Question

The drill pipes and drill collars with a combined length of 2500 m are held on the hook without rotation and mud flow. The specific gravity of the mud in the annulus is 1.5 and that inside the drill string is 1.4. The material density of the drill pipe and drill collar is 7850 kg/m\(^3\). The specifications are given below.

The overall weight acting on the hook is ____________ kN (rounded off to two decimal places).

Hint:

Always apply buoyancy correction using the correct mud density for the fluid inside and outside the drill string.

Answer 1

Correct Answer: 3870

The buoyed weight is given by:
\[ W_b = W \left( 1 - \frac{\rho_m}{\rho_s} \right) \]
Material density (steel) is:
\[ \rho_s = 7850\ \text{kg/m}^3 \]
Mud densities are:
\[ \rho_{m,a} = 1.5 \times 1000 = 1500\ \text{kg/m}^3 \]
\[ \rho_{m,i} = 1.4 \times 1000 = 1400\ \text{kg/m}^3 \]
1. Drill pipe buoyed weight
\[ W_{dp} = 2000\ \text{m} \times 30\ \text{kg/m} = 60000\ \text{kg} \]
Buoyancy factor using inside mud column:
\[ BF_{dp} = 1 - \frac{1400}{7850} = 0.8217 \]
\[ W_{dp,b} = 60000 \times 0.8217 = 49302\ \text{kg} \]
2. Drill collar buoyed weight
\[ W_{dc} = 500 \times 870 = 435000\ \text{kg} \]
Buoyancy factor using annulus mud:
\[ BF_{dc} = 1 - \frac{1500}{7850} = 0.8096 \]
\[ W_{dc,b} = 435000 \times 0.8096 = 352176\ \text{kg} \]
Total buoyed mass:
\[ M_{total} = 49302 + 352176 = 401478\ \text{kg} \]
Convert to weight:
\[ W = 401478 \times 9.81 = 3.94\times 10^6\ \text{N} \]
\[ W = 3940\ \text{kN} \]
Final Answer: 3940.00 kN