Page 1 of 17
`
`Samsung Exhibit 1021
`Samsung Electronics Co., Ltd. v. Daniel L. Flamm
`
`

`
`G.P. FORNEY
`
`the computational requirements of a general N—wall radi-
`ation model are too great for now to justify incorporating
`it into a zone fire model. By implementing the net radi-
`ation equation for particular N (two, four or ten walls),
`significant algorithmic speed increases were achieved by
`exploiting the structure of the simpler problems.
`
`THE PROBLEM
`
`N Wall Segment Radiation Exchange
`The N—wall radiation model described here consid-
`
`ers radiative heat transfer between wall segments, point-
`source fires and two gas layers. An enclosure or room
`is partitioned into N wall segments where each wall seg-
`ment emits, reflects and absorbs radiant energy. The
`interior of the enclosure is partitioned into two volume
`elements; an upper and a lower layer. The problem then
`is to determine the net radiation flux emitted by each wall
`segment and the energy absorbed by each layer given the
`temperature and emittance of each wall segment and the
`temperature and absorptance of the two gas layers.
`These calculations can be performed in conjunction
`with a zone fire model such as CFAST[12]. Typically,
`the solution (wall temperatures, gas layer temperatures
`etc) is known at a given time t. The solution is then
`advanced to a new time, t + At. The calculated ra-
`diation fluxes along with convective fluxes are used as
`a boundary condition for an associated heat conduction
`problem in order to calculate wall temperatures. Gas
`layer energy absorption due to radiation contributes to
`the energy source terms of the associated zone fire mod-
`eling differential equations. The time step, At, must be
`chosen sufficiently small so that changes in wall temper-
`atures are small over the duration of the time step.
`
`Modeling Assumptions The following assumptions
`are made in order to simplify the radiation heat exchange
`model and to make its calculation tractable.
`
`Each gas layer and each wall seg-
`iso-thermal
`ment is assumed to be at a uniform temper-
`ature. This assumption breaks down where
`wall segments meet.
`
`The wall segments and gas layers
`equilibrium
`are assumed to be in a quasi-steady state.
`The wall and gas layer temperatures are as-
`sumed to change slowly over the duration
`of the time step of the associated differen-
`tial equation.
`The fire is assumed to radiate uni-
`
`fire source
`
`formly in all directions from a single point
`giving off a fraction, X, of the total energy
`release rate to thermal radiation.
`
`radiators
`
`The radiation emitted from a wall sur-
`
`face, a gas and a fire is assumed to be
`diffuse and gray.
`In other words, the ra-
`diant fluxes emitted by these objects_are
`independent of the direction and the wave-
`length. They can depend on temperature,
`however. Since boththe emittances and the
`
`temperatures of wall segments are inputs to
`the radiation algorithms, it is assumed that
`the emittances are consistent with the cor-
`
`responding wall temperatures. Diffusivity
`implies that 6;, = a,\ for each wavelength /\
`while the gray gas/surface assumption im-
`plies that 6;. is constant for all wavelengths.
`These assumptions allow us to infer that the
`emittance, 6 and absorbance, Oz are related
`via 6 = a. A discussion of this assumption
`can be found in [13, p. 589-590].
`
`The wall surfaces are assumed to be opaque.
`When radiation encounters a surface it is
`either reflected or absorbed. It is not trans-
`
`mitted through the surface. Equation (1),
`found below, would have to be modified
`to account for the loss (or gain) of energy
`through semi-transparent surfaces.
`
`Rooms or compartments are assumed
`to be rectangular boxes. Each wall is ei-
`ther perpendicular or parallel to every other
`wall. Radiation transfer through vent open-
`ings, doors, etc is neglected.
`
`geometry
`
`The Net Radiation Equations Net radiation refers to
`the difference between outgoing and incoming radiation
`at a wall surface. As illustrated in Figure 1, incoming
`radiation consists of gray-body surface radiation emitted
`from all other surfaces, radiating point—source fires and
`emission from the two gas layers. Outgoing radiation
`consists of gray-body surface radiation and incoming
`radiation that is in turn reflected.
`Integrating the net
`radiation equation in Segel and Howell[7,, Chapter 17]
`over all wavelengths, we obtain an equation for the net
`radiation at each wall surface is given by
`
`Page 2 of 17
`
`

`
`Fire Science & Technology Vol.14 No.1 & No.2 1994
`
`Outgoing Radiation
`
`Incoming Radiation
`
`Grey body
`surface
`radiation
`
`Reflected
`radiant energy
`
`incoming radiation from
`other surfaces, fires and
`emitting, absorbing gas
`layers
`
`AkG€kT?}
`
`( 1 — €.k)qil:\
`
`qt‘
`
`Energy added to the k'th
`surface to maintain a
`
`AkAqk constant temperature
`
`Surface k
`
`R
`
`FlGURE 1: Input and Output Energy Distribution at the k‘th Wall Surface
`
`N
`A ” .
`1 — 6'
`eq I” ‘Z 5. ’A‘I"jFk—F:'—k =
`k
`jzl
`J
`N
`
`4
`4
`0'Tk - E0'Tj F],-_jTj__k —
`2:1
`
`ck
`
`(1)
`
`Other terms are defined in the nomenclature. Wall open-
`ings (vents, doors, etc) can be modeled by replacing Tj
`in equation (1) with T‘ where
`A
`A
`T4 = T; ’
`
`_ T:mb)v
`
`where Aq”k is the unknown radiative flux and ck/Ak,
`accounts for radiative flux striking the lc’th wall surface
`
`due to point source fires and gas layers and is given by
`
`q"§7‘:”if“ + q"§-1%“) .
`
`(2)
`
`A1, is the vent area and Tamb is the ambient temperature.
`Figure 2 presents a surface plot showing the effect of
`this equation.
`It plots the absolute temperature differ-
`ence, (T — T), versus relative vent area, A1,/Aj, and
`temperature, T. Note that over this broad range of tem-
`peratures and vent area fractions that the absolute change
`
`FIGURE 2: Surface plot of temperature difference (original temperature and equivalent temperature ac-
`counting for vents and doors) as a function of fractional vent area and temperature showing the effect of
`vent openings on computing radiation heat transfer.
`
`Page 3 of 17
`
`

`
`G.P. FORNEY
`
`in T required to account for vent openings is small. For
`large temperatures T or large vent fractions A,, /A,- then
`vent openings need to be taken into account.
`Subsequent sections discuss the computation of terms
`in (1) and (2).
`
`In general,
`Heat Flux Striking a Wall Segment
`every possible path between two wall segments should
`be considered in order to compute the total radiant heat
`transfer between these segments. This is not practical
`in a zone fire model due to the excessive computational
`costs. The approach taken here is to model this heat
`transfer using just one path: For a typical path there are
`four cases to consider. A path from wall segmentj to k
`can start in either the upper or lower layer and finish in
`either the upper or lower layer. A fraction, 04 = 1 — T, of
`the energy encountering a layer is absorbed. The rest, 7‘,
`passes through unimpeded. Table 1 gives formulas for
`the heat flux striking the k’th wall segment due to point
`source fires and heat emitting gas layers.
`
`Heat Flux Striking a Wall Segment Due to a Point
`Source Fire If the gas layers are transparent then the
`flux striking the k’th surface due to the f’th fire is
`
`qnfire = Xq§fotT:1Wf—k
`f_k
`47rAk
`
`where the total energy release rate ofvthe fire is qtfjl, X
`is the fraction of this energy that contributes to radiation
`and wf_k/(47rA;,.) is the fraction of the radiant energy
`leaving the f ’th fire that is intercepted by the k’th wall
`segment, ie a configuration factor. On the other hand,
`if the gas layers are not transparent then there are four
`cases to consider. The fire can be in the upper or lower
`layer and the surface can be in the upper or lower layer.
`Figure 3 shows how radiation from a fire is absorbed by
`each layer when the fire is in the lower layer and the
`surface k is in the upper layer. The other three cases are
`handled similarly. These four cases are summarized in
`the first column of Table 1. This column give formulas
`
`Radiative Heat Flux Striking the k’th Rectangular Wall Segment
`
`Fire
`
`//fire
`9 f—k
`
`Gas Layer
`
`9 ;—k
`
`[
`
`q,,U,gas
`j—k
`
`in upper
`
`from upperto lower
`
`from lowerto upper
`
`inlower
`
`><<z{J$".fM—r=
`47I'A)c
`
`X9{§eTifzWr—k
`41rAk
`
`Tf_k
`U
`L
`Tf_k Tf_ [C
`U Xqfm W4
`TfL__k Tf_k
`4fl_Ak
`L Xqjire wf_k
`Tf_k
`47rAk
`
`0
`
`L
`Fk_jUaj_kT2
`
`F1¢_jUa§J_kTg
`
`Fk__,'0'0¢]U_kTg7'JL_k
`_
`U
`4
`Fk_,o'0zj_kTU J
`
`as
`, deposited in
`’
`
`at
`
`L “ Li’
`f’
`qrl-if Tr—k°‘ik
`
`fire
`
`deposited in
`lower layer
`
`FIGURE 3: Schematic illustrating energy deposited into the lower layer, q”;-i_r;:aJ’:_,c, deposited into the
`upper layer, q”J{'f§cv}’_,grfL_k, and arriving at the Ic’th wall surface, q”j“_’§r}3_,crJE’_k due to the f’th fire.
`
`Page 4 of 17
`
`

`
`for the flux striking a surface k due to a point source fire.
`
`Heat Flux Striking a Wall Segment Due to an Emit-
`ting Gas Layer The energy emitted by the i’th layer
`(i=upper, or i=lower) along the j-k’th path is
`
`uzlyas
`j—k = Ct;-_ k0'Ti4
`
`‘I
`
`where a;._ k = 1 — 'r_7i_ k. The emittance of the gas in this
`equation is the same as the absorptance due to the gray
`gas assumption. Again four cases must be considered to
`calculate the flux striking a wall segment. The last two
`columns of Table 1 gives formulas for radiation striking
`the k’th wall segment due to lower/upper gas layer heat
`emissions for each possible path.
`
`Gas Absorbance The energy absorbed by the gas
`layers may be due to radiating wall segments, emission
`from other gas layers and radiation from fires. Tables
`2 and 3 summarize the formulas used to compute gas
`layer energy gain/loss due to these phenomena. Again,
`there are four cases to consider, since an arbitrary path
`may start in either the lower or the upper layer and end
`in the lower or upper layer. Figure 4 illustrates the heat
`
`Fire Science & Technology Vol.14 No.1 & No.2 19.94
`
`absorbed by the gas layers due to surface rectangle emis-
`sion where the “from” wall segment is in the upper layer
`and the “to” wall segment is in the lower layer. The other
`three cases are handled similarly.
`
`Configuration Factor Properties A configuration
`factor, F‘,-_,,., is the fraction of radiant energy leaving a
`surface j that is intercepted by a surface k. The following
`symmetry and additive properties (see [7, Chapter 7]) are
`used later to reduce the number of computations in the
`four-wall and ten-wall model
`
`= AkFk_j
`
`Fi—j + Fi—k
`AiFi—k + Aj-Fj—k
`
`1, 3': 1,...,N
`
`where i {B j denotes the union of two wall surfaces i and
`j. If four wall segments are configured as illustrated in
`Figure 5 then it can be shown that
`
`A1F1_4 = A2F2_3 .
`
`(7)
`
`Table 2: Radiant Heat Absorbed by the Upper Layer
`
`Path through the
`Gas
`
`Due to Heat Emitting
`Wall Surface
`
`q;’:‘,. = Air,-_,. (ar;-*-
`1*‘€1'
`u
`5]‘ Ag J‘)
`
`to Gas Layer
`Due
`Emission
`4
`'
`//z'9a.s__
`4 j’—k - ;'—k¢7T¢
`i,ga.s
`4_,__;¢
`_
`V
`/11 gas
`4 f—Ic AJFJ—k
`
`Due to Point Source
`Fire
`
`ufire _ >(‘1{”!e[“-’f—k
`q f-k _
`47rAk
`
`from the upper to
`either the lower or
`upper layer
`
`q;-’f‘kozJU_k
`
`from the lower to
`the upper layer
`
`from the lower to
`the lower layer
`
`Page 5 of 17
`
`

`
`G.P. FORNEY
`
`Table 3: Radiant Heat Absorbed by the Lower layer
`
`Path through the
`Gas
`
`Due to Heat Emitting
`Wall Surface
`4
`out _
`_
`_
`qj-_k — A,F,_—k (rrTj —
`1-é‘
`II
`___1Aq J.)
`6,’
`
`to Gas Layer Due to Point Source
`_Due
`Emission
`'
`4
`rmgas _ i
`q J-_k — oz]-_kzrT,v
`i,ga.s
`
`qjflc//2,905
`9 j—k AJ'FJ—k
`
`from the lower to
`either the lower or
`
`upper layer
`
`from the upper to
`the lower layer
`
`from the upper to
`the upper layer
`
`/deposited in qgutau
`upper layer
`J
`
`deposited in
`iowerlayer
`
`L
`out U
`qj Tj-kaj-k
`
`arriving at
`rectangle
`
`out U1!-
`qi Ti-k J'k
`
`FIGURE 4: Schematic illustrating energy deposited into the upper layer, q”}""a§’_ deposited in lower iayer,
`q”;?"‘af_kr]F’_k, and arriving at the 1c’th wall surface, q”}’“’rjL_krjU_k due to the j’th wall surface.
`
`Page 6 of 17
`
`

`
`Fire Science & Technology Vol.14 No.1 & No.2 1994
`
`FIGURE 5: Configuration factor symmetries used to reduce number of direct configuration factor calcula-
`tions.
`
`The configuration factor between two rectangles with a
`common edge of length l lying on perpendicular planes
`can be found in [7, p. 825] to be
`’
`
`¢perp(h: la 7“) :
`
`1
`
`-11
`
`—Ll_
`
`'l' HTED H
`'7?/V— {WTEH ‘VT,’
`1
`H2 + W2 Tan-1 ——— +
`\/H2 + W2
`
`4 °g
`
`1+H2+W2
`
`(1+W2)(1+H2)[ w2(1+H2+W2) V“
`11
`H2(1 + H2 + W)
`H2
`"l(1+W2)(H2+W2)l
`(9)
`
`(
`
`1 + W2)(H2 + W2
`
`)
`
`where H = h/l and W = w/ l and the two rectangles
`have dimensionsl X h andl X 11). Similarly, the config-
`uration factor between two rectangles lying on parallel
`planes, at distance c apart can be found in [7, p. 824] to
`be
`
`$110-T'(a’=bvc) =
`
`(10) are expensive to compute due to the complicated
`expressions involving log and Tan” functions. This
`portion of the work is reduced in RAD4 by noting that
`only 2 configuration factor calculations involving equa-
`tion (8) are required rather than 4 x 4 = 16. The other 14
`configuration factors are obtained using algebraic rela-
`tionships. For the RADIO case only eight configuration
`factor calculations need be calculated using equations
`(8) and (10). Again, the other 92 can be obtained using
`algebraic formulas. These formulas are detailed in [14].
`
`Solid Angles The fraction of a radiating point-source
`fire striking a wall surface is determined using solid an-
`gles. For a wall surface with sides of length :1: and y that
`lies in a plane a distance 7* from the fire the solid angle is
`
`w(w, y) =
`
`-1
`
`-
`
`{S111
`
`39
`
`(A __._W)+
`
`2
`
`(1+X2)(1+Y2)
`
`1+X2+Y2 +
`7rXY {log
`Xx/1+Y2Tan_1 X +
`1+1/2
`_
`Y
`Yx/1+X2Tan IHX2 —
`X Tan‘1X — Yran-1Y}
`
`The solid angle w(:c,y) is symmetric in ac and y. Solid
`angles are also additive, so that the solid angle of an
`arbitrary rectangle can be computed using (1 1) and
`
`(10)
`
`where X = a,/c and Y = b/c and the two rectangles
`each have dimension a x b.
`
`For a room with N wall segments, N X N = N2 con-
`figuration factors must be calculated. Equations (8) and
`
`Page 7 of 17
`
`

`
`G.P. FORNEY
`
`w($17 m27y1> 3/2)
`
`Sg11($27J2)w(l$2lv l?J2|) —
`S8H($21I1lW(l$2l»l?J1l) —
`
`S€;I1(=U1?12)W(|331|»|Z/2|)+
`
`SgI1(f61?J1)W(|~”v1lsl?J1l)
`
`ssn(z)
`
`1
`
`ifcc 2 0
`
`-1 ifa: < 0.
`
`Transmission Factors A transmission factor, 7', is
`the fraction of energy passing through a gas unimpeded.
`The transmittance of a gas depends on the absorption
`coefficient of the gas and the length of the path through
`the gas. A simple relationship for 7' can be determined
`by assuming that the absorptance of the gas (a local
`phenomena) is uniform throughout the gas layer. This
`factor, a decaying exponential, is given by
`
`1-(L) = e-"L
`
`where a, is the absorptance of the gas per unit length and
`L is the path length. The gas absorptance is not calcu-
`lated by the radiation exchange algorithms presented in
`this paper. Modak in [15] gives an algorithm for cal-
`culating gas absorptance from such information as soot
`concentration, partial pressures of C0, C02 etc. The
`emittance of the gas is the same as its absorption due
`to the gray gas assumption. The transmission factor, '7',
`in the above equation is defined for one specific path
`through a gas. We are, however considering radiation
`exchange between a pair of finite area rectangles where
`many paths of different lengths occur. Siegel and Howell
`define an average transmission factor [7, p. 603] con-
`sidering all possible paths between two surfaces through
`the gas. This form of T is defined to be
`
`fi=[[J4‘ dAjdAk/(AjFj—k)'
`
`7rL2
`
`A
`
`For a hemisphere of radius L, this integral reduces
`to e‘“L. This integral can be estimated by finding a
`characteristic path with length I, (a mean—beam length),
`such that
`
`For an optically thin gas, the mean beam length for
`radiation from an entire gas volume to the bounding
`surface (see [7]) is 4V/A where V is the volume of the
`gas and A is the surface area of the region bounding the
`gas. This formula is not applicable in these calculations
`
`since the radiation transfer of the gas to a wall surface is
`computed by splitting the entire gas volume into several
`pieces.
`For the ten-wall model, the characteristic path is
`taken to be between the centers of two rectangles. This
`length is an underestimate of Z. This approximation
`breaks down when the two wall segments are close to-
`gether or one of the wall segments is a complex shape
`(such as the union of four upper walls). The two and
`four-wall model estimates of Z are based upon an av-
`erage distance between the rectangles that make up the
`wall segments.
`For a given path between surface j and surface k we
`need to calculate the path length, LU, through the upper
`layer and the path length, LL, through the lower layer.
`Transmission factors for the upper and lower layers are
`then defined to be
`
`T]U_k
`L
`7'J_k
`
`9
`
`_ 1.1
`C LJTECLU
`L
`—Lj_,caL '
`
`C
`
`The energy fraction that passes through both layers is
`then
`
`L
`U
`TJ‘_k : Tj_kTj_k .
`
`The energy fraction absorbed by a layer is just the frac-
`tion that doesn’t pass through a layer or
`U
`_.
`1 — e‘LiJ—’°“”
`1—Tj_k—
`"“
`L _ _
`_
`k _ 1-—T]»_k—1
`
`L,’
`
`Q
`
`VJ...
`P3-a
`
`7
`
`—L1L»‘_,caL '
`
`e
`
`Two, Four, Ten Wall Segment Radiation
`Exchange
`Equation (1) for computing radiation exchange were
`specified in terms of general wall segments. This section
`discusses the radiation exchange computation in terms
`of a two-wall, four-wall and ten-wall model.
`
`Two-Wall Configuration Factors The two—wa11 model
`combines the ceiling and four upper walls into one wall
`segment and the four lower walls and the floor into the
`second wall segment. The configuration factors for these
`two surfaces are derived by Quintiere in [16, Appendix]
`and are
`
`Page 8 of 17
`
`

`
`Fire Science & Technology V0l.14 No.1 & No.2 1994
`
`Appendix]. The setup for the following derivation is
`given in Figure 6. We wish to determine the configura-
`
`Surface 1, The ceiling
`Surface 2, The
`"‘>
`upper walls
`Surface d, The layer interface
`
`Surface 3, The
`lower walls
`
`Surface 4, The floor
`
`FIGURE 6: Schematic used to derive four wall configuration
`factor formulas.
`
`tion factors
`
`F,-_j fori,j = 1,...,4.v
`
`The 16 configuration factors can be determined in terms
`of F1_4, F1_d and F4_,1. F1_4 does not change during
`a simulation since its value depends only on the height
`of the room and the area of the floor. Therefore, F1_4
`only needs to be computed once. Configuration factors,
`F1_d and F4_d depend on the layer interface height so
`need to be calculated each time the radiation exchange
`is to be calculated. Configuration factors F1_4, F1_d
`and 17.1.4 are determined using equation (10). Since
`A1 = A4 it follows that F4_1 = F1_4. The other
`14 configuration factors can be calculated using simple
`algebraic formulas.
`Since the floor and the ceiling is assumed to be a flat
`rectangular surface it follows that
`
`F1—1 = F4—4 = 0-
`
`Using the fact that configuration factors in an enclosure
`sum to l and that due to symmetry F2_1 = F2_d, it
`follows that
`
`F1—2 + F1—d
`
`1,
`
`(13)
`
`F2—1 + F2—2 + F2—d
`
`2F2—1 + F2—2 = 1 (14)
`
`Equations (13) and (14) can be solved for F1_2 and
`F2_2 respectively to obtain
`
`where A1, AD and A2 are the areas of the extended
`ceiling, layer interface and extended floor respectively.
`These configuration factors are used in the original two-
`wall radiation model in BRI [3, 4] and in CFAST [5].
`The tw0—wal1 model, RAD2, interacts with a four-
`wall heat conduction model in CFAST. The ceiling and
`upper wall temperatures may be different, so the ques-
`tion of how to represent the extended ceiling temperature
`arises. RAD2 chooses an extended ceiling temperature
`that results in the same energy contribution to the enclo-
`sure that a four-wall radiation algorithm would predict.
`The energy added to the room due to the ceiling and
`upper wall temperatures of T1,, and T11, is
`
`0 (A1a51aT14g + A1b51bT14b)
`
`-
`
`where the subscripts la and lb represents the ceiling and
`upper wall. We want to choose an effective or average
`temperature, T1109, and emittance, saw for the extended
`ceiling that matches this energy contribution, or
`energy from ceiling
`energy from extended ceiling
`/---mm-—"\
`?"——\
`
`0'(Ala '1” A1b)Ea'ugT:1,g
`
`=
`
`0'A1a.51aT14g, +
`energy from upper wall
`,.._/y?‘
`
`UA1b€1bT;1b
`
`Choosing an average emittance computed using an av-
`erage of 61,1 and 611, weighted by wall segment areas
`gives
`
`5avg :
`
`5610, + (1 - fl)€1{,.
`
`where ,6 = A1,,/(A1,, + A15). Equation (12) can now
`be solved for Tavg using this value of Envy to obtain
`
`Tang : 4 ’YT14a + (1 — ’Y)T14b
`
`where’)! = A1,,e1,,/(A1,,e1,, +A1(,e11,). A similar proce-
`dure could be used to compute an effective temperature
`and emittance for the extended floor.
`
`Four-Wall Configuration Factors. The configura-
`tion factors for four-wall radiation exchange are derived
`similarly to Quintiere’s derivation for two walls in [16,
`
`Page 9 of 17
`
`

`
`Similarly,
`
`Using the above configuration factors and equation
`(6) it follows that
`
`Ten-Wall Configuration Factors To handle the more
`general radiation exchange case, a room is split into ten
`surfaces as illustrated in Figure 7. These surfaces are the
`
`4 = upper back
`
`FIGURE 7: Schematic used for ten wall configuration factor
`formulas.
`
`ceiling, four upper walls, four lower walls and the floor.
`The radiation exchange is computed between these ten
`surfaces and the intervening gas layer(s). In general, 100
`configuration factors, F]-_k and 100 transmission factors
`T,~_k need to be determined each time this algorithm is
`invoked. Although there are 100 configuration factors
`for this room, only eight have to be calculated directly
`using equations (8) and (10). The other 92 can be com-
`
`puted in terms of simple algebraic relationships using the
`properties outlined in equations (3) to (7). This reduc-
`tion in required configuration factor calculations is due
`to the fact that the rectangle pairs 2 and 4, 3 and 5, 6 and
`8 and 7 and 9 each have equal areas. The details of these
`calculations are documented in [14].
`
`SOLVING THE NET RADIATION EQUA-
`TIONS
`
`Solving The Net Radiation Equations Ef-
`ficiently
`The net radiation equation (1) is not diagonally dom-
`inant. Therefore, iterative methods should not be used
`to solve this equation unless it is suitably transformed.
`This can be done by substituting, Aq”k = e;,A§i’ into
`equation (1) to obtain
`N
`
`4‘-iii’ - Z (1 ” 6:’) Ari3’Fk—m-re =
`3:1
`
`UT: —
`
`(15)
`
`There are two reasons for solving equation (15) instead
`of (1). First, since ck does not occur in the denominator,
`radiation exchange can be calculated when a wall seg-
`ment emittance is zero. Second and more importantly,
`the matrix corresponding to the linear system of equa-
`tions in (15) is diagonally dominant. When the number
`of wall segments is large and the wall segments have
`emittances close to one which often occurs in typical fire
`scenarios, the time required to solve this modified linear
`system can be significantly reduced due to this diagonal
`dominance by using iterative methods.
`To see this, re-write equation (1) into matrix form
`to obtain
`
`AAq” = BE — c
`
`(16)
`
`where the k, j’th components of the N x N matrices A
`and B are
`
`aka‘
`
`bkvjy
`
`6kg’ — Fk_jTj-k
`
`and the lc’th component of the vectors c’’ and E arfi
`
`Page 10 of 17
`
`

`
`Fire Science & Technology Vol. 14 No.1 & No.2 1994
`
`Substituting (21) into (22) we get the following require-
`ment for A to be diagonally dominant
`N
`
`1- Fk__k (1 —-£k)Tk._k > Z Fk_j (1 — Ej) 73-};
`2:=1
`J 76 k
`
`or equivalently
`
`N
`
`1 > ZFk_j (1 — F.j)7'j_k .
`i=1
`~
`
`The matrix A is then diagonally dominant since 1 >
`(1 — E,-)7‘,-_;, and
`
`N
`
`N
`
`1: Z Fk_J' > Z Fk_j(1—Ej)7'j_k.
`J'=1
`i=1
`
`Iterative techniques for solving linear systems such
`as Gauss-Seidel are guaranteed to converge for diago-
`nally dominant matrices [17, p. 542]. They also can be
`much more efficient than direct methods such as Gaus-
`
`sian elimination. The convergence speed depends on
`how small the right hand side of the above inequality is
`compared to 1. Physically, if the surfaces being modeled
`are approximate black bodies (6 close 1) or the gas lay-
`ers are thick ('1' close to 0) then iterative techniques for
`solving the net radiation equations will converge rapidly.
`Typical emittances for materials used in fire simulations
`range from 5 = .85 to .95. For the limiting case when
`the wall materials are black bodies then the matrix A is
`a diagonal matrix and iterative methods will converge in
`one iteration.
`
`The advantage of using an iterative method over a
`direct method for computing radiation exchange between
`approximate black bodies increases as the number of
`wall segments increases. The cost of solving the linear
`system directly is proportional %N3 while the cost of
`using iterative techniques is proportional to kN2 where
`k is the number of iterations and N is the number of
`
`wall segments. Using Gauss-Seidel iterative methods, it
`has been found that convergence is achieved after two to
`three iterations for emittances around .9. The break-even
`
`point between iterative and direct methods for matrices
`of size 10 is about 6 or 7. The linear system for RAD2
`and RAD4 is of size 2 X 2 and 4 X 4 respectively. Iterative
`methods are not faster for problems this small. RADIO
`and problems with more wall segments can use iterative
`methods to decrease the time required to solve the linear
`system without sacrificing accuracy.
`
`(«W + ms‘) +
`
`N Z
`
`J'=1
`
`‘
`Nfire
`Z q"§‘.'":.
`f=1
`
`E1, = aT;§ ,
`
`<19)
`
`(20)
`
`and 6;” is the Kronecker delta function, F;,_,- is the con-
`figuration factor from the k’th to the j’th wall segment,
`ej is the emittance of the j‘th wall segment, T,-_k is a
`fraction ranging from O to 1 indicating the amount of
`radiation that is transmitted through a gas. Also, q”§];”,:”
`nluaas
`and q J-_k
`are radiation contributions due to the gas
`layers and q’ are radiation terms due to the f’_th fire.
`The matrix A can be transformed into a diagonally dom-
`inant matrix using the following scaling matrix,
`
`0
`
`0
`
`0
`
`0
`EN
`
`17=
`
`0
`0
`
`where 5;, is the emittance of the k’th wall segment. De-
`fine the scaled matrix A by post—multiplying A by D and
`pre-multiplying Aq” by D“1 to obtain
`
`A V: AD-
`
`Aén = D—lAqII-
`
`Equation (16) then reduces to
`
`Ana" = AAq” = BE ~ c.
`
`Once the solution A4” is found we may recover the
`solution, Aq" , to the original problem by using Aq” =
`DAQ”.
`The matrix /l is diagonally dominant which is now
`shown. Using the definition of am in equation (17) the
`kj’th element of A is
`
`LAl.;V~,j = agjej = dkd‘ — Fk_jTj_k(1 — 6]‘).
`
`(21)
`
`A matrix is diagonally dominant if for each row the
`absolute value of the diagonal element is greater than the
`sum of the absolute values of the off diagonal elements
`or equivalently
`
`N
`
`|€w|> Z lineal
`v:=1
`J 95 k
`
`(22)
`
`Page 11 of 17
`
`

`
`AM" = BE — c
`
`(23)
`
`for Ag” net radiation leaving each surface
`where Aq”k = Atj;c’e;,.. If the emittances are
`sufficiently close to 1 then use iteration to
`solve equation (23) otherwise use Gaussian
`elimination.
`
`. Calculate the energy absorbed by the upper
`and lower gas layers due to the total energy ,
`q,‘;”‘ leaving each rectangle k.
`
`COMPUTATIONAL RESULTS
`
`Checks
`
`Several simple checks can be made to verify a por-
`tion of the radiation calculation. First, no heat transfer
`occurs when all wall segments and both gas layers are at
`the same temperature. Therefore, the net radiation flux,
`Aq”k given off by each surface and the energy absorbed
`by the gas should be zero under these uniform temper-
`ature conditions. Second, when there is no fire, the net
`energy absorbed by the gas must be the same as the net
`energy given off by the wall segments or equivalently
`
`qlower + qupper = energy absorbed by interior gases
`
`AI; AW}.-
`
`N 2
`
`k=1
`
`When the layers are transparent then the above equa-
`tion sums to zero even though the individual wall fluxes
`Aq”,, will in general be non-zero. The gas absorbance
`terms, q;m,,e,. and qupper, are computed by RAD2, RAD4
`and RADIO. These values can be summed to verify that
`above equation is satisfied.
`
`Timings
`Configuration calculations are one of the major bot-
`tle necks in the radiation exchange calculation. Tech-
`niques to reduce the number of these calculations will
`improve the algorithms efficiency. A preliminary ver-
`sion of RAD4 was based on RADIO, a ten wall segment
`model.
`It computed 45 configuration factors directly.
`Subsequent versions of RAD4 computed eight and then
`two configuration factors directly. Table 4 summarizes
`the time required by these three different versions of
`RAD4. This table shows that the first version of RAD4
`
`used approximately 70% of the time setting up the linear
`system and 30% solving it. Reducing the setup over-
`head by computing fewer configurations factors reduced
`
`The radiation exchange equations can be solved an-
`alytically using Cramer’s rule for the two wall segment
`case. This is how the radiation exchange equations were
`derived in [3] and [5]. Cramer’s rule is not a good numer-
`ical technique to use for the solution of linear systems
`(even for 2 X 2 systems) due to cancellation error that
`can be introduced when solving equations that are ill-
`conditioned.
`
`Algorithm for Calculating Four Wall Ra-
`diation Exchange
`The strategy for computing the radiation exchange
`between four wall segments is outlined below. RAD4
`performs these steps directly or calls subroutines that
`performs them. RAD2 and RADIO follow the same
`logic.
`
`Input
`
`Temperatures Ceiling, Upper Wall, Lower
`Wall, Floor
`Emissivities Ceiling, Wall, Floor
`
`Absorptivities Upper, Lower Layer
`Fire Size, Location, number
`
`Room room number, dimensions, layer height
`
`u
`Output
`Flux ceiling, upper wall, lower wall, floor
`
`Energy Absorption Rate upperlayer, lower
`layer
`
`Steps 1. Calculate configuration factors, solid an-
`gles.
`
`2. Determine the effective length between each
`pair of wall segments. From these lengths
`and inputted layer absorptivities calculate trans-
`mission factors for surface j to surface k
`
`. Calculate transmission factors and gas layer
`absorptions for each fire f to surface k.
`
`. Calculate the energy absorbed by each gas
`layer due to upper/lower gas layer emission
`and due to the fire(s) following Tables 2 and
`3.
`
`. Set up the linear algebra
`
`(a) Define vector E using equation (20)
`(b) Define matrix fl using equation (21)
`(c) Define matrix B using equation (18)
`(d) Define vector c, using equation (19) and
`Table 1.
`
`. Solve the linear system
`
`Page 12 of 17
`
`

`
`Fire Science & Technology Vol.14 No.1 & No.2 1994
`
`Table 4: Four Wall Radiation Algorithm Timings
`
`Total Time (5)
`
`Linear Solve Time (s)
`
`45 configuration factors and direct lin-
`ear solve
`
`Eight configuration factors and itera-
`tive linear solve
`
`Two configuration factors and direct
`linear solve
`
`0.06
`
`J2“
`
`
`
`J
`
`the computation time required by a factor of 17. Why
`quibble over the timings of a subroutine that only took .2
`seconds to execute to begin with? The relative impact of
`RAD4 on the zone fire model CFAST was measured by
`comparing the time required to execute DSOURC with
`and without RAD4. DSOURC is the subroutine CFAST
`
`uses to calculate the right hand side of the modeling dif-
`ferential equations. Most of the work is performed by
`this routine or routines that DSOURC calls. For a six
`room CFAST test case DSOURC took about .06 seconds.-
`
`All times were measured on a Compaq 386/20 Deskpro.
`This computer has a 20mhz clock and,uses a floating
`point accelerator (math co—processor). The actual times
`will be different on different computers. But the relative
`times and hence the conclusions should be the same. A
`
`routine that takes .2 seconds per room used in each room
`will result in a 21-fold increase in computer time since
`the ratio of the time in DSOURC with RAD4 to the time
`
`in DSOURC without RAD4 is (.2 * 6 + .06)/.06 w 21.
`Even the fastest version of RAD4 will cause an in-
`crease of execution time of 2.25 if it is used in each
`room.
`
`Comparisons of RAD2 with RAD4
`The predictions of a two wall radiation exchange
`model, RAD2, are compared with a four-wall model,
`RAD4. One of the assumptions made about N wall
`segment radiation models is that the temperature distri-
`bution of each wall segment is approximately uniform.
`The zone fire model CFAST models the temperature of
`four wall segments independently. Therefore, a two wall
`model for radiation exchange can break down when the
`temperatures of the ceiling and upper walls differ sig-
`nificantly. This could happen in CFAST, for example,
`when different wall materials are used to model the ceil-
`
`ing, walls and floor. To demonstrate this consider the
`following example.
`To simplify the comparison between the two and
`
`four wall segment models, assume that the wall seg-
`ments are black bodies (the emissivities of all wall seg-
`ments are one) and the gas layers are transparent (the gas
`absorptivities are zero) . This is legitimate since for this
`example we are only interested in comparing how a two
`wall and a four wall radiation algorithm transfers heat to
`wall segments. Let the room dimensions be 4 X 4 x 4
`[m], the temperature of the floor and the lower and up-
`per walls be 300 [K]. Let the ceiling temperature vary
`from 300 [K] to 600 [K]. Figure