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Consider first the integral on the left side of Eq. (21.40). Interchanging the order of integration and differentiation* results in
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where ;(x ,s) is the Laplace transform of T(x ,t). + It should be noted that the presence of x has no effect on the second integral of Eq. (21.41) because the integration is with respect to t. Also note that the derivative on the right side of Eq. (21.41) is taken as an ordinary derivative because T(x ,s) will later be seen to be a function of only one independent variable x and a parameter s. Next consider the integral on the right side of Eq. (21.40). Again, the presence of x has no effect on the integration with respect to t, and the rule for the transform of a derivative may be applied directly to yield
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= s ;(x,s) - T(x,O)
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where T(x ,0) is the initial temperature distribution in the solid. Introducing the results of the transformation into Eq. (21.40) gives
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*This interchange is allowed for most functions of engineering interest. See R. V. Churchill (1972). + In this chapter the overbar will often be used to indicate the Laplace transform of a function of two variables.
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APPLICATTONS
K d2 ;(x,s)
= s ;(x,s) - T(x,O)
The partial differential equation has now been reduced to an ordinary differential equation, which can usually be solved without difficulty. It should be clear that s in Eq. (21.43) is merely a parameter, with the result that this equation is an ordinary second-order differential equation in the independent variable n. This follows because there are no derivatives with respect to s in Eq. (21.43). Since we have taken T(x,O) = 0 for the example under consideration, Eq. (21.43) becomes (21.44) Equation (21.44) is a linear differential equation and can be solved to give T = Ale- Gx + A2e @ (21.45)
The arbitrary coefficients A1 and A2 may be evaluated as follows: In order that ; may be finite as x --, a, it is necessary that A2 = 0. Equation (21.45) then becomes T = Ale-@x (21.45a) The transformed forcing function at x = 0 is ;(O,s), which can be substituted into Eq. (21.4%) to determine Al; then ;(O, s) = Aleo or Al = T(O, s) Substituting Al into Eq. (21.45~) gives (21.46) By specifying a particular value of x, say x = L, the transfer function is (21.47)
STEP RESPONSE. To illustrate the use of this transfer function, consider a forcing
function that is the unit-step function; thus T(0, t) = u(t) for which case T(O,s) = l/s. Substituting this into Eq. (21.47) gives
;(L,s) = Ie-flL
(21.48)
THEORETICAL
ANALYSIS OF COMPLEX PROCESSES
To obtain the response in the time domain, we must invert Eq. (21.48). A table of transforms* gives the following transform pair:
where erfc x is the error-function complement of x defined as erfc x = 1 - - ox e- du ; 5 This function is tabulated in many textbooks+ and mathematical tables. Using this transform pair, l&. (21.48) becomes
T(L, t) = erfc-
A plot of T versus the dimensionless group K t/L2 is shown in Fig. 21.7.
SINUSOIDAL RESPONSE. It is instructive to consider the response in temperature
at x = L for the case where the forcing function is a sinusoidal variation; thus T(0, t) = Asinot Using the substitution rule of Chap. 16, in which s is replaced by jo, Eq. (21.47) becomes (21.51)
*Tables of transforms that include transform pairs frequently encountered in the solution of partial differential equations may be found in many textbooks on heat conduction and applied mathematics. For example, see Mickley, Sherwood, and Reed (1957). Inversion of complicated transforms such as that of Eq. (21.48) can be achieved systematically by the method of complex residues, which is also discussed in the above reference. t&e Carslaw and Jaeger (1959), p. 485.
Response of temperature in the interior of a solid to a unit-step change in temperatum at the surface.
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