Lecture 10_Unsteady Conduction II_CHBE 351

8/13/2019 Lecture 10_Unsteady Conduction II_CHBE 351

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2-D Unsteady-State Conduction X-1

THE SEMI-INFINITE SOLID

In this section we will examine three important cases for a semi-infinite wall which is initially

at a uniform temperature, T i. The exposed surface is suddenly subject to either a new constant

temperature, or a constant heat flux or a convection heat transfer. In all three cases we would

like to determine how the temperature changes in the solid.

The equation for this case is

t

T

x

T

12

2

subject to:

Initial condition: iT xT )0,( (the temperature is constant at T i at t =0 at all x locations)

Boundary conditions: x iT T (same for all three cases)

0 x sT T or ""

0

os

x

qq x

T k

or ),0(0

t T T h x

T k

x

The solution can be found by a method called a similarity variable method. According to this,

the two independent variables, t and x, are combined to result a single and the PDE is




transformed into an ODE. The similarity variable in this case is 2/1)4/( t x . Using this the

PDE becomes

T T 2

2

2

The solutions in these cases are

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

t

xerf

T T

T t xT

si

s

2

),(

t

T T k t q is

s

)()("

Case 2 Constant Surface Heat Flux: ""os qq

t

xerfc

k

xq

t

x

k

t qT t xT oo

i

24exp

)/(2),(

"22/1"

"" )( os qt q

Case 3 Surface Convection: ),0(0

t T T h x

T k

x

k t h

t xerfc

k t h

k hx

t xerfc

T T T t xT

i

i

2exp

2),(

2

2

),0()("t T T ht qs

Where )( erf is the Gaussian error function defined as

o

duuerf )exp(2

)( 2

)( erfc is the complementary error function defined as

)(1)( erf erfc

Appendix B.2 in your textbook includes the values of the )( erf and )( erfc functions.

Note: For case 1 the temperature at the surface is constant and the heat flux decreases with

time, 2/1" )( t t qs . In case 2, the flux is constant while the temperature at the surface increases




monotonically with 2/1t T s . Finally in case 3, the flux decreases with time since T s

approaches T . Selected profiles of temperature for this case (surface convection) are plotted

below.

An interesting application of case 1 is when 2 semi-infinite solids, initially at uniform

temperatures i AT , and i BT , are placed into contact. If the contact resistance is negligible, the

requirement of thermal equilibrium requires both T’s and fluxes at the interface to be the same.

Thus,

"

,

"

, Bs As qq or 2/1

,

2/1

, )()(

t

T T k

t

T T k

B

ibs B

A

i As A

Solving for T s.

2/12/1

,

2/1

,

2/1

B A

i B Bi A A

sck ck

T ck T ck T

Since the surface temperature is constant upon contact, the equations for case one can be used to

determine the profiles that are plotted in the figure below.




EXAMPLE: In laying water mains, utilities must be concerned with the possibility of freezing

during cold periods. Although the problem of determining the temperature in soil as a function

of time is complicated by changing surface conditions, reasonable estimates can be based on the

assumption of constant surface temperature over a prolonged period of cold weather. Consider

the water main in the schematic below which is buried in soil initially at 20oC and is suddenly

subjected to a constant surface temperature of -15oC for 60 days.

(a.) Calculate and plot the temperature history at the burial depth of 0.68m for thermal

diffusivity of smand /0.3,38.1,0.110 27

(b.) For sm /1038.1 27 , plot the temperature distribution over the depth m x 0.10

for times of 1,5,10,30 and 60 days.

(c.) For sm /1038.1 27 , show that the heat flux from the soil decreases with increasing

time by plotting ),0(" t q x as a function of time for the 60-day period. On this graph, also

plot the heat flux at the depth of the buried main, ),68.0(" t mq x










FINITE-DIFFERENCE METHODS

Analytical methods are restricted to simple geometries and boundary conditions. Thus in most

cases we have to use numerical methods to obtain solutions. One such method is the finite-

difference.

DISCRETIZATION OF THE HEAT EQUATION: THE EXPLICIT METHOD

Consider a 2-D system under transient conditions with constant properties and no internal

generation. The heat diffusion equation reduces to

t

T

y

T

x

T

12

2

2

2

To obtain the finite-difference form of this equation, we may use the same approximations as

before to the spatial derivatives. Again m, n subscripts are used to designate the x and y locations

of the discrete nodal points.

To discretize this equation in time, we use the integer p, so that

t pt

Thus the total time is subdivided into time intervals of t at which we would like to perform the

calculations. These time interval are t pt pt t t )1(,,....3,2,

Then the finite-difference approximation to the time derivative can be expressed as




t

T T

t

T p

nm

p

nm

nm

,

1

,

,

Where (p+1) refers to the new time and p to the previous time interval.

This finite-difference scheme is called a forward-difference and the method is called explicit

method because all temperatures are evaluated at the previous time ( p).

The overall equation for an interior node can be written

2

,1,1,

2

,,1,1,

1

,

)(

2

)(

21

y

T T T

x

T T T

t

T T p

nm

p

nm

p

nm

p

nm

p

nm

p

nm

p

nm

p

nm

Assuming ( ) y x and 2

x

t Fo

, then

p

nm

p

nm

p

nm

p

nm

p

nm

p

nm T FoT T T T FoT ,1,1,,1,1

1

, )41(

Thus we can calculate ),( y xT at the new time t p )1( in a straightforward manner.

If the system is 1-D unsteady

t

T

x

T

12

2

The finite-difference equation becomes

p

m

p

m

p

m

p

nm T FoT T FoT )21(11

1

,

It is now evident why this method is called explicit. Since the initial conditions is known

),(0 y xT t

It becomes easy to calculate ),( y xT at later times. This method is not very accurate, but you can

improve the accuracy if you use small values for t and y x , . However, in this case the

computational time increases.

This explicit method frequently becomes unstable (the solution diverges). To prevent this

instability from happening, the x and t should satisfy a requirement, that is: The coefficient

of the temperature p

nm

T ,

of the equation for that specific node (m,n) should be positive, thus

1-D: 021 Fo or 2/1Fo or

2/12

x

t

2-D: 041 Fo or 4/1Fo or

4/12

x

t




The above finite-difference equations can be derived by using the energy method (see the

example below for a boundary node with convection)

Assumptions: 1-D conduction, no heat generation, unsteady-state and constant physical

properties.

The energy equation for the control volume reduces to

st in E E

or t

T T xcAT T

x

kAT T Ah

p

o

p

o p

o

p p

o

1

12

or solving for 1 p

oT we get

p

o

p

o

p p

o

p

o T T T x

t T T

xc

t hT

12

1 22

However, BiFo x

t

k

xh

xc

t h22

22

p

o

p p

o T BiFoFo BiT T FoT 2212 1

1

The finite-difference Biot number is

k

xh Bi

The stability criterion defined before requires

0221 BiFoFo or 2/1)1( BiFo




From each equation you may get a different criterion for stability. You compare all of them and

take the most restrictive. Table 5.3 of your textbook gives you the forward finite-difference

equation for various cases.




EXAMPLE 5.6 : A fuel element of a nuclear reactor is in the shape of a plane wall of thickness

2 L = 20 mm and is convectively cooled at both surfaces, with h = 1100 W/m2 • K and T =

250°C. At normal operating power, heat is generated uniformly within the element at a

volumetric rate of1

q = 107W/m3. A departure from the steady-state conditions associated with

normal operation will occur if there is a change in the generation rate. Consider a sudden change

to 2q = 2 107 W/m3, and use the explicit finite-difference method to determine the fuel

element temperature distribution after 1.5 s. The fuel element thermal properties are k = 30 W/m

• K and a = 5 106m2/s.

Assumptions: 1-D conduction, no heat generation, unsteady-state and constant physical

properties.

Using a space increment of mm x 2 we can write the explicit finite difference equation










EXAMPLE (Textbook 5.94): A one dimensional slab of thickness 2 L is initially at a uniform

temperature Ti. Suddenly, electric current is passed through the slab causing a uniform

volumetric heating q (W/m3). At the same time, both

outer surfaces ( L x ) are subjected to a convection

process at T with a heat transfer coefficient.

Write the finite difference equation expressing

conservation of energy for node 0 located on the outer surface at L x . Rearrange your

equation and identify any important dimensionless coefficients.

Lecture 10_Unsteady Conduction II_CHBE 351

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