Transcript PETROLEUM ENGINEERING 405

Hoisting System
1.3.3-1
1.3.3-2
T
W
W
• FIG 1-1 Simple Pulley System
T=W
LD = 2W
(no friction in sheave)
W
n = number
of lines
1.3.3-3
W = weight
(hook load)
LD = load
on derrick
• FIG 1-2 Block and Tackle System
Assuming no friction
W=4T
T = W/4
LD = 6 T = 6 W/4
 n 2 
LD  
W
 n 
1.3.3-4 Example 1.1
(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.
1.3.3-5 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
1.3.3-6 Solution, cont.
Example 1.1 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.
1.3.3-7
A Rotary Rig
Hoisting System
1.3.3-8
Projection of
Drilling Lines
on Rig Floor
TOTAL
1.3.3-9 Load on Derrick
(considering friction in sheaves)
Derrick Load = Hook Load
+ Fast Line Load
 Line Load
+ Dead
Fd = W + Ff + Fs
W
W
 1  E  En 
Fd  W 

=
W


En
n
En
E = overall efficiency, e.g., E = en = 0.98n
1.3.4-1 Rig Power
Mechanical (Power)
Diesel Electric
Electric