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ChE 516 Numerical Methods in Chemical

Engineering Shuguang Deng Department of

Chemical Engineering New Mexico State

University Fall 2009

Lecture 1. Introduction

- Course syllabus
- Suggested study method
- Methods for solving ChE problems
- MATLAB basics
- Numerical solution of nonlinear equations

Introduction

- Welcome!
- Shuguang Deng
- Professor, Chem. Eng. Dept., NMSU
- Ph.D., Chem. Eng., 1996 (Univ. Cincinnati)
- 10 industrial experience
- Research adsorption, nanostructured materials,

fuel cells - 7th year at NMSU

????????

Course Objectives

Course Objectives

- To study numerical analysis methods and their

applications in chemical engineering - To solve chemical engineering problems with

numerical analysis techniques - To learn the basics of MATLAB and to write simple

MATLAB codes

Course Topics

- Numerical solution of nonlinear equations
- Numerical solution of simultaneous linear

algebraic equations - Finite difference methods and interpolation
- Numerical differentiation and integration
- Numerical solution of ordinary differential

equations - Numerical solution of partial differential

equations

Course Syllabus

- 24 lectures
- Exams 50
- Homework 20
- Project 20
- Quizzes 10
- A 90-100, B 80-90, C70-79, D 60-69

Course Features

- A core course in ChE graduate program
- A mathematical subject in engineering
- Labeled as MATLAB ( ? ? )
- Suggested study method
- Review calculus, differential equations and other

ChE courses - Self-study MATAB
- Take class notes and practice examples in class
- Do your HWs independently

Mathematical Models in Chem. Eng.

It is difficult to obtain analytical solutions

for many models used in chemical engineering,

numerical solutions are preferred in these cases.

Numerical analysis is a branch of mathematics, it

studies algorithms for the problems of continuous

mathematics.

- Non-linear equation(s)

Ordinary equation(s)

Partial differential equation(s)

Common Software in ChE

Simulation software

Math Software

- PRO/II (SimSci)
- AspenPlus
- ChemCAD
- Flowtran
- Superpro Designer
- Fluent
- CHEMKIN

- Matlab
- Mathematica
- Mathcad
- Maple
- Staticstica

A Typical ChE Problem

A long rod 40 mm in diameter, fabricated from

sapphire (aluminum oxide) and initially at a

uniform temperature of 800 K, is suddenly cooled

by a fluid at 300 K having a heat transfer

coefficient of 1600 W/m2 K. After 35 s, the rod

is wrapped in insulation and experiences no heat

losses. What will be the temperature of the rod

after a long period of time?

Problem-Solving Process

Problem-Solving Process

Radial System with Convection

Infinite Cylinder (L/r0 gt10)

i.c.

b.c.s

Infinite Cylinder

Exact Solution

Infinite Cylinder

Definition of Bessel functions (J0 J1) of the

first Kind

Infinite Cylinder

Approximate Solution (Fogt0.2)

C1 and ?1 can be found in Table 5.1, page 274.

Problem-Solving Process

Problem-Solving Process

1. Define the problem. 2. Create a mathematical

model. 3. Develop an analytical or computational

method for solving the problem. 4. Implement

the computational method. 5. Test and assess the

solution.

Mathematics

Computer Programming

MATLAB

MATrix LABoratory

A computer programming language developed to

perform numeral computations encountered in

science and engineering operations.

MATLAB

MATLAB was invented by Prof. Moler of University

of New Mexico. It was originally written in

FORTRAN and improved by C.

MathWorks was established to promote MATLAB in

1984. MATLAB has become a very popular

computational software that is widely used in

mathematics, science and engineering.

Why MATLAB ?

- Computers dominate the industrial workplace.
- The fundamental programming concepts learned

using MATLAB will enable you to adapt to other

programming environments. - You will be required to use MATLAB in this course

and future courses. - MATLAB is widely used in industry.

MATLAB Resources

Etters, D.M., D.C. Kuncicky, Introduction to

MATLAB, Prentice-Hall. Appendix A of your

textbook

MATLAB Resources

- Getting Started html document on
- Mathworks web site
- http/www.mathworks.com/access
- /helpdesk/help/helpdesk.shtml
- Click on Getting Started
- Click on This manual in PDF
- Open or save to disk
- Help from MATLAB program
- Course Webpage
- Practice and learn from each other

MATLAB Basics

Please download getstart.pdf from the following

link http//www.mathworks.com/access/helpdesk/help

/pdf_doc/matlab/getstart.pdf Let lean the basics

of MATLAB from the a demo video

Main Features of MATLAB

- Calculator
- Mathematic functions
- Vector, matrices operations
- Symbolic operations
- Programming
- Graphics

MATLAB Basics

- Desktop Tools and Development Environment
- Matrices and Arrays
- MATLAB Functions
- M-files (Script and Function)
- Graphics
- Programming

MATLAB Basics

Lets Practice MATLAB

Chapter 1

Numerical Solution of Nonlinear Equations

Needs for Solving Nonlinear Equations

Soave-Redlich-Kwong Equation (P-V-T)

Needs for Solving Nonlinear Equations

Colebrook Equation

Needs for Solving Nonlinear Equations

Differential Equation

Types of Roots

- Real and distinct
- Real and repeat
- Complex conjugates
- A combination of any or all of the above

Real and Distinct Roots

gtgt f'x46x37x2-6x-8' gtgt

fplot(f,-5,2) gtgt fv1 6 7 -6 -8 gtgt f1

roots(fv) f1 -4.0000 1.0000 -2.0000

-1.0000

Real and Repeated Roots

gtgt g'x47x312x2-4x-16' gtgt

fplot(g,-5,2) gtgt gv 1 7 12 -4 -16 gtgt g1

roots(gv) g1 -4.0000 1.0000

-2.0000 0.0000i -2.0000 - 0.0000i

Complex Conjugates

gtgt h'x4-6x318x2-30x25' gtgt

fplot(h,0,3) gtgt hv 1 -6 18 -30 25 gtgt h1

roots(hv) h1 1.0000 2.0000i 1.0000 -

2.0000i 2.0000 1.0000i 2.0000 - 1.0000i

Combination of types

gtgt j'x4x3-5x223x-20' gtgt

fplot(j,-5,2) gtgt jv 1 1 -5 23 -20 gtgt j1

roots(jv) j1 -4.0000 1.0000

2.0000i 1.0000 - 2.0000i 1.0000

Real and complex conjugates

Verify roots of nth degree polynomial

- Newton's first relation
- Newton's second relation
- Newton's third relation
- Newton's nth relation

where i ? j ? k ? for all relationships

Descartes' rule of sign

- Described by René Descartes in La Geometrie, the

rule of sign" is a technique for determining the

number of positive or negative roots of a

polynomial. The rule states if the terms of a

single-variable polynomial with real coefficients

are ordered by descending variable exponent, then

the number of positive roots of the polynomial is

either equal to the number of sign differences

between consecutive nonzero coefficients, or less

than it by a multiple of 2. Multiple roots of the

same value are counted separately. - As a corollary of the rule, the number of

negative roots is the number of sign changes

after negating the coefficients of odd-power

terms (otherwise seen as substituting the

negation of the variable for the variable

itself), or less than it by a multiple of 2.

Descartes' rule of sign

- As an example, consider the polynomial
- x3 x2 x - 1
- Which has one sign change between the second and

third terms. Therefore it has exactly 1 positive

root. - Negating the odd-power terms gives
- -x3 x2 x - 1, this polynomial has two sign

changes, so the original polynomial has 2 or 0

negative roots. - The polynomial factors easily as
- ( x 1 )2 ( x 1 )
- so the roots are -1 (twice) and 1.

Roots Finding Techniques

General approach to devise iterative algorithms

that start at an initial estimate of a root and

converge to the exact value of the desired root

in a finite number of steps. Main methods

Fixed point iteration (Successive substitution,

Wegstein method) Linear interpolation

(Method of false position, BISECTION, SECANT)

Newton-Raphson, MATLAB functions fzero, roots,

solve

MATLAB Functions

MATLAB functions fzero, roots, solve fzero

(fun, a,b), f(a)f(b)lt0 fzero(_at_fun,

x0) fzero(fun, x0) roots is a built-in

MATLAB function for solving linear algebraic

equations. You need to define the coefficient

matrix P first, then rootsP to find the

roots. solve is another built-in MATLAB

function for getting symbolic solution of

algebraic equations.

1. Bisection Iteration

We assume we are given a function f (x) and in

addition, we assume we have an interval a, b

containing the root, on which the function is

continuous. We also assume we are given an error

tolerance e gt 0, and we want an approximate root

x in a, b for which x - x e. (x is

the real root) We further assume the function f

(x) changes sign on a, b, with f (a)f (b) lt 0

1. Bisection Iteration

Algorithm Bisect(f, a, b, e). Step 1 Define c

1/2 (a b) Step 2 If b - c e, accept c as

our root, and then stop. Step 3 If b - c gt e,

then check compare the sign of f (c) to that of f

(a) and f (b). If sign(f (b)) sign(f (c))

0 then replace a with c and otherwise, replace b

with c. Return to Step 1.

1. Bisection Iteration

Denote the initial interval by a1, b1, and

denote each successive interval by aj, bj. Let

cj denote the center of aj, bj. Then Since

each interval decreases by half from the

preceding one, we have by induction,

2. Fixed Point Iteration

f(x)0, x g(x), xn1 g(xn)

Convergence

Divergence

2. Fixed Point Iteration

- x g(x)
- select initial value x1
- evaluate x2 g(x1)
- iterate as xn1 g(xn)
- convergence occurs if x3 x2 lt x2 x1
- continue until a specified tolerance is met

2. Fixed Point Iteration

function y fss1(x) y sqrt(2x) gtgt

x00.1 gtgt y0fss1(x0) y0 0.4472 gtgt x1 y0 gtgt

y1fss1(x1) gtgt y1fss1(x1) y1 0.9457 gtgt x2

y1 gtgt y2fss1(x2) y2 1.3753 gtgt x3 y2 gtgt x3

y2 x3 1.3753 gtgt y3fss1(x3) y3 1.6585

gtgt x4 y3 gtgt y4fss1(x4) y4 1.8213 gtgt x5

y4 gtgt y5fss1(x5) y5 1.9085 gtgt x6 y5 gtgt

y6fss1(x6) y6 1.9537 gtgt x7 y6 gtgt

y7fss1(x7) y7 1.9767 gtgt x8 y7 gtgt

y8fss1(x8) y8 1.9883 gtgt x9 y8 gtgt

y9fss1(x9) y9 1.9942

2. Fixed Point Iteration

function y fss1(x) y x.2x./2 gtgt

x(1)1.95 gtgt y(1)fss1(x(1)) gtgt x(2)y(1) gtgt

y(2)fss1(x(2)) gtgt x(3)y(2) gtgt

y(3)fss1(x(3)) gtgt x(4)y(3) gtgt

y(4)fss1(x(4)) gtgt x(5)y(4) gtgt

y(5)fss1(x(5)) gtgt x(6)y(5) gtgt

y(6)fss1(x(6)) gtgt x(7)y(6) gtgt

y(7)fss1(x(7)) ...etc

3. The Wegstein Method

xg(x)

3. Wegstein Method

- x g(x)
- select initial value x1, evaluate x2 g(x1)
- estimate function g(x) by line passing through

points (x1, g(x1)) and (x2, g(x2)) - next estimation found by intersection of this

line and the y x line

4.1 Newton-Raphson Method

Given an estimate of a, say a x0, approximate f

(x) by its linear Taylor polynomial at (x0, f

(x0)) p(x) f (x0) (x - x0) f(x0) If x0 is

very close to a, then the root of p(x) should be

close to a. Denote this approximating root by x1

repeat the process to further improve our

estimate of a.

4.1 Newton-Raphson Method

For a general equation f (x) 0, we assume we

are given an initial estimate x0 of the root x

or a. The iterates are generated by the formula

4.2 Newton Raphson 2nd Order Method

- Increased accuracy obtained by retaining

additional term in Taylor Series expansion - quadratic in ?x1, of general solution

4.2 Newton Raphson 2nd Order Method

- Choice between solutions x and x- determined by

which one results in the function f(x) or f(x-)

being closer to the zero value

4.2 Newton Raphson 2nd Order Method

- One may also treat the original Taylor Series

expansion as another nonlinear equation in ?x and

apply Newton-Raphson for its solution

5. The SECANT Method

5. The SECANT Method

Example 1

- Practice the example in class
- Try to understand the command in each line
- Three different methods XGX.m, LI.m and NR.m
- Discussion of results

Example 1

Example 2

Example1_2.m This program solves the problem

posed in Example 1.2. It calculates the real

gas specific volume from the SRK equation of

state using the Newton-Raphson method for

calculating the roots of a polynomial.

clear clc clf Input data P input(' Input

the vector of pressure range (Pa) ') T

input(' Input temperature (K) ') R 8314

Gas constant (J/kmol.K) Tc input(' Critical

temperature (K) ') Pc input(' Critical

pressure (Pa) ') omega input(' Acentric

factor ')

Example 2

Constants of Soave-Redlich-Kwong equation of

state a 0.4278 R2 Tc2 / Pc b 0.0867

R Tc / Pc sc -0.15613, 1.55171, 0.48508 s

polyval(sc,omega) alpha (1 s (1 -

sqrt(T/Tc)))2 A a alpha P / (R2

T2) B b P / (R T) for k 1length(P)

Defining the polynomial coefficients coef

1, -1, A(k)-B(k)-B(k)2, -A(k)B(k)

v0(k) R T / P(k) Ideal gas specific

volume vol(k) NRpoly(coef , 1) R T /

P(k) Finding the root end

Example 2

Show numerical results fprintf('\nRESULTS\n')

fprintf('Pres. 5.2f Ideal gas vol.

7.4f',P(1),v0(1)) fprintf(' Real gas vol.

7.4f\n',vol(1)) for k1010length(P)

fprintf('Pres. 5.2f Ideal gas vol.

7.4f',P(k),v0(k)) fprintf(' Real gas vol.

7.4f\n',vol(k)) end plotting the

results loglog(P/1000,v0,'.',P/1000,vol) xlabel('P

ressure, kPa') ylabel('Specific Volume,

m3/kmol') legend('Ideal','SRK')

Example 2

function x NRpoly(c,x0,tol,trace) NRPOLY Finds

a root of polynomial by the Newton-Raphson

method. NRPOLY(C,X0) computes a root of the

polynomial whose coefficients are the

elements of the vector C. If C has N1

components, the polynomial is C(1)XN ...

C(N)X C(N1). X0 is a starting point.

NRPOLY(C,X0,TOL,TRACE) uses tolerance TOL

for convergence test. TRACE1 shows the

calculation steps numerically and TRACE2

shows the calculation steps both numerically

and graphically. See also ROOTS,

NRsdivision, NR

Example 2

(c) N. Mostoufi A. Constantinides January

1, 1999 Initialization if nargin lt 3

isempty(tol) tol 1e-6 end if nargin lt 4

isempty(trace) trace 0 end if tol 0

tol 1e-6 end if (length(x0) gt 1)

(isfinite(x0)) error('Second argument must be

a finite scalar.') end

Example 2

iter 0 fnk polyval(c,x0) Function if

trace header ' Iteration x

f(x)' disp(header) disp(sprintf('5.0f

13.6g 13.6g ',iter, x0 fnk)) if trace

2 xpath x0 x0 ypath 0 fnk

end end x x0 x0 x .1 maxiter 100

Example 2

Solving the polynomial by Newton-Raphson

method while abs(x0 - x) gt tol iter lt maxiter

iter iter 1 x0 x fnkp

polyval(polyder(c),x0) Derivative if fnkp

0 x x0 - fnk / fnkp Next

approximation else x x0 .01 end

Example 2

fnk polyval(c,x) Function Show the

results of calculation if trace

disp(sprintf('5.0f 13.6g 13.6g ',iter, x

fnk)) if trace 2 xpath

xpath x x ypath ypath 0 fnk

end end end

Example 2

if trace 2 Plot the function and path to

the root xmin min(xpath) xmax

max(xpath) dx xmax - xmin xi xmin -

dx/10 xf xmax dx/10 yc for

xc xi (xf - xi)/99 xf ycyc

polyval(c,xc) end

Example 2

xc linspace(xi,xf,100) ax

linspace(0,0,100) plot(xc,yc,xpath,ypath,xc,ax

,xpath(1),ypath(2),'',x,fnk,'o') axis(xi xf

min(yc) max(yc)) xlabel('x')

ylabel('f(x)') title('Newton-Raphson The

function and path to the root ( initial guess

o root)') end if iter maxiter

disp('Warning Maximum iterations reached.') end

Synthetic Division Algorithm

Coefficients in the above two equations can be

related and used for solving bi.

Synthetic Division Algorithm

- Method for removing each real root from an

nth-order polynomial, thereby reducing the

polynomial to degree (n-1). - Each successive application reduces polynomial by

one more degree. - Subsequent algorithm by Lapidus.

Synthetic Division Algorithm

- Consider the fourth-order polynomial
- With x identified to be a root of f(x), the root

can be factored from the polynomial

Synthetic Division Algorithm

- To determine the coefficients (bi) of the

third-degree polynomial, multiply out and group

in descending powers of x

Synthetic Division Algorithm

- Equate coefficients of like powers of x in the

two polynomials forms (original and

factored/expanded), solve for bi

Synthetic Division Algorithm

- Or, in general
- where r 1, 2, , (n-1)
- Whereby the polynomial is reduced one degree to

each jth iteration

The Eigenvalue Method

Eigenvalues are a special set of scalars

associated with a linear system of equations

(i.e., a matrix equation) that are sometimes

also known as characteristic roots,

characteristic values, proper values, or latent

roots. Root-finding methods discussed so far are

not efficient for finding eigenvalues.

The Eigenvalue Method

- A square matrix has a characteristic polynomial

whose roots are called eigenvalues of the matrix.

- Consider the characteristic polynomial
- for which we define the n?n companion matrix A

which contains the coefficients of the original

polynomial

The Eigenvalue Method

The Eigenvalue Method

MATLAB has its own function, roots.m, for

calculating all roots of a polynomial equation in

the above equation. It first converts the

polynomial to the companion matrix A shown above.

It then uses the built-in function eig.m to

calcualte the eigenvalues of the matrix A, which

are also the roots of the polynomial equation.

Newtons Method for Simultaneous Nonlinear

Equations

f1(x1, x2)0, f2(x1, x2)0

Newtons Method for Simultaneous Nonlinear

Equations

Newtons Method for Simultaneous Nonlinear

Equations

Newtons Method for Simultaneous Nonlinear

Equations

Newtons Method for Simultaneous Nonlinear

Equations

f1(x1, xk)0 .. fk(x1, . xk)0

J d -f

Newtons Method for Simultaneous Nonlinear

Equations

J is the Jacobian matrix, d is the correction

vector, and f is the vector functions.

Strongly nonlinear equations likely to diverge,

therefore relaxation is used to stabilize the

iterative process (? typically 0.5)

Example 3

Example on synthetic division or eigenvalue

methods.

Example 4

Example on solution of two equations using

Newtons method.

Problem 1.6

Carbon monoxide from a water gas plant is burned

with air in an adiabatic reactor. Both the

carbon monoxide and air are being fed to the

reactor at 25?C and atmospheric pressure. For

the reaction CO ½O2 ? CO2 the following

standard free energy and enthalpy changes have

been determined ?G?To -257 kJ/(gmol CO)

?H?To -283 kJ/(gmol CO) The standard states

for all components are pure gases at 1 atm

pressure. Calculate the adiabatic flame

temperature and the conversion of CO for the

following cases (a) 0.4 mole O2 per mole CO

(b) 0.8 mole O2 per mole CO

Problem 1.6

The solution of this problem is described in

standard thermodynamic texts under the title

"adiabatic flame temperature. Base all

calculations on one gmole of CO in the feed

gases. Let no represent the number of moles of

O2 per mole of CO in the feed gases. Then the

number of moles of the various constitutents of

the feed and product gases are as follows

Problem 1.6

For an adiabatic reaction, the reacting system

exchanges no heat with the surroundings. The

enthalpy change for the overall process

(Hproducts Hreactants) must therefore equal

zero. The process can be viewed as taking place

in two stages (1) reaction of the entering gases

at the inlet temperature T, and (2) elevation of

the temperature of the product stream from

reaction temperature Tr to the final adiabatic

flame temperature, T.

Problem 1.6

The enthalpy change of reaction ?HR at Tr per

mole of CO in the feed at conversion fraction z

is given by The enthaply change for the

products ?Hp between temperature Tr and T is

given by

ni is the number of moles of the ith component

gas in the product mixture

Problem 1.6

The heat capacity is given by Combine

with energy balance

Problem 1.6

A pressure-based equilibrium constant can be

defined as At equilibrium, total moles of

gas is Therefore, mole fractions are

Problem 1.6

Substitute mole fractions in KP The

thermodynamic equilibrium constant is

Problem 1.6

The quantity ?GT?/T can be evaluated for any

temperature if it's value at To is known, by

integrating the thermodynamic relationship

Integrate both sides

Problem 1.6

Substituting known numerical values

Problem 1.6

Rewrite with constant of integration, C1

where C1 is given by Substitute into

where

Problem 1.6

One may now evaluate ?GT? As such, it is

not possible to evaluate the thermodynamic

equilibrium constant Ka at any value of T

Problem 1.6

For each constituent a f/fo, where a is the

thermodynamic activity, f is the fugacity of the

component, and fo is the fugacity of the

component in the standard state

Problem 1.6

Ka and Kp must be identical, such that

coupled with energy balance from earlier

Problem 1.6

Note that once values for parameters T0, Tr,

?H0Tr , ?G0T0, P and n0 have been supplied, the

above two equations contain only two unknowns z

and T. The problem then is to solve this set of

two simultaneous equations for z and T, given the

following values from the problem statement

T0 Tr298.15 K (25 C) ?H0T -283000J/gmol

CO ?G0T -257000J/gmol CO

n0 , the O2/CO ratio, and P are variables but

they assume fixed values for a solution.

Problem 1.6

Newtons method for simultaneous nonlinear

algebraic equations is used for the solution of

the two equations. The MATLAB function Newton.m

(developed in Example 1.4) can be used for the

solution of this problem. Program Probl_6.m is

used to get the input from the keyboard and

display the results. The set of equations to be

solved are also introduced in the MATLAB function

P1_6func.m.

Problem 1.6

Probl_6.m Probl6.m This program solves

Problem 1.6 using Newton.m. clear clc fnazne

input( Name of the function containing the set

of equations ) disp( ) repeat 1 while

repeat Initial data nO input( Ratio of

moles of 02 to moles of CO zO input(

Initial guess of conversion (z) TO input(

Initial guess of flame temperature (K) 1 end

Solution x, m Newton(fname, zO, TO), ,

, nO) disp( ) fprintf( z 5.4f \n,x(l))

fprintf( T 5.lf K\n,x(2)) fprintf(

Solution reached after 2d iterations.\n,m)

disp( 9 repeat input( Do you want to try

again (0/1)? ) disp( )

Problem 1.6

function f P1_6_func(x, n0) z x(1)

Conversion T x(2) Temperature (K) R 8.314

Gas constant CJ/gmol.K) P 1 Pressure

(atm) dGo -257e3 dHo -283e3 To

25273.15 C1 dHo10.32To-10.355e-3To22.29

2e-6To3 C2 -C1/To-10.32log(To)10.355e-3To-

1.146e-6To2 C3 C2dGo/To dGoT

C1/T10.32log(T)-10.355e-3T1.146e-6T2C3

a 26.16(T-To)8.75e-3(T2-To2)/2-1.92e-6(T

3-To3)/3 b 25.66(T-To)12.52e-3(T2-To2)/2-

3.37e-6(T3-To3)/3 c 28.67(T-To)35.72e-3(T

2-To2)/2-10.39e-6(T3-To3)/3 d

26.37(T-To)7.61e-3(T2-To2)/2-1.44e-6(T3-To

3)/3 f(1) z/(1-z)sqrt((1-z/2100/21n0)/((n0

-z/2)P))-exp(-dGoT/R) f(2) zdHo(1-z)a(n0-z

/2)bzc79/21n0d f f' It has to be a

column vector.

Problem 1.6

Input and Results gtgt Probl_6 Name of the

function containing the set of equations

P1_6_funcRatio of moles of O2 to moles of CO

0.4 Initial guess of conversion (z) 0.7

Initial guess of f lame temperature (K) 2000 z

0.7577 T 2433.8 K Solution reached after 9

iterations Do you want to try again (0/1)? 1

Ratio of moles of O2 to moles of CO 0.8

Initial guess of conversion (z) 0.99 Initial

guess of flame temperature (K) 2000 z 0.9901

T 2017.6 K Solution reached after 6

iterations. Do you want to try again (0/1)? 0

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For a small fee you can get the industry's best online privacy or publicly promote your presentations and slide shows with top rankings. But aside from that it's free. We'll even convert your presentations and slide shows into the universal Flash format with all their original multimedia glory, including animation, 2D and 3D transition effects, embedded music or other audio, or even video embedded in slides. All for free. Most of the presentations and slideshows on PowerShow.com are free to view, many are even free to download. (You can choose whether to allow people to download your original PowerPoint presentations and photo slideshows for a fee or free or not at all.) Check out PowerShow.com today - for FREE. There is truly something for everyone!

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For a small fee you can get the industry's best online privacy or publicly promote your presentations and slide shows with top rankings. But aside from that it's free. We'll even convert your presentations and slide shows into the universal Flash format with all their original multimedia glory, including animation, 2D and 3D transition effects, embedded music or other audio, or even video embedded in slides. All for free. Most of the presentations and slideshows on PowerShow.com are free to view, many are even free to download. (You can choose whether to allow people to download your original PowerPoint presentations and photo slideshows for a fee or free or not at all.) Check out PowerShow.com today - for FREE. There is truly something for everyone!

presentations for free. Or use it to find and download high-quality how-to PowerPoint ppt presentations with illustrated or animated slides that will teach you how to do something new, also for free. Or use it to upload your own PowerPoint slides so you can share them with your teachers, class, students, bosses, employees, customers, potential investors or the world. Or use it to create really cool photo slideshows - with 2D and 3D transitions, animation, and your choice of music - that you can share with your Facebook friends or Google+ circles. That's all free as well!

For a small fee you can get the industry's best online privacy or publicly promote your presentations and slide shows with top rankings. But aside from that it's free. We'll even convert your presentations and slide shows into the universal Flash format with all their original multimedia glory, including animation, 2D and 3D transition effects, embedded music or other audio, or even video embedded in slides. All for free. Most of the presentations and slideshows on PowerShow.com are free to view, many are even free to download. (You can choose whether to allow people to download your original PowerPoint presentations and photo slideshows for a fee or free or not at all.) Check out PowerShow.com today - for FREE. There is truly something for everyone!

For a small fee you can get the industry's best online privacy or publicly promote your presentations and slide shows with top rankings. But aside from that it's free. We'll even convert your presentations and slide shows into the universal Flash format with all their original multimedia glory, including animation, 2D and 3D transition effects, embedded music or other audio, or even video embedded in slides. All for free. Most of the presentations and slideshows on PowerShow.com are free to view, many are even free to download. (You can choose whether to allow people to download your original PowerPoint presentations and photo slideshows for a fee or free or not at all.) Check out PowerShow.com today - for FREE. There is truly something for everyone!

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