Legacy · Thermal validation

Homogeneous material

A $1\,\mathrm{m}\times 1\,\mathrm{m}$ body, insulated on the whole boundary. The isothermal formulation, with the $Q(t_{eq})$ below, must return the analytical adiabatic rise. Temperature stays uniform.

Paper: Analytical and numerical validation of the thermochemical problem of early-age concrete · DOI 10.70271/rirc.v1n2a008

Input data

Item	Value
Geometry	$1\,\mathrm{m}\times 1\,\mathrm{m}$ square
Boundary	$q=0$ on every edge
$T_0$	$25\,^{\circ}\mathrm{C}$
$T_{\mathrm{ref}}$	$25\,^{\circ}\mathrm{C}$
$E_a/R$	$4000\,\mathrm{K}$
$C$	$300\,\mathrm{kg/m^3}$
$k$	$9.36\,\mathrm{kJ/(m\cdot h\cdot^{\circ}C)}$
$\rho$	$2300\,\mathrm{kg/m^3}$
$c_e$	$0.900\,\mathrm{kJ/(kg\cdot^{\circ}C)}$
Adiabatic rise	$T(t)=25+30\,t^{2}/(12^{2}+t^{2})$ , $t$ in hours
Series horizon	$0$ to $480\,\mathrm{h}$ , step $1\,\mathrm{h}$

The source imposed on the model is isothermal, two Hill terms, with $a$ in hours:

$\Delta T^{Q}(t_{eq})=16.4\,\frac{t_{eq}^{1.93}}{(0.43\cdot 24)^{1.93}+t_{eq}^{1.93}}+13.6\,\frac{t_{eq}^{1.57}}{(1.35\cdot 24)^{1.57}+t_{eq}^{1.57}}.$

$C\cdot Q=\rho c_e\,\Delta T^{Q}$ . The sum $16.4+13.6$ is the $30\,^{\circ}\mathrm{C}$ of the adiabatic curve.

Reference result

Tmax_C in the series is the analytical adiabatic temperature. The body is uniform, so any point is the maximum. Q_kJ_kg is the isothermal generation at the equivalent time integrated along that temperature.