Section 2 Multiphase Flow, Flowing Well Performance

Multiphase Vertical Flow When producing an oil or gas well, the flow of the fluids up the tubing will be in most cases be 2 phase, liquid and gas. The pressure at any point in the tubing string is influenced by many factors. Such as type of flow at that point, the viscosity’s of the oil and gas, the PVT characteristics of the fluid which are pressure and temperature dependent, the bottom hole and tubing pressures. In the tubing there is a pressure drop and a temperature variation which causes several of the fluid properties to vary along with the pressure drop that allows the solution gas to come out of the oil at different rates in various parts of the tubing, causing holdup behavior. The drag of the flowing fluids (friction) also plays a part along with gravity in resisting the flow. Flow Regimes As the oil travels up the tubing different types of flow can occur as the gas is liberated from the liquid phase. If the well is producing above the bubble point pressure the flow is single phase but as it rises up the tubing the pressure will drop below this pressure and the multiphase regimes will start to form. Bubble Flow The gas starts to come out of solution and form in the liquid medium in an even dispersion.

Slug or Plug Flow As the fluid moves upward the gas bubbles move faster than the liquid, because of the buoyancy effect, and collect in large bubbles. These bubbles grow to a size where extend across the diameter of the tubing separating slugs of liquid containing the smaller bubbles

Churn Flow With increasing gas velocity, the larger bubbles become unstable and collapse, resulting in a highly turbulent flow pattern. Churn flow is characterized by oscillatory, up and down motions of the liquid. Annular Flow As the pressure is lowered these large pockets of case break through the slugs and form a continuos phase in the center of the tubing carrying droplets of the liquid with a film of oil flowing along the walls of the tubing.

Mist Flow Eventually the gas volume is so large that the film of liquid on the wall of the tubing disappears and the only liquid moving is the droplets or mist in the gas phase.

Not all of these regimes are necessarily present in a well but usually more than one is present. As can be seen the flow of oil and gas up a string of tubing has many complex components that must be considered in designing a flowing well. For a given bottom hole pressure there is a certain amount of reservoir fluids that will enter the well bore, what is needed to be known is whether the pressure drop in the tubing will allow flow to the surface. Several methods have been developed to describe this flow so pressure curves of the tubing can be drawn to use in the planning of the well and its equipment. Poettmann & Carpenter Method Based on the energy equation with the assumptions that the difference in the kinetic energy at the top and bottom is negligible and that external work done by the flowing fluid is also negligible. The energy loss Wf is caused by the slippage and frictional effects.

gDhfuWf 2

4 2 Δ= (2-1)

where u is the average mixture velocity over Δh and D is the inside diameter of the tubing, g is gravity constant and f is found empirically.

With the given assumptions and field units equation 2-1 can be reduced to

144ΔΔ

ph

K= +

−

ρρ

_

_ (2-2)

where Δp = pressure drop over Δh, psi

ρ = average density of the fluid, lb/cu ft

Kfq M

x D_

( . )=

2 2

10 57 413 10 (2-3)

q = liquid production rate, STB/day M = Total mass of gas and liquid associated with 1 STB, lb D = ID of tubing, ft f = energy loss factor defined by equation 2-1

The way that this is used is to start at the bottom of the tubing with the known pwf which can produce a certain production of reservoir fluids that is known from the productivity index of the well. All the PVT and reservoir parameters such as FVF, water-cut, gas solubility are known or can be estimated. Divide the tubing into equal parts, H1, H2, H3, H4, each a length of Δh and let the pressures, p1, p2, p3, p4, at the Hn points. At H1 the pressure is pwf, so this pressure can be used for H1H2 for the first approximation for the factors needed for equation 2-2, such as the density and Mass. If a better approximation is needed use the average between of pwf and the first approximation for the next calculations. This process is repeated for all the intervals in the tubing, to the surface or till a negative pressure is calculated. Doing this for several starting pwf a plot can be generated for any size tubing.

Figure 2-1 This graph can be used to find the bottom hole pressure and production rates for any producing tubing head pressure. The maximum rate that the well can flow is indicated by point E on the graph, this point must always be lower than q’ since there is a positive pressure loss in the tubing.

Figure 2-2 Plotting the tubing head pressures for several sizes of tubing is used to find the optimal tubing size for different flow rates. In this case at 200 psi THP the 2” tubing is better but at 100 psi the 3 ½” tubing produces at a higher rate. The only factor in equation 2-3 that is not easily found is the energy loss factor f. In an effort to establish values for f as a function of the variables involved in the flow of oil wells a reverse calculation was made. This involved taking field data in the form of production, PVT data, and pressure and temperature transverses. It was found that f correlated best as a function of tubing ID and mass velocity of the fluid in the tubing. In terms of field units

DMqxDv )104737.1( 5−=ρ (2-4)

From this the following plot is used to find f for different flow rates and tubing sizes.

Figure 2-3 It is noted here that this plotted on log-log paper and that the scatter of the points is greater that its appears on this graph. Also Baxendell extended this work to in include flow in an annulus, and with Thomas plotted f over higher flow rates. The weakness of this approach is the inaccuracy of finding f. Work by others have been to overcome this problem. Using the Poettman & Carpenter Method

Calculating M, mass of the fluid in STB M = wt of Oil + wt of Water + wt of Gas in lbm/STB wt of Oil = Oil cut x γo x 350 lb/bbl wt of Water = Water cut x γw x 350 lb/bbl wt of gas = GLR cf/bbl x γg x .0764 lb/cf w = Mass flow rate, = Mq, lbm / day Note: oo API+

=5.131

5.141γ γw and γg and GOR should be given &

the oil and water cut calculated from the WOR. Calculating ave. Density, ρ, lbm/cf ρ = M/volume Volume = vol of oil + vol of water + vol of free gas in cf Using the pressure of interval being calculated, lab or correlations of

PVT data needed. βo, βg, Rs Volume bbls = Oil cut x βo + Water cut + βg{ GLR – Oil cut x Rs} Volume cf = Volume bbls x 5.614 cf/bbl ρ = M/(5.614[oil cut x βo + water cut] + βg{GLR – Oil cut x Rs}) With values of M and ρ the Δp/Δh can be calculated using equations 2-2 and 2-3 by finding f from figure 2-3. By dividing the pressure increment by this pressure gradient gives the length of tubing over which the increment occurs. This way a plot of the pressure can be obtained.

Example: Given Tubing 2” qo. = 400 bpd 22o API oil qw.= 600 bpd γw = 1.07 GLR = 500 cf/bbl γg = .65 γo = 141.5/131.5 + 22 = .922 M = .4*.922*350 + .6*1.07*350 + 500*.65*.0764 = 378.6 lb/bbl w = M*q = 378.6*1000 = 378610 lb/day Dρv = 1.4737E-5 x w/D = 1.4737E-5 * 378610/1.995/12 = 33.56 Form fig. 2-3 f = .008 K = f w2 /(7.413E 10 D5) = .008*378610^2 / (7.413E+10*.166^5) = 122.7 For pressure point 1, 500 psi βo = 1.043 Rs = 59 scf/bbl z = .93 ρ = M/5.614[oil cut x βo + water cut] + βg{GLR – Oil cut x Rs} 378/(5.614[.4*1.043 + .6] + .0298{500 - .4*59}]) = 19 lb/cf For pressure point 2, 1000 psi βo = 1.083 Rs = 120 scf/bbl z = .9 ρ = 378.6/(5.614[.4*1.083 + .6] + .0153{500 - .4*120}) = 29.8 lb/cf ave ρ = ρ1 + ρ2 / 2 = 24.4 lb/cf Use this average density to find Δp/Δh in equation 2-2 Δp/Δh = [ρ + K/ρ]/144 = [24.4 + 122.7/24.4]/144 = .204 psi/ft

Now solve for /Δh for the Δp Δh = 1000 − 500 / .204 = 2447 ft. Plot this point, say THP is 500 psi then it 1000 psi at a depth of 2447. Then repeat the process for the next interval. A large interval was used here in the example, the interval used when doing a pressure profile of a well the pressure intervals will be much smaller, like about 25 psi.

Modified Hagedron and Brown Method (mH-B) This is an empirical two-phase flow correlation, the core of which is correlation for liquid hold-up. Griffith correlation is used for flow in the bubble flow region. The mechanical energy balance equation used for the mH-B correlation

( )

( )h

gu

DE

fwdhdp cm

ΔΔ

++=2/

10413.7144

2_

_5

2_ρ

ρρ (2-5)

in oilfield units.- Bubble flow regime exists if λg < LB where

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

DuL m

B

2

2218.071.1 (2-6)

if LB <.13 set LB to .13. The input fraction of gas λg is

m

sgg u

u=λ

To find the liquid hold-up that is needed to calculate the pressure gradient from correlations require the following dimensionless numbers

Liquid velocity 4938.1σρ l

slvl uN = (2-7)

Gas velocity 4938.1σρ l

sgvg uN = (2-8)

Pipe diameter σρ l

D DN 872.120= (2-9)

Liquid viscosity 4 3115726.σρ

μl

lLN = (2-10)

Where superficial velocities us in ft/sec, density ρ in lbm/ft3, surface tension σ in dynes/cm, diameter D in feet. Using mH-B method First find the flow regime. Calculate mass flow rate w and density ρ as before. Calculate the dimensionless numbers. Calculate the velocities ul & ug

86400

614.5

t

ll A

qu = 86400t

glg A

GLRqu

β= glm uuu +=

Using NL find CNL from fig. 2-3

Then find the group ( )Davg

Lvl

NpNCNpN

1.0575.0

1.0

find yl/ψ from fig 2-4

p here is the absolute pressure at the location of interest.

Compute 14.2

38.0

D

Lvg

NNN from fig 2-5 we get ψ

Get the liquid hold-up yl yl =(yl / ψ)/ψ Note: For low viscosity fluid ψ will generally be 1. For f a 2 phase Reynolds number is be used

)1(Re22.2

ll yg

ylD

wEN −

−=

μμ

f from Moody diagram. Get pressure gradient from, assuming no kinetic energy

( ) _

5

2_

10413.7144

ρρ

DE

fwdhdp

+=

The Griffith correlation For bubble flow Griffith uses a different hold-up correlation, bases the frictional pressure gradient on the in-situ average liquid velocity.

( ) 25

2_

10413.7144

ll

l

yDEfw

dhdp

ρρ += (2-11)

The liquid hold-up is

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛+−+−=

s

sg

s

m

s

ml u

uuu

uuy 411

211

2

(2-12)

where slip velocity us is .8 ft/sec. Reynolds number is based on in-situ liquid velocity

l

lluDNμ

ρ_

Re = or l

l

DwEN

μ22.2

Re−

= (2-13)

figure 2-3

figure 2-4

figure 2-5 Equations to replace figure 2-3

( ) YLCN 10=

41

31

211 12195.054785.055100.015841.069851.2 XXXXY −+−+−=

( )[ ]3log1 += LNX

to replace figure 2-4

( ) ( )[ ] ( )[ ] ++−++−= 2

22 6log63295.06log61777.010307.0/ XXyL ψ

( )[ ] ( )[ ]4

23

2 6log0401.06log29598.0 +−++ XX ( )

DavG

LvL

NpNCNpNX 1.0575.0

1.0

2 =

for figure 2-5

43

33

233 3.1161746.2225325.123282176.491163.0 XXXX +−+−=ψ

14.2

38.0

3D

LvG

NNNX =

if X3>.01 for X3<.01 then ψ = 1

and for f

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−−=

8981.0

Re

1098.1

Re

149.78257.2

log0452.57065.3

log41NNf f

εε

Find the pressure gradient at 500 psi Tubing 2” qo. = 400 bpd 22o API oil qw.= 600 bpd γw = 1.07 GLR = 500 cf/bbl γg = .65 σo = 30 dynes/cm σw = 70 dynes/cm μo = 3 cp μg = .015 cp

γo = 141.5/131.5 + 22 = .922

M = .4*.922*350 + .6*1.07*350 + 500*.65*.0764 = 378.6 lb/bbl w = M*q = 378.6*1000 = 378610 lb/day Find the flow regime

86400614.5

t

ll A

qu = 86400t

glg A

GLRqu

β= glm uuu +=

sec/98.286400022.

614.51000 ftul =⋅⋅

= sec/84.786400022.

0298.5001000ftug =

⋅

⋅⋅=

sec/82.1084.798.2 ftuuu glm =+=+=

3.155166.82.102218.071.12218.071.1

22

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

Du

L mB

LB<.13 so LB =.13 725.82.10/84.7 ===

m

sgg u

uλ

λg is larger than LB so the flow regime is not bubble. ρl = Ml/5.614[oil cut x βo + water cut] 353.8/5.614[.4*1.043 + .6] = 61 lb/cf ρg = Mg/ βg{GLR – Oil cut x Rs} 24.8/ .0298{500 - .4*59}=1.75 lb/cf

89.6546198.2938.1938.1 44 =⋅==

σρ l

slvl uN

66.15546184.7938.1938.1 44 =⋅==

σρ l

sgvg uN

4.215461166.872.120872.120 =⋅==

σρ l

D DN

0084.5461198.215726.115726. 4

34 3 =⋅

⋅==σρ

μl

lLN

From chart 2-3 CNL = .0022

( )41.2

4.217.1466.150022.50089.61.575.

1.

1.0575.0

1.0

−=⋅

= ENpN

CNpN

Davg

Lvl

From chart 2-4 yl/ψ = .47

0036.4.210084.66.15

14.2

38.

14.2

38.0

=⋅

=D

Lvg

NNN

From chart 2-5 Ψ =1 so yl = .47

( ) 2777288015.3166.37861022.222.2

47.147.)1(Re =⋅⋅⋅−

=−

= −−

ED

wENll y

gyl μμ

From the graph f = .005

( )

ftpsiEDE

fwdhdp /028.144/63.3

19166.10412.7378610005.19

10413.7144 5

2

_5

2_==

⋅⋅

+=+=ρ

ρ

Homework Use the data from the previous homework and solve the pressure gradient at 1000 psi. Additional data μo = 4 cp μg = .02 cp σo = 30 dynes/cm σw = 70 dynes/cm

Ros Method Uses a pressure balance rather than a energy balance equation. So that in a single phase flow the pressure gradient is the sum of the static, friction and acceleration gradients,

D

vfg

dhdp

2

21

4ρ

ρ += (2-14)

here with the acceleration gradient neglected. For 2 phase flow the static term should be modified to allow for gas slippage though the liquid. Ros introduced the a static gradient gg ggll ρερε + (2-15) Where ε is the liquid and gas holdup. Holdup is defined as the volume of that phase actually present in a certain length of pipe. By substituting equation 2-6 into 2-5

termfrictionldh

dpg

gll

l

−+++=ρρ

εερ

)1(1 (2-16)

The density of gas is small compared to the oil so 2-7 can be rewritten; termfriction

dhdp

g ll

−+= ερ1 (2-17)

Ros then conducted experiments to determine the influence of varies factors, resulting in an empirical correlation’s for liquid holdup and wall friction which are very complex. A computer program based on these correlation’s calculate pressure gradients that compare favorably with Baxendell & Thomas field data. Gilbert Method

Empirical based method, based on measured values of tubing flow pressure losses, a families of curves were derived that can be used for extrapolation and interpolation purposes. The following measurements were taken on a large number of wells; Tubing depth, ft Flowing BHP, tubing in-take pressure, psi THP, psi Flow rate, gross liquid, bbl/day GLR, mcf/bbl Tubing size Assuming that the BHP depends only on the other five listed variables, group the wells with the same tubing size, production rate, and GLR to get a correlation. Plot the BHP as a function of depth for these groups and a plot like figure 2-4 is obtained showing the pressure curves for several THP.

figure 2-6

Take a point on the curve of a THP A at a selected depth and place a valve there that can be opened in such a way that the pressure and production rate will be unchanged when the valve at the tubing head is closed. It can be seen that pressure curve below this point will also be unchanged and that this point can now be viewed as the tubing head. So actually all the curves are just extensions of one curve. This curve is given in figure 2-6.

figure 2-7 To use this curve to find the BHP at a given depth, the equivalent depth of the THP is found on the curve and added to the total length of the tubing. For example if the THP is 200 psi, and the BHP at 10,000’ is wanted, the depth on the curve of 200 psi is noted, say is 1200’, this is added to the 10,000’, making a depth of 11,200’ which is used on the curve to find the BHP. It should be noted that at low pressures it found that the correct shape of the curve is actually concave downward or the reverse of the higher pressure portion of the curve. This is seen if Poettmann and Carpenter method is used to calculate the curves. Baxendell and Carpenter ran very carefully controlled well tests and confirmed this shape at the lower pressures.

The Gilbert Curves use GLR as a parameter and there is one family of curves for each tubing size and flow rate. Two approaches to solving flowing well problems. 1) Find the IPR curve of the well, then find the pwf for different flow rates for a given THP. The intersection of the plot of these two lines will be the expected flow rate for the given THP.

IPR & 200 THP

qo bpd

0 100 200 300 400 500 600 700

Pw

f ps

i

600

700

800

900

1000

1100

1200

1300

1400

1500

IPRthp 200

2) After calculating the IPR curve, calculate the pressure loss in the tubing at various flow rates. This will give a plot like figure 2-1, it can be seen that for any flow rate a THP and BHP (pwf) can be found. Also the flow rate for and THP can be found.