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Manuals/FDS_Technical_Reference_Guide/Aerosol_Chapter.tex

@@ -84,9 +84,9 @@ \subsection{Turbulent Deposition}
 for these regimes is given below~\cite{McCoy_Hanratty}.
 \be
 u_{\rm dt} = \left\{ \begin{array}{r@{\quad \quad}l}
-	0.086 \; \text{Sc}^{-0.7} \; u_{\tau}        &  \tau^+ < 0.2 \\
-	3.5 \times 10^{-4} \; {\tau^+}^2 \; u_{\tau} &  0.2 < \tau^+ < 22.9 \\
-	0.17 \; u_{\tau}                             &  \tau^+ > 22.9
+   0.086 \; \text{Sc}^{-0.7} \; u_{\tau}        &  \tau^+ < 0.2 \\
+   3.5 \times 10^{-4} \; {\tau^+}^2 \; u_{\tau} &  0.2 < \tau^+ < 22.9 \\
+   0.17 \; u_{\tau}                             &  \tau^+ > 22.9
 \end{array} \right.
 \ee
 where Sc is the particle Schmidt number, or the ratio of the kinematic viscosity to the
@@ -114,8 +114,8 @@ \section{Aerosol Agglomeration}
 where $\eta$, shown below, is a function for apportioning mass between adjacent particle bins.
 \be
 \eta = \begin{cases}
-	\frac{x_{i+1}-m_i}{x_{i+1}-x_i} & x_i< m_i \leq x_{i+1} \\
-	\frac{m_i-x_{i-1}}{x_i-x_{i-1}} & x_{i-1}< m_i \leq x_i
+   \frac{x_{i+1}-m_i}{x_{i+1}-x_i} & x_i< m_i \leq x_{i+1} \\
+   \frac{m_i-x_{i-1}}{x_i-x_{i-1}} & x_{i-1}< m_i \leq x_i
 \end{cases}
 \ee
 The agglomeration kernel is given by the sum of the Brownian (Br), gravitational (Gr), shear (Sh), and inertial (In) agglomeration  terms.
@@ -178,14 +178,14 @@ \section{Aerosol Scrubbing}
 \ee
 \be
 \epsilon_{im,vis} = \left\{ \begin{array}{ll}
-	0  & {\rm Stk} \leq 1.214  \\[0.1in]
-	\left(1+\frac{0.75 \ln (2 {\rm Stk}) }{{\rm Stk}-1.214} \right)^{-2} & {\rm Stk} > 1.214
+   0  & {\rm Stk} \leq 1.214  \\[0.1in]
+   \left(1+\frac{0.75 \ln (2 {\rm Stk}) }{{\rm Stk}-1.214} \right)^{-2} & {\rm Stk} > 1.214
 \end{array} \right.  \\[0.1in]
 \ee
 \be
 \epsilon_{im,pot} = \left\{ \begin{array}{ll}
-	0  & Stk \leq 0.0834  \\[0.1in]
-	\left(\frac{{\rm Stk}}{{\rm Stk}+0.5} \right)^2 & {\rm Stk} > 0.2
+   0  & Stk \leq 0.0834  \\[0.1in]
+   \left(\frac{{\rm Stk}}{{\rm Stk}+0.5} \right)^2 & {\rm Stk} > 0.2
 \end{array} \right.  \\[0.1in]
 \ee
 Where ${\rm Stk}$ is the Stokes number. For other values of ${\rm Stk}$, $\epsilon_{im,pot}$ is interpolated. The viscous and potential values are combined as:

Manuals/FDS_Technical_Reference_Guide/Appendices.tex

@@ -875,11 +875,11 @@ \subsection*{Moments of the PDF Transport Equation}
         fixed, fixed zerofill, precision=1,
         /tikz/.cd  % Change back to the normal key directory
     }
-    	]
-	\addplot graphics [xmin=-2.10, xmax=-0.1, ymin=-1, ymax=1] {FIGURES/transport_vs_mixing_0461_b}
-	node [above,white] at (30,15) {\Large(a)};
-	\addplot graphics [xmin= 0.10, xmax= 2.1, ymin=-1, ymax=1] {FIGURES/transport_vs_mixing_0461_a}
-	node [above,white] at (250,15) {\Large(b)};
+       ]
+   \addplot graphics [xmin=-2.10, xmax=-0.1, ymin=-1, ymax=1] {FIGURES/transport_vs_mixing_0461_b}
+   node [above,white] at (30,15) {\Large(a)};
+   \addplot graphics [xmin= 0.10, xmax= 2.1, ymin=-1, ymax=1] {FIGURES/transport_vs_mixing_0461_a}
+   node [above,white] at (250,15) {\Large(b)};
 \end{axis}
 \end{tikzpicture}
 \end{center}
@@ -1756,7 +1756,7 @@ \chapter{Development of an Implicit Solution for Droplet Evaporation}
 \frac{\d T_{\rm w}}{\d t} & = -\frac{A_{\rm p,s} \, h_{\rm w}}{m_{\rm w} \, c_{\rm w}} (T_{\rm w}-T_{\rm p})  \label{app_droplet_solid_temp}
 \end{align}
 
-	Below we will consider the time integration for a single substep in the evaporation routine, which updates the local computational cell from time $t^n$ to $t^n + \Delta t^n$.  Usually, $\Delta t^n = \Delta t_{\mathrm{\tiny LES}}$.  However, it is possible that subtimesteps may be required to prevent spurious mass and temperature overshoots.  In what follows, unless otherwise specified RHS quantities are evaluated at the beginning of the substep, $t^n$ (hence the designation ``semi-implicit'' method).  This limits the stability range of the method (since the method is not fully implicit it is not unconditionally stable), but it greatly simplifies the solution procedure.
+   Below we will consider the time integration for a single substep in the evaporation routine, which updates the local computational cell from time $t^n$ to $t^n + \Delta t^n$.  Usually, $\Delta t^n = \Delta t_{\mathrm{\tiny LES}}$.  However, it is possible that subtimesteps may be required to prevent spurious mass and temperature overshoots.  In what follows, unless otherwise specified RHS quantities are evaluated at the beginning of the substep, $t^n$ (hence the designation ``semi-implicit'' method).  This limits the stability range of the method (since the method is not fully implicit it is not unconditionally stable), but it greatly simplifies the solution procedure.
 
 To begin the development, we expanded Eq.~(\ref{app_droplet_gas_fraction}) with the mass loss rate (to be determined) evaluated at the midpoint of the time interval:
 \begin{equation}
@@ -1868,7 +1868,7 @@ \section{Mass Conservation}
 
 Mass conservation for reaction $\beta$ for solid material $\alpha$ is given by:
 \be
-	\sum_{\alpha'=1}^{N_{\rm m}} \nu_{\alpha',\alpha \beta} = 0
+   \sum_{\alpha'=1}^{N_{\rm m}} \nu_{\alpha',\alpha \beta} = 0
 \ee
 where $N_{\rm m}$ is the number of all types of substances in FDS (solid, gas, or particle) that the reaction produces and $\nu_{\alpha',\alpha \beta}$ is the yield of $\alpha'$ in the reaction where $\alpha'$ could be a new material, a gas, or a type of Lagrangian particle. In FDS input processing, the yield of the original material, $\nu_{\alpha,\alpha \beta}$, is taken as -1, and the prior equation written in terms of the yields as specified in an input file becomes:
 \be
@@ -1899,9 +1899,9 @@ \subsection{Solid Phase Heat of Reaction}
 In this equation $h_{\alpha,ref}$ is the enthalpy at the temperature $T_{ref}$. This is the value being solved for. In many tabulations of solid enthalpies the value of $T_{ref}$ is 298.15~K; however, since it can be determined at any reference temperature, for convenience a temperature of 0~K is picked. The specific heat of a material, $c_{p,\alpha}(T)$, is defined in the input file. Evaluating the integral requires picking the upper bound temperature, T$_R$. This is taken as the reference temperature for the reaction. This may be user defined as part of the input. If not, it can be determined from the reaction $A_{\alpha\beta}$ and $E_{\alpha\beta}$ by performing a TGA simulation for material $\alpha$, reaction $\beta$. The value for $T_R$ is set to the temperature of the peak reaction rate. With $T_R$ defined, the various $\nu_{\alpha',\alpha \beta} h_{\alpha,ref}$ terms can be collected on the left hand side, and the heats of reaction and the various $\nu_{\alpha',\alpha \beta} \int_{T_{ref}}^{T_{R}} c_{p,\alpha}(T) \, \mbox{d}T$ terms can be collected on the right hand side. The result is a system of linear equations $A \bar{x} = \bar{b}$ where the unknowns, $\bar{x}$, are the values of $h_{\alpha,ref}$; however, it is not guaranteed that there are an equal number of reactions and unknown material reference enthalpies. Solving the system of equations depends on the relative numbers of equations and unknowns as follows:
 
 \begin{description}
-	\item[Reactions = unknowns]{Solution can be found by a simple solution of the linear system: $\bar{x} = A^{-1} \bar{b}$}
-	\item[Reactions < unknowns]{In this case there are an infinite number of solutions; however, any single solution will suffice. The solution that gives the minimum vector magnitude of $\bar{x}$ is $\bar{x}= A^T (A A^T)^{-1} \bar{b}$.}
-	\item[Reactions > unknowns]{In this case the system is overdetermined, and there may not be a solution vector $\bar{x}$ that will perfectly conserve energy. A least-squares solution is given by $\bar{x}=  (A^T A)^{-1} A^T \bar{b}$.}
+   \item[Reactions = unknowns]{Solution can be found by a simple solution of the linear system: $\bar{x} = A^{-1} \bar{b}$}
+   \item[Reactions < unknowns]{In this case there are an infinite number of solutions; however, any single solution will suffice. The solution that gives the minimum vector magnitude of $\bar{x}$ is $\bar{x}= A^T (A A^T)^{-1} \bar{b}$.}
+   \item[Reactions > unknowns]{In this case the system is overdetermined, and there may not be a solution vector $\bar{x}$ that will perfectly conserve energy. A least-squares solution is given by $\bar{x}=  (A^T A)^{-1} A^T \bar{b}$.}
 \end{description}
 The first two cases are typically expected. The first case is where all materials decomposes into one or more gas species and there is no material residue. The second case is where one or more materials produce one or more residues including the possibility that the residue has a reaction. For example, wood pyrolyzing into a fuel gas plus char where the char reacst with oxygen to produce and ash remnant and CO$_2$. The third case would be where multiple solid materials are reactants in the same reaction. This case, while possible, is not typical for pyrolysis reactions.
 

Manuals/FDS_Technical_Reference_Guide/Complex_Geometry_Chapter.tex

@@ -138,7 +138,7 @@ \subsection{FV Discretization in Cut-Cells} \label{sec:cc}
       \put(-350,-10){(a)}
       \put(-120,-10){(b)}
       \caption{Cut-cell FV discretization in 2D: (a) \texttt{GASPHASE} Cut-cell  $ii$ is bounded by: $\mathbf{1,2}$ \texttt{INBOUNDARY} faces, $\mathbf{3,6}$ \texttt{GASPHASE} cut-faces, and $\mathbf{4,5}$  \texttt{GASPHASE} regular faces.  (b) Geometrical elements used in the discretization of the diffusive term for \texttt{GASPHASE} cut-face $\mathbf{3}$, and \texttt{INBOUNDARY} face $\mathbf{1}$.}
-	\label{Fig:FVdiscCC}
+   \label{Fig:FVdiscCC}
 \end{figure}
 %
 \subsection{CC Discretization of Diffusive Term}  \label{sec:CCdiff}
@@ -292,7 +292,7 @@ \subsection*{A. Faces $k=\mathbf{4},\mathbf{5}$ are \textit{regular} \texttt{GAS
       \put(-330,-10){(a)}
       \put(-110,-10){(b)}
       \caption{Flux limited interpolation: (a)  Regular and cut-cell species densities are available for 4 point stencil fluc limited interpolation to face $RC$.  (b) Lower cell in stencil is within solid region (cell S), therefore inner cell value $\rho Y_2$ is used for this extreme point.}
-	\label{Fig:FVFlxLimCC}
+   \label{Fig:FVFlxLimCC}
 \end{figure}
 
 \subsection*{B. Faces $k=\mathbf{3},\mathbf{6}$ are \texttt{GASPHASE} \textit{Cut-Faces}:}

Manuals/FDS_Technical_Reference_Guide/HVAC_Chapter.tex

@@ -173,28 +173,28 @@ \section{Fan Curves}
 This approach cannot be relied upon if there are multiple fans in a duct run as the approach will not correctly account for the combined effect. This approach can also not be relied upon if large pressure differences occur across the endpoints of the duct run. This might occur with a rapidly growing fire in a relatively well sealed space. In these cases one needs to define a system curve and find its intersection with the fan curve. This is the second approach in FDS.
 
 \begin{figure}[ht!]
-	\begin{center}
-		\scalebox{0.8}{\includegraphics{FIGURES/operationPoint}}
-		\caption[Illustration of the fan curve, system curve, and operation point for an HVAC fan.]{\label{hvac_curves} Illustration of the fan curve, system curve, and operation point for an HVAC fan.}
-	\end{center}
+   \begin{center}
+      \scalebox{0.8}{\includegraphics{FIGURES/operationPoint}}
+      \caption[Illustration of the fan curve, system curve, and operation point for an HVAC fan.]{\label{hvac_curves} Illustration of the fan curve, system curve, and operation point for an HVAC fan.}
+   \end{center}
 \end{figure}
 
 A challenge in fire simulation is that the system curve is a function of the pressures seen at the end points of a duct run and the flow rate through the duct run. With a fire whose size changes over time, the pressures will also change. Therefore, a pre-calculated system curve cannot be used. Instead an alternate method~\cite{Ralph:2} is used. This method relies on the observation that pressure and flow in an HVAC system have a quadratic relationship. Pressure drop in a duct is a function of the duct velocity squared. The following steps are taken at each time step to find the operation point for each fan:
 
 \begin{enumerate}
-	\item Locate all duct runs that contain fans. A duct run is any collection of directly interconnected HVAC components. See Fig.~\ref{HVAC_ductrun} for two examples. The figure shows two sets of two compartments connected by HVAC systems. In each set, each compartment has a supply and exhaust duct. On the left side, the two exhaust ducts connect to duct containing a fan which splits into a supply duct for each compartment. In this case there is one duct run. Every HVAC component is connected to every other HVAC component by a path containing only HVAC components. On the right side, the exhaust for each compartment is connected to its own fan which becomes the supply for the other compartment. In this case there are two duct runs. There is no direct path through only HVAC components that join the two ducts.
-	\item Determine the current pressures at each inlet and outlet for each duct run containing a fan.
-	\item For each fan, using just the duct run containing the fan, solve for the flow in the duct run given the current endpoint pressures.
-	\item For each fan, using just the duct run containing the fan, solve for the flow in the duct run given the current endpoint pressures and assuming the fan is applying its maximum pressure to the duct run.
-	\item For each fan, fit a quadratic between the two system points for that fan.
-	\item Use Newton's method to find the intersection between the user specified fan curve and the system curve. The fan pressure, $\Delta p_j$, for the time step is set to the pressure at the intersection. If there is no intersection, the fan pressure is set to either the maximum pressure for the fan or zero pressure. It is set to zero pressure if the system curve flow for zero fan pressure exceeds the maximum flow defined in the fan curve. It is set to maximum pressure if there is reverse flow in the system curve at maximum pressure.
+   \item Locate all duct runs that contain fans. A duct run is any collection of directly interconnected HVAC components. See Fig.~\ref{HVAC_ductrun} for two examples. The figure shows two sets of two compartments connected by HVAC systems. In each set, each compartment has a supply and exhaust duct. On the left side, the two exhaust ducts connect to duct containing a fan which splits into a supply duct for each compartment. In this case there is one duct run. Every HVAC component is connected to every other HVAC component by a path containing only HVAC components. On the right side, the exhaust for each compartment is connected to its own fan which becomes the supply for the other compartment. In this case there are two duct runs. There is no direct path through only HVAC components that join the two ducts.
+   \item Determine the current pressures at each inlet and outlet for each duct run containing a fan.
+   \item For each fan, using just the duct run containing the fan, solve for the flow in the duct run given the current endpoint pressures.
+   \item For each fan, using just the duct run containing the fan, solve for the flow in the duct run given the current endpoint pressures and assuming the fan is applying its maximum pressure to the duct run.
+   \item For each fan, fit a quadratic between the two system points for that fan.
+   \item Use Newton's method to find the intersection between the user specified fan curve and the system curve. The fan pressure, $\Delta p_j$, for the time step is set to the pressure at the intersection. If there is no intersection, the fan pressure is set to either the maximum pressure for the fan or zero pressure. It is set to zero pressure if the system curve flow for zero fan pressure exceeds the maximum flow defined in the fan curve. It is set to maximum pressure if there is reverse flow in the system curve at maximum pressure.
 \end{enumerate}
 
 \begin{figure}[ht!]
-	\begin{center}
-		\scalebox{0.8}{\includegraphics{FIGURES/HVAC_ductrun}}
-		\caption[Illustration of determining duct runs.]{\label{HVAC_ductrun} Illustration determining duct runs. Left side has a single duct run. Right side has two duct runs.}
-	\end{center}
+   \begin{center}
+      \scalebox{0.8}{\includegraphics{FIGURES/HVAC_ductrun}}
+      \caption[Illustration of determining duct runs.]{\label{HVAC_ductrun} Illustration determining duct runs. Left side has a single duct run. Right side has two duct runs.}
+   \end{center}
 \end{figure}
 
 \section{Duct Transport Delay}
@@ -205,4 +205,4 @@ \section{Duct Transport Delay}
 \be (\rho \, Y_{\alpha})_c^{n+1}=  \rho_j \, u_j^n - \frac{\Delta t}{\Delta x} \, (\rho \, u)_j^n left(Y_{\alpha ,c}^n - Y_{\alpha ,c-1}^n ) \ee  
 \be (\rho \, h)_c^{n+1}=  \rho_j \, u_j^n - \frac{\Delta t}{\Delta x} \, (\rho \, u)_j^n left(h_c^n - h_{c-1}^n ) \ee  
 
-Updating duct conditions begins at the upwind duct node and marches over the duct length to the downwind duct node. Cell temperatures are extracted from the cell enthalpy $h_c$. This solution is done over groups of ducts that form a duct network basis. A duct network is a set of ducts and nodes that are interconnected via a flow path through HVAC components. The time step $\Delta t$ is taken as the minimum of the current simulation time step and the minimum value of $\frac{\Delta x_j}{2 u_j}$ over all ducts in the netwok. In the summation terms in Eq.~(\ref{HVACmass}) and Eq.~(\ref{HVACenergy}), the mass flow and enthalpy flow into a node from a discretized duct are obtained by integrating those flows using the values in the last cell over each time step taken.
+Updating duct conditions begins at the upwind duct node and marches over the duct length to the downwind duct node. Cell temperatures are extracted from the cell enthalpy $h_c$. This solution is done over groups of ducts that form a duct network basis. A duct network is a set of ducts and nodes that are interconnected via a flow path through HVAC components. The time step $\Delta t$ is taken as the minimum of the current simulation time step and the minimum value of $\frac{\Delta x_j}{2 u_j}$ over all ducts in the netwok. In the summation terms in Eq.~(\ref{HVACmass}) and Eq.~(\ref{HVACenergy}), the mass flow and enthalpy flow into a node from a discretized duct are obtained by integrating those flows using the values in the last cell over each time step taken.