Tutorial Lecture 1- 5

Kanad Kulkarni 05/11/2013

Hydrostatic Pressure HP = g x ρf x D Pressure Gradient HG = HP / D … (psi/ft)

Calculate the hydrostatic pressure for the following wells: a.  mud weight = 9 ppg, hole depth = 10100 ft MD

(measured depth), 9900 ft TVD (truevertical depth)

Solution: HP (psi) = 0.052 x ρf (ppg) x D (ft) = 0.052 x 9 x 9900 = 4632 psi

b. mud gradient = 0.468 psi / ft, hole depth = 10100 ft MD (measured depth), 9900 ft TVD (true vertical depth) Solution = 0.468 (psi /ft) x 9900(ft) = 4633 psi

The overburden pressure can therefore be expressed as the hydrostatic pressure exerted by all materials overlying the depth of interest:

σov = 0.052 x ρb x D where σov = overburden pressure (psi) ρb = formation bulk density (ppg) D = true vertical depth (ft)

And similarly as a gradient (EMW) in ppg: σov= 0.433xρb/0.052 Where, σovg = overburden gradient, ppg ρb = formation bulk density (gm/cc) (the factor 0.433 converts bulk density from gm/cc to psi/ft)

•  A useful equation for calculating the overburden gradient under field conditions of varying lithological and pore fluid density is given by:

σovg= 0.433[(1 – φ)ρma + (φxρf)]

where σovg= overburden gradient, psi/ft φ= porosity expressed as a fraction ρf= formation fluid density, gm/cc ρma= matrix density, gm/cc

Power  System  Performance  Characteris3cs  Ø The   overall   power   efficiency   determines   the  rate  of  fuel  consump7on,  wf,  at  a  given  speed.  

Ø The  hea7ng  value  of  a  fuel  for  internal  combus7on  is  H  

Ø The  input  power  is  expressed  in  terms  of  wf  and  H:    

TABLE 1 - HEATING VALUE OF VARIOUS FUELS

Fuel Type

Density (lbm/gal)

Heating Value (Btu/lbm)

diesel gasoline butane methane

7.2 6.6 4.7 ---

19,000 20,000 21,000 24,000

Engine power output

P = F . V

Power = Force * Velocity

Power = Ang.Vel. * Torque

Power  System  Performance  Characteris3cs  Ø The   overall   efficiency   of   power-­‐genera7ng  systems   may   be   defined   as   the   energy  output  per  energy  input:  

Efficiency = (Power Out / Power in)

Example 1. A diesel engine gives an output torque of 1,740 ft-lbf at an engine speed of 1,200 rpm. If the fuel consumption rate was 31.5 gal/hr, what is the output power and overall efficiency of the engine?

Solution: The angular velocity, ω , is given by

ω = 2π (1,200) = 7,539.8 rad/min.

The power output can be computed using Eq.1 ( )

hp5.397/hplbf/min-ft 33,000

lbf/min-ft (1,740) 7,539.8T P ===ω

Since the fuel type is diesel, the density is 7.2 lbm/gal and the heating value H is 19,000 Btu/lbm (Table 1). Thus, the fuel consumption rate w f is:

wf = 3.78 lbm/min. The total heat energy consumed by the engine is given by Eq. 2:

⎟⎠

⎞⎜⎝

⎛=minutes 60hour 1 lbm/gal) (7.2gal/hr 31.5 wf

Qi = w f H

Thus, the overall efficiency of the engine at 1,200 RPM given by Eq. 3 is

( ) ( )lbf/min/hp-ft 33,000

lbf/Btu-ft 779lbm19,000Btu/lbm/min 3.78=iQ

Efficiency = (Power Out / Power in)

23.4%or 0.2341695.4

397.5===

it Q

PE

Calculate the overburden gradient for the following: Formation type: sandstone, density = 2.65 gm/cc Formation water: 1.03 gm/cc For porosities 5%, 20% and 35%. Solutions •  For Sandstone For = 5% σovg = 0.433 x [(1 – 0.05)x2.65 + (0.05 x 1.03)] = 1.11 psi/ft For = 20% σovg =1.01 psi/ft For = 35% σovg = 0.90 psi/ft-

Water Depth= 500 ft RKB/MSL= 65 ft Specific gravity of sea water= 1.03 gm/cc Rock density= 1.9 gm/cc from seabed to 1000ft, and 2.1gm/cc from 1000-3000 ft Calculate the overburden gradient of the formations: At seabed, 200 ft, 500 ft, 1000 ft and at 3000 ft below seabed.

Example 1 (no friction)

The total weight of 9,000 ft of 9 5/8-inch casing for a deep well is determined to be 400,000 lbs. Since this will be the heaviest casing string run, the maximum mast load must be calculated. Assuming that 10 lines run between the crown and the traveling blocks and neglecting buoyancy effects, calculate the maximum load.

Solution:

The tension, T, will be distributed equally between the 10 lines. Therefore,

T = 400,000/10 = 40,000 lbf The tension in the fast line and dead line will also be 40,000 lbf, so the total load is

40,000 X 12 = 480,000 lbf

Solution, cont.

Example 1 demonstrates two additional points.

1. The marginal decrease in mast load decreases with additional lines.

2. The total mast load is always greater than the load being lifted.

Example 2 A rig must hoist a load of 300,000 lbf. The drawworks can provide an input power to the block and tackle system as high as 500 hp. Eight lines are strung between the crown block and traveling block. Calculate 1. The static tension in the fast line

when upward motion is impending, 2. the maximum hook horsepower

available,

Solution 1. The power efficiency for n = 8 is given as 0.841 in Table 1.2. The tension in the fast line is given by Eq. 1.7.

Tension in the Fast Line,

lbnE

WF 590,448*841.0

000,300===

( or: 0.988 = 0.851 )

Solution

2. The maximum hook horsepower available is

Ph = E•pi = 0.841(500) = 420.5 hp. •••

Solution

3. The maximum hoisting speed is given by

vPWbh=

=

⎛

⎝⎜

⎞

⎠⎟

hp

ft - lbf / minhp

300,000 lbf = 46.3 ft / min

420 533 000

.,

Solution to 3., cont.

To pull a 90-ft stand would require

t = =90

1 9 ft

46.3 ft / min . min.

Solution 4. The actual derrick load is given by Eq.1.8b:

FE EnEn

Wd =+ +⎛

⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

1

=1+0.841+0.841(8)

0.841(8)(300,000)

= 382,090 lbf.

Solution 5. The maximum equivalent load is given by Eq.1.9:

lbfF

WnnF

de

de

000,450

000,300*8484

=

⎟⎠

⎞⎜⎝

⎛ +=⎟

⎠

⎞⎜⎝

⎛ +=

nW

4WFDLL +=

Wn44nFDLL ⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

Wn4nFde ⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

Solution

6. The derrick efficiency factor is:

000,450090,382

FFEde

dd ==

84.9% or 849.0E d =

29

30

Example 3

Compute the pump factor in units of barrels per stroke for a double-acting duplex pump having:

6.5-inch liners (dL) 2.5 inch rods (dr) 18-inch strokes (LS) and a volumetric efficiency of 90%. (EV)

31

Solution: The pump factor for a duplex pump can be determined using Equation 1.10

( )

( )( ) ( ) ( )[ ]stroke

ddELF rLVSp

/in 991,1

5.25.629.0182

22

3

22

22

=

−=

−=

π

π

32

There are 231 in.3 in a U. S. gallon and 42 U.S. gallons in a U.S. barrel. Thus converting to the desired field units yields:

1,991 in.3/stroke * gal/231 in.3 * bbl/42 gal.

= 0.2052 bbl/stroke.

Thus: Pump Factor = 0.2052 bbl/stroke

33 Pump Factor = 3 * π/4 dL2 LS EV / (231 * 42)

Example:

Pump Factor for Triplex Pump

( )( ) ⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛∗⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛=

∗⎟⎠

⎞⎜⎝

⎛∗=

3

32

VS 2

L

inbbl

422311

strokein0.90126

4π3

ELd 4π3

bbl/stroke 0.09442FactorPump =

Example: Pump Rate

= Pump Factor * Strokes/min

= 0.09442

= 7.554 bbl/min = 317.3 gal/min

min

stks80stkbbl

∗

Pump Rate = 317 gal/min

Drilling Pipe size and grade table

37

Cutaway View and Dimensions for

Example Tool Joint

PIN BOX TJ Shoulder

Rotary  System  Ø Capacity   is   used   to   refer   to   the   cross-­‐sec7onal   area   of   the   pipe   or   annulus  expressed   in   units   of   contained   volume   per  unit  length  

Ø Displacement   is   o@en   used   to   refer   to   the  cross-­‐sec7onal   area   of   steel   in   the   pipe  expressed  in  units  of  volume  per  unit  length  

39 Capacity = Area * Length

What is the capacity of 10,000 ft of 5” OD, 19.50 lb/ft drillpipe?

Capacity = Area * Length

Area = π/4 d2 = π/4 * 4.2762 = 14.36 in2

Length = 10,000 ft = 120,000 in

Capacity = 14.36 *120,000 in3 /(231*42 in3 /bbl)

Capacity = 177.6 bbls

What is the displacement of 10,000 ft of 5” OD, 19.50 lb/ft drillpipe?

Capacity = Area * Length

Area = π /4 (od2 - id2 ) = π /4 * (52 - 4.2762)

= 5.275 in2

Length = 10,000 ft = 120,000 in

Displ. = 5.276 * 120,000 in3 /(231 * 42 in3 /bbl)

Displacement = ? bbls

Pressure at the shoe after drillers rotation can be calculated by Px = Pf - Pg - (TD - H - CSD) xpm Where Pf= Formation Pressure next to TD Pg= Pressure in gas Bubble= HXG H = height of gas bubble at casing shoe, ft G = gradient of gas = 0.05 to 0.15 psi/ft TD = next hole total depth, ft CSD = casing setting depth, ft pm = maximum mud weight for next hole section, ppg

Rearranging in terms of H & replacing Px with FG (fracture Gradient) In vertical and near-vertical holes the FBG is invariably greater than the FG. In highly inclined holes the FBG is usually smaller than the FG. For kick tolerance calculations, it is recommended to reduce the value recorded during leak-off tests in vertical wells by 100 psi and to use the resulting value as an approximate value of FG.

The volume of influx at the casing shoe is V1= H x Ca bbl

where Ca = capacity between pipe and hole, bbl/ft At bottom hole conditions the volume of influx (V2) is given by:

P2 V2 = P1 V1 (The effects of T and Z are ignored for the moment) V2=P1V1/P2 where P1 =fracture pressure at shoe, psi P2 =Pf, psi The value of V2 is the circulation kick tolerance in bbls.

The maximum allowable drillpipe shut-in pressure (DPSIP) is given by:

DPSIP = (FG - m) x CSD x 0.052 And in terms of additional mud weight, Kick Tolerance= (FG- pm)

Calculate the kick tolerance for the following well: 9 5/8" casing =14,500 ft Next TD = 17000 ft FG at 9 5/8" shoe = 16 ppg Temperature gradient = 0.02 FÆ/ft Max. mud weight for next hole =14.5 ppg Max formation pressure at next hole= 14 ppg Assume next hole 8 ó" and there is 5" drillpipe from surface to TD

In case of Exploration well where the fracture gradient is calculated constantly it is advisable to change the mud weight accordingly and also the Kick Tolerance should be calculated repeatedly. Revised calculateions for the above problem

Given that the formations pressure at 5000 ft is 2400 psi and the overburden stress is 1 psi/ft(determined from bulk density logs), estimate the formation fracture gradient at 5000 ft.

(1) The required hydrostatic pressure of mud is taken as equal to pore pressure + 200 psi, where 200 psi is the magnitude of overbalance. Any reasonable value of overbalance may be used depending on company policy. (2) Calculate the pore pressure and mud pressure gradients by simply dividing pore pressure and mud pressure by depth to obtain the gradient in psi/ft.

Taking σv/D = 1psi/ft, and using Table 2.2, the fracture gradient at 5000 ft, for example, is calculated as follows: The same equation can be used to calculate FG at other depths