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Question 1 Consider a triplex pump having 6-in. Liners and 11-in. Strokes

operating at 120 cycles/min and a discharge pressure of 3000psig. a) Compute the pump factor in units of gal/cycles at 100% volumetric efficiency.  b) Compute the flow rate in gal/min. c) Compute the energy expended by each piston during each cycle and pump  power developed. Given:

dL = 6in; L S= 11in; N=120cycles/min and

∆ = 3000psig

Find:

a) FP  b) Q c) E and PH d)

 ×  ×  ×    (Triplex)       ) = 4.039 gal/ cycle FP= × 1 1 × 1 × 6   = 933.05 in 3/ cycle(  

a) FP=

 b) Q=N*FP Q= (120 cycle/min)(4.039gal/cycle) = 484.68 gal/ cycle

c) E=

∆×

      E= (3000psi)(484.68gal/min)               E= 77 750.75 ft-lbf/cycle/cylinder

∆× 

PH=

×. = 848.33hp 

PH=

Question 2 How many sacks of barite are required to rise the mud weight of

775bbl from 77lb/ft3 to 92 lb/ft3. Calculate the new mud density when 126bbl of oil added to the new system. Given:

Mud weight: 750bbl W1=77lb/ft3 = 10.3lb/gal W2=92lb/ft3= 12.3lb/gal Find:

Sucks of barite and new mud density when 126 bbl of oil is added

Sacks of barite = 1490× −  =1490× .−. = 129     .− .−. Sacks of barite = 129×7.75 = 999.8 sx    

×   V ×  = V ×  775×12.3+ 126×6.8=901×  = 11.5lb/ft VM

O

F

3

Question 3:  A 1000 bbl unweighted fresh water mud system has a density of 9.5

lbm/gal. What mud treatment would be required to reduce the solids content to 4% by volume? The total mud must be maintained at 1000 bbl and the minimum allowable mud density is 8.8 lbm/gal. Given:

Vm = 1000 bbl W1 = 9.5 lbm/gal f sf  = final volume fraction of solids = 4 % Ww = 8.33 ppg Ws = 21.7 lb/gal Find: What mud treatment would be required to reduce the solids content to 4%

 by volume? The initial density of mud can be found using the following formula:

×

×

W1 = Ws f si -Ww f wi f si + f wi = 1 f si = (W1 - Ww)/(Ws - Ww) = (9.5 - 8.33)/(21.7 - 8.33) f si = 8.7%; f wi = 91.3% The Vm contents to 4% can now be found:

×

×

V = Vm  ( f sf  /f si) = 1000  (0.04/0.087) V = 457 bbl So the discarded volume is; V discarded  = 1000 - 457 = 543 bbl Yet the same amount of water has to be added to keep a V TotalMud  of 1000 bbl and reduce to 4% the content of solids by volume as stated in the question. So; Vwater-discarded = 543 bbl

Question 4 The density of 600bbl of 12lbm/gal mud must be increased to

14lbm/gal using API barite. One gallon of water per sack of barite will be added to maintain an acceptable mud consistency. The final volume is not limited. How much barite is required? Given:

V1=600bbl

 = 12blm/gal  = 14blm/gal Find: How much barite is requered

A=

+× = 0.802444; + ×

V2= V1(

× ×

V2= 1.142

mB= (

A

×  = 28.09

) = V1(.− .− =1.142×1

×600= 685.17bbl

 )(85.17×42) = 92740.71 +.+

Question 5 The density of 800 bbl of 14lbm/gal must be increased to

14.5lbm/gal using API barite. The total mud volume is limited to 800bbl. Compute the volume of old volume that should be discarded and the weight of API barite required. Given:

V1=800bbl

 = 14/  = 14.5/ Find: Volume of old mud discarded and the weight of API barite required.

V2= V1(

  

.− ) )= 800(.−.  =819.14bbl

Volume discarded= 819.14bbl-800bbl= 19.14bbl Weight=(V2-V1)  = (19.14)(35.4)(42) = 28457.35bbl