.MCAD 304020000 1 79 172 0
.CMD PLOTFORMAT
0 0 1 1 1 0 0 1 1 
0 0 1 1 1 0 0 1 1 
0 1 0 0 1 1 NO-TRACE-STRING
0 2 1 0 1 1 NO-TRACE-STRING
0 3 2 0 1 1 NO-TRACE-STRING
0 4 3 0 1 1 NO-TRACE-STRING
0 1 4 0 1 1 NO-TRACE-STRING
0 2 5 0 1 1 NO-TRACE-STRING
0 3 6 0 1 1 NO-TRACE-STRING
0 4 0 0 1 1 NO-TRACE-STRING
0 1 1 0 1 1 NO-TRACE-STRING
0 2 2 0 1 1 NO-TRACE-STRING
0 3 3 0 1 1 NO-TRACE-STRING
0 4 4 0 1 1 NO-TRACE-STRING
0 1 5 0 1 1 NO-TRACE-STRING
0 2 6 0 1 1 NO-TRACE-STRING
0 3 0 0 1 1 NO-TRACE-STRING
0 4 1 0 1 1 NO-TRACE-STRING
0 1 1 21 15 0 0 3 
.CMD FORMAT  rd=d ct=10 im=i et=3 zt=15 pr=3 mass length time charge temperature tr=0 vm=0
.CMD SET ORIGIN 0
.CMD SET TOL 0.001000000000000
.CMD SET PRNCOLWIDTH 8
.CMD SET PRNPRECISION 4
.CMD PRINT_SETUP 1.000000 1.000000 1.000000 1.000000 0
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.CMD DEFAULT_TEXT_PARPROPS 0 0 0
.CMD DEFINE_FONTSTYLE_NAME fontID=0 name=Variables
.CMD DEFINE_FONTSTYLE_NAME fontID=1 name=Constants
.CMD DEFINE_FONTSTYLE_NAME fontID=2 name=Text
.CMD DEFINE_FONTSTYLE_NAME fontID=4 name=User^1
.CMD DEFINE_FONTSTYLE_NAME fontID=5 name=User^2
.CMD DEFINE_FONTSTYLE_NAME fontID=6 name=User^3
.CMD DEFINE_FONTSTYLE_NAME fontID=7 name=User^4
.CMD DEFINE_FONTSTYLE_NAME fontID=8 name=User^5
.CMD DEFINE_FONTSTYLE_NAME fontID=9 name=User^6
.CMD DEFINE_FONTSTYLE_NAME fontID=10 name=User^7
.CMD DEFINE_FONTSTYLE fontID=0 family=Times^New^Roman points=10 bold=0 italic=0 underline=0 colrid=-1
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.CMD UNITS U=1
.CMD DIMENSIONS_ANALYSIS 0 0
.CMD COLORTAB_ENTRY 0 0 0
.CMD COLORTAB_ENTRY 128 0 0
.CMD COLORTAB_ENTRY 0 128 0
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.TXT 3 23 1 0 0
Cg a42.000000,42.000000,17
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}{\f1\fcharset0\fnil Times New Roman;}}\plain\cf1
\fs20 \pard {\f1\fs36\b HWK}{\f1\fs36\b  }{\f1\fs36\b #5, }{\f1\fs36\b 
Problem 1}}
.EQN 12 -22 3 0 0
{0:xes}NAME:({8,1}ö1ö0.78ö0.581ö0.411ö0.258ö0.13ö0.017ö0)
.EQN 0 12 4 0 0
{0:yes}NAME:({8,1}ö1ö0.9ö0.777ö0.632ö0.456ö0.261ö0.039ö0)
.EQN 14 -12 7 0 0
{0:SplineFit}NAME({0:xArray}NAME,{0:yArray}NAME,{0:x}NAME):{0:interp}NAME({0:pspline}NAME({0:xArray}NAME,{0:yArray}NAME),{0:xArray}NAME,{0:yArray}NAME,{0:x}NAME)
.EQN 4 0 5 0 0
{0:yeq}NAME({0:x}NAME):{0:SplineFit}NAME({0:xes}NAME,{0:yes}NAME,{0:x}NAME)
.EQN 4 0 6 0 0
{0:xeq}NAME({0:y}NAME):{0:SplineFit}NAME({0:yes}NAME,{0:xes}NAME,{0:y}NAME)
.EQN 4 0 8 0 0
{0:ie}NAME:0;{0:last}NAME({0:xes}NAME)
.EQN 0 14 9 0 0
{0:zi}NAME:0,.02;1
.EQN 5 -14 10 0 0
1&&(_n_u_l_l_&_n_u_l_l_)&({0:yes}NAME)[({0:ie}NAME),{0:yeq}NAME({0:zi}NAME),{0:zi}NAME,{0:zi}NAME@1&&(_n_u_l_l_&_n_u_l_l_)&({0:xes}NAME)[({0:ie}NAME),{0:zi}NAME,{0:zi}NAME,{0:xeq}NAME({0:zi}NAME)
0 0 1 1 1 0 0 1 1 
0 0 1 1 1 0 0 1 1 
3 0 0 0 1 1 tabulated data
0 1 0 0 1 1 y(x) interpolation
0 1 6 0 1 1 y=x
2 0 1 0 1 1 x(y) interpolation
0 1 4 0 1 1 NO-TRACE-STRING
0 2 5 0 1 1 NO-TRACE-STRING
0 3 6 0 1 1 NO-TRACE-STRING
0 4 0 0 1 1 NO-TRACE-STRING
0 1 1 0 1 1 NO-TRACE-STRING
0 2 2 0 1 1 NO-TRACE-STRING
0 3 3 0 1 1 NO-TRACE-STRING
0 4 4 0 1 1 NO-TRACE-STRING
0 1 5 0 1 1 NO-TRACE-STRING
0 2 6 0 1 1 NO-TRACE-STRING
0 3 0 0 1 1 NO-TRACE-STRING
0 4 1 0 1 1 NO-TRACE-STRING
0 0 1 37 36 10 0 3 
.TXT 9 49 11 0 0
Cg a24.000000,24.000000,73
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard This plot shows how good 
or bad yeq(x) and xeq(y) fit the tabular data.  }
.TXT 56 -49 2 0 0
Cg a73.000000,73.000000,7
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\fs36\ul Part a}{\fs36
\ul )}}
.EQN 3 0 103 0 0
{   23    52 34460 0 8 191 418}{66}
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!!!!(_#g!!+u!!!'!1#e5i!!!#+x+%!_"!.]+U1)!E#jH_'1"!<9,w1-!E6pIQ'7""!#-y14!EDLJs':""6m.+19!EDLKE':""9S-y1;!EAfKU'9""<9-i1;!E?"KU'8""<9-Y1;!E9VKU'5""9S-)19!E4,K5'3""6m,g19!E+zK5'/""9S,'1:!E!&KU',"">}+U1=!DG2L7'+""Is+U.[!TG2!A)p!q)4+u._!U&P!q)t!q.^,g.a!U4,
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*bKfJf<6?</7LR&@+4KeJfDF?,/:LQLL+TKcJfG,>Z/;LQD<+dK_JfLV=h/<LQ16+tK[Jg)0L&/?LV+j'\LGJe4&J2/%LV&@(>LHJeDFJ2/+LV&@)PLGJf+tIp/1LUIf*BLCJf9PI0/6LU<,+$L>JfA`H./8LU)&+DL4JfDFEx/9LS)&+DJ2JtA`5H1\LN6a$VJ2Jt.Z&J1\LHLM!0K,Js#d5H4!L8LMKdGLJs#d!a!!!#DH!A!%!!!!(_#i
!!+u!!!N!1)1!a!!!#DH!A!%!!!!"##f!'1A!!!E!T#e@N1xLT)',fL,Jw.ZEH3kLD17G\IFK"LUEH3aLD3{FJIGK">yEh3^LD9GExIJK"6iFJ3\LDG#EhIQK"4%Gl3[LE+kE(I+K"+s@n3[LBG#EXI1K"6iB23\LC.QEhI6K"9OBr3^LC3{F:I8K"A_C$3aLC9GG,I9K#)/C43kLC9G&zK}Ju6jC41mLS6a)0K|Jv)0C$1tLS3{*"K{Jv1@
Bb1wLS.Q*BKxJv6jB21xLS&A*RKtJv6jA@1wLRAW*2KnJv.Z@N1xLG+u!!!N!1&K!a!!!#DH!a!$!!!!!!!Q

.TXT 27 36 163 0 0
Cg a38.000000,38.000000,99
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Enter the total number 
of flash units.  This was adjusted until the goal of reaching y=0.9 
was met.}
.EQN 1 -36 15 0 0
{0:N}NAME:7
.EQN 4 0 19 0 0
{0:i}NAME:1;{0:N}NAME
.TXT 0 36 20 0 0
Cg a38.000000,38.000000,42
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Repeat calc's for each 
flash ({\i i} = 1,2,...)}
.EQN 3 -36 16 0 0
({0:y}NAME)[(0):0.3
.TXT 0 36 21 0 0
Cg a38.000000,38.000000,40
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Enter composition of 
feed to first stage}
.EQN 5 -36 22 0 0
{0:f}NAME:0.5
.TXT 0 36 23 0 0
Cg a38.000000,38.000000,80
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Enter fraction of feed 
to leave as vapor (assumed to be the same for each stage)}
.EQN 7 -36 27 0 0
{0:yop}NAME({0:x}NAME,{0:xF}NAME):-((1-{0:f}NAME)/({0:f}NAME))*{0:x}NAME+({0:xF}NAME)/({0:f}NAME)
.TXT 0 36 14 0 0
Cg a32.250000,32.250000,124
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Mole balances about the {
\i i}{\fs16\up  th} flash unit yields the "operating line": a 
relationship between x and y of the two output steams}
.EQN 8 -36 33 0 0
{0:\x}NAME:0.5
.TXT 0 36 39 0 0
Cg a38.000000,38.000000,95
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard This is an initial guess 
for the value of x{\fs16\dn i} required for the "root" function in the 
next step.}
.TXT 8 0 40 0 0
Cg a38.000000,38.000000,178
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard "root" finds the value 
of x{\fs16\dn i} at which the operating line for the i{\fs16\up th} 
stage intersects with the equilibrium curve.  Note that the feed to 
stage i is the vapor leaving stage i-1.  }
.EQN 3 -36 42 0 0
({2,1}ö({0:y}NAME)[({0:i}NAME)ö({0:x}NAME)[({0:i}NAME)):({2,1}ö{0:yeq}NAME({0:root}NAME({0:yop}NAME({0:\x}NAME,({0:y}NAME)[({0:i}NAME-1))-{0:yeq}NAME({0:\x}NAME),{0:\x}NAME))ö{0:root}NAME({0:yop}NAME({0:\x}NAME,({0:y}NAME)[({0:i}NAME-1))-{0:yeq}NAME(
{0:\x}NAME),{0:\x}NAME))
.TXT 7 36 166 0 0
Cg a42.000000,42.000000,168
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard There is no flash 
corresponding to "i=0" (first row), but y{\dn 0}=0.3 represents the 
feed to the first flash.  x{\dn 0} was never assigned a value but 
takes on 0 value by default.}
.EQN 9 -36 36 0 0
{0:x}NAME={0}?_n_u_l_l_
.EQN 0 13 37 0 0
{0:y}NAME={0}?_n_u_l_l_
.TXT 13 -13 96 0 0
Cg a71.625000,71.625000,121
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Place results in another 
array (having twice as many elements) showing steps between 45-degree 
line and equilibrium curve}
.EQN 5 0 50 0 0
({0:xs}NAME)[(0):0.3
.EQN 0 12 51 0 0
({0:ys}NAME)[(0):0.3
.TXT 0 25 97 0 0
Cg a36.000000,36.000000,24
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Feed to first flast unit}
.EQN 5 -37 54 0 0
({0:xs}NAME)[(2*{0:i}NAME-1):({0:x}NAME)[({0:i}NAME)
.EQN 0 12 55 0 0
({0:ys}NAME)[(2*{0:i}NAME-1):({0:y}NAME)[({0:i}NAME)
.TXT 0 25 99 0 0
Cg a36.000000,36.000000,46
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Step along operating 
line to equilibrium curve}
.EQN 5 -37 56 0 0
({0:xs}NAME)[(2*{0:i}NAME):({0:ys}NAME)[(2*{0:i}NAME-1)
.EQN 0 12 58 0 0
({0:ys}NAME)[(2*{0:i}NAME):({0:ys}NAME)[(2*{0:i}NAME-1)
.TXT 0 25 101 0 0
Cg a36.000000,36.000000,61
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Horizontal step from 
equilibrium curve back to 45 degree line}
.EQN 5 -37 60 0 0
{0:j}NAME:0,1;{0:last}NAME({0:ys}NAME)
.EQN 5 0 59 0 0
1&&(0.3&0.9)&{0:yeq}NAME({0:zi}NAME),{0:zi}NAME,({0:ys}NAME)[({0:j}NAME)@1&&(_n_u_l_l_&_n_u_l_l_)&{0:zi}NAME,{0:zi}NAME,({0:xs}NAME)[({0:j}NAME)
0 0 1 1 1 0 0 1 1 
0 0 1 1 1 0 1 1 1 
0 1 0 0 1 1 equilibrium curve
0 1 6 0 1 1 y = x
0 1 2 0 1 1 stepping off stages
0 4 3 0 1 1 NO-TRACE-STRING
0 1 4 0 1 1 NO-TRACE-STRING
0 2 5 0 1 1 NO-TRACE-STRING
0 3 6 0 1 1 NO-TRACE-STRING
0 4 0 0 1 1 NO-TRACE-STRING
0 1 1 0 1 1 NO-TRACE-STRING
0 2 2 0 1 1 NO-TRACE-STRING
0 3 3 0 1 1 NO-TRACE-STRING
0 4 4 0 1 1 NO-TRACE-STRING
0 1 5 0 1 1 NO-TRACE-STRING
0 2 6 0 1 1 NO-TRACE-STRING
0 3 0 0 1 1 NO-TRACE-STRING
0 4 1 0 1 1 NO-TRACE-STRING
1 0 1 42 35 10 0 3 
.EQN 52 0 61 0 0
{0:N}NAME={0}?_n_u_l_l_
.TXT 0 32 62 0 0
Cg a41.000000,41.000000,33
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard number of equilibrium 
stages used}
.EQN 5 -32 63 0 0
(({0:y}NAME)[({0:N}NAME)*(0.5)^({0:N}NAME))/(({0:xs}NAME)[(0))={84532}?_n_u_l_l_
.TXT 0 32 64 0 0
Cg a41.000000,41.000000,53
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard fractional recovery of 
toluene in vapor of last stage}
.TXT 6 -32 65 0 0
Cg a73.000000,73.000000,7
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\fs36\ul Part b}{\fs36
\ul )}}
.EQN 4 0 151 0 0
{0:x}NAME:0
.EQN 0 9 152 0 0
{0:y}NAME:0
.TXT 0 23 154 0 0
Cg a41.000000,41.000000,34
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Clear these arrays also 
used above}
.EQN 4 -32 167 0 0
{0:R}NAME:8
.TXT 0 32 168 0 0
Cg a46.000000,46.000000,60
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Assign values to the 
reflux ratio and distillate composition}
.EQN 3 -32 71 0 0
{0:x.D}NAME:0.9
.EQN 5 0 69 0 0
{0:yop}NAME({0:x}NAME):({0:R}NAME)/({0:R}NAME+1)*{0:x}NAME+({0:x.D}NAME)/({0:R}NAME+1)
.TXT 0 32 169 0 0
Cg a46.000000,46.000000,45
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Define general equation 
of the operating line}
.TXT 5 0 170 0 0
Cg a46.000000,46.000000,132
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Declare the number of 
steps to take (adjusted until bottoms composition is less than 0.3).  
Define index "i" to be the stage number.}
.EQN 2 -32 72 0 0
{0:N}NAME:5
.EQN 0 9 92 0 0
{0:i}NAME:1;{0:N}NAME
.EQN 6 -9 73 0 0
({0:x}NAME)[(0):{0:x.D}NAME
.TXT 0 32 171 0 0
Cg a46.000000,46.000000,77
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Seed the recursion 
formula with the composition of the reflux stream as x{\dn 0}.  }
.TXT 6 0 158 0 0
Cg a41.000000,41.000000,123
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Apply the recursion 
formula for the N stages.  (x{\fs16\dn i},y{\fs16\dn i}) are the 
compositions of the liquid and vapor streams leaving stage i}
.EQN 1 -32 132 0 0
({2,1}ö({0:x}NAME)[({0:i}NAME)ö({0:y}NAME)[({0:i}NAME)):({2,1}ö{0:xeq}NAME({0:yop}NAME(({0:x}NAME)[({0:i}NAME-1)))ö{0:yop}NAME(({0:x}NAME)[({0:i}NAME-1)))
.EQN 6 0 155 0 0
({0:y}NAME)[({0:N}NAME+1):{0:yop}NAME(({0:x}NAME)[({0:N}NAME))
.TXT 11 32 160 0 0
Cg a41.000000,41.000000,17
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard y{\fs16\dn 0} has no meaning}
.EQN 1 -32 140 0 0
{0:x}NAME={0}?_n_u_l_l_
.EQN 0 12 141 0 0
{0:y}NAME={0}?_n_u_l_l_
.TXT 12 -12 172 0 0
Cg a71.625000,71.625000,114
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Place results in another 
array (twice as many elements) showing steps between operating line 
and equilibrium curve}
.EQN 5 0 143 0 0
({0:xs}NAME)[(0):({0:x}NAME)[(0)
.EQN 0 12 144 0 0
({0:ys}NAME)[(0):({0:x}NAME)[(0)
.EQN 5 -12 145 0 0
({0:xs}NAME)[(2*{0:i}NAME-1):({0:x}NAME)[({0:i}NAME)
.EQN 0 12 146 0 0
({0:ys}NAME)[(2*{0:i}NAME-1):({0:y}NAME)[({0:i}NAME)
.TXT 0 23 148 0 0
Cg a38.000000,38.000000,40
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Horizontal step to the 
equilibrium curve}
.EQN 5 -35 149 0 0
({0:xs}NAME)[(2*{0:i}NAME):({0:x}NAME)[({0:i}NAME)
.EQN 0 12 150 0 0
({0:ys}NAME)[(2*{0:i}NAME):({0:y}NAME)[({0:i}NAME+1)
.TXT 1 23 147 0 0
Cg a38.000000,38.000000,35
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Vertical step to the 
operating line}
.EQN 3 -35 106 0 0
{0:j}NAME:0;2*{0:N}NAME
.EQN 6 0 109 0 0
{0:xR}NAME:({2,1}ö0ö{0:x.D}NAME)
.EQN 0 11 110 0 0
{0:yR}NAME:({2,1}ö{0:yop}NAME(0)ö{0:x.D}NAME)
.TXT 0 24 111 0 0
Cg a41.000000,41.000000,17
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Two points on ROL}
.EQN 6 -35 112 0 0
{0:i2}NAME:0;1
.EQN 1 0 162 0 0
1&&(0.3&0.9)&{0:yeq}NAME({0:zi}NAME),{0:zi}NAME,({0:ys}NAME)[({0:j}NAME),({0:yR}NAME)[({0:i2}NAME)@1&&(_n_u_l_l_&0.9)&{0:zi}NAME,{0:zi}NAME,({0:xs}NAME)[({0:j}NAME),({0:xR}NAME)[({0:i2}NAME)
0 0 1 1 1 0 1 1 1 
0 0 1 1 1 0 1 1 1 
0 1 0 0 1 1 equilibrium curve
0 1 6 0 1 1 y = x
0 1 2 0 1 1 stepping off stages
0 1 3 0 1 1 operating line
0 1 4 0 1 1 NO-TRACE-STRING
0 2 5 0 1 1 NO-TRACE-STRING
0 3 6 0 1 1 NO-TRACE-STRING
0 4 0 0 1 1 NO-TRACE-STRING
0 1 1 0 1 1 NO-TRACE-STRING
0 2 2 0 1 1 NO-TRACE-STRING
0 3 3 0 1 1 NO-TRACE-STRING
0 4 4 0 1 1 NO-TRACE-STRING
0 1 5 0 1 1 NO-TRACE-STRING
0 2 6 0 1 1 NO-TRACE-STRING
0 3 0 0 1 1 NO-TRACE-STRING
0 4 1 0 1 1 NO-TRACE-STRING
1 0 1 42 35 10 0 3 
.EQN 58 0 114 0 0
{   25    18 7410 0 8 200 145}{66}
!1!*!!!$5YG-!!1AF;#e!!!!!q!!!!>}E9/@L'.[!!!-!ND>Gm.b!!!!Ku&L!!!!!!!!!!+u!!!N!1!!"S!!!0<9!!!!!!!!!!!!!#&K!a!!!#DH!1!%!!!!!a#r!!4'!!#a!A!!!!#d!!!!!a!!!#DH!A!*!!!!KU&P!!!!!!#dLVIq&k!%!!!!(_#h!!+u!!!'!1#e,w!!!#+x&[#_!&IvG=!}!}9Q/M0S!%<<<W![!z&K'm0&!")66k#`
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GM+vLU6n#$L?!QA`GM+{LTDK"2L>!a&JH?+vLU&O#e!!!#+x"#"+!/&O2c#L!a4&H?+{LTDK2c#B!g6kGM+x!!!!&+!%!!!!"##f!%Aa!!!E!S+u6;#G!aDG!a.h!!+z#E!#!r!!!!.l!0G1$'#a!b)1KE,+!04+$W#Z!b1AJC,.!0#i$w#T!b6kIA,/!/?!%9#M!b9QH/,0!/+y%9#G!b9QG-,0!.G1%9#A!b6kEy,/!.4+$w#:!b4'Dw,-
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.EQN 3 25 115 0 0
{0:x.F}NAME:({0:y}NAME)[({0:N}NAME+1)
.EQN 0 14 116 0 0
{0:x.F}NAME={0}?_n_u_l_l_
.TXT 4 -14 113 0 0
Cg a47.125000,47.125000,194
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Recovery is the fraction 
of the benzene in the feed which ends up in the distillate: x{\fs16\dn D}D/x{
\fs16\dn F}V.  The flowrate of product is D and the feed flowrate is V 
= L+D = RD+D = (R+1)D.  Thus D/V = 1/(R+1).}
.EQN 10 0 118 0 0
({0:x.D}NAME)/({0:x.F}NAME)*(1)/({0:R}NAME+1)={0}?_n_u_l_l_
.EQN 0 20 90 0 0
{0:N}NAME={0}?_n_u_l_l_
.TXT 6 -20 119 0 0
Cg a48.000000,48.000000,139
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Alternatively, you might 
recognize that the last step takes us past the feed composition, so 
that the "actual" number of stages required is}
.EQN 9 0 120 0 0
5-(0.3-({0:y}NAME)[({0:N}NAME+1))/(({0:y}NAME)[({0:N}NAME)-({0:y}NAME)[({0:N}NAME+1))={84529}?_n_u_l_l_
.TXT 6 0 128 0 0
Cg a48.000000,48.000000,50
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard For this smaller number 
of stages, the recovery is}
.EQN 5 0 130 0 0
({0:x.D}NAME)/(0.3)*(1)/({0:R}NAME+1)={0}?_n_u_l_l_
.TXT 6 -25 131 0 0
Cg a72.000000,72.000000,186
{\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard The overall lession from 
this problem is that a countercurrent cascade is better than a series 
of flash units: The recovery is an order of magnitude better and fewer 
stages are required.}