Lecture 9ChE 333 1

Heat Transfer in a Slab

Consider a large planar solid whose thickness (y-direction) is L. What isthe temperature history of the slab if it is suddenly brought into contactwith a fluid at temperature T? The transient conduction equation is

Let’s make the problem dimensionless.

The temperature can be expressed as

so that the problem reposed is

∂T∂t

= α∂2T

∂y2

at t = 0,T = T0

at y = 0 , T = T1 for t > 0at y = L , T = T1

θ =

T – T1

T0 – T1

; η =yL

; τ =αt

L2

∂θ∂t

= α∂2θ∂y2

θ = 1 at τ = 0θ = 0 at η = 0θ = 0 as η = 1

Lecture 9ChE 333 2

How do we solve the equation ?

Suppose z has the form Θ = Y(τ)G(η)

The equation is separable in the form

Integrating each of the equations we obtain

The solution for y(θ,η) has the form

We can construct the exact solution using the boundary conditions

It follows that B must be 0 if the condition is true for all θ > 0

∂θ∂τ –

∂2θ∂η2 =

∂YG∂τ –

∂2YG

∂η2 = GdYdτ

– Yd2Gdη2 = 0

1Y

dYdτ

=1G

d2Gdη2 = – λ2

dYdτ

= – λ2Y ;d2Gdη2 = – λ2G

Y(τ) = Ke–λ2τ and G(η) = Asin(λη) +Bcos(λη)

θ(η, τ) = Asin(λη) +Bcos(λη) e–λ2τ

θ(0, τ) = Asin(0) +Bcos(0) e –λ2τ = 0

Lecture 9ChE 333 3

Now the other boundary condition

Now this is true for all τ > 0 if and only if sin(λ) = 0but sin(λ) = 0 only where λ = nπ where n = 0, 1, 2, .....

This means there are a countable infinity of solutions so that

To obtain the coefficients An , we need to use the initial condition.

To determine the coefficients, we can use the orthogonality properties ofthe sine and cosine functions. (See Appendix)

sin(nπξ)sin(mπξ)dξ

–1

1

=0 for m ≠ nπ for m = n

θ(1, τ) = Asin(λ) e–λ2τ = 0

θ(η, τ) = e–λ2τ A n sin(nπη)Σ

n= 1

∞

θ = 1 at τ < 0

z(η, 0) = A n sin(nπη) = 1Σn= 1

∞

Lecture 9ChE 333 4

We integrate

You might remember that the first sine integral is non-zero if and only ifn = m. Now the equation for An is

The result for the definite integrals follow from what I gave above . Itfollows that

We saw earlier that the solution can be described as:

θ(η, 0)sin(mπη)dξ

0

1

= sin(mπη)dη0

1

A n sin(nπη)sin(mπη)dη

0

1

Σn= 1

∞

= sin(mπη)dη0

1

A n =– sin(nπη)dη

0

1

sin2(nπη)dη0

1= –

sin(x)dx0

nπ

sin2(x)dx0

nπ

A n =

4π

– 1n

n +12

Lecture 9ChE 333 5

We have already examined how the sum converges. For τ > 0.2, onlyone term suffices to describe the solution. We can look at many differentclasses of problems. The general problem for transient heat transfer in aslab is one posed as

The dimensionless form is:

The solution is of the form

θ(η, τ) = 4

π1

2n + 1e– 2n +1

2π 2 τsin( 2n + 1 πη)Σn = 0

∞

∂T∂t = α∂ 2T

∂y2

at t = 0, T = T0

at y = 0 , – k ∂T∂y = h T – T1 for t > 0

at y = L2 , ∂T

∂y = 0

Cn =

4 sin ζn

2ζn + sin 2 ζ n

and ζn tan ζn = Bi

θ(η, τ) = Cne–ζ n

2 τsin(ζn2η)Σ

n = 0

∞

∂θ∂t = α∂2θ

∂y2

θ = 1 at τ= 0∂θ∂η = – Bi θ at η = 0

∂θ∂η = 0 as η = 1

Lecture 9ChE 333 6

Again the approximate solution is the one-term solution

This argument is the same for any transient 1-dimensional heat transferproblems involving cylinders, planes or spheres.

Examples

Infinite Cylinder

The solution is

An approximate one-term solution is

Note that along the center line

θ(η, τ) ≈ C1e–ζ 12 τsin(ζ1

2η)

θ(η, τ) = Cne–ζ n

2 τJ 0(ζnη)Σn = 0

∞

Cn = 2

ζn

J1 ζn

2 J02 ζn + J1

2 ζn

and ζn

J1 ζn

J0 ζn

= Bi

∂θ∂ τ = 1

η∂

∂η η∂θ∂η

θ = 1 at τ= 0 in η ∋ 0, 1∂θ∂η = – Bi θ at η = 1

∂θ∂η = 0 at η = 0

θ(η, τ) = C1e–ζ 12 τJ0(ζ1η)

θ 0(0, τ) = C1e–ζ 12 τ

Lecture 9ChE 333 7

So that the simpler representation is θ(η, τ) = θ 0 J0(ζ1η)

Lecture 9ChE 333 8

Sphere

The solution is

The Approximate Solution

The center temperature is

So that the temperature can be expressed as

∂θ∂ τ = 1

η2∂

∂η η2 ∂θ∂η

θ = 1 at τ= 0 in η ∋ 0, 1∂θ∂η = – Bi θ at η = 1

∂θ∂η = 0 at η = 0

Cn =

4 sin ζ n – ζncos ζn

2 ζn + sin 2ζn

and 1 – ζ n cot ζn = Bi

θ(η, τ) = Cne–ζ n

2 τsin(ζnη)ζnη

Σn = 0

∞

θ(η, τ) = C1e–ζ 1

2 τsin(ζ1η)ζ1η

θ 0(0, τ) = C1e–ζ 12 τ

θ(η, τ) = θ 0

sin(ζ1η)ζ1η

Lecture 9ChE 333 9

Short Time Solutions

Consider a large planar solid whose extent (y-direction) is very large.What is the temperature history of the slab if it is suddenly brought intocontact with a fluid at temperature Ta? The transient conduction equationis

Let’s make the problem dimensionless.

The temperature can be expressed as

θ = T – Ta

T0 – Ta

so that the problem reposed is

∂T∂t = α∂ 2T

∂y2

at t = 0, T = T0

at y = 0 , – k ∂T∂y = h T – Ta Ta

as y → ∞ , T = T0

∂θ∂t = α∂2θ

∂y2

θ = 1 at t = 0∂θ∂y = Bi θ at y = 0

θ = 1 as y → ∞

Lecture 9ChE 333 10

Solutions

We noted earlier that the equation can be solved by a combination ofvariables supposing that Τ = Τ(η,t) and we saw that the the appropriatechoice for η is

η =y4αt

The solutions for a number of different cases are as follows:

Case 1 – Constant Surface Temperature (T = Ts)

Case 2 – Constant Surface Heat Flux (q”s = q”0)

Case 3 –Surface Convection

θ = erf y

4t+ exp y + t erfc y

4t+ t

T – Ts

Ti – Ts

= erf yαt

q"

s t =k Ts – Ti

παt

T – Ts =2 q"

0αtπ

ke–

y 2

4αt – q"0yk

erfc y4αt

Lecture 9ChE 333 11

where y = hy

k s

and t = hk s

2

αt

Lecture 9ChE 333 12

Surface temperature of a Cooling Sheet

Polyethylene is extruded and coated onto an insulated substrate, moving at20 cm/sec. The molten polymer is coated at a uniform temperature T0 of400°F. Cooling is achieved by blowing air at a temperature Ta of 80°F.Earlier heat transfer studies determined that the heat transfer coefficient, h,is 0.08 cal/cm-sec-°F. The coating thickness B is 0.1 cm.

At what point downstream does the surface temperature, T(0) fall to 144°F ?

Data

T0 = 400°F h = 3.35 kW/m2-°K B = 0.1 cm.Ta = 80°F ks = 0.33 W/m-°K α = 1.3 10-7 m2/sec

The Biot number can be estimated as:

Bi = hB

k s

=3350 0.001

0.33= 10.15

The dimensionless surface temperature ratio is

θ s =

T 0 – Ta

T0 – Ta

= 144 – 80400 – 80

= 0.2

The Gurney-Lurie Chart 11.4c yields for Bi ≈ 10, the ratio of the surfacetemperature to the mid-plane temperature

However, since θ

θ10 = 0.15

, we can calculate the mid-plane temperature

from the relation for θs which is θ s = θ1

0 θθ1

0 = 0.2

This gives a midplane-temperature of θ10 = 0.2/0.15 > 1........Nonsense

What’s wrong ???We did a lot of things wrong.

Lecture 9ChE 333 13

First of all the solution we used involved only 1 term of an infinite series...

θ 1 = A 1e–β 12xFosin β 1ξ = θ1

0sin β 1ξ

We also get into trouble if we use such an equation for a “short” timesolution. Therefore avoid the charts for small xFo and large Biot numbers.

The “short time” solution we presented in the last lecture had the form.

θ = erf y

4t+ exp y + t erfc y

4t+ t

where y = hB

k s

yB and t = hB

k s

2 αtB2

Now for this case, y = 0 and we can use figure 11.3.2. We can determinethat the value of t at which Θ = 0.2.

We observe that t1/2 = 2.65 and consequently t = 7.65

Recall that ρCp = k/α = 2.5 MJ/m2-°K.

This leads to

t = hB

k s

2 αtB2 = 7.65 = h

k s

2

αt

We calculate that the time passed is t = 0.52 seconds and since d = Vt, thedistance is d = (20 cm/s) 0.52 sec = 10.4 cm.

Lecture 9ChE 333 14

An alternative method

We can use the complete Fourier expansion, not just one term.

θ =

4 sin λ n

2λn+ sin 2λn

cos λnξ e – λn2xFoΣ

n =1

∞

λntan λn = Bi

The first set of eigenvalues are

n λn1 1.4292 4.3063 7.228

If we calculate the first three terms of the Fourier expansion, we obtain

θ(1) = 0.178e–2.04x Fo + 0.155e –18.5xFo

For Θ = 0.2, by trial and error, we obtain xFo = 0.608. If we calculate thetime, we get 0.52 sec. The same as the short time solution.

This allows us a measure of “short time”. as for a slab

4 αtB2 ≤ 1 or x Fo ≤ 1

16