HEMISKI Zbirka re{eni zada~i REAKTORI 3

Y =
1 7,
077 ( 1 0,
45)
( 1 0,
45)
5
1,
75 7,
077 150 2,
782 10
1,
9625 1057,
65 6,
618 0,
0075 3
0,
45
0,
0075
538
2
1 kg/(m s ) Pa
Y 5,
259 .
2
m s
kg kat kg kat
Ovaa vrednost za Y konvertirana vo atm/kgkat: Y
5 5
5,
259 /( 1,
01310
) 5,
19110
atm/kgkat
se zamenuva vo ravenkata (17):
dP 5
FT
5,
19110
dW
FTo
T Po
To
P
atm/kgkat
. (19)
Re{enieto na sistemot diferencijalni ravenki (11), (18)
i (19), so temperatura na mediumot za ladewe Ta = 490 K, e slednoto
(samo del od izve{tajot i grafi~ki prikaz):
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
W 0 0 2.0E+04 2.0E+04
Fa 500 346.44914 500 346.44914
Fb 250 122.27095 250 122.27095
Fc 0 0 133.16941 133.16941
Fd 0 0 40.762894 40.762894
Fw 0 0 40.762894 40.762894
T 533 496.50185 646.84954 496.50185
P 10 8.9043767 10 8.9043767
X 0 0 0.3071017 0.3071017
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(W) = -ra
[2] d(Fb)/d(W) = -rb
[3] d(Fc)/d(W) = rc
[4] d(Fd)/d(W) = rd
[5] d(Fw)/d(W) = rw
[6] d(T)/d(W) = (500*0.07564*(490-T)+A)/B
[7] d(P)/d(W) = -5.191*(10^(-5))*(Po/P)*(T/To)*(Ft/Fto)