HEMISKI Zbirka re{eni zada~i REAKTORI 3

Od izve{tajot i od rezultatite od primena na solverot
za diferencijalni ravenki koi{to sledat, se gleda deka za da se
postignat 19,8698% konverzija na etilenot na temperatura
od 271 o C = 544 K, }e bidat potrebni W =2000 kg katalizator.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
W 0 0 2000 2000
X 0 0 0.198698 0.198698
Fao 2.268 2.268 2.268 2.268
T 544 544 544 544
Ka 0.0091526 0.0091526 0.0091526 0.0091526
k 0.0062932 0.0062932 0.0062932 0.0062932
K 0.0046371 0.0046371 0.0046371 0.0046371
P 140 140 140 140
pa 70 62.27844 70 62.27844
pr 0 0 15.44312 15.44312
B 2.2813598 2.1400155 2.2813598 2.1400155
r 4.963E-04 6.311E-05 4.963E-04 6.311E-05
D 0.6063593 0.6063593 0.6063593 0.6063593
R 0.2213093 0.2213093 0.2213093 0.2213093
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = r/Fao
Explicit equations as entered by the user
[1] Fao = 2.268
[2] T = 544
[3] Ka = exp(-44.061+(21415.8/T))
[4] k = exp(20.474-(13895/T))
[5] K = exp(-14.37+(4894/T))
[6] P = 140
[7] pa = P*(1-X)/(2-X)
[8] pr = P*X/(2-X)
[9] B = (1+(2*Ka*pa))
[10] r = k*(Ka^2)*((pa^2)-(pr/K))/(B^2)
[11] D = 1-((2*K*P)/((2*K*P)+2))
[12] R = 1-(D^0.5)
Zabele{ka: Ako se insistira na konverzija od to~no 20%,
toga{ to~nata vrednost za potrebnata koli~ina katalizator e
W = 2050 kg. Za ponatamo{nite presmetki se usvojuvaat 2000 kg.
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