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Application of Series in Heat Transfer: transient heat conduction

By

Alain Kassab

Mechanical, Materials and Aerospace EngineeringUCF EXCEL Applications of Calculus

Application of Series in Heat Transfer-

transient heat conduction


Part I –

background and review of series (Monday 14, April 2008)

1. Taylor and Maclaurin series (section 12.10).2. Fourier series.

Part II –

applications (Monday 21, April 2008)

1. Transient heat conduction: a. Application of Maclaurin

series to heat transfer: quenching of a metal bar at early times.

b. Application of Fourier series to heat transfer:quenching of a metal bar at later times.

2. Finite Difference: computing the heat flux at the wall.3. Applications to computational fluid dynamics and heat

transfer.


1. Transient Heat Conduction: introduction andbackground

Heat conduction: mode of energy transport in solids.

energy is transported by:

1. free electrons in electrical conductors: metals.2. atomic lattice vibrations (phonons) in electrical

insulators.3. free electrons and phonons electrical

semi-conductors.


First explained and characterized mathematically by J.B. Fourier

in his famous 1822 treatise Théorie Analytique de la Chaleure (the Analytical Theory of Heat).

1. Fourier “law”

of heat conduction:

heat (q) flow (energy/time) in a conductingmedium is proportional to the temperature (T) gradient and the area (A) through which heat flows

2. the proportionality constant is called the thermal conductivity, k [W/mK], and isparticular to the material and often depends on temperature at which it ismeasured

3. the relationship has a negative sign so that heat flows from high to low temperature (the gradient

points in the direction of maximum increase

of a function)

Note:

the partial derivativeis used here to account fortransient

behavior T(x,t)where t denotes time.

][WdxdTAq∝

0⟩dxdT

X=0 X=L

Area: A

qT1 T(x)

T2

Watt:

W = Joule/s

x=0 x=Lx

T(x)

x=0 x=L

x

T(x)

q >0 q<0

0⟨dxdT

http://en.wikipedia.org/wiki/Image:Fourier2.jpg


1.a

Application of Maclaurin

series to heat transfer: quenching of a metal bar at early times

quenching: a rapid cooling by immersion in water or oil. In metallurgy it is used to prevent phase transformations or diffusion and “lock in”

a phase of interest.

quenching is used in metallurgy to harden steel by rapidly cooling austenite phase of steel, which is a non-magnetic solution of the alloy of iron (Fe) and carbon (C) that forms above 1000K or 727oC,in order to produce what is called the martensite

phase

(hardened steel).

quenching effectively locks in the carbon so that it does not diffuse out of the crystalline structure.

by repeated tempering and quenching skilled blacksmiths control the amount of martensite to craft high quality swords throughout history.

Japanese mastered this process to produce their famousKatana swords that were the weapon of choice of the Samurai.

We wish to study the temperature history in the metal bar that is being worked into what may eventually become such a sword.


Problem statement: Suppose then that we take a long metal barof width, L=10 cm, that is initially at an elevated temperaturewe denote by Ti

=1000°C, and we plunge this extremely hot metal into a large bath of water. The water in contact withthe metal surface will boil at a constant temperature of

100°C,we denote as To

=100°C . The temperature of the bar, we denote as, T(x,t) , will decrease as a function of time, we denote as, t ,and will also vary as a function of space, x , as the heat leaves the metal.

Solution: Using principles of conservation of energy and analyticalmethods you will be exposed to later in courses such as EML 4142, the temperature distribution for the early part of the quenching process is given by:

α

is a property of the solid, it is called the thermal diffusivity, and itcontrols the rate at which heat penetrates or leaves a solid.

Let us now investigate the temperature in the bar.


Computing the temperature with the error function erf(x):

1. Let us take a typical value for the thermal diffusivity, α

, for steel which is aboutα=10-5

[m2/s], and a location close to the surface, say xo

=1 cm.

2. What is the temperature there 1 s after the metal

bar was plunged into the water bath?

3. We compute the argument of the error function for

this combination:

4. Using the Maclaurin

series for erf(x) we compute the error function of 1.581 as:

Let’s compute this with MATHCAD


5. With the value of the error function for x=1.581, we now compute the temperature

at xo

=1cm and after 1s as:

6. Examining the temperature there (at xo

=1cm) after 5 s, we compute the argument of the error function and evaluate the error function:

Note:

as the argument, x, of erf(x) gets closer to the expansion point of series (zero), we needfewer terms to converge!



0 cm 1 cm 2 cm 3 cm 4 cm 5 cm 6 cm0

500oC

1000oC

T x 1s,( )

T x 5s,( )

T x 15s,( )

T x ,( )

7 cmx

25s

x=0

To

=100oC

x

100oC

steelbar

In such a manner, we can determine the temperature at various locations in the bar for various times and a plot of these results is:

δ(5 s)Note: the effect of the cooling from the water bath takes time to befelt at the interior of the bar. That is called the penetrationdepth δ(t),

and it is controlled by the parameter δ(1 s)



The solution we have been considering is valid until the penetration depth reaches the midpoint of the bar, L= 5cm. When does this happen? We have to agree how to define

that the midpoint has felt the cooling effect of the water. One way to do this is to use behavior of the error function:

1) we know that for x = 2.576 the value of erf(2.576) = 0.999.2) then at this value T(x=L/2,t) ~ 0.999 Ti ; that is the cold

temperature at x=0 is just being felt

at the centerline.3) the corresponding time is then found by solving for

t at that location:

After the penetration depth reaches the centerline, we then have

to use another solution for the temperature, and this is in terms of what is called a Fourier series. This part of the solution will then comprise our second example of series applications in heat transfer.

or, we find that the time for the penetration depth*

to reach the midpoint of the bar is

erf 0.5( ) 0.52= erf 2( ) 0.995= erf 2.5( ) 1=

0 0.5 1 1.5 2 2.5 3 3.5 40

0.2

0.4

0.6

0.8

erf x( )

x

*Note:

we could relax this definition and use a lowervalue for x. For instance, for x=1.575 then erf(1.575)=0.974 which corresponds to a time of 25s at which the centerlinetemperature is ~98% of the initial temperature according tothe error function solution (not a good idea ~ see later).


1.b Application of Fourier series to heat transfer: quenching of

a metal bar at later times

By "later times," we mean after the middle of the bar has felt the coolingeffects of the water bath. The solution for the temperature at any location

in the bar and at any time after this is given by the infinite series,

X=0 X=L

T=100oCT=100oC

x

steel bar

T(x,t)

X=0 X=L

T=100oCT=100oC

x

steel bar

T(x,t)

where the coefficients are given by

Integrating and using the fact that ,

n13579

1113151719212325

= cn1.2730.4240.2550.1820.1410.1160.0980.0850.0750.0670.0610.0550.051

=

0 5 10 15 20 250

0.5

1

1.5

cn

n


This Fourier series for the temperature converges slowly when the non-dimensional value of time that appears in the exponential called the Fourier number, Fo, is less than 0.2, where

Fo

< 0.2 slow convergenceFo

> 0.2 fast convergence

In our case, our bar is L=10cm and α=10-5m2/s, then a Fourier number of Fo=0.2 corresponds to a time of:

For times larger than Fo=0.2, or 200s in our case, the series converges rather quickly, and you can obtain the solution pretty accurately by using only one term!

Example:

the temperature at the midpoint at 200s is using one term


N 1 3, M..:=

0 10 20 30259.180424

259.180426

259.180428

259.18043

259.180432

259.180434

T1 xo 200, N,( )

N

n13579

1113151719212325

= T1 xo 200, N,( )259.18259.18259.18259.18259.18259.18259.18259.18259.18259.18259.18259.18259.18

= Terms xo t1, n,( )0.995

-0.0465.333·10 -4

-1.021·10 -6

2.957·10 -10

-1.251·10 -14

0000000

=

Example:

the temperature at the midpoint at 200s is using one term

Taking more terms does not improve the solution much at all


N 1 3, M..:=

10 20 30950

960

970

980

990

1000

T1 xo t1, N,( )

N

n13579

1113151719212325

= T1 xo t1, N,( )995.354953.896954.376954.375954.375954.375954.375954.375954.375954.375954.375954.375954.375

= Terms xo t1, n,( )0.995

-0.0465.333·10 -4

-1.021·10 -6

2.957·10 -10

-1.251·10 -14

0000000

=

Example:

the temperature at the midpoint at 25s which corresponds to a

Fourier number of Fo=0.025


N 1 3, M..:=

10 20 30900

1000

1100

1200

T1 xo t1, N,( )

N

n13579

1113151719212325

= T1 xo t1, N,( )1.191·10 3

945.751.012·10 3

997.9061·10 3

999.9791·10 3

999.999999.999999.999999.999999.999999.999

= Terms xo t1, n,( )1.212

-0.2720.074

-0.0162.598·10 -3

-2.953·10 -4

2.339·10 -5

-1.279·10 -6

4.794·10 -8

-1.228·10 -9

2.145·10 -11

-2.546·10 -13

2.052·10 -15

=

Example:

the temperature at the midpoint at 5s which corresponds to a Fourier number of Fo=0.005


0 2 cm 4 cm 6 cm 8 cm 10 cm0

200oC

400oC

600oC

800oC

1000oC

1200oC

T x 1s,( )

T x 5s,( )

T x 15s,( )

T x 25s,( )

T x 50s,( )

T x ,( )

x

100s

0 2 cm 4 cm 6 cm 8 cm 10 cm0

200oC

400oC

600oC

800oC

1000oC

1200oC

T x 1s,( )

T x 5s,( )

T x 15s,( )

T x 25s,( )

T x 50s,( )

T x ,( )

x

100s

Doing this for various times and x-values we can develop the temperature history in the bar

X=0 X=L

T=100oCT=100oC

x

steel bar

T(x,t)

X=0 X=L

T=100oCT=100oC

x

steel bar

T(x,t)



Check solution: at time t=0

the temperature should be the initial temperature, Ti

. Let’s check this,

t=0

T(x,0)=To

+ ( Ti

–To ) = Ti

Note:

series representsx=1 on x Є

[0,0.1] and does so periodically

Checks!!

0 0.02 0.04 0.06 0.08

1

0

1

1

500

n

C n( ) sinn π⋅Lx

xx⋅⎛⎜⎝

⎞⎟⎠

⋅⎛⎜⎝

⎞⎟⎠

∑=

xx

sine series for 1 on the interval [0,0.1]

C n( )2

n π⋅1 1−( )n−⎡⎣ ⎤⎦⋅:=


0 m 0.02 m 0.04 m 0.06 m 0.08 m 0.1 m0

500oC

1000oC

1500oC

T (x,0)

T(x,0)

T(x,0)

T(x,0) N=400

x

N=200

N=50N=20

0 m 0.02 m 0.04 m 0.06 m 0.08 m 0.1 m0

500oC

1000oC

1500oC

T (x,0)

T(x,0)

T(x,0)

T(x,0) N=400

x

N=200

N=50N=20


2. Finite Difference: computing the heat flux at the wall.

Taylor series and their use in finite difference methods (FDM). a method to approximate derivatives and why we can't take limits to zero on the computer.

FDM is utilized routinely in research and commercial computer programs to approximate various derivatives that appear in the equations that govern the behavior of the dependentvariable(s) of the problem of interest; in our case, the dependent variable is the temperature.

Such equations are derived from the conservation principle (balance) typically applied to mass, linear momentum, energy, and species, and they are called differential equations.

You will learn how to solve such differential equations in MAP 3032 and in most of yourengineering courses, and you will encounter finite differences in EML 3034 and EML 4142.

The reason that finite differences are used is that although we can solve certain differentialequations analytically, we cannot do so in many cases especially when the geometry is complicated or the problem is non-linear.

We simply focus on one of the basic devices utilized in FDM: approximation of the first derivative using one sided differences to compute the heat flux at the wall.


develop and apply a finite difference method to estimate the heat flux (heat flow per unit area), qf , through the walls of our quenched bar.

X=0 X=L

T=100oCT=100oC

x

steel bar

T(x,t)

X=0 X=L

T=100oCT=100oC

x

steel bar

T(x,t)

qf qf

Point of departure for the finite difference method is the Taylor series

Which is re-arranged to solve for the 1st

derivative

approximate this derivativeusing FDM

location at which we take x can be anywhere to the right or to the left of the point xo .


Approximate the derivative of temperatureat the left wall of the bar, that is at the point xo

=0.

Since there is no solid to the left of xo

=0, we then take x=xo

+ Δx with Δx

denoting a distance(usually small) to the right of xo

x=0 x=L

xxo

+Δxxo

Δx

qf

Note as we learned in Calculus, if we take the limit Δx 0 we retrieve exactly the definition of the first derivative of f(x) at xo

! We will see what happens when we try to take this limit on the computer

Dropping (truncating) the terms as indicated we find the following approximate formula for the first derivative:

Forward Finite DifferenceFirst order

(truncation error (TE) is proportional to Δx)

truncation error is order of

Δxbig “O”

notation introduced byLev Landau

Lev Davidovich

Landau1908-1968 (Nobel Prizein Physics 1962)


Applying the forward difference to estimate the heat flux at the

left wall,

Let us choose a time to

=250s and estimate the heat flux at the left wall utilizing the above first order forward finite difference, using various values

We expect the estimate for the heat flux to approach the exact value as Δx approaches zero.

We can compute the exact solution using a one term solution since Fo

> 0.2, and

And we can computer the error in the finite difference as a function of Δx as


Plot of the temperature at t=250s Δxi0.01

1·10 -3

1·10 -4

1·10 -5

1·10 -6

1·10 -7

1·10 -8

1·10 -9

1·10 -10

1·10 -11

1·10 -12

1·10 -13

1·10 -14

1·10 -15

0

= T Δxi( )130.030067588828103.052476838824100.305297398701100.030529789587100.003052979008100.000305297901100.00003052979

100.000003052979100.000000305298100.00000003053

100.000000003053100.000000000305100.000000000031100.000000000003

100

=

0 0.02 0.04 0.06 0.08100

150

200

T x( )

x

xo

+Δxxo

Δx

T(xo

+Δx)

T(xo

=0)=

First order Forward Finite Difference



Result:

what happened to my limit Δx -> 0 ????


Observations:

error is reduced until about Δx ~ 10-6

error increases

until about Δx ~ 10-15

error is 100% for Δx < 10-15


Why can’t take the limit Δx->0 on the computer ? Answer:

Round Off (RO) error.

The computer makes very small mistakes every time it stores and computes with numbers due to the fact that it utilizes finite arithmetic: there is no way to get around it.

Round off will always exist due to the binary nature of representation of data on the computer

(one and zeroes, called bits) on the computer and the limited amount of bits the computer

uses to represent a number (in our case we used 64 bits).

Where does RO occur in our computation:1. when we compute

Δx and xo

+Δx.2. when we use that value to compute T(xo

) and T(xo

+Δx).3. when we subtract these two values.

Suppose we denote that small error by ε.

Then the total error is: Note:

at some points, calledmachine precision, the computerdoes not recognize the differencebetween xo+Δx and xo. In our case this is Δx ~ 10-15


Finite difference equations are not unique.

If we wished to evaluate the heat flux at the

right wall, xo

=L ,then we could not take

x=xo

+Δx

as there is no solid to the right.

Rather, we take a step back and choose

x=xo-Δx , and we find that

x=0 x=L

xxo

-Δx xo

Δx

qf

First order backwardFinite difference


There is yet another way to approximate the first derivative, and this methodis called a central difference and is of second order accuracy TE ~ O(Δx2).

In light of our discussions on the limits of the computer, this is a big advantage over first order forward and backward finite differences.

x=0 x=L

xxo

-Δx xo

Δ

x

qf

Δ

x

xo

+Δx


3. Applications to computational fluid dynamics and heat transfer:

Finite difference methods and the closely related finite volume method are used widely to solve problems in a variety of fields:

Power generationAerospaceDefenseHeating, ventilations and refrigerationOil and Gas industrySemiconductorPolymer processingBiomedical engineeringNuclear Marine and coastal engineering

…


How?• We start from equations that describe the physical phenomenon: what causes

the temperature to vary, the stresses to exist etc…

-

these are usually ordinary

or more often partial differential equations-

all derived from the conservation principle

Control volume

in

out

out• mass• linear momentum• energy• species• angular momentum

+ fundamental relation between first principleand measurable quantity –

constitutive lawe.g. heat flows due to a temperature difference,

stress is linearly related to displacement,…


• Why can’t we solve the differential equation(s) exactly?

• geometry is too complicated (number one reason in many cases)

• problem is non-linear (often the case but not always)

•Solution:

solve the problem approximately on the computer by approximating the derivatives by finite differences and producing algebraic equations that we can solve.

Steps in the Finite Difference Method:1. mesh generation:

discretization

of continuous space

(and time) into a given number of points in space

(and time)

x

X=0 X=L

x

X=0 X=LExact: the temperature

T(x) is continuous

i =1 i=2 i=3 …

i = IL

Numerical:

the temperature Ti

only at discrete locations –

mesh points, xi


2.

develop an algebraic equations:

apply finite difference approximationsto derivatives in governing differential equation(s), boundary, and initial condition(s)

on the grid.

Differential Equation(s)

+ BC’s (and IC)

Set of SimultaneousAlgebraic Equations

Finite DifferenceFor derivatives

Exact Physicseverywhere in space

and time

Approximate Physicson the grid

3. Solution of the algebraic equations:

-

number of algebraic equations often in the order of several millions of simultaneous equations.

-

computational efficiency of solution algorithm and computing power.


EXAMPLE:

FDM modeling blood flow

mesh

velocity magnitude at various times in cardiac cycle:

6 (mm)

4 (mm)45º

Outflow

Blockage

Inflow

Floor IH

Toe IHHeel IH occluded artery

0.00 0.07 0.14 0.21 0.28 0.35 0.42 0.49 0.56 0.63 0.70 0.77 0.84

0.00 0.02 0.05 0.07 0.09 0.11 0.14 0.16 0.18 0.20 0.23 0.25 0.27

0.00 0.02 0.04 0.06 0.08 0.10 0.13 0.15 0.17 0.19 0.21 0.23 0.25

0.00 0.01 0.02 0.03 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.14


Movie of FDM model of blood flow in artery through the cardiac cycle


EXAMPLE:

Natural convection in a slender cavity

Thotair

g

insulated

insulated

Tcold

Convection cells as at several progressive times colored by magnitude.


Conclusions

Part I – background and review of series

1. Taylor and Maclaurin

series (section 12.10).

2. Fourier series.

Part II – applications

1. Transient heat conduction: a. Application of Maclaurin

series to heat transfer: quenching of a metal bar at early times.

b. Application of Fourier series to heat transfer: quenching of a metal bar

at later times.

2. Finite Difference: computing the heat flux at the wall.

3. Applications to computational fluid dynamics and heat transfer.

application of series in heat transfer transient heat ... · pdf fileapplication of series in...

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