2.2 three approaches to solve for normal … · 27/02/2003 · • with the assumptions made, the...
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CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.1
2.2 THREE APPROACHES TO SOLVE FOR NORMAL (UNIFORM) FLOW DEPTH IN A WIDE RECTANGULAR CHANNEL
Derivation of Normal Flow Depths by Applying the Shallow Water Equations
• Consider the stretch of channel shown:
• The geoid is defined at z=0, the free surface (water-air interface) at z= , and the bottom(water-sediment interface) at z=-h.
η
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.2
• We assume the following:
• Steady state flow
• A wide channel of flow depth over a channel bed with slope
• Steady flow in the cross channel averaged variables in the x (almost identical to the s)direction. Therefore
• Uniform flow in the x-direction: constant in the x-direction and
• No cross channel gradient or flow: ;
• No applied free surface stress
• Flow depth is constant and set to along the entire section ofinterest.
∂∂t---- 0→
d0 θ0
u 0>
u∂u∂x------ 0=
∂η∂y------ 0= v 0=
τxs
0; τys
0==
H η h+= H d0=
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.3
• We now consider the Shallow Water equations:
(2.2.1)
(2.2.2)
(2.2.3)
(2.2.4)
(2.2.5)
∂η∂t------
∂ uH( )∂x
---------------∂ vH( )
∂y--------------- 0=+ +
∂u∂t------ u
∂u∂x------ v
∂u∂y------ g
∂η∂x------– +=+ +
1H----
∂∂x-----
τxxt m⁄
ρ---------- u
2–� �� �� �
zd
h–
η
�1H----
∂∂y-----
τyxt m⁄
ρ---------- uv–� �� �� �
z1H----
τxs
ρ----
1H----
τxb
ρ-----–+d
h–
η
�+
∂v∂t----- u
∂v∂x----- v
∂v∂y----- g
∂η∂y------– +=+ +
1H----
∂∂x-----
τxyt m⁄
ρ---------- uv–� �� �� �
zd
h–
η
�1H----
∂∂y-----
τyyt m⁄
ρ---------- v
2–� �� �� �
z1H----
τys
ρ----
1H----
τyb
ρ-----–+d
h–
η
�+
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.4
• With the assumptions made, the Shallow Water equations simplify to
(2.2.6)
(2.2.7)
(2.2.8)
• We are going to assume that the lateral dispersion/diffusion terms simplify as follows:
• No cross channel depth averaged currents or variations in cross channel currents;
(2.2.9)
• Since all the laminar-turbulent-depth averaged shear stress terms will all be depen-dent on gradients in the mean flow field in the x and y directions (e.g. like the EddyDispersion relationships in equations (1.5.73) through (1.5.75), the shear stresses inthe xx, xy and yy planes/directions are negligible.
0 0=
0 g∂η∂x------–
1H----
∂∂x-----
τxxt m⁄
ρ---------- u
2–
� �� �� �
z1H----
∂∂y-----
τyxt m⁄
ρ---------- uv–� �� �� �
z1H----
τxb
ρ-----–d
h–
η
�+d
h–
η
�+=
01H----
∂∂x-----
τxyt m⁄
ρ---------- uv–� �� �� �
z1H----
∂∂y-----
τyyt m⁄
ρ---------- v
2–
� �� �� �
zd
h–
η
�+d
h–
η
�=
v v 0= =
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.5
• With these assumptions the governing equations simplify to:
(2.2.10)
(2.2.11)
(2.2.12)
• Thus the free surface gradient (gravity) term balances the bottom friction term:
(2.2.13)
• We note that for steady uniform flow:
(2.2.14)
• Also we showed that for uniform steady flow:
. (2.2.15)
0 0=
0 g∂η∂x------
1H----
τxb
ρ-----––=
0 0=
g∂η∂x------
1H----
τxb
ρ-----–=
∂η∂x------ free surface gradient = = Sw–
Sw S0=
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.6
• Thus
(2.2.16)
• The x-direction momentum equation becomes:
(2.2.17)
• Re-arranging
(2.2.18)
• Noting that and that the total water column remains constant at the normal depth:
(2.2.19)
∂η∂x------ S0–=
gS0–1H----
τxb
ρ-----–=
τxb ρgHS0=
γ ρg≡H d0=
τxb γd0S0=
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.7
• Now we must apply a constitutive relationship to relate bottom stress to the depth aver-aged flow field.
(2.2.20)
• Thus the momentum statement becomes:
(2.2.21)
• Solving for
(2.2.22)
• For a Darcy-Weisbach bottom friction closure:
(2.2.23)
• Thus
(2.2.24)
τxb
ρ----- cfu
2=
gS0d0 cfu2
=
u
ugd0S0
cf--------------=
cf
fDW
8---------=
u8gd0S0
fDW------------------=
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.8
• In terms of flow rate per unit width:
(2.2.25)
• Given a flow per unit width , the normal flow depth can be computed as:
(2.2.26)
• Thus from a force balance perspective, we can characterize steady uniform flow asfollows:
• For a given flow rate per unit width, , the normal depth will have a velocityassociated with it which causes sufficient boundary shear resistance together withbottom roughness to balance the gravity forcing component.
qx ud0
8gd0S0
fDW------------------d0= =
qx
d0
qx2fDW
8gS0---------------� �� �� �
23---
=
qx d0
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.9
Derivation of Normal Flow Depth by Applying the Bernoulli Equation
• We consider the same stretch of channel with steady uniform flow as in the previousderivation between 2 sections.
• Based on these assumptions in the previous section listed on page 2.2.2:
(2.2.27)
(2.2.28)
• Also, we note that pressure at the free surface is atmospheric:
(2.2.29)
d1 d2 d0= =
u1 u2 u= =
p1 p2 0= =
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.10
• Applying the Bernoulli equation between sections 1 and 2 along a streamline on the freesurface:
(2.2.30)
(2.2.31)
(2.2.32)
• Dividing by L, the distance between points 1 and 2 in the x-direction
(2.2.33)
• We note that
(2.2.34)
p1
γ----- z1
u12
2g------
p2
γ----- z2
u22
2g------ hL1 2–
+ + +=+ +
0 zb1d0
u2
2g------+ + + 0 zb2
d0u
2
2g------ hL1 2–
+ + + +=
hL1 2–zb1
zb2–=
hL1 2–
L------------
zb1zb2
–
L-------------------=
S0
zb1zb2
–
L-------------------=
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.11
• Thus
(2.2.35)
• If we consider the head loss formula for pipe flow (a constitutive-type relationship):
(2.2.36)
• We now modify this relationship to apply to open channel flow by replacing by theHydraulic Diameter :
(2.2.37)
• For a wide open channel . In this case since the depth is the normaldepth.
(2.2.38)
hL1 2–
L------------ S0=
hL1 2–
fDWu2Ls
Dpipe2g--------------------=
Dpipe
DH
hL1 2–
fDWu2Ls
DH2g--------------------=
DH 4d= DH 4d0=
hL1 2–
fDWu2Ls
8d0g--------------------=
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.12
• Substituting for into our simplified Bernoulli equation:
(2.2.39)
• Noting that cos , we solve for .
(2.2.40)
• This solution is identical to that obtained from the Shallow Water equation derivation.
• This is logical since both the Shallow Water equation and the Bernoulli equationwere derived from the Navier-Stokes equation.
• Compatible constitutive relationships were used in both the Shallow Water andBernoulli equations. Both were derived from formulae developed for pipe flow.
• Thus, from a mechanical energy perspective, we can characterize steady uniform flow asfollows:
• The drop in potential energy due to the fall in elevation of the channel bottom mustbe exactly consumed by the energy dissipation through bottom friction and turbu-lence.
hL1 2–
fDWu2Ls
8d0g--------------------
1L--- S0=⋅
θ0LLs----- 1≅= u
u8gd0S0
fDW------------------=
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.13
Derivation of Normal Flow Depth by Applying Conservation of Momentum to a Fi-nite-Sized Control Volume
• We consider a finite sized control volume oriented along the channel bottom:
• Again we consider the same stretch of channel with steady uniform flow as in theprevious 2 derivations using the same assumptions listed on page 2.2.2.
• Let’s apply momentum conservation in integral form in the s-direction:
(2.2.41)∂∂t---- usρ V us ρU n⋅ Ad( ) Ts Ad
cs�� Bsρ Vd�
cv��+=
cs��+d�
cv��
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.14
• The vanishes since the problem is steady state.
• The net momentum flux into the CV equals and assuming cross sectionally averagedflow:
(2.2.42)
(2.2.43)
(2.2.44)
• Noting that where W = the width of the channel and = normal flow depth:
(2.2.45)
∂∂t----
usρU n⋅ A usρus Ad
A1
��–=d
A1
��
u– s2ρ Ad
A1
��=
u– s2ρA1=
A1 d0W= d0
usρU n⋅ A u– s2ρ=d d0W
A1
��
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.15
• The net momentum flux out of the CV equals
(2.2.46)
(2.2.47)
(2.2.48)
• Since :
(2.2.49)
• Total surface forces in the s-direction consist of pressure on both sides.
(2.2.50)
(2.2.51)
usρU n⋅ A usρus Ad
A2
��=d
A2
��
us2ρ Ad
A2
��=
us2ρA2=
A2 d0W=
usρU n⋅ A us2ρ=d d0W
A2
��
F1 s–
ρgd1A1
2------------------=
F2 s– ρgd2A2
2------------------–=
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.16
• Noting that
(2.2.52)
(2.2.53)
where W=width of channel (2.2.54)
• Thus pressure forces are:
(2.2.55)
(2.2.56)
• A total frictional surface force in the s-direction also exists:
(2.2.57)
(2.2.58)
d1 d2 d0= =
A1 A2 d0W= =
F1 s–
ρgd02W
2-----------------=
F2 s– ρgd0
2W
2-----------------–=
Ff s– τ0LsPw–=
Ff s– τ0Ls W 2d0+( )–=
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.17
• A total body force in the s-direction consists of
(2.2.59)
where (2.2.60)
• Thus
(2.2.61)
• Substituting into our momentum statement:
(2.2.62)
• This leads to
(2.2.63)
FB s– ρgALs θ0sin=
A Wd0=
FB s– ρgWd0Ls θ0sin=
0 us2ρd0W us
2ρd0Wρgd0
2W
2-----------------
ρgd02W
2-----------------– τ0Ls W 2d0+( )– ρgWd0Ls θ0sin+=–+
τ0Ls W 2d0+( ) ρgWd0Ls θ0sin=
CE 344 - Topic 2.2 - Spring 2003 - February 27, 2003 9:07 pm
p. 2.2.18
• Noting that sin and re-organizing:
(2.2.64)
• Again for a wide channel
(2.2.65)
• Also for modest slope channels ; thus the momentum conservation statementsimplifies to:
(2.2.66)
• Again this solution is identical to that obtained from the Shallow Water equations.
• To complete the solution, we would substitute in the appropriate constitutive relation-ship for and relate and and a friction factor.
• Thus we clearly see that we can use a variety of approaches, all rooted in the same basicphysics and using related constitutive relationships, to solve the problem.
θ0 S0 θ0cos=
τ0 γS0 θ0
Wd0
W 2d0+--------------------cos=
W d0»
Wd0
W 2d0+--------------------
Wd0
W---------- d0=≅
θ0 1≅cos
τ0 γd0S0=
τ0 us u≅ d0 to S0
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