Question 1 of 5: Production Casing Design

Question 1: Production Casing Design (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $6\,5/8$ in. production casing to $D=15{,}000$ ft in an $8\,1/2$ in. open hole; setting mud weight $16$ ppg; pore-pressure gradient $15.5$ ppg; fracture gradient $17.6$ ppg; normal formation pressure gradient $0.465$ psi/ft; design factors burst $1.1$, collapse $1.0$, tensile $1.6$; three candidate casings (table above); gas is to be produced.

Find. A casing program (grade, weight and length of each section) satisfying burst, collapse and tensile design.

Approach. Collapse governs where the casing sees the largest net external pressure (an evacuated string against the full mud column, worst at the shoe); burst governs where the internal pressure from a produced-gas kick exceeds the external backup the most (worst at surface); find the depth ranges each candidate casing satisfies for both, then check tension on the resulting tapered string.

1. Collapse load (worst case: evacuated casing, full mud column outside). $P_{collapse}(D)=0.052\,MW\,D=0.052(16)D=0.832D$ psi, maximum at the shoe: $P_{collapse}(15{,}000)=12{,}480$ psi. With $DF_{collapse}=1.0$, required rating $\ge12{,}480$ psi: N-80 24# (5,760) and N-80 32# (10,320) both fail; only $\boxed{\text{P-110, 32 lb/ft (13,220 psi) survives at the shoe}}$.
2. Burst load (produced-gas kick to surface, worst case: surface). Formation pressure at $15{,}000$ ft: $P_f=0.052(15.5)(15{,}000)=12{,}090$ psi. Assuming a standard methane gradient of $0.1$ psi/ft for the internal gas column (no PVT data given), the shut-in surface pressure is $SITP=P_f-0.1(15{,}000)=10{,}590$ psi, and internal pressure at depth $D$ is $P_{int}(D)=SITP+0.1D$. Using the given normal formation-water gradient as the worst-case external backup, $P_{ext}(D)=0.465D$. The burst differential $\Delta P_{burst}(D)=P_{int}(D)-P_{ext}(D)=SITP-0.365D$ is largest at $D=0$: $\Delta P_{burst}(0)=10{,}590$ psi. With $DF_{burst}=1.1$, required rating $\ge11{,}649$ psi at surface: N-80 32# (10,040) fails; $\boxed{\text{P-110, 32 lb/ft (13,800 psi) is required at surface}}$.
3. Where can the cheaper N-80, 32 lb/ft be used? N-80 32# satisfies burst once $\Delta P_{burst}(D)\le10{,}040/1.1=9{,}127$ psi, i.e. for $D\ge4{,}008$ ft, and satisfies collapse while $0.832D\le10{,}320$ psi, i.e. for $D\le12{,}404$ ft. So N-80, 32 lb/ft is adequate over the middle interval $\boxed{4{,}008\ \text{ft} \le D \le 12{,}404\ \text{ft}}$ (length $8{,}396$ ft), bracketed top and bottom by P-110, 32 lb/ft (lengths $4{,}008$ ft and $2{,}596$ ft). N-80, 24 lb/ft never satisfies both criteria at any depth in this well — it is not usable for this string.
4. Tensile check (buoyed weight, $DF_{tensile}=1.6$). Buoyancy factor $BF=1-MW/65.5=1-16/65.5=0.7557$. The critical joint is the top of the N-80 middle section, which carries the buoyed weight of the N-80 and P-110 (shoe) sections below it: $W_{below}=(8{,}396+2{,}596)(32)=351{,}744$ lbf air weight, $\times BF=265{,}835$ lbf buoyed; required $=1.6(265{,}835)=425{,}336$ lbf, well below N-80 32#'s joint strength of $814{,}000$ lbf. At the very top of the string, the full string's buoyed weight is $15{,}000(32)(0.7557)=362{,}748$ lbf, requiring $1.6(362{,}748)=580{,}397$ lbf against the P-110 joint strength of $1{,}040{,}000$ lbf. $\boxed{\text{Tensile is satisfied everywhere with ample margin; it does not govern this design.}}$