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The density of an oil-based drilling mud must be 1700 kg / m3 at the surface. You are only told that the oil water ratio should be 4 and you must calculate up to what proportion of barite must be added. Barite has a density of 4200 kg / m3. The base oil has a density of 830 kg / m3. Water has a density of 1000 kg / m3. We have the following 3 equations to solve the problem. Note that x stands for volume fraction (0-1) and vm is an abbreviation for weight material. Xo/Xw = 4 Xo + Xw + Xvm = 1 XoPo + XwPw+ XvmPvm = 1700 Find out what the fractions of each component are.

The density of an oil-based drilling mud must be 1700 kg / m3 at the surface. You are only told that the oil water ratio should be 4 and you must calculate up to what proportion of barite must be added. Barite has a density of 4200 kg / m3. The base oil has a density of 830 kg / m3. Water has a density of 1000 kg / m3. We have the following 3 equations to solve the problem. Note that x stands for volume fraction (0-1) and vm is an abbreviation for weight material. Xo/Xw = 4 Xo + Xw + Xvm = 1 XoPo + XwPw+ XvmPvm = 1700 Find out what the fractions of each component are.

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
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The density of an oil-based drilling mud must be 1700 kg/ m3 at the surface. You are only told that the oil water ratio should be 4 and you must calculate up to what proportion of barite must be added. Barite has a density of 4200 kg / m3. The base oil has a density of 830 kg/ m3. Water has a density of 1000 kg / m3. We have the following 3 equations to solve the problem. Note that x stands for volume fraction (0-1) and vm is an abbreviation for weight material. Xo/Xw = 4 Xo + Xw + Xvm = 1 XoPo + XwPw + XvmPvm = 1700 Find out what the fractions of each component are.
Transcribed Image Text:The density of an oil-based drilling mud must be 1700 kg/ m3 at the surface. You are only told that the oil water ratio should be 4 and you must calculate up to what proportion of barite must be added. Barite has a density of 4200 kg / m3. The base oil has a density of 830 kg/ m3. Water has a density of 1000 kg / m3. We have the following 3 equations to solve the problem. Note that x stands for volume fraction (0-1) and vm is an abbreviation for weight material. Xo/Xw = 4 Xo + Xw + Xvm = 1 XoPo + XwPw + XvmPvm = 1700 Find out what the fractions of each component are.
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