1. Hoisting System 1.3.3-1

2. W FIG 1-1 Simple Pulley System T = W L D = 2W (no friction in sheave) TW 1.3.3-2

3. W = 4 T T = W/4 L D = 6 T = 6 W/4 n = number of lines W = weight (hook load) L D = load on derrick Assuming no friction FIG 1-2 Block and Tackle System 1.3.3-3

4. 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.

5. 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

6. 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.

7. A Rotary Rig Hoisting System 1.3.3-7

8. Projection of Drilling Lines on Rig Floor TOTAL 1.3.3-8

9. 1.3.3-9 Load on Derrick ( considering friction in sheaves) Derrick Load = Hook Load + Fast Line Load + Dead Line Load F d = W + F f + F s  E = overall efficiency, e.g., E = e n = 0.98 n

10. 1.3.4-1 Rig Power zMechanical (Power) zDiesel Electric zElectric