drilling engineering year 2 tutorial questions for exam
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Tutorial Lecture 1 5
Kanad Kulkarni 05/11/2013

Hydrostatic Pressure HP = g x f 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 f (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:
ov = 0.052 x b x D where ov = overburden pressure (psi) b = formation bulk density (ppg) D = true vertical depth (ft)

And similarly as a gradient (EMW) in ppg: ov= 0.433xb/0.052 Where, ovg = overburden gradient, ppg b = 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:
ovg= 0.433[(1 )ma + (xf)]
where ovg= overburden gradient, psi/ft = porosity expressed as a fraction f= formation fluid density, gm/cc ma= matrix density, gm/cc

Power System Performance Characteris3cs The overall power eciency determines the rate of fuel consump7on, wf, at a given speed.
The hea7ng value of a fuel for internal combus7on is H
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 The overall eciency of powergenera7ng systems may be dened as the energy output per energy input:
Efficiency = (Power Out / Power in)

Example 1. A diesel engine gives an output torque of 1,740 ftlbf 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, , is given by
= 2 (1,200) = 7,539.8 rad/min.
The power output can be computed using Eq.1 ( )
hp5.397/hplbf/minft 33,000
lbf/minft (1,740) 7,539.8T P ===

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/hpft 33,000
lbf/Btuft 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% ovg = 0.433 x [(1 0.05)x2.65 + (0.05 x 1.03)] = 1.11 psi/ft For = 20% ovg =1.01 psi/ft For = 35% ovg = 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 10003000 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 5/8inch 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 = Epi = 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 90ft 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 doubleacting duplex pump having:
6.5inch liners (dL) 2.5 inch rods (dr) 18inch 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
=
=
=

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 * /4 dL2 LS EV / (231 * 42)

Example:
Pump Factor for Triplex Pump
( )( )
=
=
3
32
VS 2
L
inbbl
422311
strokein0.90126
43
ELd 43
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 Capacity is used to refer to the crosssec7onal area of the pipe or annulus expressed in units of contained volume per unit length
Displacement is [email protected] used to refer to the crosssec7onal 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 = /4 d2 = /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 = /4 (od2  id2 ) = /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 nearvertical 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 leakoff 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 shutin 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 F/ft Max. mud weight for next hole =14.5 ppg Max formation pressure at next hole= 14 ppg Assume next hole 8 " 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 v/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