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Case A: Case B

The document contains three calculation problems related to fluid dynamics: 1) Calculating hydrostatic pressure at the bottom of fluid columns of varying depths and densities. 2) Determining the mud density required to fracture a stratum at a given depth and fracture pressure. 3) Finding the density of an ideal gas at a given molecular weight, pressure, and temperature.

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caroline
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555 views2 pages

Case A: Case B

The document contains three calculation problems related to fluid dynamics: 1) Calculating hydrostatic pressure at the bottom of fluid columns of varying depths and densities. 2) Determining the mud density required to fracture a stratum at a given depth and fracture pressure. 3) Finding the density of an ideal gas at a given molecular weight, pressure, and temperature.

Uploaded by

caroline
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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1 Calculate the hydrostatic pressure at the bottom of the fluid column for each case shown in Fig 4.52.

(Ans. 5200 psig for


Case A)

Given: D = 10,000 ft ρ = 10 ppg Required: PA, PB, PC

Solution:

CASE A: 𝑃𝐴 = 0.052 × 𝜌 × 𝐷 CASE B: 𝑃𝐵 = 0.052 × 𝜌 × 𝐷

𝑃𝐴 = 0.052 × 10𝑝𝑝𝑔 × 10,000𝑓𝑡 𝑃𝐵 = 0.052 × 10𝑝𝑝𝑔 × 10,000𝑓𝑡

𝑷𝑨 = 𝟓, 𝟐𝟎𝟎 𝒑𝒔𝒊 𝑷𝑩 = 𝟓, 𝟐𝟎𝟎 𝒑𝒔𝒊

CASE C: 𝑃𝐶 = 0.052 × 𝜌 × 𝐷

𝑃𝐶 = 0.052 × 10𝑝𝑝𝑔 × 10,000𝑓𝑡

𝑷𝑪 = 𝟓, 𝟐𝟎𝟎 𝒑𝒔𝒊

2 Calculate the mud density required to fracture stratum at 5,000 ft. if the fracture pressure is 3,800 psig. (Ans. 14.6
lbm/gal)

Given: D = 5,000 ft PF = 3,800 psig

Required: ρmud

Solution:

𝑃𝐹
𝜌=
0.052 × 𝐷
3,800 𝑝𝑠𝑖𝑔
𝜌=
0.052 × 5,000𝑓𝑡

𝝆 = 𝟏𝟒. 𝟔𝟏𝟓𝟑𝟖𝟓 𝒑𝒑𝒈

3 An ideal gas has an average molecular weight of 20. What is the density of the gas at 2,000 psia and 600⁰R? (Ans. 0.8
Ibm/gal)

Given: MW = 20 Solution:

P = 2,000 psia 𝑃 × 𝑀𝑊
𝜌=
80.3 × 𝑧 × 𝑇
T = 600⁰R

Z=1 2,000 psia × 20


𝜌=
80.3 × 1 × 600⁰R
Required: ρ
𝝆 = 𝟎. 𝟖𝟑𝟎𝟐𝟐𝟎 𝒍𝒃𝒎 ⁄𝒈𝒂𝒍

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