Extraction. - separation of a liquid mixture by adding a liquid solvent forming a second phase enriched by some component(s) Notation:
|
|
- Tobias Shelton
- 5 years ago
- Views:
Transcription
1 Extraction - separation of a liquid mixture by adding a liquid solvent forming a second phase enriched by some component(s) Notation: A extracted component B added solvent C original solvent F feed R rafinate S solvent E extract Two cases: 1) immiscible solvents B and C (only A is transported) 2) partially miscible solvents (A is transported, but B and C also a little) Example: extraction of acetic acid from water solution by adding an ether; after extraction, ether is regenerated by distillation; this is cheaper than distilling water solution of acetic acid directly Extractors 1) staged a. cascade of mixers and separators (settlers) b. column with plates (vibration is used to facilitate separation) 2) continuous a. packed or spray columns (slow) b. centrifuge (fast) General extractor is described as a general equilibrium stage (see earlier chapter), only enthalpy balance is often neglected (no heat effects). For correction to non-equilibrium efficiency - only A crosses the interface - F and R contain only A+C (phase x) - S and E contain only A+B (phase y) It is convenient to use relative mass fractions: Special case immiscible solvents X A = m A = x A ; x 1 x A = X A ; m A 1 + X A = x A (m A + ) A
2 Y A = m A = y A ; y m B 1 y A = Y A ; m A 1 + Y A = y A (m A + m B ) A m B and are constant. Single stage For mass balances we use finite time intervals no flows only masses - mass balance of A at steady state: X AF + m B Y AS = X A + m B Y A or - equilibrium relation for A: y A = ψ A x A or Y A = φ A X A By using the definition of relative fractions we obtain: (X AF X A ) = m B (Y A Y AS ) Y A = ψ A 1 (ψ A 1)X A X A Remark: for low concentrations x A = X A and y A = Y A and thus ψ A = φ A = const To find X A, Y A we solve the balance and equilibrium equations simultaneously either graphically or algebraically. Graphical solution: We rewrite the balance of A: Y A = m B (X A X AF ) + Y AS which is an operational line with slope equal to m B passing through point [X AF, Y AS ].
3 Algebraic solution: By combining balance of A and equilibrium relation X AF + m B Y AS = X A ( + ψ A m B ) / 1 X AF + m B Y AS = X A (1 + ψ Am B ) we introduce an extraction factor ζ x = ψ A m B and ζ y = ψ A m B = 1 ζ x Thus: X A = (1 + ζ x ) 1 (X AF + Y AS m B ) In general, solution of this equation is by iterations, because ζ x may depend X A on through ψ A = ψ A (X A ). That is, we estimate X A (0), then calculate XA (1) from the equation, and then use XA (1) to calculate XA (2), etc. until X A (n+1) XA (n) < ε where ε is small. For a real stage we have to combine mass balance, equilibrium relation and efficiency. We denote equilibrium composition by stars: Y A = φ A X A And efficiency is: E A = Y A Y AS Y A Y AS = X AF X A X AF X A
4 Graphical solution Algebraic solution 1) solution for equilibrium case: 2) from efficiency we have: X A = (1 + ζ x ) 1 (X AF + Y AS m B ) X A = X AF E A (X AF X A ) Repeated extraction (or cross-flow extraction) - mass balance of A in the entire cascade: or X AF + m B,n Y AS = X A + m B,n Y A (X AF X A ) = m B,n ( Y A Y AS ) = m A where m A is the mass of A being transferred from phase x to phase y. This equation serves for determining m A or Y A
5 - mass balance of A in the stage n X A,n 1 + m Bn Y AS = X A,n + m B,n Y A,n ; n = 1 N or Y A,n = m B,n (X A,n X A,n 1 ) + Y AS which are operational lines with slopes m B,n passing through points [X A,n 1, Y AS ]; n = 1 N Equilibrium condition (in equilibrium stages): Y A,n = φ A X A,n ; n = 1 N Solution of mass balance and equilibrium conditions leads to all N unknown pairs X A,n, Y A,n. Graphical solution - intersection of operational lines and equilibrium curve - either repeat N times (if N is given) or do until a given is reached Algebraic solution - the use of combined balance and equilibrium equations with extraction factor ζ x,n - in analogy to one extractor: where X A,n = (1 + ζ x,n ) 1 (X A,n 1 + Y AS m B,n ) ζ x,n = ζ A,n m B,n simulation (or control) calculation
6 o N is given; X A,n is calculated consecutively, n = 1 N design calculation o X A,n is given, N is calculated either for n = 1,2 until X A,N is reached or for n = N, N 1 until X A,0 is reached (more convenient if φ A,n depends on X A,n ) For a real cascade efficiency must be used in addition to mass balance and equilibrium condition. - for equilibrium we use asterisks: - the efficiency is Graphical solution (analogous to one stage): Y A,n = φ A,n X A,n E A = Y A,n Y AS = X A,n 1 X A,n Y A,n Y AS X A,n 1 X A,n Algebraic solution 1) solution for the equilibrium stage n: 2) the use of efficiency: X A,n We can introduce overall efficiency: and use it to calculate N real. Special case: E C = = (1 + ζ x,n ) 1 (X A,n 1 + Y AS m B,n ) X A,n = X A,n 1 E A,n (X A,n 1 X A,n ) N number of equilibrium stages = N real number of real stages a) m B,n = const = m B b) φ A,n = const = φ A linear equilibrium line then also ζ x,n = const = ζ x = φ A m B
7 Combined mass balance and equilibrium equation reads: X A,n = (1 + ζ x ) 1 (X A,n 1 + Y AS m B ) let us assume for simplicity that the solvent B is pure Y AS = 0. Then X A,n X A,n 1 = (1 + ζ x ) 1 or X A,n 1 X A,n = 1 + ζ x For all stages together: X A,0 X A,1 X A,N 1 = (1 + ζ X A,1 X A,2 X x ) N A,N X A,0 X A,N A more general form with Y AS > 0 is: X A,0 Y AS φa X A,N Y AS φa = (1 + ζ x ) N Remark: A similar formula with efficiency for real stages can be derived. Counter-current cascade (for flow systems) Mass balance for A in the entire cascade or m CX AF + m BY AS = m CX A,N + m BY A,1 m C(X AF X A,N ) = m B(Y A,1 Y AS ) = m A where m A is flow of A from phase x to phase y. Mass balance of A in stage n: m B(Y A,n Y A,n+1 ) = m C(X A,n 1 X A,n ); n = 1 N or
8 (Y A,n Y A,n+1 ) (X A,n 1 X A,n ) = m C m B operational line with slope m C passing through the points [X m A,n 1, Y A,n ]; n = 1 N B Equilibrium condition: Y A,n = φ A,n X A,n ; n = 1 N Graphical solution - slope is constant, all points [X A,n 1, Y A,n ] lie on one line - stages are found as rectangular steps Minimal consumption of the solvent (B) - two cases o no tangent o tangent
9 Minimal consumption can be calculated from: m C m B,min = (Y A,max Y AS ) (X AF X A,N ) m (X AF X A,N ) B,min = m C (Y A,max Y AS ) Remark: if m B = m B,min then N ; in practice m B = γm B,min where γ = Algebraic solution The mass balance in stage n can be rewritten as m CX A,n 1 + m BY A,n = m CX A,n + m BY A,n+1 = m A where m A is constant for all stages and can be determined from balance of all stages. By using equilibrium condition Y A,n = φ A,n X A,n we get: X A,n 1 = φ A,nm B m A + m C m C ζ x,n This formula can be used repeatedly to calculate N if is X A,N given or X A,N if N is given. For real stages a new efficiency must be defined Murphree efficiency where X A,n E X,n = X A,n 1 X A,n X A,n 1 X A,n = Y A,n φa,n (i.e. Y A,n is for real stage, not Y A,n for equilibrium stage). E Y,n = Y A,n Y A,n+1 Y A,n+1 Y A,n
10 where Y A,n = φ A,n X A,n (i.e. X A,n is for real stage, not X A,n for equilibrium stage). The two efficiencies E X,n and E Y,n are not equal. Graphical representation Graphical solution of the cascade Special case - φ A,n = const = φ A linear equilibrium then ζ X = φ Am B m C = const we introduce effectiveness factors Graphical representation η X = X AF X AN X AF Y AS φ A ; η Y = Y A1 Y AS φ A X AF Y AS
11 For η X = 1 or η Y = 1 the cascade would have N. It can be shown that then for ζ Z = 1: 1 η X = ( 1 N ) = ζ N 1 η Y ζ Y and η X = η Y ; η X ζ Y = η X Y ζ X N = ln (1 η Z ζz 1 η ) Z ln ζ Z ; where Z is either X or Y η Z N = 1 η Z General case partially miscible solvents In addition to A, also B and C cross the interface. Therefore, all three mass balances must be used now relative mass fractions are not useful ordinary mass fractions should be used together with masses of the streams, not just. Equilibrium: - formally y k = ψ k x k ; k = A,B,C - ψ k depends on composition and temperature - graphical representation in a triangle diagram is more convenient x C = 1 x A x B ; y C = 1 y A y B
12 - data for the equilibrium curve are typically given in tables Mass balance of A, B, C is graphically represented in the triangle diagram by the lever rule. Extractor is thought of as a combination of a mixer and settler. - overall mass balance: m F + m S = m M = m R + m E - component k: m F x kf + m S y ks = m M z km = m R x k + m E y k Combination yields: m F (x kf y ks ) = m M (z km y ks ) in analogy the lever rule for the settler is or m F m M = y ks z km y ks x kf = SM SF lever rule for mixer m R = ME m M RE Other variants of the lever rule (less convenient for calculations)
13 - mixer: - settler: m F = MS m S FM m R = ME m E RM Repeated extraction (cross-flow) mass balance of stage n (assume constant mass of solvent) - overall: m R,n 1 + m S = m M,n = m R,n + m E,n - components: m R,n 1 x k,n 1 + m S y ks = m M,n z k,n = m R,n x k,n + m E,n y k,n This implies lever rules for points - mixer (R n 1, M n, S); n = 1 N - settler (R n, M n, E n ); n = 1 N Equilibrium in stage n: y k,n = ψ k,n x k,n ; n = 1 N or via graph Graphical solution a) lever ruler for stage 1: o mixer: m F = SM 1 determine M m M 1 SF 1 o then tie line and lever rule for settler: m R1 = M 1E 1 m E R 1 1 E 1 b) for stage 2 determine m R1
14 c) etc. o o mixer: settler: m R 1 m R 1 +m S m R 2 m R 1 +m S = SM 2 SR 1 determine M 2 = M 2E 2 R 2 E 2 determine m 2 Counter current cascade of extractors (or counter-current staged column) Mass balance of all stages summing form: - overall: m F + m S = m M = m R N + m E 1 - components: m Fx kf + m Sy ks = m Mz km = m R N x k,n + m E 1 y k,1 These equations imply lever rules connecting points (F, M, S) and (R N, M, E 1 ) Mass balance of stages 1 to n difference form: - overall: m F m E 1 = m R n m E n+1 = m P = const - components: m Fx kf m E 1 y k,1 = m R n x k,n m E n+1 y k,n+1 = m Pz kp = const Here the point P with composition z kp is hypothetical but useful; lever rule applies on the lines connecting (F, E 1, P) and (R n, E n+1, P) for n = 1 N in particular (R N, E N+1 =S, P ) Equilibrium in stage n - formally: y k,n = ψ k,n x k,n ; n = 1 N - in graph: equilibrium curve + tie lines Graphical solution 1) Summing form of mass balances serves to find the point M and then the point E 1 provided that the compositions of F, S and R N are given. Lever rule: m F m F+m S = SM SF determine M, then connect R N and M to find E 1 2) Difference form of mass balances serves first to find the point P (=pole) as intersection of line (F, E 1 ) and (R N, S). The tie line from E 1 determines R 1 and connection of P with R 1 determines E 2 (see the difference form of balances). This is repeated until R N is reached. In the figure below six step were needed to reach R N N = 6.
15
Liquid Liquid Extraction
Liquid Liquid Extraction By Dr. Salih Rushdi 1 Introduction Liquid-liquid extraction (sometimes abbreviated LLX) is a mass transfer operation in which a solution (called the feed which is a mixture of
More informationLiquid-liquid extraction
Liquid-liquid extraction Basic principles In liquid-liquid extraction, a soluble component (the solute) moves from one liquid phase to another. The two liquid phases must be either immiscible, or partially
More informationERT 313 BIOSEPARATION ENGINEERING EXTRACTION. Prepared by: Miss Hairul Nazirah Abdul Halim
ERT 313 BIOSEPARATION ENGINEERING EXTRACTION Prepared by: Miss Hairul Nazirah Abdul Halim Definition of Extraction Liquid-Liquid extraction is a mass transfer operation in which a liquid solution (the
More informationChemical reactors. H has thermal contribution, pressure contribution (often negligible) and reaction contribution ( source - like)
Chemical reactors - chemical transformation of reactants into products Classification: a) according to the type of equipment o batch stirred tanks small-scale production, mostly liquids o continuous stirred
More informationExamples Liquid- Liquid- Extraction
Examples Liquid- Liquid- Extraction Lecturer: Thomas Gamse ao.univ.prof.dipl.-ing.dr.techn. Department of Chemical Engineering and Environmental Technology Graz University of Technology Inffeldgasse 25,
More informationMass Transfer Operations I Prof. Bishnupada Mandal Department of Chemical Engineering Indian Institute of Technology, Guwahati
Mass Transfer Operations I Prof. Bishnupada Mandal Department of Chemical Engineering Indian Institute of Technology, Guwahati Module - 4 Absorption Lecture - 3 Packed Tower Design Part 2 (Refer Slide
More informationLIQUID-LIQUID EQUILIBRIUM FOR THE DESIGN OF EXTRACTION COLUMN
LIQUID-LIQUID EQUILIBRIUM FOR THE DESIGN OF EXTRACTION COLUMN Magdah Abdelbasit Nory Salih Faculty of Engineering, Red sea University, Port Sudan- SUDAN ABSTRACT Liquid-Liquid Extraction is a mass transfer
More informationDistillation. This is often given as the definition of relative volatility, it can be calculated directly from vapor-liquid equilibrium data.
Distillation Distillation may be defined as the separation of the components of a liquid mixture by a process involving partial vaporization. The vapor evolved is usually recovered by condensation. Volatility
More informationMODULE 5: DISTILLATION
MOULE 5: ISTILLATION LECTURE NO. 3 5.2.2. Continuous distillation columns In contrast, continuous columns process a continuous feed stream. No interruptions occur unless there is a problem with the column
More informationVapor-liquid Separation Process MULTICOMPONENT DISTILLATION
Vapor-liquid Separation Process MULTICOMPONENT DISTILLATION Outline: Introduction to multicomponent distillation Phase Equilibria in Multicomponent Distillation (Pg. 737) Bubble-point and dew-point calculation
More informationSolution of Algebric & Transcendental Equations
Page15 Solution of Algebric & Transcendental Equations Contents: o Introduction o Evaluation of Polynomials by Horner s Method o Methods of solving non linear equations o Bracketing Methods o Bisection
More informationSolutionbank Edexcel AS and A Level Modular Mathematics
Page of Exercise A, Question The curve C, with equation y = x ln x, x > 0, has a stationary point P. Find, in terms of e, the coordinates of P. (7) y = x ln x, x > 0 Differentiate as a product: = x + x
More informationLecture 9 Laminar Diffusion Flame Configurations
Lecture 9 Laminar Diffusion Flame Configurations 9.-1 Different Flame Geometries and Single Droplet Burning Solutions for the velocities and the mixture fraction fields for some typical laminar flame configurations.
More informationMODULE 5: DISTILLATION
MODULE 5: DISTILLATION LECTURE NO. 5 Determination of the stripping section operating line (SOL): The stripping section operating line (SOL) can be obtained from the ROL and q- line without doing any material
More informationSeparation Processes: Liquid-Liquid Extraction
Separation Processes: Liquid-Liquid Extraction ChE 4M3 Kevin Dunn, 2014 kevin.dunn@mcmaster.ca http://learnche.mcmaster.ca/4m3 Overall revision number: 314 (October 2014) 1 Copyright, sharing, and attribution
More informationSeparation Processes: Liquid-liquid extraction
Separation Processes: Liquid-liquid extraction ChE 4M3 Kevin Dunn, 2013 kevin.dunn@mcmaster.ca http://learnche.mcmaster.ca/4m3 Overall revision number: 256 (November 2013) 1 Copyright, sharing, and attribution
More informationPhase Transitions in Multicomponent Systems
Physics 127b: Statistical Mechanics Phase Transitions in Multicomponent Systems The ibbs Phase Rule Consider a system with n components (different types of molecules) with r s in equilibrium. The state
More informationMass Transfer Operations I Prof. BishnupadaMandal Department of Chemical Engineering Indian Institute of Technology, Guwahati
Mass Transfer Operations I Prof. BishnupadaMandal Department of Chemical Engineering Indian Institute of Technology, Guwahati Module -5 Distillation Lecture - 8 Fractional Distillation: Subcooled Reflux,
More information3.012 PS 7 Thermo solutions Issued: Fall 2003 Graded problems due:
3.012 PS 7 Thermo solutions 3.012 Issued: 11.17.03 Fall 2003 Graded problems due: 11.26.03 Graded problems: 1. Analysis of equilibrium phases with a binary phase diagram. Shown below is the phase diagram
More informationThe distribution of a solute between the raffinate and the extract can be expressed in terms of the partition coefficient K:
Extraction Introduction An extraction process makes use of the partitioning of a solute between two immiscible or partially miscible phases. For example the antibiotic penicillin is more soluble in the
More informationSeparationsteknik / Separation technology
Separationsteknik / Separation technology 424105 1. Introduktion / Introduction Page 47 was added Nov. 2017 Ron Zevenhoven Åbo Akademi University Thermal and Flow Engineering Laboratory / Värme- och strömningsteknik
More informationUNC Charlotte 2004 Algebra with solutions
with solutions March 8, 2004 1. Let z denote the real number solution to of the digits of z? (A) 13 (B) 14 (C) 15 (D) 16 (E) 17 3 + x 1 = 5. What is the sum Solution: E. Square both sides twice to get
More informationEXAMPLE 2: Visualization and Animation of Enthalpy Method in Binary Distillation
EXAMPLE 2: Visualization and Animation of Enthalp Method in Binar Distillation A graphical method that includes energ balances as well as material balances and phase equilibrium relations is the Ponchon-Savarit
More informationROOT FINDING REVIEW MICHELLE FENG
ROOT FINDING REVIEW MICHELLE FENG 1.1. Bisection Method. 1. Root Finding Methods (1) Very naive approach based on the Intermediate Value Theorem (2) You need to be looking in an interval with only one
More informationL/O/G/O 單元操作 ( 三 ) Chapter 23 Leaching and Extraction 化學工程學系李玉郎
L/O/G/O 單元操作 ( 三 ) Chapter 23 Leaching and Extraction 化學工程學系李玉郎 Leaching Solid extraction, dissolve soluble matter from its mixture with an insoluble solid solid (solute A+ inert B) solvent (C) concentrated
More informationCurve Fitting. Objectives
Curve Fitting Objectives Understanding the difference between regression and interpolation. Knowing how to fit curve of discrete with least-squares regression. Knowing how to compute and understand the
More informationMSE 201A Thermodynamics and Phase Transformations Fall, 2008 Problem Set No. 8. Given Gibbs generals condition of equilibrium as derived in the notes,
MSE 201A hermodynamics and Phase ransformations Fall, 2008 Problem Set No. 8 Problem 1: (a) Let a homogeneous fluid with composition, {x}, be surrounded by an impermeable wall and in contact with a reservoir
More informationMass Transfer Operations I Prof. Bishnupada Mandal Department of Chemical Engineering Indian Institute of Technology, Guwahati
Mass Transfer Operations I Prof. Bishnupada Mandal Department of Chemical Engineering Indian Institute of Technology, Guwahati Module - 5 Distillation Lecture - 5 Fractional Distillation Welcome to the
More informationMass Transfer Operations I Prof. Bishnupada Mandal Department of Chemical Engineering Indian Institute of Technology, Guwahati
Mass Transfer Operations I Prof. Bishnupada Mandal Department of Chemical Engineering Indian Institute of Technology, Guwahati Module - 4 Absorption Lecture - 4 Packed Tower Design Part - 3 Welcome to
More informationLecture 6 Asymptotic Structure for Four-Step Premixed Stoichiometric Methane Flames
Lecture 6 Asymptotic Structure for Four-Step Premixed Stoichiometric Methane Flames 6.-1 Previous lecture: Asymptotic description of premixed flames based on an assumed one-step reaction. basic understanding
More informationx is also called the abscissa y is also called the ordinate "If you can create a t-table, you can graph anything!"
Senior Math Section 6-1 Notes Rectangular Coordinates and Lines Label the following 1. quadrant 1 2. quadrant 2 3. quadrant 3 4. quadrant 4 5. origin 6. x-axis 7. y-axis 8. Ordered Pair (x, y) at (2, 1)
More informationLaminar Premixed Flames: Flame Structure
Laminar Premixed Flames: Flame Structure Combustion Summer School 2018 Prof. Dr.-Ing. Heinz Pitsch Course Overview Part I: Fundamentals and Laminar Flames Introduction Fundamentals and mass balances of
More informationAssignment No. 1 RESULTANT OF COPLANAR FORCES
Assignment No. 1 RESULTANT OF COPLANAR FORCES Theory Questions: 1) Define force and body. (Dec. 2004 2 Mks) 2) State and explain the law of transmissibility of forces. (May 2009 4 Mks) Or 3) What is law
More informationSolving Systems of Linear Equations
Section 2.3 Solving Systems of Linear Equations TERMINOLOGY 2.3 Previously Used: Equivalent Equations Literal Equation Properties of Equations Substitution Principle Prerequisite Terms: Coordinate Axes
More informationDRAFT - Math 101 Lecture Note - Dr. Said Algarni
2 Limits 2.1 The Tangent Problems The word tangent is derived from the Latin word tangens, which means touching. A tangent line to a curve is a line that touches the curve and a secant line is a line that
More informationDr. Hatem Alsyouri Heat and Mass Transfer Operations
Dr. Hatem Alsyouri Heat and Mass Transer Operations Chemical Engineering Department The University o Jordan Reerences 1. Wankat: 10.6 10.9 and 15.1 15.6 2. Coulson & Richardson Vol 6: 11.14 3. Seader and
More informationMass Transfer Fundamentals. Chapter#3
Mass Transfer Fundamentals Chapter#3 Mass Transfer Co-efficient Types of Mass Transfer Co-efficient Convective mass transfer can occur in a gas or liquid medium. Different types of mass transfer coefficients
More informationSOLUTION OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS BISECTION METHOD
BISECTION METHOD If a function f(x) is continuous between a and b, and f(a) and f(b) are of opposite signs, then there exists at least one root between a and b. It is shown graphically as, Let f a be negative
More informationAn Efficient Design of Multi Component Distillation Column by Approximate & Rigorous Method
An Efficient Design of Multi Component Distillation Column by Approximate & Rigorous Method Syed Mujahed Ali Rizwan Senior Lecturer in Chemistry Challenger College, Moinabad, Hyderabad. Abstract: In this
More informationTheory. Humidity h of an air-vapor mixture is defined as the mass ratio of water vapor and dry air,
Theory Background In a cooling tower with open water circulation, heat is removed from water because of the material and heat exchange between the water and the ambient air. The cooling tower is a special
More informationSections 2.1, 2.2 and 2.4: Limit of a function Motivation:
Sections 2.1, 2.2 and 2.4: Limit of a function Motivation: There are expressions which can be computed only using Algebra, meaning only using the operations +,, and. Examples which can be computed using
More information1 Functions and Graphs
1 Functions and Graphs 1.1 Functions Cartesian Coordinate System A Cartesian or rectangular coordinate system is formed by the intersection of a horizontal real number line, usually called the x axis,
More informationINTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES
INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES You will be expected to reread and digest these typed notes after class, line by line, trying to follow why the line is true, for example how it
More informationUnit operations are ubiquitous in any chemical process. Equilibrium-Staged Separations using Matlab and Mathematica. ChE class and home problems
ChE class and home problems The object of this column is to enhance our readers collections of interesting and novel problems in chemical engineering. Problems of the type that can be used to motivate
More informationUNIT 3 CIRCLES AND VOLUME Lesson 1: Introducing Circles Instruction
Prerequisite Skills This lesson requires the use of the following skills: performing operations with fractions understanding slope, both algebraically and graphically understanding the relationship of
More informationDIFFERENTIATION RULES
3 DIFFERENTIATION RULES DIFFERENTIATION RULES We have: Seen how to interpret derivatives as slopes and rates of change Seen how to estimate derivatives of functions given by tables of values Learned how
More informationSection 3.1. Best Affine Approximations. Difference Equations to Differential Equations
Difference Equations to Differential Equations Section 3.1 Best Affine Approximations We are now in a position to discuss the two central problems of calculus as mentioned in Section 1.1. In this chapter
More informationAssignment 1, SOLUTIONS
MAT 10: Calculus for the Life Sciences I 01.10.2008 Pawel Lorek Assignment 1, SOLUTIONS University of Ottawa Problem 1: [4 points] Suppose that every morning a patient receives the same dose of drug. From
More informationTaylor and Maclaurin Series. Approximating functions using Polynomials.
Taylor and Maclaurin Series Approximating functions using Polynomials. Approximating f x = e x near x = 0 In order to approximate the function f x = e x near x = 0, we can use the tangent line (The Linear
More informationSection 1.6. Functions
Section 1.6 Functions Definitions Relation, Domain, Range, and Function The table describes a relationship between the variables x and y. This relationship is also described graphically. x y 3 2 4 1 5
More informationPRINCIPLES AND MODERN APPLICATIONS OF MASS TRANSFER OPERATIONS
PRINCIPLES AND MODERN APPLICATIONS OF MASS TRANSFER OPERATIONS Jaime Benitez iwiley- INTERSCIENCE A JOHN WILEY & SONS, INC., PUBLICATION Preface Nomenclature xiii xv 1. FUNDAMENTALS OF MASS TRANSFER 1
More information3.012 PS 6 THERMODYANMICS SOLUTIONS Issued: Fall 2005 Due:
3.012 PS 6 THERMODYANMICS SOLUTIONS 3.012 Issued: 11.28.05 Fall 2005 Due: THERMODYNAMICS 1. Building a binary phase diagram. Given below are data for a binary system of two materials A and B. The two components
More information. For each initial condition y(0) = y 0, there exists a. unique solution. In fact, given any point (x, y), there is a unique curve through this point,
1.2. Direction Fields: Graphical Representation of the ODE and its Solution Section Objective(s): Constructing Direction Fields. Interpreting Direction Fields. Definition 1.2.1. A first order ODE of the
More informationMATH The Derivative as a Function - Section 3.2. The derivative of f is the function. f x h f x. f x lim
MATH 90 - The Derivative as a Function - Section 3.2 The derivative of f is the function f x lim h 0 f x h f x h for all x for which the limit exists. The notation f x is read "f prime of x". Note that
More information3.012 PS 7 3.012 Issued: 11.05.04 Fall 2004 Due: 11.12.04 THERMODYNAMICS 1. single-component phase diagrams. Shown below is a hypothetical phase diagram for a single-component closed system. Answer the
More informationSemester 2 Final Review
Name: Date: Per: Unit 6: Radical Functions [1-6] Simplify each real expression completely. 1. 27x 2 y 7 2. 80m n 5 3. 5x 2 8x 3 y 6 3. 2m 6 n 5 5. (6x 9 ) 1 3 6. 3x 1 2 8x 3 [7-10] Perform the operation
More informationTaylor and Maclaurin Series. Approximating functions using Polynomials.
Taylor and Maclaurin Series Approximating functions using Polynomials. Approximating f x = e x near x = 0 In order to approximate the function f x = e x near x = 0, we can use the tangent line (The Linear
More informationCore Mathematics C12
Write your name here Surname Other names Pearson Edexcel International Advanced Level Centre Number Candidate Number Core Mathematics C12 Advanced Subsidiary Tuesday 10 January 2017 Morning Time: 2 hours
More informationHomework 2. Due Friday, July We studied the logistic equation in class as a model of population growth. It is given by dn dt = rn 1 N
Problem 1 (10 points) Homework Due Friday, July 7 017 We studied the logistic equation in class as a model of population growth. It is given by dn dt = rn 1 N, (1) K with N(0) = N 0. (a) Make the substitutions
More information1.2. Direction Fields: Graphical Representation of the ODE and its Solution Let us consider a first order differential equation of the form dy
.. Direction Fields: Graphical Representation of the ODE and its Solution Let us consider a first order differential equation of the form dy = f(x, y). In this section we aim to understand the solution
More informationMass Transfer Operations I Prof. Bishnupada Mandal Department of Chemical Engineering Indian Institute of Technology, Guwahati
Mass Transfer Operations I Prof. Bishnupada Mandal Department of Chemical Engineering Indian Institute of Technology, Guwahati Module - 5 Distillation Lecture - 6 Fractional Distillation: McCabe Thiele
More informationCryogenic Engineering Prof. M. D. Atrey Department of Mechanical Engineering Indian Institute of Technology, Bombay. Lecture No. #23 Gas Separation
Cryogenic Engineering Prof. M. D. Atrey Department of Mechanical Engineering Indian Institute of Technology, Bombay Lecture No. #23 Gas Separation So, welcome to the 23rd lecture, on Cryogenic Engineering,
More informationCH.7 Fugacities in Liquid Mixtures: Models and Theories of Solutions
CH.7 Fugacities in Liquid Mixtures: Models and Theories of Solutions The aim of solution theory is to express the properties of liquid mixture in terms of intermolecular forces and liquid structure. The
More information(a 1. By convention the vector a = and so on. r = and b =
By convention the vector a = (a 1 a 3), a and b = (b1 b 3), b and so on. r = ( x z) y There are two sort of half-multiplications for three dimensional vectors. a.b gives an ordinary number (not a vector)
More informationReflections on the use of the McCabe and Thiele method
From the Selectedorks of João F Gomes January 2007 Reflections on the use of the McCabe and Thiele method Contact Author Start Your Own Selectedorks Notify Me of New ork Available at: http://works.bepress.com/joao_gomes/42
More informationDISTILLATION. Keywords: Phase Equilibrium, Isothermal Flash, Adiabatic Flash, Batch Distillation
25 DISTILLATION Keywords: Phase Equilibrium, Isothermal Flash, Adiabatic Flash, Batch Distillation Distillation refers to the physical separation of a mixture into two or more fractions that have different
More informationMATHEMATICAL METHODS UNIT 1 Chapter 1 Reviewing Linear Equations Chapter 2 Coordinate geometry & linear relations
REVIEWING LINEAR EQUATIONS E da = q ε ( B da = 0 E ds = dφ. B ds = μ ( i + μ ( ε ( dφ 3 MATHEMATICAL METHODS UNIT 1 Chapter 1 Reviewing Linear Equations Chapter 2 Coordinate geometry & linear relations
More informationChapter 4: Column Distillation: Internal Stage-by-Stage Balances
Chapter 4: Column Distillation: Internal Stage-by-Stage Balances In Chapter 3 (Introduction to Column Distillation), we performed the external balances or the balances around the distillation column To
More informationUNC Charlotte 2005 Comprehensive March 7, 2005
March 7, 2005 1 The numbers x and y satisfy 2 x = 15 and 15 y = 32 What is the value xy? (A) 3 (B) 4 (C) 5 (D) 6 (E) none of A, B, C or D 2 Suppose x, y, z, and w are real numbers satisfying x/y = 4/7,
More informationThermodynamics revisited
Thermodynamics revisited How can I do an energy balance for a reactor system? 1 st law of thermodynamics (differential form): de de = = dq dq--dw dw Energy: de = du + de kin + de pot + de other du = Work:
More informationASSIGNMENT COVER SHEET omicron
ASSIGNMENT COVER SHEET omicron Name Question Done Backpack Ready for test Drill A differentiation Drill B sketches Drill C Partial fractions Drill D integration Drill E differentiation Section A integration
More informationConcept of the chemical potential and the activity of elements
Concept of the chemical potential and the activity of elements Gibb s free energy, G is function of temperature, T, pressure, P and amount of elements, n, n dg G G (T, P, n, n ) t particular temperature
More informationA2 HW Imaginary Numbers
Name: A2 HW Imaginary Numbers Rewrite the following in terms of i and in simplest form: 1) 100 2) 289 3) 15 4) 4 81 5) 5 12 6) -8 72 Rewrite the following as a radical: 7) 12i 8) 20i Solve for x in simplest
More informationThe Coordinate Plane; Graphs of Equations of Two Variables. A graph of an equation is the set of all points which are solutions to the equation.
Hartfield College Algebra (Version 2015b - Thomas Hartfield) Unit TWO Page 1 of 30 Topic 9: The Coordinate Plane; Graphs of Equations of Two Variables A graph of an equation is the set of all points which
More informationAnnouncements. Related Rates (last week), Linear approximations (today) l Hôpital s Rule (today) Newton s Method Curve sketching Optimization problems
Announcements Assignment 4 is now posted. Midterm results should be available by the end of the week (assuming the scantron results are back in time). Today: Continuation of applications of derivatives:
More informationDEPARTMENT OF CHEMICAL ENGINEERING University of Engineering & Technology, Lahore. Mass Transfer Lab
DEPARTMENT OF CHEMICAL ENGINEERING University of Engineering & Technology, Lahore Mass Transfer Lab Introduction Separation equipments account for a major part of the capital investment in process industry.
More informationA New Modification of Newton s Method
A New Modification of Newton s Method Vejdi I. Hasanov, Ivan G. Ivanov, Gurhan Nedjibov Laboratory of Mathematical Modelling, Shoumen University, Shoumen 971, Bulgaria e-mail: v.hasanov@@fmi.shu-bg.net
More informationTEST CODE: MMA (Objective type) 2015 SYLLABUS
TEST CODE: MMA (Objective type) 2015 SYLLABUS Analytical Reasoning Algebra Arithmetic, geometric and harmonic progression. Continued fractions. Elementary combinatorics: Permutations and combinations,
More informationMATH 2331 Linear Algebra. Section 1.1 Systems of Linear Equations. Finding the solution to a set of two equations in two variables: Example 1: Solve:
MATH 2331 Linear Algebra Section 1.1 Systems of Linear Equations Finding the solution to a set of two equations in two variables: Example 1: Solve: x x = 3 1 2 2x + 4x = 12 1 2 Geometric meaning: Do these
More informationUnit #3 Rules of Differentiation Homework Packet
Unit #3 Rules of Differentiation Homework Packet In the table below, a function is given. Show the algebraic analysis that leads to the derivative of the function. Find the derivative by the specified
More informationEGMO 2016, Day 1 Solutions. Problem 1. Let n be an odd positive integer, and let x1, x2,..., xn be non-negative real numbers.
EGMO 2016, Day 1 Solutions Problem 1. Let n be an odd positive integer, and let x1, x2,..., xn be non-negative real numbers. Show that min (x2i + x2i+1 ) max (2xj xj+1 ), j=1,...,n i=1,...,n where xn+1
More informationExam 3 Solutions. ClO g. At 200 K and a total pressure of 1.0 bar, the partial pressure ratio for the chlorine-containing compounds is p ClO2
Chemistry 360 Dr. Jean M. Standard Fall 2016 Name KEY Exam 3 Solutions 1.) (14 points) Consider the gas phase decomposition of chlorine dioxide, ClO 2, ClO 2 ( g) ClO ( g) + O ( g). At 200 K and a total
More informationMASS TRANSPORT Macroscopic Balances for Multicomponent Systems
TRANSPORT PHENOMENA MASS TRANSPORT Macroscopic Balances for Multicomponent Systems Macroscopic Balances for Multicomponent Systems 1. The Macroscopic Mass Balance 2. The Macroscopic Momentum and Angular
More informationSolutions. .5 = e k k = ln(.5) Now that we know k we find t for which the exponential function is = e kt
MATH 1220-03 Exponential Growth and Decay Spring 08 Solutions 1. (#15 from 6.5.) Cesium 137 and strontium 90 were two radioactive chemicals released at the Chernobyl nuclear reactor in April 1986. The
More informationCorrelation of WNCP Curriculum to Pearson Foundations and Pre-calculus Mathematics 10
Measurement General Outcome: Develop spatial sense and proportional reasoning. 1. Solve problems that involve linear measurement, using: SI and imperial units of measure estimation strategies measurement
More informationInfinite series, improper integrals, and Taylor series
Chapter 2 Infinite series, improper integrals, and Taylor series 2. Introduction to series In studying calculus, we have explored a variety of functions. Among the most basic are polynomials, i.e. functions
More informationR.2 Number Line and Interval Notation
8 R.2 Number Line and Interval Notation As mentioned in the previous section, it is convenient to visualise the set of real numbers by identifying each number with a unique point on a number line. Order
More informationOQ4867. Let ABC be a triangle and AA 1 BB 1 CC 1 = {M} where A 1 BC, B 1 CA, C 1 AB. Determine all points M for which ana 1...
764 Octogon Mathematical Magazine, Vol. 24, No.2, October 206 Open questions OQ4867. Let ABC be a triangle and AA BB CC = {M} where A BC, B CA, C AB. Determine all points M for which 4 s 2 3r 2 2Rr AA
More informationIntegration Techniques for the AB exam
For the AB eam, students need to: determine antiderivatives of the basic functions calculate antiderivatives of functions using u-substitution use algebraic manipulation to rewrite the integrand prior
More information(a) Show that there is a root α of f (x) = 0 in the interval [1.2, 1.3]. (2)
. f() = 4 cosec 4 +, where is in radians. (a) Show that there is a root α of f () = 0 in the interval [.,.3]. Show that the equation f() = 0 can be written in the form = + sin 4 Use the iterative formula
More informationOutline of the Course
Outline of the Course 1) Review and Definitions 2) Molecules and their Energies 3) 1 st Law of Thermodynamics Conservation of Energy. 4) 2 nd Law of Thermodynamics Ever-Increasing Entropy. 5) Gibbs Free
More informationNote: items marked with * you should be able to perform on a closed book exam. Chapter 10 Learning Objective Checklist
Note: items marked with * you should be able to perform on a closed book exam. Chapter 10 Learning Objective Checklist Sections 10.1-10.13 find pure component properties on a binary P-x-y or T-x-y diagram.*
More informationCHAPTER 7. An Introduction to Numerical Methods for. Linear and Nonlinear ODE s
A SERIES OF CLASS NOTES FOR 2005-2006 TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 1 A COLLECTION OF HANDOUTS ON FIRST ORDER ORDINARY DIFFERENTIAL
More informationModel Answers Attempt any TEN of the following :
(ISO/IEC - 70-005 Certified) Model Answer: Winter 7 Sub. Code: 17 Important Instructions to Examiners: 1) The answers should be examined by key words and not as word-to-word as given in the model answer
More informationIntroduction. Chapter Points, Vectors and Coordinate Systems
Chapter 1 Introduction Computer aided geometric design (CAGD) concerns itself with the mathematical description of shape for use in computer graphics, manufacturing, or analysis. It draws upon the fields
More informationSystems of Ordinary Differential Equations
Systems of Ordinary Differential Equations Systems of ordinary differential equations Last two lectures we have studied models of the form y (t) F (y), y(0) y0 (1) this is an scalar ordinary differential
More informationUNM - PNM STATEWIDE MATHEMATICS CONTEST XLIX. February 4, 2017 Second Round Three Hours
UNM - PNM STATEWIDE MATHEMATICS CONTEST XLIX February 4, 2017 Second Round Three Hours 1. What are the last two digits of 2017 2017? Answer: 77. From Round 1, we now that (2017) 4 7 4 1 mod 10. Thus 7
More informationSome Results Concerning Uniqueness of Triangle Sequences
Some Results Concerning Uniqueness of Triangle Sequences T. Cheslack-Postava A. Diesl M. Lepinski A. Schuyler August 12 1999 Abstract In this paper we will begin by reviewing the triangle iteration. We
More informationChapter 11 section 6 and Chapter 8 Sections 1-4 from Atkins
Lecture Announce: Chapter 11 section 6 and Chapter 8 Sections 1-4 from Atkins Outline: osmotic pressure electrolyte solutions phase diagrams of mixtures Gibbs phase rule liquid-vapor distillation azeotropes
More informationSolving mass transfer problems on the computer using Mathcad
Solving mass transfer problems on the computer using Mathcad E. N. Bart, J. Kisutcza NJIT, Department of Chemical Engineering, University Heights, Newark NJ 712-1982 Tel 973 596 2998, e-mail: Bart@NJIT.edu
More information