Document on basics and fundamentals of Hydraulic fracturing
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Research essay about the pros and cons of hydraulic fracturing. This essay discusses the positives and the consequences of the use of hydraulic fracturing to make oil wells more productive. …Full description
it is good to check out.Full description
Jika copy filenya tolong masukin ke Daftar Pustakanya ya teman, Thanks best regards yogie mohammadDeskripsi lengkap
Hydraulic fracturing is widely accepted and applied to improve the gas recovery in unconventional reservoirs. Unconventional reservoirs to be addressed here are with very low permeability, compli...
Descripción: today in Petrovietnam University, I uploaded this file to share with everyone on the worldwide.
Jika copy filenya tolong masukin ke Daftar Pustakanya ya teman, Thanks best regards yogie mohammadFull description
today in Petrovietnam University, I uploaded this file to share with everyone on the worldwide.
brief effort to help newcomer in hydraulic hydrant designDescrição completa
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Hydraulic Calculation
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CALCULATION PROCEDURE 1. Bottom hole fracture treatment pressure p fr = g f × D (psi) The differential pressure across fracture face : ∆ p = p fr − p R (psi) The fracturing coefficient K e × Qe × C R K c = 0.0374 × ∆ p (ft/min) µ R 2. The total pumping time t =
V i
42(Qi )
x =
(minutes)
2 K c π t W f
determine The erfc (x) value from 4 6 ⎤ 2 x ⎡ x 2 x x 1 .... erfc( x ) = 1 − − + − + ⎢ ⎥ or figure 6 π ⎢⎣ 1!×3 2!×5 3!×7 ⎥⎦
1 ⎡ x2 2 x ⎤ e erfc( x) + − 1⎥ (%) ⎢ 2 π ⎦ x ⎣ 3. Fracture Area Eff × qit ft2 At = ⎛ W f ⎞ ⎟ 7.48⎜ ⎜ 12 ⎟ ⎝ ⎠ Eff =
4. Weight of sand required ⎛ W f ⎞ ⎟ At (1 − φ sand )× γ sand (62.4) lbs S = ⎜⎜ ⎟ 12 ⎝ ⎠ 5. Total injection rate S qit + (8.34) γ sand gal/min or bbls/min qT = t 6. Density of fluid plus sand 141.5 S (γ o )60 = ppg X = 131.5 + API V i the specific gravity should be adjusted to the reservoir temperature, assume the coefficient of thermal expansion β is 0.0005 (γ o )T = (γ o ) 60 [1 − β (T R − 60)] ρ =
8.34 × (γ o )T + X 1 + 0.0456 X
(ppg)
7. The hydroststic pressure ph = 0.052 × ρ × D (psi) If assumed laminar flow (frictional losses ignored) 8. Surface injection pressure p wh = p fr − ph (psi) 9. Horsepower requirement H h = 0.0245 × p wh × qT (hp) >>> qt in bbl/min
10. The mean fluid velocity 17.16(qT ) v= (ft/sec) 2 d Determine the resulting Reynolds number 928 × ρ × v × d NRe = µ Since NRe is greater than 2.000, the flow is turbulent. Therefore the friction factor should be calculated. • From chart ε/d vs. d 16 (laminar flow) • f = N Re
•
f =
c
where, c =
b N Re
' log .n + 2.5
50
and b =
1.4 − log .n 7
'
(turbulent flow)
11. The friction pressure drop 2
∆ p f =
f × ρ × v × L
(25.8) × D
(psi)
12. The Surface injection pressure p wh = p fr − ph + ∆p f (psi) 13. Horsepower requirement H h = 0.0245 × p wh × qT