Question 2 of 5: Bit Hydraulics — Maximum Bit Hydraulic Horsepower

Question 2: Bit Hydraulics — Maximum Bit Hydraulic Horsepower (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. Two parasitic (surface + annular friction) pressure-loss data points at different flow rates, mud weight $12$ ppg, pump rated at $1{,}800$ HP with $80\%$ volumetric efficiency, $P_{s,max}=4{,}500$ psi, $Q_{min}=350$ gpm, current bit nozzles $13$–$13$–$13$ (in $1/32$ in units).

Find. The parasitic loss, flow rate, bit nozzle pressure drop and optimum nozzle sizes at maximum bit hydraulic horsepower.

Approach. Fit the parasitic-loss power law $P_p=CQ^m$ from the two given data points, find the pump's available hydraulic horsepower, check whether the pump is horsepower-limited or pressure-limited at the optimum, apply the correct maximum-bit-hydraulics condition for that regime, then size the nozzles from the resulting bit pressure drop and flow rate.

1. Fit the parasitic pressure-loss law $P_p=CQ^m$. Taking the ratio of the two data points, $m=\dfrac{\ln(2{,}173/1{,}388)}{\ln(500/390)}=1.804$, and $C=P_{p1}/Q_1^{\,m}=2{,}173/500^{1.804}=0.02937$ (psi per gpm$^{1.804}$).
2. Available hydraulic horsepower. Only the volumetric-efficiency share of the rated pump horsepower is delivered as usable hydraulic power: $HP_{avail}=1{,}800(0.80)=1{,}440$ HP.
3. Check which pump limit governs. If the pump were purely horsepower-limited, the unconstrained optimum ($P_{p,opt}=\tfrac{m}{m+2}P_s$ with $P_s=1{,}714\,HP_{avail}/Q$) solves to $Q\approx513$ gpm at $P_s\approx4{,}807$ psi — but that exceeds the $4{,}500$ psi pump rating, so the pump is actually pressure-limited at the optimum, not horsepower-limited. The correct condition for a pressure-limited pump ($P_s=P_{s,max}$ fixed) is found by maximizing bit horsepower $\propto (P_s-CQ^m)Q$, which gives $\boxed{P_{p,opt}=\dfrac{P_{s,max}}{m+1}}$.
4. Optimum parasitic pressure loss (part a). $P_{p,opt}=\dfrac{4{,}500}{1.804+1}=\dfrac{4{,}500}{2.804}$, so $\boxed{P_{p,opt}=1{,}605\ \text{psi}}$.
5. Optimum flow rate (part b). From $P_{p,opt}=CQ_{opt}^{\,m}$: $Q_{opt}=\left(\dfrac{1{,}605}{0.02937}\right)^{1/1.804}$, so $\boxed{Q_{opt}=422.7\ \text{gpm}}$ — above the $350$ gpm minimum, so the optimum is achievable. (Check: horsepower used $=P_{s,max}Q_{opt}/1{,}714=4{,}500(422.7)/1{,}714=1{,}110$ HP, below the $1{,}440$ HP available, confirming the pump is pressure- not horsepower-limited here.)
6. Bit nozzle pressure drop (part c). $\Delta P_{bit}=P_{s,max}-P_{p,opt}=4{,}500-1{,}605$, so $\boxed{\Delta P_{bit}=2{,}895\ \text{psi}}$.
7. Optimum nozzle sizes (part d). Using $\Delta P_{bit}=\dfrac{MW\,Q_{opt}^2}{12{,}032\,C_d^2A_t^2}$ with the standard discharge coefficient $C_d=0.95$: $A_t=\sqrt{\dfrac{MW\,Q_{opt}^2}{12{,}032\,C_d^2\,\Delta P_{bit}}}=\sqrt{\dfrac{12(422.7)^2}{12{,}032(0.95)^2(2{,}895)}}=0.2611\ \text{in}^2$. The closest practical set of three near-equal, whole-$1/32$-in nozzles is $10$–$11$–$11$ ($A_t=\tfrac{\pi}{4}\left[(10/32)^2+2(11/32)^2\right]=0.2623\ \text{in}^2$), which back-calculates to $\Delta P_{bit}\approx2{,}869$ psi, within $1\%$ of the target — so $\boxed{\text{use }10\text{-}11\text{-}11\ (1/32\ \text{in})\ \text{nozzles}}$, a slight downsizing from the current $13$-$13$-$13$ set.