fourier’s law and the heat equation

7/29/2019 Fouriers Law and the Heat Equation

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Fouriers Law

and the

Heat Equation

Chapter Two


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Fouriers Law

A rate equation that allows determination of the conduction heat flux

from knowledge of the temperature distribution in a medium

Fouriers 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.

Fouriers Law serves to define the thermal conductivity of the

medium /k q T


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Heat Flux Components

(2.18)T T Tq k i k j k k r r z

rq q zq

Cylindrical Coordinates: , ,T r z

sin

T T Tq k i k j k k

r r r

(2.21)

rq q q

Spherical Coordinates: , ,T r

Cartesian Coordinates: , ,T x y z

T T T

q k i k j k k x y z

xq yq zq

(2.3)


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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 ofC/m and not C/deg.

or ,

Heat rate forone-dimensional, radial conduction in a cylinder or sphere:

Cylinder

2r r r r q A q rLq

or,

2r r r r q A q rq

Sphere

24r r r r q A q r q


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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 transferofthermal energy into the

control volume (inflow-outflow)

p

T T T T k k k q c

x x y y z z t

(2.13)

Thermal energy

generationChange in thermal

energy storage


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Heat Equation (Radial Systems)

2

1 1

p

T T T T

kr k k q cr r r z z t r

(2.20)

Spherical Coordinates:

Cyli

ndrical Coordinates:

2

2 2 2 2

1 1 1sin

sin sinp

T T T T kr k k q c

r r tr r r

(2.33)


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Heat Equation (Special Case)

One-Dimensional Conduction in a Planar Medium with Constant Properties

andNo Generation

2

2

1T T

tx

thermal diffu osivit f the medy iump

k

c


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Boundary Conditions

Boundary and Initial Conditions Fortransient 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, twoboundary conditionsmust be specified. Some common cases:

Constant Surface Temperature:

0, sT t T

Constant Heat Flux:

0|x s

T

k qx

Applied F lux I nsulated Surface

0| 0x

T

x

Convection

0| 0,xT

k h T T t x


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Properties

Thermophysical Properties

Thermal Conductivity: A measure of a materials ability to transfer thermal

energy by conduction.

Thermal Diffusivity: A measure of a materials ability to respond to changes

in its thermal environment.

Property Tables:

Solids: Tables A.1A.3

Gases: Table A.4

Liquids: Tables A.5A.7


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Conduction Analysis

Methodology of a Conduction Analysis

Solve appropriate form of heat equation to obtain the temperature

distribution.

Knowing the temperature distribution, apply Fouriers 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


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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 oftime 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 1T

x

T

t2

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 theconditions 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 thatthe gradient at x = L decreases with time.

fourier’s law and the heat equation

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