Chuong 7cb more casing design

Petroleum Engineering 405 Drilling Engineering * Well Drilling Engineering More Casing Design Dr DO QUANG KHANH * Casing Design cont’d Casing Threads Using the Halliburton Cementing Tables Yield Stren[.]

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   y ield * A s Pipe Body Yield Strength  2 Py  (D  d )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 Pipe Body Yield Strength Example What is yield strength of body of 7”, 26 #/ft, P-110 casing?  Py  (D  d2 )Yp Internal Yield Pressure for Pipe (Burst)  2Yp t  P 0.875    D  where P internal yield pressure, psi FT FP FP = DLP FT = 2tLYP Yp minimum yield strength, psi DLP = 2tLYP t nominal wall thickness, in  2Yp t  P   D   D O.D of pipe, in 10 Ellipse of Plasticity 12 COLLAPSE TENSION 13 14 L Length of arc, L = Rrad L = (R + r) - R R  dn L r     L dn   dn       L L 212  100 180 30 * 10  E  2,400   218  dn  dn 180 R+r 218dn F  218  dn A s (7.14a) 15 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  d n w ( 14b ) where Fab , , dn , and w have units of lbf, degrees/10 ft, in., and lbf/ft, respective ly 16 Example   deg/ 100 ft dn,  in w  35 lbf / ft Fab  64  dn w ( 14b ) Fab = 64 * * * 35 = 74,400 lbf Fab = 74,400 lbf 17 Rc = 22 ( Rc = 22 corresponds to a yield strength of about 90,000 psi ) 18 19 20