Bernatz R. Fourier Series and Numerical Methods for Partial Differential Equations

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102 HEAT EQUATION

y + Ay —

y —

x x-\- Ax

Figure 4.3 An arbitrary control volume for energy transfer in 2D.

The time rate of change of the total energy is

g rx+Ax çy+Ay

time rate of change of energy = — / / a6Hu(r,s,t)drds

Jχ Jy

çx+Ax ry+Ay

= j / aSHu

(r,s,t)drds

Jx Jy

(4.13)

Next, the net energy flux in the x direction is expressed based on Fourier's law

net energy flux = —KHu

(x, s, t)Ay + KHu

(x 4- Δχ, s, t)Ay

= KH[u

(x + Ax,s,t) - u

(x,s,t)]Ay (4.14)

where K is the thermal conductivity [W/(ms)] of the material. A similar expression

for the y direction is

net energy flux = —KHu

(r, y, t)Ax + KHu

(r, y -f Ay, t)Ax

= KH[uy(r,y + Ay,t)-u

(r,y,t)]Ax (4.15)

The total energy source within the control volume may be expressed as an integral

of the form

x+Ax ¡>y+Ay

rate of internal energy production = / / HQ(r,s,t)drds (4.

J x J v

16)

where Q(r,s, t) is the rate of internal heat production.

The principle of conservation of energy applied to the control region using the

appropriate terms above gives

rx+Ax ry+Ay

j I a8Hu

{r,s,t)drds = KH[u

(x + Δχ, s,t)

—

(x, s,t)]Ay

J x Jy

HEAT EQUATION IN 3D 103

+ KH[u

(r, y + Ay, t) - u

(r, y, t)]Ax

px+Ax py+Ay

+ / / HQ{r,s,t)drds

J x Jy

(4.17)

The Mean Value theorem for integrals is applied to each of the integral expressions

above to write

px+Ax py+Ay

I l aôHu

(r,s,t)drds = aSHu

(ri,si,t)AyAx

J x J y

px+Ax py+Ay

I I HQ(r, s, t)drds = HQ(r

, s

,t)AyAx

Jx J y

where the points {r\,s\) and (r2, S

) are points within the differential control region

at which the Mean Value theorem is satisfied for the respective integrals. Substituting

these expression for the corresponding integrals in Equation (4.17) gives

oSHut^i^si.^AyAx = KH[u

{x + Ax,s,t)

—

(x,s,t)]Ay

+ KH[uy(r,y + Ay,t)-Uy(r,y,t)]Ax

+ HQ{r

,t)AyAx (4.18)

Dividing both sides of Equation (4.18) by H Ax Ay and taking the limit

(Δχ, Ay)

—>

(0,0) gives

ut(x,y,t) = k{u

(x,y,t) +u

(x,y,t)) +q(x,y,t) (4.19)

where k = Κ/(σδ) and q = Q/(aS) are the thermal diffusivity and rate of internal

heat production per specific heat and density of the material. Note also that, because

the

continuity of u

and Q, u

(ri,s\,t) andQ(r

,t) have limits u

(x,y,t) and

Q{x, y, t), respectively, as both Ax and Ay approach zero.

4.4 HEAT EQUATION IN 3D

The heat equation in three spatial dimensions results when a solid medium may have

energy transfer in each of the three directions. The case for 3D Cartesian coordinates

will be given in this section.

Figure 4.4 shows an arbitrary differential volume from within a solid medium.

The conservation of energy principle used in the ID case is the basis in the 3D case

as well. In the 3D case, the net flux must be considered in each of the three Cartesian

directions. The resulting equation is

nzArAz py+Ay px+Ax

¡¡I a6ut(x,y, z,t)dxdydz =

J z Jy Jx

K[u

(x + Δχ, y, z, t) - u

(x, y, z, t)]AyAz

104 HEAT EQUATION

z + Az

—

+ Ay

Figure 4.4 An arbitrary volume for energy transfer in 3D.

K[u

(x, y + Ay, z, t)

—

(x, y, z, t)]AxAz

K[u

(x, y,z + Az, t)

—

(x, y, z, t)]AxAy

f>z+Az py+Ay px+Ax

Q{x,y,z,t)dxdydz

pz-tnz py-\-¿\y px

Jz Jy Jx

(4.20)

where Fourier's law is used to express the flux at each of the six sides of the differential

volume. The Mean Value theorem for integrals in 3D is used to write

a5ut(a\,bi,ci,t)AxAyAz =

K[u

(x + Ax, y, z, t) - u

(x, y, z,t)]AyAz

K[u

(x, y + Ay, z, t) - u

(x, y, z, t))AxAz

K[u

(x, y,z + Az, i)

—

(x, y, z, t)]AxAy

+Q(a

, b

, c

,t)AxAyAz (4.21)

where a\ and a

are real numbers between

[x,

x + Ax], b\ and b

are real numbers

between

[y,

y + Ay], and c\ and c

are real numbers between \z,z

-V

Az]. Dividing

both sides of Equation (4.21) by aSAxAyAz gives

{ai,bi,ci,t)

(x + Ax,y,z,t) -

^[η

(χ^

+ Ay,z,t) -

(x,y,z + Az,t) -

-u

(x,y,z,t)]

-u

{x,y,z,t)]

-u

(x,y,z,t)]

+q(a

,t) (4.22)

where, as in the ID case, k = ^ is the constant thermal diffusivity of the solid, and

x,y,z

σδ

q(x, y, z, t) = Q(

¿>. Now, taking the limit of both sides of Equation (4.22) as

POLAR-CYLINDRICAL COORDINATES 105

Ax, Ay and Az all go to zero gives

ut(x, y, z, t) = kV

u{x, y, z, t) + q(x, y, z, t)

(4.23)

which is the 3D heat equation in Cartesian coordinates. The first term on the right-

hand side of Equation (4.23) is

u(x, y, z, t) =

(x,

y, z, t) +

{x,

y, z, t) +

(x,

y, z, t) (4.24)

and represents the Laplacian of u in Cartesian coordinates.

4.5 POLAR-CYLINDRICAL COORDINATES

There are numerous PDE applications that require polar-cylindrical coordinates for

a more efficient solution process. Therefore, the Laplacian in polar-cylindrical form

is derived in this section.

Figure 4.5 Relationship between x, y, and 2, and

φ,

and z.

Referring to Figure

4.5,

the relationships between Cartesian variables (x, y, z) and

polar-cylindrical variables (p,

</>,

z) are

x —

p cos

y = p sin φ

z — z

and, for polar-cylindrical variables in terms of x, y and z,

p = \Jx

+ y

φ = tan

z = z

•(i)

(4.25)

(4.26)

(4.27)

(4.28)

(4.29)

(4.30)

106

HEAT EQUATION

We will begin by determining an expression for u

in polar-cylindrical coordi-

nates.

Applying the chain rule to u(p, 0, z), we have

du du dp du do du dz , _

— = -A -H (4.31)

dx dp dx

dcf)

dx dz dx

Using Equation (4.28), we have

dp _ d_

dx dx

\Jx

+ y

Λ/Χ

+ y

pcos(/>

= cos0 (4.32)

after employing the relationships in Equations (4.25) and (4.28). Using Equation

(4.29),

we have

dx dx

tan

■©

-JL

rr-2

i + (f)

+ y

—

sin(/>

(4.33)

Substituting the results in Equations (4.32) and (4.33) into Equation (4.31), and

knowing that —- = 0, we have

du sinö!) du ,

„

— = cos φ- ^ — (4.34)

ox op ρ οφ

Next, substitute

——

for u in Equation (4.34)

d /du\ _ d f du\ sin0 d Ídu\

dx \dxJ dp \dxj p

d<fi

\dxJ

Then, subbing for —- on the right-hand side of Equation (4.35) using Equation (4.34)

results in

d (du\ . d ( ßu sinφ du

-— = cos©-—- cos

dx \dxj dp \ dp p

δϊηφ d ( du smódu

. ,

„^

COS0- -τττ)

(

)

ρ dφ \ dp p

POLAR-CYLINDRICAL COORDINATES 107

Executing the derivative in Equation (4.36) and simplifying results in

u 2

sin φ cos φ du sin

φ du

8φ p dp

2 sin φ cos φ d

u sin

φ d

u ,,

Ύ_ Ύ_ ! Ύ_ (4.37)

ρ δρδφ ρ

δφ

Now, a similar sequence of manipulations will be used to express u

in terms of

polar-cylindrical variables.

du du dp du dφ du dz

dy dp dy άφ dy dz dy

Finding proper expressions for -^- and -^, substituting them in Equation (4.38), and

simplifying gives

du .

du cos φ du ,

„^

— = sin φ— + ^ — (4.39)

dy dp ρ οφ

Using Equation (4.39) in the same way as Equation (4.34), the following result for

Uyy

is found.

u .

sin 0 cos φ du cos

φ du

άφ ρ dp

2sinφcos φ d

u cos

φ d

u ,

ΛΛ

L· L· 1 Z (4.40)

p dpdφ p

dφ

Adding the results from Equations (4.37) and (4.40) gives

u d

u 1 du 1 d

p dp ρ

άφ

Because the z coordinate is the same for Cartesian and polar-cylindrical coor-

dinates, it follows that the Laplacian of u in polar-cylindrical coordinates is given

2 _ 9

u 1 du 1 d

u d

p dp p

άφ

Using the identity

ld_/ du\ _ldu (Pu

pdp \ dp J p dp dp

Equation (4.42) may be written as

--—( —\ —— —

pdp\dpj p

άφ

u = -(pUp)p + -^ηψφ + u

(4.45)

108

HEAT EQUATION

4.6 SPHERICAL COORDINATES

The relationship between Cartesian coordinate variables and spherical coordinate

variables is shown in Figure 4.6. The process for determining an expression for the

Laplacian in spherical coordinates can be made shorter

establishing

the

relationship

Figure 4.6 Relationship between x, y, and z, and r,

Θ,

and φ.

between variables in the cylindrical and spherical coordinate systems. These are

z = p cos θ p = r sin

φ = φ. (4.46)

The relations in Equations (4.46) are of the same form as those between Cartesian

and cylindrical system. Therefore, one may begin with the Laplacian in cylindrical

coordinates and transform it to its spherical form through the relations in Equations

(4.46) and the equations developed in Section 4.5.

Without providing the details, various forms of the Laplacian in spherical coordi-

nates are shown below

u =

u 2 du

sin

<9d</>

' r

3Θ

u 1 d

u cot

Vu =

-(ru)

-\-

sin

/ 2 x 1

-ö(r u

)

ηψφ

sin

-ν>ΦΦ

sin

-(sin 6^0)0

(sinö u

(4.47)

(4.48)

(4.49)

EXERCISES

4.1 Derive the following ID version of the heat equation

1 /-)

ut =

—z

— [K(x)u

] +q(x,t)

σο ox

where the thermal conductivity K depends on x.

EXERCISES

109

4.2 Derive the ID form of the heat equation for the case where the thermal con-

ductivity K is a function of temperature u.

4.3 By making the substitution τ = kt, show that the 2D heat Equation (4.19),

without internal heat sources or sinks, can be written as

ΊΙχχ i tLyy

showing this change of variable makes it possible to remove the explicit dependency

of u on the thermal diffusivity k.

4.4 A solid slab occupies the region between 0 < x < c and — oo < y < oo. The

slab's face at x = 0 is maintained at a constant temperature of

while the face at

x — c has a constant temperature of zero. Write the steady-state boundary value

problem (BVP) for this case. Then, find the solution for u(x) in terms of

To,

c, and

k, the given thermal diffusivity of the material.

4.5 A solid slab has faces at x = 0 and x — c. The face is kept at x = c is kept

at a constant temperature of zero. There is a constant flux of heat into the slab at

x = 0 of Φο· Write down the steady-state BVP for this case. Then solve the BVP

for temperature u(x).

4.6 In this exercise, we consider a solid slab with faces at x = 0 and x = c. There

is surface heat transfer at both faces with the same surface conductance of h. The

surrounding medium for x < 0 has a constant temperature of

and the surrounding

for x > c has a constant temperature of T

a) Show that the steady-state BVP for this case is

(x) =0 0 < x < c

(0) = hu(0) KU

{C) — h[T

— u(c)]

when Newton's law of cooling and Fourier's law of heat transfer are applied

to the respective boundaries.

b) Solve the BVP derived above to show that

u(x)=

° Jh*x + 1)

where h* = h/κ.

4.7 Show that Equations (4.48) and (4.49) follow from Equation (4.47).

4.8 Suppose u{r) is the temperature as a function of radius r in the region bounded

by two concentric spheres, the inner sphere has radius a and the outer sphere has

radius

The surface of

the

inner sphere is kept at temperature zero while the surface

at r = b is maintained at temperature

T&.

a) Show that the spherical form of the Laplacian given in Equation (4.48)

reduces to

(ru)

= 0

110 HEAT EQUATION

b) Solve

the

BVP

give

/ a\

u(r)

= I a <r <b

- a \ r

h*b

s Λ

αχ

wir)

= —— 1 a < r < b

4.9 Suppose

the

boundary condition

at r = 6 of

Exercise

4.8 is

replaced with

that

surface heat transfer, obeying Newton's

law of

cooling, with

the

surrounding

temperature

of T

Show that the temperature

u(r) in

this case

given

h*b

h*b(b-a)

where h*

—

h/κ.

4.10 Consider

the

hollow cylinder

(a < p < b)

whose horizontal cross section

is shown

Figure

4.7. The

temperature

at p = a is

held constant

at T

, and the

temperature

at p = b is

held

at a

constant value

of T

These boundary conditions

and

the

relative dimensions (length

diameter)

of the

hollow cylinder allow

to assume

steady-state temperature that

is a

function

of p

only. Determine

the

Figure 4.7 The hollow cylinder.

temperature profile

u(p) for

this case.

4.11 Consider

the

hollow cylinder

described

Exercise

4.10 and

shown

Figure 4.7. Suppose the surface

at p

—

a is

insulated, while the heat transfer

at p = b

obeys

Robin boundary condition with

the

constant exterior temperature given

. Determine

the

steady-state temperature

u(p) for

this case.

4.12 Consider

the

hollow cylinder

described

Exercise

4.10 and

shown

Figure

4.7.

Suppose

the

surface

at p = a is

maintained

at a

constant temperature

of zero, while

the

heat transfer

at p =

b obeys

Robin boundary condition with

the

constant exterior temperature given

by T

Determine

the

steady-state temperature

u(p)

for

this case.

4.13 Suppose

wire lies along

the x

axis.

The

surrounding medium

has a

constant

uniform temperature of

Suppose surface heat transfer takes place along the length

of the wire. The wire

slender enough

one may assume temperature

is a

function

t and x

only. Using

approach similar

that

Section

4.1,

derive

the PDE

(x,t)

b[T

u(x,t)]

EXERCISES 111

where b is a constant that is given by material characteristics including the surface

heat conductance represented by H. What is the formula for 6?

4.14 For each of the following descriptions, give a complete and accurate statement

of the IB VP (PDE, IC, and BCs). In each case, let the heat diffusion coefficient = Λ,

the constant thermal conductivity = κ, and thermal diffusivity = k.

a) A thin wire is stretched from x = 0 to x = c. The wire is insulated on the

lateral boundaries and at the left endpoint. At the right endpoint, the wire

is exposed to a medium at a constant temperature of 20 °C, and the system

obeys Newton's law of cooling at the boundary. The wire has an initial

temperature of 10 °C.

b) A thin wire is stretched from x

—

0 to x = c. The wire is insulated on the

lateral boundaries. Both the left and right boundaries are insulated. The

(

7TX

c) A long pipe has an inner radius of 2 cm and an outer radius of

cm. The

outside of the pipe is insulated, and a fluid with constant temperature of 10

°C.

flows through the pipe. The pipe and the fluid obey Newton's law of

cooling, and has an initial uniform temperature of 30 °C.

d) A solid steel ball with radius of 5 cm is dropped into a medium with

constant temperature of 14 °C. The ball's interaction with the medium

obeys Newton's law of cooling, and the initial temperature distribution in

the ball is proportional to its distance from the center of the ball.