Lecture Notes for BBE 5513: Watershed Engineering
CHAPTER FOUR
DETENTION PONDS AND HYDRAULIC ROUTING
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-1
Components of reservoirs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-1
Fundamental hydrologic relationship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-2
Additional relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-3
ELEVATION-STORAGE CURVE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-4
Computations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-4
Example problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-6
Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-6
ELEVATION-OUTFLOW CURVE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-7
Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-7
Weir flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-8
Orifice flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-9
Pipe flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-9
Example problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-11
RESERVOIR ROUTING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-14
Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-14
Water balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-15
Storage-indication curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-16
Solution procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-17
Occurrence of the peak outflow rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-21
Solution to example problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-21
CHANNEL ROUTING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-24
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-24
Conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-25
Muskingum method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-26
SEDIMENTATION PROCESSES IN DETENTION PONDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-27
Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-27
Trap efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-27
Characteristics of deposited sediment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-28
PHYSICAL PROPERTIES OF SEDIMENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-29
Density and specific gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-29
Primary particle size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-29
Particle size distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-30
SETTLING VELOCITY OF SEDIMENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Physical setting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Terminal velocity in quiescent water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Predictive relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
DISCRETE PARTICLE SEDIMENTATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Key relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Adjustment for dead space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Trapping of smaller particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
TRAP EFFICIENCY EXAMPLE PROBLEM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solution with no dead space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solution with 50% dead space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
PREDICTION OF SEDIMENTATION IN SEDCAD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
DEPOSITS Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
CSTRS Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
APPENDIX 4-A: MUSKINGUM METHOD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
APPENDIX 4-B: ADDITIONAL THEORY ON TRAP EFFICIENCIES . . . . . . . . . . . . . . . . . . . . . . .
APPENDIX 4-C: IMPACT OF TURBULENCE ON TRAP EFFICIENCY . . . . . . . . . . . . . . . . . . . . . .
Pond Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4-30
4-30
4-31
4-32
4-34
4-34
4-35
4-36
4-37
4-37
4-37
4-39
4-40
4-40
4-40
4-41
4-42
4-44
4-49
4-53
4-57
CHAPTER FOUR1
DETENTION PONDS AND HYDRAULIC ROUTING
For this reason a settling reservoir was built upstream from the intake, so that in it and between the river and the
conduit the water might come to rest and clarify itself. But in spite of this construction the water reaches the city in
a discolored condition, whenever there are heavy rains.
Water Supply of Rome, Frontinus (. 100 A.D.)
INTRODUCTION
Components of reservoirs
Reservoir schematic
EMERGENCY
SPILLWAY
f
D
P
ADDITIONAL
PERMANENT
POOL
STORAGE
IMPACT BASIN
S
Spillway components
Principal spillway
The principal spillway is used to provide for outflow for most runoff events.
The principal spillway is designed (1) to reduce peak flow rate for flood control and/or
(2) to enhance trap efficiencies of detention ponds.
Emergency spillway
The emergency spillway is used to provide for outflow for extreme runoff events.
The emergency spillway is designed to prevent overtopping of the dam or pond
embankment.
1
© 2011 Bruce N. Wilson and the Regents of the University of Minnesota. All Rights Reserved.
4-1
4-2
Storage components
Sediment storage
Sediment storage is the volume of the reservoir (or pond) that is used to store deposited
sediment.
Permanent pool or active storage
The permanent pool (also called active storage) is the volume of the reservoir that is used
for water supply and recreation or to enhance trap efficiency. It is equal to the difference
of the volume corresponding to the elevation of the principal spillway and the volume for
sediment storage.
Detention storage or flood control storage
Detention storage (also called flood control storage) is the volume of the reservoir that is
used to temporarily store water for reducing the peak outflow rate and/or for trapping
sediment. It is the difference between the cumulative volume at the crest of the
emergency spillway and the crest of the principal spillway.
Surcharge and freeboard storage
Surcharge and freeboard storage is the volume that lies between the top of the dam and
the crest of the emergency spillway.
Fundamental hydrologic relationship
Conservation of mass
The conservation of mass requires that the mass of water cannot be created or destroyed. For
a constant density fluid, this principle requires that the difference between the inflow and
outflow volumes must be stored in the system. For a reservoir, this concept is illustrated
below.
4-3
The water balance for a reservoir can be written as
where
I - Inflow hydrograph determined using hydrologic principles in Chapter 3
O - Outflow hydrograph (unknown)
S - Storage volume (unknown)
Closure problem
The water balance for a reservoir is one equation with two unknowns. A solution cannot be
obtained until an additional relationship(s) is provided.
Additional relationships
Level pool assumption
Additional relationships can be obtained if the water surface is assumed to be horizontal in
the reservoir as shown below.
4-4
For level pool routing, both S and O can be related to water surface elevation in the reservoir.
Therefore an elevation-storage curve can be defined corresponding to the elevation-flow rate
curve, or
Summary of equation set
To summarize, we have the following unknown variables:
* Outflow
* Storage
* Elevation
and following equations or relationships to solve for the three unknowns
* Water balance equation
* Elevation-storage curve
* Elevation-outflow or elevation-discharge curve
ELEVATION-STORAGE CURVE
Computations
Views of a reservoir
A contour map of a hypothetical reservoir is shown below.
4-5
The side view of the pond is shown below.
Mathematics
The cumulative volume, or storage, at elevation 1005 is simply defined as
and the volume at elevation 1010 as
or
4-6
In general, the volume for any ith elevation is then
Example problem
Let’s compute the elevation-storage values for the following surface area measurement for
the contour lines previously given.
Elevation
(feet)
Area
(acre)
1000
1
Ave
Area
(acre)
5
1010
10
1015
1020
)vol
(ac-ft)
5.0
15.0
15.0
7.5
5.0
37.5
20.0
5.0
100.0
65.0
5.0
325.0
52.5
30
100
Storage
Volume
(ac-ft)
0
3.0
1005
)EL
(feet)
152.5
477.5
Applications
Spillway elevations
Let's review the definitions of sediment storage, active storage and flood storage volumes.
For known sediment storage, active storage, and flood storage volumes, the elevations of the
crest of the principal and emergency spillways are be obtained as
4-7
Storage (ac-ft)
600
400
200
0
1000
1005
1010
Elevation (ft)
1015
1020
Solution of water balance equation
As discussed in greater depth later, the storage-elevation allows us to convert a difference in
water surface elevation to a change in storage in the reservoir. This concept is shown on the
previous graph.
ELEVATION-OUTFLOW CURVE
Background
Drop-inlet and emergency spillways
Consider the following reservoir with a drop-inlet principal spillway and an grassed
waterway emergency spillway.
For the drop-inlet spillway the diameter of riser is usually 1.5 times larger than the diameter
of the barrel. Computation procedures for determining the elevation-discharge for the
drop-inlet spillway are given in this section.
The emergency spillway is designed to move large flow rates for small change in water
4-8
surface elevation. Procedures to compute elevation-discharge for a grassed-waterway
emergency spillway are given in Chapter 6.
Types of flow in drop-inlet spillways
There are three possible flows in drop-inlet spillways:
* Weir flow
* Orifice flow
* Pipe flow
Equations for each of these flow conditions are given below.
Computational procedures
For selected water surface elevations in the reservoir,
* Calculate weir, orifice and pipe flow rates for each elevation
* Select design flow rate using the lowest flow rate
Weir flow
Physical flow condition
H
Equation
As discussed in Chapter 6, the general form of the weir flow equation can be derived using
specific energy concepts. Flow rate is predicted as
where
Qw =
Lw =
Cw =
H=
Weir flow in cfs
Weir length in feet = π Dr (circular pipe)
Weir coefficient . 3.0
Head above spillway inlet in feet
4-9
Orifice flow
Physical flow condition
H
Equation
The general form of the orifice flow equation can be obtained using Bernoulli's equation.
Orifice flow is predicted as
where
Qo =
Ar =
Co =
g=
H=
Orifice flow in cfs
Area of riser (ft2) = πDr2/4
Orifice coefficient . 0.6
Acceleration of gravity = 32.2 ft/s2
Head above the principal spillway in feet
Pipe flow
Physical conditions for free fall outlet
H
D
H’
H*
0.6D
Bernoulli's equation
4-10
Let's consider a point located at the water surface in the pond and at the exit point.
Bernoulli's equation can be written as
where
V1 =
P1 =
z1 =
V2 =
z2 =
P2 =
hf =
H* =
Surface velocity in the pond . 0
Pressure equal to atmospheric pressure
Elevation of water surface
Velocity at the outlet (barrel velocity)
Elevation of the outlet
Pressure at outlet, equal to the atmospheric pressure for free-fall condition
or the hydrostatic pressure for submerged condition (i.e., ρgy).
Friction losses
Elevation difference from the crest of the principal spillway to z2.
By canceling terms and evaluating elevation differences, we obtain
The main components of the friction losses are shown below.
ELEVATION OF WATER IN RESERVOIR
EN
ERG
Y G
RA
DE
LIN
E
V2
2g
V2
2g
L
ELBOW AND TRANSITION
Ke . 1 (entrance coefficient)
Kb
H’
0.6D
Entrance losses:
V2
2g
V2
KCL
2g
D
Friction losses
Ke
4-11
Bend losses:
Kb . 0.5 (bend coefficient)
Friction losses:
L=
Kc =
Db =
n=
Length of pipe in feet
Friction loss coefficient
Diameter of barrel in inches
Manning’s n (given by manufactures of pipes)
Solution
By substituting the relationships for friction losses into Bernoulli's equation, we obtain
or
Since Qp = V2 A2 = V2 Ab ,
Example problem
Problem statement
Determine the elevation-discharge curve for the dam with the contour lines previously given.
We will assume that the reservoir has the following principal spillway characteristics.
Elevation of inlet = 1010 ft
Elevation of invert of pipe at outlet = 1000 ft
Dr = 24 inches
Db = 18 inches
Total length = 50 feet
Cw = 3.0, Co = 0.60, Ke = 1.0, Kb = 0.50, n = 0.012
Outflow rates will be determined for water surface elevations of 1000, 1010, 1010.5, 1012,
1015, 1020, 1021 and 1022 feet.
4-12
A graphical illustration of the example problem is shown below.
Weir flow equation
As previously given,
By using the characteristics of the principal spillway, we obtain
Orifice flow equation
As previously given,
By using the characteristics of the principal spillway, we obtain
Pipe flow equation
As previously given,
where for the characteristics of the principal spillway we obtain the fixed elevation distance
as
4-13
and the friction loss coefficient as
and therefore the pipe flow equation can be written as
Computations
We obtain the following design discharges for the specified elevations.
Elevation
(feet)
1000.0
1005.0
Head
(feet)
-
Weir
flow
(cfs)
-
Orifice
flow
(cfs)
-
Pipe
flow
(cfs)
-
Design
flow
(cfs)
0.0
0.0
1010.0
0.0
0.0
0.0
23.6
0.0
1010.5
0.5
6.7
10.7
24.3
6.7
1012.0
1015.0
1020.0
1021.0
1022.0
2.0
5.0
10.0
11.0
12.0
53.3
210.7
596.1
687.7
783.6
21.4
33.8
47.8
50.2
52.4
26.1
29.4
34.2
35.1
36.0
21.4
29.4
34.2
35.1
36.0
Graphical representation is shown below.
4-14
Outflow (cfs)
40
Orifice
Weir
30
20
Pipe
Design Flow Rate
10
0
1010
1015
1020
1025
Elevation (ft)
RESERVOIR ROUTING
Background
Objectives
The objectives of the reservoir routing is to determine:
* Outflow hydrograph
* Water surface elevation
from the following information:
* Inflow hydrograph
* Elevation-storage curve
* Elevation-outflow curve
Example Problem Information
Data from previous problems will be used to illustrate the concepts. The inflow hydrograph
will be taken from the convolution solution given in Chapter 3 using a time step of Δt=0.25
h. The first ten points are shown below.
Time
(h)
Inflow
(cfs)
Time
(h)
Inflow
(cfs)
0.00
0.25
0.0
2.9
1.50
1.75
1831.2
1992.3
4-15
0.50
0.75
1.00
27.9
232.8
659.0
2.00
2.25
2.50
1867.7
1569.9
1154.0
We will use the elevation-storage curve computed in the example given earlier in the chapter.
Storage values at additional elevations were obtained from this curve and are shown below.
Elevation
(ft)
1000.0
1010.0
1010.5
1012.0
Storage
(ac-ft)
0.0
52.5
57.6
82.9
Elevation
(ft)
1015.0
1020.0
1021.0
1022.0
Storage
(ac-ft)
152.5
477.5
586.9
711.1
We will use the elevation-outflow curve obtained in an earlier example problem. An
emergency spillway has been placed at elevation 1020 and flow rate values are given below.
These values are from an example problem in Chapter 6. The total outflow for each
elevation is shown below.
Elevation
(ft)
1000.0
1010.0
1010.5
1012.0
1015.0
1020.0
1021.0
1022.0
P.S. Flow
(cfs)
0.0
0.0
6.7
21.4
29.4
34.2
35.1
36.0
E.S. Flow
(cfs)
0.0
0.0
0.0
0.0
0.0
0.0
77.3
276.1
Total Flow
(cfs)
0.0
0.0
6.7
21.4
29.4
34.2
112.4
312.1
To perform the routing, we need to also specify the water surface location at the start of the
runoff. For the example problem, we will use an initial condition of
Initial elevation of water surface = 1010 ft
Water balance
As previously given,
or
Physical interpretation of water balance equation:
4-16
By using the above numerical approximation, we obtain
which can be rearranged into known and unknown terms,
where 2S/t+O is called the storage indication value.
The above equation is of the following general form
If the storage indication and outflow values are known then
Storage-indication curve
Example problem values
For the example problem elevation-storage and elevation-outflow curves, the storageindication values are computed as
4-17
Elevation
(ft)
1000.0
Storage
(ac-ft)
0.0
Outflow
(cfs)
0.0
2S/Δt + O
(cfs)
1010.0
52.5
0.0
5082.0
1010.5
57.6
6.7
5581.9
1012.0
1015.0
1020.0
1021.0
1022.0
82.9
152.5
477.5
586.9
711.1
21.4
29.4
34.2
112.4
312.1
8046.1
14791.4
46256.2
56924.3
69146.6
0.0
The storage-indication value for elevation 1010.5 is computed as
The outflow-versus-storage-indication curve for the example problem is shown below.
250
Outflow (cfs)
200
150
100
50
0
0
20000
40000
Storage-Indication (cfs)
60000
Solution procedure
Step #1: Determine inflow hydrograph
The inflow hydrograph is calculated by methods discussed in previous chapters. The inflow
hydrograph is usually known at constant time increments as shown below.
Subscript
1
2
Time
0
t
Inflow
I1 ( 0.0)
I2 ( 2.9)
Outflow - Unknown
O1
O2
4-18
Subscript
3
4
:
:
i
:
:
Time
2t
3t
:
:
(i-1)t
:
:
Inflow
I3 ( 27.9)
I4 ( 232.8)
:
:
Ii
:
:
Outflow - Unknown
O3
O4
:
:
Oi
:
:
Step #2: Specify initial conditions
The initial (at t=0, i=1) water surface in the reservoir needs to be specified or known. It is
frequently taken as the elevation corresponding to the crest of the principal spillway. This
concept is illustrated below.
From a specified water surface elevation at t=0, the storage-indication value can be
determined from the above graph and the outflow can be determined from the following
graph.
250
Outflow (cfs)
200
150
100
50
0
1010
1015
Elevation (ft)
1020
4-19
The initial storage term can be estimated as
250
Outflow (cfs)
200
150
100
50
0
0
20000
40000
Storage-Indication (cfs)
60000
For the example problem, the initial water surface is at an elevation of 1010 feet. Therefore
we obtain an initial outflow of
O1 = 0.0
and an initial storage-indication value of
Step #3: Determine storage-indication value
At t=Δt (i=2), the storage-indication value can be determined from the water balance
equation as
The right-hand side of the equation. The values for I1 and I2 are known from the inflow
hydrograph. The last term can be computed from the initial storage-indication value and the
initial outflow, that is,
For the example problem, the storage-indication value is then obtained as
Step #4: Determine outflow rate
4-20
The outflow rate for a known storage-indication method can now be estimated from the plot,
Outflow (cfs)
40
30
20
10
0
5000
15000
25000
35000
Storage-Indication (cfs)
45000
For the example problem, we estimate an outflow value of
O2 = 0.02 cfs
Step #5: Determine next storage value
The storage value for the next time increment can then be calculated as
For the example problem, we obtain
Repeat steps for subsequent times
Steps 3, 4 and 5 are repeated until the inflow hydrograph is routed through the pond. For
example, at t=2Δt (i=3) the storage indication value can be determined using the water
balance equation as
and the outflow rate from the above curve as
O3 = 0.2 cfs
4-21
and the storage value for the next time step as
Occurrence of the peak outflow rate
Let’s consider the reservoir as at water surface increases with time as shown below.
Therefore the peak outflow rate corresponds to
and therefore the peak outflow rate can be defined as
or the peak outflow must hit the falling limb of the inflow hydrograph. This concept is
shown graphically below.
Solution to example problem
Review
We will solve the routing problem introduced in the previous section. The inflow
4-22
Outflow (cfs)
40
30
20
10
0
5000
15000
25000
35000
Storage-Indication (cfs)
45000
hydrograph is based on the convolution solution given in Chapter 3. The elevation-storage
and elevation-outflow are represented by the following storage-indication curve.
Solution
We will use the following table to solve for the outflow values.
Time
(hr)
Ii
(cfs)
Ii-1
(cfs)
2Si-1/Δt - Oi-1
(cfs)
2Si/Δt + Oi
(cfs)
Oi
(cfs)
0.00
0.25
0.50
0.0
2.9
27.9
0.0
0.0
2.9
0.0
5082.0
5084.9
5082.0
5084.9
5115.6
0.0
0.0
0.2
0.75
232.8
27.9
5115.3
5375.9
2.3
1.00
659.0
232.8
5371.3
6263.1
14.3
1.25
1309.4
659.0
6234.5
8202.8
21.8
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
1831.2
1992.3
1867.7
1569.9
1154.0
809.5
579.4
424.9
304.3
220.4
159.3
1309.4
1831.2
1992.3
1867.7
1569.9
1154.0
809.5
579.4
424.9
304.3
220.4
8159.2
11243.9
15008.3
18807.2
22182.2
24842.5
26742.0
28066.4
29005.9
29670.2
30129.9
11299.7
15067.4
18868.3
22244.7
24906.0
26806.0
28130.8
29070.6
29735.1
30194.9
30509.5
27.9
29.5
30.6
31.3
31.7
32.0
32.2
32.4
32.4
32.5
32.5
4-23
4.25
4.50
4.75
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.75
7.00
114.1
82.5
59.7
42.8
31.2
22.7
15.6
9.9
4.6
1.6
0.2
0.0
159.3
114.1
82.5
59.7
42.8
31.2
22.7
15.6
9.9
4.6
1.6
0.2
30444.4
30652.7
30784.2
30861.3
30898.6
30907.5
30896.2
30869.3
30829.6
30778.9
30719.9
30656.6
30717.9
30849.4
30926.5
30963.8
30972.7
30961.4
30934.5
30894.8
30844.1
30785.1
30721.7
30656.8
32.6
32.6
32.6
32.6
32.6
32.6
32.6
32.6
32.6
32.6
32.6
32.6
Computational values for i=4 are given below.
O4 = 2.3 cfs (from storage-indication graph)
and the computational values for i = 5 are
O5 = 14.3 cfs (from storage-indication graph)
Determine water surface elevation
The water surface elevation at any time (such as the value corresponding to the peak outflow
rate) can be determined from the elevation-outflow curve for a known outflow. This concept
is illustrated below.
P.S. Flow Rate (cfs)
4-24
40
36
32
28
24
20
16
12
8
4
0
1010
1014
Elevation (ft)
1018
The maximum water surface elevation in the reservoir is approximately 1018.3 ft for this
storm.
Effects of storage on outflow hydrograph
The effects of storage on the outflow hydrograph is shown schematically below.
CHANNEL ROUTING
Introduction
Definition
Channel routing is the prediction of the change in the hydrograph shape as it moves through
4-25
a channel or river.
Schematic illustration
Let’s consider the shape of the hydrograph between Points A and B along a river.
The hydrograph at Point B can be represented as
Conservation of mass
Consider the control volume of the entire reach
4-26
Mass Balance Equation
As previously given, the mass balance for a channel reach can be written as
where lateral flow is added at the end of the reach length. Similar to reservoir routing, we
have one equation and two unknowns (outflow and rate of change in storage).
Muskingum method
Let us consider the following idealized view of storage in a river
where
Prism storage = f(O) = KO
Wedge storage = f(I-O) = Kθ(I - O)
Total storage is the sum of prism and wedge storage, or
where K and θ are parameters that are assumed to be independent of time.
A lower limit of θ is usually taken as zero, corresponding to the simple linear reservoir
condition of S = KO.
If parameters K and θ are known, we have one equation with one unknown (outflow).
Numerical procedures for solving the combined solution of the mass balance and the
Muskingum equations are given in Appendix 4-A.
SEDCAD procedures for K and θ
4-27
For ungaged streams and rivers, simple SCS relationships are given in Appendix 4-A to
obtain the following estimates of K and θ
where L is the channel length in feet and V is the mean velocity in ft/s.
Procedures for estimating K and θ are given in Appendix 4-A. In Appendix 4-A, alternative
equations are given to estimate K and θ for ungaged streams. Although these equations are
superior, the above SCS equations still represent a reasonable first approximation.
SEDIMENTATION PROCESSES IN DETENTION PONDS
Applications
Sediment storage volume
Let’s consider the sediment storage volume required for the life of a structure.
Water quality
Sediment is an important contaminant influencing the quality of surface waters.
Sedimentation ponds are frequently designed to trap sediment. Here the performance of the
pond for a single storm is usually considered.
Trap efficiency
Definition
Trap efficiency is defined as
or
4-28
Uses
If the inflow mass of sediment and trap efficiency are known, the amount of mass deposited
can be computed as
and outflow mass estimated as
Empirical curve of trap efficiency
For long-term estimate of sediment storage, Brune's curve or similar curves are sometimes
used.
Characteristics of deposited sediment
Density
The density of deposited sediment (ρd) can be computed from the following equation
where ΔFc, ΔFm, and ΔFs are the fractions of clay, silt, and sand, respectively, and WC, Wm,
and WS are the equivalent densities (lbm /ft3) of clay, silt, and sand, respectively, as defined
below.
Type of Reservoir Operation
Sediment always submerged
Moderate to considerable drawdown
Reservoir normally empty
Riverbed sediment
Wc
26
35
40
60
Wm
70
71
72
73
Ws
97
97
97
97
4-29
Volume
Let's review how the principal spillway elevation is determined.
The deposited mass can be estimated as
and the volume of deposited sediment
PHYSICAL PROPERTIES OF SEDIMENT
Density and specific gravity
Range: 2.3 (coal) to 7.6 (galen).
Usual situation: ρs = 2.65 gm/cm3 and SG =2.65
Primary particle size
Definition of diameter
Sieve diameter: The sieve diameter is defined as the diameter of a sphere that is equal to the
length of the side of a square sieve opening.
Sedimentation diameter: The sedimentation diameter is the diameter of a sphere of the same
fall velocity under the same conditions (including density) of the particle.
USDA Classifications
4-30
Millimeters
> 2.0
2.00 to 1.000
1.00 to 0.500
0.50 to 0.250
0.25 to 0.100
0.10 to 0.050
0.05 to 0.002
< 0.002
Particle Size
Microns
> 2000
2000 to 1000
1000 to 500
500 to 250
250 to 100
100 to 50
50 to 2
<2
US Sieve
10
18
35
60
Class
Gravel
Very Course Sand
Course Sand
Medium Sand
Fine Sand
Very Fine Sand
Silt
Clay
Particle size distribution
Determination
Particle size distributions are usually defined using the fraction finer values defined as
Typically large particles are determined using sieve procedures and small particles are
determined using pipette analysis.
Fraction finer curve
Fraction finer values are frequently summarized using the following curve.
SETTLING VELOCITY OF SEDIMENT
Physical setting
Let's consider an isolated spherical particle in a homogenous fluid.
4-31
Terminal velocity in quiescent water
Newton's second law of motion
By evaluating the forces and momentum of the particle, we are able to determine
where ωs is the settling velocity. Drag force is defined as
where CD is the drag coefficient, U is the velocity of the fluid, Ap is the projected area of the
sphere equal to πd2/4, and gravity force is defined as
Simplification
At terminal velocity, the particle is no longer accelerating and therefore
For this condition,
or
4-32
where Ap = πd2/4.
By rearranging terms, we are able to obtain
Evaluation of drag coefficient
To evaluate the settling velocity, we need an estimate of the drag coefficient. Trends in
observed drag coefficients are given below.
Predictive relationships
Small particles (small Reynolds number)
A theoretical solution is possible for a sphere if the inertia terms are neglected in the
Navier-Stokes equation. This solution results in a drag coefficient defined as
where Re is the Reynolds number defined here as ωsd/ν, where ν is the kinematic viscosity.
Therefore the drag coefficient can be written as
By substituting this relationship for the drag coefficient into the drag-gravity force balance
equation previously given, we obtain
4-33
The settling velocity can be obtained as
By using SG =2.65 and the viscosity at 68EF, one obtains
where ωs is in ft/sec and d is in mm. In the WATER model, Stokes equation is used to
compute settling velocities for particles smaller than or equal to 0.01 mm.
Medium-sized particles
Brown developed a curve to determine the settling velocity for medium sized particles using
the observed drag coefficients. In the WATER model, the following equation is used to
approximate Brown’s curve:
where d is in units of mm and settling velocity in units of mm/s. The above equation was
developed for particle diameters between 0.01 mm < d < 1 mm. The settling velocity can be
computed in ft/s as
The settling velocity from the above equation is approximately equal to that obtained from
Stokes equation for d=0.01 mm.
If the settling velocity is known (ft/s), the above equation can be rearranged to solve for
log10(d) as
and therefore the particle diameter (mm) as
Large particles (large Reynolds number)
For large Reynolds numbers, the previous graph showed that the drag coefficient is
approximately constant which for a sphere CD . 0.40 and for a disk CD . 1.1.
Let's repeat the force balance at terminal velocity,
4-34
For a constant CD, terminal velocity can be calculated as
Rubey's solution is often used for large particles greater than 1 mm. This solution
corresponds to a CD . 2.1. Rubey's solution is
For SG = 2.65 and g = 32.2 ft/sec2
where ωs has units of ft/s and d has units of mm.
To allow a smooth transition between settling velocities from medium to large sized
particles, the WATER model uses the Brown curve at d=1 mm to set the coefficient for
settling velocity. We then obtain the following equation for particles larger than or equal to
1 mm
where settling velocity is units of ft/s and d in units of mm.
DISCRETE PARTICLE SEDIMENTATION
Overview
Introduction
We will discuss the simplest approach to predict trap efficiency using physically-based
processes. Although simple, it is illustrative of the processes of sedimentation in reservoirs
and ponds. The approach discussed below is frequently called the overflow rate method.
Assumptions
We will make the following assumptions:
* Steady-state flow
* Rectangular reservoir
* Discrete particle settling-no turbulence
* No resuspension
* Particles uniformly distributed at inlet
* Plug flow
4-35
The above assumptions are shown schematically below.
A theoretical framework for including turbulence and other factors is given in Appendix 4-C.
Key relationships
Critical fall velocity
The critical fall velocity is defined as the velocity of particle falling a height of h within the
detention time of the tank, or
All particles with settling velocities greater than ωc are trapped.
Detention time
Detention time is defined as the time that particles are in the basin, which for plug flow is
equal to the time for a particle to travel from the inlet to the outlet of the pond, or
Since
the detention time is therefore also defined as
where As is the surface area (i.e. As =LW). The above definition of detention time is
physically related to the time to displace the volume of the tank for a steady flow rate of Q.
Overflow rate
4-36
Let's review the definition of critical settling velocity previously given as
By using the above definition of detention time, we obtain
Overflow rate is defined as
which is equal to the critical fall velocity for rectangular basin with constant flow rate.
Calculation of critical particle diameter
From Stokes equation (ωs = 2.81d2), the diameter of the particle corresponding to the critical
fall velocity can be computed as
For critical velocities greater than 0.000282 ft/s, the inverse relationship corresponding to
medium-sized particles can be used to compute dc.
Adjustment for dead space
Definition
Dead space is the fraction or percentage of pond volume that does not mix with the inflow
concentration. Let’s consider the profile and top view of a detention pond shown below.
Adjustment in computations
The detention time of sediment for a pond with dead space would be computed as
4-37
where fd is the fraction of dead space.
The critical settling velocity is then calculated as
This equation can be used to compute the corresponding diameter using Stokes or similar
equation.
Dead space value
The amount of dead space is a function of the shape of the pond and the size of the pond
relative to the inflow volume. Typical values are between 10 and 30%.
Trapping of smaller particles
Thus far, equations have been developed to determine the critical fall velocity and
corresponding diameter. All particles larger than the critical diameter will be trapped.
However, smaller particles located nearer to the bed will also be trapped. This concept is
shown below.
Relationships and an example problem are given in Appendix 4-B to compute the trap
efficiency corresponding to the trapping of these smaller particles.
TRAP EFFICIENCY EXAMPLE PROBLEM
Problem statement
Determine fraction mass trapped associated with the critical fall velocity (1) for no dead
space and (2) for 50% dead space.
4-38
We are given the following inflow and basin characteristics:
Flow rate, Q= 5 cfs
Surface area, A = 1.0 acre
Rectangular reservoir as shown below.
The particle size distribution at the inlet is shown below.
100
Percent Finer
80
60
40
20
0
0.0001
0.001
Solution with no dead space
Calculate overflow rate
The overflow rate is defined as
0.01
Particle Diameter (mm)
0.1
1
4-39
Determine critical particle diameter
By using Stokes equation, the critical fall diameter can be computed as
Fraction finer value
By using the particle size distribution at the inlet, the fraction finer value corresponding to dc
is obtained
Trap efficiency
By assuming that all particles larger than d3 are trapped (and all smaller particles are
discharged), the trap efficiency of the basin is determined as
Solution with 50% dead space
Overflow rate, critical diameter and fraction finer
We can compute overflow rate, critical diameter, and fraction finer using the following
equations.
Trap efficiency
The trap efficiency is then computed as
4-40
PREDICTION OF SEDIMENTATION IN SEDCAD
Introduction
SEDCAD Options:
* Plug flow model (DEPOSITS model)
* Series of continuous stirred reactors model (CSTRS model)
Both use time-consuming algorithms that would be (practically) impossible to do with hand
computations.
DEPOSITS Model
Introduction
Plug flow model that extends the previous concepts to irregular-shaped pond and unsteady
flow rates.
Let's first review the plug flow assumption.
Only key components of the model can be discussed.
Irregular geometry
Let's consider the following pond profile.
4-41
For each elevation of the water surface, the average "fall" distance is obtained using a
volume-weighted depth. As a general rule, the average fall depth is approximately
two-thirds the maximum depth.
Unsteady flow
Let's consider a pond with and without a permanent pool volume.
We can compute detention time for each plug using inflow and outflow hydrographs.
After the detention time and fall distance have been determined for each plug, the trap
efficiency is computed using concepts given for the overflow rate.
CSTRS Model
Introduction
4-42
This approach allows mixing within the pond. A schematic illustrating the modeling
approach is given below.
The CSTRS model requires that the user enter the number of reactors in series. The number
of reactors is typically two for most detention ponds.
Mathematical approach
In the CSTRS model, the following mass balance is used for each reactor.
The most difficult term to estimate in the above equation is the deposition rate. In contrast to
DEPOSITS, the CSTRS model cannot use a single detention time for all particles within a
given inflow plug. A sophisticated algorithm has been developed that maintains a mass
balance for each plug.
4-43
APPENDIX 4-A: MUSKINGUM METHOD
Numerical approximation
Water balance
Let's review the water balance equation evaluated between ti-1 and ti previously given for
reservoir routing
which was evaluated numerically as
Muskingum formulation
The Muskingum relationships allow storage terms to be evaluated as
By substituting these relationships, we obtain
By rearranging terms, the Muskingum method can be evaluated as
where
Selection of time step
To avoid negative flows, Δt should be approximately equal to K, which, as shown later, is
often viewed as the travel time throughout the reach. The impact of a small value of Δt is that
the downstream point is evaluated before the upstream flow has reached it.
If we set K = Δt, the numerical coefficients can be evaluated by
4-44
Clearly for any positive value of θ less than 0.5, the routing coefficients are positive, and
negative flows are not possible. For θ =0.5, we obtain
that is, the outflow hydrograph is equal to the inflow hydrograph lagged by the travel time
through the reach. The upper limit of θ is usually taken as one-half.
K and θ estimates using observed data
Observed data requirements
*A
Gaging station - inflow hydrograph
* B Gaging station - outflow hydrograph
Review storage relationship
The Muskingum storage relationship is defined as
where
is a weighted discharge defined as
Possible computational steps
Determine observed storage:
From the mass balance, we know that
where inflow and outflow values are known, and therefore we can estimate Si from
observed data as follows
Compute weighted discharge:
4-45
We need to assume a θ value. With this assumed value, we can compute the weighted
discharge for storage Si as
Plot different Si versus for observed
values:
If data plots as a straight line, determine K from slope. If data does not plot as a straight line,
guess a new θ and repeat computational steps.
SEDCAD estimates of K and θ
Geometry assumptions
Let's consider the following water surface profile.
where the geometry of the upstream and downstream channels are assumed to be
approximated by rectangular channel of width W.
Evaluation of K
4-46
The Muskingum method uses the following relationship for prism storage
For the geometry given above, prism storage would be defined as
By setting the prism storage of Muskingum method equal to the prism storage for the SCS
geometry, we obtain
where V is the uniform flow velocity defined as
Evaluation of θ
The Muskingum method uses the following relationship for wedge storage
For the SCS's geometry previously given, wedge storage would be defined as
By setting the wedge storage of Muskingum method equal to the wedge storage for the SCS
geometry, we obtain
We will defined a wedge velocity, U, as
which is conceptually similar to the kinematic wave velocity defined as dQ/dA. By using the
wedge velocity and the definition of K previously derived, we are able to determine θ as
By using empirical data, the SCS recommends that the ratio of uniform and wedge velocities
be evaluated as
Muskingum parameter θ can then be estimated as
4-47
Alternate estimates of K and θ
Muskingum parameters can also be estimated using linearized forms of the equation of
motion using appropriate numerical approximations. These parameters have been derived
from the dynamic wave and diffusion wave models. In the interest of simplicity, the diffusion
wave results are given below.
The Muskingum parameter K is estimated as travel time through the reach for a velocity
equal to the kinematic wave speed, or
where cko is the kinematic wave speed (ck = mQ/A), Qo and Ao are representative flow rates
and cross-sectional areas, respectively, and m is a rating coefficient that is equal to 5/3 for
Manning's equation.
The parameter θ is defined as
where So is the bed slope and
is the flow depth.
4-48
APPENDIX 4-B: ADDITIONAL THEORY ON TRAP EFFICIENCIES
Trapping of smaller particles
As discussed in the chapter, the critical velocity is computed as
Since smaller particles closer to the bed will also be trapped, additional relationships need to
be developed to consider these particles.
"Critical velocity" for smaller depths
Let's consider the trapping of smaller particles by subdividing the entire depth into smaller
layers and computing the equivalent critical velocity for each of these layers as shown below.
where h1 through hn correspond to fall distances less than the total distance h. These layers
are illustrated below.
For each velocity, a diameter can be estimated by Stokes equation,
Smaller depths can be related to these "critical velocities" as
Influence on trap efficiency
Trap efficiency definition
Let's review the definition of trap efficiency
4-49
where inflow mass will be taken as
where C is the effective concentration of sediment at the inlet.
Mass deposited for critical velocity
As previously given:
Total mass deposited including smaller particles
Let's consider the smaller particles trapped for a fall depth of h1. This mass would be defined
as
The additional mass, not included in the critical velocity computation, smaller than critical
diameter is obtained as
This logic can be applied to other depths to obtain the following amount of trapped mass as
Graphic illustration of this concept is shown below.
By rearrange terms, we obtain
By using the relationship that
, we can further simplify as
4-50
The trap efficiency can then be calculated as
where ωc is defined for rectangular basin as
For a continuous curve, the trap efficiency is obtained as
Example problem
Problem Statement
Let's resolve the trap efficiency problem given in the chapter accounting for the trapping of
sediment that has a settling velocities less than ωc.
Solution obtained in chapter problem
The critical fall velocity, diameter and fraction finer mass were obtained as
Trapping of smaller particles
Let's divide attached particle size distribution into 0.04 ΔFF and used the midpoint of each
range to determine settling velocity for the particle size distribution used in the trapefficiency example problem.
4-51
100
Percent Finer
80
60
40
20
0
0.0001
0.001
0.01
0.1
Particle Diameter (mm)
1
For this particle size distribution, we obtain the following values.
Diameter
(mm)
)FF
0.00075
0.0014
0.0021
0.0030
0.0042
0.0056
G
0.04
0.04
0.04
0.04
0.04
0.04
0.24
Ti
(ft/sec)
0.00000158
0.00000551
0.0000124
0.0000253
0.0000496
0.0000881
Estimated trap efficiency
By using the following equation,
we obtain
Hence the trapping efficiency is approximately 82 percent.
0.0005
0.0019
0.0043
0.0088
0.0173
0.0306
0.0634
4-52
APPENDIX 4-C: IMPACT OF TURBULENCE ON TRAP EFFICIENCY
Mass balance
Let’s consider a control volume moving with a velocity of the mean velocity for onedimensional flow.
s A 
 (s A)
z
z
 t
C 
C
 (  t
)z
z z
z
A
Mean Velocity 
z
Change in Mass Storage 
Settling Flux  s A
Turbulent Flux   t
Q
HW
 (CAz)
t
C
z
For an observer moving with the mean flow, the conservation of mass for one-dimensional
flow can be written as
(C-1)
where C is the concentration of sediment (time-averaged), ωs is the settling velocity of
sediment, and εt is the turbulent diffusion coefficient, assumed to be independent of depth.
As discussed by Wilson and Barfield (1986) and others, the turbulent diffusion coefficient is
independent of depth for a parabolic velocity profile.
There are two boundary conditions, corresponding to the flux at the surface and the bed.
Since there is no sediment movement at the surface, the flux is defined as
(C-2a)
At the bed, the flux is equal to the detachment rate, or
(C-2b)
where e is the detachment or erosion rate written as a function of time. The initial conditions
can be written as
4-53
(C-2c)
where Co is the initial condition, which can be a function of depth.
Dobbins (1944) and Camp (1944) obtained an analytical solution to the conservation of mass
for the boundary and initial conditions using separation-of-variables techniques. This
solution results in a powerful solution using dimensionless parameters. The sediment
concentration can be written as the product of arbitrary functions of z and t, or
(C-3)
where Z and T are function to be determined. For the assumptions of negligible bed erosion
(i.e., e = 0) and uniform initial concentrations (i.e., Co … f(z)), the solution is
(C-4)
where Np is the dimensionless sedimentation Peclet number defined as
(C-5)
which is a measure of the movement of particles by settling relative to that by turbulent
diffusion. In Equation C-4, the Zi and Ji are defined as
(C-6a)
and
(C-6b)
The parameter βi is obtained from the separation-of-variable technique as
(C-7)
which can be solved using iterative or graphical techniques. There is an infinite number of
positive values of βi. By using the above equation, the last term of Ji can be written as
(C-8)
where the following trigonometric identity was also used
(C-9)
where x is defined for this problem as 2βiNp/(β2i -N2p). Positive values are used for x in the
quadrants of -π/2 < x < π/2 and negative values for x in the quadrants defined by π/2 < x <
3π/2. The variable Ji is then simplified as
4-54
(C-10)
An analytical solution for trap efficiency can now be derived from Equation C-4 by using the
difference in the concentrations, averaged over depth, from when the sediment enters the
pond (t = 0) and when it leaves (t = Td). The initial mass of sediment per unit area is simply
Co H. The final mass of sediment per unit area can be obtained from Equation C-4 using t =
Td and integrating over the flow depth. Mathematically, it is defined as
(C-11)
where Nh is the dimensionless Hazen number defined as
(C-12)
which is a measure of the settling velocity of a particle of interest relative to that of a particle
that, under quiescent settling, would fall the entire flow depth within Td. This latter velocity
is called the critical fall velocity (ωc).
The trap efficiency is defined as the difference of inflow and outflow mass divided by the
inflow mass, or 1 - Mo/CoH. By evaluating the integral in Equation C-11, trap efficiency is
then obtained as
(C-13)
A graphical solution to Equation C-13 is shown in the following figure.
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1.0
0.8
Trap Efficiency
Np = 1
Np = 10
0.6
0.4
Np = 0.1
0.2
0.0
0.0
0.5
1.0
1.5
2.0
Hazen Number
Trap efficiency is uniquely defined by the Hazen number and the sedimentation Peclet
number. Let’s consider two basins, both of which have flows that are well approximated by
the assumptions leading to Equation C-13. The trap efficiencies of both basins can then be
represented by the above results and are identical for the same Hazen number and
sedimentation Peclet number:
4-56
Pond Assignment
Due Date:
Format
All homework assignments must be done in a neat and organized manner. Clearly identify
the solution in the problems with several mathematical steps by underlining or boxing in
your answer. Points will be deducted from problems that are sloppy and difficult to follow.
Problems
(15 points)
1. Assume that the following information has been determined from a topographic map.
Elevation
(ft)
1592
1595
1600
1605
1607
1608
1610
Area
(acres)
0.0
0.8
3.2
4.8
6.5
8.9
12.0
Calculate and plot the elevation-storage curve using this information.
(5 points)
2. For the pond in Problem #1, assume that the crest of the principal spillway is located at an
elevation of 1600 feet and the crest of the emergency spillway is located at an elevation of
1608 feet. How much of the pond volume is available for deposition and active storage
(water supply, recreation, etc.)? How much of the pond volume is available for flood
storage? Indicate these values on your elevation-storage plot.
(20 points)
3. For the pond in Problem #1, assume that a principal spillway will be installed with the
following characteristics:
Driser = 27 inches;
Dbarrel = 24 inches
Co = 0.6
Cw =3.0;
Ke = 1.0; Kb = 0.5; n = 0.025; L = 120 feet
Elevation drop from crest to invert of the outlet pipe is 12 feet.
Calculate the elevation-discharge curve using the same elevation values of Problem #1. The
elevation of the principal spillway is 1600 feet.
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(20 points)
4. Further assume that the flow rate through the emergency spillway has been calculated by a
colleague in your firm as
Elevation
ft
1608
1610
EmergencySpillway
Flow Rate - cfs
0.0
50.0
Add this flow to the discharge obtained in Problem #3 and calculate storage-indication
values using a time increment of 0.5 hr. Plot the storage indication values (with respect to
discharge).
(20 points)
5. Using the information obtained in Problem #4, calculate the first ten points of the outflow
hydrograph using the storage-indication method. Assume the following inflow hydrograph:
Time
(hr)
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
Inflow
(cfs)
0
100
300
750
300
100
50
0
0
0
Assume that the initial water surface elevation is at the crest of the principal spillway.
(20 points)
6. Consider a rectangular basin of a depth of 4 feet, a width of 50 feet and a length of 130 feet.
Assume that 15% of the volume is dead space. For a constant flow rate of 4 cfs, estimate the
trap efficiency for each of the two particle size distributions shown below.
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100
Size #1
Percent Finer
80
60
40
Size #2
20
0
0.0001
0.001
0.01
0.1
Particle Diameter (mm)
1
10