$y_w$ mole fraction of water in the gas, fraction
$p_s(T)$ vapour pressure of pure water at the temperature of the gas, MPa
$p$ absolute pressure of the gas, MPa
$\beta(T)$ correction for non-ideality: an addition to the mole fraction of water that depends only on temperature, fraction
$T$ gas temperature, K
$W$ water content: the mass of water in one cubic metre of gas at 20 °C and 0.101325 MPa, g/m³

Units: absolute pressure in MPa (in the ratio $p_s/p$ any identical units), temperature in kelvins

Holds when

Sweet natural gas, without noticeable hydrogen sulfide and carbon dioxide (they increase the water content), in equilibrium with fresh water (salty water gives a little less).
The tested range of the correlation is from 15.5 to 238 °C and from 0.1 to 70 MPa. Below 15.5 °C, and still more in the hydrate region, the result is an estimate.
We take $p_s(T)$ from an accurate equation for water (in the models of the lesson, the international equation for the vapour pressure of water). Any error in it goes straight into the result.
748.9 g/m³ is the mass of a kilomole of water (18.015 kg) divided by the volume of a kilomole of gas at 20 °C and 0.101325 MPa (24.055 m³).

The first term is the ideal part, the second is the correction. At 5–7 MPa the correction is about a quarter of the result, and at 1 MPa a few per cent. The water content is almost proportional to $p_s(T)$ : cooling the gas by 10 °C almost halves it, and the “extra” water condenses.

In other unit systems (1)

Oilfield units

W\,[\text{lb/MMscf}] = 47{,}484\,\frac{p_s}{p} + B,\qquad \log_{10} B = -\frac{3083.87}{459.6 + t} + 6.69449

$W$ in pounds of water per million standard cubic feet (60 °F, 14.696 psia); $p_s$ and $p$ in psia; $t$ in °F.

The original form of the correlation. 47,484 is the mass of a pound-mole of water divided by the volume of a pound-mole of gas at 60 °F (about 379.5 ft³), multiplied by $10^6$ . So $\beta = B/47{,}484$ and $2.0179 = 6.69449 - \log_{10} 47{,}484$ . Also $459.6 + t = 1.8T - 0.07$ if $T$ is in kelvins. 1 lb/MMscf ≈ 15.78 mg/m³ at 20 °C.

CalculatedWater content of natural gas saturated with water against temperature at 1, 5.5 and 12 MPa (Bukacek correlation, logarithmic scale). Below 15.5 °C the curves leave the tested range (dashed). The points are the gathering line of the example below: 30 and 6 °C at 5.5 MPa.

Worked exampleHow much water drops out in a gathering line

Given (assumed data)

Gas: 1.5 million m³/d (at 20 °C and 0.101325 MPa), sweet, saturated with water at the inlet of the gathering line
Gathering line: pressure 5.5 MPa (absolute); the gas enters at 30 °C and in winter cools to the ground temperature of 6 °C (assumed)

Solution

The vapour pressure of water at 30 °C is $p_s = 4.247$ kPa. The ideal part is $4.247/5500 = 7.722\cdot 10^{-4}$ .
The correction is $\beta = 10^{2.0179 - 3083.87/(1.8\cdot 303.15 - 0.07)} = 2.321\cdot 10^{-4}$ . The mole fraction of water is $y_w = (7.722 + 2.321)\cdot 10^{-4} = 10.043\cdot 10^{-4}$ , and the water content is $W = 748.9\cdot 10.043\cdot 10^{-4} = 0.752$ g/m³.
At 6 °C: $p_s = 0.9353$ kPa, $y_w = (1.701 + 0.758)\cdot 10^{-4} = 2.459\cdot 10^{-4}$ and $W = 0.184$ g/m³ (an estimate, because 6 °C is below the tested range of the correlation).
The water that drops out is $(0.752 - 0.184)\cdot 1{,}500{,}000/1000 = 852$ kg/d.
For comparison, the ideal part alone at 30 °C would give $748.9\cdot 7.722\cdot 10^{-4} = 0.578$ g/m³, which is 23% less.

Answer: About 852 kg of water a day condenses in the gathering line: the gas brings 0.752 g/m³, and at 6 °C it holds only 0.184 g/m³.

What it means. The dew point of the gas at the inlet is 30 °C at 5.5 MPa: anything colder already condenses water. About 0.85 t of water a day at 5.5 MPa and 6 °C leads straight to hydrates. By the estimate from the lesson on chokes, gas of this density forms them at this pressure below about 14 °C (next section). Without the Bukacek correction, the water content at 30 °C would be 23% less, and we would plan too little inhibitor. The value at 6 °C comes from outside the tested range, but its error is small compared with the water that drops out. The data in this example are assumed for illustration.

Hydrates and inhibitors

Gas hydrates are crystals in which a framework of water molecules holds gas molecules. They are stable at high pressure and low temperature, including above the freezing point of water. What they are, and how to estimate the hydrate formation temperature from the specific gravity of the gas, is covered in the lessons on complex fluids and on chokes and temperature. Here we look at how they are prevented.

Keep the gas warm: heaters before the chokes, thermal insulation, burying the gathering lines.
Remove the water before cooling: dry the gas so that its water dew point is below the lowest temperature in the pipe (the section on drying).
Keep the pressure below the hydrate formation pressure. This is not always possible, but it helps during shutdowns and when plugs are removed.
Thermodynamic inhibitors: methanol, monoethylene glycol (MEG) and diethylene glycol (DEG). Dissolved in water, they lower the water activity (the effective concentration of the water) and shift the hydrate curve (the line that separates the hydrate zone from the safe zone on a pressure–temperature chart) to lower temperatures. They are needed at concentrations of tens of per cent of the mass of the water.
Low-dosage inhibitors: kinetic inhibitors slow down the start and growth of crystals, and anti-agglomerants keep them from sticking together into a plug. The doses are from fractions of a per cent to a few per cent of the mass of the water. But they work only in a limited range of supercooling (how far the temperature is below the hydrate formation temperature) and for a limited time.

Lowering of the hydrate formation temperature by an inhibitor (Hammerschmidt formula)

\Delta T = \frac{K\,W}{M\,(100 - W)}

where:

$\Delta T$ how much lower the hydrate formation temperature becomes, °C
$W$ concentration of the inhibitor in the water phase (water plus inhibitor). The letter is the one used in the Hammerschmidt formula; it is not the water content of the gas $W$ from the previous section, % by mass
$M$ molar mass of the inhibitor: methanol 32.04, MEG 62.07, kg/kmol
$K$ constant: 1297 for $\Delta T$ in °C (2335 in °F), °C·kg/kmol

Units: $\Delta T$ in °C, $W$ in per cent by mass

Holds when

A dilute (weak) solution: for methanol up to about 20–25% by mass. At higher concentrations the formula overestimates $\Delta T$ , that is, it gives too little inhibitor (the advanced part of the lesson has a logarithmic form for methanol).
For monoethylene glycol, reference books give its own constant, a few per cent smaller. With the same constant, about twice as much MEG by mass is needed as methanol, because its molar mass is twice as large.
The hydrate formation temperature without inhibitor must be known in advance, from the gas composition, laboratory tests or an estimate from the specific gravity.

The lowering grows with concentration faster than linearly, and for each 1% of inhibitor, methanol gives almost twice the effect of MEG. The inverse form $W = 100\,\Delta T M/(K + \Delta T M)$ gives the concentration needed. The mass of inhibitor in the water is $m_{\text{i}} = m_{\text{w}}W/(100 - W)$ .

In other unit systems (1)

Oilfield units

\Delta T\,[{}^\circ\text{F}] = \frac{2335\,W}{M\,(100 - W)}

$\Delta T$ in degrees Fahrenheit (a temperature difference), the rest as above.

$2335/1.8 = 1297$ : a difference of 1 °C equals 1.8 °F.

Worked exampleHow much methanol a gathering line needs

Given (assumed data)

Gathering line: from the previous example: 5.5 MPa, gas of specific gravity 0.60, coldest point 6 °C, 852 kg of water drops out a day
Margin: keep the hydrate formation temperature 3 °C below the lowest temperature (an operating rule, assumed)
Inhibitor: methanol, $M = 32.04$ kg/kmol; $K = 1297$

Solution

The hydrate formation temperature without inhibitor (the Towler and Mokhatab formula from the lesson on chokes): $p = 5.5\cdot 145.04 = 797.7$ psia, $\ln p = 6.6817$ , $\ln 0.60 = -0.5108$ . Then $T_h = 13.47\cdot 6.6817 + 34.27\cdot(-0.5108) - 1.675\cdot 6.6817\cdot(-0.5108) - 20.35 = 57.86$ °F, or 14.37 °C.
The lowering needed: $\Delta T = 14.37 - 6 + 3 = 11.37$ °C.
The methanol concentration in the water: $W = \dfrac{100\cdot 11.37\cdot 32.04}{1297 + 11.37\cdot 32.04} = 21.93$ %, which is within the range where the formula applies.
Methanol in the water phase: $852\cdot 21.93/(100 - 21.93) = 239$ kg/d, or 0.16 kg per thousand cubic metres of gas.
For comparison, monoethylene glycol by the same formula with the same constant: $W = 35.2$ %, about 464 kg/d. This is a rough estimate.

Answer: There must be about 21.93% of methanol in the water of the gathering line, which is 239 kg a day for the water phase alone.

What it means. In practice we must feed in more. Methanol is volatile (it evaporates easily): part of it goes into the gas and dissolves in the hydrocarbon condensate. At these pressures and temperatures these losses are comparable to the amount in the water. We estimate them from reference data on how methanol is shared between the phases, or by modelling. We inject methanol before the point where the gas cools, at the wellhead or before the choke, through nozzles so that it spreads through the flow. At the plant we recover it from the water phase and return it. About twice as much MEG by mass is needed, but it hardly goes into the gas. So in long offshore pipelines it circulates in a closed loop with regeneration. If the lowering needed is more than 13.5 °C, the methanol concentration goes beyond 25%, and we calculate with the logarithmic form or with an equation of state. The data in this example are assumed for illustration.

CalculatedThe hydrate formation curve of gas of specific gravity 0.60 (Towler and Mokhatab formula) and the same curve with 21.9% of methanol in the water (a shift of 11.4 °C by the Hammerschmidt formula). The point is the coldest place of the gathering line of the example. Without inhibitor it is in the hydrate region (where hydrates can form); with inhibitor it is to the right of the curve, with a margin of 3 °C.

Drying gas: glycols and adsorbents

An inhibitor protects the pipe only while it is being fed. It is more reliable to remove the water from the gas altogether, that is, to dry it (gas dehydration) to a dew point below the lowest temperature that the gas will meet in the pipeline. Two methods are used: absorption by a liquid absorbent, glycol, and adsorption on a solid adsorbent. Cooling with the injection of inhibitor in low-temperature separation also dries the gas, but not as deeply (the next sections).

SchematicA glycol dehydration unit (schematic): in the absorber the wet gas rises against the glycol that flows down. The glycol loaded with water (rich glycol) is degassed, heated and boiled in the regenerator to drive off the water, and the regenerated (lean) glycol is cooled and fed to the absorber again.

Glycol dehydration. In the absorber (contactor) the gas rises through trays or packing against a concentrated glycol, usually triethylene glycol (TEG), and at CIS plants also diethylene glycol (DEG). The glycol absorbs water eagerly, and the solution loaded with water (rich glycol) goes down. It is degassed in a flash tank (a vessel in which the pressure is lowered and the dissolved gas comes out), heated, and the water is boiled off in a regenerator with a reboiler (a heater that boils the glycol). Two parameters set the depth of drying. The first is the concentration of the regenerated (lean) glycol: the less water in it, the lower the dew point that can be reached. TEG starts to break down at a temperature of about 204 °C (400 °F), so we keep the reboiler at not more than about 200 °C. We get a purer glycol by stripping (blowing dry gas through the glycol) or with a vacuum. The second is the temperature in the absorber: warm gas carries more water, while a glycol that is too cold becomes thick and foams.

We choose the glycol rate from the amount of water to be removed. The industry rule is 2 to 5 US gallons of TEG per pound of water, most often about 3 (17–42 L per kilogram, about 25 L/kg). With less, there is not enough absorbent. With more, the load on the reboiler and the losses of glycol with the gas grow.

Worked exampleDrying gas with triethylene glycol: how much water and how much glycol

Given (assumed data)

Gas: 5 million m³/d, saturated with water at the absorber inlet at 6.0 MPa and 30 °C
Requirement at the outlet: not more than 0.10 g of water per cubic metre (a contract condition, assumed)
TEG rate: 25 L per kilogram of water removed (the industry rule, assumed)

Solution

Water content at the inlet: $p_s(30\ ^\circ\text{C}) = 4.247$ kPa, the ideal part is $4.247/6000 = 7.078\cdot 10^{-4}$ , and the correction is $2.321\cdot 10^{-4}$ (as in the gathering line example). Then $W = 748.9\cdot(7.078 + 2.321)\cdot 10^{-4} = 0.704$ g/m³.
The water to be removed is $(0.704 - 0.10)\cdot 5{,}000{,}000/1000 = 3020$ kg a day, or $3020/24 = 125.8$ kg/h.
The glycol rate is $125.8\cdot 25 = 3145$ L/h, about 3.1 m³/h.

Answer: The absorber removes about 3020 kg of water a day, and we pump about 3.1 m³ of TEG an hour through it for this.

What it means. The amount of glycol is set not by the gas but by the water: with the same gas rate but 40 °C at the inlet there would be almost twice as much water. So we cool the gas before the absorber and separate the water that drops out on cooling in a separator. The requirement of 0.10 g/m³ is only an assumption: the gas transport standards set not the water content but the dew point at a stated pressure (the table at the start of the lesson). To convert from one to the other at low temperatures, we use a standard method, not the Bukacek correlation, which is tested only above 15.5 °C. The data in this example are assumed for illustration.

Adsorption drying. The gas passes through a bed of a solid adsorbent with an enormous inner surface: silica gel, activated alumina or zeolites (molecular sieves). The water molecules are held on that surface. There are two or more adsorbers. While one adsorber dries the gas, we purge another with hot gas at a temperature of several hundred degrees. The water evaporates from it and condenses in a cooler, and then we cool the bed. Zeolites leave a residual water content of the order of fractions of a ppm (part per million) and give dew points around minus 100 °C. This is needed before deep cooling, in ethane recovery and gas liquefaction, where any water would freeze.

Ways of removing water from gas

Method: Inhibitor + cooling (low-temperature separation)
Depth of drying: down to the separator temperature
Where it fits: gas condensate fields, while a pressure drop is available

Method: Absorption by glycol (TEG, DEG)
Depth of drying: a dew point tens of degrees below the absorber temperature
Where it fits: the main method for sales gas

Method: Adsorption by zeolites
Depth of drying: fractions of a ppm of water, a dew point of about −100 °C and below
Where it fits: before cryogenic processing (at very low temperatures) and gas liquefaction

Method	Depth of drying	Where it fits
Inhibitor + cooling (low-temperature separation)	down to the separator temperature	gas condensate fields, while a pressure drop is available
Absorption by glycol (TEG, DEG)	a dew point tens of degrees below the absorber temperature	the main method for sales gas
Adsorption by zeolites	fractions of a ppm of water, a dew point of about −100 °C and below	before cryogenic processing (at very low temperatures) and gas liquefaction

Approximate values for comparing the methods. The dew point that glycol drying can reach depends on the concentration of the regenerated glycol and on the absorber temperature.

Hydrogen sulfide and carbon dioxide

Hydrogen sulfide (H₂S), carbon dioxide (CO₂) and mercaptans are called acid components: in water they form acid solutions (sour gas). Hydrogen sulfide is deadly poisonous. Together with water it causes sulfide stress cracking (cracking of steel under stress in the presence of hydrogen sulfide) and hydrogen embrittlement of steels (hydrogen makes the steel brittle). Carbon dioxide with water corrodes carbon steel, and as a non-combustible component it lowers the heating value. How dangerous the environment is for metal is decided not by the percentage of the component but by its partial pressure: its share in the gas multiplied by the pressure.

Partial pressure of a gas component

p_i = y_i\,p

where:

$p_i$ partial pressure of component $i$ , kPa
$y_i$ mole fraction of the component in the gas, fraction
$p$ absolute pressure of the gas, kPa

Units: any identical units of pressure

Holds when

An ideal mixture of gases. At high pressures it is more accurate to use fugacity (a corrected pressure that accounts for non-ideal behaviour), but the standards for materials use this simple measure.

The international standard for materials for hydrogen sulfide environments (ISO 15156, also known as NACE MR0175) requires steels resistant to sulfide stress cracking if the partial pressure of hydrogen sulfide in the gas reaches 0.3 kPa (0.05 psia). At high pressure this is very little hydrogen sulfide: at 8 MPa it is 37.5 ppm by volume.

Mole fraction of a component from its mass concentration

y = \frac{c}{M\,n_{\text{sc}}},\qquad n_{\text{sc}} = \frac{p_{\text{sc}}}{R\,T_{\text{sc}}} = 41.57\ \text{mol/m}^3

where:

$y$ mole fraction (for an ideal gas, also the volume fraction) of the component, fraction
$c$ mass concentration of the component at standard conditions, g/m³
$M$ molar mass of the component: hydrogen sulfide 34.08, g/mol
$n_{\text{sc}}$ number of moles of gas in one cubic metre at 20 °C and 101.325 kPa, mol/m³

Units: g/m³, g/mol; standard conditions 20 °C and 101.325 kPa

Holds when

The gas at standard conditions is ideal.

The standards for gas are written in grams per cubic metre, while analysers and process calculations often work with mole fractions and ppm. The formula converts one into the other.

Chemical solvents: amines (amine gas treating). These are solutions in water of monoethanolamine (MEA), diethanolamine (DEA) and methyldiethanolamine (MDEA). In the absorber they take up H₂S and CO₂ in a reversible chemical reaction, and in the regenerator they release them again on heating. The scheme is the same as for glycol drying. MDEA absorbs hydrogen sulfide faster than carbon dioxide, and we use this for selective treating (removing the hydrogen sulfide and leaving most of the carbon dioxide).
Physical solvents absorb acid gases without a chemical reaction. The higher the partial pressure of the acid gases, the more they absorb. They pay off for gas with a high content of H₂S and CO₂ under high pressure.
Solid absorbents, iron and zinc oxides, are for small amounts of hydrogen sulfide; we replace the spent absorbent. Membranes (thin films that let some gases through more easily than others) mainly remove carbon dioxide.
What to do with the acid gas. A mixture of H₂S and CO₂ (the acid gas) leaves the regenerator. We either turn it into elemental sulfur in a Claus unit (a plant that turns hydrogen sulfide into sulfur), or compress it and inject it into a reservoir. At Tengiz, Karachaganak and Kashagan, what goes back into the reservoir is not even the acid gas but part of the raw gas as a whole. This keeps up the reservoir pressure, and less gas has to be treated (gas injection in Kazakhstan).

Reactions of the Claus process

2\,\mathrm{H_2S} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{SO_2} + 2\,\mathrm{H_2O},\qquad 2\,\mathrm{H_2S} + \mathrm{SO_2} \rightarrow 3\,\mathrm{S} + 2\,\mathrm{H_2O}

where:

$\mathrm{H_2S}$ hydrogen sulfide from the acid gas, —
$\mathrm{SO_2}$ sulfur dioxide, made by burning about a third of the hydrogen sulfide, —
$\mathrm{S}$ elemental sulfur, a sales product, —

Units: reaction equations; stoichiometry in moles

Holds when

The first reaction takes place in a furnace, and the second in the furnace and in catalytic reactors (vessels with a catalyst that speeds up the reaction). How completely the sulfur is recovered depends on the number of stages, and the remainder is cleaned up further.
Overall, each mole of hydrogen sulfide gives one mole of sulfur, so the mass of sulfur is the moles of H₂S multiplied by 32.06 g/mol.

Sulfur is a significant sales product of the sour fields of the Pre-Caspian Basin (sulfur of Tengiz and Kashagan).

Worked exampleSour gas: materials, the limit and sulfur

Given (assumed data)

Gas: 1 million m³/d at 8.0 MPa (absolute), 1.5% (mole) of hydrogen sulfide
Limit: hydrogen sulfide not more than 0.007 g/m³; in Kazakhstan until 2030, for the domestic market, 0.020 g/m³

Solution

The partial pressure of hydrogen sulfide is $p_{\text{H}_2\text{S}} = 0.015\cdot 8000 = 120$ kPa. This is 400 times the threshold of 0.3 kPa, so steels resistant to sulfide stress cracking are needed.
The threshold at 8 MPa is $0.3/8000 = 37.5\cdot 10^{-6}$ . So 37.5 ppm of hydrogen sulfide is already enough to require such steels.
The limit in mole fractions is $0.007/(34.08\cdot 41.57) = 4.94\cdot 10^{-6}$ (4.94 ppm). The temporary Kazakhstan limit is $0.020/(34.08\cdot 41.57) = 14.1\cdot 10^{-6}$ .
The gas treatment must remove $1 - 4.94\cdot 10^{-6}/0.015 = 99.97$ % of the hydrogen sulfide.
Sulfur: $1{,}000{,}000\cdot 0.015\cdot 41.57 = 623{,}550$ mol of H₂S per day. With full conversion this gives $623{,}550\cdot 32.06/10^6 = 20.0$ t of sulfur per day.

Answer: The gas is sour and needs special steels: 120 kPa of hydrogen sulfide against the threshold of 0.3 kPa. To deliver it into the gas pipeline, 99.97% of the hydrogen sulfide must be removed, and about 20.0 t of sulfur a day can be made from it.

What it means. The limit of 0.007 g/m³ is about 4.94 ppm. Hydrogen sulfide must be removed almost completely, and the most expensive units of gas processing plants do this work. Sulfur comes out in tonnes a day even from a small stream. If there is nowhere to sell it, or the process is too expensive, we inject the acid gas into the reservoir. The threshold of 0.3 kPa refers to sulfide stress cracking. Hydrogen cracking of steels is possible with even smaller amounts of hydrogen sulfide, so in sour environments we choose materials according to the whole standard. The data in this example are assumed for illustration.

Hydrocarbon dew point and the recovery of liquid hydrocarbons

Natural gas with an admixture of propane, butanes, and especially pentanes and heavier hydrocarbons gives off liquid when it is cooled and when its pressure falls within a certain range. Gas condensate systems behave like this (retrograde condensation: liquid drops out of the gas as the pressure falls). In a gas pipeline this liquid collects in low places, narrows the cross-section and forms liquid plugs (slugs). So we cool the gas in advance below the required hydrocarbon dew point and separate the liquid that condenses. On the way, this gives valuable products: stable condensate, liquefied propane and butanes, the wide fraction of light hydrocarbons, and at large plants also ethane.

Low-temperature separation (LTS). We cool the high-pressure gas in a recuperative heat exchanger (one in which the cold gas from the outlet of the unit cools the incoming gas), and then throttle it. The Joule–Thomson effect (the cooling of a gas as it expands) lowers the temperature (cooling by throttling). In the low-temperature separator the condensate and water are separated. We inject methanol or glycol before the cooling, or hydrates will block the unit. The reservoir itself provides the energy, as long as there is a free pressure drop.
Refrigeration unit: the gas is cooled to minus 20–40 °C by evaporating propane in a heat exchanger (an evaporator). It does not need a pressure drop, but it uses power for the compressor of the refrigeration cycle.
Turboexpander: the gas expands in a turbine and does work, so it cools more than in throttling with the same pressure drop. The work of the turbine drives a compressor, which partly restores the pressure. Turboexpander plants cool the gas below minus 70 °C and recover most of the ethane and almost all the propane.
Stabilisation of condensate and fractionation of NGL: the liquid that has been separated is saturated with methane and ethane. It is stabilised in columns, as oil is (oil stabilisation), and the wide fraction is split into separate products: propane, butanes and pentanes.

Worked exampleIs the pressure drop enough for low-temperature separation?

Given (assumed data)

Inlet gas: 11.0 MPa and 20 °C (after the inlet separator)
Low-temperature separator: 6.5 MPa (the pipeline pressure); −15 °C is needed to provide the dew point with a margin
Cooling by throttling: on average 3.5 °C per 1 MPa (as for methane in the lesson on chokes, assumed)

Solution

The pressure drop is $11.0 - 6.5 = 4.5$ MPa. Throttling alone would cool the gas to $20 - 3.5\cdot 4.5 = 4.25$ °C, which is far from −15 °C.
To have −15 °C after the choke, the gas before it must be at $-15 + 3.5\cdot 4.5 = 0.75$ °C. The recuperative heat exchanger must cool the incoming gas from 20 to 0.75 °C with the cold gas from the separator.
How the recovery of cold works. We assume a counter-current heat exchanger (the two gas streams flow in opposite directions) with equal flow rates and heat capacities of the streams. Then the temperature difference is the same along its whole length and equals the cooling at the choke, here $3.5\cdot 4.5 = 15.75$ °C. The heat that the exchanger takes from the incoming gas is proportional to this difference. So with the same heat transfer surface and the same gas rate, the total cooling of the unit ( $20 - (-15) = 35$ °C) is proportional to the pressure drop at the choke.
When the inlet pressure falls to 8.0 MPa, throttling gives only $3.5\cdot(8.0 - 6.5) = 5.25$ °C, and the whole unit cools the gas by $35\cdot 5.25/15.75 = 11.7$ °C, to 8.3 °C instead of −15 °C.

Answer: While the inlet pressure is 11 MPa, the unit works on the energy of the reservoir: the pressure drop of 4.5 MPa and the recovery of cold give −15 °C. At an inlet pressure of 8 MPa the separator will be at about 8.3 °C, and the dew point can no longer be met.

What it means. This is how gas condensate fields age: low-temperature separation, which pays at the start, loses its cold as the reservoir pressure falls. Then we either place a booster compressor station before the unit to create the pressure drop again, or add a refrigeration unit or a turboexpander. The simplifications of the calculation work in both directions. The Joule–Thomson coefficient (the cooling per unit of pressure drop) at low temperatures and for heavier gas is larger than 3.5 °C/MPa, so the cooling at the choke will be stronger than the estimate. But heat is released when gas condenses in the exchanger, and there is less cold gas after the separator than incoming gas. These two effects make the recovery of cold worse. An accurate calculation uses an equation of state with an enthalpy balance (a balance of the heat content). The data in this example are assumed for illustration.

Gas compression

Compressors raise the gas pressure where it is not enough. The gas of the last stages of oil separation is raised to the pressure of the gas pipeline or of processing. The gas of depleting reservoirs is raised to the pressure of the treatment unit (booster stations). The gas for gas lift (the gas lift system on the surface) and for injection into the reservoir is raised to tens of megapascals. A compressor unit is a compressor together with its driver: a gas turbine, a gas engine or an electric motor.

Reciprocating and centrifugal compressors

Property: Principle
Reciprocating: a piston pushes the gas out of a cylinder
Centrifugal: impellers (spinning wheels with blades) speed the gas up, and a diffuser (a widening passage) slows it down

Property: Compression ratio of a stage
Reciprocating: high, usually up to 3–4
Centrifugal: small per impeller; a casing holds several impellers

Property: Flow rate
Reciprocating: small and medium; easy to change
Centrifugal: large; a narrow working range

Property: Features
Reciprocating: pulsations, more maintenance
Centrifugal: surge (unstable flow reversals) at low flow, so anti-surge protection is needed

Property	Reciprocating	Centrifugal
Principle	a piston pushes the gas out of a cylinder	impellers (spinning wheels with blades) speed the gas up, and a diffuser (a widening passage) slows it down
Compression ratio of a stage	high, usually up to 3–4	small per impeller; a casing holds several impellers
Flow rate	small and medium; easy to change	large; a narrow working range
Features	pulsations, more maintenance	surge (unstable flow reversals) at low flow, so anti-surge protection is needed

Screw compressors are used for low-pressure gas, for example in units that recover vapours from tanks. Liquid in the gas is dangerous for any machine: in reciprocating compressors it breaks the valves, and in centrifugal ones it wears the impeller blades. So we place a gas scrubber (a vessel that removes liquid droplets from gas) at the inlet of each stage.

Gas heats up when it is compressed: the work spent on compression turns into its internal energy. Without heat exchange (adiabatically) the temperature grows as a power of the compression ratio, and a real compressor heats the gas even more because of losses. A high discharge temperature (the temperature of the gas leaving the compressor) breaks down the lubricating oil and damages the seals and the valves. So industry standards and manufacturers limit it, as a rule to values of about 135–150 °C. This is where stages come from. We compress the gas in one stage, cool it in an intercooler (in fields, usually an air cooler: a heat exchanger cooled by fans), separate the liquid that dropped out, and compress the gas further. Intercooling also reduces the work: cold gas takes a smaller volume.

Compression ratio of a stage when the ratios are equal

r = \left(\frac{p_{\text{out}}}{p_{\text{in}}}\right)^{1/n}

where:

$r$ compression ratio of one stage: the ratio of the discharge and suction pressures, —
$p_{\text{out}},\ p_{\text{in}}$ absolute discharge pressure of the station and suction pressure of the first stage, MPa
$n$ number of stages, —

Units: absolute pressures in identical units

Holds when

The gas is cooled after each stage to the same temperature. The pressure losses in coolers and separators are small (in practice the ratios are raised slightly to cover them).

Equal compression ratios give the smallest total work at the same suction temperature of the stages (the derivation is in the advanced part of the lesson), and the same discharge temperature in all stages. The same rule of thumb, equal pressure ratios, gives a first estimate of the pressures of separation stages (number of separation stages). We usually compress the gas of each separation stage to the pressure of the previous one.

Discharge temperature and specific work of a stage

T_2 = T_1\left[1 + \frac{r^{(k-1)/k} - 1}{\eta}\right],\qquad w = \frac{k}{k-1}\cdot\frac{Z\,R\,T_1}{M}\cdot\frac{r^{(k-1)/k} - 1}{\eta}

where:

$T_1,\ T_2$ absolute suction and discharge temperature, K
$k$ adiabatic exponent of the gas, $c_p/c_v$ (the ratio of the heat capacities at constant pressure and at constant volume): about 1.2–1.3 for natural gas, —
$\eta$ adiabatic (isentropic) efficiency of the stage: the ideal work divided by the real work, —
$w$ specific work of compression, per kilogram of gas, J/kg
$Z$ gas deviation factor at suction (the correction for the difference between real and ideal gas), —
$R$ universal gas constant, 8.314, J/(mol·K)
$M$ molar mass of the gas, kg/mol

Units: SI; the power of a stage is $w$ multiplied by the mass flow rate of the gas

Holds when

An ideal gas with a constant $k$ and the correction $Z$ at suction. At high pressures, and for rich gas, it is more accurate to calculate with an equation of state.
The losses are covered by one efficiency, referred to adiabatic compression. Mechanical losses and losses in the driver are added separately.
The gas is single-phase: the liquid is separated before the stage.

Work and heating grow with the compression ratio not in proportion but as its power $(k-1)/k$ , about 0.2. In the ideal case, doubling the compression ratio raises the absolute discharge temperature by $2^{0.2} \approx 1.15$ times, and the work of a stage grows more slowly than the compression ratio. The power is proportional to the mass flow rate and to the absolute suction temperature: warm gas is more expensive to compress.

CalculatedDischarge temperature of a stage against its compression ratio: gas entering the stage at 40 °C,

k

= 1.25, efficiency 0.78. The dashed line is the limit of 150 °C assumed in the example. The points are two, three and four equal stages for a total compression ratio of 25.

Worked exampleHow many stages and how much power are needed to deliver associated gas

Given (assumed data)

Gas: 300,000 m³/d (20 °C, 0.101325 MPa), specific gravity 0.80 ( $M = 23.18$ kg/kmol)
Pressures: suction 0.3 MPa, discharge 7.5 MPa (absolute)
Conditions: the gas is cooled to 40 °C before each stage; $k = 1.25$ , adiabatic efficiency 0.78; the discharge temperature is not more than 150 °C (assumed)

Solution

The total compression ratio is $7.5/0.3 = 25$ ; the exponent is $(k-1)/k = 0.2$ ; $T_1 = 313.15$ K.
One stage: $25^{0.2} = 1.9037$ , $T_2 = 313.15\cdot[1 + (1.9037 - 1)/0.78] - 273.15 = 403$ °C. This is not possible.
Two stages: $r = \sqrt{25} = 5$ , $5^{0.2} = 1.3797$ , $T_2 = 192$ °C. This is above the limit.
Three stages: $r = 25^{1/3} = 2.924$ , $2.924^{0.2} = 1.2394$ , $T_2 = 313.15\cdot[1 + 0.2394/0.78] - 273.15 = 136$ °C. This is acceptable. The suction pressures of the stages are 0.3, $0.3\cdot 2.924 = 0.877$ and $0.877\cdot 2.924 = 2.564$ MPa.
Mass flow rate: the gas density at standard conditions is $0.80\cdot 1.2043 = 0.9634$ kg/m³ (1.2043 kg/m³ is the density of air at 20 °C and 0.101325 MPa), and $300{,}000\cdot 0.9634/86{,}400 = 3.345$ kg/s.
Specific work: $RT_1/M = 8.314\cdot 313.15/23.18 = 112.32$ kJ/kg. With $Z$ at suction of 0.9908, 0.9729 and 0.9199, the stages give $5\cdot Z\cdot 112.32\cdot 0.2394/0.78$ = 170.8, 167.7 and 158.6 kJ/kg, in total 497.1 kJ/kg.
The compression power is $3.345\cdot 497.1 = 1663$ kW. The power at the shaft is larger by the mechanical losses of the compressor.

Answer: Three stages are needed, with a compression ratio of 2.924 and a discharge temperature of 136 °C. The compression power is about 1.66 MW, without mechanical losses and losses in the driver.

What it means. Here the number of stages is set by the temperature, not by savings: two stages would give 192 °C, and a fourth stage would save only about 3.4% more power (the table below). In the intermediate separators, after cooling, condensate drops out: propane, butanes and heavier. It is collected and stabilised, and it is part of the value of the associated gas. The calculation with a constant $k$ and an ideal gas is approximate. For rich gas at high pressures, the work and the temperatures are refined with an equation of state. The data in this example are assumed for illustration.

Number of stages: the same station

Stages: 1
Compression ratio of a stage: 25.00
Discharge temperature, °C: 403
Power, kW: 2157

Stages: 2
Compression ratio of a stage: 5.00
Discharge temperature, °C: 192
Power, kW: 1779

Stages: 3
Compression ratio of a stage: 2.92
Discharge temperature, °C: 136
Power, kW: 1663

Stages: 4
Compression ratio of a stage: 2.24
Discharge temperature, °C: 110
Power, kW: 1606

Stages	Compression ratio of a stage	Discharge temperature, °C	Power, kW
1	25.00	403	2157
2	5.00	192	1779
3	2.92	136	1663
4	2.24	110	1606

Calculated with the model of the lesson: equal compression ratios, cooling to 40 °C before each stage, $k$ = 1.25, efficiency 0.78, and $Z$ at the suction of each stage from a correlation for natural gas. One stage is only for comparison: no compressor can stand such a temperature. The data are assumed for illustration.

Gas treatment and processing in Kazakhstan

According to the data presented at the meeting of the Government of the Republic of Kazakhstan on 7 July 2026, the country produced 68.2 billion m³ of gas in 2025, and the production of sales gas was 27.4 billion m³. For 2026 the plan is 64.5 billion m³ of production and keeping sales gas at 27.4 billion m³. The difference, about 60%, is not a loss. Much of it is raw gas that the largest fields inject back into the reservoir to keep up the pressure. In 2024 Karachaganak alone injected about 14.2 billion m³ and Kashagan about 5.5 billion m³ (facilities of Tengiz and Karachaganak, gas injection). In addition, part of the raw gas of Karachaganak is processed at the plant in Orenburg, and part of the gas is used for the fields' own needs.

Gas processing projects of Kazakhstan as of July 2026

Project: Gas processing plant (GPP) at the Kashagan field (QazaqGaz)
Capacity: 1 billion m³ a year
Status: overall progress 61.4%, 98% of the main equipment delivered; construction and installation work is planned to finish by the end of 2026

Project: Second GPP at Kashagan
Capacity: 2.5 billion m³ a year
Status: the contractor has been chosen; the design and construction contract is planned to be signed in the fourth quarter of 2026

Project: GPP at Karachaganak
Capacity: 5 billion m³ a year
Status: QazaqGaz was appointed operator of the project on 13 April 2026; an agreement for early engineering work was signed on 2 July

Project: New GPP in Zhanaozen
Capacity: 900 million m³ of raw gas a year
Status: start-up is planned for 2027; it has been decided not to close the existing Kazakh GPP

Project	Capacity	Status
Gas processing plant (GPP) at the Kashagan field (QazaqGaz)	1 billion m³ a year	overall progress 61.4%, 98% of the main equipment delivered; construction and installation work is planned to finish by the end of 2026
Second GPP at Kashagan	2.5 billion m³ a year	the contractor has been chosen; the design and construction contract is planned to be signed in the fourth quarter of 2026
GPP at Karachaganak	5 billion m³ a year	QazaqGaz was appointed operator of the project on 13 April 2026; an agreement for early engineering work was signed on 2 July
New GPP in Zhanaozen	900 million m³ of raw gas a year	start-up is planned for 2027; it has been decided not to close the existing Kazakh GPP

As reported by the Ministry of Energy, QazaqGaz and KazMunayGas for the Government meeting of 7 July 2026. The plans and dates of the projects may change.

It is more profitable to connect small fields to plants that already work than to build new plants for them. According to the same data, the gas of the West Prorva and Central Urikhtau projects and of the Rozhkovskoye field is planned to be processed at existing plants: the Borankol and Zhanazhol GPPs and the plant of Zhaikmunai LLP. KazMunayGas plans to raise its production of sales gas from 2.4 to 4.5 billion m³ a year by 2030, and of liquefied petroleum gas from 1.3 to 1.5 million t.

Terms and symbols

Gas treatment units, gas processing plant, booster compressor station: The integrated gas treatment unit at a field brings the gas to the sales requirements. The preliminary gas treatment unit does the first operations near the wells. The gas processing plant (GPP) does deep processing. The booster compressor station raises the pressure of the gas when the reservoir pressure falls.
Water dew point: The temperature at which water drops out of the gas at a given pressure.
Hydrocarbon dew point: The temperature at which hydrocarbons condense from the gas at a given pressure.
Water content: The mass of water vapour in one cubic metre of gas at standard conditions.
Bukacek correlation: A formula for the water content of gas saturated with water: the ideal part $p_s/p$ plus a correction that depends on temperature.
Hydrate inhibitor: A substance that lowers the temperature at which hydrates form (methanol, glycols) or hinders the growth and sticking together of their crystals.
Hammerschmidt formula: A formula for the lowering of the hydrate formation temperature from the concentration of the inhibitor in the water and its molar mass.
Glycol dehydration: Absorption of water vapour from gas by triethylene or diethylene glycol, with the glycol regenerated by heating.
Adsorption drying: Uptake of water by solid adsorbents: silica gel, alumina and zeolites.
Amine gas treating: Removal of hydrogen sulfide and carbon dioxide with solutions of amines in water. The amines are then regenerated.
Low-temperature separation: Cooling of gas by throttling with recovery of cold, and separation of condensate and water.
Compression ratio: The ratio of the absolute discharge and suction pressures of a compressor stage.
Surge: Unstable operation of a centrifugal compressor at low flow: the flow periodically changes direction.

Common mistakes and misconceptions

Wrong

“The dew point of the gas is −10 °C, so water will not drop out of it at any temperature above −10 °C.”

In fact

The dew point depends on pressure: at a higher pressure the same water drops out at a higher temperature. That is why the standards state the pressure (3.92 MPa for the water dew point), and we check the gas at the highest pressure and the lowest temperature on its way.

Wrong

“The amount of methanol needed is what the Hammerschmidt formula gives.”

In fact

The formula gives the concentration in the water phase. Methanol also goes into the gas and the condensate, so we must feed in more. And at concentrations above 25% the formula underestimates the need.

Wrong

“The more compression stages, the better.”

In fact

Each further stage saves less power, but costs as much as a separate cylinder or casing with a cooler and a separator. The minimum number of stages is set by the discharge temperature, and after that economics decides.

Wrong

“There is little hydrogen sulfide in the gas, hundredths of a per cent, so ordinary steels will do.”

In fact

What decides is the partial pressure: at 8 MPa, 40 ppm of hydrogen sulfide already gives more than 0.3 kPa, and steels resistant to sulfide stress cracking are needed.

In depth: advanced detail

Where the Hammerschmidt formula comes from and where it stops working

An inhibitor works in the same way as salt on ice. Dissolved in water, it lowers the chemical potential (the activity) of the water, and it becomes harder for the solid phase, ice or hydrate, to exist. For an ideal solution the lowering of the equilibrium temperature is proportional to $-\ln x_w$ , where $x_w$ is the mole fraction of water in the solution. This gives the logarithmic form proposed for methanol by Nielsen and Bucklin:

Lowering of the hydrate formation temperature (the logarithmic form of Nielsen and Bucklin)

\Delta T = -72.0\,\ln x_w,\qquad x_w = \frac{1}{1 + \dfrac{18.015\,W}{M\,(100 - W)}}

where:

$\Delta T$ lowering of the hydrate formation temperature, °C
$x_w$ mole fraction of water in the water solution of the inhibitor, fraction
$W,\ M$ concentration of the inhibitor (% by mass) and its molar mass, %, kg/kmol
$\ln$ natural logarithm, —

Units: °C (in °F the constant is 129.6)

Holds when

It was proposed for methanol. Reference books recommend it for concentrations from about 25 to 50% by mass, where the Hammerschmidt formula is already inaccurate.
The solution is treated as ideal. For glycols, and for mixed inhibitors with salts, a calculation with an equation of state and a water activity model is more accurate.

At small concentrations $-\ln x_w \approx 18.015\,W/(M(100 - W))$ , and the formula turns into the Hammerschmidt formula with $K = 72.0\cdot 18.015 \approx 1297$ . Both describe the same physics. The empirical Hammerschmidt formula is the same as the first term of the series for the logarithm, without the higher-order terms.

The difference grows quickly with concentration. At 21.93% of methanol from the example of the lesson, the Hammerschmidt formula gives a lowering of 11.4 °C and the logarithmic one 10.6 °C. At 40% they give 27.0 and 22.9 °C. At high concentrations the Hammerschmidt formula promises more protection than the solution gives, so the dose it gives is too small. When a large lowering is needed, we calculate with the logarithmic form or with an equation of state, and we choose the temperature margin with the inaccuracy of the method in mind.

Why stages with equal compression ratios are better

Take two stages with cooling between them to the same temperature $T_1$ , and treat the gas as ideal. The work of compression from $p_1$ to the intermediate pressure $p_2$ and then to $p_3$ is proportional to the sum $(p_2/p_1)^{e} + (p_3/p_2)^{e} - 2$ , where $e = (k-1)/k$ . The sum is smallest where its derivative with respect to $p_2$ (its rate of change) is zero. This happens when $p_2^{2e} = (p_1 p_3)^{e}$ , that is, at $p_2 = \sqrt{p_1 p_3}$ . Both stages compress the gas by a factor of $\sqrt{p_3/p_1}$ . For $n$ stages the same derivation gives $r = (p_{\text{out}}/p_{\text{in}})^{1/n}$ .

The minimum is flat: a small departure from equal ratios hardly changes the power. So in practice we choose the ratios also for other reasons: so that the stages discharge into the headers (common pipes) of separators at the required pressure, to equalise the discharge temperatures when the cooling temperatures differ, and to cover the pressure losses in the coolers. If the gas after a cooler is warmer than before the first stage, it pays to give that stage a smaller compression ratio.

Connections to other topics

Oil treating: separators, dehydration, desalting and stabilisation — where the associated gas of the separation stages comes from, and how the liquid is stabilised.
Chokes, surface pressure losses and well temperature — cooling by throttling and the hydrate formation temperature.
Field development concepts: surface facilities — where gas treatment units and compressors stand in the facilities of a field.
Gas lift: principles and choice of operating conditions — compressors and gas distribution for gas lift.
Displacement of oil by gas — gas injection into the reservoir in the fields of Kazakhstan.
Retrograde condensation, wax, asphaltenes and high-pressure fluids — retrograde condensation, hydrates and complex fluids.
State regulation of field development — the ban on flaring and the gas processing programmes.
Ideal and real gas: the gas deviation factor (Z) — the gas deviation factor in calculations of volumes and of the work of compression.

Check yourself

1
Gas is saturated with water at 7 MPa and 40 °C. It is cooled to 20 °C at the same pressure. What happens to the water content, and where does the “extra” water go?
Show the answer
The water content is almost proportional to the vapour pressure of water, and on cooling from 40 to 20 °C this falls by a factor of about 3.2 (from 7.38 to 2.34 kPa). The gas can hold only about a third of its former water (a little more, with the Bukacek correction). The rest condenses. In a pipe it becomes liquid water, which can form hydrates together with the gas. In a plant it condenses in a separator, where it is drawn off.
2
To lower the hydrate formation temperature by 8 °C, what methanol concentration is needed in the water, by the Hammerschmidt formula?
1. A.About 40%
2. B.About 25%
3. C.About 16%
4. D.About 8%
Show the answer
Correct answer: C. About 16%
$W = 100\cdot 8\cdot 32.04/(1297 + 8\cdot 32.04) = 25{,}632/1553.3 \approx 16.5$ %. This is within the range where the formula applies. For the same lowering, monoethylene glycol by the same formula would need about 28%, almost twice as much by mass.
3
Why does low-temperature separation, which worked well in the first years, in time stop meeting the dew point requirement at a gas condensate field?
Show the answer
The cold in it comes from throttling, that is, from the difference between the inlet pressure and the pipeline pressure. As the accumulation is depleted, the inlet pressure falls, and the pressure drop and the cooling become smaller (at 3.5 °C per megapascal, 1.5 MPa gives only about 5 °C). Then we place a booster compressor station before the unit, or add a refrigeration unit or a turboexpander.
4
Gas at 10 MPa contains 30 ppm of hydrogen sulfide. Are steels resistant to sulfide stress cracking needed?
1. A.Yes: the partial pressure is 3 kPa
2. B.No: 30 ppm is only traces
3. C.Yes: the partial pressure is 0.3 kPa, at the threshold
4. D.No: the threshold applies only to oil
Show the answer
Correct answer: C. Yes: the partial pressure is 0.3 kPa, at the threshold
$p_{\text{H}_2\text{S}} = 30\cdot 10^{-6}\cdot 10{,}000 = 0.3$ kPa, exactly the threshold of the standard for materials for hydrogen sulfide environments. We treat such an environment as sour and choose the materials according to the standard. At 15 MPa the same 30 ppm would already give 0.45 kPa.
5
Gas must be compressed from 0.5 to 6.0 MPa under the same conditions as in the example of the lesson: 40 °C at the suction of each stage, $k = 1.25$ , adiabatic efficiency 0.78, and a discharge temperature of not more than 150 °C. Will two stages be enough?
Show the answer
The total compression ratio is $6.0/0.5 = 12$ . For two equal stages $r = \sqrt{12} \approx 3.46$ , $3.46^{0.2} \approx 1.282$ , and $T_2 = 313.15\cdot[1 + 0.282/0.78] - 273.15 \approx 153$ °C, a little above the limit. There are two ways out: a third stage ( $r = 12^{1/3} \approx 2.29$ , about 112 °C), or colder gas at suction; at 30 °C two stages give about 140 °C. Which way is better is decided by cost and climate: in hot summers air coolers cool the gas less well.

Updated: 19 October 2026