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• Dhafin Rizqi

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TM3104.nb

������������� Wellbore flow can be divided into several broad categories, depending on the flow geometry, the fluid properties, and the flow rate. The flow geometry of interest in the wellbores is usually flow through a circular pipe, though flow in an annular space, such as between tubing and casing, sometimes occurs. ◼ Single Phase Flow (gas or liquid only, e.g. injection wells) ◼ Multiphase Flow (at least two phases) ◼ Oil + Gas Flow ◼ Gas + Water Flow ◼ Oil + Gas + Water Flow ◼ Liquid/Gas + Solid?

Our objectives: predict the pressure as a function of position between the bottomhole location and the surface.

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������������������������������������������������������� Incompressible? Newtonian Fluid?

��������������������������������������������� In circular pipe, generally characterized by Reynold’s Number, NRe = (ρ u D) / μ (that is the ratio between inertial forces to the viscous forces) Laminar Flow : Fluid moves in distinct laminae, with no fluid motion transverse to the bulk flow direction. Turbulent flow : Characterized by eddy currents that cause fluctuating velocity components in all directions. Both will strongly influence the overall pressure drop, e.g. frictional pressure drop. Transition between laminar and turbulent flow occured at NRe = 2100

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Example 1: For the injection of 1.03 - specific gravity water (ρ = 64.3 lbm/ft3) in an injection well with 7 - in., 32 - lb/ft casing, construct a graph of Reynolds number versus volumetric flow rate (in bbl/d). The viscosity of the water at bottomhole conditions is 0.6 cp. At what volumetric flow rate will the transition from laminar to turbulent flow occur? Note that in this example I consistently calculate everything in SI unit! SG = 1.03; OD = 7 inch;(* casing weight of 32 lbft, we get ID = 6.094 inch from API casing table *) ID = 6.094 inch; μ = 0.6 cp; ρ = SG ρw; 1 π ID2 ; A= 4

Recall: NRe = ρ u D / μ, rearranging: u =

NRe μ ρD

If u = q / A, then the transition to turbulent occured at q= q=

NRe μ A ρD

where NRe = 2100

2100 μ A ρ ID

;

qlaminar-turbulent

80.9804

bbl/d

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������������� The pressure drop over a distance, L, of single-phase flow in a pipe can be obtained by solving the mechanical energy balance equation: ⅆp ρ

+

u ⅆu gc

g

+g + c

2 ff u2 ⅆL gc D

+ ⅆWs = 0

If the fluid is incompressible (ρ = constant), and there is no shaft work device in the pipeline (a pump, compressor, turbine, etc.):

Δp = p1 - p2 =

g gc

ρ

ρΔz + 2 g Δu2 +

or

Δp = ΔpPE + ΔpKE + ΔpF

c

2 ff ρu2 L gc D

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TM3104.nb

ΔpPE - Pressure drop due to Potential Energy change

Potential energy pressure drop:

ΔpPE =

g gc

ρ Δz, where

Δz = L sin θ Δz : The difference in elevation between positions 1 and 2, with z increasing upward. L : Pipe length θ : The angle between horizontal and the direction of flow. (e.g.: θ is +90° for upward, vertical flow, 0° for horizontal flow, and –90° for downward flow in a vertical well) Example 2: Suppose that 1000 bbl/d of brine (γw = 1.05) is being injected through 2 7/8-in., 8.6-lbm/ft tubing in a well that is deviated 50° from vertical. Calculate the pressure drop over 1000 ft of tubing due to the potential energy change.

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ΔpKE - Pressure drop due to Kinetic Energy change ΔpKE is the pressure drop resulting from a change in the velocity of the fluid between positions 1 and 2. It will be zero for an incompressible fluid unless the cross-sectional area of the pipe is different at the two positions of interest. ρ 2 gc ρ = 2 gc

ΔpKE =

Δu2

ΔpKE

u22 - u12

The velocity only varies only with the cross-sectional area of the pipe. Since A = π D2 4, then 4q , therefore π D2 8 ρ q2 1 1   ΔpKE = 2 4 D41 π g c D2

u=

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TM3104.nb

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ΔpKE - Pressure drop due to Kinetic Energy change Example 3: Suppose that 2000 bbl/d of oil with SG of 0.93 is flowing in a horizontal pipeline having a diameter reduction from D1 to D2 where D2 is one-half of D1 where D1 is 4 inch. What is the pressure drop (in psi) caused by diameter change?

Note that in this example I consistently calculate everything in SI unit! SG = 0.93; D1 = 4 inch; D2 = 0.5 D1; qo = 2000 bbl  day; ρo = SG ρw; u2 = 4 qo  π D22 ; u1 = 4 qo  π D12 ; ρo u22 - u12 ; ΔpKE = 2 u1 1.48931 ft/s 5.95726 ft/s u2 0.208046 psi ΔpKE

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ΔpF - Frictional pressure drop The frictional pressure drop is obtained from the Fanning equation: ΔpF =

2 ff ρ u 2 L gc D

where ff is the Fanning friction factor. For laminar flow, the frictional pressure drop is only a function of NRe , where: ff =

16 NRe

In turbulent flow, the friction factor may depend on both the Reynolds number and the relative pipe roughness, ε. The relative roughness is the amount of surface roughness that exists inside the pipe. ε = k / D, where k is the absolute roughness of the pipe

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ΔpF - Frictional pressure drop The Fanning friction factor is most commonly obtained from the Moody friction factor chart. This chart was generated from the Colebrook-White equation: 1

= - 4 log

ff

ε 3.7065

+

1.2613

NRe

ff

In an explicit form (Chen’s equation): 1

ff

= - 4 log

ε 3.7065

-

5.0452

NRe

log

ε1.1098 2.8257

+

7.149

NRe

0.8981



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TM3104.nb

ΔpF - Frictional pressure drop Example 4: Suppose that 1000 bbl/d of brine (γw = 1.05 and μw = 1.2 cp) is being injected through 2 7/8-in., 8.6lbm/ft tubing in a well that is deviated 50° from vertical. The tubing relative roughness is 0.001. Calculate the pressure drop over 1000 ft of tubing due to the friction. Solution: First, we need to calculate the Reynold’s number to determine if the flow is laminar or turbulent. From the API tubing table, we get tubing ID of 2.259”. Note that in this example I consistently calculate everything in SI unit! q = 1000 bbl  day; SG = 1.05; ρ = SG ρw; μw = 1.2 cp; ID = 2.259 inch; ε = 0.0001; L = 1000 ft; u = 4 q  π ID2 ; ρ u ID NRe = ; μw u NRe Flow is

2.33477 35 657.2 Turbulent

ft/s

Second, we calculate the Fanning friction factor using Chen’s equation: ff = - 4 Log10

ε 3.7065

-

5.0452

Then, ΔpF = ff ΔpF

2 ff ρ u2 L

; ID 0.00572125 4.67848

psi

NRe

Log10

ε1.1098 2.8257

+

7.149 NRe

0.8981

-2



;

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QUIZ

Suppose that 1100 bbl/d of brine (γw = 1.03 and μw = 1.1 cp) is being injected through 1000 ft length of 9 5/8-in., 47-lb/ft and 1000 ft length of 5-in., 20.3 lb/ft casings in a well that is deviated 50° from vertical. The casings relative roughness is 0.0015. a. Calculate the overall pressure drop in 9 5/8-in casing (psi) b. Calculate the overall pressure drop in 5-in casing (psi) c. Calculate the overall pressure drop of the system (psi) Solution: Note that in this example I consistently calculate everything in SI unit! q = 1100 bbl  day; SG = 1.03; μ = 1.1 cp; ID1 = 8.681 inch; ID2 = 4.184 inch; L1 = 1000 ft; L2 = 1000 ft; θ = - 40; ρ = SG * 1000; ε = 0.0015;

a.

Δp = ΔpPE + ΔpKE + ΔpF since there is no change in Kinetic Energy then ΔpKE = 0

ΔpPE1 = ρ gu L1 Sinπ  180  θ;

To calculate the friction factor, first we need to determine if the flow is laminar or turbulent using NRe

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TM3104.nb

u1 =

4q

; π ID12 ρ u1 ID1 ; NRe1 = μ

ff1 = - 4 Log10

ε 3.7065

-

5.0452 NRe1

Log10

ε1.1098 2.8257

+

7.149

0.8981

-2



NRe1

;

Then, ΔpF1 =

2 ff1 ρ u12 L1 ID1

;

Therefore in the first segment: Δp1 = ΔpPE1 + ΔpF1; u1 NRe-1 Flow is ff-1 ΔpPE-1 ΔpF-1 Δp1

b.

0.173912 10 944.4 Turbulent 0.00813121 - 287.025 0.00943635 - 287.015

ft/s

psi psi psi

Δp = ΔpPE + ΔpKE + ΔpF Similarly to solution a. since there is no change in Kinetic Energy, then ΔpKE = 0 ΔpPE2 = ρ gu L2 Sinπ  180  θ;

To calculate the friction factor, first we need to determine if the flow is laminar or turbulent using NRe 4q ; π ID22 ρ u2 ID2 ; NRe2 = μ

u2 =

ff2 = - 4 Log10

ε 3.7065

-

5.0452 NRe2

Then, ΔpF2 =

2 ff2 ρ u22 L1 ID1

;

Therefore in the second segment: Δp2 = ΔpPE2 + ΔpF2; u2 NRe-2 Flow is ff-2 ΔpPE-2 ΔpF-2 Δp2

0.748662 22 707.6 Turbulent 0.00708944 - 287.025 0.152466 - 286.872

ft/s

psi psi psi

Log10

ε1.1098 2.8257

+

7.149 NRe2

0.8981

-2



;

TM3104.nb

c.

Overall pressure drop of the system ΔpTotal = ΔpPE-Total + ΔpKE + ΔpF-Total Where ΔpPE-Total = ΔpPE-1 + ΔpPE-2 and ΔpF-Total = ΔpF-1 + ΔpF-2 ρ

u22 - u12 ; 2 ΔpTotal = ΔpPE1 + ΔpF1 + ΔpPE2 + ΔpF2 + ΔpKE

ΔpKE =

ΔpKE ΔpTotal

0.00367958 - 573.884

psi psi

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