HEMISKI Zbirka re{eni zada~i REAKTORI 3

330
Fio Fi() Fi(X) pi = yiP = (Fi/FT)P
A FAo FAo– FAo(1–X)
B FBo FBo–0,5 FBo–0,5FAoX
C 0 FAoX
I
(N2)
FI =
= 1,881FAo
FTo=
FAo+FBo+FI
= 3,381FAo
FI =
= 1,881FAo
FT =
FTo–0,5
FI =
= 1,881FAo
FT =
FTo–0,5FAoX =
=FTo(1-0,15X)
p
A
y
FAo
( 1
X ) P
F ( 1
0,
15X
)
Ao
To
( 1
X ) P
( 1
0,
15X
)
0,
5FAo
( 1
X ) P
pB
FTo
( 1
0,
15X
)
y Ao ( 1
X ) P
2 ( 1
0,
15X
)
p
C
y
p
I
F
Ao
( 1
To
F
Ao
( 1
0,
15X
)
X P
X P
0,
15X
)
FAo1,
881P
F ( 1
0,
15X
)
To
y Ao P
1,
881
( 1
0,
15X
)
pi
So kombinirawe na brzinskiot izraz i stehiometrijata
(soglasno so stehiometriskata tablica) se dobiva:
(
1/
3 2/
3
rA ) k p A pB
(kmol/kgkat
1/
3
( r
( r
A
A
) k y
) k y
2 / 3
Ao
Ao
( 1
X )
P
( 1
0,
15X
)
1
2
1
k y P
2
'
k Ao
(
r A
) k
'
2
/ 3
0,
0141
)/h ; p
y
2
( 1
X )
P;
( 1
0,
15X
)
0,
296
Ao
( 1 X )
( 1 X )
0,
0263
( 1 0,
15X
) ( 1 0,
15X
)
A
, p
B
( atm);
( 1
X )
P
( 1
0,
15X
)
2/
3
0,
63 10
0,
0263 (kmol/kg
(kmol/kg
kat
;
)/h .
P
kat
)/h ;