Module 1 Section 4

PETR4303 PETROLEUM PRODUCTION METHODS SYNOPSIS OF CLASS PRESENTATIONS MODULE-1 PRODUCTION METHODS UNDER NATURAL FLOW

Section -4: Estimating Wellbore Flow Performance  Concept of Tubing Performance Relationship (TPR)  Concept of Surface Flowline Performance Relationship (FPR)  Concept of Choke Flow Performance (CPR)  Estimating Frictional Pressure Drop in Tubing: Single-phase flow  Estimating Frictional Pressure Drop in Tubing: Multi-phase flow

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RECAPITULATION: In our last discussion, we understood IPR—Inflow Performance Relationship. We know how much oil will enter (qo) into the wellbore. The oil flow rate, qo depends on the drawdown, (𝑝–𝑝𝑤𝑓 ). But 𝑝 is constant for some time, therefore the only way we can increase (or decrease) the oil flow rate, qo, is by decreasing (or increasing) the flowing bottomhole pressure, 𝑝𝑤𝑓 . We are still waiting to know how much oil will come out of the well to the surface! To know that we have to understand the flow behavior in the vertical pipe, which is the tubing string.

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Concept of Tubing Performance Relationship (TPR) Now that certain amount of oil is available at the bottom of the wellbore—thanks to Mother Nature, our concern would be: How to LIFT this oil up to the Surface through the tubing string? For sure, we need a certain amount of energy (i.e., pressure) at the entrance of the tubing (i.e., at the bottom of the wellbore) in order to lift the oil to the surface.  For naturally flowing wells, the reservoir itself has enough energy (pressure) to provide for the lift.  If the reservoir does not have enough energy, we will have to install a pump or any other appropriate device (e.g., Gas-Lift!)

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In either case, though, we have to figure out how much energy (pressure) we need to obtain certain oil flow rate, qo. In other words, we are now looking for an equation for pipe flow in terms of the input pressure (at the entrance of the tubing), and the output flow rate (at the end of the tubing, which is at the surface). This equation is called the Tubing Performance Relationship (TPR). It is also called sometimes as the Vertical Lift Performance (VLP). We will use both term interchangeably. TPR: Basically, the TPR is nothing but the Darcy-Weisbach Pipe Flow equation that is used in physics and fluid dynamics:

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Consider a fluid flowing from point 1 to point 2 in a tubing string of length L and height z (Fig.1). z = Vertical height, traversed by the pipe of length L. Volumetric flow rate = q = q1 = q2 for a single-phase liquid flow. Note:  For gas, which is compressible, q2> q1, but the mass flow rate is unchanged.  For two-phase oil & gas gas, which is compressible, q2> q1, (remember the oil FVF, Bo?) Page 6 of 28

For an isothermal system, the law of conservation of energy (i.e., the first law of thermodynamics) yields the following equation for pressure drop, P: … … … (1)

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The Eqn-1 shows that the total pressure drop, Dp, for the fluid to 𝑢

flow at a flow rate, 𝑞 = ( ) has 3 components, expressed in FPS 𝐴

system of units (or SI): 𝑔

1.

∆𝑝1 = 𝜌∆𝑧 = Pressure loss due to decrease in elevation, z

2.

∆𝑝2 =

3.

∆𝑝3 =

𝑔𝑐 𝜌

2𝑔𝑐 2𝑓 𝐹 𝑔𝑐 𝐷

∆𝑢2 = Pressure loss due to acceleration 𝜌𝑢2 𝐿 = Pressure loss due to viscous friction.

Out of these 3 components, p1 is the major component, while p2 is usually negligible.

p1 can be calculated using the ‘hydrostatic pressure’ formula: p1 (psi) = 0.052 x L (lbm/gal) x D (ft), Or,

= 0.433 x L (sp.gr.) x D (ft) Page 8 of 28

Example-1: Suppose a well is flowing only oil (sp.gr. = 0.85 @ 60 oF) at 200 STB/d from a directional well with 10,000-ft MD and 8,500-ft TVD. Tubing ID = 2.441 in. Calculate the hydrostatic pressure loss. The average flowing temperature (bottom to top) is 160 oF. Solution: First, we need to calc. oil sp.gr at 160 oF, by using some published chart or correlation. Suppose (o)150F = 0.80. Then,

p1 (psi) = 0.433 x 0.80 x 8,500 = 3,128 psi.

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Pressure Loss due to Acceleration/De-acceleration: p2 This happens if the pipe cross-sectional area increases (p2 is +ve) or decreases (p2 is –ve). This is illustrated in the following diagram, where the pipe diameter (ID) decreases downstream from D1 to D2 (D2 < D1):

where,

pKE = p2 is in psi, q STB/d,   lbm/cuft, D1, D2  in.

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Example-2: In Example-1, the tubing diameter (ID) changes from 2.441-in. to 2.25-in. at the SSSS Valve (650-ft below the surface, Tf = 130 oF). Calculate the pressure drop due to this change of diameter. Is this pressure drop positive, or negative? Solution: We need to know the temperature of the flowing oil at the SSSV. If the sp.gr of the flowing oil is 0.85 at 60 oF, then we need to find its sp.gr. at the temperature at the SSSV (o) TSSSV, then convert to density in lbm/cuft = 0.85 x (H2O) at TSSSV [Use either a published chart, or Correlation!] Suppose these data are (o)130F = 0.82 and (H2O)130F = 62.1 lbm/cuft, respectively. Page 11 of 28

Therefore, (O)130F = 0.82 x 62.1 = 50.92 lbm/cuft

= 1.3 x 10-8 x (200)2 x 50.92 (1/2.254 – 1/2.4414)

p2

= 2.87 x 10-4 psi  very very small!! [Compare to p1 (psi) = 3,128 psi]

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The viscous friction loss component, p3, = pf is significant as long the length of the pipe, L, is significant (in thousand feet), although it arises from viscosity () and the pipe’s internal surface roughness,  , expressed as roughness factor,(/D): 𝑓𝐹 =

16 𝑁𝑅𝑒

𝑁𝑅𝑒 =

, for laminar flow …..

1.48𝜌 𝑞 𝜇𝐷

…..

…. (2a)

, for turbulent flow …..

…. (2b)

= Reynold’s Number, dimensionless, where

 = fluid density, lbm/cuft  = fluid viscosity, cp q = fluid flow rate, bbl/d D = Pipe inner diameter, in. Page 13 of 28

For laminar flow, i.e., NRe < 2,000 For turbulent flow, i.e., NRe > 2,100, use Chen’s correlation to estimate the Fanning friction factor, fF:

……. ……(3a) If the pipe roughness factor is ignored, then 1 𝑓

= −4 log⁡ {−

5.0452 𝑁𝑅𝑒

𝑙𝑜𝑔

7.149 0.8981 𝑁𝑅𝑒

} ……..

….

…. (3b)

Eqn.(3b) shows that a pipe will have non-zero friction factor, fF, even if the pipe’s roughness is nil, i.e., the pipe’s inner surface, which is in contact with the fluid, is frictionless ( = 0, or ‘smooth’)! Page 14 of 28

Where does the friction loss (pf) arise from, then, if the pipe’s inner surface is frictionless? --The friction loss, (pf), due to flow of a viscous fluid ( > 0) in a frictionless pipe (=0) arises from the ‘interlayer’ frictions within the fluid itself! Can you calculate the average velocity, vavg ? Flow direction

Laminar flow:

Turbulent flow: 1/2

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Example-3. For the problem in Example 1-2, calc. the viscous friction pressure loss, p3. Given o =1.5 cp at 60 oF. Solution: First, find o @ avg. flowing temperature of 150 oF. Suppose (o)150F= 50.2, and (o)150F =1.1 cp Step-1: Calc. u = q/A =4q/(D ), 2

q = 200 STB/d = 200 x5.615/(24x60x60) = 0.013 cuft/s u = q/A =4q/(D2) *144 = 4x0.013 /(x2.4412) *144=0.4 ft/s Step-2: Calc. 𝑵𝑹𝒆 =

𝟏.𝟒𝟖𝝆 𝒒 𝝁𝑫

= 1.48x50.2x200/(32.7x2.441)

= 186.2 < 2,100  Flow is Laminar. Therefore, fF = 16/NRe = 16/186.2 = 0.0859 Page 16 of 28

Step-3: Calc. ∆𝑝3 =

2𝑓 𝐹 𝑔𝑐 𝐷

𝜌𝑢2 𝐿, in lb/ft2

= 2*0.859x50.2x0.42x10,000 / (32.7x2.441/12) = 20,745 lb/ft2 =144 psi Summary: In the Example (1,2,3), we found that the Total pressure loss, pf, as 200 STB/d oil flows along a 10,000-ft long directional well (TVD=8,500 ft) = p1 + p2 + p3 =(3,128 + 2.87 x 10-4 + 144) psi = 3,272 psi. 1. Hyd. Pressure loss, % = 3,128/3,272 = 96% 2. Kinetic energy pressure loss, % = 0% 3. Viscous friction loss, % = 4% Page 17 of 28

Concept of Flowline Performance Relationship (FPR) • The surface flowline is also pipe flow, therefore, the same equations as in TPR apply: pFL = p1 + p2 + p3 • The surface flowline is approximately horizontal, since it is laid over flat terrain—from the wellhead to the nearest separator/storage tank. In that case, p1 = 0 • Since p2 = 0, therefore only p3 = pf need to be calculated.

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Concept of Choke Performance Relationship (CPR) The choke is a narrow cross-section short (about 2-3 inches) tube inserted in a pipe adapter inserted between the wellhead and the flowline. It’s purpose is to cause a high K.E. pressure loss, p2, so that the flow rate (qo) from the well can be reduced if so desired.

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Fig. Flow regimes in a Fixed Choke. Page 20 of 28

Example-4. Repeat Example-2, assuming that the choke diameter is (5/64)-in. [Example-2: In Example-1, the tubing diameter (ID) changes from 2.441-in. to 2.25-in. at the SSSS Valve (650-ft below the surface, Tf = 130 oF). Calculate the pressure drop due to this change of diameter. Is this pressure drop positive, or negative?]

Solution: We need to know the temperature of the flowing oil at the SSSV. If the sp.gr of the flowing oil is 0.85 at 60 oF, then we need to find its sp.gr. at the temperature at the SSSV (o) TSSSV, then convert to density in lbm/cuft = 0.85 x (H2O) at TSSSV [Use either a published chart, or Correlation!] Suppose these data are (o)130F = 0.82 and (H2O)130F = 62.1 lbm/cuft, respectively. Page 21 of 28

Therefore, (O)130F = 0.82 x 62.1 = 50.92 lbm/cuft

= 1.3 x 10-8 x (200)2 x 50.92 (1/(5/64)4 – 1/2.4414)

p2

= 710.8 psi  quite significant !!

Since the above p2 is +ve, therefore, the FBHP, pwf, will be increased by this amount. That means, the reservoir pressure drawdown, (pR – pwf) will decrease by 711 psi  net flow rate will be decreased. Thus, the choke can be used for flow rate control, if need be, esp. in case of a new well. Page 22 of 28

Graphical Form of Friction Factor, fF and fD: We’ve seen that for Laminar flow, the Fanning friction factor is given by: fF = 16/NRe, which clearly shows a non-linear relationship between fF (y-axis) and NRe (x-axis). So take logarithm and we have: log(fF) = log(16) – log(NRe),  Y = C - X Which shows that a graph of Y = Log(fF) and X = log(NRe) is a straight line, with intercept C = log(16), and s Page 23 of 28

lope m = -1.

This is traditionally shown as the Moody Chart (1944)! But wait a minute! In the Moody Chart, the Y-axis is not Log(fF) but Log(fD), where fD =Darcy friction factor, which is 4-times our Fanning friction factor, so that, fD = 64/NRe

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Therefore, if you ever see the term “friction factor,” DO ASK Which one it is: fF , or fD

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The graphical form of the Friction Factor equation is shown in the following figure.

Turbulent flow

Laminar flow

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Estimating Frictional

Pressure Drop in Pipe flow:

Multi-phase flow Our discussion was confined so far to a Single-phase flow (liquid). In an oil well, the flow inside the tubing string, surface flow line, and the choke (if present) is Two-phase: liquid oil and gas.  The combined flow of oil and gas in the same physical domain is complex, as far as the dispersion of gas in the liquid, and vice versa, is concerned. Page 27 of 28

 Even there is a marked difference of two-phase flow in VERTICAL, HOROZONTAL and INCLINED pipes!!  We have NO ‘law-based’ equation to calc. pf : instead, we have to rely on one of many CORRELATIONS! Some popular correlations are:  Poetmann & Carpenter (1952)  Duns & Ross (1963)  Hagedorn & Brown (1965)  Beggs & Brill (1973)  Orkiszewski (1967)  Aziz, Govier, and Fogarasi (1972)  Ansari et al. (1994) Page 28 of 28