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by DRILLINGFORMULAS.COM on SEPTEMBER 22, 2013
The drilling line is
reeved over a set of crow block and down to another set of sheaves known as
travelling block. The hook connected to travelling block is used to suspend the
drilling load. One end of drilling line is wound onto the drawworks and this
line is called “Fast Line”. Other end of the drilling line is tied into an
anchor point on the rig floor and this line is named as “Dead Line”. The
drilling line is reeved around the blocks several times in order to meet
required load.
Static Derrick Load
W = N x Tf
Tf = W/N
Where,
N – Number of drilling lines in
a travelling block
Tf = Td (Tf and Td are the same
value because the same tension in the drilling line)
Static derrick load equates to
summation of fast line tension, dead line tension and hook load. We can describe
into the following equation.
FD = Tf + W + Td
Where,
FD –
Static derrick load
Tf – Fast
line tension
Td – dead
line tension
W –
hookload
Note: Neglect a small effect of
small angle of the fast line and the dead line.
FD = W/N +
W + W/N
FD = (N+2)
x W ÷ N
Dynamic
Derrick Load
Under dynamic condition,
friction in sheave bearings and block lines make the fast tension higher than
the dead line tension. It means that the fast line tension will increase under
a dynamic condition; however, the dead line tension will remain the same
because it is still in static condition.
The fast line tension under the
dynamic environments can be described as the equation below;
Tf = W ÷ (
E x N )
Where,
Tf – The
fast line tension.
W – Block
weight
N – Number
of lines
Derrick load under dynamic
condition is also equal to summation of hook load, dynamic fast line tension
and dead line tension as described in the equation below
FD = Tf +
W + Td
Exercise
for Derrick Load Calculation
Buoyed weight of the drill
string is 260,000 lb which will be pulled out of hole. Weight of travelling
block and hook is 40,000 lb. The rig has 10 lines strung in crown block and
travelling block.
Solution:
Efficiency of 10 lines = 0.811
Total hook load = 260,000 +
40,000 = 300,000 lb
The
fast line tension:
Tt = 300,000 ÷ (10 x 0.811)
Tt = 36,991 lb
The
dead line tension:
Td = 300,000 ÷ 10
Td = 30,000 lb
Derrick load under the dynamic
condition:
FD = Tf + W + Td
FD = 36,991 + 30,000 + 300,000
FD = 366,991 lb
Derrick
load in dynamic condition equates to 366,991 lb
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