PETE 411 Well Dri rill llin ing g
Lesson Lesso n 11 Lami La mina narr Flo low w
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Lesson Le sson 11 - La Lamina minarr Flow Flow
Rheo Rh eolo logi gica call Mo Mode dels ls • Newtonian • Bingham Plastic • Power-Law
Rotational Viscometer
Laminar Flow in Wellbore • Fluid Flow in Pipes • Fluid Fl Flow in in An Annuli
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Read ADE A DE Ch Ch.. 4 to t o p . 138
HW #5 ADE A DE Pr Pro o b l em ems s 4.3, 4.4, 4.5, 4.6 Due Frid Fri d ay, Sep Sep t . 27, 27, 2002 2002
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New to ton n ian Flu Fluid id Mod ode el
Experiment ally ,
F A
= µ
V L
τ = µ γ Shear stress = viscosity * shear rate
4
F A
= µ
V L
Lami La mina narr Flo Flow w of o f Ne Newt wton onia ian n Flu Fluid ids s 5
Newt wton onian ian Flu Fluid id Mo del In a Newtonian fluid the shear stress is directly proportional to the shear rate (in laminar flow):
.
τ = µ γ
1 = µ 2 cm sec
dyne
i.e., The constant of proportionality, is the viscosity of the fluid and is independent of shear rate. 6
Newt wto o ni nia an Flu Fluid id Mod ode el µ
=
τ
.
γ
dyne • sec cm 2
Viscosity may be expressed in poise or centipoise.. centipoise dyne - s g 1 poise = 1 =1 2 cm cm − s 1 centipoise
= 0.01 poise 7
Sh ear St r es s v s . Sh ear Rat e f o r a Newt wton onia ian n Flu luid id
.
τ = µ γ
Slo lope pe of lin l ine e=µ 8
Exa xample mple - Newtonia wtonian n Fluid
9
Ex amp ampll e 4. 4.16 16
Area of upper plate = 20 cm2 Distance between plates = 1 cm Force req’d to move upper plate at 10 cm/s = 100 dynes. What is fluid viscosity? 10
Ex amp ampll e 4. 4.16 16 τ = µ γ 2
shear stress F / A 100/ 20 dynes/cm µ = = = shear rate V / L 10/1 sec-1
=
5 10
= 0 .5
dyne − s cm
2
= 0 .5 poise
µ = 50 cp 11
Bing Bi ngha ham m Pla Plast stic ic Mod ode el
12
Bing Bi ngha ham m Pla Plast stic ic Mod ode el ⋅
τ = µp γ + τ y
if τ > τ y
⋅
γ = 0
if - τ y ≤ τ ≤ τ y ⋅
τ = µp γ − τ y
if τ < -τ y
τ τ and τ τ y are often expressed in lbf/100 sq.ft 13
Bing Bi ngha ham m Pla last stic ic Mod ode el g cm 454 980 2 lbf sec lbf lbf = 1 * 2 2 2 100 ft 100 ft cm 30 . 48 ft 1
lbf 2
100 ft
= 4.79 dyne/cm
2
(p.134)
1 dyne is the force that, if applied to a standard 1 gram body, would give that body an acceleration of 1 cm/sec2 14
Ex am p l e 4.17 {pa {p ar all lle el plate p lates s again again!} !} Bingh Bin gha am Pla Plasti stic c Flu Fluid id Area of upper plate
= 20 cm2
Distance between plates 1. Min. force to cause plate to move
= 1 cm = 200 dynes
2. Force req’d to move plate at 10 cm/s cm /s = 400 dynes
Calc lcul ula ate yi yie eld poi p oint nt and and pla pl asti stic c visc vi scosi osity ty 15
Ex amp ampll e 4. 4.17 17 Yield point,
⋅
τ = τ y + µ p γ
Fy
200 dynes dynes τy = = = 10 2 A 20 cm cm 2
but 1
∴τ y =
lbf 100 ft
10 4.79
=
2
= 4 . 79
dynes cm
2
2.09 lbf/100 ft
2
16
Ex amp ampll e 4. 4.17 17 Plastic viscosity,
p
is given by ⋅
τ = τ y + µ p γ 400 dynes 20 cm
∴ µ p =
2
=
200 dynes 20 cm
20 − 10 10
=1
2
10 cm/s + µ p 1 cm
dyne − s cm
2
= 1 poise
i.e. µp = 100 cp 17
Po w er -L aw Mode Mod el
18
Po w er -L aw Mode Mod el
⋅ n
τ = K γ ⋅
⋅
if γ ≥ 0 ⋅ n −1
τ = K γ γ
⋅
if γ < 0
n = flow behavior index K = consistency index 19
Po w er -L aw Mode Mod el lbf − sn lbf − sn 1 = * 2 2 ft ft
1
lbf − sn ft
2
g 454 lbf
cm 980 2 sec
cm 30.48 ft n
2
2
= 479 dyne − s / cm = 479 eq. poise.
1
lbf − sn ft
2
= 47,900 eq. cp. 20
Ex am p l e 4.18 Po w er -L aw Fl Fl u i d
Area of upper plate
= 20 cm2
Distance between plates
= 1 cm
Force on upper plate = 50 dyne if v = 4 cm/s Force on upper plate = 100 dyne if v = 10 cm/s Calculate consistency index (K) and flow flo w be beha havi vior or in inde dex x (n) 21
Ex am p l e 4.18 v = 4 cm/s ⋅
Area of upper plate = 20 cm2
n
τ 4 = K γ 4
Distance between plates = 1 cm
50 4 = K 20 1 2 .5 = K
n
(4 )
n
Force on upper plate = 50 dy dyne ne if v = 4 cm cm/s /s
(i) 22
Ex am p l e 4. 4.1 18 v = 10 cm/s
Area of upper plate = 20 cm2
⋅
τ 10
= K γ 10
n
Distance between plates = 1 cm Force on upper plate
100 20
10 = K 1
5 = K (10
n
)
n
= 100 dy dyne if v = 10 cm/s
(ii) 23
Ex am p l e 4.18
n
2 . 5 = K (4 ) 5 = K (10
n
)
(i) (ii)
Combining Eqs. (i) & (ii): 5 2 .5
=
K 10 n K 4n
= 2 .5 n
log 2 = n log 2 . 5
n = 0.7565 24
Ex amp ampll e 4. 4.18 18
5 = K (10
n
) (ii)
From Eq. (ii):
∴ K =
5 10
n
=
5 10
0 .7565
= 0 .8760 eq. poise
K = 87.6 eq. cp.
25
Ap A p p ar aren entt Vi Vis scosity
Apparent viscosity = (τ / γ ) is the slope at each shear rate,
⋅
⋅
⋅
γ 1 , γ 2 , γ 3 . 26
Ap A p p ar aren entt Vi Vis scosity Is no nott con const stan antt for for a ps pseu eudo dopl plas asti tic c fl flui uid d The apparent viscosity decreases with increasing shear rate (for a power-law fluid) (and also for a Bingham Plastic fluid) 27
Typ ypic ica al Dri Drill llin ing g Flu Fluid id Vs. (Plotted on linear paper) New to toni nia an , B in ingh gha am and and Pow owe er Law La w Flu Fluid ids s
0
28
Rheolo heologi gical cal Mod ode els 1. Newtonian Fluid:
= shear stress µ = absolute viscosity
⋅
τ = µ γ
⋅
γ = shear rate
2. Bingham Plastic Fluid: ⋅
τ = τ y + ( µ p ) * γ
τ y = yield point µ p = plastic viscosity
What if τ τ y = 0?
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Rheolo heologi gical cal Mod ode els 3. Power Law Fluid: ⋅
τ = K (γ )
n
K = consistency index n = flow behavior index
When n = 1, fluid is Newtonian and
We shall use power-law model(s) to calculate pressure losses (mostly).
K = µ
30
Rotating Viscometer
We determine rheological properties of drilling fluids in this device
Fi g u r e 3. 3.6
Rheometer
Infinite parallel plates
31
Rh eo m et er (Ro (Ro tati tatio o n al Viscometer)
sleeve BOB
fluid
τ = f ( ) Shear Stress = f (Dial Reading) Shear Rate = f (Sleeve RPM) Shear Stress = f (Shear Rate)
τ( TAU ), the Shear Stress depends on the
value of γ (GAMMA), the Shear Rate 32
Rhe heom ome ete terr - ba base se ca case se RP M 3 6 100 200 300 600
sec-1 5.11 10.22 170 340 511 1022 RPM * 1.703 = sec-1 33
Example A rotational viscometer containing a Bingham plastic pla stic fluid gives a dial reading of 12 at a rotor speed of 300 RPM and a dial reading of 20 at a rotor speed of 600 RPM Comput ompute e pla plastic stic visc viscosit osity y and and yie yi eld point p oint p
= θ 600 − θ 300
θ600 = 20 θ300 = 12
= 20 - 12
µ p = 8 cp
See Appendix A 34
θ600 = 20 θ300 = 12
Example
y
= θ 300 −
(See Appendix A)
p
= 12 - 8 τ y = 4 lbf/100 ft
2
35
Rot ota ati tion ona al Vis isco comete meter, r, Pow Powe er-La r-Law w Mod ode el Example: Exampl e: A rotat rotation ional al viscom viscomete eterr contai containin ning g a non-Newtonian fluid gives a dial reading of 12 at 300 RPM and 20 at 600 RPM.
Assuming pow powe err-la law w flu fluid id,, calculate the flow behavior index and the consistency index.
36
θ600 = 20 θ300 = 12
Example
θ 600 20 = 3 .322 log n = 3 . 322 log 12 θ 300 n = 0 . 7370
K =
510 θ 300 511
n
=
510 * 12 511
0 . 7372
= 61 .67 eq. cp 37
Gel St St r en g th
38
Gel St St r en g th = shear stress at which fluid movement begins
The yield strength, extrapolated from the 300 and 600 RPM readings is not a good representation of the gel strength of the fluid
Gel strength may be measured by turning the rotor at a low speed and noting the dial reading at which the gel structure is broken (usually at 3 RPM) 39
Gel St r en g th The gel strength is the maximum dial reading when the viscometer is started at 3 rpm.
In field units, τ = 1.06 θ g
lbf / 100 ft 2
In practice, this is often approximated to
τg = θmax,3
lbf / 100 ft
2
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Velo loci city ty Pro rofi file les s (la lami mina narr flow fl ow))
Fig ig.. 4-26. Velo loci city ty pr prof ofil ile es for f or laminar flow fl ow:: (a) pi pipe pe flow flo w and (b) ann nnul ula ar flo f low w 41
3D View o off La L ami minar nar Flo Flow w in i n a pipe pi pe - Newtonia wtonian n Fluid Fluid
“ It looks like conce concentric ntric rings of flui d tele te lesc scopi oping ng down d own the pipe at diff di ffe ere rent nt ve v eloc lociti itie es”
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Tabl ble e 4.4 - Sum umma mary ry of La Lami mina narr Flow Equ qua ati tions ons for Pip ipe es and and Annu A nnuli li Fictional Pressure Loss
Shear Rate at Pipe Well
Newtonian
_
_
dp f
Pipe
dL
=
⋅
µ v
Pipe
γ w =
1,500 d 2
Annulus
dL
=
d
Annulus _
dp f
96 v
_
µ v
1,000 (d 2 − d 1 )
⋅ 2
γ w =
144 v (d 2 − d 1 ) 43
Tabl ble e 4.4 - Summ umma ary of La Lami mina narr Flow Equ qua ati tion ons s for fo r Pip Pipe es and and Ann A nnul ulii Fictional Pressure Loss
Shear Rate at Pipe Wall
Bingham Plastic Pi p e dp f dL
Pi p e
_
=
_
µ p v 2
1, 500 d
+
τ y
⋅
225 d
Annulus
γ w =
96 v d
dL
=
µ p
Annulus _
dpf
+ 159 .7
τ y
µ p v
1,000(d 2 − d 1 )
+
τ y
200(d 2 − d 1 )
_ ⋅
γ w =
144 v (d 2 − d 1 )
+ 239.5
τ y µ p 44
Tabl ble e 4.4 - Summ umma ary of La Lami mina narr Flow Equ qua ati tion ons s for fo r Pip Pipe es and and Ann A nnul ulii Fictional Pressure Loss
Shear Rate at Pipe Well
Power-Law Pipe dp f dL
Pipe _ n
=
K v
1+ n
144,000 d
3 + 1 / n 0.0416
_
n
⋅
γ w =
Annulus dp f dL
d
(3 + 1 / n )
Annulus _ n
=
24 v
K v
_
2 +1/ n
1+ n 144,000 (d 2 − d 1 ) 0.0208
n
⋅
γ w =
48v d 2 − d 1
(2 +1/ n) 45
Tabl ble e 4.3 - Sum umma mary ry of Equ qua ati tion ons s forr Rot fo ota ati tion ona al Vis Visco come mete ter r
µa = θ300
Newtonian Model
or
µa = ⋅
γ =
300
θN
N
5 . 066 2
r
N 46
Tabl ble e 4.3 - Summ umma ary of o f Equ Equa atio tions ns for f or Rot ota ati tion ona al Vis isco come meter ter Bingham Plastic Model
oror p
= θ 600 − θ 300
µ p =
300 N 2 − N1
(θ N 2 − θ N1 )
or
τ y = θ 300 −
or p
τ y = θ N1 − µ p
N1 300
τ g = θ max at 3 rpm
47
Tabl ble e 4.3 - Summ umma ary of o f Equ Equa atio tions ns for f or Rot ota ati tion ona al Vis isco come meter ter Po wer-Law Mode Mod el or
θ 600 n = 3.322 log( ) θ 300
K =
510 θ 300 ( 511 )
n
or
θ N log θ N n= N 2 log N1 2
1
or
or K =
510 θ N (1 .703 N ) n 48
