17.) Fracking: Surface Pressure; Pump Power. subsurface zone is to be fractured

Question

17.) Fracking: Surface Pressure; Pump Power. subsurface zone is to be fractured by high-pressure- high-rate injection of an 8-ppg (lbm per gallon), 10- cP fracturing fluid (Newtonian) through a vertical casing (4.892-in ID; 0.00065-in roughness). Designed bottomhole treating pressure is 5600 psi at 8000-ft depth (i.e., 0.7-psi/ft fracture gradient) and the injection rate is 40 bpm (barrel per minute). How much surface (wellhead) pressure is required? What pump power (ideal) is required? (Ppump.discharge wellhead pressure andp 32.174 lbm; 1 barrel 5.6146 ft; 1 cP 1 mPa s; 1 psi 6894.757 Pa; 1 hp 550 ft.lbf/s; g 32.174 ft/s2.] Ans. 4182 psi, 4098 hp. n 0.) [1 slug - Pump (well head)

Explanation / Answer

Fluid density, ? = 8 ppg = 8/(0.134*123) lb mass/ in3 = 0.035 lb mass/ in3

Fluid viscosity ? = 10cP = 10 mPa.s =10*10-3Pa.s = 10*10-3/6894.757 lbf.s/ in2 = 1.450*10-3 lbf.s/ in2

Casing ID, d = 4.892 in

Internal area of casing = (?/4)*d2 = (?/4)* (4.892)2 = 18.796 in2 = 0.131 ft2

Casing roughness = 0.00065 in

Well height, h = 8000 ft

Bottomhole treating pressure = 5600 psi

Injection rate = 40 bpm = 40*5.6146 ft3 /m = 224.584 ft3 /m

Injection velocity, V = 224.584/0.131 = 1714.382 ft /m = 28.573 ft /s

Coefficient of due to surface roughness is given by 1/ (?4f) = 2*log10*(r/R) +1.74 = 2*log10*(2.446/0.00065) +1.74 => f = 0.266 ft

Head loss due to friction = (4fLV2)/(2dg) = (4*0.266*28.5732)/(2*4.892*32.174) = 2.766 ft

Surface pressure = Bottomhole treating pressure – hydrostatic pressure

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.