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Copy file name to clipboardExpand all lines: Manuals/FDS_User_Guide/FDS_User_Guide.tex
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\subsection{Specified Internal Heat Source}
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\label{info:INTERNAL_HEAT_SOURCE}
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\label{internal_heating}
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The condensed phase heat conduction equation has a source term that describes the internal sources and sinks of energy.
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There are three types of sources that contribute to this term:
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heats of reaction for the pyrolysis (see Sec.~\ref{info:solid_pyrolysis}),
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internal absorption and emission of radiation (see Sec.~\ref{info:liquid_fuels}), and the source specified by the user.
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An example of the case where specified heat source could be needed is the heating of electrical cables due to internal current.
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The internal source term for each layer of the surface is specified using \ct{INTERNAL_HEAT_SOURCE} on the \ct{SURF} line. Its units are \unit{kW/m^3} and the default value is zero. An optional time ramp can be specified for each layer's heat source using \ct{RAMP_IHS}. In the example below, the cylindrical surface describing a cable consists of an outer plastic layer and inner core of metal. The metal core is heated with a power of 300~\unit{kW/m^3}.
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The internal source term for each layer of the surface is specified using \ct{INTERNAL_HEAT_SOURCE} on the \ct{SURF} line. Its units are \unit{kW/m^3} and the default value is zero. An optional time ramp can be specified for each layer's heat source using \ct{RAMP_IHS}. In the example below, the cylindrical surface describing a 10~cm long cable segment consists of an outer plastic layer and inner core of metal. The metal core is heated with a power of 300~\unit{kW/m^3}.
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\begin{lstlisting}
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&SURF ID = 'Cable'
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THICKNESS = 0.002,0.008
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MATL_ID(1,1) = 'PLASTIC'
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MATL_ID(2,1) = 'METAL'
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GEOMETRY = 'CYLINDRICAL'
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LENGTH = 0.1
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INTERNAL_HEAT_SOURCE = 0.,300. /
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&SURF ID = 'Cable'
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THICKNESS = 0.002,0.008
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MATL_ID(1,1) = 'PLASTIC'
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MATL_ID(2,1) = 'METAL'
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GEOMETRY = 'CYLINDRICAL'
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LENGTH = 0.1
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INTERNAL_HEAT_SOURCE = 0.,300. /
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\end{lstlisting}
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Figure~\ref{fig:internal_heating} displays the heat generated by 10 of these cable segments. The exact value is 300~\unit{kW/m^3} multiplied by the volume of the metal within the cable segment, $2 \times 10^{-5}$~\unit{m^3}, multiplied by the number of segments, 10, which equals approximately 0.06~kW.
\caption[Results of the \ct{internal_heating} test case]{Heating rate of a set of 10 cable segments. The dashed line represents the heat generated by the cable and the dotted line the heat flowing out of the computational domain.}
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