Duhamel Example1[1]

8/11/2019 Duhamel Example1[1]

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Example from Chapter 5, Ozisik, M.N., 1980.Heat Conduction John Wiley and Sons.

Duhamels Theorem

Example (1)

A semi infinite solid is initially at zero temperature. For time t > 0 the boundarysurface at x = 0 is kept at temperature f (t). Obtain an expression for the temperature distribution

T(x,t) in the solid for times t > 0.

Solution: The mathematical formulation of this problem is given as

The auxiliary problem is taken as

Then the solution of the problem (1-1) is given in term of the solution of the problem (1-2) by

the Duhamels theorem (1-3 ) as:

The solution of the auxiliary problem (1-2) is obtainable from the solution T(x,t) givenby equation


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by the relation and setting the equation(1-4) T0=1. Thus we obtain

Then

* +

Introducing equation (1-6) into equation (1-3) the solution of the problem (1-1) becomes

*

+

This result can be put into a different form by defining a new variable as

Introducing equation (1-8) into equation (1-7), we obtain

We now consider a special case of solution (1-9); if the surface temperature is a periodic function

of time in the form

The solution (1-9) becomes

* +

Or


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* +

* +

The first definite integral can be evaluated, then

*

+* +

* +

Here the second term on the right represents the transients that die away as t , and the firstterm represent the steady oscillation of temperature in the medium after the transient have

passed.

Example (2)

A slab, is initially at zero temperature. For times t > 0 the boundary at the surfaces atx = 0 and x = L are kept at temperatures f1(t) and f2(t) , respectively. Obtain an expression for the

temperature distribution T(x,t) in the slab for times t > 0.

Solution:

The mathematical formulation of this problem is given as



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And by Duhamels theorem

the solution of problem (2-1) is given as

Introducing equation (2-5) into equation (2-7) we obtain

Where . This solution seems to vanish at x = 0 and x = L, instead of converging tothe boundary condition function f1(t) and f2(t) at these locations. The reason for this is that the

term associated with the boundary-condition functions are in the form of Fourier series that are

not uniformly convergent at these locations. This difficulty can be alleviated by integrating

equation (2-8) by parts and replacing such series by their equivalent closed-form expression asnow described.

We write equation (2-8) in the form

Where

The integral terms is evaluated by parts as


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( )

, [

]

-

* +

Equation (2-11) is introduced into equation (2-8)

* + *

+

Closed form expression can readily be obtained for the first two series on the right hand side of

the equation (2-12) as:

Introducing equation (2-13) into equation (2-12), the solution becomes

*

+

* +

This solution given in this form clearly shows that at x = 0 and x = L this solution reduces to f1(t)

and f2(t), respectively.


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Example (3)

A solid cylinder is initially at zero temperature. For times t > 0 boundary surfaces atr = b is kept at temperatures T = f(t), which varies with time. Obtain an expression for the

temperature distribution T(r, t) in the cylinder for time t > 0

Solution:



The solution of the problem (3-1) can be written in the term of the solution of the auxiliary

problem (3-2) by Duhamels theorem as

If is the solution of the problem for a solid cylinder , initially at temperatureunity and for times t > 0, the boundary surface at r = b is kept at zero temperature, then the

solution for is obtainable from the solution (3-4)


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by setting T0=1 in that equation ; we find

Where are positive roots of The solution of the auxiliary problem (3-2) is obtainable from the solution given by equation (3-5) as

Introducing equation (3-6) into equation (3-3) , the solution of the problem (3-1) becomes

Where are the roots of The solution for the T(r, t) given by equation (3-7) does not explicitly show that . This result can be expressed in alternative form by integrating integral term byparts as has been done in the previous example (2). We obtain:

*

+

We note that the solution (3-5) for t=0 should be equal to the initial temperature thus:


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Which gives the desired closed-form expression for the first series on the right hand side of

equation (3-8). Then the solution (3-8) is written as

*

+

The solution given in this form clearly shows that T(r, t) = f(t) at r = b.

Example (4)

A solid cylinder is initially at zero temperature. For times t > 0 heat is generated inthe solid at rate of g(t) per unit volume and boundary surfaces at the surfaces at r = b is kept at

zero temperature. Obtain an expression for the temperature distribution T(r, t) in the cylinder for

times t > 0

Solution:




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The the solution of the problem (4-1) is related to the solution of the auxiliary problem (4-2) by

Duhamels theorem as

The solution auxiliary problem (4-2) is obtainable from equation (4-4) by setting g0=1 and

F(r)=0; we find

Where are the positive roots of Introducing equation (4-5) into (4-3) we obtain

Duhamel Example1[1]

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