Chokes, surface pressure losses and well temperature

Quick refresher

When the pressure downstream of the choke (after it) falls to about half of the upstream pressure (before it), the flow in the narrowest part reaches the speed of sound. From then on the flow is critical. The flow rate depends only on the pressure and composition of the flow upstream of the choke, and pressure swings downstream do not reach the well.
For a gas the critical pressure ratio is $(2/(k+1))^{k/(k-1)}$ , about 0.55 at $k = 1.3$ . ( $k$ is the ratio of the specific heats of the gas.) A liquid without gas has no critical flow in this sense. Its flow rate grows as the square root of the pressure drop.
For a gas–liquid flow, engineers use empirical correlations of the form $p_{wh} = b\,q\,\text{GLR}^{\,c}/S^a$ (Gilbert and his followers). In critical flow the pressure upstream of the choke is proportional to the rate. The correlations were built from the data of different fields, and they differ from one another by tens of per cent. So check them against measurements.
With the choke fully open, the flowline and the separator set the wellhead pressure. Losses in the line grow faster than the rate. Toward the separator the gas expands and speeds up the mixture. So a drop in separator pressure does not reach the wellhead in full.
The greater the mass flow rate, the less the flow cools on its way to the wellhead. By Ramey's method, the amount by which the flow temperature exceeds the rock temperature tends to $G_T A$ , where $G_T$ is the geothermal gradient and the relaxation distance $A$ is proportional to the mass flow rate. A well that is choked back is colder, and wax comes out of the oil over a greater length of pipe.
After the choke the gas cools (the Joule–Thomson effect). For methane at pressures of 5–13 MPa the cooling is roughly 2.5–4.5 °C for each megapascal of pressure drop. If the temperature after the choke is below the hydrate formation temperature, you need heating before the choke, an inhibitor (a chemical that prevents hydrates), or another way to lower the pressure.

Key relationships

r_c = \left(\dfrac{2}{k+1}\right)^{k/(k-1)}

critical ratio of the pressures downstream and upstream of the choke for a gas

q = C_d A \sqrt{\dfrac{2\,\Delta p}{\rho}}

liquid through a choke (subcritical flow)

p_{wh} = \dfrac{b\,q\,\text{GLR}^{\,c}}{S^{a}}

Gilbert-type correlations for critical flow of a gas–liquid mixture

T_f(z) = T_e(z) + G_T A\left[1 - e^{-(H - z)/A}\right]

flow temperature profile by Ramey's method

A = \dfrac{w\,c_p\,(k_e + r U f)}{2\pi\,r U\,k_e}

relaxation distance

Common pitfalls

Using Gilbert and similar correlations in subcritical flow, that is, when the pressure downstream of the choke is more than about 0.5–0.6 of the upstream pressure.
Putting gauge pressures (measured above atmospheric pressure) instead of absolute ones, and degrees Celsius instead of kelvins, into the gas formulas.
Taking the producing gas–oil ratio instead of the gas–liquid ratio (GLR), or mixing up the GLR units in Gilbert's correlation. The constant 10 goes with scf/STB (standard cubic feet per stock-tank barrel), and 435 goes with thousands of cubic feet per barrel.
Assuming that a 0.4 MPa drop in separator pressure will lower the wellhead pressure by 0.4 MPa too.
Checking the hydrate margin only at steady state. At start-up the wellhead and the flowline are cold, and that is when hydrates most often form.

After this lesson you can

explain how critical flow through a choke (the flow reaches the speed of sound in the narrowest part) differs from subcritical flow, and calculate the critical pressure ratio for a gas
calculate the flow rate of gas and liquid through a choke, and choose the diameter of the choke bean (the insert with the hole in it)
estimate the rate of a well from the pressure upstream of the choke with Gilbert-type correlations (formulas fitted to field measurements), and understand their scatter and limits
find the operating point of a well with the node at the wellhead, both with the choke in place and with the flowline fully open, and estimate how much the line and the separator contribute
calculate the temperature profile of the flow by Ramey's method, and from it the margin above the wax appearance temperature (the temperature at which wax starts to come out of the oil) and above the hydrate formation temperature

The surface part of the system: from the wellhead to the separator

After leaving the tubing, the flow passes through the Christmas tree: the valves and the cross above the tubing head. Then it goes through the choke on the working (flow) wing, and along the flowline to the gathering point. In Russian-language practice the flowlines of several wells usually meet at an automated group metering station, which measures the rate of each well in turn. From there the production goes through a common gathering header (a large collecting pipe) to a booster pump station, or straight to a treatment plant with separators.

The pressure upstream of the choke is the wellhead pressure $p_{wh}$ . In Russian-language practice it is called the buffer pressure, because the gauge on the “buffer” of the tree shows it. The pressure downstream of the choke, at the start of the flowline, is called the line pressure. We write it $p_{ch}$ . On the surface, as in the well, there is a pressure balance (pressure balance of the production system): $p_{wh} - p_{sep} = \Delta p_{\text{choke}} + \Delta p_{\text{line}}$ . Everything spent at the surface presses back on the well as back pressure. The higher the wellhead pressure, the higher the bottomhole pressure that is needed, and the lower the rate.

Elements of the surface system and the pressure losses in them

Element: Choke
Its purpose: to create a set pressure drop and limit the rate
What the loss depends on: the size of the narrowing and the rate; in critical flow, only the state of the flow upstream of it

Element: Valves, tees, bends
Its purpose: to shut off and direct the flow
What the loss depends on: small for open ones; a partly open valve acts as a choke

Element: Flowline
Its purpose: to bring the production to the gathering point
What the loss depends on: the length, diameter and profile of the route; friction grows faster than the rate

Element: Metering station, header
Its purpose: to measure the rate and collect the production of the wells
What the loss depends on: the total flow of all the wells connected to the header

Element: Separator
Its purpose: to separate gas from liquid
What the loss depends on: a regulator holds its pressure; this is the lower limit of the whole chain

Element	Its purpose	What the loss depends on
Choke	to create a set pressure drop and limit the rate	the size of the narrowing and the rate; in critical flow, only the state of the flow upstream of it
Valves, tees, bends	to shut off and direct the flow	small for open ones; a partly open valve acts as a choke
Flowline	to bring the production to the gathering point	the length, diameter and profile of the route; friction grows faster than the rate
Metering station, header	to measure the rate and collect the production of the wells	the total flow of all the wells connected to the header
Separator	to separate gas from liquid	a regulator holds its pressure; this is the lower limit of the whole chain

The resistance of the tree and the fittings is usually counted as an extra length of pipe, called the equivalent length.

The roles of the elements change as the well ages. While the reservoir pressure is high, the well can deliver more than the reservoir and the equipment allow. Then the choke “eats” most of the wellhead (buffer) pressure. As the reservoir is depleted, the choke is opened up and later removed altogether. Then the flowline and the separator pressure set the rate. This lesson looks at both stages on one and the same example well, the lesson's well. The lesson Well performance sensitivity analysis is built on the same well.

The lesson's well (assumed data)

Parameter: Depth of the middle of the perforated interval and of the tubing shoe
Value: 2000 m, vertical well

Parameter: Tubing
Value: 73 mm (2⅞″), inner diameter 62 mm

Parameter: Reservoir
Value: $\bar p_R = 22$ MPa, $p_b = 12$ MPa; $k = 80$ mD, $h = 12$ m, $\mu_o = 1.4$ mPa·s, $B_o = 1.2$ ; $r_e = 300$ m, $r_w = 0.1$ m, $S = 3$

Parameter: Productivity index
Value: $J = 29.85$ m³/(d·MPa), a straight line above $p_b$ , a Vogel curve below it

Parameter: Production
Value: oil 850 kg/m³, water 1050 kg/m³, water cut $\text{WC} = 10$ %, producing gas–oil ratio $R = 90$ m³/m³

Parameter: Gas–liquid ratio (GLR)
Value: $\text{GLR} = R\,(1 - \text{WC}) = 90\cdot(1 - 0.10) = 81$ m³/m³, gas of specific gravity 0.70

Parameter: Flowline and separator
Value: 3 km of 114 × 7 mm pipe (100 mm inside), level terrain; separator at 0.8 MPa

Parameter	Value
Depth of the middle of the perforated interval and of the tubing shoe	2000 m, vertical well
Tubing	73 mm (2⅞″), inner diameter 62 mm
Reservoir	$\bar p_R = 22$ MPa, $p_b = 12$ MPa; $k = 80$ mD, $h = 12$ m, $\mu_o = 1.4$ mPa·s, $B_o = 1.2$ ; $r_e = 300$ m, $r_w = 0.1$ m, $S = 3$
Productivity index	$J = 29.85$ m³/(d·MPa), a straight line above $p_b$ , a Vogel curve below it
Production	oil 850 kg/m³, water 1050 kg/m³, water cut $\text{WC} = 10$ %, producing gas–oil ratio $R = 90$ m³/m³
Gas–liquid ratio (GLR)	$\text{GLR} = R\,(1 - \text{WC}) = 90\cdot(1 - 0.10) = 81$ m³/m³, gas of specific gravity 0.70
Flowline and separator	3 km of 114 × 7 mm pipe (100 mm inside), level terrain; separator at 0.8 MPa

Pressures are absolute. Gas volumes are at 0.101325 MPa and 20 °C. The inflow performance relationship is composite (composite IPR). The flow in the tubing follows the drift-flux model of the lesson Multiphase flow in tubing. In that model all the gas is treated as free, and the temperature in the pipes as constant, 50 °C.

Why a well needs a choke and how it is built

Limit the drawdown on the formation, so as not to produce sand, break up weak rock or draw up a cone of water or gas (Sand production: types, prediction and cost of the problem, Water and gas coning).
Protect the equipment: reduce the pressure to what the flowline, the metering station and the separator can take.
Distribute production: keep to the operating mode set by the field development plan, and share the processing capacity between wells.
Keep the operating mode steady: in critical flow, pressure surges in the gathering network do not reach the well.

The simplest is the fixed (positive) choke. It is a replaceable bean of cemented carbide (a very hard metal) with a calibrated hole. To change the operating mode, you replace the bean with another one. An adjustable choke changes its opening without being taken apart. In a needle choke a conical stem enters a seat. In a disc choke two discs with holes are turned relative to each other. The size of a bean is given as the diameter of the hole in millimetres or, in the American tradition, in sixty-fourths of an inch: 1/64″ = 0.397 mm, so a 16/64″ bean is 6.35 mm. The jet in the choke is fast, and sand wears the bean and the body. That is why they are made of cemented carbides and ceramics, and why the size of the bean is checked from time to time: a worn bean passes more than was calculated.

SchematicFlow through a choke bean and the pressure along the axis (schematic, not to scale): acceleration in the orifice, the contracted section of the jet with the lowest pressure, eddies behind the bean and incomplete pressure recovery.

Inside the choke the flow speeds up sharply. Just after the entrance the jet goes on narrowing and becomes narrower than the orifice. This is the vena contracta (the contracted section), where the velocity is highest and the pressure lowest. Further on, the jet expands into the pipe, eddies at the walls use up its energy, and the pressure recovers only partly. The pressure difference between the two sides of the choke is exactly the irreversible loss (pressure lost for good). The choke is installed to create this loss.

Critical and subcritical flow

Pressure disturbances travel through a flow at the speed of sound relative to the flow itself. As long as the velocity in the narrowest part is below the speed of sound, a drop in the pressure downstream of the choke “reaches” the entrance. The flow speeds up, and the rate grows. This is subcritical flow. When the velocity in the contracted section becomes equal to the speed of sound, the signal about the pressure drop can no longer travel against the flow. The flow rate reaches its greatest value, and from then on it does not depend on the pressure downstream of the choke. This is critical, or sonic, flow.

For a gas, the critical pressure ratio $r_c$ marks the moment of transition. It is the ratio of the pressure downstream of the choke to the pressure upstream of it at which the speed of sound is reached in the contracted section. It comes from the energy equation for an ideal gas expanding without heat exchange or friction (the advanced section gives the derivation).

Critical pressure ratio for a gas

r_c = \left(\frac{p_2}{p_1}\right)_c = \left(\frac{2}{k+1}\right)^{\frac{k}{k-1}}

where:

$p_1,\ p_2$ absolute pressure upstream and downstream of the choke, MPa
$k$ adiabatic exponent — the ratio of the heat capacities of the gas at constant pressure and at constant volume, $c_p/c_v$ , —

Units: dimensionless

Holds when

An ideal gas, expansion without heat exchange or friction (isentropic); $k$ is constant.
The velocity of the gas upstream of the choke is small compared with the velocity in the narrowest part.
At $p_2/p_1 \le r_c$ the flow is critical and the flow rate does not depend on $p_2$ . At $p_2/p_1 > r_c$ it is subcritical.

For air ( $k = 1.40$ ) $r_c = 0.528$ . For methane ( $k \approx 1.30$ ) it is 0.546. For heavier natural gases ( $k \approx 1.25$ ) it is 0.555. So gas passes through a choke in critical flow if the pressure downstream is less than about 0.53–0.55 of the pressure upstream.

CalculatedGas flow rate through a choke against the ratio of the pressures downstream and upstream of it, at a constant pressure upstream of the choke (calculated for the gas well of the example below: 13 MPa, 38 °C, 16 mm bean). To the left of the critical ratio the flow rate is constant.

For a gas–liquid mixture, the speed of sound is much lower than in the gas or the liquid alone. The mixture is almost as compressible as a gas and almost as heavy as a liquid (the advanced section gives an estimate). So the mixture reaches critical flow at moderate velocities. Its critical pressure ratio is not constant, because it depends on the gas fraction. In practice engineers put the boundary at about 0.5–0.6. W. Gilbert wrote the first field correlation for chokes. He regarded the flow as critical if the pressure upstream of the choke was at least 1.7 times higher than downstream, that is, if the ratio $p_2/p_1$ was no more than 0.59.

Critical pressure ratio for a gas

Gas: Air
$k$: 1.40
$r_c$: 0.528

Gas: Methane, dry gas
$k$: 1.30
$r_c$: 0.546

Gas: Heavier natural gas
$k$: 1.25
$r_c$: 0.555

Gas	$k$	$r_c$
Air	1.40	0.528
Methane, dry gas	1.30	0.546
Heavier natural gas	1.25	0.555

The values of $k$ are for an ideal gas at moderate temperature. For a real gas at high pressure the adiabatic exponent is different, and the ratio is corrected using the composition of the gas.

Single-phase flow: gas and liquid

The same energy equation gives the gas flow rate through a choke. The gas speeds up as it expands. The mass flow rate equals the density of the gas in the contracted section, multiplied by the velocity and the area. A real orifice passes less than an ideal nozzle, because the jet contracts and there is friction. The discharge coefficient $C_d$ allows for this. It is the ratio of the actual flow rate to the ideal one. The manufacturer of the choke gives it, or you find it from measurements.

Gas flow rate through a choke

\dot m = C_d A \sqrt{\frac{2k}{k-1}\,p_1 \rho_1 \left[r^{2/k} - r^{(k+1)/k}\right]},\qquad r = \max\!\left(\frac{p_2}{p_1},\ r_c\right)

where:

$\dot m$ mass flow rate of the gas, kg/s
$C_d$ discharge coefficient of the choke, —
$A$ area of the bean orifice, $\pi d^2/4$ , m²
$p_1,\ \rho_1$ absolute pressure and density of the gas upstream of the choke; $\rho_1 = p_1 M/(Z_1 \mathcal{R} T_1)$ , Pa, kg/m³
$r$ ratio of the pressures downstream and upstream of the choke, but not less than the critical $r_c$ , —
$M,\ Z_1,\ T_1$ molar mass of the gas, gas deviation factor and absolute temperature upstream of the choke, kg/mol, —, K
$\mathcal{R}$ universal gas constant, 8.314 J/(mol·K), J/(mol·K)

Units: SI

Holds when

Isentropic expansion of an ideal gas, corrected for the gas deviation factor at the conditions upstream of the choke; $k$ is constant.
The velocity of the gas upstream of the choke is small (the orifice is much smaller than the pipe); there is no liquid in the flow.
Below the critical ratio, put $r_c$ into the formula. The flow rate then does not depend on the pressure downstream of the choke.

In the critical region the flow rate is proportional to the pressure upstream of the choke and to the area of the orifice. To double the rate, double the inlet pressure or the area of the bean (the diameter by a factor of $\sqrt 2$ ). The flow rate depends less on the temperature of the gas. It depends on it through the density, as $1/\sqrt{Z_1 T_1}$ .

In other unit systems (2)

Metric oilfield units, critical flow

q_{\text{sc}} = 3.326\,C_d\,d^2 p_1 \sqrt{\frac{k}{\gamma_{\text{g}} Z_1 T_1}}\left(\frac{2}{k+1}\right)^{\frac{k+1}{2(k-1)}}

$q_{\text{sc}}$ in thousand m³/d at 0.101325 MPa and 20 °C; $d$ in mm; $p_1$ in MPa (absolute); $T_1$ in K; $\gamma_{\text{g}}$ is the specific gravity of the gas relative to air.

$3.326 = (\pi/4)\cdot(T_{\text{sc}}/p_{\text{sc}})\cdot\sqrt{\mathcal{R}/M_{\text{air}}}\cdot 86{,}400/1000$ at $T_{\text{sc}} = 293.15$ K, $p_{\text{sc}} = 101{,}325$ Pa, $M_{\text{air}} = 0.02897$ kg/mol; the factors mm² → m² and MPa → Pa cancel each other.

Oilfield units, critical flow

q_{\text{sc}} = 879\,C_d\,A\,p_1 \sqrt{\frac{k}{\gamma_{\text{g}} Z_1 T_1}}\left(\frac{2}{k+1}\right)^{\frac{k+1}{2(k-1)}}

$q_{\text{sc}}$ in thousand standard cubic feet per day (Mscf/d) at 60 °F and 14.696 psia; $A$ in in²; $p_1$ in psia; $T_1$ in degrees Rankine.

$879$ is obtained from the same SI formula by converting in² → m², psi → Pa, °R → K and m³/s → thousand ft³/d (more precisely 879.0).

Worked exampleCritical flow rate and bean diameter for a gas well

Given (assumed data)

Upstream of the choke: 13.0 MPa (absolute), 38 °C
Downstream of the choke, in the flowline: 6.0 MPa (absolute)
Gas: dry natural gas — mainly methane with a small admixture of ethane and propane, without hydrogen sulfide or carbon dioxide: $\gamma_{\text{g}} = 0.58$ , $k = 1.30$
Discharge coefficient of the bean $C_d$: 0.85 (assumed)
Required rate: 500 thousand m³/d (0.101325 MPa, 20 °C)

Solution

Flow type: $p_2/p_1 = 6.0/13.0 = 0.462$ , while the critical ratio at $k = 1.30$ is $(2/2.3)^{1.3/0.3} = 0.546$ . The ratio is below the critical one, so the flow is critical, and the flowline pressure does not affect the flow rate.
Gas deviation factor upstream of the choke (from a correlation for natural gas). The pseudo-critical values are $T_{pc} = 195.7$ K and $p_{pc} = 4.64$ MPa. The pseudo-reduced values are $T_{pr} = 311.15/195.7 = 1.590$ and $p_{pr} = 13.0/4.64 = 2.80$ . So $Z_1 = 0.8266$ .
The factors of the formula: $\sqrt{k/(\gamma_{\text{g}} Z_1 T_1)} = \sqrt{1.30/(0.58\cdot 0.8266\cdot 311.15)} = 0.09335$ and $(2/2.3)^{2.3/0.6} = 0.5852$ .
Rate for each square millimetre of bean diameter: $q/d^2 = 3.326\cdot 0.85\cdot 13.0\cdot 0.09335\cdot 0.5852 = 2.008$ thousand m³/d.
Diameter: $d = \sqrt{500/2.008} = 15.8$ mm. We install the nearest bean, 16 mm. It will pass $2.008\cdot 16^2 = 514$ thousand m³/d.
Margin for the flow type: the flow stays critical as long as the flowline pressure is below $0.546\cdot 13.0 = 7.1$ MPa.

Answer: A bean of about 15.8 mm is needed. A 16 mm bean will give about 514 thousand m³/d, whatever the flowline pressure, as long as it is below 7.1 MPa.

What it means. The rate of a gas well on a choke is proportional to the pressure upstream of it and to the area of the bean. If wear enlarges the bean from 16 to 17 mm, the flow rate rises by about 13%. An uncertainty in the discharge coefficient passes into the rate one to one. That is why engineers tune the formula against a measurement. Without the correction for the gas deviation factor ( $Z_1 = 1$ ), the flow rate would come out 9% lower, because a real gas at 13 MPa is denser than an ideal one. The data in this example are assumed for illustration.

A liquid is practically incompressible (its volume hardly changes with pressure). Bernoulli's equation for discharge through an orifice describes its flow rate through a choke. A liquid has no critical flow like a gas: the flow rate grows as the square root of the pressure drop. A limit is reached only if the pressure in the contracted section falls below the vapour pressure of the liquid, or below the bubble point pressure of the oil. Then the liquid boils or releases gas, and the flow becomes two-phase.

Liquid flow rate through a choke

q = C_d A \sqrt{\frac{2\,(p_1 - p_2)}{\rho}}

where:

$q$ volumetric flow rate of the liquid, m³/s
$C_d$ discharge coefficient: allows for the contraction of the jet and for friction, —
$A$ area of the bean orifice, $\pi d^2/4$ , m²
$p_1 - p_2$ pressure drop across the choke (the difference can also be taken in gauge pressures), Pa
$\rho$ density of the liquid, kg/m³

Units: SI

Holds when

An incompressible single-phase liquid. The pressure in the contracted section is above the pressure at which it boils or releases gas.
The velocity upstream of the orifice is small compared with the velocity in it. If the orifice is close in size to the pipe, divide the flow rate by $\sqrt{1 - (d/D)^4}$ .
$C_d$ is empirical (found from experiments). In classical hydraulics it is about 0.6 for an orifice in a thin wall and about 0.8 for a short cylindrical nozzle. For a particular choke the manufacturer gives it.

The flow rate is proportional to the square root of the pressure drop. To double the rate through the same bean, the pressure drop must be four times as large. So a choke on a liquid restricts the flow strongly at high rates, and hardly at all at low rates.

In other unit systems (1)

Metric oilfield units

q = 95.97\,C_d\,d^2 \sqrt{\frac{\Delta p}{\rho}}

$q$ in m³/d, $d$ in mm, $\Delta p$ in MPa, $\rho$ in kg/m³.

$95.97 = 86{,}400\cdot(\pi/4)\cdot 10^{-6}\cdot\sqrt{2\cdot 10^{6}}$ : the seconds in a day, the area of a circle in m² from the diameter in mm, and the conversion MPa → Pa under the root.

Worked exampleA choke on an injection well

Given (assumed data)

Pressure in the water line before the well: 16.0 MPa
Pressure needed at the wellhead to inject 450 m³/d: 13.5 MPa (from the injectivity of the well, that is, how easily it takes water)
Water: density 1060 kg/m³
Discharge coefficient of the bean $C_d$: 0.82 (assumed)

Solution

The pressure drop the choke has to “eat”: $\Delta p = 16.0 - 13.5 = 2.5$ MPa.
The square root: $\sqrt{\Delta p/\rho} = \sqrt{2.5/1060} = 0.04856$ .
Diameter: $d^2 = 450/(95.97\cdot 0.82\cdot 0.04856) = 117.8$ mm², $d = 10.85$ mm. The nearest bean is 11 mm.
An 11 mm bean at a pressure drop of 2.5 MPa will pass $95.97\cdot 0.82\cdot 11^2\cdot 0.04856 = 462$ m³/d. The water leaves it at a velocity of $0.82\sqrt{2\cdot 2.5\cdot 10^{6}/1060} = 56.3$ m/s.

Answer: A bean of about 10.85 mm is needed. With an 11 mm bean the flow rate will be about 462 m³/d.

What it means. In reality the pressure drop will not stay at exactly 2.5 MPa. A slightly higher rate needs a slightly higher wellhead pressure, so the rate will settle somewhat below 462 m³/d. This is the same node logic as in a producing well. A velocity of about 56.3 m/s explains why beans are made of cemented carbides: solid particles in the water quickly wear out the orifice, and the flow rate grows. The data in this example are assumed for illustration.

Gas–liquid flow: Gilbert and similar correlations

A gas–liquid mixture flows through the chokes of oil wells, and there is no simple theory for it. Gas and liquid move at different velocities, and gas comes out of solution and expands right in the narrowest part. In 1954 W. Gilbert analysed 268 measurements on flowing wells of one field in California. The wells ran on beans from 6/64 to 18/64 inch (2.4–7.1 mm). Gilbert found a simple link between the pressure upstream of the choke, the liquid rate, the gas–liquid ratio and the bean size. Later, other authors found their own constants for the same form. Ros used his own theoretical model of critical flow of a mixture. Baxendell and Achong used measurements in other fields.

Gilbert-type correlations (critical flow)

p_{wh} = \frac{b\,q\,\text{GLR}^{\,c}}{S^{a}}

where:

$p_{wh}$ pressure upstream of the choke, gauge, psig
$q$ liquid rate (oil and water) at standard conditions, STB/d
$\text{GLR}$ gas–liquid ratio of the produced fluid, scf/STB
$S$ bean diameter in sixty-fourths of an inch, 1/64″
$b,\ c,\ a$ constants of the correlation (table below); for Gilbert 10.0, 0.546, 1.89, —

Units: oilfield units: psig, STB/d, scf/STB, 1/64″

Holds when

Critical flow only. The pressure downstream of the choke is no more than about 0.5–0.6 of the pressure upstream (for Gilbert, less than 1/1.7 ≈ 0.59).
The constants come from measurements in particular fields and with particular beans. Outside these conditions the accuracy falls, so check the correlation against rate measurements.
In Gilbert's original the constant is 435, with the GLR in thousands of cubic feet per barrel (Mscf/bbl). This is the same formula: $10.0\cdot 1000^{0.546} \approx 435$ . Different publications give the pressure as gauge or as absolute. At pressures of a few megapascals the difference is one to three per cent.

In critical flow the pressure upstream of the choke is directly proportional to the rate. On a graph of pressure against rate, this is a straight line from zero. It is steeper when the bean is smaller and when there is more gas in the flow. The exponent of $S$ is close to 2, so the flow rate grows roughly in proportion to the area of the bean. The more gas, the less liquid the same pressure passes, because the gas takes up part of the cross-section and speeds up the flow.

In other unit systems (1)

Metric oilfield units (Gilbert)

p_{wh} = 0.1924\,\frac{q\,\text{GLR}^{\,0.546}}{d^{1.89}}

$p_{wh}$ in MPa (gauge), $q$ in m³/d, GLR in m³/m³ (gas at 20 °C and 0.101325 MPa), $d$ in mm.

$0.1924 = 10.0\cdot 6.2898\cdot 5.529^{0.546}\cdot 0.3969^{1.89}/145.04$ : 6.2898 barrels in a cubic metre, 5.529 scf/STB in 1 m³/m³ (gas at 20 °C against 60 °F), 0.3969 mm in 1/64″ and 145.04 psi in a megapascal.

Constants of Gilbert-type correlations

Correlation: Gilbert
$b$: 10.0
$c$: 0.546
$a$: 1.89
Metric constant: 0.1924

Correlation: Ros
$b$: 17.40
$c$: 0.500
$a$: 2.00
Metric constant: 0.2795

Correlation: Baxendell
$b$: 9.56
$c$: 0.546
$a$: 1.93
Metric constant: 0.1772

Correlation: Achong
$b$: 3.82
$c$: 0.650
$a$: 1.88
Metric constant: 0.0886

Correlation	$b$	$c$	$a$	Metric constant
Gilbert	10.0	0.546	1.89	0.1924
Ros	17.40	0.500	2.00	0.2795
Baxendell	9.56	0.546	1.93	0.1772
Achong	3.82	0.650	1.88	0.0886

The oilfield constants are for $p$ in psig, $q$ in STB/d, GLR in scf/STB and $S$ in 1/64″; the metric ones are for $p$ in MPa, $q$ in m³/d, GLR in m³/m³ and $d$ in mm. The last column is worked out in the same way as for Gilbert's formula above.

Worked exampleRate from the pressure upstream of the choke: four correlations

Given (assumed data)

Well: the lesson's well on a 6 mm bean; $\text{GLR} = 81$ m³/m³
Wellhead (buffer) pressure: 5.82 MPa gauge (read on the pressure gauge), that is 5.92 MPa absolute
Line pressure downstream of the choke: 0.85 MPa (absolute)

Solution

Flow type: $0.85/5.92 = 0.144$ , which is much less than 0.59. So the flow is critical, and the correlations apply.
The factors: $\text{GLR}^{\,0.546} = 81^{0.546} = 11.016$ and $d^{1.89} = 6^{1.89} = 29.56$ .
Gilbert: $q = \dfrac{p_{wh}\,d^{1.89}}{0.1924\,\text{GLR}^{\,0.546}} = \dfrac{5.82\cdot 29.56}{0.1924\cdot 11.016} = 81.2$ m³/d.
Ros: $q = 5.82\cdot 6^{2}/(0.2795\cdot 81^{0.5}) = 83.3$ m³/d. Baxendell: $5.82\cdot 6^{1.93}/(0.1772\cdot 11.016) = 94.7$ m³/d. Achong: $5.82\cdot 6^{1.88}/(0.0886\cdot 81^{0.65}) = 109.6$ m³/d.
Scatter: the largest estimate is 35% above the smallest.
Sensitivity to gas: if the true $\text{GLR} = 100$ m³/m³, Gilbert's formula gives a rate of $(81/100)^{0.546} = 0.8913$ of the calculated one, that is, 72.4 m³/d.

Answer: Gilbert gives about 81.2 m³/d. The other three correlations give 83.3 to 109.6 m³/d.

What it means. The wellhead pressure of 5.82 MPa was calculated for this well in the next section with Gilbert's correlation, so Gilbert's formula “guessed” the rate. The other correlations, built from other data, give up to 35% more. Hence a practical rule: choose the correlation for a field and tune it (fit the constant $b$ ) against rate measurements at the metering station or the test separator. Only then estimate well rates from the wellhead pressure. You must know the gas–liquid ratio well, because an error in it passes into the rate to a power of almost 0.5. The data in this example are assumed for illustration.

When the pressure downstream of the choke rises above the critical fraction, Gilbert-type correlations do not apply, because the flow rate starts to depend on the pressure drop. For subcritical flow of a mixture, engineers use mechanistic models, for example those of Sachdeva and co-authors (Sachdeva, 1986) and of Perkins (Perkins, 1993). From the conservation equations, these models calculate both the critical pressure ratio for the given mixture and the flow rate at any pressure drop. Well-modelling programs usually contain both correlations and such models. As with the correlations for flow in pipes, tune the chosen one against measurements (choice of calculation method).

The choke as a node of the system

It is convenient to put the node at the wellhead, just before the choke (where to put the node). The inflow side is the reservoir and the tubing. At each rate the well delivers a certain pressure to the choke. To draw this tubing performance curve, find the bottomhole pressure from the inflow performance relationship. Then calculate the lift up the pipes. This gives the pressure that is left in the flow at the wellhead. In critical flow the outflow side is the choke itself. By Gilbert's correlation its “curve” is a straight line from the origin. (Add atmospheric pressure to the gauge pressure from the formula, because the curve of the well is drawn in absolute pressures.) The crossing point is the operating mode on that bean.

CalculatedThe node at the wellhead of the lesson's well: the tubing performance curve (the pressure the well delivers to the choke) and the lines for 5, 6 and 7 mm beans from Gilbert's correlation. The dashed line is the lowest wellhead pressure at which the flow is still critical: 1.7 times the line pressure. Calculated with the teaching models of the lesson.

The picture shows how a choke works. A small bean gives a steep line. The well then operates at a low rate and a high wellhead pressure, close to the highest pressure it can deliver. The larger the bean, the flatter the line, the higher the rate and the lower the wellhead pressure. At the same pressure, the flow through a bean grows roughly with its area. But the rate of the well grows more slowly. On a 12 mm bean the well would give about 189 m³/d, against 81 on a 6 mm bean. That is 2.3 times as much, although the area is four times larger. The reason is that the pressure the well delivers falls together with the rate. A 12 mm bean is already beyond the beans from which Gilbert's correlation was built, so the number is an estimate.

Worked exampleWhich bean will keep to the drawdown limit

Given (assumed data)

Well: the lesson's well, $\bar p_R = 22$ MPa, $J = 29.85$ m³/(d·MPa), $\text{GLR} = 81$ m³/m³
Sand production limit: drawdown no more than 3.5 MPa (from the calculation of rock stability, assumed)
Beans: 6, 7 and 8 mm
Line pressure: from the calculation of the flowline to a separator at 0.8 MPa

Solution

Permissible rate: the bottomhole pressure must not be below $22.0 - 3.5 = 18.5$ MPa. This is above the bubble point, so the inflow curve is a straight line, and $q \le 29.85\cdot 3.5 = 104.5$ m³/d.
The operating points are where the tubing performance curve crosses the lines of the beans. The 6 mm bean gives 81.1 m³/d at a wellhead pressure of 5.92 MPa, the 7 mm bean gives 102.0 at 5.56 MPa, and the 8 mm bean gives 122.0 at 5.18 MPa.
Check of the 7 mm point against the correlation: $0.1924\cdot 102.0\cdot 11.016/7^{1.89} = 5.46$ MPa gauge, or 5.56 MPa absolute. This is the same pressure that the well delivers to the choke.
Drawdowns: $q/J$ = 2.72, 3.42 and 4.09 MPa. The 8 mm bean breaks the limit, and the 7 mm bean keeps within it.
Flow type with the 7 mm bean: the line pressure is 0.884 MPa, and the ratio is 0.884/5.56 = 0.159. The flow is critical, with a large margin.
Wear: a 7 mm bean worn by sand to 7.5 mm will give 112.1 m³/d and a drawdown of 3.76 MPa.

Answer: Choose the 7 mm bean: about 102.0 m³/d at a drawdown of 3.42 MPa. An 8 mm bean would give a drawdown of 4.09 MPa.

What it means. The choke turns a limit set by the reservoir into a simple operating setting: the diameter of the bean. The margin is small. Half a millimetre of wear takes the well beyond the limit. So inspect the beans on a schedule, and watch the wellhead pressure and the rate: if they grow on the same bean, the bean is worn. Critical flow with a ratio of 0.159 means that pressure swings in the header will not change the operating mode. The 7.5 and 8 mm beans are slightly larger than the beans from which Gilbert's correlation was built (up to 7.1 mm), so their points are estimates. The data in this example are assumed for illustration.

The larger the bean, the lower the wellhead pressure. The line pressure grows with the rate, because there are more losses in the flowline. At a certain rate the wellhead pressure falls to 1.7 times the line pressure. This is the limit of critical flow by Gilbert's condition. In the figure it is where the tubing performance curve crosses the dashed curve: about 261 m³/d at a wellhead pressure of 2.14 MPa. At higher rates the flow through the choke is subcritical. Pressure swings in the gathering network then reach the well again, and Gilbert's correlations do not apply. The choke, the flowline and the separator together set the operating mode. A bean of about 19 mm would give such a mode. That is far beyond the beans of up to 7 mm from which Gilbert's correlation was built, so the boundary itself is an estimate here.

The flowline and the gathering network

In the flowline, pressure is spent on friction, on climbing rises in the terrain, and on fittings. Friction is calculated in the same way as in the tubing (components of the pressure gradient). It takes the simplest, homogeneous form, $\lambda\,\rho_{ns} v_m^2/(2d)$ , built from the no-slip density of the mixture and the velocity of the mixture. The special feature of the line is its low pressure. Toward the separator the gas expands, so the velocity of the mixture grows, and the gradient grows with it. So the losses in the line depend strongly on the pressure at its end. At the same rate but a lower pressure, the gas takes up more volume, and the mixture moves faster.

A horizontal line has no weight of a liquid column (no hydrostatic head), but the route is seldom level. On rises the liquid is held back, so the liquid holdup there is high. The column of mixture creates back pressure. On descents the mixture flows in layers: the liquid “slips” down in a thin layer, and the pressure hardly recovers. A rise of 10 m, at a liquid holdup of 0.5 and a liquid density of 870 kg/m³, creates a back pressure of about 0.04 MPa, and the descent hardly gives it back. A route with several such hills costs tenths of a megapascal, even if the beginning and end of the line are at the same elevation. Liquid collects in the low spots and slugs form. This is an unstable mode that makes the pressure swing at the wellhead and in the separator.

Equivalent diameter of two parallel lines

d_{\text{eq}} = \left(d_1^{2.5} + d_2^{2.5}\right)^{0.4}

where:

$d_1,\ d_2$ inner diameters of the lines laid in parallel (a second pipe laid beside the first is called looping), m
$d_{\text{eq}}$ diameter of a single pipe of the same length with the same losses, m

Units: any identical units of length

Holds when

Single-phase turbulent flow with the same friction factor in both lines. The lines have equal length between common points.
For a gas–liquid mixture this is only an estimate. The phases need not divide between the lines in the same proportion, and the real gain is sometimes smaller than the calculated one.

The Darcy–Weisbach pressure drop is proportional to $q^2/d^5$ . So at a given pressure drop, the flow rate of each line is proportional to $d^{2.5}$ , and the flow rates add up. A second identical line gives $d_{\text{eq}} = 2^{0.4} d \approx 1.32\,d$ . It cuts the losses by a factor of four, because the flow is divided in half and the losses are proportional to the square of the flow rate.

Worked exampleChoke fully open: what looping and a lower separator pressure give

Given (assumed data)

Well: the lesson's well, choke removed; the wellhead pressure is set by the flowline
Flowline: 3 km, inner diameter 100 mm, level route; friction factor 0.025; temperature 20 °C, $Z = 0.98$
Separator: 0.8 MPa; alternative 0.4 MPa
Looping: a second line of the same diameter along the whole flowline

Solution

The operating point is where the tubing performance curve crosses the pressure that the flowline needs: $q = 292.6$ m³/d at a wellhead pressure of 1.355 MPa. The line takes $1.355 - 0.8 = 0.555$ MPa.
Why the gradient grows toward the separator: the velocity of the mixture is 2.99 m/s at the wellhead end and 4.77 m/s at the separator, because the gas has expanded.
Looping: $d_{\text{eq}} = (100^{2.5} + 100^{2.5})^{0.4} = 132.0$ mm. The wellhead pressure falls to 0.980 MPa, and the rate grows to 306.0 m³/d (+4.6%).
Separator at 0.4 MPa (without looping): the wellhead pressure is 1.174 MPa. That is lower by only 0.18 MPa, not by 0.4. The rate is 299.4 m³/d (+2.3%).

Answer: Without a choke the well gives about 292.6 m³/d. Looping adds 4.6%, and halving the separator pressure adds 2.3%.

What it means. Only a little less than half of the 0.4 MPa reduction in separator pressure reached the wellhead. The gas in the line expanded more, so the velocity and the friction grew. Looping removed about two thirds of the losses in the line. (At the previous rate it removed 70%, which is less than the three quarters that an incompressible liquid would have given, because the pressure in the line became lower and the gas expanded more.) Looping also gave the larger gain of the two. The choice between them is a matter of economics: looping means steel pipe, while a low separator pressure means a compressor for the gas and sometimes pumps for the liquid. Both gains are modest, because this well is limited mainly by the reservoir and the tubing: the line accounts for only 2.6% of the pressure drop from the reservoir to the separator. The data in this example are assumed for illustration.

CalculatedThe node at the wellhead with the choke fully open: the tubing performance curve of the lesson's well, and the pressure the flowline needs in three cases: a single line with the separator at 0.8 MPa, looping, and the separator at 0.4 MPa. Calculated with the teaching models of the lesson.

In a gathering network the wells are linked to one another. The pressure in the common header depends on the total flow, so more production from some wells raises the back pressure on the others. Wells on a critical choke do not notice this. Wells working without a choke lose rate: on the lesson's well each 0.1 MPa at the wellhead costs about 4 m³/d (wellhead pressure). So the gathering network is calculated together with the wells. The gathering scheme is chosen at the design stage of the field facilities (Field development concepts: surface facilities).

Flow temperature in the well

The temperature of the flow affects almost everything we calculate in a well. It changes the volume of gas and the viscosity of oil, and so the pressure gradient. It changes the opening pressures of gas lift valves (Unloading a gas lift well: valves and mandrels), the precipitation of wax, hydrates and salts, and the operation of packers (seals that close the space around the tubing) and other seals. The geothermal gradient $G_T$ (how fast the rock temperature rises with depth) sets the reservoir temperature (reservoir temperature). In a shut-in well the fluid is at the temperature of the surrounding rocks. In a producing well the picture is different. The fluid arrives from the bottom warm. On the way up it gives up heat to the colder rocks, through the tubing wall, the annulus (the space between the tubing and the casing), the casing and the cement.

Two things decide how much the flow cools. The first is the flow rate. The more mass passes through the pipes each second, the more heat it carries. About the same amount of heat escapes through the walls, and this is set by the design of the well and the temperature difference. In the same way, hot water in a long garden hose comes out barely warm if the flow is weak, and almost hot if the flow is strong. The second is time. In the first days the rocks near the well are cold and take away a lot of heat. With time they warm up, and the flow cools less.

H. Ramey (1962) reduced the problem to a simple formula. He took the heat exchange in the wellbore as steady, and in the rocks as unsteady radial heat conduction (heat spreading outward from the well and changing with time). The dimensionless time function $f(t)$ describes this. The result is an equation in which the whole influence of the flow rate, the design of the well and the rocks is gathered into one length: the relaxation distance $A$ .

Relaxation distance (Ramey's method)

A = \frac{w\,c_p\,(k_e + r U f)}{2\pi\,r U\,k_e}

where:

$A$ relaxation distance — the length over which the difference between the temperatures of the flow and of the rocks falls from its starting value by a factor of $e$ (about 2.7), m
$w\,c_p$ mass flow rate of the flow multiplied by its specific heat capacity (the heat needed to warm 1 kg by 1 K), summed over oil, water and gas, W/K
$k_e$ thermal conductivity of the rocks, W/(m·K)
$r U$ product of the radius and the overall heat transfer coefficient from the flow to the wall of the well (tubing, annulus, casing, cement); it does not depend on the choice of radius, W/(m·K)
$f$ dimensionless time function of heat conduction in the rocks, —

Units: SI

Holds when

Steady flow in the wellbore. Heat capacity and thermal conductivities are constant. The heat resistances of the well design do not change with depth.
The rocks are uniform. Far from the well the temperature equals the geothermal one.
Not included: cooling of the gas as it expands, the work of the gas against gravity, and heating by friction. For an oil well the corrections are usually small, but because of the gas the flow is in fact slightly colder than calculated. For a gas well they are significant (see the advanced section).

$A$ is proportional to $w c_p$ . Twice the mass flow rate means twice as long a “memory” of the reservoir heat in the flow. The denominator holds the ability of the well design and the rocks to carry heat away. The better the insulation (the smaller $rU$ ), the greater $A$ .

Flow temperature in a producing well

T_f(z) = T_0 + G_T z + G_T A\left[1 - e^{-(H - z)/A}\right]

where:

$T_f(z)$ temperature of the flow at depth $z$ , °C
$T_0 + G_T z$ rock temperature: $T_0$ is the temperature of the neutral layer near the surface (the layer where the temperature no longer follows the seasons), $G_T$ is the geothermal gradient, °C, °C/m
$H$ depth of inflow, where the fluid enters the well at the reservoir temperature, m
$A$ relaxation distance, m

Units: any consistent units (°C and m)

Holds when

A vertical well and a linear geothermal gradient. The fluid enters the well at the reservoir temperature.
The same assumptions as for the relaxation distance.

The amount by which the flow temperature exceeds the rock temperature grows from zero at the bottom of the hole and tends to $G_T A$ at distances greater than $A$ . The temperature at the wellhead is $T_{wh} = T_0 + G_T A\,(1 - e^{-H/A})$ . If $A$ is much greater than the depth, the flow hardly cools. If $A$ is much smaller, the flow arrives at the wellhead at a temperature close to the geothermal one.

Time function at large times

f(t) \approx \tfrac{1}{2}\ln\left(4 t_D\right) - \tfrac{\gamma_E}{2},\qquad t_D = \frac{\alpha\,t}{r_{wb}^{2}}

where:

$t_D$ dimensionless time, —
$\alpha$ thermal diffusivity of the rocks, m²/s
$t$ time the well has been producing, s
$r_{wb}$ radius of the wellbore (from the bit size), m
$\gamma_E$ Euler's constant, 0.5772, —
$\ln$ natural logarithm, —

Units: dimensionless

Holds when

Large dimensionless times, which in practice means after the first days or weeks of operation. This is the same line-source asymptotic formula (the long-time approximation) as in reservoir flow (Line-source solution and the pressure function).
At small times the long-time formula underestimates $f$ and so overestimates the heat loss. The calculated flow temperature is then lower than the true one. For small times use the full time function, for example the approximation of Hasan and Kabir, which is valid for all times.

The function grows with the logarithm of time. With time the rocks near the well warm up, the heat loss decreases, $A$ grows, and the flow arrives at the wellhead warmer.

Worked exampleWellhead temperature and the zone of wax deposition

Given (assumed data)

Rocks: $T_0 = 10$ °C, $G_T = 3.0$ °C/100 m (70 °C at the bottom of the hole); $k_e = 2.0$ W/(m·K), $\alpha = 1.0\cdot 10^{-6}$ m²/s
Well: producing for 30 days; wellbore radius 0.108 m; the product $rU$ comes from the well design (advanced section)
Heat capacities: oil 2.0, water 4.18, gas 2.3 kJ/(kg·K)
Operating modes: on a 6 mm bean, 81.1 m³/d; without a choke, 292.6 m³/d (examples above)
Oil: wax appearance temperature 40 °C (from a laboratory determination, assumed)

Solution

Time function: $t_D = 10^{-6}\cdot 30\cdot 86{,}400/0.108^2 = 222$ , $f = \tfrac12\ln(4\cdot 222) - 0.2886 = 3.106$ .
The well design gives $rU = 0.6576$ W/(m·K). The factor is $(k_e + rUf)/(2\pi\,rU\,k_e) = (2.0 + 0.6576\cdot 3.106)/(2\pi\cdot 0.6576\cdot 2.0) = 0.4892$ m·K/W.
The flow at 81.1 m³/d consists of oil 0.7181 kg/s, water 0.0986 kg/s and gas 0.0641 kg/s. So $w c_p = 1996$ W/K and $A = 1996\cdot 0.4892 = 976$ m. Without a choke, $w c_p = 7201$ W/K and $A = 3523$ m.
At the wellhead: $T_{wh} = 10 + 0.03\cdot A\,(1 - e^{-2000/A})$ , which gives 35.5 °C on the choke and 55.8 °C without it.
Wax: on the choke the flow is colder than 40 °C above a depth of 174 m (from the profile). Without a choke it is warmer than 40 °C all the way to the wellhead. In a shut-in well the fluid cools to the temperature of the rocks, and wax may come out of the oil from the wellhead down to a depth of $(40 - 10)/0.03 = 1000$ m.
For the pressure calculation: without a choke, the mean temperature in the tubing is 65.0 °C instead of the 50 °C assumed in the flow model. So the required bottomhole pressure at the same rate is lower by 0.15 MPa.

Answer: On a 6 mm bean the flow reaches the wellhead at 35.5 °C, and wax deposits are possible in the top 174 m of the tubing. Without a choke it arrives at 55.8 °C, and there are no deposits in the pipes.

What it means. A choke that cut the rate by a factor of about 3.6 made the flow 20 °C colder at the wellhead. For the pressure in the tubing, an error in temperature hardly matters: 0.15 MPa out of a bottomhole pressure of 12 MPa. But for wax the temperature is decisive. Deposits narrow the pipes, and they grow fastest where the wall is colder than the wax appearance temperature. The calculation is a guide for a wax-control programme. Temperature measurements in the producing well show where deposition really begins. The data in this example are assumed for illustration.

CalculatedRock temperature and flow temperature in the lesson's well after 30 days of operation on a 6 mm bean (81 m³/d) and without a choke (293 m³/d). The vertical dashed line is the wax appearance temperature, 40 °C. Calculated by Ramey's method.

The time of operation makes a clear difference. At the same rate of 81 m³/d the flow reaches the wellhead at 29.9 °C one day after start-up, 33.3 °C after a week, 35.5 °C after a month and 38.8 °C after a year. For the first day the long-time formula for the time function underestimates the temperature. So this is an estimate with a margin on the cold side.

Temperature margin: wax and hydrates

Wax comes out of oil when the oil cools below the wax appearance temperature (WAT; in Russian-language literature, the temperature of saturation of the oil with paraffin). The laboratory finds it on an oil sample. The lesson Retrograde condensation, wax, asphaltenes and high-pressure fluids covers what paraffins are and how they are studied. Crystals grow first of all on the pipe wall, which is colder than the flow. So deposition begins somewhat deeper than the point where the mean temperature of the flow equals the WAT. The temperature profile gives this boundary and shows what shifts it: the rate, the choke, insulation and time. In Kazakhstan this is especially important for the high-wax oils of Mangyshlak (the Uzen field example).

Keep the flow warm: avoid unnecessary throttling, insulate the tubing (tubing with a vacuum gap, insulating fluids in the annulus), and heat the production.
Remove deposits: use scrapers on a wireline (a steel cable run into the well), flush with hot oil or water, or use solvents.
Stop the crystals from growing: use wax inhibitors fed into the annulus or through a capillary tube (a very thin pipe).
Remember shut-ins: in a shut-in well the fluid cools to the rock temperature. At start-up, wax may then be found along the whole length above the depth where the geotherm crosses the WAT.

Gas hydrates are solid crystals that look like ice. Water molecules form a framework with cavities, and molecules of light gases sit in the cavities: methane, ethane, propane, carbon dioxide, hydrogen sulfide. To form, hydrates need free water, gas, high pressure and low temperature. The higher the pressure, the higher the temperature at which a hydrate is still stable. Hydrate plugs block chokes, flowlines and the tree. The most dangerous place is immediately after the choke of a gas well, because there the gas cools sharply while the pressure is still high.

The cooling on throttling is the Joule–Thomson effect. A gas that expands without heat exchange and without doing work (at constant enthalpy, the total heat content) cools, because part of its internal energy goes into pulling the molecules apart against the forces of attraction. From reference data, for methane the Joule–Thomson coefficient at pressures of 5–13 MPa and temperatures of 15–40 °C is about 2.5–4.5 °C/MPa. Throttling methane from 12 MPa and 40 °C to 5 MPa cools it by 24 °C, on average almost 3.5 °C per megapascal. Heavier gases cool more strongly (ethane about three times as much). Liquids warm slightly on throttling: water by about 0.2 °C per megapascal. That is why the cooling after the chokes of oil wells, where there is a lot of liquid, is much smaller than for gas wells.

Temperature downstream of the choke

T_2 \approx T_1 - \bar\mu_{JT}\,(p_1 - p_2),\qquad \mu_{JT} = \left(\frac{\partial T}{\partial p}\right)_{h}

where:

$T_1,\ T_2$ temperature upstream and downstream of the choke, °C
$p_1,\ p_2$ pressure upstream and downstream of the choke, MPa
$\mu_{JT}$ Joule–Thomson coefficient: the change of temperature with pressure at constant enthalpy $h$ , °C/MPa
$\bar\mu_{JT}$ its mean value along the expansion path, °C/MPa

Units: °C and MPa

Holds when

Throttling without heat exchange with the surroundings. The kinetic energy before and after the choke is small, so the enthalpy is constant.
The coefficient varies with pressure and temperature. At large pressure drops it is more accurate to calculate with an equation of state, keeping the enthalpy constant. For a gas with liquid, also allow for the heat capacity of the liquid.

For a gas the coefficient is positive, so the gas cools. For liquids it is negative, so they warm up slightly. A pressure drop of 7 MPa across the choke of a gas well cools the gas by twenty to thirty degrees.

The best way to find the hydrate formation temperature is to calculate it from the composition of the gas and water, or to measure it in the laboratory. For a quick estimate for a sweet natural gas (one without hydrogen sulfide and carbon dioxide), there is a chart by gas specific gravity, and there are formulas that approximate it. One convenient formula is that of Towler and Mokhatab (2005).

Hydrate formation temperature of a sweet natural gas (Towler and Mokhatab)

T_h = 13.47\ln p + 34.27\ln\gamma_{\text{g}} - 1.675\ln p\,\ln\gamma_{\text{g}} - 20.35

where:

$T_h$ temperature below which hydrates form at pressure $p$ , °F
$p$ absolute pressure of the gas, psia
$\gamma_{\text{g}}$ specific gravity of the gas relative to air, —
$\ln$ natural logarithm, —

Units: oilfield: $T_h$ in °F, $p$ in psia; degrees Celsius are $(T_h - 32)/1.8$ , pressure in psia is the pressure in MPa multiplied by 145.04

Holds when

A sweet natural gas (without appreciable hydrogen sulfide or carbon dioxide) with a specific gravity of about 0.55–1.0, in the presence of fresh water.
The formula reproduces the chart by gas specific gravity. But gases of the same density and different composition may form hydrates under clearly different conditions (for a gas close to pure methane the estimate is too high by several degrees). Salts in the water and inhibitors lower the hydrate formation temperature. For design decisions, calculate from the composition.

The hydrate formation temperature grows with pressure roughly logarithmically: for a gas of specific gravity 0.58 it is about 9.1 °C at 3 MPa, 14.6 °C at 6 MPa and 20.8 °C at 13 MPa.

Worked exampleHydrate margin downstream of the choke of a gas well

Given (assumed data)

Well and choke: from the bean diameter example: 13.0 MPa and 38 °C upstream of the choke, 6.0 MPa downstream, 500 thousand m³/d of gas of specific gravity 0.58 with water
Mean Joule–Thomson coefficient: 3.5 °C/MPa (assumed from reference data for methane)
Required margin: 3 °C above the hydrate formation temperature (an operating rule, assumed)
Heat capacity of the gas at 13 MPa: 3.1 kJ/(kg·K) (assumed)

Solution

Temperature downstream of the choke: $T_2 = 38 - 3.5\cdot(13.0 - 6.0) = 13.5$ °C.
Hydrate formation temperature at 6.0 MPa: $p = 6.0\cdot 145.04 = 870.2$ psia, $\ln p = 6.7688$ and $\ln 0.58 = -0.54473$ . So $T_h = 13.47\cdot 6.7688 + 34.27\cdot(-0.54473) - 1.675\cdot 6.7688\cdot(-0.54473) - 20.35 = 58.33$ °F, or 14.63 °C.
Margin: $13.5 - 14.63 = -1.13$ °C. This is negative: after the choke the gas is in the hydrate region.
Required temperature upstream of the choke: $T_1 = 14.63 + 3 + 3.5\cdot 7.0 = 42.13$ °C. That is, the gas must be heated by 4.13 °C.
Heater duty (the heat the heater must supply): the mass flow rate is $500{,}000\cdot 0.6985/86{,}400 = 4.042$ kg/s ( $0.6985$ kg/m³ is the density of the gas at standard conditions). Then $Q = 4.042\cdot 3.1\cdot 4.13 = 52$ kW.
At start-up the wellhead is cold. At 13 MPa hydrates are stable below 20.8 °C, so a plug may form even before the choke, until the well has warmed up.

Answer: Without heating, the temperature after the choke, 13.5 °C, is 1.1 °C below the hydrate formation temperature. To keep a margin of 3 °C, heat the gas to 42.13 °C. This needs about 52 kW.

What it means. Even a small negative margin means that hydrates will grow after the choke, and the bean or the flowline will sooner or later be blocked. There are several solutions. One is a heater before the choke (as in the example). Another is to inject methanol or glycol (hydrate inhibitors) before the choke. A third is a downhole choke, so that the expansion takes place at depth, where it is warm. A fourth is to throttle in two stages, with heating between them. Methanol lowers the hydrate formation temperature, but a lot of it is needed, and it has to be separated from the production. The calculation of inhibition belongs to gas processing (Gas processing and compression). The data in this example are assumed for illustration.

CalculatedHydrate formation curve of a gas of specific gravity 0.58 (by the formula of Towler and Mokhatab) and the path of the gas through the choke of the example: without heating the gas enters the hydrate region, and after heating to 42 °C it stays to the right of the curve with a margin of 3 °C.

Terms and symbols

Choke, choke bean: A narrowing in the flow path that creates a set pressure drop. The bean is a replaceable part with a calibrated orifice.
Wellhead (buffer) pressure and line pressure: The pressure upstream of the choke (on the “buffer” of the Christmas tree; in Russian-language practice, the buffer pressure) and the pressure downstream of it, at the start of the flowline.
Critical flow: The mode in which the velocity in the narrowest part equals the speed of sound. The flow rate then depends only on the state of the flow upstream of the choke.
Critical pressure ratio $r_c$: The ratio of the pressure downstream of the choke to the pressure upstream of it, below which the flow is critical. For a gas it is $(2/(k+1))^{k/(k-1)}$ , about 0.55.
Discharge coefficient $C_d$: The ratio of the actual flow rate through an orifice to the ideal one. It allows for the contraction of the jet and for friction.
Gilbert-type correlations: Empirical relationships $p_{wh} = b\,q\,\text{GLR}^{\,c}/S^a$ for critical flow of a gas–liquid mixture through a choke.
Tubing performance curve: The pressure the well delivers to the wellhead at each rate. It is the inflow curve for a node at the wellhead.
Equivalent length: The length of straight pipe with the same pressure loss as a valve, a bend or another fitting.
Relaxation distance $A$: The length in Ramey's method over which the temperature difference between the flow and the rocks falls by a factor of $e$ . It is proportional to the mass flow rate.
Joule–Thomson coefficient $\mu_{JT}$: The change in temperature with a change in pressure, without heat exchange (at constant enthalpy). It is positive for a gas, so a gas cools on throttling.
Hydrate formation temperature: The temperature below which, at a given pressure, gas with water forms hydrates.
Wax appearance temperature (WAT): The temperature below which wax crystals appear in oil. Below it, deposits grow on cold walls.

Common mistakes and misconceptions

Wrong

“If we double the bean diameter, the rate will double, or even grow fourfold with the area.”

In fact

As the rate grows, the pressure that the well delivers to the choke falls. On the lesson's well a 12 mm bean instead of a 6 mm one would give about 189 m³/d instead of 81. That is 2.3 times as much (by the same correlation, which is already beyond its data). Find the new operating mode where the tubing performance curve crosses the line of the new bean.

Wrong

“If the pressure downstream of the choke is lowered, the rate will always grow.”

In fact

In critical flow it will not. The signal about the pressure drop cannot pass through the narrowest part against the sonic flow. The rate will grow only when the flow is subcritical, or when the choke is removed.

Wrong

“Gilbert's correlation has been tested for decades — it is accurate for any field.”

In fact

It was built from measurements in one field and with beans of up to 7 mm. The correlations of other authors for the same operating mode differ from it by tens of per cent. For a field, choose the correlation and tune it against rate measurements.

Wrong

“The wellhead temperature is a property of the well; it hardly depends on the operating mode.”

In fact

It depends strongly on the rate and the time of operation. In the example of this lesson, a choke that cut the rate by a factor of about 3.6 cooled the flow at the wellhead by about 20 °C.

Wrong

“Hydrates are a concern of gas producers; they do not occur on oil wells.”

In fact

They occur wherever there are gas, free water, high pressure and cold: in the flowlines and trees of oil wells in winter, in gas lift gas lines, and in shut-in wells. After the choke of an oil well the cooling is smaller than for a gas well, but it is not zero.

In depth: advanced detail

Derivation of the gas formula and the critical ratio

The gas speeds up in the narrowest part at the expense of its own enthalpy (its heat content). Take expansion without heat exchange, from a low velocity before the choke. The energy equation then gives $v_2^2/2 = h_1 - h_2$ . For an ideal gas $h_1 - h_2 = c_p (T_1 - T_2)$ . For isentropic expansion (no friction and no heat exchange) $T_2/T_1 = r^{(k-1)/k}$ with $r = p_2/p_1$ , and $c_p T_1 = \tfrac{k}{k-1}\,p_1/\rho_1$ . So the velocity in the narrowest part is:

Velocity of the gas in the narrowest part

v_2 = \sqrt{\frac{2k}{k-1}\,\frac{p_1}{\rho_1}\left[1 - r^{(k-1)/k}\right]}

where:

$v_2$ velocity of the gas in the contracted section, m/s
$p_1,\ \rho_1$ pressure and density of the gas upstream of the choke, Pa, kg/m³
$r$ pressure ratio $p_2/p_1$ , —
$k$ adiabatic exponent, —

Units: SI

Holds when

An ideal gas and isentropic expansion. The velocity upstream of the choke is small.

The density in the narrowest part is $\rho_2 = \rho_1 r^{1/k}$ . The mass flow rate $\rho_2 v_2 A$ then gives the lesson's formula for the gas flow rate, so far without the discharge coefficient. The factor $r^{2/k} - r^{(k+1)/k}$ in it is zero at $r = 1$ and at $r = 0$ , and it has a maximum between them. Set its derivative with respect to $r$ to zero. This gives $\tfrac{2}{k}\,r^{2/k - 1} = \tfrac{k+1}{k}\,r^{1/k}$ , so $r_c^{(k-1)/k} = 2/(k+1)$ . This is the critical ratio. At the maximum, the velocity in the narrowest part is exactly equal to the local speed of sound $\sqrt{k p_2/\rho_2}$ . This is the physical meaning of critical flow.

Below $r_c$ the calculated “nozzle” would give a smaller flow rate. This does not happen in fact: the narrowest part passes the greatest flow rate, and the gas expands further beyond it, in the jet. Put $r_c$ and $\rho_1 = p_1 M/(Z_1 \mathcal{R} T_1)$ into the formula. This gives the critical flow rate:

Critical mass flow rate of a gas

\dot m_c = C_d A\,p_1 \sqrt{\frac{k M}{Z_1 \mathcal{R} T_1}}\left(\frac{2}{k+1}\right)^{\frac{k+1}{2(k-1)}}

where:

$\dot m_c$ mass flow rate in critical flow, kg/s
$C_d,\ A$ discharge coefficient and area of the orifice, —, m²
$p_1,\ T_1,\ Z_1$ absolute pressure and temperature and gas deviation factor upstream of the choke, Pa, K, —
$M,\ \mathcal{R}$ molar mass of the gas and the universal gas constant, kg/mol, J/(mol·K)

Units: SI

Holds when

The same as for the gas flow rate formula; $p_2/p_1 \le r_c$ .

Divide by the density of the gas at standard conditions, $p_{\text{sc}} M/(\mathcal{R} T_{\text{sc}})$ , and write $M = \gamma_{\text{g}} M_{\text{air}}$ . This gives the lesson's formulas in metric and oilfield units, with the constants 3.326 and 879.

Why a mixture “locks up” at low velocities

The compressibility of a medium and its density set the speed of sound: $c^2 = 1/(\rho\,\beta)$ , where $\beta$ is the compressibility (how much the volume shrinks under pressure). In a uniform, finely dispersed mixture, almost all the compressibility comes from the gas, and the density from the liquid. So sound in the mixture is slower than in either phase alone. Wood's formula gives an estimate.

Speed of sound in a uniform gas–liquid mixture (Wood's formula)

\frac{1}{\rho_m c_m^{2}} = \frac{\alpha}{\rho_G c_G^{2}} + \frac{1-\alpha}{\rho_L c_L^{2}},\qquad \rho_m = \alpha\rho_G + (1-\alpha)\rho_L

where:

$c_m,\ c_G,\ c_L$ speed of sound in the mixture, in the gas and in the liquid, m/s
$\alpha$ volume fraction of gas in the mixture, —
$\rho_m,\ \rho_G,\ \rho_L$ densities of the mixture, the gas and the liquid, kg/m³

Units: SI

Holds when

The phases are finely mixed and move together. The length of the sound wave is much greater than the size of the bubbles.
The gas in the bubbles is compressed isothermally (at constant temperature, because it has time to exchange heat with the liquid), so that $\rho_G c_G^2 \approx p$ . There is no mass exchange between the phases.

Example: take oil (850 kg/m³, $c_L$ = 1300 m/s) and gas in equal volumes at 6 MPa and 30 °C (a gas of specific gravity 0.70, with a density of 57 kg/m³). Then $c_m \approx 162$ m/s. That is half as fast as in the gas and eight times slower than in the oil. For air and water in equal volumes at atmospheric pressure, it is about 20 m/s. That is why a gas–liquid flow in the bean of a choke easily reaches the critical velocity.

What the heat transfer coefficient is made of

Heat passes from the flow to the wall of the well through several layers one after another: the film at the inner wall of the tubing, the steel wall of the tubing, the liquid or gas in the annulus, the wall of the casing, and the cement sheath. Their resistances add up, like resistors connected in series.

Thermal resistance of the well design

\frac{1}{2\pi\,rU} = \frac{1}{2\pi r_{ti} h_f} + \sum_i \frac{\ln(r_{i+1}/r_i)}{2\pi k_i}

where:

$rU$ product of the radius and the heat transfer coefficient from Ramey's formula, W/(m·K)
$r_{ti},\ h_f$ inner radius of the tubing and the film heat transfer coefficient from the flow to its wall, m, W/(m²·K)
$r_i,\ r_{i+1},\ k_i$ inner and outer radii of the $i$ -th annular layer (tubing wall, annulus, casing, cement) and its thermal conductivity, m, W/(m·K)

Units: SI; resistance per metre of length, K·m/W

Holds when

Steady radial heat flow through coaxial layers (layers one inside another, around the same axis).
The annulus is treated as a layer that conducts heat. Natural convection of the liquid in it (and radiation, if it holds gas) increases heat transfer. Without them $rU$ is too low and the calculated flow temperature is too high.

For the lesson's well (73 × 5.5 mm tubing, water in the annulus at 0.6 W/(m·K), 168 mm casing, cement at 0.9 W/(m·K), 216 mm hole), the resistance is 0.242 K·m/W. Of this, the annulus accounts for 0.192, the cement 0.044, the film 0.005, and the steel only thousandths. So $rU = 0.658$ W/(m·K). The annulus is the main barrier, which is why insulation works there. Replace the water with an insulating fluid with an effective thermal conductivity of 0.06 W/(m·K). Then $rU$ falls to $0.081$ W/(m·K), and on a 6 mm bean the flow reaches the wellhead at 58 °C instead of 36 °C. That is above the wax appearance temperature.

Gas wells: what to add to Ramey's method

In Ramey's formula for a liquid, the change in pressure does not affect the temperature, because the work against gravity comes from the pressure and not from heat. For a gas it is different. As it rises, the gas does work against gravity at the expense of its enthalpy, and it cools by $g/c_p$ per metre of lift. As it expands with falling pressure, it also cools by the Joule–Thomson effect, by $\mu_{JT}\,\dd p/\dd z$ . In the balance equation, both terms are subtracted from the geothermal gradient. The long-distance excess of the flow temperature over the rock temperature then becomes $(G_T - g/c_p - \mu_{JT}\,\dd p/\dd z)\,A$ .

An estimate for a gas well: at $c_p$ = 2.6 kJ/(kg·K), the work against gravity gives 0.377 °C per 100 m. The Joule–Thomson effect, at 3.5 °C/MPa and a pressure gradient of 1.1 kPa/m, adds another 0.385 °C per 100 m. Together this is 0.762 °C per 100 m, against the geothermal 3 °C per 100 m. So the gas flow cools noticeably more than the formula for a liquid gives. Engineers use generalised methods with these terms (for example, those of Alves and co-authors and of Hasan and Kabir) for gas and gas-condensate wells and for gas injection.

Connections to other topics

Nodal analysis of the production system — the pressure balance of the system, the node at the wellhead and functional nodes are the basis of this lesson.
Multiphase flow in tubing — the pressure gradient, slip and the model of flow in the tubing on which the examples are built.
Inflow performance relationship (IPR) — the composite IPR of the lesson's well.
Well performance sensitivity analysis — the same well: how the rate depends on the tubing, the wellhead pressure, the water cut and depletion.
Ideal and real gas: the gas deviation factor (Z) — the gas deviation factor of the gas upstream of the choke.
Reservoir pressure and temperature — the geothermal gradient and the reservoir temperature.
Sand production: types, prediction and cost of the problem — where the drawdown limit that the choke holds comes from.
Retrograde condensation, wax, asphaltenes and high-pressure fluids — paraffins, asphaltenes and hydrates: what these phases are and how they are studied.
Gas lift: principles and choice of operating conditions — regulating chokes on lift gas and hydrates in gas lines.

Check yourself

1
Gas flows through a choke at 9 MPa upstream and 4 MPa downstream. The flowline pressure is lowered to 3 MPa. How will the flow rate change?
1. A.It will grow by about a quarter
2. B.It will decrease
3. C.It will not change: the flow is already critical
4. D.It will double
Show the answer
Correct answer: C. It will not change: the flow is already critical
The pressure ratio $4/9 = 0.44$ is already below the critical one (about 0.55 for natural gas). So the flow is critical, and only the pressure and temperature upstream of the choke set the flow rate. Lowering the pressure downstream of the choke to 3 MPa ( $3/9 = 0.33$ ) does not change the flow rate. The gas simply expands more in the jet beyond the narrowest part.
2
On a 5 mm bean the lesson's well gave 60 m³/d at a wellhead (buffer) pressure of 6.09 MPa (gauge). The bean was replaced by a 6 mm one. What would the wellhead pressure have been if the rate had not changed? Why did the rate in fact grow, and the pressure fall by less?
Show the answer
By Gilbert's formula, at the same rate and gas–liquid ratio the pressure is inversely proportional to $d^{1.89}$ : $6.09\cdot(5/6)^{1.89} \approx 4.3$ MPa. But at a lower back pressure the well gives more. The tubing performance curve crosses the flatter line of the new bean farther to the right, at 81 m³/d and 5.82 MPa. The pressure fell by less than the calculation for an unchanged rate gives, because the rate grew, and the pressure that the bean needs grew with it.
3
Two identical wells produce from the same reservoir: one at a rate of 300 m³/d, the other at 80 m³/d. In which is the flow temperature at the wellhead higher?
1. A.The same: the temperature is determined by depth alone
2. B.In the well producing 80 m³/d
3. C.In the well producing 300 m³/d
4. D.The same: at the wellhead the flow always takes the temperature of the neutral layer
Show the answer
Correct answer: C. In the well producing 300 m³/d
The flow carries heat in proportion to the mass flow rate, and it loses heat in proportion to the temperature difference between the flow and the rocks. The relaxation distance is proportional to the flow rate. So at the higher rate the flow cools less and reaches the wellhead warmer: in the lesson's example, 56 °C against 36 °C.
4
The wax appearance temperature of an oil is 45 °C, the temperature of the neutral layer is 12 °C, the geothermal gradient is 2.8 °C per 100 m. At what depth can wax begin to come out of the oil in a well that has stood idle for several days? Why is the zone of deposition shorter during production?
Show the answer
In a shut-in well the fluid cools to the rock temperature: $12 + 0.028z = 45$ gives $z = 33/0.028 \approx 1180$ m. Above this depth the oil is colder than the WAT. During production the flow carries the heat of the reservoir. It is warmer than the rocks by $G_T A(1 - e^{-(H-z)/A})$ , so the boundary rises, and the higher the rate, the higher it rises. After a shut-in well is started up, the first hours are dangerous, because the pipes and the oil in them are cold.
5
The gas upstream of the choke is at 15 MPa and 45 °C, downstream of the choke at 7 MPa. The mean Joule–Thomson coefficient of this gas is 4 °C/MPa, and the hydrate formation temperature at 7 MPa is 16 °C. Will there be hydrates downstream of the choke, and what must the temperature upstream of it be for the margin to be 3 °C?
Show the answer
After the choke, $T_2 = 45 - 4\cdot 8 = 13$ °C. That is 3 °C below the hydrate formation temperature, so there will be hydrates. For the temperature after the choke to be $16 + 3 = 19$ °C, it must be $19 + 32 = 51$ °C before the choke. So the gas must be heated by 6 °C. The other options are to inject an inhibitor, or to throttle in two stages with heating between them.
6
The separator pressure is lowered by 0.4 MPa, and the choke on the well has been removed. How will the wellhead pressure change?
1. A.It will fall by exactly 0.4 MPa
2. B.It will not change
3. C.It will fall by less than 0.4 MPa
4. D.It will fall by more than 0.4 MPa
Show the answer
Correct answer: C. It will fall by less than 0.4 MPa
At a lower pressure the gas in the flowline expands more. The velocity of the mixture and the friction losses grow, and the line takes up part of the gain. Also, at a lower back pressure the well gives a higher rate, which raises the losses further. In the lesson's example the wellhead received only about half of the reduction.

Updated: 19 October 2026