Casing Design Example

First step is to find the casing that meets the ID required of this string. 7” in this case.

Then calculate the bottom hole hydrostatic pressure. 10,000 feet of 9.9 mud with a safety factor of 1.2 is 6178 psi

Find the casing that meets the collapse criteria. N80 26# has a collapse of 5410 psi – too low N80 29# collapse is 7020 psi, this is the lowest grade that meets the spec’s.

Using the collapse of the next lowest pipe calculate the length of the bottom grade. The collapse pressure of this pipe has to be adjusted for the weight of the lower pipe. Starting with the 5410 psi of the 26# pipe a length of 1475’ with an adjusted PC of 5267 psi with a setting depth of 8525’ from the surface. The section of 29# pipe weights 42747#.

Next check the joint strength of the two pipes verifying that this meets the criteria. With a SF of 1.9 the joint strength of the 29# is 314,000# and the 26# JS is 273,000#, so both meet the specs at the change over.

Repeat the process using the next grade of pipe. The next pipe is the N80 23# with a PC of 3830 psi. A length of 2595’ is calculated with a corrected PC of 3664 psi, and a weight of the 26# pipe is 67466#. Depth of the top of 26# is 5930 ft. For a total string weight of 110213#. Check the JS at this change over, the JS of the 23# is 232,000# so it holds.

Repeat the process using the next grade of pipe. The next pipe is the K55 29# with a PC of 4320 psi. This is higher than the PC of the previous pipe.

Joint Strength is now the controlling factor. Take the JS of the top pipe and subtract the weight of the string below this point and the this the length of the section. Here for the N80 23# is 5295 ft. This does not reach the surface, the next section is N80 26# with a JS 273,000 #. Since this last section is less than 500 ft, make 500 ft. The casing design is

7” N80 26# ltc 500 ft 7” N80 23# ltc 5430 ft 7” N80 26 # ltc 2595 ft 7” N80 29 # ltc 1475 ft