HEMISKI Zbirka re{eni zada~i REAKTORI 3

POLYMATH Results
Calculated values of the DEQ variables
^etvrti del. Idealen ceven reaktor (PFR)
Variable initial value minimal value maximal value final value
V 0 0 1 1
X 0 0 0.1272604 0.1272604
To 953 953 953 953
FAo 1.31E-04 1.31E-04 1.31E-04 1.31E-04
T 953 883.96325 953 883.96325
CAo 0.0128 0.0128 0.0128 0.0128
k 0.0154871 3.721E-04 0.0154871 3.721E-04
R 1.982E-04 4.091E-06 1.982E-04 4.091E-06
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = R/FAo
Explicit equations as entered by the user
[1] To = 953
[2] FAo = 0.000131
[3] T = (To-406*X)/(1+0.1544*X)
[4] CAo = 0.0128
[5] k = 10^((-18963/T)+(2*log(T))+12.130)
[6] R = k*CAo*(1-X)*To/T/(1+0.75*X)
Kako {to se gleda od rezultatot, so adijabatska rabota na
reaktorot so volumen od 1 m 3 , so vlezna temperatura od 680 o C i
pritisok od 1 atm, izleznite konverzija i temperatura se:
‡ izlezna konverzija, X = 0,127,
‡ izlezna temperatura, T = 884 K (611 o C).
Ako pak se zeme deka odnosot na temperaturite vo izrazot
za molskata koncentracija e edinica i ako toplinskiot bilans
(6) se uprosti na T = To – 406 X, zemaj}i deka 0,15447