Environmental Engineering Reference
In-Depth Information
The energy balance now is (no heat supply nor extraction and no work related to
the system) U
products
−
U
reactants
=0:
2
4
3
5
0=
5
3
h
CO
2
−
+2
h
H
2
O
−
52
h
N
2
−
ð
R
u
T
ad
Þ
ð
R
u
T
ad
Þ
+7
:
ð
R
u
T
ad
Þ
2
3
+7
+
2
4
h
CH
4
−
+2
h
O
2
−
52
h
N
2
−
3
h
CO
2
−
5
−
ð
R
u
T
0
Þ
R
u
T
0
:
ð
R
u
T
0
Þ
ð
R
u
T
0
Þ
2
4
n
o
+2
h
f
,
H
2
O
+
n
o
0=
5
3
h
f
,
CO
2
+
c
p
,
CO
2
T
ad
−
ð
T
0
Þ
c
p
,
H
2
O
T
ad
−
ð
T
0
Þ
o
#
n
h
o
+7
:
52
f
,
N
2
+
c
p
,
N
2
T
ad
−
ð
T
0
Þ
2
4
3
5
−
0
@
1
A
R
u
T
ad
−
h
f
,
N
2
+
2
1
2
h
f
,
CH
4
+2
3
h
f
,
CO
2
h
f
,
O
2
+7
:
:
−
52
3
+2+7
52
ð
T
0
Þ
T
0
) to the left-hand side
of the equation, one obtains an equation that is easy to solve with T
ad
−
Now, bringing all temperature-dependent terms (T
ad
−
T
0
as the
only unknown. The numerical result is
9K, so that T
ad
= 2645 K= 2372
C
T
ad
−
T
0
= 2346
:
:
At constant volume, the end pressure can be calculated to be
p
2
=T
2
:
p
1
=
T
1
= 2645 1
ð
=
298
Þ
=8
:
9 atm
:
c
p
value constant at 1200 K as the
average temperature, but this does not cause a very large error. In fact, assuming
that no endothermic dissociation of CO
2
takes place (forming CO and ½ O
2
)
causes a larger discrepancy (Turns, 2000). If no CO
2
were present in this gas, then
a significantly higher temperature and associated pressure would result.
There is a slight discrepancy when taking the
3.4 CONSERVATION OF MOMENTUM
Momentum, unlike mass or energy, has a direction, and therefore, it is a vector
quantity and so is momentum density,
f
=
!
. For each directional component of this
vector, a conservation equation can be derived. However, the vector notation is used
here. The external source (or force) is usually only gravity, so then
s
f
=
ρ
!
.As
f
is
ρ
!
f
=
!!
+
pI
already a vector, the momentum flux becomes a tensor:
φ
ρ
−
τ
.
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