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Chapter2Introduction to Conduction

• Conduction is the transfer of energy from

OverviewCo duct o s t e t a s e o e e gy othe more energetic to the less energetic particles of a substance due to interactions between the particles.– There must be a thermal gradient in a stationary

(no moving parts) medium (solid/liquid/gas). – There are no moving parts in conductive heat

transfer. Most problems dealing with conduction involve solids.

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• Conduction is governed by Fourier’s Law: (heat rate)ddTkAqx −=

The Conduction Rate EquationConduction is governed by Fourier s Law: (heat rate)

• For the heat flux:

where k = the thermal conductivity (W/m·K), A = the area of the object (m2), dT/dx = temperature gradient of the object.

• Since heat flux is a vector quantity, a more general statement of Fourier’s Law (conduction rate equation) is:

dxqx

dxdTkqx −=′′

⎟⎟⎞

⎜⎜⎛

++−=∇−=′′dTkdTjdTikTkq( q )

– where is the three-dimensional del operator and T(x,y,z) is the scalar temperature field.

• From the previous equations it follows that:

⎟⎟⎠

⎜⎜⎝

++∇dz

kdy

jdx

ikTkq

∇

dxdTkqx −=′′

dydTkqy −=′′

dzdTkqz −=′′

Heat Flux Components

• Cartesian Coordinates: ( ), ,T x y zT T Tq k i k j k kx y z

→ → → →∂ ∂ ∂′′ = − − −∂ ∂ ∂

xq′′ yq′′ zq′′

(2.3)

(2.22)T T Tq k i k j k kr r zφ

→ → → →∂ ∂ ∂′′ = − − −∂ ∂ ∂

rq′′ qφ′′ zq′′

• Cylindrical Coordinates: ( ), ,T r zφ

• Spherical Coordinates: ( ), ,T r φ θ

sinT T Tq k i k j k kr r rθ θ φ

→ → → →∂ ∂ ∂′′ = − − −∂ ∂ ∂

(2.25)

rq′′ qθ′′ qφ′′

Spherical Coordinates: ( ), ,φ

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Heat Flux Components (cont.)

In angular coordinates , the temperature gradient is still based on temperature change over a length scale and hence has units of °C/m and not °C/deg.

( ) or ,φ φ θ

• Heat rate for one-dimensional, radial conduction in a cylinder or sphere:

– Cylinder2r r r rq A q rLqπ′′ ′′= =

or,2r r r rq A q rqπ′ ′ ′′ ′′= =

– Sphere– Sphere24r r r rq A q r qπ′′ ′′= =

Properties

Thermophysical PropertiesThermal Conductivity: A measure of a material’s ability to transfer thermal

energy by conduction.

Thermal Diffusivity: A measure of a material’s ability to respond to changesin its thermal environmentin its thermal environment.

•Property Tables:• Solids: Tables A.1 – A.3• Gases: Table A.4• Liquids: Tables A.5 – A.7

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• Physical Meaning

Thermal Conductivityy g

– property of the material, how quickly heat can be delivered• How to Determine k

– experimentally, determined from equations (example: k = -qx’’/ (∂T/∂x) )• How to Use It

– use k to find temperature distribution, which can then find the heat transfer in any direction

• For a given material: ks > kl >> kgh lid l li idwhere s: solid; l: liquid; g: gas.

• This is only true assuming that they (solid, liquid, gas) are all stationary and not moving.

• k is usually assumed to be constant but it can be a function of temperature, k = f(T)– If the material is non-isotropic, heat transfer is directional (direction dependent, such as a

composite), k = f(T,x,y,z). – Therefore for an isotropic material, k= f(T)

Properties (Micro- and Nanoscale Effects)

Micro- and Nanoscale Effects• Conduction may be viewed as a consequence of energy carrier (electron or

phonon) motion.

• For the solid state:1k Cc λ= (2 7)

• Energy carriers also collide with physical boundaries, affecting their propagation.External boundaries of a film of material

average energy carrier velocity, c < ∞.

3 mfpk Cc λ=

energy carrierspecific heat perunit volume.

mean free path → averagedistance traveled by an energycarrier before a collision.

(2.7)

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Properties (Micro- and Nanoscale Effects)

( )( )

For 1

1 2 3

1 3

mfp

x mfp

y mfp

L

k k L

k k L

λ

λ π

λ

<

= −

= −

/ ,/ // /

Grain boundaries within a solid

(2.9a)

(2.9b)

Measured thermal conductivity of a ceramic material vs. grain size, L. at 300K 25nm.mfp Tλ ≈ =

• Fourier’s law does not accurately describe the finite energy carrier propagation velocity. This limitation is not important except in problems involving extremely small time scales.

• Thermal diffusivity α is the ratio of the thermal conductivity to

Thermal Diffusivity Thermal diffusivity, α, is the ratio of the thermal conductivity to the heat capacity.

– where ρ=density, Cp = specific heat (heat required to raise one gram of material 1 K), and k = thermal conductivity

• Physical MeaningeHeatStorag

tionHeatConducT

tT

Ck

=Δ

Δ•=

ρα

• If k increases and Cp decreases, the thermal diffusivity of the material will increase

eHeatStoragtTCp

Δρ

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• Physical Meaning

Thermal Diffusivity (continued) Physical Meaning

– Several factors affect how fast heat will transfer through a material• Thermal Conductivity, k: the higher the value of a material’s heat conductivity, k,

the faster that heat will transfer through the material.• Heat Storage: when heat transfers through a material, some of the heat will be

stored in the material. Materials that have a large heat capacity will let more heat through than materials with a low heat capacity.

– Low Cp ↑Heat Transfer, ↓Heat Storage– High Cp ↓Heat Transfer, ↑ Heat Storage

• How Can We Determine It?

• How can We Use It?– The heat diffusion equation:

dtdTcq

dzdTk

dzd

dydTk

dyd

dxdTk

dxd

pρ=+⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎠⎞

⎜⎝⎛

• The heat diffusion equation: dtdTcq

dzdTk

dzd

dydTk

dyd

dxdTk

dxd

pρ=+⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎠⎞

⎜⎝⎛

The Heat Diffusion Equation q

• How did we get this equation?Derivation:T(t,x,y,z) = Temperature Distribution

dtdzdzdydydxdx ⎠⎝⎠⎝⎠⎝

Imagine a small cube of material as shown at right

Substituting gives

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The Heat Diffusion Equation (continued)

The Heat Diffusion Equation (continued)

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• 1

Simplifications: 1-D, stead state, constant properties

• 1.

• 2.

• 3.

Heat Equation (Radial Systems)

• Cylindrical Coordinates:

21 1

pT T T Tkr k k q c

r r r z z trρ

φ φ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞+ + + =⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

•(2.24)

• Spherical Coordinates:

22 2 2 2

1 1 1 sinsin sin p

T T T Tkr k k q cr r tr r r

θ ρφ φ θ θθ θ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞+ + + =⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

•(2.27)

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Heat Equation (Special Case)

One Dimensional Conduction in a Planar Medium with Constant Properties• One-Dimensional Conduction in a Planar Medium with Constant Propertiesand No Generation

2

21T T

tx α∂ ∂

=∂∂

thermal diffu osivit f the medy iumkα ≡ → ypcρ

Boundary Conditions

Boundary and Initial Conditions• For transient conduction, heat equation is first order in time, requiring

specification of an initial temperature distribution: ( ) ( )0, ,0tT x t T x= =

• Since heat equation is second order in space, two boundary conditionsmust be specified. Some common cases: Constant Surface Temperature:Constant Surface Temperature:

( )0, sT t T=

Constant Heat Flux:

T∂

Applied Flux Insulated Surface

T∂0|x s

Tk qx =

∂ ′′− =∂

0| 0xTx =

∂=

∂

Mixed surface condition:

( )0| 0,xTk h T T tx = ∞

∂− = −⎡ ⎤⎣ ⎦∂

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• For the 3-Dcase, 6 BCs and 1 IC are required to get the complete solution.

Boundary Conditions (BCs) and Initial Conditions (ICs)

For the 3 Dcase, 6 BCs and 1 IC are required to get the complete solution. However, a unique solution may still not be arrived at.

• E.g., the 1-Dimensional case:

Boundary Conditions (continued)( )1(0, )T t T t ⎫=

⎪

However:

1(0, ) ( )T t T t= ⎫⎪

0

( ) ill-defined w ith no unique solution

( , 0) ( )x

dT w tdx

T x f x=

⎪⎪= ⎬⎪⎪= ⎭

1

2

( , ) ( )( , ) ( ) well defined with a unique solution( ,0) ( )

T L t T tT x f x

⎫⎪= ⎬⎪= ⎭

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Conduction Analysis

Methodology of a Conduction Analysis

• Solve appropriate form of heat equation to obtain the temperature

• Consider possible micro- or nano-scale effects in problems involving verysmall physical dimensions or very rapid changes in heat or cooling rates.

• Solve appropriate form of heat equation to obtain the temperaturedistribution.

• Knowing the temperature distribution, apply Fourier’s Law to obtain theheat flux at any time, location and direction of interest.

• Applications:Applications:

Chapter 3: One-Dimensional, Steady-State ConductionChapter 4: Two-Dimensional, Steady-State ConductionChapter 5: Transient Conduction

Problem: Thermal Response of Plane Wall

Problem 2.46 Thermal response of a plane wall to convection heat transfer.

KNOWN: Plane wall, initially at a uniform temperature, is suddenly exposed to convective heating.

FIND: (a) Differential equation and initial and boundary conditions which may be used to find the temperature distribution, T(x,t); (b) Sketch T(x,t) for the following conditions: initial (t ≤ 0), steady-state (t → ∞), and two intermediate times; (c) Sketch heat fluxes as a function of time at the two surfaces; (d) Expression for total energy transferred to wall per unit volume (J/m3).

SCHEMATIC:

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Problem: Thermal Response (Cont).

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal heat generation.

ANALYSIS: (a) For one-dimensional conduction with constant properties, the heat equation has the form,

2

2T 1 T

t x∂ ∂

α ∂∂=

( ) i

0

L

Initial, t 0 : T x,0 T uniform temperature

Boundaries: x=0 T/ x 0 adiabatic surface

x=L k T/ x = h T L

∂ ∂

∂ ∂

≤ =

=

− ( ),t T surface convection∞

⎧⎪⎪⎨⎪

⎡ − ⎤⎪ ⎣ ⎦⎩

and the conditions are:

(b) The temperature distributions are shown on the sketch.

Note that the gradient at x = 0 is always zero, since this boundary is adiabatic. Note also that the gradient at x = L decreases with time.

c) The heat flux, ( )xq x,t ,′′ as a function of time, is shown on the sketch for the surfaces x = 0 and x = L.

Problem: Thermal Response (Cont).

in conv s0E q A dt

∞′′= ∫

d) The total energy transferred to the wall may be expressed as

( )( )in s 0E hA T T L,t dt

∞∞= −∫

0

Dividing both sides by AsL, the energy transferred per unit volume is

( ) 3in0

E h T T L,t dt J/mV L

∞∞ ⎡ ⎤= −⎡ ⎤⎣ ⎦ ⎣ ⎦∫

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Problem 2.28 Surface heat fluxes, heat generation and total rate of radiationabsorption in an irradiated semi-transparent material with a

prescribed temperature distribution.

Problem: Non-uniform Generation due to Radiation Absorption

KNOWN: Temperature distribution in a semi-transparent medium subjected to radiative flux.

FIND: (a) Expressions for the heat flux at the front and rear surfaces, (b) The heat generation rate ( )q x , and (c) Expression for absorbed radiation per unit surface area.

SCHEMATIC:

Problem : Non-uniform Generation (Cont.)

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in medium, (3) Constant properties, (4) All laser irradiation is absorbed and can be characterized by an internal volumetric heat generation term ( )q x .

ANALYSIS: (a) Knowing the temperature distribution, the surface heat fluxes are found using Fourier’s law,

( ) -axx 2

dT Aq k k - a e Bdx k

⎡ ⎤⎡ ⎤′′ = − = − − +⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ 2dx ka⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

Front Surface, x=0: ( )xA Aq 0 k + 1 B kBka a

⎡ ⎤ ⎡ ⎤′′ = − ⋅ + = − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ <

Rear Surface, x=L: ( ) -aL -aLx

A Aq L k + e B e kB .ka a

⎡ ⎤ ⎡ ⎤′′ = − + = − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ <

(b) The heat diffusion equation for the medium is

d dT q d dT0 or q=-k

dx dx k dx dx⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) -ax -axd Aq x k e B Ae .dx ka

⎡ ⎤= − + + =⎢ ⎥⎣ ⎦

( c ) Performing an energy balance on the medium, in out gE E E 0− + =

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On a unit area basis

( ) ( ) ( )-aLg in out x x

AE E E q 0 q L 1 e .a

′′ ′′ ′′ ′′ ′′= − + = − + = + − <

Alternatively, evaluate gE′′ by integration over the volume of the medium,

Problem : Non-uniform Generation (Cont.)

( ) ( )LL L -ax -ax -aLg 0 0 0

A AE q x dx= Ae dx=- e 1 e .a a⎡ ⎤′′ = = −⎣ ⎦∫ ∫