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39 changes: 34 additions & 5 deletions Manuals/FDS_Verification_Guide/FDS_Verification_Guide.tex
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Expand Up @@ -5022,6 +5022,39 @@ \section{SFPE Verification Cases}

The Society of Fire Protection Engineers (SFPE) has developed a standard entitled {\em S.02 -- Calculation Methods to Predict the Thermal Performance of Structures \& Fire Resistive Assemblies}~\cite{SFPE_S.02} that contains an appendix with verification cases to benchmark basic heat transfer calculations. This section contains several of these cases.

\subsection{Case 1: Lumped Mass Subjected to Standard Fire}
\label{SFPE_Case_1}

A plate ($\rho=7850$~kg/m$^3$, $c=0.52$~kJ/(kg~K), $\epsilon=0.7$) that has a thickness of 4~cm and an initial temperature of 20~°C is heated on the top and bottom surfaces according to the standard ISO~834 fire curve
\be
T(t) = T_\infty + 345 \, \ln \left( 8t/60 + 1 \right) \label{ISO_834}
\ee
where the time, $t$, is in seconds. If the thermal conductivity of the material is relatively large, the temperature in the section can be taken as uniform. For the convection heat transfer coefficient, $h=25$~W/(m$^2$~K), calculate the temperature of the plate as a function of time (Fig.~\ref{fig:SFPE_Case_1}).

\begin{figure}[ht]
\centering
\includegraphics[height=2.2in]{SCRIPT_FIGURES/SFPE_Case_1}
\caption[The SFPE heat transfer verification Case 1]{Temperature of a 4~cm thick plate that is heated top and bottom by the standard fire curve.}
\label{fig:SFPE_Case_1}
\end{figure}

\FloatBarrier

\subsection{Case 2: Lumped Mass Subjected to Incident Flux}
\label{SFPE_Case_2}

A 1~cm thick horizontal flat plate ($\rho=7850$~kg/m$^3$, $c=0.56$~kJ/(kg~K), $\epsilon=0.9$) with an initial temperature of 20~°C is exposed from above with a radiant heater set to an incident flux of ̇50~kW/m$^2$. The gas temperature is 20~°C and $h=12$~W/(m$^2$~K). Assuming that the bottom and sides of the plate are perfectly insulated, and that the thermal conductivity of the material is sufficiently large to assume a uniform temperature with depth, calculate the temperature of the plate as a function of time (Fig.~\ref{fig:SFPE_Case_2}).

\begin{figure}[ht]
\centering
\includegraphics[height=2.2in]{SCRIPT_FIGURES/SFPE_Case_2}
\caption[The SFPE heat transfer verification Case 2]{Temperature of a 1~cm thick plate that is heated on top via an incident flux and insulated below.}
\label{fig:SFPE_Case_2}
\end{figure}

\FloatBarrier


\subsection{Case 6: 2-D Heat Transfer with Cooling by Convection}
\label{SFPE_Case_6}

Expand All @@ -5039,11 +5072,7 @@ \subsection{Case 6: 2-D Heat Transfer with Cooling by Convection}
\subsection{Case 7: 2-D Heat Transfer by Convection and Radiation}
\label{SFPE_Case_7}

A 0.2 m by 0.2 m square column ($k=1$~W/(m~K), $\rho=2400$~kg/m$^3$, $c=1$~kJ/(kg~K), $\epsilon=0.8$) is heated according to the ISO~834 time-temperature curve
\be
T(t) = T_\infty + 345 \, \ln \left( 8t/60 + 1 \right) \label{ISO_834}
\ee
where the time, $t$, is in seconds. Assuming that $h=10$~W/(m$^2$~K) and that the initial temperature is $T_\infty=273$~K, calculate the temperature at the column center, corner and middle side surface as a function of time (Fig.~\ref{fig:SFPE_Case_7}).
A 0.2 m by 0.2 m square column ($k=1$~W/(m~K), $\rho=2400$~kg/m$^3$, $c=1$~kJ/(kg~K), $\epsilon=0.8$) is heated according to the ISO~834 time-temperature curve, Eq.~(\ref{ISO_834}). Assuming that $h=10$~W/(m$^2$~K) and that the initial temperature is $T_\infty=273$~K, calculate the temperature at the column center, corner and middle side surface as a function of time (Fig.~\ref{fig:SFPE_Case_7}).

\begin{figure}[ht]
\centering
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1 change: 0 additions & 1 deletion Source/read.f90
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Expand Up @@ -8338,7 +8338,6 @@ SUBROUTINE READ_SURF(QUICK_READ)
SF%DUCT_PATH = DUCT_PATH
SF%DT_INSERT = DT_INSERT
SF%E_COEFFICIENT = E_COEFFICIENT
IF (SF%HT_DIM>1 .AND. EMISSIVITY>=0._EB .AND. EMISSIVITY_BACK<0._EB) EMISSIVITY_BACK = EMISSIVITY
SF%EMISSIVITY = EMISSIVITY
SF%EMISSIVITY_BACK = EMISSIVITY_BACK
IF (SF%EMISSIVITY>=0._EB) THEN
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4 changes: 3 additions & 1 deletion Source/wall.f90
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Expand Up @@ -2116,7 +2116,9 @@ SUBROUTINE SOLID_HEAT_TRANSFER(NM,T,DT_BC,PARTICLE_INDEX,WALL_INDEX,CFACE_INDEX,

! Compute back side emissivity

IF (SF%EMISSIVITY_BACK_SPECIFIED) THEN
IF (SF_BACK%EMISSIVITY_SPECIFIED) THEN
EMISSIVITY_BACK = SF_BACK%EMISSIVITY
ELSEIF (SF%EMISSIVITY_BACK_SPECIFIED) THEN
EMISSIVITY_BACK = SF%EMISSIVITY_BACK
ELSEIF (BACKING /= INSULATED) THEN
CALL GET_EMISSIVITY(ONE_D,NWP,EMISSIVITY_BACK)
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2 changes: 2 additions & 0 deletions Utilities/Matlab/FDS_verification_dataplot_inputs.csv
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Expand Up @@ -612,6 +612,8 @@ d,realizable_mass_fractions,Flowfields/realizable_mass_fractions_git.txt,Flowfie
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