Question: DD 1 0 # 5 ( 8 ) 2 A B adiabatic CSTR A 3 0 0 K 4 2 mol / s

DD10"\# 5(8)2AB adiabatic CSTR A 300 K 42 mol/s 25 mol/m
3
. CSTR 50%
\Delta
=8000
mol
E=15000
mol
=50
mol.K
=100
mol.K
1
=5\times 10
6
e
-
mol.s
m
3
c
=2.5
mol
m
3
@300 K
0
1x
dx
=ln
1x
1
1
2
(1x)
2
dx
=
1x
2
1
-
1x
1
1
0
(1x)
2
dx
=
1x
0
1+\epsi x
dx
=
\epsi
1
ln(1+\epsi x)
0
1x
(1+\epsi x)dx
=(1+\epsi )ln
1x
1
\epsi x
(A4)(A5)4.
4
f(X)dX=
3
h
(v
0
+4f
1
+2f
2
+4f
3
+f
4
) h=
4
4
x
0
"

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