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‫روندیابی‬
‫مخزن‬ ‫سیل‬
‫عمران‬ ‫مهندسی‬ ‫دانشکده‬
-
‫فنی‬ ‫پردیس‬
‫تهران‬ ‫دانشگاه‬
Reservoir Flood Routing
List of content
• Introduction
• Upstream boundary conditions
• Downstream boundary conditions
• Flood routing
• Summary and Conclusion
• Resources
Reservoir Flood Routing List of content 1
Introduction
• Statement of the problem
• Research objectives
• Report structure
Hydrologic river routing (Muskingum
Method)
Wedge storage in reach
KQ
S 
Prism
)
(
Wedge Q
I
KX
S 

K = travel time of peak through the reach
X = weight on inflow versus outflow (0 ≤ X ≤
0.5)
X = 0  Reservoir, storage depends on outflow,
no wedge
X = 0.0 - 0.3  Natural stream
)
( Q
I
KX
KQ
S 


]
)
1
(
[ Q
X
XI
K
S 


Muskingum Method (Cont.)
]
)
1
(
[ Q
X
XI
K
S 


t
Q
Q
t
I
I
S
S
j
j
j
j
j
j 









2
2
1
1
1
6
]}
)
1
(
[
]
)
1
(
{[ 1
1
1 j
j
j
j
j
j Q
X
XI
Q
X
XI
K
S
S 





 


j
j
j
j Q
C
I
C
I
C
Q 3
2
1
1
1 

 

t
X
K
t
X
K
C
t
X
K
KX
t
C
t
X
K
KX
t
C



















)
1
(
2
)
1
(
2
)
1
(
2
2
)
1
(
2
2
3
2
1
Recall:
Combine:
If I(t), K and X are known, Q(t) can be calculated using
above equations
Muskingum - Example
• Given:
 Inflow hydrograph
 K = 2.3 hr, X = 0.15, t = 1
hour, Initial Q = 85 cfs
• Find:
 Outflow hydrograph using
Muskingum routing method
Period Inflow
(hr) (cfs)
1 93
2 137
3 208
4 320
5 442
6 546
7 630
8 678
9 691
10 675
11 634
12 571
13 477
14 390
15 329
16 247
17 184
18 134
19 108
20 90
7
5927
.
0
1
)
15
.
0
1
(
3
.
2
*
2
1
)
15
.
0
1
(
*
3
.
2
*
2
)
1
(
2
)
1
(
2
3442
.
0
1
)
15
.
0
1
(
3
.
2
*
2
15
.
0
*
3
.
2
*
2
1
)
1
(
2
2
0631
.
0
1
)
15
.
0
1
(
3
.
2
*
2
15
.
0
*
3
.
2
*
2
1
)
1
(
2
2
3
2
1



































t
X
K
t
X
K
C
t
X
K
KX
t
C
t
X
K
KX
t
C
Muskingum – Example (Cont.)
j
j
j
j Q
C
I
C
I
C
Q 3
2
1
1
1 

 

Period Inflow C1Ij+1 C2Ij C3Qj Outflow
(hr) (cfs) (cfs)
1 93 0 0 0 85
2 137 9 32 50 91
3 208 13 47 54 114
4 320 20 72 68 159
5 442 28 110 95 233
6 546 34 152 138 324
7 630 40 188 192 420
8 678 43 217 249 509
9 691 44 233 301 578
10 675 43 238 343 623
11 634 40 232 369 642
12 571 36 218 380 635
13 477 30 197 376 603
14 390 25 164 357 546
15 329 21 134 324 479
16 247 16 113 284 413
17 184 12 85 245 341
18 134 8 63 202 274
19 108 7 46 162 215
20 90 6 37 128 170
8
0
100
200
300
400
500
600
700
800
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Time (hr)
Discharge
(cfs)
C1 = 0.0631, C2 = 0.3442, C3 =
0.5927
Unsteady Flow Routing in Open
Channels
• Flow is one-dimensional
• Hydrostatic pressure prevails and vertical accelerations
are negligible
• Streamline curvature is small.
• Bottom slope of the channel is small.
• Manning’s equation is used to describe resistance effects
• The fluid is incompressible
Continuity Equation
dx
x
Q
Q



x
Q


t
Adx

 )
(

 


.
.
.
.
.
0
s
c
v
c
dA
V
d
dt
d


Q = inflow to the control volume
q = lateral inflow
Elevation View
Plan View
Rate of change of flow
with distance
Outflow from the C.V.
Change in mass
Reynolds transport theorem
Momentum Equation
• From Newton’s 2nd Law:
• Net force = time rate of change of momentum


 


.
.
.
.
.
s
c
v
c
dA
V
V
d
V
dt
d
F 

Sum of forces on
the C.V.
Momentum stored
within the C.V
Momentum flow
across the C. S.
Forces acting on the C.V.
• Fg = Gravity force due to
weight of water in the C.V.
• Ff = friction force due to
shear stress along the
bottom and sides of the C.V.
• Fe = contraction/expansion
force due to abrupt changes
in the channel cross-section
• Fw = wind shear force due
to frictional resistance of
wind at the water surface
• Fp = unbalanced pressure
forces due to hydrostatic
forces on the left and right
hand side of the C.V. and
pressure force exerted by
banks
Elevation View
Plan View
Momentum Equation


 


.
.
.
.
.
s
c
v
c
dA
V
V
d
V
dt
d
F 

Sum of forces on
the C.V.
Momentum stored
within the C.V
Momentum flow
across the C. S.
0
)
(
1
1 2



















f
o S
S
g
x
y
g
A
Q
x
A
t
Q
A
Momentum Equation(2)
0
)
(
1
1 2



















f
o S
S
g
x
y
g
A
Q
x
A
t
Q
A
0
)
( 










f
o S
S
g
x
y
g
x
V
V
t
V
Local
acceleration
term
Convective
acceleration
term
Pressure
force
term
Gravity
force
term
Friction
force
term
Kinematic Wave
Diffusion Wave
Dynamic Wave
Momentum Equation (3)
f
o S
S
x
y
x
V
g
V
t
V
g











1
Steady, uniform flow
Steady, non-uniform flow
Unsteady, non-uniform flow
Applications of different forms of momentum
equation
• Kinematic wave: when gravity forces and friction forces
balance each other (steep slope channels with no back
water effects)
• Diffusion wave: when pressure forces are important in
addition to gravity and frictional forces
• Dynamic wave: when both inertial and pressure forces
are important and backwater effects are not negligible
(mild slope channels with downstream control, backwater
effects)
0
)
( 










f
o S
S
g
x
y
g
x
V
V
t
V
16
Kinematic Wave
• Kinematic wave celerity, ck is the speed of movement of
the mass of a flood wave downstream
 Approximately, ck = 5v/3 where v = water velocity
Muskingum-Cunge Method
• A variant of the Muskingum method that has a more physical hydraulic
basis
• This is what Dean Djokic has used in the Brushy Creek HEC-HMS
models
• 𝐾 =
Δ𝑥
𝑐𝑘
, where Δx = reach length or an increment of this length
• 𝑋 =
1
2
1 −
𝑄
𝐵𝑐𝑘𝑆0Δ𝑥
, where B = surface width, S0 is the bed slope
Reach SBR_080 Downstream
of Dam 7
How do we route the
flow through Reach
SBR_080?
Longitudinal profile for reach
SBR_080
0.000
8 1
1545 ft
Cross-Section for SBR_080
Station Elevation
0 797.6057
118.1 790.0711
236.2 781.6702
284 777.0652
304 777.0652
323.42 783.5712
344.26 789.859
365.1 795.4788
775.00
780.00
785.00
790.00
795.00
800.00
0.00 100.00 200.00 300.00 400.00
Elevation
above
datum
(ft)
Distance (ft)
Cross-Section
Routing in stream reach downstream
of Dam 7

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ReservoirFlowRouting.pptx

  • 1. ‫روندیابی‬ ‫مخزن‬ ‫سیل‬ ‫عمران‬ ‫مهندسی‬ ‫دانشکده‬ - ‫فنی‬ ‫پردیس‬ ‫تهران‬ ‫دانشگاه‬
  • 3. List of content • Introduction • Upstream boundary conditions • Downstream boundary conditions • Flood routing • Summary and Conclusion • Resources Reservoir Flood Routing List of content 1
  • 4. Introduction • Statement of the problem • Research objectives • Report structure
  • 5. Hydrologic river routing (Muskingum Method) Wedge storage in reach KQ S  Prism ) ( Wedge Q I KX S   K = travel time of peak through the reach X = weight on inflow versus outflow (0 ≤ X ≤ 0.5) X = 0  Reservoir, storage depends on outflow, no wedge X = 0.0 - 0.3  Natural stream ) ( Q I KX KQ S    ] ) 1 ( [ Q X XI K S   
  • 6. Muskingum Method (Cont.) ] ) 1 ( [ Q X XI K S    t Q Q t I I S S j j j j j j           2 2 1 1 1 6 ]} ) 1 ( [ ] ) 1 ( {[ 1 1 1 j j j j j j Q X XI Q X XI K S S           j j j j Q C I C I C Q 3 2 1 1 1      t X K t X K C t X K KX t C t X K KX t C                    ) 1 ( 2 ) 1 ( 2 ) 1 ( 2 2 ) 1 ( 2 2 3 2 1 Recall: Combine: If I(t), K and X are known, Q(t) can be calculated using above equations
  • 7. Muskingum - Example • Given:  Inflow hydrograph  K = 2.3 hr, X = 0.15, t = 1 hour, Initial Q = 85 cfs • Find:  Outflow hydrograph using Muskingum routing method Period Inflow (hr) (cfs) 1 93 2 137 3 208 4 320 5 442 6 546 7 630 8 678 9 691 10 675 11 634 12 571 13 477 14 390 15 329 16 247 17 184 18 134 19 108 20 90 7 5927 . 0 1 ) 15 . 0 1 ( 3 . 2 * 2 1 ) 15 . 0 1 ( * 3 . 2 * 2 ) 1 ( 2 ) 1 ( 2 3442 . 0 1 ) 15 . 0 1 ( 3 . 2 * 2 15 . 0 * 3 . 2 * 2 1 ) 1 ( 2 2 0631 . 0 1 ) 15 . 0 1 ( 3 . 2 * 2 15 . 0 * 3 . 2 * 2 1 ) 1 ( 2 2 3 2 1                                    t X K t X K C t X K KX t C t X K KX t C
  • 8. Muskingum – Example (Cont.) j j j j Q C I C I C Q 3 2 1 1 1      Period Inflow C1Ij+1 C2Ij C3Qj Outflow (hr) (cfs) (cfs) 1 93 0 0 0 85 2 137 9 32 50 91 3 208 13 47 54 114 4 320 20 72 68 159 5 442 28 110 95 233 6 546 34 152 138 324 7 630 40 188 192 420 8 678 43 217 249 509 9 691 44 233 301 578 10 675 43 238 343 623 11 634 40 232 369 642 12 571 36 218 380 635 13 477 30 197 376 603 14 390 25 164 357 546 15 329 21 134 324 479 16 247 16 113 284 413 17 184 12 85 245 341 18 134 8 63 202 274 19 108 7 46 162 215 20 90 6 37 128 170 8 0 100 200 300 400 500 600 700 800 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Time (hr) Discharge (cfs) C1 = 0.0631, C2 = 0.3442, C3 = 0.5927
  • 9. Unsteady Flow Routing in Open Channels • Flow is one-dimensional • Hydrostatic pressure prevails and vertical accelerations are negligible • Streamline curvature is small. • Bottom slope of the channel is small. • Manning’s equation is used to describe resistance effects • The fluid is incompressible
  • 10. Continuity Equation dx x Q Q    x Q   t Adx   ) (      . . . . . 0 s c v c dA V d dt d   Q = inflow to the control volume q = lateral inflow Elevation View Plan View Rate of change of flow with distance Outflow from the C.V. Change in mass Reynolds transport theorem
  • 11. Momentum Equation • From Newton’s 2nd Law: • Net force = time rate of change of momentum       . . . . . s c v c dA V V d V dt d F   Sum of forces on the C.V. Momentum stored within the C.V Momentum flow across the C. S.
  • 12. Forces acting on the C.V. • Fg = Gravity force due to weight of water in the C.V. • Ff = friction force due to shear stress along the bottom and sides of the C.V. • Fe = contraction/expansion force due to abrupt changes in the channel cross-section • Fw = wind shear force due to frictional resistance of wind at the water surface • Fp = unbalanced pressure forces due to hydrostatic forces on the left and right hand side of the C.V. and pressure force exerted by banks Elevation View Plan View
  • 13. Momentum Equation       . . . . . s c v c dA V V d V dt d F   Sum of forces on the C.V. Momentum stored within the C.V Momentum flow across the C. S. 0 ) ( 1 1 2                    f o S S g x y g A Q x A t Q A
  • 14. Momentum Equation(2) 0 ) ( 1 1 2                    f o S S g x y g A Q x A t Q A 0 ) (            f o S S g x y g x V V t V Local acceleration term Convective acceleration term Pressure force term Gravity force term Friction force term Kinematic Wave Diffusion Wave Dynamic Wave
  • 15. Momentum Equation (3) f o S S x y x V g V t V g            1 Steady, uniform flow Steady, non-uniform flow Unsteady, non-uniform flow
  • 16. Applications of different forms of momentum equation • Kinematic wave: when gravity forces and friction forces balance each other (steep slope channels with no back water effects) • Diffusion wave: when pressure forces are important in addition to gravity and frictional forces • Dynamic wave: when both inertial and pressure forces are important and backwater effects are not negligible (mild slope channels with downstream control, backwater effects) 0 ) (            f o S S g x y g x V V t V 16
  • 17. Kinematic Wave • Kinematic wave celerity, ck is the speed of movement of the mass of a flood wave downstream  Approximately, ck = 5v/3 where v = water velocity
  • 18. Muskingum-Cunge Method • A variant of the Muskingum method that has a more physical hydraulic basis • This is what Dean Djokic has used in the Brushy Creek HEC-HMS models • 𝐾 = Δ𝑥 𝑐𝑘 , where Δx = reach length or an increment of this length • 𝑋 = 1 2 1 − 𝑄 𝐵𝑐𝑘𝑆0Δ𝑥 , where B = surface width, S0 is the bed slope
  • 19. Reach SBR_080 Downstream of Dam 7 How do we route the flow through Reach SBR_080?
  • 20. Longitudinal profile for reach SBR_080 0.000 8 1 1545 ft
  • 21. Cross-Section for SBR_080 Station Elevation 0 797.6057 118.1 790.0711 236.2 781.6702 284 777.0652 304 777.0652 323.42 783.5712 344.26 789.859 365.1 795.4788 775.00 780.00 785.00 790.00 795.00 800.00 0.00 100.00 200.00 300.00 400.00 Elevation above datum (ft) Distance (ft) Cross-Section
  • 22. Routing in stream reach downstream of Dam 7