PETE 411
Well Drilling
Lesson 14
Jet Bit Nozzle Size Selection
1
14. Jet Bit Nozzle Size Selection
 Nozzle Size Selection
for Optimum Bit Hydraulics:
 Max. Nozzle Velocity
 Max. Bit Hydraulic Horsepower
 Max. Jet Impact Force
 Graphical Analysis
 Surge Pressure due to Pipe Movement
2
Read:
Applied Drilling Engineering, to p.162
HW #7:
On the Web - due 10-09-02
Quiz A
Thursday, Oct. 10, 7 - 9 p.m. Rm. 101
Closed Book
1 Equation sheet allowed, 8 1/2”x 11” (both sides)
{ Quiz A_2001 is on the web }
3
Jet Bit Nozzle Size Selection
 Proper bottom-hole cleaning
• will eliminate excessive regrinding of drilled
solids, and
• will result in improved penetration rates
 Bottom-hole cleaning efficiency
• is achieved through proper selection of bit
nozzle sizes
4
Jet Bit Nozzle Size Selection
- Optimization Through nozzle size selection,
optimization may be based on
maximizing one of the following:
 Bit Nozzle Velocity
 Bit Hydraulic Horsepower
 Jet impact force
• There is no general agreement on which of
these three parameters should be maximized.
5
Maximum Nozzle Velocity
Nozzle velocity may be maximized consistent with
the following two constraints:
 1. The annular fluid velocity needs to be high
enough to lift the drill cuttings out of the hole.
- This requirement sets the minimum
fluid circulation rate.
 2. The surface pump pressure must stay within the
maximum allowable pressure rating of the
pump and the surface equipment.
6
Maximum Nozzle Velocity
From Eq. (4.31)
i.e.
v n  Cd
Pb
8.074 *10  4 
v n  Pb
so the bit pressure drop should be maximized in
order to obtain the maximum nozzle velocity
7
Maximum Nozzle Velocity
This (maximization) will be achieved when
the surface pressure is maximized and the
frictional pressure loss everywhere is
minimized, i.e., when the flow rate is
minimized.
 v n is maximized when 1& 2 above are satisfied,
at the minimum circulatio n rate
and the maximum allowable surface pressure.
8
Maximum Bit Hydraulic Horsepower
The hydraulic horsepower at the bit is
maximized when (p bit q) is maximized.
ppump  pd  pbit
p bit  p pump  p d
where p d may be called the parasitic pressure
loss in the system (friction).
9
Maximum Bit Hydraulic Horsepower
The parasitic pressure loss in the system,
p d  ps  p dp  p dc  p dca  p dpa  cq
1.75
if the flow is turbulent .
In general,
p d  cq
m
where 0  m  2
10
Maximum Bit Hydraulic Horsepower
p bit  p pump  p d
 PHbit
p d  cq
pbit q p pumpq  cq


1714
1714
dPHbit

 0 when
dq
m
m 1
p pump  c(m  1)q  0
m
11
Maximum Bit Hydraulic Horsepower
p pump  c(m  1)q  0
m
i.e., when p pump  ( m  1) pd
 1 
i.e., when p d  
p pump
 m 1
 PHbit is maximum when
pd
 1 

 p pump
 m 1
12
Maximum Bit Hydraulic Horsepower
- Examples In turbulent flow, m = 1.75
1
p d 
pp
m 1
1


 p d  
p pump *100%
 1.75  1 
 36% of p pump
 p bit  64% of p pump
13
Maximum Bit Hydraulic Horsepower
Examples - cont’d
In laminar flow, for Newtonian fluids,
m=1
 1 
 p d  
p pump *100%
 11
 50% of p pump
 p b  50% of p pump
14
Maximum Bit Hydraulic Horsepower
 In general, the hydraulic horsepower is not
optimized at all times
 It is usually more convenient to select a
pump liner size that will be suitable for
the entire well
 Note that at no time should the flow rate be
allowed to drop below the minimum
required for proper cuttings removal
15
Maximum Jet Impact Force
The jet impact force is given by Eq. 4.37:
F j  0.01823 cd q  pbit
 0.01823 c d q  (p pump  pd )
16
Maximum Jet Impact Force
Fj  0.01823 cd q  (p pump  pd )
But parasitic pressure drop,
pd  cq
 F j  0.01823 cd
m
 p p q   cd q
2
m2
17
Maximum Jet Impact Force
Upon differentiating, setting the first derivative
to zero, and solving the resulting quadratic
equation, it may be seen that the impact
force is maximized when,
2
p d 
p p
m2
18
Maximum Jet
Impact Force
- Examples Thus, if m  1.75,
2
p d 
p p
m2
p d  53% of p p
and p b  47% of p p
Also, if m  1.00
p d  67% of p p
and p b  33% of p p
19
Nozzle Size Selection
- Graphical Approach -
20
21
22
1. Show opt. hydraulic path
2. Plot pd vs q
3. From Plot, determine
optimum q and pd
p bit  p pump  p d
4. Calculate
5. Calculate
2
5
8.311*10  qopt
Total Nozzle Area: ( At ) opt 
2
Cd (pb ) opt
(TFA)
6. Calculate Nozzle Diameter
With 3 nozzles:
4A tot
dN 
3
23
Example 4.31
Determine the proper pump operating
conditions and bit nozzle sizes for max.
jet impact force for the next bit run.
Current nozzle sizes: 3 EA 12/32”
Mud Density = 9.6 lbm.gal
At 485 gal/min, Ppump = 2,800 psi
At 247 gal/min, Ppump = 900 psi
24
Example 4.31 - given data:
Max pump HP (Mech.) = 1,250 hp
Pump Efficiency
= 0.91
Max pump pressure
= 3,000 psig
Minimum flow rate
to lift cuttings
= 225 gal/min
25
Example 4.31 - 1(a), 485 gpm
Calculate pressure drop through bit nozzles:
Eq.(4.34) : pb 
pb 
8.311*10 5  q 2
2
cd At
8.311(10 -5 )(9.6)( 485)2
2

  12  
2
(0.95) 3 
 
 4  32  
2
2
 1,894 psi
 parasitic pressure loss  2,800 - 1,894  906 psi
26
Example 4.31 - 1(b), 247 gpm
pb 
5
8.311(10 )(9.6)( 247)
   12 
(0.95) 3 

 4  32 
2
2



2
2
 491 psi
 parasitic pressure loss  900 - 491  409 psi
(q1, p1) = (485, 906)
(q2, p2) = (247, 409)
Plot these two
points in Fig. 4.36
27
28
Example 4.31 - cont’d
3
2
2. For optimum hydraulics:
1
(a ) Interval 1,
1,714 PHp E 1,714(1,250)(0.91)
q max 

 650 gal/min
Pmax
3,000
(b) Interval 2,
 2 
 2 
p d  
 Pmax  
(3,000)
 m2
 1.2  2 
 1,875 psi
(c) Interval 3,
q min  225 gal/min
29
Example 4.31
3. From graph, optimum point is at
gal
q  650
, p d  1,300 psi  pb  1,700 psi
min
8.311*10 qopt
5
 ( At ) opt 
A opt  0.47 in
Cd (pb ) opt
2
2

2
8.311*10-5 * 9.6 * (650) 2

(0.95) 2 * (1,700)
d N opt  14
32
nds
in
30
gal
q  650
, p d  1,300 psi  pb  1,700 psi
min
31
Example 4.32
Well Planning
It is desired to estimate the proper pump
operating conditions and bit nozzle sizes for
maximum bit horsepower at 1,000-ft
increments for an interval of the well
between surface casing at 4,000 ft and
intermediate casing at 9,000 ft. The well
plan calls for the following conditions:
32
Example 4.32
Pump: 3,423 psi maximum surface pressure
1,600 hp maximum input
0.85 pump efficiency
Drillstring: 4.5-in., 16.6-lbm/ft drillpipe
(3.826-in. I.D.)
600 ft of 7.5-in.-O.D. x 2.75-in.I.D. drill collars
33
Example 4.32
Surface Equipment: Equivalent to 340
ft. of drillpipe
Hole Size: 9.857 in. washed out to 10.05 in.
10.05-in.-I.D. casing
Minimum Annular Velocity: 120 ft/min
34
Mud Program
Depth
(ft)
Mud
Density
(lbm/gal)
Plastic
Yield
Viscosity
Point
(cp)
(lbf/100 sq ft)
5,000
9.5
15
5
6,000
9.5
15
5
7,000
9.5
15
5
8,000
12.0
25
9
9,000
13.0
30
12
35
Solution
The path of optimum hydraulics is as
follows:
Interval 1
q max 
1,714 PHp E
p max
1,714(1,600)(0.85)

3,423
 681 gal/min.
36
Solution
Interval 2
Since measured pump pressure data are not
available and a simplified solution technique
is desired, a theoretical m value of 1.75 is
used. For maximum bit horsepower,
 1 
 1 
pd  
 pmax  
 3,423
 m 1
 1.75  1 
 1,245 psia
37
Solution
Interval 3
For a minimum annular velocity of
120 ft/min opposite the drillpipe,
qmin  2.448 10.05  4.5
2
2

 120 


 60 
 395 gal/min
38
Table
The frictional pressure loss in other
sections is computed following a
procedure similar to that outlined above for
the sections of drillpipe. The entire
procedure then can be repeated to
determine the total parasitic losses at
depths of 6,000, 7,000, 8,000 and 9,000 ft.
The results of these computations are
summarized in the following table:
39
Table
Depth ps p dp p dc p dca p dpa p d
5,000
6,000
7,000
8,000
9,000
38
38
38
51
57
490
601
713
1,116
1,407
320
320
320
433
482
20
20
20
28
27*
20
25
29
75*
111*
888
1,004
1,120
1,703
2,084
* Laminar flow pattern indicated by
Hedstrom number criteria.
40
Table
The proper pump operating conditions
and nozzle areas, are as follows:
(l)Depth (2)Flow Rate (3) p d (4) p b
(ft )
(gal/min)
5,000
6,000
7,000
8,000
9,000
600
570
533
420
395
(psi)
1,245
1,245
1,245
1,245
1,370
(5)A t
(psi) (sq in.)
2,178
2,178
2,178
2,178
2,053
0.380
0.361
0.338
0.299
0.302
41
Table
The first three columns were read directly
from Fig. 4.37. (depth, flow rate and pd)
Col. 4 (pb) was obtained by subtracting p d
shown in Col.3 from the maximum pump
pressure of 3,423 psi.
Col.5 (Atot) was obtained using Eq. 4.85
42
43
Surge Pressure due to Pipe Movement
When a string of pipe is
being lowered into the
wellbore, drilling fluid is
being displaced and forced
out of the wellbore.
The pressure required to
force the displaced fluid out
of the wellbore is called the
surge pressure.
44
Surge Pressure due to Pipe Movement
An excessively high surge pressure can
result in breakdown of a formation.
When pipe is being withdrawn a similar
reduction is pressure is experienced. This
is called a swab pressure, and may be
high enough to suck fluids into the wellbore,
resulting in a kick.
For fixed
v pipe ,
Psurge  Pswab
45
Figure 4.40B
- Velocity profile for laminar flow pattern when closed
pipe
is being run into hole
46