Drill Bit Hydraulics
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- Claire Mosley
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1 Drill i yraulic Aumpion ) Change of preure ue o elevaion i negligible. ) Velociy upream i negligible compare o nozzle. 3) reure ue o fricion i negligible. Δ Δ E ρvn 0 reure rop acro bi, vn nozzle velociy Solving for nozzle velociy v n Δ 8.074E 4ρ In he fiel i ha been hown ha velociy preice by hi equaion i off. So i ha been moifie, v n C Δ 8.074E 4ρ he recommene valve for C i.95. If 3 nozzle are preen
2 q q v A A 3 he velociy i equal in all he e. q A 3 q q + q + q3 vn A + vn A + vn A3 Tha give u v q n In fiel uni A v n q 3.7A q in gpm, A in inche, v n in f/ec olving for he preure rop Δ 8.3E 5ρq C A ρ i #/gal Flow Exponen I can be euce ha f CQ C i a conan log f logc + logq So he log log plo of hi equaion i a raigh line wih a lope of. can foun if wo f an Q are known, hi can be achieve by meauring he anpipe or urface preure for pumping rae. f + o by uing he above equaion can be calculae an ubrace from o fin f. f 8.3E 5ρq afer fining f, can be foun by C A
3 f log log Q f Q Maximum Drill i yraulic orepower Crierion aume ha opimum hole cleaning i achieve if he hyraulic horepower acro he bi i imize wih repec o he flow rae Q. Q Sub in + Take he fir erivaive of wih repec o Q e he reul o 0. g Q ( + ) CQ 0 f CQ ( ) 0 f or f S + hi i he roo ha make a imum. ence he opimum bi hyraulic will be achieve if fricion preure lo in he yem i mainaine a an opimum value of + acro he nozzle op +
4 Calculae or meaure a ome Q a hen knowing a Q op can be calculae by Q op Qaani log log fqa Wih Q op known he op can be rewrien op 8.3E 5ρQ A C op op olve for A op 8.3E 5ρQ Aop C op op π if all nozzle are he ame ize A op n nop n i he number of nozzle 4 olve for nop nop Aop nπ
5 Example: D 4/ 0#/f, Collar 7 0.3#/f 000 Mu θ 300, θ 600 9, ρ 5.5 #/gal ump 5440 pi 600hp 80% TD,000 V amin 85 f/min i 8 7/ ole ize 9 7/8 Rae aa Q 300 f 966 pi Q 400 f 4883 pi Fin 8.3E5ρQ C A 8.3E E pi pi f pi f pi f log log Q Q f ( ) log( 400 ) log Fin Q an Q min ( ) gpm Q ae on pump 5440 ( 4.5 ) 85 gpm Qmin ae on velociy 60 Opimum fricion preure p pi
6 Opimum preure rop a he bi p pi Opimum flow rae 047 Qop Qaani log log 300ani log log 7gpm fqa Thi i lower han he an higher han min flow rae. Opimum nozzle area A 8.3E 5 ρ Q op 8.3E op C op in For 3 equal ize e nop.5. 5in 3π omework 0/3/08 The rae in gpm o have urbulen flow in he annulu of he previou homework. Wha will he fricion rop be in he rill ring an annulu if he flui i waer.
7 The imum e impac force crierion aume ha he boom-hole cleaning i achieve by imizing he e impac force wih repec o he flow rae. The impac force a he boom of he hole can be erive form Newon econ law of moion F Q. 083C ρ Q in gpm ρ in #/gal f F Q limiaion ) imum pump horepower ) imum urface preure For he hallow porion of he well f i mall an he flow rae requiremen i large he impac force i limie only by he pump horepower, herefore, he allowable urface preure, expree a p Q ubiuing F Q p Q p Q Differeniae an e o 0 F.5 p Q + [ ( + ) CQ ] p Q + 0 For a vali oluion he numeraor mu be equal o zero.
8 Solve for he opimum fricion preure op + hen olve for he opimum bi preure op op + + op In he eeper ecion of he well he fricion preure lo increae, while he flow rae requiremen ecreae. Therefore he impac force will limie by he imum allowe pump preure,. Q Differeniae an e o 0 F.5 Q + [ ( + ) CQ ] Q + 0 For a vali oluion he numeraor mu be equal o zero. Give + op +
9 Example Same aa a yraulic example So.66 Q 4.3 gpm Q min 68 gpm A,000 fee he pump preure i he limiing facor pi op pi 975 Qop Qaani log log 300ani log log 347gpm fqa I i boune by he min an flow rae, o A 8.3E 5 ρ Q op 8.3E op C op in nop.6.33in 0.6 / 3" 3π 3 - e have an area of.7in. Secion 4.3 in ex, page 56, 57
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