Petroleum Engineering 405 Drilling Engineering 1 Well Drilling Engineering More Casing Design Dr DO QUANG KHANH 2 Casing Design cont’d Casing Threads Using the Halliburton Cementing Tables Yield[.]
Well Drilling Engineering More Casing Design Dr DO QUANG KHANH Casing Design - cont’d Casing Threads Using the Halliburton Cementing Tables Yield Strength of Casing (in tension) Burst Strength Effect of Axial Tension on Collapse Strength Effect of Pipe Bending Effect of Hydrogen Sulfide Selection of Casing Settling Depths Read: Applied Drilling Engineering, Ch.7 HW # Rounded Threads * per inch ~ Square Threads * Longer * Stronger Integral Joint * Smaller ID, OD * Costs more * Strong < - BURST -> < - TENSION -> Tensile force balance on pipe body Example 7.1: Compute the bodyyield strength for 20in., K-55 casing with a nominal wall thickness of 0.635 in and a nominal weight per foot of 133 lbf/ft Ften = σ y ield * A s Tensile force balance on pipe body K55 Solution: This pipe has a minimum yield strength of 55,000 psi and an ID of: Ften = σ y ield * A s d = 20.00 − 2(0.635) = 18.730in Tensile force balance on pipe body Thus, the cross-sectional area of steel is As = π ( 20 − 18.732 ) = 38.63sq.in and a minimum pipe-body yield is predicted by Eq 7.1 at an axial force of: Ften = σ y ield * A s Ften = 55,000 (38.63) = 2,125,000 lbf Pipe Body Yield Strength π Py = (D − d2 )Yp where Py = pipe body yield strength, lbf Yp = specified minimum yield strength, psi D = outside diameter of pipe, in d = inside diameter of pipe, in 10 Ellipse of Plasticity 14 TENSION 15 16 L Length of arc, L = R∆θrad ∆L = (R + r)∆θ - R∆θ R ∆θ R+r dn ∆L = r ∆θ = ∆θ ∆L dn ∆θ α π dn ∆ε = = = L L 2(12) 100 180 30 * 10 ∆σ = E∆ε = 2,400 ∆σ = 218 α dn π αdn 180 = 218αdn F = 218 α dn A s (7.14a) 17 Figure 7.14 - Incremental stress caused by bending of casing in a directional well The area of steel, As, can be expressed conveniently as the weight per foot of pipe divided by the density of steel For common field units, Eq 7.14a becomes Fab = 64 α d n w (7.14b) where Fab , α, dn , and w have units of lbf, degrees/10 ft, in., and lbf/ft, respectively 18 Example α = deg/ 100 ft dn, = in w = 35 lbf / ft Fab = 64 α dn w ( 7.14b) Fab = 64 * * * 35 = 74,400 lbf Fab = 74,400 lbf 19 Rc = 22 ( Rc = 22 corresponds to a yield strength of about 90,000 psi ) 20