Civil Engineering Reference
In-Depth Information
2
2
ρ
fv
−
1
∂
∂
z
s
∂
∂
z
s
f
D
V
=
1
dV
dt
−
−
(6.1)
τ
=
,
,
0
8
γ
2
g
g
For flow direction changes:
(6.2)
2
VVV
=
dV
dt
+
1
∂
∂
p
s
dz
ds
f
D
VV
+
+
(
) (
g
=
0
,Eulerequation
(6.3)
ρ
2
Continuity equation and for fluid fine element:
1
dd
d A
ρρ
1
dA
Dt
∂
∂
V
s
1
d
dt
−
+
(6.4)
(()
=
+
ds
0 214
ρ
ds
-fluid elasticity property,
ρρ
1
d
=
0
ρ
d
-Pipe elasticity property
+
1
dA
dt
−
∂
∂
v
s
1
dp
dt
=
(
)
(6.5)
2
c
0
,Continuityequation
,
A
ρ
dV
dt
+
1
∂
∂
p
s
dz
ds
f
D
VV
+
+
(
)
g
=
0
,EulerEquation
,
(6.6)
ρ
2
Using the MOC, the two partial differential equations can be transformed to the
following equations:
g
c
dH
dt
dV
dt
ds
dt
+
(
)
=⇒ =
fv vd
/
20
c
+
,
(6.7)
g
c
dH
dt
dV
dt
fVVD
ds
dt
+
(
)
=⇒ =
(6.8)
−
−
/
2
0
c
,
Method of characteristic solution for partial differential equation:
The method of characteristics (MOC) is a finite difference technique where pressures
are computed along the pipe for each time step:
dp
dt
∂
∂
p
t
∂
∂
p
s
ds
dt
=
+
(6.9)
Search WWH ::
Custom Search