Biomedical Engineering Reference
In-Depth Information
In the flue gas,
F
N
2
;
e
¼
F
N
2
;0
¼ 3573:8
mol
=
h;
F
CO
2
;
e
¼
F
CO
;0
¼ 1000
mol
=
h;
F
O
2
;
e
¼
F
O
2
;
0
F
O
2
;
min
¼ 0:45
F
CO
;0
¼ 450
mol
=
h;
Energy balance over the system:
⎡
N
⎤
⎡
N
⎤
d
(
U
⋅
n
)
S
S
∑
=1
∑
=1
Q
−
W
F
H
=
+
F
H
−
⎣
⎦
⎣
⎦
j
j
j
j
dt
j
j
in
out
0
0
Steady state
no work done
Therefore:
F
j;
out
H
j;
out
F
j;
in
H
j;
in
X
N
S
Q
¼
j
¼1
Since the pressure is constant (at standard pressure), we have:
Z
T
H
j;T
¼ DH
fj;298
þ
¼ DH
fj;298
þ
C
P
j
d
T
C
P
j
ð
T
298Þ
298
Remark: if a table of C
P
vs T is tabulated, numerical integration will be needed. If a correlation
equation is known, then direct integration can be applied.
F
j;
out
H
j;
out
F
j;
in
H
j;
in
¼
P
N
S
Q
j
¼1
n
F
j;
e
h
H
j;298
þ
C
P
j;
e
T
j;
e
298
i
h
H
j;298
þ
T
j;0
298
io
¼
P
N
S
F
j;0
C
P
j;0
j
¼1
F
j;
e
H
j;298
þ
P
h
F
j;
e
C
P
j;
e
T
j;
e
298
T
j;0
298
i
¼
P
N
S
N
S
Q
F
j;0
F
j;0
C
P
j;0
j
¼1
j
¼1
h
F
j;
e
C
P
j;
e
T
j;
e
298
T
j;0
298
i
F
CO
;0
DH
R
;298
þ
P
N
S
¼
F
j;0
C
P
j;0
j
¼1
¼ 1000
282:990 10
3
þ
3573:8 31:47 þ 1000 49:91 þ 450 33:25
1273 298
3573:8 31:47
373 298
þ 1000 39:38
473 298
þ 950 29:66
373 298
J
Q
=
h
¼127523350
J
=
h
¼1:2752 10
8
J
=
h
¼35:42
kW
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