Appendices 10.A & 10.B: - PowerPoint Presentation

Slide1

Appendices 10.A & 10.B: An Educational Presentation

Presented By:

Joseph Ash

Jordan Baldwin

Justin Hirt

Andrea LanceSlide2

History of Heat Conduction

Jean Baptiste Biot

(1774-1862)

French PhysicistWorked on analysis of heat conductionUnsuccessful at dealing with the problem of incorporating external convection effects in heat conduction analysisSlide3

History of Heat Conduction

Jean Baptiste Joseph Fourier

(1768 – 1830)

Read Biot’s work1807 determined how to solve the problemFourier’s LawTime rate of heat flow (Q) through a slab is proportional to the gradient of

temperature differenceSlide4

History of Heat Conduction

Ernst Schmidt

German scientist

Pioneer in Engineering ThermodynamicsPublished paper “Graphical Difference Method for Unsteady Heat Conduction”

First to measure velocity and temperature field in free convection boundary layer and large heat transfer coefficients

Schmidt Number

Analogy between heat and mass

transfer that causes a dimensionless

quantitySlide5

Derivation of the Heat Conduction Equation

A first approximation of the equations that govern the conduction of heat in a solid rod.Slide6

Consider the following:

A uniform rod is insulated on both lateral ends.

Heat can now only flow in the axial direction.

It is proven that heat per unit time will pass from the warmer section to the cooler one.The amount of heat is proportional to the area, A, and to the temperature difference T2-T1, and is inversely proportional to the separation distance, d.Slide7

The final consideration can be expressed as the following:

is a proportionality factor called the thermal conductivity and is determined by material properties Slide8

Assumptions

The bar has a length L so x=0 and x=L

Perfectly insulated

Temperature, u, depends only on position, x, and time, tUsually valid when the lateral dimensions are small compared to the total length.Slide9

The differential equation governing the temperature of the bar is a physical balance between two rates:

Flux/Flow term

Absorption termSlide10

Flux

The instantaneous rate of heat transfer from left to right across the cross sections x=x

0

where x0 is arbitrary can be defined as:

The negative is needed in order to show a positive rate from left to right (hot to cold)Slide11

Flux

Similarly, the instantaneous rate of heat transfer from right to left across the cross section x=x

0

+Δx where Δx is small can be defined as:Slide12

Flux

The amount of heat entering the bar in a time span of

Δ

t is found by subtracting the previous two equations and then multiplying the result by Δt:Slide13

Heat Absorption

The average change in temperature,

Δ

u, can be written in terms of the heat introduced, Q Δt and the mass Δm of the element as:

where s = specific heat of the material

ρ

= densitySlide14

Heat Absorption

The actual temperature change of the bar is simply the actual change in temperature at some intermediate point, so the above equation can also be written as:

This is the heat absorption equation.Slide15

Heat Equation

Equating the Q

Δ

t in the flux and absorption terms, we find the heat absorption equation to be:Slide16

If we divide the above equation by

Δ

xΔt and allow both Δx and

Δ

t to both go to 0, we will obtain the

heat conduction

or

diffusion equation

: where

and has the dimensions of length^2/time and called the thermal diffusivitySlide17

Boundary Conditions

Certain boundary conditions may apply to the specific heat conduction problem, for example:

If one end is maintained at some constant temperature value, then the boundary condition for that end is

u = T.If one end is perfectly insulated, then the boundary condition stipulates ux =

0.Slide18

Generalized Boundary Conditions

Consider the end where x=0 and the rate of flow of heat is proportional to the temperature at the end of the bar.

Recall that the rate of flow will be given, from left to right, as

With this said, the rate of heat flow out of the bar from right to left will be

Therefore, the boundary condition at x=0 is

where h

1

is a proportionality constant

if h

1=0, then it corresponds to an insulated end

if h

1

goes to infinity, then the end is held at 0 temp.Slide19

Generalized Boundary Conditions

Similarly, if heat flow occurs at the end x = L, then the boundary condition is as follows:

where, again, h

2 is a nonzero proportionality factorSlide20

Initial Boundary Condition

Finally, the temperature distribution at one fixed instant – usually taken at t = 0, takes the form:

occurring throughout the barSlide21

Generalizations

Sometimes, the thermal conductivity, density, specific heat, or area may change as the axial position changes. The rate of heat transfer under such conditions at x=x

0

is now:The heat equation then becomes a partial differential equation in the form: orSlide22

Generalizations

Other ways for heat to enter or leave a bar must also be taken into consideration.

Assume

G(x,t,u) is a rate per unit per time.SourceG(

x,t,u

) is added to the bar

G

(

x,t,u) is positive, non-zero, linear, and u does not depend on tG(x,t,u) must be added to the left side of the heat equation yielding the following differential equationSlide23

Generalizations

Similarly,

Sink

G(x,t,u) is subtracted from the barG(x,t,u) is positive, non-zero, linear, and u does not depend on t

G

(

x,t,u

) then under this sink condition takes the form:Slide24

Generalizations

Putting the source and sink equations together in the heat equation yields

which is commonly called the

generalized heat conduction equationSlide25

Multi-dimensional space

Now consider a bar in which the temperature is a function of more than just the axial x-direction. Then the heat conduction equation can then be written:

2-D:

3-D:Slide26

Example 1: Section 10.6, Problem 9

Let an aluminum rod of length 20 cm be initially at the uniform temperature 25

C. Suppose that at time t=0, the end x=0 is cooled to 0C while the end x=20 is heated to 60C, and both are thereafter maintained at those temperatures. Find the temperature distribution in the rod at any time t

Slide27

Example 1: Section 10.6, Problem 9

Find the temperature distribution, u(x,t)



2

u

xx

=u

t

, 0<x<20, t<0

u(0,t)=0 u(20,t)=60, t<0

u(x,0)=25, 0<x<20

From the initial equation we find that:

L=20, T

1

=0, T

2

=60, f(x)=25

We look up the Thermal Diffusivity of aluminum

→

2

=0.86Slide28

Example 1: Section 10.6, Problem 9

Using Equations 16 and 17 found on page 614, we find that

whereSlide29

Example 1: Section 10.6, Problem 9

Evaluating c

n

, we find thatSlide30

Example 1: Section 10.6, Problem 9

Now we can solve for u(x,t)Slide31

Example 1: Section 10.6, Problem 9Slide32

Derivation of the Wave Equation

Applicable for:

One space dimension, transverse vibrations on elastic string

Endpoints at

x

= 0 and

x

= L along the

x

-axis

Set in motion at

t

= 0 and then left undisturbed Slide33

Schematic of String in TensionSlide34

Equation Derivation

Since there is no acceleration in the horizontal direction

However the vertical components must satisfy

where is the coordinate to the center of mass and the weight is neglected

Replacing

T

with

V

the and rearranging the equation becomesSlide35

Derivation continued

Letting , the equation becomes

To express this in terms of only terms of u we note that

The resulting equation in terms of

u

is

and since

H(t)

is not dependant on x

the resulting equation isSlide36

Derivation Continued

For small motions of the string, it is approximated that

using the substitution that

the wave equation takes its customary form ofSlide37

Wave Equation Generalizations

The telegraph equation

where c and k are nonnegative constants

cu

t

arises from a viscous damping force

ku arises from an elastic restoring force F(x,t) arises from an external force

The differences between this telegraph equation and the customary

wave equation are due to the consideration of internal elastic

forces. This equation also governs flow of voltage or current in a

transmission line, where the coefficients are related to the electrical

parameters in the line. Slide38

Wave Equations in Additional Dimensions

For a vibrating system with more than on significant space coordinate it may be necessary to consider the wave equation in more than one dimension.

For two dimensions the wave equation becomes

For three dimensions the wave equation becomesSlide39

Example 2: Section 10.7, Problem 6

Consider an elastic string of length L whose ends are held fixed. The string is set in motion from its equilibrium position with an initial velocity g(x). Let L=10 and a=1. Find the string displacement for any time t.Slide40

Example 2: Section 10.7, Problem 6

From equations 35 and 36 on page 631, we find that

whereSlide41

Example 2: Section 10.7, Problem 6

Solving for k

n

, we find:Slide42

Example 2: Section 10.7, Problem 6

Now we can solve for u(x,t)Slide43

THE END