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MATLAB%20Ex%2010_5

# MATLAB%20Ex%2010_5 - %EX1D_e e Single reaction Incomplete...

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Unformatted text preview: ::%EX1D_e e Single reaction. Incomplete reaction. close all; clear all; clc; 0 Fractional convereion of limiting reactant. f = 0.6T; % Basie ie 100 kmol feed. 0 Inlet kmol. ni =.[40 % CH4 50 e 012 10]: % N2 % Set up matrix A for system of equations, Ax = h. e Outlet kmol are given by x = Alb. % no - column vector of unknown product kmcl. e b = column vector giving right hand side of equat % noilj = CH4. % noizl = 012. % noi3] = HCl. e notd] - CHSCl. % noiSJ - H2 {does not react}. .'\$ = [ni{1} % carbon, 0 gﬁ 2 * niizl % chlorine; Cl 4 * niil} % hydrogen, H 2 * ni{3] % nitrogen, N {1 - f}*ni{11]; % extent of reaction A = [1 0 0 1 0 % carbon,r C 0 2 l 1 0 % chlorine, Cl 4 0 1 3 0 % hydrogen. H 0 0 0 0 2 % nitrogen,r H 1 0 0 0 0]: % extent of reaction no = Alb; ' %Reeulte, kmol no- 13.2000 CH4 23.2000 012 26.0000 HC1 26.8000 CHSCl 10.0000 N2 (does not react} EPEFﬂFdFWEP J 3EX10_5 % Two competing reactions. Incomplete reaction. close all; clear all; clc; % Fractional conversions. f1 - 0.90: %fraction of inlet CHBOH that reacts. f2 = 0.75; %fraction of inlet CHBOH that yields CHEO. 0 Basis is 1 kmol CHEOH. - % Inlet kmol- ni - [1 % CHEOH 1 % 02 is twice that required for stoichionetric reaction. 0.?9!0.21]; % N2 % Set up matrix A for system of equations, Ax - b. 0 Outlet kmol are given by x = AKb. '-4L——i—n-h\$1rdw4-meeeéxrw % no = column vector of unknown productIr kmol. % h = column vector giving right hand side of equation. % noill - CHBOH. % no12} - 02. % noiBl = N2 (does not react}. '1% noldl = Cﬂzﬂ. W e nolS) = CO. % noiﬁl = H20. b = [{l-f1]*ni{l} % extent of reaction f2*ni{l} % extent of reaction nitl) % carbon, C 4 * nitl} % hydrogen, H ni{1) + 2*nif2} % oxygen, 0 2*nii3ll; % nitrogen, N A.= [1 0 0 0 0 0 0 0 0 l D 0 1 0 0 l l 0 4 0 0 2 0 2 1 2 0 1 1 1 0 0 2 0 0 0]; no = AXb; #5 Calculate INDIE fractions. y = no.fsum[noﬂ; e Results 0 no = % 0.1000 CH30H % 0.4150 02 % 3.7619 N2 % 0.1500 CHEO :)% 0.1500 CD % 1.0500 H20 EPWﬂPdelcPOF 0.ﬂ159 3.0T55 0.5934 H.1193 D.U239 D.lE?ﬂ ...
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