# Scaling

Thermal penetration depth for conduction in a semi-infinite solid.

Scaling, or scale analysis, is a process that uses the basic principles of heat transfer (or other engineering disciplines) to provide order-of-magnitude estimates for quantities of interest. For example, scale analysis of a boundary-layer type flow can provide the order of magnitude of the boundary layer thickness. In addition, scale analysis can provide the order of magnitude of the heat transfer coefficient or Nusselt number, as well as the form of the functions that describe these quantities. Scale analysis confers remarkable capability because its result is within a few percentage points of the results produced by the exact solution (Bejan, 2004). Scaling will be demonstrated by analyzing a heat conduction problem and a contact melting problem.

The first example is a scale analysis of a thermal penetration depth for conduction in a semi-infinite solid as shown in the figure on the right. The initial temperature of the semi-infinite body is Ti. At t = 0 the surface temperature is suddenly increased to T0. At a given time t, the thermal penetration depth is δ, beyond which the temperature of the solid is not affected by the surface temperature. The temperature satisfies the following two conditions

$T(\delta ,t) = {T_i} \qquad \qquad(1)$
${\left. {\frac{{\partial T}}{{\partial x}}} \right|_{x = \delta }} = 0 \qquad \qquad(2)$

The energy equation for this problem and the corresponding initial and boundary conditions are:

$\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}\quad \quad x > 0,\quad t > 0 \qquad \qquad(3)$
$T(x,t) = {T_0}\quad \quad x = 0,\quad t > 0 \qquad \qquad(4)$
$T(x,t) = {T_i}\quad \quad x > 0,\quad t = 0 \qquad \qquad(5)$

Since temperature difference occurs only within $0 \le x < \delta$, the order of magnitude of x is the same as δ, i.e.,

$x \sim \delta \qquad \qquad(6)$

The order of magnitude of the term on the left-hand side of eq. (3) is

$\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{\partial }{{\partial x}}\left( {\frac{{\partial T}}{{\partial x}}} \right) \sim \frac{1}{\delta }\frac{{\Delta T}}{\delta } = \frac{{\Delta T}}{{{\delta ^2}}} \qquad \qquad(7)$

where ΔT = T0Ti. The order of magnitude of the right-hand side of eq. (3) is

$\frac{1}{\alpha }\frac{{\partial T}}{{\partial t}} \sim \frac{1}{\alpha }\frac{{\Delta T}}{t} \qquad \qquad(8)$

Equation (3) requires that the two orders of magnitude represented by eqs. (7) and (8) equal each other, i.e.,

$\frac{{\Delta T}}{{{\delta ^2}}} \sim \frac{1}{\alpha }\frac{{\Delta T}}{t} \qquad \qquad(9)$

The order of magnitude of the thermal penetration depth is then

$\delta \sim \sqrt {\alpha t} \qquad \qquad(10)$

Equation (10) indicates that the thermal penetration depth is proportional to the square root of time, which agrees with the results obtained by the integral approximate solution.

Heat conduction in a rectangular domain.

The additional capability of scale analysis over dimensional analysis can be demonstrated through analysis of a transient heat conduction in a rectangular domain with dimensions of L by H (see figure on the right). The condition under which the problem can be simplified to one-dimension can be determined through scaling. The governing equation for transient heat conduction in a two-dimensional domain is

$\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}} \qquad \qquad(11)$

The orders of magnitude of x and y are, respectively, the same as L and H, i.e.,

$x \sim L, y \sim H \qquad \qquad(12)$

The orders of magnitude of the two terms on the left-hand side of eq. (11) are

$\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{\partial }{{\partial x}}\left( {\frac{{\partial T}}{{\partial x}}} \right) \sim \frac{1}{L}\frac{{\Delta T}}{L} = \frac{{\Delta T}}{{{L^2}}} \qquad \qquad(13)$
$\frac{{{\partial ^2}T}}{{\partial {y^2}}} = \frac{\partial }{{\partial y}}\left( {\frac{{\partial T}}{{\partial y}}} \right) \sim \frac{1}{H}\frac{{\Delta T}}{H} = \frac{{\Delta T}}{{{H^2}}} \qquad \qquad(14)$

where ΔT = T1T2. The condition under which the problem can be simplified as 1-D is

$\frac{{{\partial ^2}T}}{{\partial {x^2}}} \gg \frac{{{\partial ^2}T}}{{\partial {y^2}}} \qquad \qquad(15)$

i.e.,

$H \gg L \qquad \qquad(16)$

## References

Bejan, A., 2004, Convection Heat Transfer, 3rd edition, John Wiley & Sons, New York.

Faghri, A., and Zhang, Y., 2006, Transport Phenomena in Multiphase Systems, Elsevier, Burlington, MA.

Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.