3. Example :
Calculatethe reel capacity
Where: Tubing Outside Diameter (D) = 1.5 in.
Flange Height (A) = 24 in.
Free Board (E) = 2 in.
Flange Width (B) = 72 in.
Core Diameter (C) = 72 in.
solution
1) Calculate N = (A-E) / D
N = 14
2) CalculateM = C / D
M = 48
3) L = π x N x M (B +DN)
L = 196237 in.
L = 16,353ft.
4. 2- Pipe Sizes,volumes & Displacement
We can get internal volume or displaced volume directly by using
tables in “FieldData Hand Book” chapter 2 or can calculatedby
following equations
The internal volume:
Vi = (π/4) di
2 = 0.7854 di
2
The displacement volume:
V0 = (π/4) d0
2 = 0.7854 d0
2
Where: di = internal diameter of the pipe
d0 = external diameter of the pipe
Vi = internal volume of the pipe
V0 = external volume of the pipe
5. 3-Annular Volumes & Capacities
Also we can directly get the annularvolumes for CT inside of tubing
and tubing inside casing by using tables in “FieldData Hand Book”
chapterthree ,or by using next equation
The annular volume between two pipes
Va = (π/4) (d2
i-o- d2
o-i) = 0.7854 (d2
i-o- d2
o-i)
Where: Va = annularvolume between two pipes
di-o = internal diameter of the outer pipe
do-I = external diameter of the inner pipe
6. Example
16,200of 1.5 in OD C.T with wall thickness 0.095in
1) What is the total volume to fill the reel for P.T
2) How many barrels can 2000 ft of this string displaced, assuming that
the well is full of water without pumping
3) If the C.T run in 3.5 in x 12.7 lb/ft tubingwhat will be the annular
volume between 5000 ft C.T and tubing
Solution
From “Field Data Hand Book” table (2-2)
1) Internal volume = 1.667 bbl/1000ft
So, the total volume = 27 bbl.
2) External displacement by C.T = 2.168 bbls/1000 ft
So, the volume displaced by 2000 ft of C.T = 4.3 bbls
From “Field Data Hand Book” table (3-3)
3) The annular volume = 5.160 bbls/ 1000ft
So, the annular volume between 5000 ft C.T and tubing is
equal = 25.8 bbls
7. C
oiledTubingSizes,Volum
es&D
isplacem
ents
A
rea InternalVolum
e ExternalD
isplacem
ent
O
utside
D
iam
eter
W
all
Thickness
(N
om
inal) W
all Internal Per1000ft
Per
m
eter
Per1000ft
Per
m
eter
in m
m in m
m in
2
m
m
2
in
2
m
m
2
ft
3
gal bbls liters ft
3
gal bbls liters
1 25.40 0.075 1.905 0.218 140.6 0.567 366.1 3.941 29.48 0.702 0.3661 5.454 40.80 0.971 0.5067
0.080 2.032 0.231 149.2 0.554 357.5 3.848 28.79 0.685 0.3575 5.454 40.80 0.971 0.5067
0.087 2.210 0.250 161.0 0.536 345.7 3.721 27.84 0.663 0.3457 5.454 40.80 0.971 0.5067
0.095 2.413 0.270 174.3 0.515 332.5 3.578 26.77 0.637 0.3325 5.454 40.80 0.971 0.5067
0.102 2.591 0.288 185.6 0.498 321.1 3.456 25.85 0.615 0.3211 5.454 40.80 0.971 0.5067
0.109 2.769 0.305 196.8 0.480 309.9 3.335 24.95 0.594 0.3099 5.454 40.80 0.971 0.5067
0.125 3.175 0.344 221.7 0.442 285.0 3.068 22.95 0.546 0.2850 5.454 40.80 0.971 0.5067
1.25 31.75 0.075 1.905 0.277 178.6 0.950 613.1 6.600 49.37 1.175 0.6131 8.522 63.75 1.518 0.7917
0.087 2.210 0.318 205.1 0.909 586.7 6.315 47.24 1.125 0.5867 8.522 63.75 1.518 0.7917
0.090 2.286 0.328 211.6 0.899 580.1 6.244 46.71 1.112 0.5801 8.522 63.75 1.518 0.7917
0.097 2.464 0.351 226.7 0.876 565.0 6.082 45.50 1.083 0.5650 8.522 63.75 1.518 0.7917
0.104 2.642 0.374 241.6 0.853 550.2 5.922 44.30 1.055 0.5502 8.522 63.75 1.518 0.7917
1.5 38.10 0.095 2.413 0.419 270.5 1.348 869.6 9.360 70.02 1.667 0.8696 12.272 91.80 2.168 1.1401
0.102 2.591 0.448 289.0 1.319 851.1 9.161 68.53 1.632 0.8511 12.272 91.80 2.168 1.1401
0.109 2.769 0.476 307.3 1.291 832.8 8.964 67.06 1.596 0.8328 12.272 91.80 2.168 1.1401
0.125 3.175 0.540 348.4 1.227 791.7 8.522 63.75 1.518 0.7917 12.272 91.80 2.168 1.1401
0.134 3.404 0.575 371.0 1.192 769.1 8.278 61.93 1.474 0.7691 12.272 91.80 2.168 1.1401
9. 4-Friction Pressure Drop
There is graphs shows the relationship between the friction pressure
drop and pump rate for various coiled tubingsizes for the common
oilfield fluids. “FieldData Hand Book” chapterfive
Example:
What is the pressure drop occurduring pumping brine water by rate 0.5
bbl/min inside 15,000 ft of 1.5 in. CT with wall thickness 0.095.
Solution
From “Field Data Hand Book” graph (5-15)
Drop in pressure = 60 psi / 1000 ft
So, drop occurs along 15,000ft = 900 psi
11. 5-Stuck Pipe/ Free Point Determination
By using next equationwe can determine the free point
L = S(E x As)/(F x 12)
Where: L = “Free” length of pipe (ft)
F = Pull force over the weight of the tubing used to cause
the stretch (lb)
AS = Cross-sectional area of the steel in the tubing (in2)
S = stretch caused by the applied over pipe weight pull
force on the tubing
E = Modulusof Elasticity (30 x 106)
12. Example
There is 10,000ft of 1¼” x 0.095”coiled tubingstuck in the well. The
tubing weight 11,720lbs. And has an 80%minimum yield strength of
19,300lbs. With no external pressure difference.
Solution steps
1- Pick up the string weight of 11,270 lbs. And establish a reference point.
2- Pull a 5,000 lbs. Over pull force “F” on the tubingby picking up to
16,720lbs.
3- Measure the stretch in inches “S”, in this case assumed to be 20”.
4- Determine the cross-sectional area of the tubing“AS” by calculatingas
follows:
AS = (1.252 – 1.062)x 0.7854 = 0.345 in2
5- Calculatethe approximate “ free point”from the equation as follows:
L = (20 x 30,000x 0.345)/(5,000x 12)
L = 3450 feet of free point
13. 6- Tubing Elongation Due To Temperature
Difference
Steel expands and contracts0.0000828inches per foot per degree(F) of
temperaturechange. A rise in temperaturewill cause expansion or
lengtheningof the steel and lowering the temperaturewill cause
contraction or shorteningof the steel.
The equation for determining the change in length due to temperature
changes is :
L = D x (0.0000828) x TC
Where: L = Change in length (in.)
D = Depth (ft)
TC = Average temperaturechange
14. Example:
A 15,000 ft string of tubing is being run into the well. The bottomhole
temperatureis 250o F and it is 90o F on surface. Determine the increase
length in the pipe due to temperaturechange.
Solution
1- Calculatethe average well temperatureby following eq.
Average well Temperature = (BHT + Surface Temp.)/2
= (250+90)/2
= 170o F
2- Temperature change = Avg. well temp. – surface temp.
= 170 – 90
= 80o F
3- L =15,000x 0.0000828x 80
= 99.36 in
= 8.28
15. 7- Buoyancy Factors Of Steel In Fluid
When the volume of steel in the tubulardisplaces the same volume of
fluid in the well a buoyancy effect is created due to the difference in
densities between the steel and fluid
The buoyancyfactor can be calculatedby following eq.
B = (DS – DF) / DS
Where: B =Buoyancyfactor
DS= Density of steel
DF= Density of fluid
The weight of the tubing string in the well due to buoyancyeffect
calculated as following:
WW = Wa x B
Where: WW = Weight of tubing in well fluid
Wa = Weight of tubingin air
B = Buoyancy factor
16. Determine the buoyancy factor for tubing suspended in 8.4 lb/gal.fluid.
Also determine the weight of 15,000ft of 1 ¼ “ x 0.083”1.081
lb/ft.coiled tubingin 8.4 lb/gal.fluid.
Solution
1- Density of well fluid (Df) = 8.4 lb/gal.
2- Density of the steel (Ds) = 65.5lb/gal.
3- Buoyancyfactor (B) = ( Ds – Df) / Ds
(B) = (65.5-8.4)/ 65.5
(B) = 0.872
4- String weight in air ( Wa) = 1.081x15,000
( Wa) =16,215lb.
5- String weight in well ( WW) =16,215x 0.872
( WW) =14,140 lb.
Example:
