Massoud M. Engineering Thermofluids: Thermodynamics, Fluid Mechanics, and Heat Transfer

3. Numerical Solution of Partial Differential Equations 1003

Solution of One-Dimensional Wave Equation by Method of D’Alembert

The D’Alembert method, like the Laplace transform and the method of separation

of variables, provides an analytical solution to the one-dimensional wave equation.

It can be easily shown that a function in the form of y(x, t) = f

1

(x + ct) + f

2

(x – ct)

is a solution to Equation VIIe.3.15. For this equation to be the final solution, it

must satisfy the initial and the boundary conditions. For the initial conditions of

y(x, 0) = f(x) and

)(/)0,( xgtxy =∂∂

, we find the solution as:

³

+

−

+−++=

ctx

dssg

c

ctxfctxftxy )(

2

1

)]()([

2

1

),( VIIe.3.17

where s is a dummy variable. For example, consider a string, fixed at the end

points, having a length of 4 feet. The wire density and the tension force are so that

c = 2. Choosing a length increment of 1 ft, gives the optimum time step of 0.5

seconds. The string is now pulled to vibrate with an initial velocity of:

L

x

t

xy

π

sin

)0,(

=

∂

We want to find the displacement of a point located 1/5 of the length from the left

end, 1 second after the string is disturbed. To solve the problem, we use Equa-

tion VIIe.3.17 as follows:

»

¼

º

«

¬

ª

»

¼

º

«

¬

ª

+−

¸

¹

·

¨

©

§

−=

µ

¶

´

=

+

−

L

t

L

x

L

t

L

xL

ds

L

s

c

y

ctx

ππππ

π

2

cos

2

cossin4

2

1

From here, we find y for x = L/5 ft and t = 1 second as:

()

fty 296.1104.0913.0

4

15

8

cos

15

2

cos

4

35

cos

35

cos

4

=+=

¸

¹

·

¨

©

§

−=

»

¼

º

«

¬

ª

¸

¹

·

¨

©

§

+−

¸

¹

·

¨

©

§

−=

π

ππ

π

ππππ

π

The reader should use the finite-difference method and compare the results.

Solution of the Two-Dimensional Wave Equation

Displacement of a vibrating membrane can be predicted by the solution to the

two-dimensional wave equation:

2

22

2

),,(1),,(),,(

t

tyxu

cy

tyxu

x

tyxu

∂

=

∂

+

∂

This equation can be solved explicitly by substituting for the derivative terms from

the central-difference approximation.

2

1

,,

1

,

22

1,,1,

2

,1,,1

)(

2

1

)(

2

)(

2

t

uuu

cy

uuu

x

uuu

k

ji

k

ji

k

ji

k

ji

k

ji

k

ji

k

ji

k

ji

k

ji

∆

+−

=

∆

+−

+

∆

+−

+−+−

1004 VIIe. Engineering Mathematics: Numerical Analysis

If the increments along the x and the y-axis are equal, we get:

k

ji

k

ji

k

ji

k

ji

k

ji

k

ji

k

ji

u

x

tc

uuuuu

x

tc

u

,

2

22

1

,1,1,,1,1

2

22

1

,

]

)(

42[)(

)(

∆

−+−+++

∆

=

−

+−+−

+

To further simplify this relation, we may choose the spatial and the temporal in-

crements so that the last term vanishes and we simply get:

1

,1,1,,1,1

1

,

)(

2

1

−

+−+−

+

−+++=

k

ji

k

ji

k

ji

k

ji

k

ji

k

ji

uuuuuu

This equation applies when 2/1)/()(

222

=∆∆ xtc . Similar to the one-dimensional

wave problem, to find displacement corresponding to k – 1 when we begin the

first time step, we take advantage of the specified initial velocity. If the initial ve-

locity is zero, then displacement corresponding to the first time step is obtained

from:

)(

4

1

0

1,

0

1,

0

,1

0

,1

1

,

+−+−

+++=

jijijijiji

uuuuu

4. The Newton–Raphson Method

As was discussed in Chapter VIId, the mathematical modeling of most physical

phenomena reduces to a set of simultaneous differential equations the solution of

which would determine the parameters of interest. The solution to such a set of

equations involves the approximation of the differential terms by finite difference,

for example, and then linearization of the nonlinear terms. The Newton-Raphson

method is most often used for the transformation of a set of differential equations.

Consider the following set of first order non-linear differential equations, con-

sisted of N equations and N unknowns:

()

¸

¹

·

¨

©

§

=

¸

¹

·

¨

©

§

NN

N

yyytF

y

dt

d

,...,,

.

,...,,

.

21

212

211

2

1

VIIe.4.1

Functions in the right side of the above set can be linearized. To do this, each

function is expanded in Taylor’s series and then approximated by using only the

first derivative term:

()

(

)

[

]

()

n

i

n

i

N

i

n

N

nnn

i

n

N

nnn

iNi

yy

y

yyytF

yyytFyyytF −

∂

+=

+

¦

1

21

2121

,...,,,

,...,,,,...,,,

4. The Newton–Raphson Method 1005

We now substitute the approximated functions in the right side of Equa-

tion VIIe.4.1. We also expand the left side of the set of differential equations by

using a discrete difference at consecutive time steps. Equation VIIe.4.1 can then

be written as:

()

¸

¹

·

¨

©

§

−

¸

¹

·

¨

©

§

∂∂∂∂∂∂

∆+

¸

¹

·

¨

©

§

∆=

¸

¹

·

¨

©

§

−

+

n

N

n

N

nn

N

n

N

n

N

n

N

nnn

N

nnn

n

N

nnn

N

n

N

nnn

n

N

nnn

n

N

n

N

nn

yy

yFyFyF

t

yyytF

t

yy

1

2

1

2

1

21

22212

12111

21

212

211

1

2

1

2

1

.

..

.

..

,...,,

.

,...,,

.

VIIe.4.2

The matrix on the right hand side of Equation VIIe.4.2, which contains the partial

derivative terms, is called the Jacobian matrix. The above set can now be solved

for the increment in each variable. To do this, the Jacobian matrix multiplied by

the time step should be deducted from the unity matrix. Final answer is obtained

as:

()

1

21

22212

12111

21

212

211

1

2

1

2

1

..

.

..

1..11

..

1..11

,...,,

.

,...,,

.

−

+

°

¿

°

¾

½

°

¯

°

®



¸

¹

·

¨

©

§

∂∂∂∂∂∂

∆−

¸

¹

·

¨

©

§

¸

¹

·

¨

©

§

∆=

¸

¹

·

¨

©

§

−

N

n

N

n

N

n

N

nnn

N

nnn

n

N

nnn

N

n

N

nnn

n

N

nnn

n

N

n

N

nn

yFyFyF

t

yyytF

t

yy

Let’s demonstrate the application of the Newton-Raphson linearization method by

solving a simple example.

Example. Find the friction factor from:

2

10

Re

7.182

log274.1

−

»

¼

º

«

¬

ª

¸

¹

·

¨

©

§

+−=

f

D

f

e

ε

Data: D

e

= 1 in,

ε

= 0.004 in, Re = 1.0E7.

Solution: Since f appears in both sides of the correlation, we use the Newton-

Raphson method where:

0

Re

7.182

log274.1

2

10

=

»

¼

º

«

¬

ª

¸

¹

·

¨

©

§

+−−=

−

f

D

fF

e

ε

we then expand F as:

0)()( =−+=

dx

dF

xxxFF

oo

Solving for x, we find:

1006 VIIe. Engineering Mathematics: Numerical Analysis

dxxdF

xF

xx

o

/)(

)(

−=

This procedure is summarized in the following FORTRAN program.

implicit real*8(a-h,o-z)

data a1,a2,eps,De,Re,e/1.74,-2.00,4.00E-3,1.0,1.0E7,1.e-8/

f=0.1

1 continue

i=i+1

a3=2.00*eps/De

a4=18.7/Re

W=a3+(a4/sqrt(f))

Wp=-a4*(f**(-1.5))/2.00

V=a1+a2*alog10(W)

Vp=0.4343*a2*Wp/W

U=1./(V*V)

Up=-2.00*Vp/(V*V*V)

Fof=f-U

dFdf=1.00-Up

error=Fof/dFdf

f=f-error

if(abs(error).le.e) go to 3

if(i.gt.30) go to 2

go to 1

2 continue

Print *,'Iteration did not converge'

go to 4

3 continue

Write(*,*) i,f

4 continue

9 format(i5,f15.9)

stop

end

Using the above data, the program finds the answer after 3 iterations as 0.0284.

5. Curve Fitting to Experimental Data

In many engineering applications, we prefer to use an equation to represent a set

of experimental data. It is therefore our goal here to represent a set of experimen-

tal data by a curve that would best fit the data. The simplest case is fitting a line

between points in a set of data. In general however, the experimental data are

such that a nonlinear curve needs to be found to best fit the data. The most widely

used technique for curve fitting is the method of least squares. A polynomial of

degree n passes exactly through a set of N data points, if n = N – 1. Hence, in fit-

ting polynomials to a set of experimental data, we require n < N –1.

5.1. Regression Analysis, the Method of Least Squares

Since polynomials can be readily manipulated, to describe the method of least

squares, we consider the case of fitting a polynomial to a specified set of data. In

5. Curve Fitting to Experimental Data 1007

this method, the goal is to find the coefficients of a function of a single variable in

the form of:

¦

=

M

i

ii

xfcxf

1

)()( VIIe.5.1

to fit the data pairs (x

1

, y

1

), (x

2

, y

2

), ! (x

N

, y

N

) so that the summation of the square

of the errors as in the following summation:

[]

2

1

)(

¦

=

−=

N

i

ii

yxfE

is minimized. Since functions f

i

(x) comprising the function f(x) are known, the

unknown coefficients c

i

should then be determined. This is accomplished by set-

ting the derivative of E with respect to the unknown coefficients (i.e. c

i

) to zero:

[]

{}

¦

=−+++=

∂

=

N

i

iiiMMii

xfyxfcxfcxfc

c

E

1

12211

1

0)()()()(2 "

[]

{}

¦

=

=−+++=

∂

N

i

iiiMMii

xfyxfcxfcxfc

c

E

1

12211

2

0)()()()(2 "

[]

{}

¦

=

=−+++=

∂

N

i

iiiMMii

M

xfyxfcxfcxfc

c

E

1

12211

0)()()()(2 "

This results in the following set of M equations and M unknowns:

»

¼

º

«

¬

ª

=

»

¼

º

«

¬

ª

»

¼

º

«

¬

ª

¦

¦¦¦

=

===

N

i

iiM

N

i

ii

N

i

ii

M

N

i

iMiM

N

i

iiM

N

i

iiM

N

i

N

i

iMi

N

i

iiii

N

i

iMi

N

i

ii

N

i

ii

yxf

c

xfxfxfxfxfxf

1

2

1

2

1

11

2

1

11

2

1

2212

1

21

1

11

)(

)()()()()()(

#

!

###

!

VIIe.5.2

This set can be solved for the unknown coefficients c

i

using the methods described

in Chapter VIId. We may now apply the set given in Equation VIIf.4.2 to the fol-

lowing simple polynomials with linear terms:

1008 VIIe. Engineering Mathematics: Numerical Analysis

n

xaxaxaay ++++= "

2

210

Functions f

i

then become f

1

= a

0

, f

2

= a

1

x, etc. Substituting into Equation VIIe.5.2,

we obtain:

»

¼

º

«

¬

ª

=

»

¼

º

«

¬

ª

»

¼

º

«

¬

ª

¦

¦¦¦¦

¦¦¦

=

==

+

=

+

=

+

===

N

i

n

i

N

i

ii

N

i

n

N

i

n

i

N

i

n

i

N

i

n

i

N

i

n

i

N

i

n

i

N

i

N

i

N

i

N

i

n

i

N

i

N

i

yx

y

a

xxxx

xxxN

1

0

1

2

1

2

1

3

1

2

1

11

2

1

#

"

#####

"

VIIe.5.3

Note that the degree of the polynomial N is not known before hand. Rather, N

should be found by iteration. The criterion is to choose N to minimize the stan-

dard deviation.

Example VIIe.5.1. Find the coefficients of a polynomial to fit a portion of the

steam tables, in the range of 100 psia

≤ P ≤ 122 psia, for saturated steam pressure

as a function of steam temperature (F):

Solution: We fit polynomials of degree 1, 2, 3, etc. The results of this polyno-

mial fit in the form of:

"+++=

2

210

TaTaaP

are tabulated below:

As the trend in this table indicates, the best fit occurs at N = 3. Hence, the answer

is found as:

32

)42181.0(01386.0438.37.305 TETTP −+−+−=

5. Curve Fitting to Experimental Data 1009

We may use this equation to find saturation pressure corresponding to T

sat

=

338.08 F. According to the above fit

P

sat

= –305.7 + 3.438 × 338.08 –0.01386 × (338.08)

2

+ (0.2181E-4) × (338.08)

3

=

115.2 psia

The corresponding pressure from the steam tables is P = 115 psia, resulting in an

error of less than 0.2%.

The most important task in curve fitting by the method of least squares is the

selection of the type of functions (i.e. f

i

(x)) comprising function f(x). The function

may be in any of the following forms:

¦

=

+=

n

i

xBAy

1

,

¦

=

+=

n

i

xBAy

1

)ln( ,

¦

=

+=

n

i

xBAy

1

)ln()ln( ,

¸

¹

·

¨

©

§

+=

¦

=

n

i

xBAxy

1

/ ,

¦

=

+=

n

i

xBAy

1

/ ,

¦

=

+=

n

i

xBAy

1

/)ln( ,

¸

¹

·

¨

©

§

+=

¦

=

n

i

xBAy

1

/1 ,

¦

=

+=

n

i

xBAy

1

.

Polynomial fits to data (similar to the first equation shown above) can be made by

using the regression analysis available on the accompanying CD-ROM.

VIII. Appendices

Table of Contents

Appendix I. Unit Systems, Constants, and Numbers ................................... 1013

Table A.I.1. Primary SI Units .......................................................................... 1014

Table A.I.2. Derived SI Units ......................................................................... 1014

Table A.I.3. Associated SI Units ..................................................................... 1014

Table A.I.4. Common Physical Quantities

in Two Systems of Dimensions................................................................ 1015

Table A.I.5. Physical Constants ....................................................................... 1016

Table A.I.6. Dimensionless Numbers .............................................................. 1017

Table A.I.7. Multiples of SI Units.................................................................... 1017

Table A.I.8. Unit Conversion Tables ............................................................... 1018

Appendix II. Thermodynamic Data.............................................................. 1023

Table A.II.1(SI). Saturated Water

and Dry Saturated Steam Properties, f(P) ................................................ 1024

Table A.II.2(SI). Saturated Water

and Dry Saturated Steam Properties, f(T)................................................. 1028

Table A.II.3(SI). Superheated Steam Properties .............................................. 1032

Table A.II.4(SI). Subcooled Water Properties ................................................. 1036

Table A.II.5(SI). Properties of Various Ideal Gases......................................... 1037

Table A.II.1(BU). Saturated Water

and Dry Saturated Steam Properties, f(T)................................................. 1038

Table A.II.2(BU). Saturated Water

and Dry Saturated Steam Properties, f(P) ................................................ 1040

Table A.II.3(BU). Superheated Steam Properties ............................................ 1042

Table A.II.4(BU). Subcooled Water Properties ............................................... 1046

Table A.II.5(BU). Properties of Various Ideal Gases....................................... 1047

Table A.II.6. Examples of Least-Square Fit to Saturated Water

and Dry Saturated Steam.......................................................................... 1047

Appendix III. Pipe and Tube Data................................................................. 1049

Table A.III.1(SI). Commercial Steel Pipe (Schedule Wall Thickness)............ 1050

Table A.III.2(SI). Commercial Steel Pipe (Nominal Pipe Size, NPS) ............. 1051

Table A.III.3(SI). Tube Data, Birmingham Gauges

to millimeter and inches........................................................................... 1053

Table A.III.1(BU). Pipe Data, Carbon & Alloy Steel ...................................... 1054

Table A.III.2(BU). Tube Data.......................................................................... 1055

1012 VIII. Appendices

Table A.III.4. Navier-Stokes Equations

in the Cylindrical Coordinate System ...................................................... 1056

Table A.III.5. Navier-Stokes Equations

in the Spherical Coordinate System ......................................................... 1056

Table A.III.6. Substantial Derivative and Flow Acceleration Components

(Cylindrical Coordinates)......................................................................... 1057

Table A.III.7. Substantial Derivative and Flow Acceleration Components

(Spherical Coordinates)............................................................................ 1057

Appendix IV. Thermophysical Data............................................................. 1059

Table A.IV.1(SI). Thermophysical Properties

of Selected Metallic Solids....................................................................... 1060

Table A.IV.2(SI). Thermophysical Properties

of Selected Nonmetallic Solids ................................................................ 1064

Table A.IV.3(SI). Thermophysical Properties

of Common Materials at 300 K................................................................ 1066

Table A.IV.4(SI). Thermophysical Properties

of Gases at Atmospheric Pressure............................................................ 1072

Table A.IV.5(SI). Thermophysical Properties

of Saturated Water and Saturated Steam.................................................. 1077

Table A.IV.6(SI). Thermophysical Properties of Liquid Metals...................... 1079

Table A.IV.4(BU). Thermophysical Properties of Gases

at Atmospheric Pressure........................................................................... 1080

Table A.IV.5(BU). Thermophysical Properties of Saturated Water ................ 1082

Table A.IV.6(BU). Thermophysical Properties of Saturated Steam ................ 1082

Table A.IV.7(BU). Thermophysical Properties of Superheated Steam............ 1083

Table A.IV.8(BU). Thermal Properties of Solid Dielectrics

at Normal Temperature ............................................................................ 1084

Table A.IV.9(BU). Normal, Total Emissivity of Metallic Surfaces................. 1086

Table A.IV.10(BU). Normal, Total Emissivity of Non-Metallic Surfaces ....... 1088

Appendix V. Nuclear Properties of Elements ............................................... 1091

Table A.V.1(SI). Absorption Coefficient of Gamma Rays.............................. 1092

Table A.V.2(SI). Cross Sections for Neutron Interaction ................................ 1093

Appendix I

Unit Systems, Constants, and Numbers

Table A.I.1. Primary SI Units .......................................................................... 1014

Table A.I.2. Derived SI Units ......................................................................... 1014

Table A.I.3. Associated SI Units ..................................................................... 1014

Table A.I.4. Common Physical Quantities

in Two Systems of Dimensions................................................................ 1015

Table A.I.5. Physical constants ........................................................................ 1016

Table A.I.6. Dimensionless Numbers .............................................................. 1017

Table A.I.7. Multiples of SI Units.................................................................... 1017

Table A.I.8. Unit Conversion Tables ............................................................... 1018

Massoud M. Engineering Thermofluids: Thermodynamics, Fluid Mechanics, and Heat Transfer

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