HEMISKI Zbirka re{eni zada~i REAKTORI 3

Petti del. Kataliti~ki reaktor so fiksen sloj katalizator (PBR)
Pp 0 0 1.2058351 1.1240547
Po 1.5 0.2798397 1.5 0.2798397
Ph 1.5 0.2798397 1.5 0.2798397
R 1.465E-05 8.597E-07 1.465E-05 8.597E-07
D 6.826E+04 6.826E+04 1.163E+06 1.163E+06
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = R/FOo
Explicit equations as entered by the user
[1] FOo = 0.00833
[2] P0 = 3
[3] k = 0.115/3600
[4] Ko = 0.4926
[5] Kh = 0.3174
[6] Kp = 0.4263
[7] P = P0-0.0008834*W
[8] Pp = X*P/(2-X)
[9] Po = (1-X)*P/(2-X)
[10] Ph = (1-X)*P/(2-X)
[11] R = k*Po*Ph/((1+Ko*Po+Kh*Ph+Kp*Pp)^2)
[12] D = 1/R
Soglasno so dobienite rezultati, so vklu~uvawe na linearnata
zavisnost za padot na pritisokot nadol` reaktorot, za
80% konverzija na izo-oktenot PBR treba da bide ispolnet so
W = 1490 kg katalizator, dodeka negoviot volumen treba da bide:
V W / 1490 / 1560 0,
955 m .
sloj
W (kg)
P(W)
pP(W)
pO(W)
3
381