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3.3 Semi-infinite approximation

75

θ

= dθ η

= dθ

x

× −

1

× (αt)3/2 ×α

(3.43)

 

t

 

2

 

dη t

dη 2

 

 

 

 

Substituting in Eq. (3.37) we get the following

 

d2θ

dθ

= 0

 

η2

+ 2η dη

(3.44)

 

 

 

 

Please note that Eq. (3.44) is an ordinary differential equation (ODE), as opposed to Eq. (3.37), which is a partial differential equation (PDE). Eq. (3.44) is a secondorder differential equation that supports two conditions. These are

η= 0; θ = 0

η→ ∞; θ =θi

The second condition is quite interesting. This actually means that, x → ∞ as well at t = 0, θ = θi. Hence, our proposal to introduce a new variable η, frequently referred to as a similarity variable has helped us to maintain the integrity of the original governing equation, as now we have only two conditions for θ, as opposed to three in the original PDE.

Let

dθ

= p

 

dη

 

 

dp

 

 

 

 

 

 

dη + 2ηp = 0

(3.45)

 

 

 

 

 

dp

 

 

 

p = −2ηdη

(3.46)

 

 

 

Integrating both sides, we have

 

 

 

ln p = −η2 +C

(3.47)

 

 

p = Aeη2

(3.48)

 

 

dθ = Aeη2

(3.49)

 

 

dη

 

 

 

dθ = Aeη2 dη

(3.50)

Integrating Eq. (3.50) between limits (0,η) in η and (0, θ) in θ, we have

 

 

 

θ dθ = A0η eη2 dη

(3.51)

 

 

0

 

In order to obtain the constant A, We need to integrate Eq. (3.50) from 0 to and the corresponding limits for θ are 0 and θi.

θi

η

2

 

(3.52)

0

dθ = A0 e

 

dη

 

 

 

76 CHAPTER 3  Conduction

A = eη2

dη

(3.53)

0

 

 

Combining Eqs. (3.51) and (3.52), we have

 

θ

=

0η eη2 dη

(3.54)

 

θi

eη2

dη

 

 

 

 

0

 

 

At this point in time, we can invoke the use of error function, erf(η) as follows,

erf (η) =

2

η eη2 dη

(3.55)

π

 

0

 

Consequent upon the invoking of erf(η), Eq. (3.54) may be rewritten as,

 

θ

= erf (η)

= erf (η)

(3.56)

θi

 

erf ()

 

 

Eq. (3.56) uses the information that erf() = 1.

The error function can be tabulated in a form that is easy to use. Values of erf(η) against η are given in Table 3.1.

Table 3.1  Error function table.

x

erf(x)

x

erf(x)

x

erf(x)

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

0.46623

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

0.95229

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

1.00

0.84270

3.00

0.99998

 

 

 

 

1.00000

 

 

 

 

 

 

 

 

3.4 The method of separation of variables

77

 

θ

 

x

 

 

 

= erf

 

 

(3.57)

θi

 

 

 

2

αt

 

Eq. (3.57) can be used to determine the temperature at any point “x” at any time “t” in the slab. Please note that nowhere does the length scale L of the problem appear in the final solution.

3.4  The method of separation of variables

Let us now move on to the more general and less restrictive case where the dimension-

less temperature φ = θ = f (ξ, Bi, Fo). Consider a one-dimensional slab of thickness

θi

2L. Let ρ,cp and k be the density, specific heat capacity, and thermal conductivity of the body respectively. The body is initially at a temperature of Ti throughout. At time t = 0, the two ends at x = ±L are suddenly brought to Twith T< Ti. The problem of transient conduction begins with the challenge being our ability to get T(x,t) in the domain for L x L and for all t 0.

Let us now try to sketch qualitatively the variation of φ across the slab at various times, as shown in Fig. 3.5

From the figure it is seen that at “early” times a substantial part of the slab is still at Ti, and it is in this regime that the semi-infinite approximation is valid. Additionally,

FIGURE 3.5

Qualitative variation of temperature time history for various time instances in an infinitely long slab of thickness 2L.

78CHAPTER 3  Conduction

for t 0, known as the late regime, all temperatures within the slab are nearly the same, and in this late regime, the lumped capacitance method is valid regardless of the Biot number. At time instants not covered by either of these two asymptotic limits and the consequent simplifications thereof, one has to necessarily solve the govern-

ing equation in its full strength to obtain T(x,t). Assume Ti > T. Let φ = T T

Ti T

0 ≤ φ ≤ 1 The governing Eq. (3.3) reduces to

2 φ = 1 φ x 2 α t

The initial condition is: φ = φi = 1 for all x, for t = 0 Boundary condition 1: At x = ±L;φ = 0 for t > 0

Boundary condition 2: At x = 0; φx = 0 for t ≥ 0

We now assume a product solution as follows.

φ(x, t) = X(x)T (t)

(3.58)

(3.59)

The objective is to see whether, by doing this,the governing equation, which is a partial differential equation (PDE), can be reduced to an ordinary differential equation (ODE). The motivation for doing this is the fact that it is a lot easier to solve an ODE rather than a PDE.

 

 

2 φ

= d2 X T

 

 

 

x2

 

 

dx2

 

 

 

 

φ

= dT

X

 

 

 

 

t

 

 

dt

 

 

 

 

d 2 X

T

=

 

1 dT

X

 

 

 

 

 

 

 

 

 

 

 

dx2

 

α dt

 

 

 

 

 

 

 

1 d 2 X

=

 

1 dT

= −λ2

 

 

 

 

 

 

 

X dx2

 

αT

dt

 

(3.60)

(3.61)

(3.62)

(3.63)

In Eq. (3.63), λ2 is a constant and the negative sign ensures that the resulting ODEs do not result in solutions inconsistent with the physics of the problem. The above is a consequence of the fact that the left hand side of eqn. 3.63 is only a function of x while the right hand side is only a function of t. For this to happen, both must be equal to a constant.

The two ordinary differential equations are

dT +αλ2T = 0

(3.64)

dt

 

 

d2 X

+ λ2 X = 0

(3.65)

dx2

 

 

 

 

 

 

 

 

 

 

 

 

3.4 The method of separation of variables

79

The general solution to Eq. (3.64) is

 

 

 

 

 

 

T = Aeαλ2t

(3.66)

 

The general solution to Eq. (3.65) is

 

 

 

 

 

 

X = Bcos λx + C sin λx

(3.67)

 

The general solution to Eq. (3.58) then becomes

 

 

 

 

φ = Aeαλ2t (B cos λx + C sin λx)

(3.68)

 

We now evaluate A, B, C, and λ using the initial and boundary conditions.

 

 

 

 

φ = Aeαλ2t (Bλ sin λx +Cλ cos λx)

(3.69)

 

 

 

x

 

 

 

 

At x = 0;

φ

= 0

 

 

 

 

 

 

 

 

 

 

 

x

 

 

 

 

Therefore

 

 

 

 

 

 

C = 0

 

 

The general solution then becomes

 

 

 

 

 

 

φ = Aeαλ2t B cos λx

(3.70)

 

Let AB = A, and Eq. (3.70) becomes

 

 

 

 

 

 

φ = Aeαλ2t

cos λx

(3.71)

 

At x = ±L; φ = 0

 

 

 

 

 

 

0 = Aeαλ2t

cos λL

(3.72)

 

From Eq. (3.72), it is clear that

 

 

 

 

 

 

cos λL = 0

(3.73)

 

Eq. (3.73) is satisfied by a succession of λs that satisfy the following equation.

 

 

 

 

nπ

 

(3.74)

 

 

 

λn = 2L

 

 

 

 

 

where n = 1, 3, 5.....

 

 

 

 

Hence, the solution to Eq. (3.70) becomes

 

 

 

 

 

 

 

 

 

 

 

 

φ = An cos(λn x)eλn2 t

(3.75)

 

 

 

n=1

 

 

 

 

An incorrect choice of λ2 instead of –λ2 would have resulted in a solution for temperature that is exponential in x and periodic in t, which completely goes against the physics of the problem.

80 CHAPTER 3  Conduction

In Eq. (3.75), An is as yet unknown. To obtain An we invoke the initial condition. At t = 0 s; φ = φi = 1

 

1 = An cos(λn x)e0

(3.76)

n=1

 

 

1 = An cos(λn x)

(3.77)

n=1

We need to use the property of a set of functions called orthogonal functions to

obtain An.

Orthogonal functions

An infinite set of functions g1 (x), g2 (x), g3 (x)…..gn (x) is said to be orthogonal in

the closed interval a x b, if ab gm (x)gn (x) = 0; for m n.

Many functions like sinx and cosx exhibit orthogonality. Hence if we multiply Eq. (3.77) by cos(λnx) on both sides and integrate from 0 to 1, we can exploit the property of orthogonal functions to evaluate An.

0L cos(λn x) dx = 0L An cos2 (λn x) dx

(3.78)

Please note that the right hand side of the equation 3.78 was originally a summa­ tion. However, in the view of the orthogonality property of cosλnx, only the term with cos2λnx will remain in the series, as all other terms will be zero.

An =

 

 

 

 

L cos(λn x) dx

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

(3.79)

 

 

L cos2 (λn x) dx

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L

 

 

 

 

 

 

 

 

 

 

sin(λn x)

 

 

An =

 

 

 

 

 

 

 

 

λn

 

0

 

(3.80)

0L

1 + 2 cos(2λn x)

dx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin(λn x) L

 

 

 

 

 

 

 

 

 

 

 

λn

 

 

 

An =

 

 

 

 

 

 

 

 

 

 

0

 

(3.81)

x

 

L

sin(2λn x) L

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

 

 

2

 

 

0

 

 

 

4λn 0

 

 

 

 

 

 

 

 

 

 

 

 

1

sin(λn L)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

An =

 

 

 

 

 

 

 

 

 

 

λn

 

 

 

 

 

 

(3.82)

 

 

L

 

+ 2sin(λn L) cos(λn L)

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

4λn

 

 

 

 

 

 

 

 

 

 

 

φi

sin(λn L)

 

 

 

 

 

 

 

 

 

 

 

 

 

An =

 

 

 

 

 

 

 

 

 

λn

 

 

 

 

 

 

(3.83)

 

 

L

 

+ sin(λn L) cos(λn L)

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

2λn

 

 

 

3.4 The method of separation of variables

81

Substituting for An, in Eq. (3.75) and rewriting φ = (T-T)/(Ti-T) as θ/θi. We get the final expression for θ (or φ) as,

θ

1

 

sin(λn L) cos(λn x)e

αλn2 t

 

 

=

 

 

 

(3.84)

θi

λn

 

 

L

+

sin(λn L) cos(λn L)

 

n=1

 

 

 

 

 

 

 

2

2λn

 

 

 

A tougher variant of this problem arises when the sides are exposed to convection with a heat transfer coefficient of “h” and free stream temperature “T”. All the conditions are the same as the previous case except that at the convection boundaries, we have

At x = ±L; k φx = hφ

φ = Aeαλn2t cos λn x

(3.85)

kAeαλn2 t (sin λn L)λn = hAeαλn2 t cos λn L

(3.86)

λn Lk sin λn L = hL cos λn L

(3.87)

ξ tanξ = Bi (Where ξ = λL)

(3.88)

Again eq. 3.88 is satisfied by a succession of ξ s that have to be obtained by numerically solving the transcendental equation 3.88

From Eq. (3.84)

 

θ

2L sin(ξn ) cos(λn x)e

αλn

2t

 

 

=

 

 

 

 

 

 

θ

i

n=1

nπ

 

 

L

+

sin(2λn L)

 

 

 

 

 

 

2

 

4λn

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

θ

2L sin(ξn ) cos(λn x)e

αλn

2t

 

 

=

 

 

 

 

 

 

θi

n=1

nπ

 

2Lλn + sin 2λn L

 

 

 

 

 

 

 

 

4λn

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

θ

2L

4 nπ sin(ξn ) cos(λn x)e

αλn2t

 

=

 

 

θi

n=1

nπ

2L

 

2ξn + sin(2ξn )

 

θ

4 sin(ξn )

 

eξn Fo cos(λn x)

 

=

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

θi

n=1

2ξn + sin(2ξn )

 

 

 

 

 

 

The first-term approximation of the solution is given as

(3.89)

(3.90)

(3.91)

(3.92)

 

θ

 

ξ2Fo

 

(3.93)

Φ = θi

= A1e

1

cos(λ1 x)

 

 

 

 

 

The location x = 0, corresponds to the mid-plane temperature. Eq. (3.93) can be simplified at x = 0 as.

θ

0

= A e

ξ 2Fo

(3.94)

 

1

θi

1

 

 

 

 

 

where θ0 is the dimensionless mid-plane temperature of the plate.

θi

for any plane. This is mainly for convenience.

82 CHAPTER 3  Conduction

We can now calculate the ratio of enthalpy transfer to the maximum enthalpy transfer possible in the one dimensional slab.

 

Q

=

 

V

 

 

ρcp (T Ti ) dV

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Qmax

 

 

 

ρcp (TTi )V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

V

 

ρcp (T T+ TTi ) dV

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ρcp (TTi )V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q

=

1

 

V

(1φ) dV

 

 

 

 

 

 

 

 

 

 

 

 

 

Q

 

V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

max

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

1

 

 

L

 

(1φ) A dx

 

 

 

 

 

 

 

 

 

 

AL

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q

 

 

 

 

 

 

1

 

 

L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= 1

 

 

 

 

 

 

0

 

φ A dx

 

 

 

 

 

 

 

 

 

 

 

 

Q

 

 

 

AL

 

 

 

 

 

 

 

 

 

 

 

 

 

 

max

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= 1

1

 

 

L φ dx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Invoking eqn. 3.93 for f, we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

L

 

 

 

 

 

 

x

 

ξ 2Fo

 

 

 

=

 

1 −

 

 

 

 

0

A1cos

ξ

 

 

 

e

 

 

 

 

 

dx

 

 

 

 

 

L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

ξ 2Fo

 

L

 

 

 

 

 

 

x

 

 

 

 

 

=

1 −

 

 

 

 

A1

e

 

 

0

cos

ξ

 

 

 

dx

 

 

 

 

L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x

 

L

 

Q

 

 

 

 

 

 

1

 

 

 

 

 

ξ 2F

 

sin ξ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

1 −

 

 

 

 

 

 

 

 

 

 

L

 

 

 

 

 

 

 

 

A1

e

 

o

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Qmax

 

 

L

 

 

 

 

 

ξ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

Q

=

1 A1e

ξ2F sin(ξ)

 

 

 

 

 

 

 

Qmax

 

 

 

 

o

ξ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where A1

=

4sin(ξ1 )

2ξ1

+ sin(2ξ1 )

 

 

(3.95)

(3.96)

(3.97)

(3.98)

(3.99)

(3.100)

(3.101)

(3.102)

(3.103)

(3.104)

Though we are able to obtain the solution analytically, Eq. (3.92) is quite tedious to use. Heisler (1947) used the first term in the series Eq. (3.94) and presented charts to solve the problem swiftly; these have now come to be known as “Heisler’s charts.” and are valid for Fo > 0.2.

Let us use the first term of Eq. (3.92) to redo these charts for the problem under consideration for convenience. We can get the mid-plane temperature first and then

develop a chart for φ

φcenter

Analytical solutions are also possible for a cylinder and a sphere. There are formidable and involve Bessel’s functions and Legendre polynomials respectively. However, these are quite involved. Charts based on the first term in the series solution for the plane wall are given in Figs. 3.6–3.7. Figs. 3.9–3.10, as

FIGURE 3.6

Mid-plane temperatures charts for a plane wall with thickness 2L initially at a uniform temperature of Ti subject to convection on both sides with a heat transfer coefficient of h and an ambient temperature of T.

83 variables of separation of method The 4.3

84 CHAPTER 3  Conduction

FIGURE 3.7

Temperature distribution chart for a plane wall with thickness 2L initially at a uniform temperature of Ti subjected to convection on both sides with a heat transfer coefficient of h and an ambient temperature of T.

FIGURE 3.8

Enthalpy chart for a plane wall with thickness 2L initially at a uniform temperature of Ti subjected to convection on both sides with a heat transfer coefficient of h and an ambient temperature of T.