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Copy file name to clipboardExpand all lines: Manuals/FDS_User_Guide/FDS_User_Guide.tex
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@@ -3452,7 +3452,7 @@ \subsubsection{Example}
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\section{Pyrolysis and Energy Conservation}
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\label{solid_phase_energy_conservation}
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The \ct{HEAT_OF_REACTION} is the energy gained or lost in converting the material to residue(s) and/or gas(es). In other words the \ct{HEAT_OF_REACTION} is the enthalpy of the products at the current solid temperature minus the enthalpy of the initial material at the current solid temperature. For total internal energy to be conserved, this must hold true. A challenge is that while the heat of reaction may be known for a solid material, detailed knowledge of the material enthalpy or the enthalpy of any residue or gaseous products is often not known. In cases where a reference enthalpy for a material is known, you can input on the \ct{MATL} line the \ct{REFERENCE_ENTHALPY} in kJ/kg and the \ct{REFERENCE_ENTHALPY_TEMPERATURE} in $^\circ$C.
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The \ct{HEAT_OF_REACTION} is the energy gained or lost in converting the material to residue(s) and/or gas(es). In other words the \ct{HEAT_OF_REACTION} is the enthalpy of the products at the current solid temperature minus the enthalpy of the initial material at the current solid temperature. For total internal energy to be conserved, this must hold true. A challenge is that while the heat of reaction may be known for a solid material, detailed knowledge of the material enthalpy or the enthalpy of any residue or gaseous products is often not known. In cases where a reference enthalpy for a material is known, you can input on the \ct{MATL} line the \ct{REFERENCE_ENTHALPY} in kJ/kg and the \ct{REFERENCE_ENTHALPY_TEMPERATURE} in $^\circ$C. All reactions require a \ct{HEAT_OF_REACTION} even if it is zero.
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FDS will attempt to adjust all material enthalpies so that internal energy is conserved. For each material reaction a linear equation is defined:
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\be
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The value of $T_{\rm ref}$ used for a reaction is the value specified by the \ct{REFERENCE_TEMPERATURE} for that reaction. If no value is given (when the reaction is defined using \ct{A} and \ct{E}), then FDS will do a virtual TGA using just the single reaction for the single material. The temperature where the peak reaction rate occurs is used for $T_{\rm ref}$. Since in most cases the specific heat of a material and the specific heats of its residues and product gases are not the same, the heat of reaction is temperature dependent. FDS will create a temperature dependent array for the heat of reaction that accounts for this where the value at $T_{\rm ref}$ is fixed to the value specified with \ct{HEAT_OF_REACTION}.
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This process of adjusting enthalpies can be skipped by setting \ct{ADJUST_H=F} on a \ct{MATL} line. This should be done when material reactions do not represent actual chemical reactions. If no \ct{HEAT_OF_REACTION} is specified for a \ct{MATL} with a reaction then \ct{ADJUST_H=F} will be set. Note that this means if a reaction has \ct{HEAT_OF_REACTION} that is actually zero, and enthalpy adjustment is desired, then set \ct{HEAT_OF_REACTION} to 0 in the input file for that reaction.
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This process of adjusting enthalpies can be skipped by setting \ct{ADJUST_H=F} on a \ct{MATL} line. This should be done when material reactions do not represent actual chemical reactions.
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If the sum of the yields is less than 1, then for the purpose of solving for the $H_{\rm adj}$ values, FDS will assume that the missing mass is a material with the same specific heat as the original material.
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@@ -5679,7 +5679,7 @@ \subsection{Example Case: Soot Deposition from a Propane Flame}
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&SPEC ID = 'NITROGEN', LUMPED_COMPONENT_ONLY = T /
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&SPEC ID = 'WATER VAPOR', LUMPED_COMPONENT_ONLY = T /
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&SPEC ID = 'CARBON DIOXIDE', LUMPED_COMPONENT_ONLY = T /
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&SPEC ID = 'SOOT', AEROSOL = T /
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&SPEC ID = 'SOOT', AEROSOL = T, MEAN_DIAMETER=1.E-6 /
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\end{lstlisting}
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If Eq.~(\ref{eq:PROPANE_depo}) is properly balanced, you can directly use the stoichiometric coefficients of the primitive species to define the lumped species:
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242 \> \ct{Spray Pattern Table ... massflux <= 0 for line ...} \> Section~\ref{info:sprinklers} \\
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243 \> \ct{PROP ... VIEW_ANGLE must be between 0 and 180} \> Section~\ref{info:sprinklers} \\
\caption[The \ct{heat\_conduction} test cases]{Comparison of heat conduction calculations with analytical solutions.}
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\label{heat_conduction}
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\end{figure}
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\section{Heat Conduction Through a Solid Slab Part 2 (\texorpdfstring{\ct{back\_wall}}{back\_wall})}
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\label{back_wall}
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A slab of thickness $L=0.01$~m with a conductivity of 10~\si{W/(m.K)} is exposed to a backside temperature of $T_g=200$~\si{\degree C}.
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The backside heat transfer conditions are $h$=50~\si{W/(m^2.K)} and $e$=0.5. The frontside heat transfer conditions are $h$=10~\si{W/(m^2.K)} and $e$=1.
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This case test that both sets of conditions are applied.
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