Fouriers Law and the Heat Equation Chapter Two Lecture 3.

Fourier’s Lawand the

Heat Equation

Chapter Two

Lecture 3

Fourier’s Law

• A rate equation that allows determination of the conduction heat flux from knowledge of the temperature distribution in a medium

Fourier’s Law

• Its most general (vector) form for multidimensional conduction is:

q k T

Implications:

– Heat transfer is in the direction of decreasing temperature

(basis for minus sign).

– Direction of heat transfer is perpendicular to lines of constant

temperature (isotherms).

– Heat flux vector may be resolved into orthogonal components.

– Fourier’s Law serves to define the thermal conductivity of the

medium

Heat Flux Components

(2.24)T T T

q k i k j k kr r z

rq q zq

• Cylindrical Coordinates: , ,T r z

sin

T T Tq k i k j k k

r r r

(2.27)

rq q q

• Spherical Coordinates: , ,T r

• Cartesian Coordinates: , ,T x y z

T T Tq k i k j k k

x y z

xq yq zq

(2.3)

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 ,

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

– Cylinder

2r r r rq A q rLq

or,

2r r r rq A q rq

– Sphere24r r r rq A q r q

Heat Equation

The Heat Equation• A differential equation whose solution provides the temperature distribution in a stationary medium.

• Based on applying conservation of energy to a differential control volume through which energy transfer is exclusively by conduction.

• Cartesian Coordinates:

Net transfer of thermal energy into the control volume (inflow-outflow)

pT T T T

k k k q cx x y y z z t

(2.19)

Thermal energygeneration

Change in thermalenergy storage

Heat Equation (Radial Systems)

2

1 1p

T T T Tkr k k q c

r r r z z tr

(2.26)

• Spherical Coordinates:

• Cylindrical Coordinates:

22 2 2 2

1 1 1sin

sin sinp

T T T Tkr k k q c

r r tr r r

(2.29)

Heat Equation (Special Case)

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

2

2

1T T

tx

2thermal diffusivit of the medium my /sp

k

c

pT T

k cx x t

becomes

Boundary Conditions

Boundary and Initial Conditions• For transient conduction, heat equation is first order in time, requiring specification of an initial temperature distribution: 0

0t=

T x,t = T x,

• Since heat equation is second order in space, two boundary conditions must be specified. Some common cases:

Constant Surface Temperature:

Constant Heat Flux:

0x= sT

-k | = qx

Applied Flux Insulated Surface

0 0x=T

| =x

Convection:

0 0x=T

-k | = h T - T ,tx

T(0, t) = Ts

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 environment.

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

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:

• Energy carriers also collide with physical boundaries, affecting their propagation.

External boundaries of a film of material. thick film (left) and thin film (right).

average energy carrier velocity, c < .

mfp

1

3k Cc

energy carrierspecific heat perunit volume.

mean free path → average distancetraveled by an energy carrier beforea collision.

(2.7)

Properties (Micro- and Nanoscale Effects)

mfp

mfp

mfp

For 1

1 3

1 2 3

x

y

L

k k L

k k L

/ ,

/ /

/ /

Grain boundaries within a solid

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

• 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.

(2.9a)

(2.9b)

mfpwhere is the average distance

traveled before experiencing a

collision with another energy carrier

or boundary (See Table 2.1 and Eq. 2.11).

Conduction Analysis

Typical Methodology of a Conduction Analysis

• Solve appropriate form of heat equation to obtain the temperature distribution.

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

• Applications:

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

• Consider possible microscale or nanoscale effects in problems involving very small physical dimensions or very rapid changes in heat or cooling rates.

Problem: Thermal Response of Plane Wall

Problem 2.57 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:

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

2

T 1 T=

a tx

0

Initial: 0 0 uniform temperature

Boundaries: 0 0 adiabatic surface

it T x, = T

x = T / x =

x = L

surface convectionL

- k T / x = h T L,t - T

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.

<

<

conv0in sE = q A dt

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

in s 0E = hA T - T L,t dt

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

3

0 J/minE h

= T - T L,t dtV L

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).

<

<

Problem 2.37 Surface heat fluxes, heat generation and total rate of radiationabsorption in an irradiated semi-transparent material with a prescribed temperature distribution.

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 due to Radiation Absorption

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 + B

dx ka

Front Surface, x=0: xA A

q 0 = -k + 1+ B = - +kBka a

<

(b) The heat diffusion equation for the medium is

0d dT q d dT

+ = or q = -kdx dx k dx dx

-ax -axd Aq x = -k + e + B = Ae .

dx ka

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

Rear Surface, x=L: -aL -aLx

A Aq L = -k + e + B = - e +kB

ka a

<

<

On a unit area basis

in out 0 -aLg x x

AE = -E + E = -q +q L = + 1- e .

a <

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

0

LL L -ax -ax -aLg 0 0

A AE = q x dx = Ae dx = - e = 1- e .

a a

Problem : Non-uniform Generation (cont.)

Example 2.3 (pages 75-76)

The temperature distribution across a wall 1m thick at a certain

instant of time is given as:

where T is in degree Celsius and x is in meters, while a=900C,

b=-300C/m, c=-50C/m2. A uniform heat generation, =1000

W/m3, is present in the wall of area 10 m2 having the property of

=1000 kg/m3, k=40W/mK, and cP=4 kJ/kgK. 1.Determine the rate of heat transfer entering the wall (x=0) and leaving the

wall (x=1m).2.Determine the rate of change of energy storage in the wall.3.Determine the time of temperature change at x =0, 0.25 and 0.5m.

2)( cxbxaxT

q


Solution

Known: Temperature distribution T(x) at an instant of time t

in a 1-D wall with uniform generation

Find: 1. Heat rates entering and leaving;

2. Rate of change of energy storage in the wall,

3. Time rate of temperature change at several locations

Schematic:

stE


Schematic:


Assumptions: 1. 1-D conduction in x-direction

2. Isotropic medium with constant

properties

3. Uniform internal heat generation, (W/m3)

Analysis:

1. For steady state 1-D conduction, Fourier’s can be applied

to calculate qin and qout. (qin= 120 kW ; qout = 160 kW)

q

dx

dTkq


Analysis:

2. The rate of change of energy storage in the wall ( ) can

be calculated by applying an overall energy balance to the

wall. Using Equation 1.1 for control volume about the wall,

stoutgin EEEE

stE

kWqALqqEEEE outinoutginst 30


Analysis:

3. The time rate of change of the temperature at any point in

the medium may be determined from the heat Equation 2.19,

rewritten as:

From the prescribed temperature distribution, it follows:

PP c

q

x

T

c

k

t

T

2

2

22

2

/100 mCx

T

xx

T


Analysis:

sCx

c

qmC

c

k

t

T

PP

/1069.4

/100

4

2

Lecture 3

Fouriers Law and the Heat Equation Chapter Two Lecture 3.

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