Unit 4

5

Governing Equations of Heat ConductionUNIT 4 GOVERNING EQUATIONS OF HEAT

CONDUCTION Structure

4.1 Introduction Objectives

4.2 General Equation of Heat Conduction 4.2.1 Rectangular Coordinate System 4.2.2 Cylindrical Coordinates 4.2.3 Spherical Coordinates

4.3 Steady State Heat Conduction in Simple Geometrical Systems 4.3.1 Plane Wall 4.3.2 Cylindrical Wall 4.3.3 Spherical Shell

4.4 Transient Conduction 4.4.1 Lumped Capacitance Method 4.4.2 General Lumped Capacitance Analysis

4.5 The Semi-infinite Solid 4.6 Multi-dimensional Effects 4.7 Summary 4.8 Key Words 4.9 Answers to SAQs

1.1 INTRODUCTION

In conduction mode heat is transferred through a complex sub-microscopic mechanism that involves flow of free electrons and lattice vibration. Conduction is predominant in case of solid or liquid metals. In case of liquids and gases, once heat begins to flow, even if no external force is applied, density gradients are set up and convective currents are set in motion. Heat is then transported on a macroscopic scale as well as on a microscopic scale with convection currents generally being the more effective. In the following sub-units conduction heat transfer has been discussed in details for steady as well as transient states.

Objectives After studying this unit, you should be able to

• formulate heat transfer by conduction in different geometries under steady state condition,

• evaluate temperature at different locations in rectangular, cylindrical and spherical coordinate system,

• differentiate the steady state and transient nature of heat conduction,

• evaluate heat transfer by lumped capacitance method and its limitations, and

• solve some problems on steady and transient heat transfer.

4.2 GENERAL EQUATION OF HEAT CONDUCTION

General equation of heat conduction is explained in three coordinate systems.

6

4.2.1 Rectangular Coordinate System Conduction

Differential Form

Let us consider an infinitesimal volume element of sides ∂x, ∂y and ∂z as shown in Figure 4.1. The considerations here will include the non-steady condition of temperature variation with time t.

Figure 4.1 : Volume Element for Determining Heat Conduction Equation

According to the Fourier heat conduction law, the heat flowing into the left most face of the element in the x-direction

xdTdQ k y z

x= − ∂ ∂

∂ . . . (4.1)

The value of the heat flow out of the right face of the element can be obtained by expanding dQx in a Taylor series and retaining only the first two terms as an approximation :

( ) . .x dx x xdQ dQ dQ x .x−

∂= + δ +

∂ . . . (4.2)

The net heat flow by conduction in the x-direction is therefore

( )x x dx xTdQ dQ dQ x k y z x

x x+∂ ∂ ⎛ ⎞

x∂

− = − δ = − − δ δ δ⎜ ⎟∂ ∂ ⎝ ⎠∂

2

2Tk x y z

x∂

= δ δ δ∂

. . . (4.3)

Similarly, the net heat flows in the y-and z-direction are 2

2y y dyTdQ dQ k x y z

y+∂

− = δ δ∂

δ . . . (4.4)

2

2z z dzTdQ dQ k x y z

z+∂

− = δ δ δ∂

. . . (4.5)

Here, the solid has been assumed to be isotropic and homogeneous with properties uniform in all directions. Let us consider that there is some heat source within the solid, and heat is produced internally as a result of the flow of electrical current or nuclear or chemical reactions. Let qG is the rate at which heat is generated

Z

δz

dQy

δy

δx

dQz

dQy + dz

dQx

x

dQx + dx

y

dQx + dy

7

Governing Equations of Heat Conductioninternally per unit volume 3

Wm

⎛⎜⎝ ⎠

⎞⎟ . Then the total rate of heat generation in the

elemental volume is qG ∂x ∂y ∂z. The net heat flow owing to conduction and the heat generated within the element together will increase the internal energy of the volume element. The rate of accumulation of internal energy IE within the control volume is

. .cTIE x y zt

∂= ρ δ δ δ

∂ . . . (4.6)

where c is the specific heat and ρ is the density of the solid. An energy balance can be achieved on the volume element as : Rate of energy storage within the solid

= Rate of heat influx – Rate of heat outflux + Rate of heat generation

or ( ) (c x y z x dx y dyTx y z dQ dQ dQ dQ dQ dQt + + +

∂ρ δ δ δ = + + − + +

∂)z dz

Gq x y z+ δ δ δ

2 2 2

2 2 2 GT T Tk x y z q x y z

x y z

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

. . . (4.7)

or 2 2 2

2 2 2c GT T T Tkt x y z

⎛ ⎞∂ ∂ ∂ ∂ρ = + + +⎜⎜∂ ∂ ∂ ∂⎝ ⎠

q⎟⎟ . . . (4.8)

2 2 2

2 21GqT T T

z kx y∂ ∂ ∂ ∂

+ + + =∂ α∂ ∂

Tt∂

. . . (4.9)

where α is the thermal diffusivity of the solid given by c

kρ

.

If the temperature of a material is not a function of time, the system is in the steady state and does not store any energy. The steady state form of a three-dimensional conduction equation in rectangular coordinates is

2 2 2

2 2 2 0GqT T Tkx y z

∂ ∂ ∂+ + + =

∂ ∂ ∂ . . . (4.10)

If the stem is in steady state and no heat is generated internally, the conduction equation simplifies to

2 2 2

2 2 2 0T T Tx y z

∂ ∂ ∂+ + =

∂ ∂ ∂ . . . (4.11)

Eq. (4.11) is known as the Laplace equation. It occurs in a number of areas in addition to heat transfer, for example, in diffusion of mass or in electromagnetic fields. The operation of taking the second derivatives of the potential in a field has therefore been given a short symbol, ∇2, called the Laplacian operator. For the rectangular coordinate system Eq. (4.11) becomes

2 2 22

2 2 2 0T T T Tx y z

∂ ∂ ∂+ + = ∇ =

∂ ∂ ∂ . . . (4.12)

Since the operator ∇2 is independent of coordinate system, the above form will be useful when we study conduction in cylindrical and spherical coordinates. The heat conduction can thus be written as

2 1Gq TTk t

∂∇ + =

α ∂ . . . (4.13)

8

Vector Method Conduction Equation can also be derived vectorially. Let us consider a control surface S

enclosing a volume V as shown in Figure 4.2. q

ds

dυ

qG

q Control surface S

Figure 4.2 : Heat Conduction through a Volume Element

The net rate of heat outflow across the surface S is given by where n is

the normal direction. Converting the surface integral to volume integral

.s

q nds∫

. divs v

q n ds q dv=∫ ∫ . . . (4.14)

Where q is the heat flux per unit area. The net rate of heat inflow to the control

volume (CV) is . divv

q dv− ∫

If there is a volumetric heat source inside the CV the rate of heat generation is

Gv

q dv∫ . . . (4.15)

The rate of energy accumulation within the CV is

v v

e dv c T dVt t

∂ ∂ρ = ρ

∂ ∂∫ ∫ . . . (4.16)

e being the specific energy (J/kg). By energy balance,

divGv v v

c T dv q dv q dVt

∂ρ = −

∂ ∫ ∫ ∫ . . . (4.17)

where q is the heat flux per unit area. Writing the energy equation for the elemental volume dV within the CV

divGTc dV q dV q dt

V∂ρ = −

∂ . . . (4.18)

Since dV is now independent, it can be removed from the above equation. Therefore,

divGTc qt

∂ρ = −

∂q . . . (4.19)

Now, div .q q= ∇ . . . (4.20)

and q k T= − ∇ . . . (4.21) 2div ( )q k T k= ∇ − ∇ = − ∇ T . . . (4.22)

for constant k.

9

Governing Equations of Heat Conduction

Substituting in Eq. (4.19),

2G

Tc q kt

∂ρ = + ∇

∂T . . . (4.23)

Therefore, 2 1Gq TTk t

∂∇ + =

α ∂ . . . (4.24)

Which is the same as Eq. (4.13).

4.2.2 Cylindrical Coordinates

For a general transient three-dimensional heat conduction problem in the cylindrical coordinates with T = T (r, q, z, t), let us consider an elementary volume as shown in Figure 4.3.

dV d r r d dz= θ . . . (4.25)

dQθ +dθ

dQr +dr

dQz

dQrdθ

dθ θ

dQθ

z

dQz +dz

r

dr

ds

n q

r

r

z

θ

θ

Figure 4.3 : Heat Conduction in a Cylindrical Volume Element (r, θ, z)

By Fourier’s law,

( )rTdQ k r d zr

∂= − θ

∂ . . . (4.26)

( )r dr r rdQ dQ dQ drr+

∂= +

∂ . . . (4.27)

r r drTdQ dQ k r d dz dr

r r+∂ ∂⎛ ⎞− = − − θ⎜ ⎟∂ ∂⎝ ⎠

2

2Tkr d dz dr k d dz dr

rrT∂ ∂

= θ + θ∂∂

. . . (4.28)

Similarly, TdQ k dz drrθ∂

= −∂θ

. . . (4.29)

dTdQ d Q k dz dr d

rθ θ + θ⎛ ⎞∂ ∂

− = − −⎜ ⎟∂θ ∂θ⎝ ⎠θ . . . (4.30)

10

Conduction ( )ddQ d Q d Q r d

rθ + θ θ θ∂

= + θ∂θ

. . . (4.31)

2

2T Tdr r d dzc kr d dr k d dz drt rr

T∂ ∂ ∂ρ θ = θ + θ

∂ ∂∂

2 2

2 2T Tk dz dr d k dr rd dz

z∂ ∂

+ θ + θ∂θ ∂

2

21 Tk dz dr dr

∂= θ

∂θ . . . (4.32)

zTdQ k dr rdz

∂= − θ

∂ . . . (4.33)

( )z dz z zdQ dQ dQ dzz+

∂= +

∂ . . . (4.34)

z z dzTdQ dQ k dr r d dz

z z+∂ ∂⎛ ⎞− = − − θ⎜ ⎟∂ ∂⎝ ⎠

2

2Tk dr r d dz

z∂

= θ∂

. . . (4.35)

Rate of heat generation from an internal heat source

Gq dr rd dz= θ . . . (4.36)

Rate of energy accumulation within the CV

( ) Tdr r d ct

∂= ρ θ

∂ . . . (4.37)

By energy balance, from Eqs. (4.28)-(4.37), 2 2

2 2T T Tdr rd d z c k r d dr k d dz dr k dz dr dt rr

T∂ ∂ ∂ρ θ = θ + θ + θ

∂ ∂∂

∂ ∂θ

2

2 GTk dr r d dz q dr r d dz

z∂

+ θ + θ∂

. . . (4.38)

2 2

2 2 21 1T T T Tc k k k k

t r rr r∂ ∂ ∂ ∂ ∂

ρ = + + +∂ ∂

2

2T

z∂ ∂θ ∂ . . . (4.39)

or 2 2 2

2 2 2 21 1 1G

GqT T T Tq

r r k tr r z∂ ∂ ∂ ∂

+ + + + =T∂

∂ α ∂∂ ∂θ ∂ . . . (4.40)

or 2 2

2 2 21 1 GqT T Trr r r k tr z

∂ ∂ ∂ ∂ ∂+ + + =

∂ ∂ α ∂1 T

∂θ ∂ . . . (4.41)

This is the general heat conduction equation in cylindrical co-ordinates. If we compare this equation with Eq. (4.13), the Laplacian is

2 22

2 2 21 1T TT rr r r r z

T∂ ∂ ∂ ∂⎛ ⎞∇ = + +⎜ ⎟∂ ∂ ∂θ ∂⎝ ⎠ . . . (4.42)

If heat flows only in radial direction, T = T (r, t), Eq. (4.41) reduces to

1 1GqT Trr r r k t

∂ ∂ ∂⎛ ⎞ + =⎜ ⎟∂ ∂ α ∂⎝ ⎠ . . . (4.43)

11


If the temperature distribution does not vary with time, then at steady state,

1 0GqTrr r r k

∂ ∂⎛ ⎞ + =⎜ ⎟∂ ∂⎝ ⎠ . . . (4.44)

In this case the equation for the temperature contains only a single variable r and is therefore an ordinary differential equation.

When there is no volumetric energy generation and temperature is a function of the radius only, the steady-state conduction for cylindrical coordinates is

0d dTrdr dr

⎛ ⎞ =⎜ ⎟⎝ ⎠

. . . (4.45)

4.2.3 Spherical Coordinates

For spherical coordinates, as shown in Figure 4.4 the temperature is a function of the three space coordinates r, θ and φ and time t, i.e. T = T (r, θ, φ, t). The general form of the conduction equation in spherical coordinates can be found as

dr

y

dφ φ

x

θ

z

dθ

Figure 4.4 : Spherical Coordinate System for the General Conduction Equation

2

22 2 2 2 2

1 1 1sinsin sin

GqT T Trr r kr r r

∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞+ θ + +⎜ ⎟ ⎜ ⎟∂ ∂ ∂θ ∂θ α ∂θ θ ∂φ⎝ ⎠ ⎝ ⎠

1 Tt

= . . . (4.46)

Where the Lapacian includes the first three terms of the above equation in the spherical coordinates.

4.3 STEADY STATE HEAT CONDUCTION IN SIMPLE GEOMETRICAL SYSTEMS

We will now derive solutions to the conduction equation as obtained in the previous section for simple geometrical systems with and without heat generation.

4.3.1 Plane Wall

In Unit 2 we saw that the temperature distribution for one-dimensional steady conduction through a wall is linear. We will verify the result by simplify the more general equation (Eq. 4.13)

2 1Gq TTk t

∂∇ + . . . (4.47) =

α ∂

12

Conduction For steady state, 0T

t∂

=∂

. Since T is only a function of x, 0Ty

∂=

∂ and 0T

z∂

=∂

. There is

no internal heat generation, qG = 0. Therefore, the above equation reduces 2

2 0d Tdx

= . . . (4.48)

Integrating the ordinary differential equation twice yields the linear temperature distribution

1 2( )T x C x C= + . . . (4.49)

For a wall as shown in Figure 4.5, at x = 0, T = T1 and at x = b, T = T2

T1

T2

b

Figure 4.5 : Heat Conduction Through a Plane Wall

1 22

T TT xb

T−= − + . . . (4.50)

Which agrees with the linear temperature distribution deduced by increasing Fourier’s law

kdTQ kAdx

= − . . . (4.51)

Let us now consider a heat source generating throughout the system. If the thermal conductivity is constant and the heat generation is uniform, Eq. (4.13) reduces to

2

2 0Gqd Tkdx

+ = . . . (4.52)

On integration, 1GqdT C

dx k= − + . . . (4.53)

A second integration gives

21 2( )

2GqT x x C x Ck

= − + + . . . (4.54)

where C1 and C2 are constants.

At x = 0, T = T1 and at x = b, T = T2 substituting in Eq. (4.52),

1T C2= . . . (4.55)

22 12

GqT b C bk 1T= − + + . . . (4.56)

13

Governing Equations of Heat Conduction2 1

1 2GqT TC b

b k−

= + . . . (4.57)

Therefore, the temperature distribution is

2 2 11( )

2 2G Gq qT TT x x x x Tk b k

−= − + + + . . . (4.58)

It may be seen that Eq. (4.49) is now modified by two terms containing the heat generation and that the temperature distribution is no longer linear. If the two surface temperatures are equal (T1 = T2), then the temperature distribution becomes

22

1( )2

Gq x xT x b Tk b b

⎡ ⎤⎛ ⎞= − − +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

. . . (4.59)

Which is a parabolic and symmetric about the central plane with a minimum Tmax

at 2bx = .

22

1 2 02

GqdT xbdx k b b

⎛ ⎞= − − =⎜ ⎟⎝ ⎠

. . . (4.60)

or 22x

bb=

1 . . . (4.61)

or 2bx = . . . (4.62)

and 2

MAX 1 8Gq b

T Tk

= + . . . (4.63)

In dimensionless form, on dividing Eq. (4.59) by Eq. (4.63)

70 xb

= . . . (4.64)

21

max 1

( ) 4 ( )T x TT T

−= ξ − ξ

− . . . (4.65)

Let us consider the case where heat is transferred from the two sides of the wall to the surrounding fluid at T∞ (Figures 4.6-4.7). For simplicity, let us assume that both the wall surfaces are at Tw (Figure 4.7). At steady state and for one-dimensional heat flow,

∞

qG = heat generated per unit volume b/2

T1

T2

∞

x

b

Figure 4.6 : Heat Flow through a Wall with Heat Generation

14

Conduction

2

2 0GqdTkdx

+ = . . . (4.66)

Let the excess temperature be

T Tαθ = − . . . (4.68)

So that 2 2

2 2d ddx dx

θ=

T . . . (4.69)

Figure 4.7 : Heat Transfer from Two Sides of a Wall having a Heat Source

Therefore, 2

2Gqdkdx

θ= − . . . (4.70)

1Gqd x C

dx kθ

= − + . . . (4.71)

and 21 22

Gq x C x Ck

θ = − + + . . . (4.72)

Which is parabolic. The central plane is the plane of symmetry where the solid

temperature is the maximum and 0ddx

θ= and is taken as the reference plane with x = 0.

At x = 0, 0ddx

θ= , C1 = 0 . . . (4.73)

At 2bx = ,

22

Gb

qd bdx k

θ⎛ ⎞ = −⎜ ⎟⎝ ⎠

Again, at wall

2

( )2 2

G Gk wb

q q bbq k k h T T hx k ∞

∂θ⎛ ⎞w= − = = = − =⎜ ⎟∂⎝ ⎠

θ . . . (4.74)

2G

wq b

hθ = . . . (4.75)

From Eq. (4.72)

222

Gqx C

kθ = − + . . . (4.76)

When ,2 wbx = θ = θ

∞

∞

Tω

o

kT∞

Tω

θo

k

h

T ∞

b/2 b/2

T∞

h

θω

∞

T∞ ∞

15


2

22 4G

wq bq C

k= − + . . . (4.77)

22

2 2 4 2 8G G G

wq q b qbC

k h k= θ + = + b . . . (4.78)

Therefore, the final temperature distribution is

2

2 2 8G G Gq q b qx Tk h k ∞θ = − + + + . . . (4.79)

2 2( 4 )8 2

G Gq qT b xk h

b T∞= − + + . . . (4.80)

This is the temperature distribution. At the mid-plane, x = 0 and T = Tmax.

2

8 2G Gq q bT bk h ∞= + + T . . . (4.81)

SAQ 1

(a) Show that the temperature profile for heat conduction through a plane wall of constant thermal conductivity is a straight line.

(b) Show that the temperature profile for heat conduction through a plane wall with a heat source and constant thermal conductivity is parabolic.

4.3.2 Cylindrical Wall Cylinder without Heat Generation

As shown in the Figure 4.8, heat is assumed to flow only radially in a hollow pipe consisting fluid inside the heat is transferred through a hollow pipe. We want to determine the temperature distribution and the heat transfer rate in a long hollow cylinder of length L if the inside and outside surface temperatures are Ti and To respectively and there is no internal heat generation.

Toqk

L

T1

ro

r1

T1T To

rro

r1dr

T = T(r) K = uniform qG = 0

(a) (b)

Figure 4.8 : Radial Heat Conduction through a Hollow Cylinder

Since the temperatures at two surfaces are constant, the temperature distribution in wall is not a function of time, and the conduction equation is given by Eq. (4.44)

16

Conduction 0d dTr

dr dr⎛ ⎞ =⎜ ⎟⎝ ⎠

. . . (4.82)

On integration

1dTr Cdr

= . . . (4.83)

or 1CdTdr r

= . . . (4.84)

A second integration yields

1 lnT C r C2= + . . . (4.85)

At ,i ir r T T= =

1 2lni iT C r C= + . . . (4.86)

At 0 0,r r T T= =

0 1 0 2 1 0ln ln lni iT C r C C r T C ir= + = + − . . . (4.87)

01

0ln

i

i

T TCrr

−=

⎛ ⎞⎜ ⎟⎝ ⎠

. . . (4.88)

and 02

0ln

ln

ii i

i

T TC T r

rr

−= −

⎛ ⎞⎜ ⎟⎝ ⎠

. . . (4.89)

Substituting C1 and C2 in Eq. (4.87),

0 0

0 0ln ln

ln ln

i ii

i i

T T T TT r Tr rr r

ir− −

= + −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

. . . (4.90)

0 0

ln( )

ln

ii

i

i

rrT r T

T T rr

⎛ ⎞⎜ ⎟

− ⎝ ⎠=− ⎛ ⎞

⎜ ⎟⎝ ⎠

. . . (4.91)

The rate of heat transfer by conduction

01

02 2

ln

ik

i

T TCdTQ kA k r L k Ldr r r

r

−= − = − π = − π

⎛ ⎞⎜ ⎟⎝ ⎠

. . . (4.92)

0

0

2 (

ln

ik

i

k L T TQrr

)π −=

⎛ ⎞⎜ ⎟⎝ ⎠

. . . (4.93)

Then the thermal resistance offered by the wall is

0

0ln

2ii

thk

rrT T

RQ k

⎛ ⎞⎜ ⎟

− ⎝ ⎠= =π L

. . . (4.94)

17


As shown in Figure 4.9, the rate of heat conduction through a composite cylindrical wall with convection at the inside and outside surfaces is given by

32

1 2

1 1 2 0

ln ln1 1

2 2 2 2

h c

th

i

T TTQR rr

r rh r L k L k L h k3 L

−Δ= =

∑ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+ + +

π π π π

. . . (4.95)

where Th and Tc are the hot and cold fluid temperatures, hi and ho are the inside and outside heat transfer coefficients and k1 and k2 are the thermal conductivities of the two walls in series.

Figure 4.9 : Temperature Distribution in a Composite Cylinder

with Convection at the Interior and Exterior Surfaces

Tc, ∞

T2

T1

r1 = n r2

r3 = r0

hc,o

Th, ∞T1

T2

T3

Th ∞ T1 T2 T3

Tc, ∞

hc,1

1 hc1 2πr1L

1 hc,02πr0L

ln(r2/r1) 2πkAL

ln(r3/r2) 2πkBL

q

B

L

T3

AL Th, ∞

Now . . . (4.96) 0 0 ( )h cQ U A T T= −

where Uo is the overall heat transfer coefficient based on outside area 2o oA r L= π given by

0 0 32

1 2

1 2 0

1 1

ln ln1 1

2 2

th

i i

RU A rr

r rh A k L k L h A

= ∑⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+ + +

π π 0

. . . (4.97)

where ri = r1 and ro = r3. To find the wall temperature T1, T2 and T3, we can use

1 1hi i

QT T Q Rh A

− = = . . . (4.98)

11 2

2

1

2

ln

Q kT Trr

π− =

⎛ ⎞⎜ ⎟⎝ ⎠

L . . . (4.99)

2 3o o

QT Th A

− = . . . (4.100)

Cylinder with Heat Generation Let us consider the Figure 4.10, a long solid cylinder of radius R with internal heat generation, such as an electric coil in which heat is generated as a result of electric current in the wire or a cylindrical nuclear fuel element in which heat is generated by nuclear fission.

18

The one-dimensional heat conduction equation in cylindrical coordinates is Conduction

1 0Gqd dTrr dr dr k

⎛ ⎞ + =⎜ ⎟⎝ ⎠

. . . (4.101)

Gq rd dTrdr dr k

⎛ ⎞ = −⎜ ⎟⎝ ⎠

. . . (4.102)

On integration, 2

12Gq rdTr

dr kC= − + . . . (4.103)

12Gq r CdT

dr k r= − + . . . (4.104)

Figure 4.10 : Temperature Distribution in a Cylindrical Rod with Heat Generation

By second integration, 2

1 2ln4Gq rT C r

k= + + C . . . (4.105)

At 0, 0dTrdr

= = (with origin at the centre line of the cylinder). But from

Eq. (4.104), dTdr

= ∞ , which is impossible.

1,2G

r R

q R CdTr Rdr k R=

⎛ ⎞= = −⎜ ⎟⎝ ⎠

+

L

. . . (4.106)

Heat generated in the cylindrical rod = π . 2Gq R

This heat is conducted to the surface and then converted away.

2 2Gr R

dTq R L k L Rdr =

⎛ ⎞π = − π ⎜ ⎟⎝ ⎠

. . . (4.107)

2G

r R

q RdTdr k=

⎛ ⎞ = −⎜ ⎟⎝ ⎠

. . . (4.108)

From Eqs. (4.104) and (4.108), C1 = 0

224

GqT rk

= − + C . . . (4.109)

qG

q

hc

Tω

T∞

q

hc

L

T∞O R

rTω

19


At , wr R T T= =

22 4

Gw

qC T Rk

= + . . . (4.110)

Substituting in Eq. (4.105)

2 2

4 4G G

wq qT r T

k k= − + + R . . . (4.111)

or 2 2

2 22( ) ( ) 1

4 4G G

w Wq q R rT r R r T T

k k R

⎛ ⎞= − + = − +⎜ ⎟⎜ ⎟

⎝ ⎠ . . . (4.112)

This is the maximum temperature variation along the wall radius. The maximum temperature occurs at r = 0

2

max 4G

wq RT

k= + T . . . (4.113)

In dimensionless form Eq. (4.112) becomes 2

max

( )1w

w

T r T rT T R

− ⎛ ⎞= − ⎜ ⎟− ⎝ ⎠ . . . (4.114)

For a hollow cylinder with uniformly distributed heat source and specified surface temperature T = Ti at r = ri and T = T0 at r = r0, Eq. (4.105) gives

21 2ln

4G

i i iqT r C r

k= − + + C . . . (4.115)

21 2ln

4G

o o oqT r C r

k= − + + C . . . (4.116)

Evaluating C1 and C2 from the above two equations and substituting in Eq. (4.105).

We can obtain the temperature distribution as

2 2 2 2ln

( ) ( ) ( )4 4

ln

oG Go o i o

o

i

rrq qT r r r r r T T T

k krr

⎛ ⎞⎜ ⎟

⎡ ⎤⎝ ⎠= − + − + − +⎢ ⎥⎛ ⎞ ⎣ ⎦⎜ ⎟⎝ ⎠

i o . . . (4.117)

If a solid cylinder is immersed in a fluid at T∞ and the convection heat transfer coefficient is hc and T = Tw at r = R, the heat conduction from the cylindrical is equal to the rate of convection at the surface, or

( )c wr R

dTk h Tdr

T∞=

⎛ ⎞− =⎜ ⎟⎝ ⎠

− . . . (4.118)

From Eq. (4.108)

( )2

c w G

r R

h T T q RdTdr k k

∞

=

−⎛ ⎞ = − = −⎜ ⎟⎝ ⎠

. . . (4.119)

2G

wc

q RT Th∞= + . . . (4.120)

From Eq. (4.112)

2 2

2( ) 14 2

G

c

q R q RrT r Tk hR ∞

⎛ ⎞= − + +⎜ ⎟⎜ ⎟

⎝ ⎠

G . . . (4.121)

20

In dimensionless form, Conduction

2( ) 2 14

G c

c

q R h RT r T rT h T k R

∞

∞ ∞

⎡ ⎤⎧ ⎫− ⎪ ⎪⎛ ⎞⎢ ⎥= + −⎨ ⎬⎜ ⎟⎝ ⎠⎢ ⎥⎪ ⎪⎩ ⎭⎣ ⎦

. . . (4.122)

and maximum temperature is

max 1 24

G c

c

T q R hT h T∞ ∞

⎛= + +⎜⎝ ⎠

Rk

⎞⎟ . . . (4.123)

There are two dimensionless parameters in the above equation which are important

in conduction, viz. G

c

q Rh T∞

the heat generation number, and ch Rk

the Biot number,

which appears in problems with simultaneous conduction and convection.

The Biot number is the ratio of conduction resistance to convection resistance or

1

ok

ic

c

rc oR h rkB

R kh

= = = . . . (4.124)

The limits of Biot numbers are

0iB → when 0ok

rRk

= → or, 1c

cR

h= → ∞

iB → ∞ when 1 0cc

Rh

= → or, ok

rRk

= → ∞

The Biot number approaches zero when the conductivity of solids very large (k → ∞) or the convection coefficient of heat transfer is very low (hc → 0), i.e. when the solid is practically isothermal and the temperature change is mostly caused in the fluid by convection at the interface. On the contrary, the Biot number approaches infinity when the thermal resistance predominates (k → 0) or the convection resistance is very low (hc → 0).

SAQ 2

(a) Show that the maximum temperature in a cylindrical rod with heat

generation 3kWmGq ⎛

⎜⎝ ⎠

⎞⎟ is given by max 1 2

4G c

c

T q R hT h T∞ ∞

⎛ ⎞= + +⎜ ⎟⎝ ⎠

Rk

.

(b) What are the limiting values of Biot number?

(c) What is heat generation number?

4.3.3 Spherical Wall As shown in Figure 4.11 a hollow sphere with uniform temperature at the inner and outer surfaces, the temperature distribution without heat generation in the steady state can be

21


obtained by simplifying Eq. (4.46). Under these conditions the temperature is only a function of the radius r, and the conduction equation in the spherical coordinates is

22

1 0d dTrdr drr

⎛ ⎞ =⎜ ⎟⎝ ⎠

. . . (4.125)

On integration, and putting at r = ri, T = Ti and r = ro, T = To

( ) ( ) 1o ii o i

o i

r rT r T T T

r r r⎛ ⎞− = − −⎜ ⎟− ⎝ ⎠

. . . (4.126)

T0r0

T1

T = T(r) K = uniform q″G = 0

r1

qk

Figure 4.11 : Heat Conduction in a Hollow Sphere without Heat Generation

The rate of heat transfer through the spherical wall is

24 ( )4

i ok

o i

o i

T TdTQ k r r rdrk r r

−= − π =

−π

. . . (4.127)

The thermal resistance for spherical wall is then

4o i

tho i

r rRk r r−

=π

. . . (4.128)

4.4 TRANSIENT CONDUCTION

Many heat transfer problems are time dependent. Such unsteady, or transient, problems typically arise when the boundary conditions of a system are changed. For example, if the surface temperature of a system is altered, the temperature at each point in the system will also begin to change. The changes will continue to occur until a steady-state temperature distribution is reached. Consider a hot metal billet that is removed from a furnace and exposed to a cool air stream. Energy is transferred by convection and radiation from its surface to the surroundings. Energy transfer by conduction also occurs from the interior of the metal to the surface, and the temperature at each point in the billet decreases until a steady-state condition is reached. Such time dependent effects occur in many industrial heating and cooling processes. In the following sub-units, transient heat transfer analysis has been discussed.

4.4.1 Lumped Capacitance Method A simple, yet common, transient conduction problem is one for which a solid experiences a sudden change in its thermal environment. Consider a hot metal forging that is initially at a uniform temperature Ti and is quenched by immersing it in a liquid of lower temperature T∞ < Ti as shown in Figure 4.12. If the quenching is said to begin at time t = 0, the temperature of the solid will decrease for time t > 0, until it eventually reaches T∞. This reduction is due to convection heat transfer at the solid-liquid interface. The essence of the lumped capacitance method is the assumption that the temperature of the solid is

22

spatially uniform at any instant during the transient process. This assumption implies that temperature gradients within the solid are negligible.

Conduction

From Fouriers law, heat conduction in the absence of a temperature gradient implies the existence of infinite thermal conductivity. Such a condition is clearly impossible. However, although the condition is never satisfied exactly, it is closely approximated if the resistance to conduction within the solid is small compared with the resistance to heat transfer between the solid and its surroundings. For now we assume that this is, in fact, the case.

In neglecting temperature gradients within the solid, we can no longer consider the problem from within the framework of the heat equation. Instead, the transient temperature response is determined by formulating an overall energy balance on the solid. This balance must relate the rate of heat loss at the surface to the rate of change of the internal energy. Thus, with respect to the Figure 4.12,

out stE E− = . . . (4.134)

or ( )sdTh A T T Vcdt∞− − = ρ . . . (4.135)

t < 0 T = Ti

Liquid

T∞ ≤ Y1

Ti

T(t)

t ≥ 0 T = T(t)

Eout = qconvenction

Esi

Figure 4.12 : Cooling of a Hot Metal Forging

Introducing the temperature difference

T T∞θ ≡ − . . . (4.136)

And recognizing that d dTdt dtθ⎛ ⎞ ⎛=⎜ ⎟ ⎜

⎝ ⎠ ⎝⎞⎟⎠

, it follows that

s

Vc dh A dtρ θ

= − θ . . . (4.137)

Separating variables and integrating from the initial condition, for which t = 0 and T (0) = Ti, we then obtain

0i

t

s

Vc d dth A

θ

θ

ρ θ= −

θ∫ ∫ . . . (4.138)

where i iT T∞θ ≡ − . . . (4.139)

Evaluating the integrals it follows that

ln i

s

Vc th A

θρ= =

θ . . . (4.140)

or exp s

i i c

h AT T tT T V

∞

∞

⎡ ⎤⎛ ⎞−θ= = −⎢ ⎥⎜ ⎟θ − ρ⎢ ⎥⎝ ⎠⎣ ⎦

. . . (4.141)

23


Eq. (4.140) may be used to determine the time required for the solid to reach some temperature T, or, conversely, Eq. (4.141) may be used to compute the temperature reached by the solid at same time t.

The foregoing results indicate that the difference between the solid and fluid temperatures must decay exponentially to zero as t approaches infinity. This behaviour is

shown in Figure 4.13. From Eq. (4.141) it is also evident that the quantity s

Vch A

⎛ ⎞ρ⎜⎝ ⎠

⎟ may

be interpreted as a thermal time constant τs. This time constant may be expressed as

1 ( )t ts

Vc R Ch A

⎛ ⎞τ = ρ =⎜ ⎟

⎝ ⎠t

dt

. . . (4.142)

τt = ρVc

hAs= Rt Ct

0.368

0 τt. 1 τt. 2 τt. 3 τt. 4

T T

T T1 1

−θ ∞=θ − ∞

1

t

Figure 4.13 : Transient Temperature Response of Lumped Capacitance Solids

for Different Thermal Time Constants τs

where Rt is the resistance to convection heat transfer and Ct is the lumped thermal capacitance of the solid. Any increase in Rt or Ct will cause a solid to respond more slowly to changes in its thermal environment and will increase the time required to reach thermal equilibrium (θ = 0). This behavior is analogous to the voltage decay that occurs when a capacitor is discharged through a resistor in an electrical RC circuit.

To determine the total energy transfer Q occurring up to some time t, we simply write

0 0

t tsQ q dt h A= = θ∫ ∫ . . . (4.143)

Substituting for θ from Eq. (4.140) and integrating, we obtain

( ) 1 expit

tQ Vc⎡ ⎤⎛ ⎞

= ρ θ − −⎢ ⎥⎜ ⎟τ⎢ ⎥⎝ ⎠⎣ ⎦ . . . (4.144)

The quantity Q is, of course, related to the change in the internal energy of the solid, and from Eq. (4.134)

stQ E− = Δ . . . (4.145)

For quenching Q is positive and the solid experiences a decrease in energy. Eqs. (4.140), (4.141) and (4.144) also apply to situations where the solid is heated (θ < 0), in which case Q is negative and the internal energy of the solid increases.

Validity of the Lumped Capacitance Method

From the foregoing results it is easy to see why there is a strong preference for using the lumped capacitance method. It is certainly the simplest and most convenient method that can be used to solve transient conduction problems. Hence,

24

it is important to determine under what conditions it may be used with reasonable accuracy.

Conduction

As shown in Figure 4.14, to develop a suitable criterion consider steady-state conduction through the plane wall of area A.

Although we are assuming steady-state condition, this criterion is readily extended to transient processes. One surface is maintained at a temperature Ts, 1 and the other surface is exposed to a fluid of temperature T∞ < Ts, 1. The temperature of this surface will be some intermediate value, Ts, 2, for which T∞ < Ts, 2 < Ts, 1. Hence, under steady-state conditions the surface energy balance is

,1 ,2 ,2( ) (s s skA T T hA T TL

)∞− = − . . . (4.146)

where k is the thermal conductivity of the solid. Rearranging, we then obtain

,1 , 2 cond

, 2 conv1s s

s

LT T R hLkA BiT T R k

hA∞

⎛ ⎞⎜ ⎟− ⎝ ⎠= = = =

− ⎛ ⎞⎜ ⎟⎝ ⎠

. . . (4.147)

XL

Ts. 1

T

Bi = 1

Bi >> 1

Bi << 1

qcond qconv

Ts. 2

Ts. 2

Ts. 2

T∞, h

Figure 4.14 : Effect of Biot Number on Steady-State Temperature Distribution

in a Plane Wall with Surface Convection

The quantity hLk

⎛⎜⎝ ⎠

⎞⎟ is a dimensionless parameter appearing in Eq. (4.147). It is

termed the Biot number, and it plays a fundamental role in conduction problems that involve surface convection effects. According to Eq. (4.147) and as illustrated in Figure 4.14, the Biot number provides a measure of the temperature drop in the solid relative to the temperature difference between the surface and the fluid. Note especially the conditions corresponding to Bi < < 1. The results suggest that, for these conditions, it is reasonable to assume a uniform temperature distribution across a solid at any time during a transient process. This result may also be associated with interpretation of the Biot number as a ratio of thermal resistances, Eq. (4.147). If Bi < < 1, the resistance to conduction within the solid is much less than the resistance to convection across the fluid boundary layer. Hence, the assumption of a uniform temperature distribution is reasonable.

We have introduced the Biot number because of its significance to transient conduction problems. Consider the plane wall of Figure 4.15, which is initially at a uniform temperature Ti and experiences convection cooling when it is immersed in a fluid of T∞ < Ti. The problem may be treated as one-dimensional in x, and we are interested in the temperature variation with position and time, T (x, t). This

25


variation is a strong function of the Biot number, and three conditions are shown in Figure 4.15.

For Bi < < 1 the temperature gradient in the solid is small and T (x, t) ≈ T (t). Virtually all the temperature difference is between the solid and the fluid, and the solid temperature remains nearly uniform as it decreases to T∞. For moderate to large values of the Biot number, however, the temperature gradients within the solid are significant. Hence, T = T (x, t). Note that for Bi > > 1, the temperature difference across the solid is much larger than that between the surface and the fluid.

Lumped capacitance method is a preferred method for solving transient conduction problems. Hence, when confronted with such a problem, the very first thing that one should do is calculate the Biot number. If the following condition is satisfied

0.1h LcBik

= < . . . (4.148)

T∞, h

Figure 4.15 : Transient Temperature Distributions for different Biot Numbers

in a Plane Wall Symmetrically Cooled by Convection

The error associated with using the lumped capacitance method is small. For convenience, it is customary to define the characteristic length of Eq. (4.148) as

the ratio of the solids volume to surface area, cs

VLA

= . Such a definition

facilitates calculation of Lc for solids of complicated shape and reduces to the

half-thickness L for a plane wall of thickness 2 L as shown in Figure 4.15, to 02r

for a long cylinder, and to 03r for a sphere. However, if one wishes to implement

the criterion in a conservative fashion, Lc should be associated with the length scale corresponding to the maximum spatial temperature difference. Accordingly, for a symmetrically heated (or cooled) plane wall of thickness 2 L, Lc would remain equal to the half-thickness L. However, for a long cylinder or sphere, Lc

would equal the actual radius r0, rather than 02r or 0

3r .

Finally, we note that, with cs

VLA

≡ , the exponent of Eq. (4.141) may be expressed

as

2 2s c

c c c

h A t h L h Lht k t tVc c L k c kL L

c α= = =

ρ ρ ρ . . . (4.149)

T∞, h

-L L x

T(x, O) = T1 T(x, O) = T1

T∞

-L L -L L -L L

T∞ T∞ T∞

Bi << 1T = T(t)

Bi = 1 T = T(x, t)

Bi >> 1 T = T(x, t)

26

or .sh A t Bi FoVc

=ρ

. . . (4.150) Conduction

where 2c

tFoLα

≡ . . . (4.151)

is termed the Fourier number. It is a dimensionless time, which, with the Biot number, characterizes transient conduction problems. Substituting Eq. (4.150) into Eq. (4.141), we obtain

exp ( . )i i

T T Bi FoT T

∞

∞

−θ= = −

θ − . . . (4.152)

SAQ 3 (a) What do you mean by Biot number and Fourier number? (b) What are the conditions for validity of lumped capacitance method?

4.4.2 General Lumped Capacitance Analysis

Although transient conduction in a solid is commonly initiated by convection heat transfer to or from an adjoining fluid, other processes may induce transient thermal conditions within the solid. For example, a solid may be separated from large surroundings by a gas or vacuum. If the temperatures of the solid and surroundings differ, radiation exchange could cause the internal thermal energy, and hence the temperature, of the solid to change. Temperature changes could also be induced by applying a heat flux at a portion, or all, of the surface and/or by initiating thermal energy generation within the solid. Surface heating could, for example, be applied by attaching a film or sheet electrical heater to the surface, while thermal energy could be generated by passing an electrical current through the solid.

Figure 4.16 depicts a situation for which thermal conditions within a solid may be influenced simultaneously by convection, radiation, an applied surface heat flux, and internal energy generation.

Figure 4.16 : Control Surface for Generalized Lumped Capacitance Analysis

It is presumed that, initially (t = 0), the temperature of the solid (Ti) differs from that of the fluid, T∞, and the surroundings, Tsur, and that both surface and volumetric heating (qs″ and q) are initiated. The imposed heat flux qs″ and the convection-radiation heat transfer occur at mutually exclusive portions of the surface, As (h) and As (c, r), respective and convection-radiation transfer is presumed to be from the surface. Moreover, although convection and radiation have been prescribed for the same surface, the surfaces may, in fact, differ (As, c ≠ As, r). Applying conservation of energy at any instant t

P, c, V, T(0) = Ti

q”s

As, h

Eg, E

As(c,r)

q”conv

q”rad

st

Surroundings

T sur

T∞, h

27

Governing Equations of Heat Conduction , ( ,( )s s h g conv rad s c r

dTq A E q q A Vcdt

′′ ′′ ′′+ − + = ρ) . . . (4.153)

or, 4 4, (( ) ( )s s h g sur s c r

dTq A E h T T T T A Vcdt∞⎡ ⎤′′ + − − + εσ − = ρ⎣ ⎦ , ) . . . (4.154)

Eq. (4.154) is a nonlinear, first-order, nonhomogenous, ordinary differential equation that cannot be integrated to obtain an exact solution. However, exact solutions may be obtained for simplified versions of the equation. For example, if there is no imposed heat flux or generation and convection is either nonexistent (a vacuum) or negligible relative to radiation. Eq. (4.154) reduces to

4 4, ( )s r s

dTVc A T Tdt

ρ = − ε σ − ur . . . (4.155)

Separating variables and integrating from the initial condition to any time t, it follows that

,4 40 i

t Ts r

Tsur

A dTdtVc T T

ε σ=

ρ −∫ ∫ . . . (4.156)

Evaluating both integrals and rearranging, the time required to reach the temperature T becomes

3,

ln ln4

sur sur i

sur sur is r sur

T T T TVctT T T TA T

⎧ + +ρ ⎪= −⎨ − −ε σ ⎪⎩

1 12 tan tan i

sur sur

TTT T

− −⎫⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎪+ −⎢ ⎥⎬⎜ ⎟ ⎜ ⎟

⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎪⎣ ⎦⎭ . . . (4.157)

This expression cannot be used to evaluate T explicitly in terms of t, Ti, and Tsur nor does it readily reduce to the limiting result for Tsur = 0 (radiation to deep space). However, returning to Eq. (4.156), its solution for Tsur = 0 yields

3 3,

1 13 s r i

VctA T T

⎛ ⎞ρ= −⎜ ⎟⎜ ⎟ε σ ⎝ ⎠

. . . (4.158)

An exact solution to Eq. (4.154) may also be obtained if radiation may be neglected and

h is independent of time. Introducing a reduced temperature θ ≡ T – T∞, where d ddt dtθ

=T ,

Eq. (4.154) reduces to a linear, first-order, non-homogenous differential equation of the form

0d a bdtθ

+ θ − = . . . (4.159)

where ,s ch Aa

Vc⎛ ⎞

= ⎜ ⎟ρ⎝ ⎠

and ,s s c gq A Eb

Vc

′′⎡ ⎤+⎛ ⎞= ⎢ ⎥⎜

ρ⎟

⎢ ⎥⎝ ⎠⎣ ⎦. Although Eq. (4.159) may be solved by

summing its homogeneous and particular solutions, an alternative approach is to eliminate the non-homogeneity by introducing the transformation

ba

′θ ≡ θ − . . . (4.160)

Recognising that d ddt dt

′θ=

θ , Eq. (4.160) may be substituted into Eq. (4.159) to yield

0d adt

′θ ′+ θ = . . . (4.161)

Separating variables and integrating from 0 to t ( to )i′ ′θ θ , it follows that

28

exp ( )i

at′θ

= −′θ

. . . (4.162) Conduction

or substituting for and θ, ′θ

exp ( )i

bT Ta atbT Ta

∞

∞

⎛ ⎞− − ⎜ ⎟⎝ ⎠ = −⎛ ⎞− − ⎜ ⎟⎝ ⎠

. . . (4.163)

Hence [exp ( ) 1 exp ( )i i

bT T aat atT T T T

∞

∞ ∞

−= − + − −

− −] . . . (4.164)

As it must, Eq. (4.164) reduces to Eq. (4.141) when b = 0 and yields T = Ti at t = 0. As

t → ∞, Eq. (4.164) reduces to ( ) bT Ta∞

⎛ ⎞− = ⎜ ⎟⎝ ⎠

, which could also be obtained by

performing an energy balance on the control surface of Figure 4.16 for steady-state conditions.

4.5 THE SEMI-INFINITE SOLID

A semi-infinite solid is characterized by a single identifiable surface as shown in Figure 4.17. In principle, such a solid extends to infinity in all but one direction. If a sudden change of conditions is imposed at this surface, transient, one-dimensional conduction will occur within the solid. The semi-infinite solid provides a useful idealization for many practical problems. It may be used to determine transient heat transfer near the surface of the earth or to approximate the transient response of a finite solid, such as a thick slab. For this second situation the approximation would be reasonable for the early portion of the transient, during which temperatures in the slab interior (well removed from the surface) are uninfluenced by the change in surface conditions.

Case (1) T(x,o) = Ti T(o,t) = Ts

Case (2) T(x,o) = Ti

-k ∂ T/∂xx=o = q”o

Case (3) T(x,o) = Ti

-k ∂/∂xx=o = h(T∞ 0 – T(O,i)

Ts q”T∞, h

x x x

Figure 4.17 : Transient Temperature Distributions in a Semi-Infinite Solid for Three Surface Conditions (a) Constant Surface Temperature, (b) Constant Surface Heat Flux,

and (c) Surface Convection

The interior boundary condition is of the form

( , ) iT x t T→ ∞ = . . . (4.165)

Closed-form solutions have been obtained for three important surface conditions, instantaneously applied at t = 0. These conditions are shown in Figure 4.17. They include application of a constant surface temperature application of a constant surface heat flux

Ts

Ti

x

t

T (x, t)

t

Ti x

T∞

Ti

tt

x x

t

x

x

29


T0q′′ and exposure of the surface to a fluid characterized by iT∞ ≠ and the convection coefficient h.

The solution for Case 1 may be obtained by recognizing the existence of a similarity variable η, through which the heat equation may be transformed from a partial differential equation, involving two independent variables (x and t), to an ordinary differential equation expressed in terms of the single similarity variable. To confirm that

such a requirement is satisfied by 12(4 )

x

t

η =

α

, we first transform the pertinent

differential operators, such that

( )1 21

4

T dT dTx d x dt

∂ ∂η= =

∂ η ∂ α η . . . (4.166)

2 2

21

4T d T d T

d x x t 2x d∂ ∂ ∂η⎡ ⎤= =⎢ ⎥η ∂ ∂ α∂ η⎣ ⎦

. . . (4.167)

122 (4 )

T dT x dTt d t d

t t

∂ ∂η= = −

∂ η ∂α

η . . . (4.168)

With assumption of no internal heat generation and constant thermal conductivity the heat equation becomes

2

2 2d T dTdd

= − ηηη

. . . (4.169)

With x = 0 corresponding to η = 0, the surface condition may be expressed as

( 0) sT η = = T . . . (4.170)

And with x → ∞, as well as t = 0, corresponding to η → ∞, both the initial condition and the interior boundary condition correspond to the single requirement that

( ) iT n T→ ∞ = . . . (4.171)

Since the transformed heat equation and the initial/boundary conditions are independent

of x and t, 12(4 )

x

t

η =

α

is indeed, a similarity variable. Its existence implies that,

irrespective of the values of x and t, the temperature may be represented as a unique function of η.

The specific form of the temperature dependence, T (η), may be obtained by separating variables in Eq. (4.169), such that

2

dtdd

ddTd

⎛ ⎞⎜ ⎟η⎝ ⎠ = − η η

⎛ ⎞⎜ ⎟η⎝ ⎠

. . . (4.172)

Integrating, it follows that

21ln dT C

d⎛ ⎞ ′= − η +⎜ ⎟η⎝ ⎠

. . . (4.173)

30

Conduction or 2

1 exp ( )dT Cd

= − ηη

. . . (4.174)

Integrating a second time, we obtain

21 0

exp ( )T C u du Cη

= −∫ 2+ . . . (4.175)

Where u is a dummy variable. Applying the boundary condition at η = 0, Eq. (4.170), it follows that C2 = T5 and

21 0

exp ( ) sT C u du Tη

= − +∫ . . . (4.176)

From the second boundary condition, Eq. (4.171), we obtain

21 0

exp ( )i sT C u du T∞

= −∫ + . . . (4.177)

or, evaluating the definite integral,

1 12

2 ( )i sT TC

−=

π

. . . (4.178)

Hence, the temperature distribution may be expressed as

21 02

2 exp ( ) erfs

s

T Tu du

T Tη

⎛ ⎞− ⎜ ⎟= − ≡⎜ ⎟− ⎜ ⎟π⎝ ⎠

∫ η . . . (4.179)

Where the Gaussian error function, erf η, is a standard mathematical function that is tabulated in Appendix A. The surface heat flux may be obtained by applying Fourier’s law at x = 0, in which case

0 0

(erf )( )s i sx

T dq k k T Tx d x= η =

∂′′ = − = − −∂ η

η ∂η∂

. . . (4.180)

12 2

12 0

2( ) exp ( ) (4 )s s iq k T T t−

η =

⎛ ⎞⎜ ⎟′′ = − − η α⎜ ⎟⎜ ⎟π⎝ ⎠

. . . (4.181)

12

(

( )

)s is

k T Tq

t

−′′ =

π α

. . . (4.182)

Analytical solutions may also be obtained for the Case 2 and Case 3 surface conditions, and results for all three cases are summarized as follows.

Case 1 : Constant Surface Temperature : T (0, t) = Ts

( , ) erf2

s

i s

T x t T xT T t

−=

− α . . . (4.183)

(( ) )s is

k T Tq tt

−′′ =πα

. . . (4.184)

Case 2 : Constant Surface Heat Flux : s′′ ′q t q0( ) = ′

31


12

0 20

2( , ) exp erf

4 2i

tqq xx xT x t T c

k t k t

α⎛ ⎞′′ ⎜ ⎟ ⎛ ⎞⎛ ⎞ ′′−π⎝ ⎠− = − ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟α α⎝ ⎠ ⎝ ⎠ . . . (4.185)

Case 3 : Surface Convection

0[ (0,

x

Tk h T Tx ∞

=

∂− = −

∂)]t

2

2( , )

erf exp2

i

i

T x t T x h x h tcT T kt k∞

⎡ ⎤⎛ ⎞ ⎛ ⎞− α= − +⎜ ⎟ ⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟− α ⎢ ⎥⎝ ⎠⎝ ⎠ ⎣ ⎦

erf2

h txckt

⎡ ⎤⎛ ⎞α⎢ +⎜⎜ α

⎥⎟⎟⎢ ⎥⎝ ⎠⎣ ⎦ . . . (4.186)

The complementary error function, erf c w, is defined as erf c w ≡ 1 – erf c w. Temperature histories for the three cases are shown in Figure 4.17, and distinguishing features should be noted. With a step change in the surface temperature, Case 1, temperatures within the medium monotonically approach Ts with increasing t, while the magnitude of the surface temperature gradient, and hence the surface heat flux, decreases

as 12t . In contrast, for a fixed surface heat flux (Case 2), Eq. (4.185) reveals that

increases monotonically as

−

(0, ) ( )sT t T t=12t . For surface convection (Case 3), the

surface temperature and temperatures within the medium approach the fluid temperature T∞ with increasing time. As Ts approaches T∞ there is, of course, a reduction in the surface heat flux, . Specific temperature histories computed from Eq. (4.186) are plotted in Figure 4.18. The result corresponding to h = ∞ is equivalent to that associated with a sudden change in surface temperature, Case 1. That is, for h = ∞, the surface instantaneously achieves the imposed fluid temperature (T

( ) [ ( )]sq t h T T t∞′′ = − s

s = T∞), and with the second term on the right-hand side of Eq. (4.185) reducing to zero, the result is equivalent to Eq. (4.183).

0.001 0 0.5 1.0 1.5

0.005

0.1

0.5

1.0

∞ 3

2 1

0.5 0.4

0.3 0.2

0.1

0.05H √αt = 0.05

k

T∞ T(x,t)h

x

2√αt 1 x

T –

T iT

∞T i

Figure 4.18 : Temperature Histories in a Semi-Infinite Solid with Surface Convection

32

An interesting permutation of Case 1 results when two semi-infinite solids, initially at uniform temperatures TA, i and TB,

Conduction

B i, are placed in contact at their free surfaces (Figure 4.19). If the contact resistance is negligible, the requirement of thermal equilibrium dictates that, at the instant of contact (t = 0), both surfaces must assume the same temperature Ts, for which TB, B i < Ts < TA, i. Science Ts does not change with increasing time, it follows that the transient thermal response and the surface heat flux of each of the solids are determined by Eqs. (4.179) and (4.182), respectively.

The equilibrium surface temperature of Figure 4.19 may be determined from a surface energy balance, which requires that

, ,s A sq q B′′ ′′= . . . (4.187)

Substituting from Eq. (4.182) for ,s Aq′′ and ,s Bq′′ and recognising that the x coordinate of Figure 4.19 requires a sign change for ,s Aq′′ , it follows that

( ) ( )

, ,1 12 2

( ) ( )A s A i B s B i

A B

k T T k T T

t t

− − −=

π α π α . . . (4.188)

or, solving for Ts,

1/ 2 1/ 2, ,

1 2 1 2

( ) ( )

( ) ( )A A i B B i

sBA

k c T k c TT

k c k c

ρ + ρ=

ρ + ρ . . . (4.189)

T

x

TA, i

q”

q”s,A

Ts

t

t

kA, pA, cA

A

TB, i

s,B

kB, pB, cB

B

Figure 4.19 : Interfacial Contact between Two Semi-Infinite Solids at Different Initial Temperatures

Hence the quantity 12( )m k c≡ ρ is a weighing factor that determines whether Ts will

more closely approach , ( )A i A BmT m > or , ( )B i B AmT m > . = =

SAQ 4

(a) What do you mean by a semi-infinite solid? What is its speciality?

(b) What are the three surface conditions in solution of a transient heat conduction problem for a semi-infinite solid?

33

Governing Equations of Heat Conduction4.6 MULTIDIMENSIONAL EFFECTS

Consider immersing the short cylinder of Figure 4.20, which is initially at a uniform temperature Ti, in a fluid of temperature T∞ ≠ Ti. Because the length and diameter are comparable, the subsequent transfer of energy by conduction will be significant for both the r and x coordinate directions. The temperature within the cylinder will therefore depend on r, x, and t.

Assuming constant properties and no generation, the appropriate form of the heat equation is

2

21 1T Trr r r tx

∂ ∂ ∂ ∂⎛ ⎞ + =⎜ ⎟∂ ∂ α ∂∂⎝ ⎠

T . . . (4.190)

where x has been used in place of z to designate the axial coordinate. A closed-form solution to this equation may be obtained by the separation of variables method. Although we will not consider the details of this solution, it is important to note that the end result may be expressed in the following form :

Plane Infinitewall cylinder

( , , ) ( , ) ( , ).i i i

T r x t T T x t T T r t TT T T T T T

∞ ∞ ∞

∞ ∞ ∞

− − −=

− − − . . . (4.191)

ror

T(r, x, t)

Midplane

-L

x

+L

L

L

(r, x)

=

ro

L

L

r

r

x x

x

T∞, h

T∞, h

T∞, h

T∞, h

i i i

θ(r, x.t) θ(r, t) θ(x, t)= ×

θ θ θ

* * * * *θ =c(r ,t )×P(x ,t )Figure 4.20 : Two-Dimensional Transient Conduction in a Short Cylinder,

(a) Geometry, (b) Form of the Product Solution

That is, the two-dimensional solution may be expressed as a product of one-dimensional solutions that correspond to those for a plane wall of thickness 2 L and an infinite cylinder of radius r0. For F0 > 0.2, these solutions are provided by Figures D.1 and D.2 for the plane wall and Figures D.3 and D.4 for the infinite cylinder. Results for others multidimensional geometries are summarized in Figure 4.21. In each case the multidimensional solution is prescribed in terms of a product involving one or more of the following one-dimensional solutions :

Semi-infinitesolid

( , )( , )i

T x t TS x tT T

∞

∞

−≡

− . . . (4.192)

Planewall

( , )( , )i

T x t TP x tT T

∞

∞

−≡

− . . . (4.193)

Infinitecylinder

( , )( , )i

T r t TC r tT T

∞

∞

−≡

− . . . (4.194)

34

The x coordinate for the semi-infinite solid is measured from the surface, whereas for the plane wall it is measured from the mid plane. In using Figure 4.21 the coordinate origins should carefully be noted. The transient, three-dimensional temperature distribution in a rectangular parallelepiped, Figure 4.21(h), is then, for example, the product of three one-dimensional solutions for plane walls of thicknesses 2L1 and 2L3. That is

Conduction

1 2 31 2 3

( , , , ) ( , ) . ( , ) . ( , )i

T x x x t T P x t P x t P x tT T

∞

∞

−=

− . . . (4.195)

The distances x1, x2 and x3 are all measured with respect to a rectangular coordinate system whose origin is at the center of the parallelepiped.

Figure 4.21 : Solutions for Multi-Dimensional Systems Expressed

as products of One Dimensional Results

The amount of energy Q transferred to or from a solid during a multi-dimensional transient conduction process may also be determined by combining one-dimensional results.

Example 4.1

(a) Semi-infinite Solid

S(x,t) P(x,t)

2L1

(b) Plane Wall (c) Infinite Cylinder

C(r,t)

S(x1,t)P(x2,t) P(x1,t)P(x2,t) C(x1,t)S(x,t)

x2

x1

2L1

2L2

x1

2L2

x2

x

ro

(d) Semi-infinite Plate (e) Infinite Rectangular Bar (f) Semi-infinite Cylinder

r

ro

S(x3,t)P(x1,t)P(x2,t)

2L2

2L1 x1

x3

x2

S(x1,t)P(x2,t)P(x3,t)

x3

x1x2

2L1

2L2

2L2

2Lt

C(r,1)P(x,t)

r

r x

(g) Semi-infinite Rectangular Bar (h) Rectangular Parallelepiped (i) Short Cylinder

x x r

35


The air inside a chamber at T∞, i = 50oC is heated convectively with hi = 20 W/m2 K by a 200-mm thick wall having thermal conductivity of 4 W/m K and a uniform heat generation of 1000 W/m3. To prevent any heat, generated within the wall, from being lost to the outside of the chamber at T∞, 0 = 25oC with h0 = 5 W/m2 K, a very thin electrical strip heater is placed on the outer wall to provide a uniform heat flux . 0q′′

(a) Sketch the temperature distribution in the wall on T – x coordinates for the condition where no heat generated within the wall is lost to the outside chamber. Identity T∞, i and T∞, 0 on the plot.

(b) What are the temperatures at the wall boundaries T0 and TL (at x = 0 and x = L) for the conditions of Part (a).

(c) Determine the value of that must be supplied by the strip heater so that all heat generated within the wall is transferred to the inside of the chamber for the temperature T

0q′′

0 as computed in part (b).

Solution 2

2 0Gqd Tkdx

+ =

1GqdT x C

dx k+ + . . . (1)

(a) 21 22

GqT x C xk

= + + C

(b) At x = 0, T = T0,

0dTdx

= . . . (2)

From Eq. (1), C1 = 0

222

GqT xk

= − + C

At x = L, T = TL

222

GL

qT Lk

= − + C

or 2

2 2G

Lq LC T

k= +

The temperature distribution in the wall is thus,

2 2( )2

GL

qT L xk

= − + T . . . (3)

At x = 0, from Eq. (3),

20 max 2

GL

qT T T L Tk

= = = +

21000 (0.2)

2 4 LT×= +

× . . . (4)

Again, at the inside wall surface, from Eq. (1),

,( ) Gi L i

x L

q LdTh T T k kdx k∞

=

⎛ ⎞− = − =⎜ ⎟⎝ ⎠

36

o,

1000 0.2 10 C20

GL i

i

q LT Th∞

×− = = =

Conduction

o50 10 60 CLT = + = C Ans. (a)

From Eq. (4),

o0 max 5 60 65 CT T= = + = Ans. (b)

(c) 0 0 0 ,0( ) 5 (65q h T T∞= − = − 25)

= 200 W/m2 Ans. (c)

The temperature profile is shown in Figure 4.22 given below :

T0x

Figure 4.22

Example 4.2

In a cylindrical fuel rod of a nuclear reactor heat is generated internally, accordingly according to the equation

2

00

1Grq qr

⎡ ⎤⎛ ⎞⎢ ⎥= − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

where qG is the local rate of heat generation per unit volume at radius r, ro is the outside radius, and q0 is the rate of heat generation per unit volume at the centre line. Calculate the temperature drop from the centre line to the surface for a 2.5 cm outer diameter rod having k = 25 W/m K, if the rate of heat removal from the surface is 1650 kW/m2.

Solution

In cylindrical coordinates, the radial variation of temperature in at steady state when there is heat generation is given by

T∞, 0 = 50°C

h1, = 20W/m2K

H0, = 5W/m2K

T∞, 0

h0

L

Outside Chamber T∞,0,h0

L = 0.2, K = 4W/mk

qo = 1 kW/m3

q

q

Air

Inside chamber

37


r0

qG

r

h0

Tr

h

0

o

Figure 2.23

1 Gqd dTrr dr dr k

⎛ ⎞ = −⎜ ⎟⎝ ⎠

or 2

0 20

1Gq rd dT rr qdr dr k kr

⎛ ⎞⎛ ⎞ = − = − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

r

or 2 4

012

02 4qdT r rr C

dr k r

⎛ ⎞= − + +⎜ ⎟⎜ ⎟

⎝ ⎠

or 2

0 12

01

2 2q r CdT rk

dr rr

⎛ ⎞= − + +⎜ ⎟⎜ ⎟

⎝ ⎠ . . . (1)

At 0, dTr kdr

⎛= ⎜⎝ ⎠

⎞⎟ cannot be defined,

∴ C1 = 0

2

02

01

2 2q rdT rk

dr r

⎛ ⎞= − −⎜ ⎟⎜ ⎟

⎝ ⎠

Heat transfer from the rod

0

0 0 0 0112 2F

r r

q r q rdTq kdr =

⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 4

Given : qF = 1650 kW/m2, r0 = 1.25 cm

20 1.25 101650

4q −× ×

=

3 30

1650 100 16 528 10 kW/m5

q × ×= = ×

This is the volumetric heat generation at the centre line of the rod. Now,

20

20

12 2q rdT r

dr k r

⎛ ⎞= − −⎜ ⎟⎜ ⎟

⎝ ⎠

38

Conduction

2 40

2202 2 8

q r rT Ck r

⎛ ⎞= − − +⎜ ⎟⎜ ⎟

⎝ ⎠

At r = 0, T = Tc the centre line temperature.

C2 = Tc

2 40

202 2 8 c

q r rT Tk r

⎛ ⎞= − −⎜ ⎟⎜ ⎟

⎝ ⎠+

At, r = r0, the temperature drop,

0

2 20 0 0 01 1 32 2 8 16c r

q r q rT T

k k⎛ ⎞− = − =⎜ ⎟⎝ ⎠

6 3 2 43 528 10 (W/m ) (1.25) 10 m

16 25 W/mK

−× × ×= ×

2

= 618.7oC.

Example 4.3

A bar of square cross-section connects two metallic structures. One structure is maintained at a temperature 200oC and the other is maintained at 50oC. The bar, 20 mm × 20 mm, is 100 mm long and is made of mild steel (k = 0.06 kW/m K). The surroundings are at 20oC and the heat transfer coefficient between the bar and the surroundings is 0.01 kW/m2K.

Derive an equation for the temperature distribution along the bar and hence calculation the total heat flow rate from the bar to the surroundings.

Solution

From Figure 2.24 below :

T∞ = 20 oC A = 0.2 cm × 0.2 cm

Q1T1 = 200o C

L = 0.1 m K = 0.06 kW/m K H = 0.01 k/Wm2 K

Q T2 = 50 oC

Figure 2.24

1dTQ kAdx

= −

12 1

d QQ Q ddx

= + x

2

1 2 2d dT d TQ Q kA dx kA dxdx dx dx

⎛ ⎞− = − − =⎜ ⎟⎝ ⎠

( )h P dx T T∞= −

Letting T T∞θ = − ,

2

2 0d hPkAdx

θ= θ =

39

Governing Equations of Heat Conduction0or 2 2( )D m− θ =

where

12hPm

kA⎛ ⎞= ⎜ ⎟⎝ ⎠

The general solution is

1 2mx mxC e C e−θ = + . . . (1)

At . . . (2) o1 1 20, 180 Cx C C= θ = θ = + =

At . . . (3) 1 12 1 21, 30 Cm mx C e C e−= θ = θ = + = o

11 122 20.01 4 0.02 4

0.06 0.02 0.02 0.12hPmkA

⎛ ⎞× ×⎛ ⎞ ⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠⎝ ⎠

10.577, 1.78mml e= = ,

1 10.561 5.77me m− −= =

1 230 1.78 0.561C C= × + ×

1 23.173 53.476C C+ =

From Eq. (2)

1 2 180.00C C+ =

On subtraction,

12.173 126.524C = −

1 58.22C = −

From Eq. (2),

2 238.22C =

The temperature distribution is

5.77 5.7758.22 238.22x xe e−θ = − +

5.77 5.771 0

0( 58.22 5.77 238.22 5.77 )x x

xx

dQ kA kA e edx

−=

=

θ⎛ ⎞= − = − − × − ×⎜ ⎟⎝ ⎠

40.06 4 10 (335.93 1374.53)−= × × +

= 0.0410 kW = 41.0 W

21

( 58.22 5.77 1.78 238.22 5.77 0.561)x

dQ kA kAdx =

θ⎛ ⎞= − = − − × × − × ×⎜ ⎟⎝ ⎠

40.06 4 10 (598 771.5)−= × × +

= 0.0328 kW = 32.8 W

Heat flow rate from the bar to the surroundings

1 2 41.0 32.8 8.2 WQ Q= − = − =

40

Example 4.4 Conduction

A load of peas at a temperature of 25oC is to be cooled down in a room at a constant air temperature of 1oC.

(a) How long the peas will require to cool down to 2oC when the surface heat transfer coefficient of the peas is 5.81 W/m2 K?

(b) What is the temperature of the peas after a lapse of 10 min from the start of cooling?

(c) What air temperature must be used if the peas were to be cooled down to 5oC in 30 min? The peas are supposed to have an average diameter of 8 mm. Their density is 750 kg/m3 and specific heat 3.35 kJ/kg K.

Solution

We have for a lumped heat-capacity system

.hAt

Bi FocV

i i

T T e eT T

−ρ∞

∞

−θ= = =

θ −

Since the diameter of the peas is only 8 mm, we can neglect any temperature variation within the pea.

3

2

4750 0.0082 1

3 2 3 23 4

2

dV dA d

⎛ ⎞ρ × π ⎜ ⎟ρ ρ ×⎝ ⎠= = =× ×⎛ ⎞× π ⎜ ⎟

⎝ ⎠

=

(a) 32 1 5.81ln25 1 3.35 10

ht tc

−= − = −

− ×

35.81ln 24 3.178

3.35 10t

= =×

1832.4 30.54 mint s= =

(b) 5.81 600

33501 0.35325 1T e

×−−

= =−

o9.48 CT =

(c) 5.81 30 60

33505 0.04425

T eT

× ×−

∞

∞

−= =

−

1.1 0.044 5T T∞ ∞− = −

0.956 3.9T∞ =

o4.08 CT∞ =

Example 4.5

A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500oC to 30oC in a large reservoir of water at 1ooC. Below 100oC the heat transfer coefficient is 1.5 kW/m2 K. Above 1000oC it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500oC and 100oC is 0.5 kW/m2 K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.

41


Solution di = 10 cm, d0 = 12 cm, l = 20 cm, T∞ = 10oC

20( ) (10 12) 20 1382 cmiA d d l= π + = π + × =

2 20( ) (144 100) 20 691 cm

4 4iV d d lπ π= − = − × = 3

Cooling from 500oC to 100oC

lni

T T hAtT T cV

∞

∞

−− = +

− ρ

4

6100 10 490 500 1382 10ln ln500 10 90 7800 470 691 10

t−

−− × × ×

− = =− × × ×

∴ t = 62.12 s Cooling from 100oC to 30oC

4

630 10 1500 1382 10ln

100 10 7800 470 691 10t−

−− × × ×

− =− × × ×

∴ t =18.38 s Total time for quenching = 62.12 + 18.38 = 80.5 s.

Example 4.6 Steel ball bearings (k = 50 W/m K, α = 1.3 × 10– 5 m2/s) having a diameter of 40 mm are heated to a temperature of 650oC and then quenched in a tank of oil at 55oC. If the heat transfer coefficient between the ball bearings and oil is 300 W/m2 K, determine : (a) the duration of time the bearings must remain in oil to reach a temperature

of 200oC, (b) the total amount of heat removed from each bearing during this time, and (c) the instantaneous heat transfer rate from the bearings when they are first

immersed in oil and when they reach 200oC. Solution

To determine if the bearings have negligible resistance, we first check the magnitude of the Biot number.

300 0.023 0.043 50

rhhLBik k

×= = = =

×

Since Bi < 0.1, internal resistance may be neglected.

5

2 2 21.3 10 0.2925

0.023 3

t t tFo tL r

−α α ×= = = =

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(a) The time required for the ball bearings to reach 200oC is

.Bi Fo

ie−θ

=θ

0.04 0.2925 0.0117200 55 145650 55 595

t te e− × −−= = =

−

∴ t = 120.67 s which corresponds to a Fourier number of 35.3.

42

(b) Total amount of heat removed from each bearing 120.67 s Conduction

.

0

( ) ( )t

Bi FoiQ hA T T hA T T e dt−

∞ ∞= − = −∫

.

0

(1 ).

hAttBi ForeV

i ithA e dt hA e

Bi Fo− −= θ = θ −∫

2 0.04 35.3 120.5300 4 (0.02) (650 55) (1 )0.04 35.31

e− ×= × π − − ××

= 57.9 × 104 W s or J = 57.9 kJ.

(c) Instantaneous heat transfer rate at t = 0 (or Fo = 0) is 2( ) 300 4 (0.02) (650 55)iQ hA T T∞= − = × π × × −

= 897 W

and at t = 120 s (Fo = 35.3),

( ) Bi FoiQ hA T T e−

∞= −

2 (0.04) (35.3)300 4 (0.02) (650 50) e−= × π × −

= 218 W.

Example 4.7

A semi-infinite aluminium cylinder (k = 237 W/m K, α 9.71 × 10– 5 m2/s) of diameter 20 cm is initially at a uniform temperature of 200oC. The cylinder is now placed in water at 15oC where heat transfer takes place by convection with h = 120 W/m2 K. Determine the temperature at the centre of the cylinder 15 cm from the end surface 5 min after the start of the cooling.

Solution

We will solve the problem using the one-term solution for the cylinder and the analytic solution for the semi-infinite medium.

0 120 0.1 0.05237

hrBik

×= = =

5

2 29.71 10 5 60 2.913 0.2

(0.1)tFo

r

−α × × ×= = = >

Thus, one-term solution is applicable for cylinder, and Bi = 0.05, A1 = 1.0124 and δ1 = 0.3126

2 21 0 (0.3126) (2.913)

1 1.0124Fc A e e− δ − ×θ = =

= 0.7616

The solution for the semi-infinite solid can be determine from

2

inf 1 22

1 ( , ) erf exp

2 ( )semi

h tx hxx t ck k

t−

⎡ ⎤⎛ ⎞⎢ ⎥ α

− θ = − +⎜ ⎟⎢ ⎥ ⎜ ⎟⎝ ⎠⎢ ⎥

α⎣ ⎦

12

12

( )erf

2 ( )

h txck

t

⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥α⎪ ⎪+⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥

α⎪ ⎪⎣ ⎦⎩ ⎭

43


Tl = 200 oC

D = 0.2 m

h

T∞L = 15 cm

T∞ = 15 oC

h

Figure 2.25

where 1 152 2

0.15 m 0.88

2 ( ) 2 (9.17 10 5 60)

x

t −

ξ = = =

α × × ×

1 152 2( ) 120 [9.71 10 5 60] 0.086

237h t

k

−α × × ×= =

120 0.15 0.0759237

hxx

×= =

2

22 (0.086) 0.0074h t

kα

= =

( ) inf 1 erf (0.88) exp (0.0759 0.0074) erf (0.88 0.086)semi c c−θ = − + + +

1 0.2133 exp (0.0833) 0.170 0.974= − + × =

inf 1 0.974 0.762 0.742csemi cy

i

T TT T

∞−

∞

−= θ × θ = × =

−

. o15 (200 15) 0.742 152.3 CcT = + − × =

Exercise 4.1

A solid sphere of radius b = 5 cm and thermal conductivity k = 20 W/moC is heated uniformly throughout the volume at a rate of 2000 W/m3, and heat is dissipated by convection from its outer surface into the ambient air at T∞ = 25oC with a heat transfer coefficient h = 20 W/m2.oC. Determine the steady state temperature at the centre and the outer surface of the sphere.

Exercise 4.2

A hollow cylinder with inner radius 30 mm and outer radius 50 mm is heated at the inner surface at a rate of 105 W/m2 and dissipates heat by convection from the outer surface into a fluid at temperature 100oC with a heat transfer coefficient of 400 W/m2 K. There is no energy generation, and the thermal conductivity of the solid is assumed to be constant at 15 W/m K. Calculate the temperatures of the inside and outside surfaces of the cylinder.

44

Conduction

Exercise 4.3

An aluminium plate (k = 160 W/m K, ρ = 2790 kg/m3, cp = 0.88 kJ/kg K) of thickness 30 mm and at a uniform temperature of 225o is suddenly immersed at time t = 0 in a well-stirred fluid at a constant temperature of 25oC. The heat transfer coefficient between the plate and the fluid is 320 W/m2 K. Determine the time required for the centre of the plate to reach 50oC.

Exercise 4.4

A cubical piece of aluminium with the same properties, as given in problem 4.5, is 10 mm on a side and is heated from 50oC to 300oC by a direct flame. How long should the aluminium remain in the flame if the flame temperature is 800oC and the convective heat transfer coefficient between the flame and aluminium is 190 W/m2 K?

Exercise 4.5

Steel cylinder of diameter 0.25 m and length 0.8 m initially at 25oC is placed in a furnace, where T∞ = 1000oC. Determine the temperature at the centre and on the surface of the cylinder after a lapse of 1 h. Assume k, ρ, cp and h.

Exercise 4.6

During quenching, a cylindrical rod made of 1080 steel, 1 cm in diameter and 20 cm in length is first heated to 750oC and then immersed in a water bath at 100oC. The heat transfer coefficient can be taken as 250 W/msqC. The density, sp heat and thermal conductivity of the steel are ρ = 7801 kg/m3, c = 473 J/kgoC and k = 43 W/moC, respectively. Calculate the time required for the rod to reach 300oC.

4.9 SUMMARY

In the present unit, both the steady state and transient heat transfer due to conduction are discussed. General method to solve a steady state conduction heat transfer problem is illustrated. All the formulations in Cartesian, cylindrical and spherical coordinates are presented. Solution of problems for simple cases, such as plane wall, long cylinder and spheres are given. Subsequently, discussion is presented for transient heat transfer problems. Lumped capacitance method is presented in details. Transient problems in semi-infinite medium is given at the end. Some solved examples and SAQs are given for better understanding of the unit.

45

Governing Equations of Heat Conduction4.10 KEY WORDS

Steady State Conduction : Parameters are independent of time

Transient Conduction : Conduction parameters are time dependent

Semi-Infinite Solid : Characterized by a single identifiable surface

Rectangular Coordinate : Coordinates in x, y and z direction.

Cylindrical Coordinate : Coordinates in radial r, circumferential θ and axial direction z of a cylinder.

Spherical Coordinate : Coordinates in radial r, circumferential θ and azumthal direction φ of a sphere.

BIOT Number : Ratio of conduction resistance to convection resistance.

4.11 ANSWERS TO SAQs

Please refer the relevant preceding text in this unit for answers to SAQs.

46

Conduction APPENDIX-A

w erf w w erf w w erf w 0.00 0.00000 0.36 0.38933 1.04 0.85865 0.02 0.02256 0.38 0.40901 1.08 0.87333 0.04 0.04511 0.40 0.42839 1.12 0.88679 0.06 0.06762 0.44 0.46622 1.16 0.89910 0.08 0.09008 0.48 0.50275 1.20 0.91031 0.10 0.11246 0.52 0.53790 1.30 0.93401 0.12 0.13476 0.56 0.57162 1.40 0.95228 0.14 0.15695 0.60 0.60386 1.50 0.96611 0.16 0.17901 0.64 0.63459 1.60 0.97635 0.18 0.20094 0.68 0.66378 1.70 0.98379 0.20 0.22270 0.72 0.69143 1.80 0.98909 0.22 0.24430 0.76 0.71754 1.90 0.99279 0.24 0.26570 0.80 0.74210 2.00 0.99532 0.26 0.28690 0.84 0.76514 2.20 0.99814 0.28 0.30788 0.88 0.78669 2.40 0.99931 0.30 0.32863 0.92 0.80677 2.60 0.99976 0.32 0.34913 0.96 0.82542 2.80 0.99992 0.34 0.36936 0.100 0.84270 3.00 0.99998

The Gaussian error function is defined as

2

0

2erfw

w dvw e−=π ∫

The complementary error function is defined as

erf 1 erfw w= −

1.0 0.8 0.5 0.4 0.3 0.2

0.1 0.07 0.05 0.04 0.03 0.02

0.01 0.007 0.005 0.004 0.003 0.002 0.001

0 1 2 3 4 6 8 10 12 14 16 18 20 24 26 28 30 40 50 60 70 80 90 110 130

0.1 0

0.2

0.3

0.4 0.5

0.7

1.0

150 300 400 500 600

0

0.1 0.3 0.5 0.8 1.0 1.4

2.0

2.5

3

6 7

9 10

20 30

100 50

1 2 3 4

0 0.05

0.1 0.2

0.3 0.4

0.5

0.6

0.7

0.8

1.0

1.2

1.4

1.6

1.8 2.0

2.5

3

4

6

7

8 9

10

12 14

16 18

20 25 30

35 40

45 50 60 70 80

90 100

Bi -1 = k/hl

D1

47


1.0

0.9

0.8

0.7

D2

D3

0.6

0.5

0.4

0.3

0.2

0.1

0 0.01 0.02 0.05 0.1 0.2 0.5 1.0 2 3 5 10 20 50

1.0

0.9

0.8

0.6

0.4

0.2

(k/hL) = Bi -

x/L

1

1.0 0.8 0.5 0.4 0.3 0.2

0.1 0.07 0.05 0.04 0.03 0.02

0.01 0.007 0.005 0.004 0.003 0.002 0.001

0 1 2 3 4 6 8 10 12 14 16 18 20 24 26 28 30 40 50 60 70 80 90 110 130 150 300 400 500 600

0.1

0.2

0.3

0.4 0.5

0.7

1.0 100

0

0 0.6 1.2 1.6

1 2 3 4

0 0.05

0.1 0.2

0.3 0.4

0.5

0.6

0.7

0.8

1.0

1.2

1.4

1.6

1.8 2.0

2.5

3

4

6

7

8 9

10

12 14

16 18

20 25

30 35 40 45

50 60

70 80

90

100

Bi -1 = k/hl

0.8 2.0 2.5

3.0

3.5

4 5

30 50

18

12

6 8

48

Conduction 1.0 0.2

0.9

D4

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00.01 0.02 0.05 0.1 0.2 0.5 1.0 2 3 5 10 20 50

1.0

0.9

0.8

0.6

0.4

x/L

(k/hro )= Bi-1

49

Governing Equations of Heat ConductionREFERENCES

F. P. Incropera and D. P. Dewitt (2004), Fundamentals of Heat and Mass Transfer, 5th Edition , John Willay and Sons. W. M. Rohsenow and H. C. Choi (1961), Heat Mass and Momentum Transfer, Prentice-Hall Englewood Cliffs, New Jersey. E. R. G. Eckert and R. M. Drake Jr. (1959), Heat and Mass Transfer, McGraw-Hill, New York. P. K. Nag (2002), Heat Transfer, Tata McGraw-Hill Publishing Company Limited, New Delhi. J. P. Holman (2002), Heat Transfer, 9th Edition, Tata McGraw-Hill, New Delhi.

Unit 4

Documents

t dqx x

y z x x x x

heat conduction equation

qg x y z

specific heat

heat source

time t

steady state heat conduction