Question 2 of 5: Mud Pump Liner Selection for Cuttings Transport

Question 2: Mud Pump Liner Selection for Cuttings Transport (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. Well at $10{,}000$ ft; $10\,3/4$ in. casing (ID $9.5$ in.) set at $4{,}000$ ft; $8\,1/2$ in. open hole below the shoe; $4$ in. OD drill pipe from surface to $9{,}000$ ft, $6$ in. OD $\times 1{,}000$ ft drill collars from $9{,}000$–$10{,}000$ ft; minimum annular velocity for cuttings transport $90$ ft/min; single-acting triplex pump, $80\%$ volumetric efficiency, $20$ in. stroke, $60$ SPM.

Find. The minimum liner diameter that delivers the required flow rate.

Approach. A single flow rate sets the annular velocity in every section simultaneously, and velocity $=Q/A$ is lowest wherever the annular area is largest — so the LARGEST annulus, not the open hole, sets the minimum required pump output. Compare the three annulus geometries, size the pump output from the governing (largest) one, then back-solve the single-acting triplex displacement equation for the liner diameter.

1. Annular areas of the three sections. $A=\dfrac{\pi}{4}\left(D_{out}^2-D_{in}^2\right)$: cased hole–drill pipe, $A_1=\dfrac{\pi}{4}(9.5^2-4^2)=58.32\ \text{in}^2$; open hole–drill pipe, $A_2=\dfrac{\pi}{4}(8.5^2-4^2)=44.18\ \text{in}^2$; open hole–drill collar, $A_3=\dfrac{\pi}{4}(8.5^2-6^2)=28.47\ \text{in}^2$. The cased-hole section is the largest, so $\boxed{A_{max}=58.32\ \text{in}^2\ \text{(cased hole)}}$ governs the required flow rate.
2. Minimum flow rate. $Q_{min}=V_{min}\,A_{max}=90\ \text{ft/min}\times\left(58.32/144\right)\ \text{ft}^2=36.45\ \text{ft}^3/\text{min}$. Converting at $7.48052$ gal/ft$^3$: $\boxed{Q_{min}=272.6\ \text{gpm}}$. (Both other sections easily clear $90$ ft/min at this rate, since they have smaller area and therefore higher velocity for the same $Q$.)
3. Pump displacement equation (single-acting triplex). Each of the three cylinders discharges once per revolution (single-acting — only the forward stroke pumps, so the rod diameter does not enter the calculation, unlike a double-acting duplex pump): $Q_{pump}=\dfrac{3\left(\pi/4\right)D_{liner}^2\,L_{stroke}\,SPM\,E_v}{231}$. With $L_{stroke}=20$ in., $SPM=60$, $E_v=0.80$: $Q_{pump}=\dfrac{3(\pi/4)(20)(60)(0.80)}{231}D_{liner}^2=9.792\,D_{liner}^2$ gpm (D in inches).
4. Solve for the minimum liner diameter. $D_{liner}=\sqrt{Q_{min}/9.792}=\sqrt{272.6/9.792}$, so $\boxed{D_{liner}=5.28\ \text{in.}}$ (minimum). Checking the nearest standard $1/4$-in. liner sizes: a $5\,1/4$ in. liner delivers only $9.792(5.25)^2=269.9$ gpm — just short of the $272.6$ gpm requirement — while a $5\,1/2$ in. liner delivers $9.792(5.5)^2=296.2$ gpm, which clears it. So $\boxed{\text{select a }5\,1/2\ \text{in. liner}}$.