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TECHNICAL PAPER

The effective thermal conductivity of the elementary unit m = +1). For the special case of m → 0 for random distribution
cells can be obtained from the effective heat flow Q X at the of cells, the suggested empirical relationship has been given by
boundaries along X-axis, namely at X = 0; and X = 1 as, earlier researchers [38,42] and corresponds to weighted geometric
average. Generally, k ec1 and k ec2 differ from each other by few
(17)
orders, especially when conductivities of solid and pores differ
where, Q X is numerically equal to the heat inflow at X = 0 or, heat by large order, hence geometric mean, is more appropriate. The
outflow at X = 1. k ec , is the effective thermal conductivity of the conductivities k ec1 and k ec2 represent lower and upper bounds of
cell under consideration. A typical 3-D view of the elementary conductivity.
cell with enclosed pore with appropriate boundary conditions
is shown in Figure 16. Closed form solution to this problem (19)
is not readily available. Hence, finite element method (FEM) The non-dimensional ratios of conductivities obtained from the
was used for solving the heat transfer problem. Standard FEM numerical solution of heat transfer problem are λ 1 and λ 2 for the
formulation with eight nodded linear iso-parametric brick cells containing enclosing and enclosed pores, and are given
elements belonging to the serendipity family and automatic by, λ 1 = k ec1 / k s and λ 2 = k ec2 / k s respectively. Replacing k ec1 and
mesh generation provided an easy solution. For pore sizes k ec2 by λ 1 and λ 2 respectively, in Equation 19, results in following
smaller than 3 mm, in dry and saturated conditions; convection, Equation 20,
radiation, and evaporation-condensation in pores are neglected.
Thus, as discussed earlier, conduction is the main mode of (20)
heat transfer in dry and saturated pores. Thermal conductivity
of air and water in the ambient conditions are adopted as The values of λ 1 and λ 2 depend on state of material, i.e., dry or
0.0258 W/m.K and 0.6051 W/m.K [36,39] . The output obtained from saturated. Using added subscripts “d” and “s” respectively for
the program is the dimensionless effective thermal conductivity dry and saturated states the equations are obtained as follows,
of the given elementary cell. The inputs are k s , porosity, and (21)
equivalent conductivity of the pore (air/water).
(22)
Let f be the fraction of enclosed pores in the total porosity p of
the material. Then (1 – f) is the fraction of enclosing pores. Let When k s and f are known, for a given porosity p, and for the
k ec1 and k ec2 are the effective thermal conductivities of the cells given state, dry or saturated, the effective thermal conductivity
containing the enclosing pores and enclosed pores, respectively. of the material can be obtained using Equations 21 and 22 once
The effective thermal conductivity of the overall material k e λ 1d , λ 2d , λ 1s , and λ 2s are obtained through the solution of heat
can be obtained by combining the above conductivities of the transfer problems. A plot of λ 1d against porosity obtained from
elementary cells using general Law of mixture as follows , solution of heat transfer problem for various k s values is given
[42]
in Figure 17. Similar Figures for λ 2d , λ 1s , and λ 2s against porosity
(18)
for various k s values respectively are available in references [36,39] .
where, m, is an empirically determined exponent in the range of Direct experimental determination of k s and f, are difficult,
–1 ≤ m ≤ +1, depending on the arrangement of cells (for series however they can be determined from the experimentally
arrangement of cells m = – 1 and that for parallel arrangement determined values of k ed and k es .

Figure 16: Heat flow direction and boundary conditions for enclosed pore Figure 17: λ 1d against porosity

18 THE INDIAN CONCRETE JOURNAL | JANUARY 2026