NATIONAL ENGINEERING IIAI~IBOOK
SECTION
4
HYDROLOGY
CHAPTER 17.
FMOD ROUTING
Victor Mockus
Hydraulic Engineer
Revisions by
Wendell Styner
1972
Revised and r e p r i n t e d , 1972
NM Notice 4-102,
AUgUSt
1972
NATIONAL ENGINEERING HANDBOOK
SECTION
4
HYDROLOGY
.
CHAPTER 17
FLOOD ROUTING
Contents
. . . . . . . . . . . . . . . . . . . . . . . . . . 17-1
. . . . . . . . . . . . . . . 17-2
. . . . . . . . . . . . . . . . . . . . . . . . . 17-2
. . . . . . . . . . . . . . . . . 17-3
Elevation-storage and Elevation-discharge relationships . . . . 17-3
Elevation-storage relationships for reservoirs. . . . . . . . 17-3
Elevation-discharge relationships for reservoirs . . . . . . 17-7
Storage-discharge relationships for reservoirs . . . . . . . 17-7
Elevation. stage. storage. discharge relationships
for streams . . . . . . . . . . . . . . . . . . . . . . . 17-7
Reservoir Routing Methods . . . . . . . . . . . . . . . . . . .
The continuity equation . . . . . . . . . . . . . . . . . . .
Methods and examples . . . . . . . . . . . . . . . . . . . .
Mass curve method: numerical version . . . . . . . . . . .
Example 17-1 . . . . . . . . . . . . . . . . . . . . .
Example 17-2 . . . . . . . . . . . . . . . . . . . . .
Mass curve method: direct version . . . . . . . . . . . .
Mass curve method: graphical version . . . . . . . . . . .
Example 17-3 . . . . . . . . . . . . . . . . . . . . .
Storage indication method . . . . . . . . . . . . . . . .
Example 17-4 . . . . . . . . . . . . . . . . . . . . .
Example 17-5 . . . . . . . . . . . . . . . . . . . . .
Storage indication method as used in the SCS
electronic computer program . . . . . . . . . . . . . .
Culp's method . . . . . . . . . . . . . . . . . . . . . .
Example 17-6 . . . . . . . . . . . . . . . . . . . . .
Short cuts for reservoir routings . . . . . . . . . . . .
Introduction
SCS electronic computer program
References
hmmary of chapter contents
L
L
.....................
.................
.........................
..........
.....................
....................
...................
......
Channel Routing Methods
Theory of the convex method
Discussion
Some useful realtionships and procedures
Determination o f K
Determination of C
Determination of At
Procedure for routing through any reach length
Variability of routing parameters. selection of velocity V
NM Notice 4.102.
August 1972
.
CONTENTS cont'd
.
. . . . . . . . . . . . . . . 17-53
Examples: Convex routing methods
Example 17-7
Example 17-8
Example 17-9
Example17-10
Example 17-11
Effects of transmission l o s s e s on routed flows
Routing through a system of channels
. . . . . . . . . . . . . . . . . . . . . . 17-54
. . . . . . . . . . . . . . . . . . . . . .17-56
. . . . . . . . . . . . . . . . . . . . . .17-57'
. . . . . . . . . . . . . . . . . . . . . 17-59
. . . . . . . . . . . . . . . . . . . . . 17-61
. . . . . . . . 17-66
. . . . . . . . . . . . 17-66
. . . . . . . . . . . .
Unit-hydrograph Routing Methods
Basic equations
Effects of storm duration and time of concentration
Elimination of Tp
Working equations f o r s p e c i a l cases
Examples
Use of equation 17-40
Example 17-12
Use of equation 17-43
Example 17-13
Use o f equation 17-43 on l a r g e watersheds
Use of equations 17.48. 17.50. and 17-52
Example 17-14
Example 17-15
Example 17-16
Example 17-17
Discussion
.....
......................
.............
. . . . . . . . . . . . . . . . . . . . . .
a
*
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
17-82
17-82
17-83
17-84
17-85
17-86
17-86
17-86
................
. . . . . . . . . . . . . . . . . . . . . 17-86
.................
. . . . . . . . . . . . .. .. .. .. .. .. .. .. .. 17-86
17-86
. . . . . . . . . 17-89
. . . . . . . . . . . . . . . . . . . . . 17-89
. . . . . . . . . . . . . . . . . . . . . 17-89
. . . . . . . . . . . . . . . . . . . . . 17-90
. . . . . . . . . . . . . . . . . . . . . 17-90
. . . . . . . . . . . . . . . . . . . . . . . . . . 17-91
Figures
17-1
17-2
17-3
17-4
17-5
17-6
17-7
17-8
17-9
17-10
17-11
17-12
17-13
Elevation. storage. discharge r e l a t i o n s h i p
forareservoir
Storage. Discharge r e l a t i o n s h i p and p l o t t e d mass
inflow curve f o r a r e s e r v o i r
Mass inflow. storage. and mass outflow curves
f o r Example 17-2
Graphical version of Mass Curve method of
r e s e r v o i r routing f o r Example 17-3
Graphical version f o r Example 17-2. 'Step 4
Working curve f o r Storage-Indication method of
r e s e r v o i r routing f o r Example 17-4
Inflow and outflow hydrograph f o r Example 17-4
P r i n c i p a l spillway hydrograph and outflow
hydrograph f o r Example 17-5
Working curves f o r Storage-Indication method
of r e s e r v o i r routing f o r Example 17-5
Culp's method of r e s e r v o i r routing f o r Example 17-6
Typical shortcut method of r e s e r v o i r flood routing
Relationships f o r Convex method of channel routing
Convex routing c o e f f i c i e n t versus v e l o c i t y
..em.....d..........
.............
...................
..........
.......
..........
.....
..............
.........
..
..
...
.......
NM Notice 4.102.
August 1972
CONTENTS cont'd.
Figures
17-14 ES-1025 rev.
Sheetlof2
S e e t 2 0 f 2 . .
17-15 Inflow and routed outflow hydrograph
f o r Example 17-7
17-16 Inflow and routed outflow hydrograph f o r
Example 17-8
17-17 Outflow and routed inflow hydrograph f o r
Example 17-9
17-18 Mass inflow, mass outflow and r a t e hydrograph
f o r Example 17-10
17-19 Inflow hydrograph and routed outflow
hydrographs f o r Example 17-11, Method 1 and 2
17-20 !&pica1 schematic diagram f o r routing through
a system of channels
17-21 Q/Q versus A f o r a t y p i c a l physiographic a r e a
......................
....................
. . '. . . . . . . . . . . . . . . .
........ ............
.. ..................
.................
...
. ... . ...........
...
Tables
L
17-1
17-2
17-3
17-4
17-5
17-6
17-7
17-8
17-9
17-10
17-11
17-12
17-13
17-14
L
17-15
17-16
17-17
17-18
.... ...
..................
..........
... ........... ...
......
..
....
............
.................
...........
...........
.......
. . . . . . . . . . . . . . . . ..
.......
.......
......
..
Equations f o r conversion of u n i t s
Elevation-storage r e l a t i o n s h i p f o r a
reservoir
Elevation-discharge r e l a t i o n s h i p f o r a
2-stage p r i n c i p a l spillway
Working t a b l e f o r a storage-discharge
relationship
Operations t a b l e f o r t h e mass-curve
method of routing f o r Example 17-1
Operations t a b l e f o r determining storage
a f t e r 10 days of drawdown f o r Example 17-2
Working t a b l e f o r t h e graphical version of
t h e mass-curve method f o r Example 17-3
Working t a b l e for preparation of t h e working
curve f o r Example 17-4
Operations t a b l e f o r t h e S-I method f o r
Example 17-4
Procedure f o r routing by t h e storage-indication
method f o r Example 17-4.
Working t a b l e f o r preparation of t h e working
curves f o r Example 17-5.
Operations t a b l e f o r Example 17-5
Working t a b l e f o r Culp method, step 1 3 of
Example 17-6
Basic operations i n t h e Convex routing method
Operations t a b l e f o r Example 17-8
Operations t a b l e f o r Example 17-9
Operations t a b l e f o r Example 17-10.
Operations t a b l e f o r Example 17-11 Method 1
NM Notice 4-102, August 1972
CONTENTS c o n t ' d .
Tables
. . . . . . . 17-67
. . . . . . . . . . . 17-71
. . . . . . . . . . . . . . 17-88
. . . . . . . . . 17-92
17-19 Operation t a b l e f o r Example 17-11 Method 2
17-20 P o r t i o n of a t y p i c a l o p e r a t i o n s t a b l e f o r
r o u t i n g through a s t r e a m s y s t e m .
17-21 Data and working t a b l e f o r use of Equation
17-43 on a l a r g e watershed
17-22 Area and s t o r a g e d a t a f o r Example 17-17
NEH Notice 4-102,
August 1972
NATIONAL ENGINEERING HANDBOOK
SECTION
4
HYDROLOGY
Introduction
I n t h e Americari Society of C i v i l Engineers' manual, "Nomenclature for
Hydraulics," flood r o u t i n g i s variously defined as follows:
.--
(1)The derivation of an outflow hydrorouting (hydraulics)
graph of a stream from known values of upstream inflow.
The procedure u t i l i z e s wave v e l o c i t y and t h e storage equat i o n ; sometimes both.
( 2 ) Computing t h e flood a t a downstream point from t h e flood inflow a t an upstream point, and
taking channel storage i n t o account.
routing. flood.-- The process of determining progressively t h e
timing and shape of a flood wave a t successive points along
a river.
routing, streamflow.--The procedure used t o derive a downstream
hydrograph from an upstream hydrograph, or t r i b u t a r y hydrographs, and from considerations of l o c a l inflow by solving
t h e storage equation.
Routing i s a l s o done with mass curves of runoff o r with merely peak
r a t e s o r peak stages of runoff, as well a s hydrographs. The routing
need not be only downstream because the process can be reversed f o r
upstream routing, which i s often done t o determine upstream hydrographs from hydrographs gaged downstream. Nor i s routing confined t o
streams and r i v e r s ; it i s r e g u l a r l y used i n obtaining inflow o r outflow hydrographs, mass curves, o r peak r a t e s i n r e s e r v o i r s , farm ponds,
tanks, swamps, and lakes. And low flows a r e routed, as well as floods.
The term "flood routing" covers all of t h e s e p r a c t i c e s .
The purpose of flood r o u t i n g i n most engineering work i s t o l e a r n what
stages or r a t e s of flow occur, without a c t u a l l y measuring them, a t
s p e c i f i c locations i n streams o r s t r u c t u r e s during passages of floods.
The stages o r r a t e s a r e used i n evaluating o r designing a watercontrol s t r u c t u r e o r p r o j e c t . Differences i n stages or r a t e s from
routings made with and without t h e s t r u c t u r e o r p r o j e c t i n place
show i t s e f f e c t s on t h e flood flows. I n evaluations, t h e differences
a r e t r a n s l a t e d i n t o monetary terms t o show b e n e f i t s on an e a s i l y compar a b l e b a s i s ; i n design, t h e differences a r e used d i r e c t l y i n developing
or modifying t h e s t r u c t u r e or p r o j e c t c h a r a c t e r i s t i c s .
L
NM Notice 4-102, AugUSt 1972
The routing process i s based on one of t h e following approaches:
1. Solution of simultaneous p a r t i a l d i f f e r e n t i a l equations of
motion and continuity. Simplified versions of t h e equations a r e
generally used i n e l e c t r o n i c computer routings; even t h e simplifications a r e t o o laborious f o r manual routings.
2. Solution of the continuity equation alone. A simplified form
of t h e equation i s t h e b a s i s f o r many routing methods.
3. Use of inflow-outflow hydrograph r e l a t i o n s h i p s .
4. Use of u n i t hydrograph theory.
5. Use of empirical r e l a t i o n s h i p s between inflow and outflow pesk
stages or r a t e s . Mostly used f o r l a r g e r i v e r s .
6. Use of hydraulic models.
.
Methods based on t h e second, t h i r d , and fourth approaches a r e presented
i n t h i s chapter. The routing operations i n t h e methods can be made
numerically by means of an e l e c t r o n i c computer, desk c a l c u l a t o r , s l i d e
r u l e , nomograph, network c h a r t , o r by mental calculations; o r graphically
by means of an analog machine, s p e c i a l c h a r t , o r by successive geometric a l drawings. Methods s p e c i f i c a l l y intended f o r e l e c t r o n i c computers
o r analog machines are n e i t h e r presented nor discussed.
A l l methods presented i n t h i s chapter are accurate enough f o r p r a c t i c a l
work i f t h e y a r e applied a s they a r e meant t o be and i f a a t a needed f o r
t h e i r proper application a r e used. Advantages and disadvantages of part i c u l a r methods a r e mentioned and s i t u a t i o n s t h a t l e a d t o g r e a t e r o r
l e s s e r accuracy of a method a r e pointed out, but t h e r e i s no presentat i o n of t e s t s f o r accuracy o r of comparisons between routed and gaged
hydrographs
.
SCS e l e c t r o r i c computer program
The e l e c t r o n i c computer program now being used i n SCS watershed evaluat i o n s contains two methods of flood routing. The S t o r q e - I n d i c a t i o n
method i s used f o r routing through r e s e r v o i r s and t h e Convex method
f o r routing through stream channels. Manual versions of both methods
a r e described i n t h i s chapter.
References
Each of t h e following references contains general m a t e r i a l on flood
routing and descriptions o f two or more methods. References whose
main subject i s not flood routing but which contain a useful example
of routing are c i t e d i n t h e chapter as necessary.
1. Thomas, H. A . , 1937, The hydraulics of flood movements i n r i v e r s :
Pittsburg, Carnegie I n s t . Tech., Eng. B u l l . Out of p r i n t but it
can be found i n most l i b r a r i e s having c o l l e c t i o n s of engineering
literature.
2.
G i l c r e s t , B. R.,
1950, Flood routing: Engineering Hydraulics
(H. Rouse, ed. ), New York, John Wiley and Sons, Chapter 10, pp.
635-710.
NEH Notice 6-102, August 1972
3. U S . Department of t h e Army, Corps of &gineers, 1960, Routing
of floods t h r o u ~ hr i v e r channels: Eng. Manual EM 1110-2-1408.
4. Carter, R.W., and R. G. Godfrey, 1960, Storage and flood
routing: U.S. Geol. Survey Water-Supply Paper 1543-B.
5. Yevdjevich, Vujica M., 1964, Bibliography and discussion of
flood-routing methods and unsteady flow i n channels: U.S. Geol.
Survey Water-Supply Paper 1690. Prepared i n cooperation with t h e
S o i l Conservation Service.
6. Lawler, Edward A., 1964, Flood routing: Handbook of Applied
Hydrology (v.T. Chow, ed. ) , New York, McGraw-Hill Book Co. , sect i o n 25-11, pp. 34-59.
Summary of chapter contents
The remainder of t h i s chapter
i s divided i n t o four p
- a r t s : elevationstorage and elevation-discharge r e l a t i o n s h i p s , r e s e r v o i r routing methods,
channel routing methods, and unit-hydrograph routing methods. I n t h e
first p a r t , some r e l a t i o n s h i p s used i n r e s e r v o i r o r channel routing a r e
discussed and e x h i b i t s of t y p i c a l r e s u l t s a r e given; i n t h e second, t h e
continuity equation i s discussed and methods of using it i n r e s e r v o i r
routings a r e shown i n examples of t y p i c a l a p p l i c a t i o n s ; i n t h e t h i r d ,
t h e theory of t h e Convex method i s presented and examples of t y p i c a l
applications i n channel routings a r e given; and i n t h e fourth, t h e u n i t
hydrograph theory i s discussed and methods of applying it i n systems
analysis a r e shown i n examples using systems of floodwater-retarding
strictures
.
Elevation-Storaae and Elevation-Discharge Relationships
In t h e examples of r o u t i n g through r e s e r v o i r s and stream channels it
w i l l be necessary t o use elevation-storage o r elevation-discharge
curves ( o r both) i n making a routing o r as a preliminary t o routing.
Preparation of such curves i s not emphasized i n t h e examples because
t h e i r construction i s described i n other SCS publications. The
r e l a t i o n s h i p s a r e b r i e f l y discussed here as preliminary material;
e x h i b i t s of t a b l e s and curves used i n routings a r e given here and i n
some of t h e examples. Conversion equations used i n preparing t h e t a b l e s
and curves s r e given i n Table 17-1.
Elevation storage r e l a t i o n s h i p s f o r r e s e r v o i r s
Table 17-2 i s a working t a b l e t h a t shows data and computed r e s u l t s f o r
an elevation-storage r e l a t i o n s h i p t o be used i n some of t h e examples
given l a t e r . Columus 1 and 7 o r 1 and 8 give t h e r e l a t i o n s h i p i n
d i f f e r e n t u n i t s of storage.
The r e l a t i o n s h i p i s developed from a contour map ( o r equivalent) of t h e
r e s e r v o i r area and t h e t a b l e i s a record of t h e computations t h a t were
made. Once t h e map i s a v a i l a b l e , t h e work goes as follows: ( 1 ) s e l e c t
contours close enough t o define t h e topography with reasonable accuracy
and t a b u l a t e t h e contour elevations i n column 1; ( 2 ) determine t h e
NEH ~ o i i c e4-102, August 1972
r e s e r v o i r surface a r e a a t each elevation; f o r t h i s t a b l e t h e a r e a s were
determined i n square f e e t as shown i n column 2 and converted t o acres
as shown i n column 3; ( 3 ) compute average surface areas a s shown i n
column 4; ( 4 ) t a b u l a t e t h e increments of depth i n column 5; ( 5 ) compute
the increments of storage f o r column 6 by multiplying an average a r e a
i n column 4 by i t s appropriate depth increment i n column 5; ( 6 ) accmul a t e t h e storage increments of column 6 t o get accumulated storage i n
column 7 f o r each elevation of column 1; ( 7 ) convert storages of column 7 t o storages i n another u n i t , i f required, and show them i n t h e
next column. The r e l a t i o n s h i p of data i n columns 1 and 8 i s p l o t t e d i n
f i g u r e 17.1 as an elevation-storage curve.
NM Notice 4-102, August 1972
Table 17-1. Equations for conversions of units
- ---
Conversion
where A
A,
AF
L
q&
Equation No.
= drainage area in square miles
= cross section end-area in square
feet.for discharge x
= kcre-feet
= reach length in feet
= discharge in acre-feet per day
= discharge in acre-feet per hour
qcfs = discharge in cfs
qid = discharge in inches per day
qih = discharge in inches per hour
Sx
SA
= reach storage in cfs-hours for
a given discharge x
= reach storage in acre-feet for a
given discharge x
N&X
Notice 4-102, August 1972
17-6
Table 17-2. Elevation-storage relationship for a reservoir.
Elevat ion
(feet)
Surface
area
Surface
area
(q.ft.1
(acres)
Average
surface
area
(acres)
A
A
depth
storage
(feet)
(AF)
Storage
(AF)
Storage
(inches)
Elevation-dischar~e r e l a t i o n s h i p s f o r r e s e r v o i r s
The elevation-discharge r e l a t i o n s h i p f o r a r e s e r v o i r i s made using elevations of t h e reservoir and discharges of t h e spillways t o be used i n a
routing. A t y p i c a l r e l a t i o n s h i p f o r a 2-stage p r i n c i p a l spillway i s
given by columns 1 and 6 of Table 17-3 f o r discharges i n c f s , and i n colThe procedure f o r developing
umns 1 and 7 f o r discharges i n in./day.
t h e r e l a t i o n s h i p w i l l not be given here because s u f f i c i e n t c h a r t s , equat i o n s , and examples f o r p r i n c i p a l spillways a r e given i n NM-5 and i n
ES-150 through 153, and f o r emergency spillways i n ES-98 and ES-124.
Table 17-3 i l l u s t r a t e s a useful way of keeping t h e work i n order: by
t a b u l a t i n g t h e d a t a f o r d i f f e r e n t types of flow i n separate columns,
and by keeping t h e two stages separate, t h e t o t a l discharges a r e more
e a s i l y summed. Note t h a t t h e t o t a l s i n c f s a r e not merely sums of a l l
c f s i n a row; t h e operation of t h e spillway must be understood when sel e c t i n g t h e discharges t o be included i n t h e sum. To combine t h e p r i n c i p a l spillway flow with emergency spillway flow a column f o r t h e emergency
spillway discharges i s added between columns 5 and 6 , and t o t a l s i n
column 6 must include those discharges where appropriate. Column 7 gives
discharges converted from those i n column 6; it i s shown because t h i s
t a b l e i s used i n examples given l a t e r and t h a t p a r t i c u l a r u n i t of flow
i s required ( s e e Figure 17-1).
Storage-discharge r e l a t i o n s h i p s f o r r e s e r v o i r s
If rhe elevation-storage and elevation-discharge r e l a t i o n s h i p s a r e To be
used f o r many routings it i s more convenient t o use them as a storagedischarge r e l a t i o n s h i p . The r e l a t i o n s h i p s a r e combined by p l o t t i n g a
graph of storage and elevation, another of discharge and e l e v a t i o n , and,
while r e f e r r i n g t o t h e f i r s t two graphs, making a t h i r d by p l o t t i n g
storage f o r a selected elevation against discharge f o r t h a t elevation;
f o r a t y p i c a l curve see Figure 17-2. The storage-discharge curve can
a l s o be modified f o r ease of operations with a p a r t i c u l a r r o u t i n g method;
f o r a t y p i c a l modification see Figure 17-6 and s t e p 4 of Example 17-4.
Elevation, stage, storage, discharge r e l a t i o n s h i p s f o r streams
It i s common p r a c t i c e t o divide a stream channel i n t o reaches ( s e e Chapt e r 6 ) and
develop storage o r discharge r e l a t i o n s h i p s f o r i n d i v i d u a l
reaches r a t h e r than t h e stream as a whole. A stream elevation- o r stagedischarge curve i s f o r a p a r t i c u l a r cross section. I f a reach has s e v e r a l
cross s e c t i o n s within it they a r e a l l used i n developing t h e working t o o l s
f o r routing. Some routing methods require t h e use of separate discharge
curves f o r t h e head and f o o t of a reach; such methods a r e not presented
i n t h i s chapter.
to
Elevation- o r stage-discharge curves f o r cross sections o r reaches a r e
prepared a s shown i n Chapter 14. They w i l l not be discussed here.
Elevation- or stage-storage curves f o r a reach can be prepared using t h e
procedure for. r e s e r v o i r s but o r d i n a r i l y a modified approach i s used a n d
t h e storage-discharge curve prepared d i r e c t l y . Table 17-4 i s a working
t a b l e f o r developing such a curve. The work is based on t h e assumption
t h a t steady flow occurs i n t h e reach a t all stages of flow.' The reach
used i n Table 17-4 has four cross s e c t i o n s s o t h a t a weighting method
NEH Notice 4-102, August 1972
Table 17-3
Elevation-discharge r e l a t i o n s h i p f o r a 2-stage p r i n c i p a l
spillway.
Elevation
F i r s t stage:
Weir
Orifice
(cfs)
(cfs)
(feet)
Second stage:
Weir
Pipe
(cfs)
(cfs)
Total
(cfs)
Total
(in./day)
XEH Notice 4-102, August 1972
i s needed; with only one o r two sections t h e weighting i s eliminated
but t h e reach storage i s l e s s well defined. Development of t h e
storage-discharge curve goes a s follows: ( 1 ) s e l e c t a s e r i e s of discharges from zero t o a discharge g r e a t e r than any t o be routed and
t a b u l a t e them i n column 1; ( 2 ) enter t h e stage-discharge curve f o r
each cross s e c t i o n with a discharge from column 1 and f i n d t h e s t a g e ;
( 3 ) e n t e r t h e stage-end-area curve f o r t h a t s e c t i o n with t h e stage
from s t e p 2 and f i n d t h e a r e a a t t h a t st&ge, t a b u l a t i n g areas f o r a l l
sections as shown i n columns 2 , 3 , 4, and 5; ( 4 ) determine t h e distances between cross s e c t i o n s and compute t h e weights a s follows:
From
cross
section
To
cross
section
Distance
(feet)
Weight
with t h e weight f o r sub-reach 1-2 being 1000/10000 = 0.10, and so on;
( 5 ) compute weighted end areas f o r columns 6, 7, and 8; f o r example, a t
a discharge of 3,500 c f s cross s e c t i o n 1 has an end area of 2,500 square
f e e t and section 2 has 640 square f e e t , and t h e weighted end a r e a i s
0.10(2500 + 640)/2 = 157 square f e e t ; ( 6 ) sum t h e weighted areas of
columns 6, 7, and 8 f o r each discharge, t a b u l a t i n g t h e sums i n column
9 ; ( 7 ) compute storages i n column 10 by use of Equation 17-8 or 17-9,
whichever i s required; f o r example, a t a discharge of 3,500 c f s t h e
storage i n cfs-hrs i s s3500= 10000(1189)/3600 = 3300 cfs-hrs, by a
slide-rule computation. The storage-discharge curve i s p l o t t e d using
data from columns 1 and 10. Data of those columns can be used i n
preparing t h e working curve f o r routing. How t h i s i s done depends on
t h e routing method t o be used. For t h e Storage-Indication method t h e
working curve i s prepared as shown i n Example 17-4.
NM Notice 4-102, August 1972
Table
Outflow
(cfs)
17-4
Working t a b l e f o r a storage-discharge r e l a t i o n s h i p
Cross s e c t i o n end-areas
1
(sq.ft)
Wei~htedend-areas
2
3
4
1-2
2-3
3-4
(sq.ft)
(sq.ft)
(sq.ft)
(sq.ft)
(sq.ft)
(sq.ft)
Avg
.
Storage
endareas
( s q . f t ) (cfs-hrs)
Reservoir Routing Methods
Reservoirs have t h e c h a r a c t e r i s t i c t h a t t h e i r storage i s closely rel a t e d t o t h e i r outflow r a t e . I n reservoir routing methods t h e storagedischarge r e l a t i o n i s used f o r repeatedly solving t h e continuity equation,
each s o l u t i o n b e i ~ ga s t e p i n delineating t h e outflow hydrograph. A
r e s e r v o i r method i s s u i t e d f o r channel routings i f t h e channel has t h e
r e s e r v o i r c h a r a c t e r i s t i c . Suitable channels are those w i t h swamps o r
other f l a t areas i n t h e routing reach and with a c o n s t r i c t i o n o r s i m i l a r
c o n t r o l a t t h e foot of t h e reach. There i s an exception t o t h i s : a
r e s e r v o i r method i s a l s o s u i t a b l e f o r routing through any stream reach
i f t h e inflow hydrograph r i s e s and f a l l s so slowly t h a t nearly steady
flow occurs and makes storage i n t h e reach closely r e l a t e d t o t h e outflow r a t e . Examples i n t h i s p a r t show t h e use of r e s e r v o i r methods f o r
both r e s e r v o i r s and stream channels.
'I'ne Continuity Equation
The continuity equation used i n r e s e r v o i r routing methods i s concerned
with conservation of mass: For a given time i n t e r v a l , t h e volume of
inflow minus t h e volume of outflow equals t h e change i n volume of
storage. The equation i s o f t e n w r i t t e n i n t h e simple form:
~t (7
where
- -O )
= AS
(EQ. 17-10)
A t = a time i n t e r v a l
-I
= average r a t e of inflow during t h e time i n t e r v a l
-0
= average r a t e of outflow during t h e time i n t e r v a l
AS = change i n volume of storage during t h e time i n t e r v a l
I n most applications of t h e continuity equation t h e flow and storage
v a r i a b l e s a r e expanded as follows:
so t h a t Equation 17-10 becomes:
where
A t = t2 - tl = time i n t e r v a l ; tl ts t h e time a t t h e beginning
of t h e i n t e r v a l and t2 t h e tlme a t t h e end of t h e i n t e r v a l
Il = inflow r a t e a t t l
I 2 = inflow r a t e a t t 2
01 = outflow r a t e a t t l
NEH Notice 4-102, August 1972
02 = outflow r a t e a t t 2
S1 = storage volume a t tl
S2 = storage volume a t t 2
When routing with Equation 17-10 t h e usual objective i s t o f i n d 5, with
Equation 17-11 f i n d 02; t h i s means t h a t t h e equations must be rearranged
i n some more convenient working form. It i s a l s o necessary t o use t h e
r e l a t i o n s h i p of outflow t o storage i n m+king a solution. Most r e s e r v o i r
routing methods now i n use d i f f e r only i n t h e i r arrangement of t h e routing
equation and i n t h e i r form of t h e storage-outflow r e l a t i o n s h i p .
It i s necessary t o use consistent u n i t s with any routing equation.
commonly used s e t s of u n i t s are:
Time
Hours
days
days
hours
days
Rates
Inflow
Outflow
Inflow
cfs
cfs
AF'/day
i n . /hr
i n . /day
cfs-hrs
cfs-days
AF
inches
inches
cfs
cfs
M/dw
i n . /hr
i n . /day
Some
Volumes
Outflow
Storage
cfs-hrs
cfs-days
cf s-hrs
cfs-days
AF'
AF
inches
inches
inches
inches
Methods and Examples
Two methods of r e s e r v o i r routing based on t h e continuity equation a r e
presented i n t h i s section, a mass-curve method and t h e Storage-Indication
method. The mass-curve nethod i s given because it i s one of t h e most
v e r s a t i l e of a l l r e s e r v o i r methods. It can be applied numerically or
graphically; examples of both versions a r e given. The Storage-Indication
method i s given because it i s t h e method used a t t h e present time i n t h e
SCS e l e c t r o n i c computer program f o r watershed evaluations and because
it i s a widely used method f o r both r e s e r v o i r and channel routings. Examples of r e s e r v o i r and channel routing a r e given.
Mass-Curve Method: Numerical Version - According t o item 52 i n reference
5 , a mass-curve method of routing through r e s e r v o i r s was already i n use
i n 1883. Many o t h e r mass-curve methods have s i n c e been developed. The
method described here i s similar t o a method given i n King's "Handbook
of Hydraulics," 3rd e d i t i o n , 1939, pages 522-527; another resembling it
i s given i n "Design of Small Dams," U. S. Bureau of Reclamation, 1960,
pages 250-252.
The method requires the use of elevation-storage and elevation-discharge
r e l a t i o n s h i p s e i t h e r separately or i n combination. The i n p u t i s t h e
mass ( o r accumulated) inflow; t h e output i s t h e mass outflow, outflow
hydrograph, and reservoir storage. The routing operation i s a trial-ande r r o r process when performed numerically, but it i s simple and e a s i l y
done. Each operation i s a s o l u t i o n of Equation 17-10 r e w r i t t e n i n t h e
form:
NM Notice 4-102, August 1972
MI2
where
-
(MO1 + 'i;; a t ) = S2
MI2 = mass inflow a t time 2
M01 = mass outflow a t time 1
0
= average discharge during t h e routing i n t e r v a l
At
= routing i n t e r v a l = time 2 minus time 1
S2 = storage a t time 2
The routing i n t e r v a l can be e i t h e r v a r i a b l e or constant. Usually it i s
more convenient t o use a variable i n t e r v a l , making it small f o r a l a r g e
change i n mass inflow and l a r g e f o r a s m a l l change. The PSMC of Chapter
21 a r e tabulated i n i n t e r v a l s e s p e c i a l l y s u i t e d f o r t h i s method of routing.
The following example shows t h e a p p l i c a t i o n of t h e method i n determining
minimum required storage f o r a floodwater-retarding s t r u c t u r e by use of
a PSMC from Chapter 21.
Example 17-1.--Determine t h e minimum required storage, by SCS
f o r a floodwater-retarding s t r u c t u r e having t h e drainage a r e a
Example 21-2 of Chapter 21. Use t h e data and r e s u l t s of t h a t
f o r t h i s s t r u c t u r e . Work with volumes i n inches and r a t e s i n
day; round off a l l r e s u l t s t o t h e nearest 0.01 inch.
criteria,
use i n
example
inches per
1. Develop an elevation-discharge curve f o r t h e s t r u c t u r e . A
curve f o r t h e p r i n c i p a l spillway discharges i s needed f o r t h i s
routing. Columns 1 and 7 of Table 17-3 w i l l be used f o r t h i s
s t r u c t u r e . The elevation-discharge curve i s p l o t t e d i n Figure
17-1.
2. Develop an elevation-storage curve f o r t h e s t r u c t u r e .
Columns 1 and 8 of Table 17-2 w i l l be used f o r t h i s s t r u c t u r e .
elevation-storage curve i s p l o t t e d i n Figure 17-1.
The
(Note: The curves of s t e p s 1 and 2 can be combined i n t o a storagedischarge curve as shown by t h e i n s e t of Figure 17-2. This curve i s a
time-saver i f more than one routing i s made.)
3. Develop and p l o t t h e curve of mass inflow (PSMC).
The PSMC developed i n Example 21-2, and given by columns 1 and 7 of
Table 21-7, w i l i be used f i r t h i s example.
he- p l o t t e d mass inflow
i s shown i n Figure 17-2. The p l o t t i n g i s used as a guide i n t h e
routing and l a t e r used t o show t h e r e s u l t s but it i s not e s s e n t i a l
t o t h e method.
4. Prepare an operations t a b l e f o r t h e routing.
Suitable headings and arrangement f o r an operations t a b l e a r e shown
i n Table 17-5.
NM Notice 4-102, August 1972
5. Determine t h e r e s e r v o i r storage f o r t h e s t a r t of t h e routing.
If t h e routing i s t o begin with some storage already occupied then
e i t h e r t h e amount i n storage i s entered i n t h e f i r s t l i n e o r column
5 of t h e operations t a b l e ( a s done i n Example 17-2) or t h e elevationstorage curve i s modified t o give a zero storage f o r t h e f i r s t l i n e .
I n t h i s example t h e sediment o r dead s t o r a g e , which i s n o t t o be
used i n t h e r o u t i n g , occupies t h e r e s e r v o i r t o elevation 580.2 f e e t
as shown i n Figure 17-1. Storage a t t h a t elevation i s 1.00 inches
and because t h i s i s a whole s c a l e u n i t t h e storage curve f o r routing
i s e a s i l y obtained by s h i f t i n g t h e point of o r i g i n as shown i n
Figure 17-1.
Ordinarily, if t h e Sediment o r dead storage i s some
f r a c t i o n a l quantity it i s b e t t e r t o re-plot t h e curve t o show zero
storage a t t h e elevation where t h e routing begins.
6.
Determine t h e spillway discharge a t t h e start of the routing.
If t h e spillway i s flowing a t t h e s t a r t of t h e routing t h e discharge
r a t e i s entered i n t h e f i r s t l i n e of column 7 of able 17-5 (see
Example 17-2). For t h i s example t h e s t a r t i n g r a t e i s zero.
7.
Do t h e routinq.
The trial-and-error
procedure goes a s follows:
a. Select a time and t a b u l a t e it i n column 1, Table 17-5. For
This example t h e times used w i l l be those given f o r t h e PSMC
i n Table 21-7, except f o r occasional omissions unimportant f o r
t h i s routing.
b.
- Compute A t and e n t e r t h e r e s u l t i n column 2.
-c.
Tabulate i n column 3 t h e mass inflow f o r t h e t i m e i n column
1. The e n t r i e s f o r t h i s example come from'column 7 of Table
21-7.
-d.
Assume a m a s s outflow amount and e n t e r , it i n column 4.
Compute t h e r e s e r v o i r storage, which i s t h e inflow of
-e.column
3 minus t h e outflow of column 4, and e n t e r it i n col-
umn 5.
-f .
Determine t h e instantaneous discharge r a t e of t h e spillway.
Using t h e elevation-storage curve of Figure 17-1, f i n d the e l e vation f o r t h e storage of column 5 ; with t h a t e l e v a t i o n e n t e r
t h e elevation-discharge curve and f i n d t h e discharge, tabulating
it i n column 6. I f a storage-discharge curve i s being used,
simply e n t e r t h e curve with t h e storage and f i n d t h e corresponding discharge.
g. Compute t h e average discharge f o r A t .
The average i s
always t h e arithmetic mean o f t h e r a t e determined i n s t e p f
and t h e r a t e f o r t h e previous time. For t h e time 0.5 days t h e
r a t e i n column 6 i s 0.03 i n . /day; f o r t h e previous time t h e
r a t e i s zero; t h e average r a t e i s ( 0 + 0.03)/2 = 0.015, which
NEH n o t i c e 4-102, August 1972
Table 17-5.
Operations table for the mass-curve method of routing
for Example 17-1.
Spillway
Acc. Assumed
Res.
discharne
Outflow Acc
Acc .
At
inflow acc. volume
Inst. A m-. for At
outflow
outflow
(days) (days) (in.)
(in.) (in.) (in./dsy)(in./day)(in.)
(in.)
.
Time
(1)
0
.5
1.0
2.0
3.0
3.5
4.0
4.4
4.8
5.0
5.1
5.2
5.3
5.4
5.6
6.0
6.5
7.0
8.0
etc.
-
etc.
etc.
etc.
etc.
etc.
etc.
etc.
etc.
Mass outflow is plotted using entries in column 4 or column 9. The outflow hydrograph is plotted using column 6, which gives instantaneous
rates at the accumulated times shown in column 1.
NEH Notice 4-102, A~gUst1972
i s rounded t o 0.02 in./day. For t h e time 1.0 days t h e average
i s (0.03 + 0.08)/2 = 0.055, which i s rounded t o 0.06 in./day;
and so on.
h. Compute the outflow f o r A t .
by t h e average r a t e of column
flow f o r column 8.
Multiply the A t of column 2
7 and get t h e increment of out-
Add t h e outflow increment of column 8 t o t h e t o t a l of column
-i.
9 f o r t h e previous time and t a b u l a t e t h e sum i n column 9.
.j-. Compare t h e mass outflow of column 9 with t h e assumed mass
outflow of column 4. I f t h e two e n t r i e s agree within t h e spec i f i e d degree of accuracy (0.01 inch, i n t h i s r o u t i n g ) t h e n
t h i s routing operation i s complete and a new one i s begun with
s t e p 5. If the two e n t r i e s do not agree well enough then assume
another mass outflow f o r column 4 and repeat s t e p s 2 through ;L.
8. Determine t h e minimum required s t o r w e .
Examine t h e e n t r i e s i n column 5 and f i n d t h e l a r g e s t e n t r y , which i s
2.82 inches a t 5.3 days.
This i s t h e minimum required storage.
The routing gives t h e reservoir storages i n column 5, outflow hydrograph
i n column 6, and mass outflow i n column 9, f o r t h e times of column 1.
Unless t h e r e s u l t s a r e t o be used i n a r e p o r t o r e x h i b i t , t h e routing i s
usually c a r r i e d only f a r enough p a s t t h e time of maximum storage t o
ensure t h a t no l a r g e r storage w i l l occur. The mass inflow and outflow
f o r t h i s example a r e p l o t t e d i n Figure 17-2, with outflow shown only t o
8.0 days. If t h e mass outflow p l o t t i n g i s made during t h e routing t h e
trend of t h e curve i n d i c a t e s t h e b e s t assumption f o r t h e next s t e p i n
column 4.
The next example shows how t h e routing proceeds when it must s t a r t w i t h
t h e r e s e r v o i r containing l i v e storage and t h e spillway discharging.
Example 17-2.--For t h e same r e s e r v o i r used i n Example 17-1, determine
t h e e l e v a t i o n and amount of storage remaining i n t h e r e s e r v o i r a f t e r
10 days of drawdown from the minimum l e v e l allowed by SCS c r i t e r i a .
The base flow used i n developing t h e PSMC ( s e e Example 21-2) i s assumed
t o continue a t t h e same r a t e throughout the routing. Round a l l work
t o the nearest 0.01 inch.
1. Determine t h e storage volume i n t h e r e s e r v o i r and t h e spillway
discharge f o r t h e s t a r t of t h e routing.
SCS c r i t e r i a s e r m i t t h e drawdown r o u t i n a t o start with storage at
t h e maximum eievation a t t a i n e d i n t h e r o u t i n g of t h e PSH o r PSMC
used i n determining t h e minimum required storage, even though t h e
s t r u c t u r e may be designed t o contain more than t h e minimum storage.
For t h i s example t h e s t a r t i n g storage of 2.82 inches i s found i n
column 5 , and t h e associated discharge of 1.69 in./day i n column 6,
of Table 17-5 i n t h e l i n e f o r 5.3 days.
REH Notice 4-102, August 1972
2. Prepare an operations t a b l e f o r t h e routing.
Ordinarily t h e s u i t a b l e headings and arrangement a r e those of Table
17-5, but- i f base flow, snowmeit, or upstream r e l e a s e s must be
included (base flow i n t h i s routing) then one o r more a d d i t i o n a l
columns a r e needed. Table 17-6 shows headings and arrangement s u i t able f o r t h i s example.
3. Do t h e routing.
The procedure of s t e p 7, Example 17-1, i s s l i g h t l y modified f o r t h i s
routing. The f i r s t l i n e of d a t a i n t h e operations t a b l e must cont a i n t h e i n i t i a l reservoir volume i n column 4 and t h e i n i t i a 1 , s p i l l way discharge i n column 7. Accumulated base flow i s added t o t h e
i n i t i a l value of column 4 t o give t h e "accumulat;d inflow" of t h a t
column. In a l l other respects t h e routing procedure i s t h a t of
s t e p 7, Example 17-1.
4. Determine t h e s t o r w e remaining a f t e r 10 days of drawdown.
The entry i n column 6 a t day 10 shows t h a t t h e remaining storage
i s 0.20 inches, which i s a t elevation 581.1 f e e t .
The routing f o r t h i s example has been c a r r i e d t o 1 4 days t o show t h a t
when t h e inflow r a t e i s steady, as it i s i n t h i s case (0.045 i n . / d a y ) ,
then t h e outflow r a t e eventually a l s o becomes steady a t t h e same r a t e .
The l a r g e r t h e steady r a t e of inflow t h e sooner t h e outflow becomes
steady. Note t h a t if the routing had been done with an accuracy t o t h e
nearest 0.001 inch, t h e outflow r a t e would be 0.045 i n . / d q , t h e base
flow r a t e .
The mass inflow, storage, and mass outflow curves f o r t h i s example a r e
shown i n Figure 17-3. Note t h a t t h e work i s accurate t o t h e nearest
0.01 inch, therefore the curves must follow the p l o t t e d points within
t h a t l i m i t . S l i g h t i r r e g u l a r i t i e s i n the smooth curves a r e due t o slope
changes i n the storage-discharge curve.
Mass-Curve Method: Direct Version.- It is easy enough t o eliminate t h e
trial-and-error process of t h e mass-curve method but t h e r e s u l t i n g
" d i r e c t version" i s much more laborious than t h e trial-and-error version.
To get a d i r e c t version t h e working equation i s obtained from Equation
17-12 a s follows.
The average discharge 5 i n Equation 17-12 la (01 + 02)/2 so t h a t t h e
equation can be w r i t t e n :
Because O2 as well as S2 i s unknown it i s necessary t o make combinations
of S and 0 t o get d i r e c t solutions i n t h e routing operation. A t any time,
mass outflow i s equal t o mass inflow minus storage, or:
NM Notice 4-102, August 1972
Table 17-6
Operations table f o r determining storage after 1 0 days
of drawdown for Ekample 17-2.
As-
Time
kc.
At
(days) (days)
Acc.
Acc.
base inflow* flow
(in.) (in.)
*
1.0
etc.
Out-
flow
for
acc.
voloutflow me
At
( i n . ) (in;) (in./dqr) ( i n . / d w ) ( i n . )
2.82
2.83
.02 2.84
-03 2.85
.01
14.0
etc.
Spillvray
discharge
Inst.
Avg.
Acc.
outflow
(in.)
(4)
(3)
0
sumed Res.
,04
.07
.o9
;ll
2.86
2.89
2.91
2.93
.14
2.96
.16
2.98
.18
.20
;22
.27
.32
-36
3.00
3.02
3.04
3.09
3.14
3.18
-40
.45
.50
.54
.58
3.22
3.27
3.32
3.36
3.40
.63 3.45
e t c etc.
.
1.50 1.39
1.80 L . l l
2.03
.go
2.01
.92
2.23
.73
2.19
.77
.68
3.30
2.32
.66
2.42
.58
2.52
.5O
2-59 .45
.36
2.73
2.85
.29
2.94
.24
2.93
.25
3.00
.22
.20
3.07
3.13
-19
3.19
-17
3.25
.l5
3.24
.16
.16
3.29
etc.
etc.
1.69
1.66
1.35
.87
.98
.66
.60
.53
.37
.38
.27
.29
.24
.23
.20
-17
.I5
.12
.O9
07
.08
-07
.07
-06
.05
.04
05
05
etc.
0
1.67
1.50
1.n
1.16
.82
.63
.56
.45
.46
.32
.34
.26
.26
-22
.18
.16
.14
.10
.08
.08
.08
.07
.06
.06
.04
.05
.O5
etc.
A t a r a t e of 0.045 inches per day.
NELI Notice 4-102, August 1972
0.33
.3O
.22
.23
.33
.32
.28
.22
.23
.16
-17
.I3
.I3
.11
.09.
.08
.14
.10
.08
.08
.08
.07
.06
.06
.04
-05
.O5
etc.
.33
.63
.85
.86
1.19
1.51
1.79
2.01
2.02
2.18
2.19
2.32
2.32
2.43
2.52
2.60
2.74
2.84
2.92
2.92
3.00
3.07
3.13
3.19
3.23
3.24
3.29
etc.
- S1 for MOl in Equation 17-13 and rearranging gives:
At
At
(Eq. 17-15)
M I 2 - MI1 + (sl - - ol) = S2 " - 02
2
2
Substituting MI1
which is the working equation for the direct version. Working curves of
01 and (Sl - (At 01)/2) and of 02 and (S2 + (At 02)/2) are needed for
routing
.
Other arrangements of working equations can also be obtained from Equation 17-12. Equation 17-15 is the mass-curve version of the StorageIndication method, which is described later in this part. Routing by
use of Equation 17-15 takes about twice as much work as routing by the
Storage-Indication method.
Examples of direct versions of the mass-curve method are n.ot given in
this chapter because the trial-and-error version is more efficient in
every respect.
Mass-Curve Method: Graphical Version.- The graphical version of the
mass-curve method is in a sense a direct version because there is no
trial-and-error involved. The graphical version is usually faster than
the trial-and-error version if the routing job is simple. For complex
jobs the trial-and-error version is more efficient and its results more
easily reviewed. For any routing it gives mass outflow, storage, and the
outflow hydrograph; the graphical version gives only the mass outflow
and storage. The following example shows the use of the graphical version with the data and problem of Example 17-1.
Example 17-3.--Use the graphical version of the mass-curve method to
determine the minimum required storage for the structure used in
Example 17-1. Use the data of that example.
1. Develop an elevation-discharge curve for the structure.
The curve used in Example 17-1 will be used here.
2. Develop an elevation-storwe curve for the structure.
The curve used in Example 17-1 will be used here.
3. Prepare a working table for the routing.
Using the curves of steps 1 and 2, select enough discharges on the
discharge curve to define the curve accurately and tabulate them in
column 2, Table 17-7. Tabulate the associated elevations in column
1 and storages at those elevations in column 4. Compute average discharges from column 2 for column 3. The designations in column 5
show which line is associated with each pair of storages shown on
Figure 17-4. Thus, line A applies when the storage is between 0
and 0.18 inches; line B when it is between 0.18 and 0.40 inches; and
so on.
4. Plot the mass inflow.
The PSMC used in Example 17-1 is used here. It is plotted in Figure
IiEH Botice 4-102, August 1972
5. Do t h e routing.
The work i s done on t h e graph of mass inflow, Figure 17-4. Table
17-7 i s used during t h e work. The procedure goes a s follows:
-a.
Draw l i n e A with i t s o r i g i n a t t h e beginning of mass inflow
and with i t s slope equal t o t h e associated average discharge
(column 3 of Table 17-7), which i s 0.025 in./day. This i s t h e
f i r s t portion of t h e mass outflow curve.
(Note: Every p a r t of t h e l i n e of mass outflow must f a l l on o r below t h e
mass inflow curve. If some p a r t i s above t h e inflow, determine t h e slope
and storage l i m i t s f o r a l i n e with a f l a t t e r slope and use it i n s t e a d . )
b.
-
Determine t h e time a t which t h e difference between mass
inflow and l i n e A i s equal t o t h e l a r g e r of t h e storage l i m i t s
f o r l i n e A, i n t h i s case 0.18 inches, which occurs a t 0.65 days.
This i s t h e p o i n t of o r i g i n f o r l i n e B.
l i n e B with i t s o r i g i n a t t h e point
-c.andDraw
with a slope of 0.09 in./day.
found i n s t e p
2
-d.
Determine t h e time a t which t h e difference between mass inflow and l i n e B i s equal t o t h e l a r g e r of t h e storage limits
f o r l i n e B, i n t h i s case 0.40 inches, which occui-s a t 1.50 days.
This i s t h e point of o r i g i n f o r l i n e C.
e.
-
Repeat t h e procedure of s t e p s 5 and d with l i n e s C, I), E,
e t c . , u n t i l t h e storage being used i s s o l a r g e it exceeds t h e
possible difference between mass inflow and mass outflow. For
t h i s example t h i s occurs with l i n e H. The p a r a l l e l l i n e above
it shows t h a t t h e associated storage of 3.44 inches f a l l s above
t h e mass inflow l i n e . When t h i s s t e p i s reached t h e required
storage i s obtained by t a k i n g t h e maximum difference between
l i n e H and the mass inflow curve. The difference occurs a t t h e
point on t h e mass inflow curve where a l i n e p a r a l l e l t o l i n e
H i s tangent t o t h e inflow curve. For t h i s example it i s 2.80
inches a t 5.33 days. This s t e p completes t h e routing.
The graphical method can a l s o be used f o r routings s t a r t i n g with some
storage occupied and with t h e spillway discharging. For t h e problem used
i n Example 17-2 t h e graphical method s t a r t s with l i n e H and continues
with l i n e s G, F, E, D , C, B, and A i n t h a t order. The r e s u l t s a r e
shown i n Figure 17-5.
The storage a f t e r 10 days of drawdown i s 0.18
inches, which i s nearly t h e same as found i n Example 17-2. Differences
between r e s u l t s of t h e two methods a r e due mainly t o t h e use of s m a l l s c a l e graphs f o r working curves; l a r g e r s c a l e s increase t h e accuracy.
Note t h a t l i n e A i'n Figure 17-5 i s f l a t t e r %ban t h e l i n e of accumulated
base flow. This i n d i c a t e s t h a t t h e flow becomes steady a t o r near 10
days and t h a t t h e dashed l i n e ( p a r a l l e l t o mass inflow) i s t h e a c t u a l
outflow.
NEB Notice 4-102, August 1972
Table 17-7
Working t a b l e f o r t h e graphical version of t h e mass-curve
method f o r Example 17-3.
Elevation
Spillway discharge
Inst.
Avg
Storage
(feet )
(in./day) (in./day)
(inches )
.
(2)
(1)
580.2
(3)
(4)
Designation
on F i g . 17-4
(5)
0
0
0.025
581.0
.05
582.0
.13
583.5
.33
584.6
52
587.0
70
587.8
1.20
588.4
1.64
591.0
1.74
592.5
1.80
line A
.18
.09
.40
.23
.42
line B
line C
.80
line D
1.09
.61
line E
1.86
.95
1.42
line F
2.16
line
G
2.38
1.69
line H
3.44
1.77
line I
4.15
NEH Notice 4-102, A m s t 1972
Storage-Indication Method.- Reservoir routing methods t h a t a r e a l s o used
f o r stream routings a r e generally discharge, not mass, methods because
it i s usually only t h e discharge hydrograph t h a t i s wanted. The StorageIndication method, which has been widely used f o r channel and r e s e r v o i r
routings, has discharge r a t e s as input and output. The method was given
i n t h e 1955 e d i t i o n of NEH-4, Supplement A. Example 17-4, below i s t h e
same example used i n t h a t publication except f o r minor changes.
The Storage-Indication method uses Equation 17-11 %n t h e form:
where 7 = (I1 + 1 2 ) / 2 . The values of
a r e e i t h e r taken from midpoints
of routing i n t e r v a l s of p l o t t e d inflow hydrographs o r computed from inflows tabulated a t regular i n t e r v a l s . A working curve of 02 p l o t t e d
against (s2/At) + (02/2) i s necessary f o r solving t h e equation.
I n channel routing t h e Storage-Indication method has t h e defect t h a t
outflow begins a t the same time inflow begins so t h a t presumably t h e inflow a t t h e head of t h e reach passes instantaneously through t h e reach
regardless of i t s length. This defect i s not serious i f t h e r a t i o T t / T p
i s about 1 / 2 or l e s s , where Tp i s t h e inflow hydrograph time t o peak and
T t i s a t r a v e l time defined as:
where
Tt =
reach t r a v e l time i n hours; t h e time it t a k e s a s e l e c t e d
steady-flow discharge t o pass through t h e reach
L
=
reach length i n f e e t
A
=
average end-area f o r discharge q i n square f e e t
q
=
s e l e c t e d steady-flow discharge i s c f s
V
=
q/A = average v e l o c i t y of discharge q i n f p s
I n determining T t the discharge q i s usually t h e bank-full discharge
under steady flow conditions ( s e e Chapter 1 5 ) .
Another defect of t h e Storage-Indication method, f o r both channel and
r e s e r v o i r r o u t i n g , i s t h a t t h e r e i s no r u l e f o r s e l e c t i n g t h e proper
s i z e of routing i n t e r v a l . T r i a l routings show t h a t negative outflows
w i l l occur during recession periods of outflow whenever A t i s g r e a t e r
than 2 S2/02 ( o r whenever 02/2 i s g r e a t e r than S2/At). This a l s o means
t h a t r i s i n g portion's of hydrographs a r e being d i s t o r t e d . I n p r a c t i c e ,
t o avoid t h e s e p o s s i b i l i t i e s , t h e working curve can be p l o t t e d a s shown
i n Figure 17-6; i f any p a r t of t h e working curve falls above t h e l i n e of
equal values then t h e e n t i r e curve should be discarded and a new one
made using a smaller value of A t . For channel routing t h e p o s s i b i l i t y
of negative outflows i s usually excluded by t a k i n g A t l e s s than Tt.
Notice 4-102, A u g u s t 1972
The following example shows t h e use of t h e Storage-Indication method i n
channel routing. The example i s t h e one used i n t h e 1955 e d i t i o n of
NEE-4, Supplement A, with some minor changes.
Example 17-4.--Use t h e Storage-Tndication method of r e s e r v o i r
routing t o route the inflow hydrograph of Figure 17-7 through the
stream reach of Table 17-4.
1. Prepare the storage-discharge r e l a t i o n s h i p f o r t h e reach.
This i s done i n Table 17-4 and t h e t e x t accompanying it.
2. Determine t h e reach t r a v e l time.
This i s done using Equation 17-17. Table 17-4 and t h e accompanying
t e x t supply t h e following data: L = 10,000 f e e t and f o r a bankf u l l discharge of 800 c f s as q t h e end-area A = 234 square f e e t .
Then by Equation 17-17, T t = 10000(23~)/3600(800)= 0.813 hours.
3. Select the routing i n t e r v a l .
The routing i n t e r v a l f o r t h i s example w i l l be 0.5 hours, which i s
l e s s than t h e t r a v e l time of s t e p 2 and which i s a convenient s i z e
f o r t h e given inflow hydrograph. (See t h e discussion i n t h e t e x t
accompanying Equation 17-30 f o r f u r t h e r information on t h e select i o n of reach routing i n t e r v a l s . )
4. Prepare the working curve.
Use the storage-discharge r e l a t i o n s h i p of s t e p 1, which i s given i n
columns 1 and-10 of able 17-4. 'Fnese two coiumk a r e reproduced
as columns 1 and 3 of Table 17-8, t h e working t a b l e ; columns 2, 4 ,
and 5 of t h e t a b l e a r e self-explanatory.
The working curve i s
p l o t t e d using columns 1 and 5. The f i n i s h e d curve i s shown i n
Figure 17-6.
5. Prepare t h e operations t a b l e .
Suitable headings and arrangement f o r an operations t a b l e a r e shown
i n Table 17-9.
6. Enter times and inflows i n t h e operations t a b l e .
Accumulated time i n s t e p s-of t h e routing i n t e r v a l is shown i n
column 1 of Table 17-9.- I values read From midintervals on t h e
inflow hydrograph of Figure 17-7 a r e shown i n column 2.
7. Do t h e routing.
The procedure i s shown i n Table 17-10. The routing r e s u l t s a r e
shown i n columns 3 and 4 of Table 17-9. The outflow hydrograph
given i n column 4 i s p l o t t e d i n Figure 17-7.
I n routing through channels it i s generally necessary t o add l o c a l inflow
t o t h e routed outflow. The method of doing t h i s i s described l a t e r i n
t h e p a r t on channel routing methods.
NEEf Notice 4-102, August 1972
The Storage-Indication procedure f o r
t h a t f o r channel routing except t h a t
t r a v e l time. The following example
problem and data of Example'l7-1 a r e
of procedures and r e s u l t s .
r e s e r v o i r routing i s i d e n t i c a l with
t h e r e i s no need t o determine a
shows t h e r e s e r v o i r procedure. The
used i n order t o a l l o w a comparison
&ample 17-5.--Use t h e Storage-Indication method t o determine t h e
minimum required storage f o r t h e s t r u c t u r e used i n Example 17-1.
Use t h e d a t a of t h a t example where applicable. Make t h e routing with
discharges i n cfs.
1. Develop an elevation-discharge curve f o r t h e s t r u c t u r e .
The curve used i n Example 17-1 w i l l be used here. That curve i s
f o r discharges i n i n . / h r .
Ordinarily when c f s a r e t o be used t h e
curve i s developed i n t h a t u n i t . The conversion t o c f s w i l l be
made i n s t e p 5.
2. Develop an elevation-storage curve f o r t h e structure.
The curve used i n Example
17-1 w i l l be used here. That curve i s
f o r storage i n inches. The conversion t o cfs-days w i l l be made i n
step 5.
3. Develop and p l o t t h e inflow hydrograph.
Because of t h e type of problem t h e inflow hydrograph must be a Princ i p a l -Sj?illway Hydrograph (PSH) taken from Chapter-21. The PSH
corresponding t o t h e PSMC of Example 17-1 i s given i n columns 1
m d 4 of Table 21-7.
The PSH i s p l o t t e d i n Figure 17-8.
4.
Select t h e rout in^ i n t e r v a l .
Examination of t h e PSH i n F i m r e 17-8 shows t h a t two routing i n t e r v a l s w i l l be needed, one of 6.5 days f o r small changes i n r&es
and one of 0.1 days f o r l a r g e changes.
5. Prepare t h e working curves..
Data and computations f o r t h e working curves a r e shown i n Table
17-11. Two curves a r e needed because two routing i n t e r v a l s w i l l be
used. The e l e v a t i o n s of column 1 and discharges of column 2 a r e
taken from t h e curve of s t e p 1with t h e discharges being converted
from i n . / h r . t o c f s i n t h e process. The discharges a r e s e l e c t e d so
t h a t they adequately define t h e elevation-dischaxge relationship.
Column 3 of Table 17-11 gives the corresponding storages from t h e
curve of s t e p 2 , converted from inches t o cfs-hrs during t h e tabulat i o n . The remaining columns contain self-explanatory computations.
Columns 2 and 6 give t h e f i r s t working curve and columns 2 and 8
t h e second; they a r e p l o t t e d i n Figure 17-9.
Note t h a t " l i n e s of
equal values" i f drawn would be w e l l above t h e working curves,
therefore t h e routing i n t e r v a l s a r e adequately small. Also note
t h a t t h e second curve i s shown only f o r t h e higher discharges i n
order t o use a l a r g e r s c a l e ; o r d i n a r i l y t h e e n t i r e curve i s plotted.
NM Notice 4-102, AugUSt 1972
Table 17-8
Working t a b l e f o r preparation of t h e
working curve f o r Example 17-4.
NM Notice 4-102, August 1972
Table 17-9 Operations table for the S-I method for
Example
17-4.
Time
(2)
0
625"
1875
3125
4375
4615
3865
3125
2375
1635
900
265
O*"
0
.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
0
7.0
0
7.5
8.0
0
0
etc.
etc.
* 625 cfs
(3)
(4)
0
0
625
2215
4310
6805
8540
8795
8210
7135
5720
4180
2635
1425
795
420
260
178
285
1030
etc.
1880
2880
3610
3710
3450
3050
2440
1810
1210
630
375
160
82
53
etc.
is the average discharge for the time from 0 to
0.5 hours, 1875 cfs the average discharge from 0.5 to 1.0
hours, and so on.
** Inflow ceases
at 5.33 hours.
NEH Notice 4-102, August 1972
Table 17-10
Procedure f o r routing by t h e Storage-Indication method
f o r Example 17-4.
Remarks
0
Given
Given
o
285
-o
+ 625 = 625
From Figure 17-6
Given
625
1030
-
285 + 1875 = 2215
From Figure 17-6
Given
2215
1880
-
1030 + 3125 = 4310
From Figure 17-6
Given
4310
etc.
etc.
etc.
- 1880 + 4375 = 6805
2880
From Figure 17-6
etc.
etc.
NEH Notice 4-102, August 1972
6.
Prepare t h e operations t a b l e .
Suitable headings
and arrangement a r e shown i n Table 17-12. Note
t h a t t h e r e i s a column f o r i n s t g t a n e o u s r a t e s of inflow. These
r a t e s w i l l be used f o r g e t t i n g I values because it i s d i f f i c u l t t o
s e l e c t I values accurately enough from some portions of t h e p l o t t e d
hydrograph.
7. Tabulate times and r a t e s of inflow and compute 7 values.
Accumulated times a r e shown i n column 1 of Table 17-12 a t i n t e r v a l s
of A t = 0.5 days f o r t h e i n i t i a l slow-rising portion of t h e PSH, a t
A t = 0 . 1 days f o r t h e f a s t - r i s i n g and - f a l l i n g portion, and again
a t A t = 0 . 5 days f o r t h e slow recession. Instantaneous r a t e s of
inflow f o r those times a r e taken from t h e PSH of Figure 17-8 ( o r
from column 4 of Table 21-7 i f they are f o r t h e s e l e c t e d times) and
shown i n column 2. The I values of column 3 a r e arithmetic averages
of e n t r i e s i n column 2.
8. Do t h e routin@;.
The procedure i s t h e same a s t h a t given i n Table 17-10 except when
a change i s made from one working curve t o another. The changes a r e
made a s follows. A t time 4.5 days t h e routing i n t e r v a l changes,
therefore, t h e working curve must be changed. The outflow r a t e a t
t h a t time i s 116 cfs. Entering t h e second working curve with t h i s
r a t e gives 2,640 c f s as t h e value of (s2/At) + (0212) i n column 4
f o r t h e same time. Once t h i s value i s entered t h e routing continues
with use of t h e second working curve. A t time 6.0 days t h e routing
i n t e r v a l chknges back t o t h e first one and t h e r e f o r e t h e first
working curve must again be used. The outflow r a t e a t t h a t time i s
357 c f s . Entering t h e f i r s t working curve with t h i s r a t e gives
1,270 c f s as t h e value of (s2/At) + (02/2) i n column 4 f o r t h a t
time. After e n t e r i n g t h i s value t h e routing continues with use of
the first working curve.
9 . Determine t h e maximum storage a t t a i n e d i n t h e routing.
The maximum storage a t t a i n e d i n a r e s e r v o i r during t h e routing of
a single-peaked hydrograph occurs a t t h e time when outflow equals
inflow. The p l o t t i n g i n Figure 17-8 shows t h a t t h i s occurs a t 5.33
days. For t h i s time, Table 17-12 shows t h a t 02 = 364 c f s and
(s2/At) + (02/2) = 6,480 c f s . solving f o r S2 gives S2 = A t 6480
(0 1 2 ) . With A t = 0 0 . days and 02 = 364 c f s , S2 = 0.1 6480
(3&/2) = 629.8 cfs-days , t h e maximum storage. To convert t o AF
use Equation 17-2, which gives 629.8/0.504 = 1,247 AF as t h e maximum storage i n AT. To convert AF t o inches use Equation 17-3 and
t h e given drainage area of 8.0 square miles ( s e e Example 17-11,
which give l247/53.3(8.0) = 2.93 inches as t h e maximum storage i n
inches. (Tote: The storage can a l s o be found by use of a storagedischarge curve o r elevation-discharge and elevation-storage curves
but with t h e Storage-Indication method it i s generally b e s t t o use
t h e above method. )
-
-
A comparison of peak r a t e s of outflow shows t h a t t h e mass-curve method
of Example 17-1 gave a peak r a t e of 1.69 in./dsy, which converts t o 363
NEH Notice 4-102, August 1972
Table 17-11
Working t a b l e f o r preparation of the working curves
f o r Example 17-5.
For A t = 0.5 days
Elevat ion
Dis-
charge
Storage
(S2)
O2
(02)
(feet)
(cfs)
(cfs-days) ( c f s )
S2
S2
"2
At
At + 2
(cfs)
(cfs)
NM Notice 4-102, August 1972
For A t = 0.1 days
s2
At
(cfs)
S2
o2
+At
2
(cfs)
Table 17-12
Operations t a b l e f o r Example 17-5.
Gutf OH
etc.
etc.
etc.
etc.
*
From f i r s t working curve.
""From second working curve.
NEH Notice 4-102, August 1972
etc.
c f s , and t h e Storage-Indication method gave 364 cfs, which i s excellent
agreement. But a comparison of maximum storage i n inches shows t h a t t h e
mass-curve method of Example 17-1 gave 2.82 inches, t h e graphical masscurve method of Example 17-3 gave 2.80 inches, and t h e Storage-Indication
method gave 2.93 inches. The discrepancy i s f o r t h e most p a r t due t o use
of small-scale graphs f o r t h e working curves. Larger graphs would reduce
t h e discrepancy.
St orage-Indication Method as Used i n t h e SCS Electronic Computer Program.
SCS e l e c t r o n i c computer program f o r watershed evaluations uses t h e
Storage-Indication method only f o r r e s e r v o i r routings. The chief d i f f e r ence between t h e manual procedure of Example 17-5 and t h e electroniccomputer procedure i s t h a t i n t h e l a t t e r no working curves a r e used. Ins t e a d , t h e working equation i s solved during a process i n which interpol a t i o n s a r e made i n t h e elevation-discharge and elevation-storage data
stored i n t h e computer. The process i s repeated during t h e routing j u s t
as t h e working curve i s repeatedly used i n manual routing. The machine
routing has a numerical accuracy g r e a t e r than t h a t of t h e manual routing,
but t h e machine cannot improve t h e accuracy of the input data. D e t a i l s
of t h e machine routing process a r e given i n pages A-61 through A-66 of
t h e r e p o r t t i t l e d "Computer Program f o r Project Formulation
Hydrology,"
by C-E-I-R,
Inc. Arlington, Va., January 1964, which w a s prepared f o r
SCS. Copies of t h i s report a r e a v a i l a b l e from t h e Washington, D. C.
o f f i c e of SCS.
-
-
Culp's Method.- Some routing methods a r e developed f o r solving special
problems, f o r which they have a high e f f i c i e n c y . One such method i s
described next.
I n t h e design of an emergency spillway of a dam it i s SCS p r a c t i c e t o
base t h e design on t h e r e s u l t s from a routing of an Bnergency Spillway
Hydrograph. Because all of t h e spillway dimensions cannot be known i n
advance, it i s necessary t o route t h e hydrograph through t h r e e o r four
d i f f e r e n t spillways w i t h assumed dimensions before t h e spillway w i t h
t h e proper dimensions can be found. M. M. Culp's r o u t i n g method eliminates much of t h a t work by giving t h e routed peak discharge without t h e
use of spillway dimensions. The following example shows an a p p l i c a t i o n
of t h e method t o t h e s t r u c t u r e used i n previous examples. The example
i s lengthy because many d e t a i l s a r e given; a f t e r t h e method i s understood
it w i l l be seen t o be f a s t and easy t o apply.
Example 17-6.--Find t h e routed peak discharge t o be used i n design of
an emergency spillway f o r t h e s t r u c t u r e of Example 17-1. The required
difference i n elevation between t h e c r e s t of t h e spillway and t h e reserv o i r water surface, H , i s 4.0 f e e t during t h e peak discharge. Watershed and s t r u c t u r e dafa a r e given i n examples 17-1 and 17-2.
=
1. Prepare t h e elevation-discharge curve f o r t h e p r i n c i p a l s p i l l -
This curve was prepared f o r Example 17-1 with t h e discharges i n
inches per hour. It w i l l be used here as shown i n Figure 17-10(a)
with discharges i n cfs.
NEII Notice 4-102, August 1972
2. Prepare t h e elevation-storage curve f o r t h e s t r u c t u r e .
This curve was prepared f o r Example 17-1. Only t h e portion above
t h e sediment storage w i l l b e used here; it is shown i n f i g u r e
17-10(a).
3. Determine t h e elevation of t h e emergency spillway c r e s t .
According t o SCS c r i t e r i a , t h e elevation of t h e emergency
spillway
c r e s t can be a t o r above t h e maximum water-surface elevation
a t t a i n e d i n t h e r e s e r v o i r during t h e routing of t h e P r i n c i p a l
Spillway Hydrograph (PSH) o r i t s mass curve (PSMC) The watersurface e l e v a t i o n found i n Example 17-1 w i l l be used here as t h e
c r e s t elevation. This elevation i s 589.5 f e e t with floodwater
storage of 2.82 inches.
.
4. Determine t h e water-surface elevation of t h e floodwater remaini n g i n t h e r e s e r v o i r after LO days of drawdown from storage a t t h e
water-surface elevation a t t a i n e d i n routing t h e PSH o r PSMC.
This s t e p i s required by SCS c r i t e r i a . The determination i s made
i n Fxample 17-2 and those r e s u l t s w i l l be used here. The watersurface elevation a f t e r 10 days of drawdown i s 581.1 f e e t with
floodwater storage a t 0.20 inches.
5. Prepare t h e Emergency Spillway Hydrograph (ESH) and i t s mass
curve (ESMC).
The ESH f o r t h i s example was prepared using t h e method of Example
21-5 and t h e following d a t a : drainage a r e a = 8.0 square miles,
time of concentration = 2.0 hours, runoff curve number = 75, design
storm r a i n f a l l = 9 . 1 inches, storm duration = 6.0 hours, r u n o f f =
6.04 inches, hydrograph family = 2, To = 5.05 hours, i n i t i a l Tp =
1 . 4 hours, To/Tp = 3.61, s e l e c t e d To/Tp = 4, revised Tp = 1.26 hours,
qp = 3,073 c f s , and Q(q ) = 18,560 c f s . The ESMC was prepared using
Table 21-17 and t h e folEowing d a t a : hydrograph family = 2, To/Tp =
4, Tp = 1.26 hours, and Q = 6.04 inches. The hydrograph i s shown
i n Flgure 17-10(b) and t h e mass curve i n Figure 17-10(c).
(Note: The above s t e p s a r e taken, i n much t h e same way, regardless of
which manual method of r o u t i n g i s used f o r t h i s kind of problem. The
following s t e p s apply t o t h e Culp method. )
6.
Determine t h e time a t which t h e emergency spillway begins t o
flow during passage of t h e ESH or ESMC.
For t h i s example t h e time was found by routing the ESMC of s t e p 5
by t h e method of Example 17-1, using t h e curves of Figure 17-10(a)
as working c w e s . The routing w a s s t a r t e d with 0.20 inches of
floodwater i n t h e r e s e r v o i r (SCS c r i t e r i a r e q u i r e t h e ESH o r ESMC
routing t o s t a r t a t t h e elevation f o r t h e floodwater remaining
a f t e r the 10-day drawdown period; see s t e p 4). The emergency s p i l l way began t o flow a t 2.9 hours, a t which time t h e mass outflow was
0.06 inches. The time and outflow a r e indicated by point c l on
Figure 17-10(c )
iWH Notice 4-102, August 1972
Determine the average discharge of t h e p r i n c i p a l s p i l l w q
during, passage of t h e ESH or ESMC through t h e emergency spillway.
The p r i n c i p a l spillway average discharge i s f o r t h e period during
which t h e reservoir storage r i s e s from t h e elevation of t h e emergency spillway c r e s t t o t h e c r e s t elevation plus
the elevation-discharge curve of Figure 17-10(a) t o f i n d he discharges
a t t h e two elevations. These discharges a r e 361 and 392 c f s respectively; t h e i r average i s 376 c f s .
7.
2.
8. Locate a reference point i n t h e ESH f o r use i n l a t e r steps.
The reference point, shown as point b l i n Figure 17-10(b), i s
located a t t h e time determined i n s t e p 6 and a t t h e average discharge determined i n s t e p 7. A second p o i n t , not a c t u a l l y necessary
i n t h e work, i s shown as b2 on t h e recession s i d e . A s t r a i g h t l i n e
connecting points b l and b2 represents the p r i n c i p a l spillway outflow r a t e during t h e period used i n s t e p 7.
9.
Compute t h e slope of t h e p r i n c i p a l spillway mass outflow l i n e
f o r use on the mass inflow graph.
Tne mass outflow t o be used i s f o r t h e period considered i n s t e p 7.
F u l l pipe flow occurs and t h e mass outflow i s adequately represented
by a s t r a i g h t l i n e . The slope of t h e l i n e f o r t h i s example must be
i n inches per hour because t h e mass inflow s c a l e s a r e f o r inches
and hours. To get t h e slope, convert t h e average discharge of
s t e p 7 by use of Equation 17-5, which gives 376/645(8.0) = 0.073
inches per hour.
10. P l o t a reference l i n e and a working l i n e of p r i n c i p a l spillway
mass outflow on t h e graph f o r mass inflow.
The l i n e s a r e f o r t h e period considered i n s t e p 7 but f o r working
convenience they a r e extended beyond t h e limits of t h e period. To
p l o t t h e reference l i n e , f i r s t l o c a t e point c2 on t h e mass inflow
curve of Figure 17-10(c) a t t h e time determined i n s t e p 6, then
through c2 draw a s t r a i g h t l i n e having t h e slope determined i n
s t e p 9 ; t h i s gives l i n e A as shown. To p l o t t h e working l i n e ,
f i r s t determine t h e storage associated with EIp, which i s 1.84 inches
as shown i n Figure 17-10(a), then draw l i n e B p a r a l l e l t o l i n e A
and 1.84 inches of runoff above it as shown i n Figure 17-10(c).
11. Find t h e period within which t h e emergency spillway peak d i s charge w i l l occur.
Point c3 i s a t t h e i n t e r s e c t i o n of t h e mass inflow curve and l i n e B
i n Figure 17-10(c). Locate point b3 on t h e ESH of Figure 17-10(b)
a t t h e time found f o r c3. Points b3 and b2 are t h e end p o i n t s f o r
t h e period within which t h e emergency spillway peak discharge w i l l
occur.
2 . Select s e v e r a l working discharges between p o i n t s b3 and b2.
Four s e l e c t e d working discharges a r e indicated by p o i n t s b4, b5,
b6, and b7 i n Figure-17-10(b); t h e discharges a r e 4,750, 3,500,
2,200, and 920 c f s respectively. These discharges represent t h e
peak discharges of outflow hydrographs.
NEH Notice 4-102, August 1972
( ~ o t e : After some experience with t h i s method, it may be found e a s i e r
t o s e l e c t only two working discharges i n t h i s s t e p , t o work through
s t e p s 1 3 t o 15, and i f t h e r e s u l t s a r e u n s a t i s f a c t o r y t o r e t u r n t o s t e p
12 again by s e l e c t i n g a t h i r d working discharge, working through s t e p s
13 through 1 5 f o r t h a t discharge, and s o on.1
13. Compute a volume-to-peak f o r each working discharge of step 12.
I n the Culp method t h e r i s i n g s i d e of the outflow hydrograph f o r
a trapezoidal spillway i s taken a s being nearly parabolic s o t h a t
the volume from t h e beginning of r i s e t o t h e peak r a t e , o r t h e
volume-to-peak, i s :
where Qe i s t h e volume i n cfs-hrs, qe i s t h e working discharge of
s t e p 12 i n c f s , qps i s t h e p r i n c i p a l spillway r a t e of s t e p 7 i n
c f s , and Te i s t h e time i n hours from point b l t o t h e peak time.
The volume Qe must be converted t o a u n i t usable with t h e mass
inflow curve, i n t h i s case, inches. The summary of work f o r t h i s
s t e p i s given i n Table 17-13. I n t h e columns f o r points b4 through
b7, t h e items i n l i n e 1 a r e from s t e p 12; items i n l i n e 2 a r e from
s t e p 7; items i n l i n e 3 a r e obtained by s u b t r a c t i n g qps from qe;
items .in l i n e 4 a r e obtained by inspection of Figure 17-10(b);
qps) x Te; items i n l i n e 6
items i n l i n e 5 a r e products of (qe
a r e products of ( ~ e j 0 . 6 2 )x 0.62; items i n l i n e 7 a r e Qe's of l i n e
6 divided by t h e drainage a r e a of 8.0 square miles; items of l i n e
8 a r e Qe' s of l i n e 7 divided by 645. Each Qe of l i n e 8 applies
only at t h e time indicated by i t s point on the ESH.
-
14.
P l o t a curve of mass inflow minus mass outflow.
This i s a working curve, not t h e complete curve of inflow minus
outflow. Subtract each Qe of l i n e 8, Table 17-13, from t h e inflow
amount a t t h e i d e n t i c a l time on t h e mass inflow curve of Figure
17-10(c) and p l o t t h e r e s u l t as shown f o r p o i n t s c4, c5, c6, and
c7. Connect t h e p o i n t s with a curve, l i n e C.
15.
Determine t h e time and r a t e f o r the emergency spillway peak
discharge.
The i n t e r s e c t i o n of l i n e s B and C , a t point c8 i n Figure 17-10(c),
gives t h e time a t which t h e emergency spillway peak discharge occurs
The t o t a l discharge r a t e a t t h a t time i s 3,050 c f s a s shown by t h e
corresponding point b8 on t h e ESE of Figure 17-10(b). The emergency spillway discharge r a t e i s 3050
376 = 2,674 c f s , which
occurs when t h e r e s e r v o i r water surface i s a t t h e given elevation
of 593.5 f e e t ( c r e s t elevation plus Hp). This s t e p completes t h e
routing. Design of t h e emergency spillway now follows with use of
ES-98, ES-124, and spillway c r i t e r i a .
-
I f Kp is not known i n advance, t h e Culp method can be used with assumed
values of Hp t o get associated discharges from which t h e s u i t a b l e combination of Hp and discharge can be s e l e c t e d . For e a r t h spillways % can
be closely approximated from permissible v e l o c i t i e s and t h e appropriate
IiEH Notice 4-102, August 1972
17-35
Working t a b l e f o r Culp method s t e p 13 of Example 17-6.
Table 17-13
Point :
Line
It em
qe
¶€'s
qe
-
Unit
b4
b5
cfs
cfs
q~s
C ~ S
Te
hrs
Qel0.62
cfs-hrs
Qe
cfs-hrs
Qe
csm-hrs
Qe
in.
NJB Notice 4-102, August 1972
b6
b7
length and chosen p r o f i l e of t h e i n l e t channel. A close approximation
of t h e emergency spillway discharge r a t e can be obtained i n t h i s way f o r
an ELp ~ a l u enear t h e middle of t h e desired range t o get a " C curve" ( l i n e
C on Flgure 17-10(c)). The average discharge i n t h e conventional drop
i n l e t under f u l l pipe flow conditions v a r i e s only s l i g h t l y as H?, v a r i e s
r e l a t i v e l y g r e a t l y , t h u s t h e discharge through t h e emergency splllway
can be closely approximated from such an average C curve. If refinement i s j u s t i f i e d , then t r i a l adjustments on t h e slope of l i n e B w i l l
give t h e required accuracy. The correction process converges rapidly.
For preliminary layouts or comparative cost s t u d i e s such refinement i s
seldom j u s t i f i e d .
Short-Cuts f o r Reservoir R o u t i n ~ s . - Various equations and c h a r t s have
been developed f o r quickly estimating t h e required storage i n a reserv o i r o r t h e required capacity of a spillway, such estimates being used
i n preliminary s t u d i e s of s t r u c t u r e s or p r o j e c t s . The equations and
c h a r t s a r e usually based on t h e r e s u l t s of routings so t h a t using t h e
equation o r c h a r t i s i n e f f e c t a form of routing.
A t y p i c a l short-cut i s t h e graph, Figure 17-11.
The curve through t h e
c i r c l e d points i s based on information i n t a b l e 2 on page 39 of "Low
D a m s , " a design manual prepared by the Subcommittee on Small Water
Storage P r o j e c t s , National Resources Committee, Washington, D. C . , 1938
( t h e manual i s out of p r i n t and no longer a v a i l a b l e f o r ~ u r c h a s e ) .
Relationships of t h i s kind a r e developed from routings made through a
p a r t i c u l a r type of spillway and they apply only t o t h a t type. The form
of standard inflow hydrograph used f o r routing a l s o a f f e c t s t h e r e l a t i o n s h i p and t h e same form must be applicable when t h e short-cut i s used.
With such a r e l a t i o n s h i p i f any t h r e e of t h e four v a r i a b l e s a r e known
t h e f o u r t h can be estimated. Usually e i t h e r t h e r e s e r v o i r storage or
t h e r e s e r v o i r discharge r a t e i s t h e unknown.
The t r i a n g u l a r point on Figure 17-11 i s f o r t h e routing made i n Example
17-6. For t h a t example t h e outflow/inflow r a t i o i s 3050/10200 = 0.30
and t h e storage/inflow-volume r a t i o i s 2.82/(2.62 + 1.84) = 0.63. Note
t h a t t h e emergency spillway "surcharge" storage is included when computing t h e volume r a t i o . Thi: cross p o i n t s , f o r "miscellaneous routings",
a r e f o r routings of s e v e r a l kinds ofhydrographs through emergency s p i l l ways of t h e SCS type. The "Low Dams" curve appears t o be an enveloping
curve f o r t h e points. As such it can be used f o r making conservative
estimates. Thus, i f t h e inflow volume i s 8.15 inches of runoff and t h e
t o t a l a v a i l a b l e storage i s 5.7 inches then the storage r a t i o i s 0.7; a t
t h a t r a t i o t h e discharge r a t i o i s 0.4, which means t h a t t h e peak outflow
r a t e w i l l be not more than 0.4 of t h e peak inflow. Such estimates a r e
often u s e f u l i n preliminary work.
NM Notice 4-102, August 1972
I
I-w
Z(3
w a
I
DISCHARGE, INCHES PER DAY,
OF PRINCIPAL SPILLWAY WITH TWO-STAGE INLET
--zg
gk
1
2
3
4
5
6
7
4
5
6
TOTAL STORAGE- INCHES
0
I
2
3
STORAGE USED IN ROUTING- INCHES
Figure 17-2.
Storage, discharge relationship and plotted mass
inflow e w e for a reservoir.
NEB Notice 4-102, August 1972
PLUS ACCUMULATED BASE
TIME, DAYS
Figure 17-3.
Mass inflow, storage, and mass outflow curves for Example 17-2.
NM Notice 4-102, August 1972
I
MASS INFLOW OR
:ONTENTS PLUS I
dFTER 10 DAYS DRAM
I
-INE OF ACTUAL FL(
O
Y
:LINE IS PARALLEL '
KCUMULATED BASE LOW)
2
4
6
8
TIME, DAYS
Figure 17-5.
Graphical version for Example 17-2, Step 4.
NEH Notice 4-102, August 1972
10
Figure 17-6. Working curve for Storage-Indication method of
reservoir routing for Example 17-4.
NEH Notice 4-102, August 1972
TIME, HOURS
Figure 17-7.
Inflow and outflow hydrograph f o r
Example 17-4.
NEH Notice 4-102, August 1972
Figure 17-8. Principal spillway hydrograph and outflow
hydrograph f o r Example 17-5.
mEII Notice 4-102, August 1972
,
FOR At = 0.1
DAYS
(SECOND WORKING CURVE)
I
0
FOR At = 0 . 5 DAYS 0
FOR At = 0. l DAYS 2 0 0 0
3000
1000
4000
5000
2000
6000
7000
Figure 17-9. Working curves for Storage-Indication method
Example 17-5.
3000
8000
9000
of reservoir routing for
I
C
vl
Figure 17-10.
Culp's method of reservoir routing for Example 17-6.
NM Notice 4-102, August 1972
OUTFLOW PEAK RATE
INFLOW PEAK RATE
Figure 17-11.
Typical shortcut method of reservoir flood
routing.
NEH Notice 4-102, August 1972
Channel Routina Methods
The Convex method of routing through stream channels i s presented i n
t h i s p a r t . The method i s derived from inflow-outflow hydrograph r e l a tionships and, because of t h i s , t h e method has some f e a t u r e s not possessed by channel routing methods derived from consideration of t h e
continuity equation. The Storage-Indication method of channel routing,
presented i n Example 17-4, w i l l not be discussed here, but discussions
of procedures f o r adding l o c a l inflows, deducting transmission l o s s e s ,
and routing through stream systems a l s o apply t o t h a t method.
Theory of t h e Convex Method
The Convex method i s based on t h e following p r i n c i p l e : When a n a t u r a l
flood flow passes t h r o u b a n a t u r a l stream channel having n e g l i g i b l e
l o c a l inflows o r transmission l o s s e s , t h e r e i s a reach length L and a
time i n t e r v a l A t such t h a t 02 i s not more than t h e l a r g e r nor l e s s than
A t i s considered as both t h e
t h e smaller of t h e two flows 11 and O1.
t r a v e l time of t h e flood wave through t h e reach measured a t t h e beginning
of t h e r i s i n g portion of t h e hydrograph a t both ends of t h e reach; and
t h e required routing time i n t e r v a l .
The p r i n c i p l e requires t h a t :
If I1 2 01, t h e n I1 2 O2 2 O1
If 11 5 01,
then I1 5 02 5 01
(Eq. 17-19)
(Eq. 17-20)
I n general, inequality Equation 17-19 a p p l i e s t o r i s i n g portions of hydrographs and Equation 17-20 t o f a l l i n g portions. Note t h a t 12 does not
e n t e r i n t o t h e p r i n c i p l e ; t h i s makes t h e Convex method a forecasting
method (see under "Discussion" )
.
The routing p r i n c i p l e can be extended t o include l o c a l inflows and
transmission l o s s e s but t h i s unnecessarily complicates t h e working equat i o n . It i s common p r a c t i c e t o add l o c a l inflows t o t h e routed outflow
hydrographs t o get t h e t o t a l outflows, and t h i s p r a c t i c e w i l l be f o l lowed here. There may be s i t u a t i o n s , however, i n which t h e l o c a l inflow
i s added t o t h e inflow hydrograph and t h e n routed. Small transmission
l o s s e s are generally deducted a f t e r t h e routing, l a r g e ones during t h e
routing; f o r a discussion of transmission l o s s e s see t h e heading "Effects
of transmission l o s s e s on routed flows."
The routing o r working equation i s formed a f t e r examination of t y p i c a l
inflow and outflow hydrographs such as t h o s e i n Figure 17-12. Typical
flood wave combinations o f 11, O1 and O2 a r e shown on t h e r i s i n g and
f a l l i n g s i d e s of t h e hydrographs. The r o u t i n g p r i n c i p l e s t a t e s t h a t
f o r a properly selected reach l e n g t h L , hence A t , O2 K i l l f a l l somewhere on or between I1 and O1 i n magnitude but not above or below them.
This i s evident on Figure 17-12 d e s p i t e t h e displacement of 02 i n time;
it i s t h e magnitudes t h a t a r e of concern here.
R& Notice 4-102, August 1972
I
l
L
The next s t e p i s t o recognize t h a t 11, 01 and O2.are members of a
Convex s e t i / . For such a s e t , i f p o i n t s A and B a r e i n t h e s e t then
a l l points on a s t r a i g h t l i n e connecting A and B a r e a l s o i n t h e s e t .
Because t h e concern i s with magnitudes and not with time it i s not
necessary f o r O2 t o be physically on the l i n e between 11 and 01. The
routing equation can now be w r i t t e n based on t h e theory of convex s e t s .
For the s i t u a t i o n j u s t described, and using proportions as shown on t h e
i n s e t o f Figure L7-12, t h e routing o r working equation is:
where C i s a parameter with t h e range:
zero
<
C
5 one
( E ~ 17-22)
.
Given Equation 17-22, Equation 17-21 meets t h e requirements of Equations
17-19 and 17-20 and therefore of t h e routing p r i n c i p l e .
The routing method based on Equation 17-21 i s c a l l e d t h e Convex method
t o c a l l a t t e n t i o n t o t h e equation's background.
It follows from Equation 1 7 - 2 1 t h a t :
I n the i n s e t of Figure 17-12 t h e r e l a t i o n s h i p s between 11, 01,
make similar t r i a n g l e s , so t h a t :
I
and O2
where K i s considered t h e reach t r a v e l time f o r a s e l e c t e d steady flow
discharge of a water p a r t i c l e through the given reach. From Equation
17-24 it follows t h a t :
Combining Equations 17-23 and 17-25 gives:
'/ Enough of t h e theory of convex s e t s f o r t h e purposes of t h i s chapt e r i s given i n pages 41-42 of "An Introduction t o Linear Programming," by A. Charnes, W. W. Cooper, and A. Henderson; John Wiley
and Sons, Inc., New York, 1953.
NEH Notice 4-102, August 1972
I
I
from which comes t h e equation t h a t defines A t , t h e wave t r a v e l time and
a l s o t h e required r o u t i n g i n t e r v a l :
I
Discussion
This much of t h e t h e o m i s enouah
- f o r makina- a workable routina method.
The emphasis i n t h i s chapter i s on working examples, not on theory,
t h e r e f o r e t h e a d d i t i o n a l r e s u l t s from t h e theory a r e summarized i n t h e
next section without giving derivations or proofs. Further work can
be done on some aspects qf t h e Convex routing method but even i n i t s
present s t a t e t h e method i s highly useful f o r most types of problems of
r o u t i n g flood flows through stream channels.
-
The theory a s given s o f a r can be used f o r exploratory routings by
assuming magnitudes f o r any two of t h e v a r i a b l e s i n Equation 17-27 computing the t h i r d , and using Equation 17-21 with various inflow hydrographs. Such routings show t h e features of t h e Convex method. I n
Figure 17-12, f o r example, note t h a t outflow begins a t one routing i n t e r v a l , A t , a f t e r inflow begins, which i s t o be expected f o r a stream reach
because it t a k e s water waves time t o t r a v e l through t h e reach. It i s
c h i e f l y t h i s c h a r a c t e r i s t i c t h a t distinguishes t h e Convex method from
channel methods based on the continuity equation. I n Convex routing
t h e peab r a t e of the outflow hydrograph does not f a l l on t h e recession
limb of the inflow hydrograph, as i n r e s e r v o i r methods. But, as i n
all routing methods, t h e maximum storage i n t h e reach i s a t t a i n e d when
outflow equals inflow ( a t point A i n Figure 17-12). The maximum storage
i s represented by t h e a r e a under t h e inflow hydrograph t o t h e l e f t of
point A minus t h e a r e a of t h e outflow hydrograph t o t h e l e f t of point A .
Also note t h a t inflow I 2 does not appear i n t h e working equation though
it does appear i n equations f o r other channel methods. This f e a t u r e
makes the Convex method a forecasting method. For example, i f t h e rout i n g i n t e r v a l i s one day, today's inflow and outflow a r e known and l o c a l
~ n e g l i g i b l e , then tomorrow's outflow can be predicted
inflow i s k n 0 or
accurately without knowing tomorrow's inflow. The p r e d i c t i v e f e a t u r e
i s more important f o r l a r g e r i v e r s than f o r small streams because t h e
r o u t i n g i n t e r v a l f o r reaches of such streams i s usually short.
Some UsefU Relationships and Procedures
Of t h e equations so far given, only Equations 17-21 and 17-27 a r e needed
i n p r a c t i c a l a p p l i c a t i o n s of t h e Convex method. The f i r s t i s t h e working
equation and t h e second an a u x i l i a r y equation used once before a routing
begins. Several other r e l a t i o n s h i p s and procedures a l s o - u s e f u l i n applications follow
.
Determination o f K. - K i s t h e reach t r a v e l time f o r a s e l e c t e d steadyflow discharge and can be computed using Equation 17-7 s u b s t i t u t i n g K
f o r Tt. Example 17-8 shows a preferred method fok s e l e c t i n g t h e discharge. The K used i n t h e Muskingum routing method ( r e f s . 2 and 3) may
a l s o be used as the K f o r t h e Convex method.
NM Notice 4-102, August 1972
i/
L
I Determination
-
of C.
From Equations 17-17 and 17-26 the parameter o r
routing coefficient C can be derived as t h e r a t i o of two v e l o c i t i e s :
t h a t i s , C = V/U, where V i s t h e steady-flow water v e l o c i t y r e l a t e d t o
i t h e reach t r a v e l time f o r steady flow discharge, K, and U i s considered
t h e wave v e l o c i t y r e l a t e d t o t h e t r a v e l time of t h e wave through t h e
i reach, A t . For p r a c t i c a l purposes C may be estimated from an empirical
r e l a t i o n s h i p between C and V shown i n Figure 17-13. The dashed l i n e i n
t h e Figure i s represented by t h e equation:
I
I
!
I n some applications it i s more convenient t o use Equation 17-28 than
Figure 17-13. The "x" used i n t h e Muskingum routing method ( r e f s . 2
and 3 ) may a l s o be used t o approximate C. The approximation i s :
I n t h e Muskingum procedure t h e x i s sometimes determined only t o t h e
nearest t e n t h ; if t h i s i s done then C i s approximated t o t h e nearest two
t e n t h s and accurate routing r e s u l t s should not be expected.
-
'U
,
,
)
I f C and K a r e known, from Equation 17-27, t h e r e
Determination of A t .
i s only one permissible routing i n t e r v a l . This permissible i n t e n r s l
may be an inconvenient magnitude because it i s e i t h e r an unwieldy fract i o n of an hour o r does not fit t h e given hydrograph. I n s e l e c t i n g a
s u i t a b l e routing i n t e r v a l keep i n mind t h a t t o o l a r g e an i n t e r v a l w i l l
not accurately define t h e inflow hydrograph and t h a t t o o s m a l l an i n t e r v a l w i l l needlessly increase t h e e f f o r t required f o r t h e routing. A
generally s u i t a b l e r u l e of thumb t o follow i s t h a t t h e s e l e c t e d r o u t i n g
time i n t e r v a l , A t * , should be no g r e a t e r than 115 of t h e time from t h e
beginning of r i s e t o t h e time of t h e peak discharge of t h e inflow hydrograph, or:
At*
j
T
p
5
,
'I where
Tp i s t h e time t o peak (Chapter 16). I f t h e hydrograph has more
t h a n one peak the i n t e r v a l should be s e l e c t e d using the 3
the
/
/
s h o r t e s t of t h e r i s e periods of t h e b p o r t a n t peaks. It is important
t h a t an end-point of a time i n t e r v a l f a l l a t o r near t h e inflow peak
time and any other l a r g e change i n r a t e .
Procedure f o r routing through any reach length.- The r e l a t i o n s h i p of K,
C, and A t i s v a l i d f o r one and only one routing reach length f o r a given
time i n t e r v a l and inflow hydrograph. I f A t i s t o be changed t o At*
I (desired routing time i n t e r v a l ) it follows from Equation 17-27 t h a t e i t h e r
( 1 ) C or K must be changed (Method 1) o r , ( 2 ) routing through a s e r i e s
/ of subreaches, L*, (Equation 17-32) must be made u n t i l t h e sum of t h e
t r a v e l time of t h e A t ' s f o r each subreach, L*, equal t h e desired t r a v e l
: time, At*, f o r t h e t o t a l reach, L (Method 2). Selection of e i t h e r method
/
/
L
NER Notice 4-102, August 1972
depends on t h e manner of computation and t h e consistency of the answers
desired. Method 1 may be used when rough approximations of t h e routing
e f f e c t a r e desired and manual computation i s used. Method 2 is used
when consistency of t h e routing i s important o r a computer i s used. Consistency, as used here, r e f e r s t o t h e changes i n t h e outflou hydrograph
(Tp and q ) caused by varying At*. If t h e r e i s l i t t l e change i n t h e
hydrograp8 when A t * i s changed t h e routing i s considered consistent.
I
In Method 1, t h e reach length i s fixed, hence, K i s f i x e d (Equation
17-17) and C must be modified by t h e empirical r e l a t i o n s h i p :
where C* i s t h e modified routing c o e f f i c i e n t required f o r use with A t W ,
C i s the coefficient determined from Figure 17-13 o r computed by Equat i o n 17-28, At* i s t h e desired routing i n t e r v a l , and A t i s t h e r o u t i n g
i n t e r v a l determined from Equation 17-27. After s e l e c t i n g At* t h e coeffic i e n t C* i s found by using e i t h e r Equation 17-31 o r Figure 17-14 (ES1025 r e v . )
Method 2 assumes t h a t C and t h e desired routing i n t e r v a l A t * a r e f i x e d
and t h e routing i s made f o r a reach length L*. From Equation 17-27, t h e
' desired t r a v e l time i s :
I
i
From Equation 17-17 t h e proper routing reach l e n g t h t o match C and A t *
, i s then:
I f L* i s l e s s than t h e given reach length, L , t h e inflow hydrograph i s
r e p e t i t i v e l y routed u n t i l t h e difference between t h e sum of t h e L*'s
The l a s t r o u t i n g i n t h e reach i s a
and L becomes l e s s than t h e next L*.
The A t used i n
f r a c t i o n a l routing using C* computed by Equation 17-31.
Equation 17-31 i s t h e time i n t e r v a l f o r routing through t h e f r a c t i o n a l
length increment of L, L**. (see Example 17-11 Method 2).
If L* i s g r e a t e r than t h e given reach length, L, t h e , i n f l o w hydrograph
is routed once using Method 1. Example 17-11 i l l u s t r a t e s t h e use of
Methods 1 and 2.
V a r i a b i l i t y of routing parameters; s e l e c t i o n of v e l o c i t y , V.
As shown by preceding r e l a t i o n s h i p s , t h e magnitudes of t h e routing parameters C and K (and therefore of At) depend on t h e magnitude of t h e
velocity V. For steady flow i n n a t u r a l streams this v e l o c i t y v a r i e s
with stage but t h e v a r i a t i o n i s not t h e same f o r a l l seasons of a y e a r
or f o r a l l reaches of a stream, nor does t h e v e l o c i t y c o n s i s t e n t l y increase o r decrease with stage. For unsteady flow, v e l o c i t y v a r i e s not
only w i t h s t a g e but a l s o with t h e r a t e of change of t h e stream flow.
NM n o t i c e 4-102, A w t 1972
L
These f a c t s would appear t o require a change i n routing parameters f o r
each operational s t e p i n a routing. But exploratory routings with t h e
Convex method show t h a t constant parameters must be used t o conserve
mass, t h a t i s , t o make t o t a l outflow equal t o t a l infiow. The necessity
f o r t h e use of constant parameters i s a c h a r a c t e r i s t i c of c o e f f i c i e n t
routing equations, including not only Equation 17-21 but a l s o with t h e
Muskingum routing equation ( r e f s . 2 and 3) and t h e Storage-Indication
equations. Therefore all of t h e examples i n t h i s p a r t show a use of cons t a n t parameters. I n p r a c t i c e t h e parameters need not be constant f o r
a l l steps of a routing but t h e more often they a r e changed t h e more l i k e l y
t h a t t h e t o t a l outflow w i l l not equal t o t a l inflow.
The average, dominant, and peak v e l o c i t i e s of one inflow hydrograph w i l l
nearly always d i f f e r from t h e corresponding v e l o c i t i e s of another hydrograph. Even though a s i n g l e value of V i s used t o get t h e constant values of C , K , and A t f o r a routing, t h i s V w i l l nearly always be d i f f e r e n t
f o r d i f f e r e n t inflow hydrographs t o a reach. Each inflow hydrograph w i l l
need i t s own routing parameters determined from i t s own s e l e c t e d velocity.
There are various methods of s e l e c t i n g the v e l o c i t y .
One method, useful when a computer i s used, computes t h e v e l o c i t y as t h e
average of v e l o c i t i e s f o r all given discharges of t h e inflow hydrograph
2 50 percent of t h e peak discharge.
L
A manual method with t h e same objective as t h e machine method w i l l be
used i n t h i s chapter t o make manual routings comparable t o machine rout i n g s . I n t h i s method t h e dominant v e l o c i t y of t h e inflow hydrograph i s
used t o determine t h e parameters t o be used i n t h e routing. I f t h e inflow
hydrograph has a s i n g l e peak t h e v e l o c i t y i s f o r a discharge equal t o 314
of t h e peak inflow r a t e . If t h e inflow hydrograph has two o r more peaks
t h e v e l o c i t y i s f o r t h e discharge with t h e l a r g e s t value of Tq, where:
Some a d d i t i o n a l
The use of Equation 17-34 i s i l l u s t r a t e d i n Example 17-8.
remarks concerning t h e s e l e c t e d velocity a r e given i n t h e paragraph preceding Example 17-7.
Examples.- The Convex method i s generally used f o r routing hydrographs
through stream reaches. It can a l s o be used, without any change i n procedure f o r routing mass curves through reaches. Examples of both uses
w i l l be given. The method can be used f o r routing through r e s e r v o i r s but
f o r t h i s it i s not as e f f i c i e n t as t h e mass-curve method of Example 17-1;
therefore no examples of r e s e r v o i r routing a r e given i n t h i s p a r t . Examp l e s are given showing various aspects of Convex routing.
Example 17-7
Example 17-8
-
Basic routine using assumed parameters.
Routing with parameters determined from reach
d a t a and with l o c a l inflow added a t bottom of
.reach.
NEH Notice 4-102, August 1972
Example 17-9
Example 17-10
Example 17-11
- "Reverse Routing" o r determining t h e inflow
hydrograph f o r a given outflow hydrograph.
- Routing of Mass Curve and method of g e t t i n g
t h e outflow hydrograph.
- Routing any hydrograph through any reach.
Method 1 and Method 2 are. compared.
For the following examples it i s assumed t h a t stage-discharge and
stage-end-area curves a r e a v a i l a b l e for t h e routing reach. These curves
a r e used f o r determining t h e v e l o c i t y , V, a f t e r t h e dominant discharge
of t h e inflow hydrograph i s obtained. I n preliminary work such curves
may not be a v a i l a b l e , i n which case the v e l o c i t y can be estimated during
a f i e l d t r i p t o t h e stream area, o r a s u i t a b l e v e l o c i t y assumed, and t h e
routing made as a t e n t a t i v e study; such routings need v e r i f i c a t i o n by
routings based on reach d a t a before making firm decisions about a p r o j e c t .
I n t h e f i r s t example'the values o f C and A t are assumed; t h e r e f o r e t h e
reach length and K do not d i r e c t l y e n t e r i n t o t h e work:
Example 17-7.--Route the t r i a n g u l a r inflow hydrograph of Figure 17-15 by
t h e Convex computational method. Use assumed values of C = 0.4 and A t =
0.3 hours. There i s no l o c a l inflow i n t o t h e reach.
1. Prepare t h e operations t a b l e .
Suitable headings and &rangement a r e shown f o r t h e f i r s t t h r e e columns i n Table 17-14.
The "remarks" column i s used here t o e m l a i n
t h e s t e p s ; it i s not needed i n routine work.
d
2. Tabulate t h e inflow r a t e s a t accumulated times, using i n t e r v a l s
of A t .
The accumulated times a t i n t e r v a l s of A t = 0.3 hours a r e shown i n
column 1 of Table 17-14. The inflow r a t e s a t these times a r e taken
from t h e inflow hydrograph of Figure 17-15 and shown i n column 2.
-
3. Prepare t h e working equation.
is
Since C = 0.4 then (1- C ) = 0.6 and t h e working- eauation
o2 = ( 1 C ) 0 1 t C 11 = 0.6 o1 + 0.4 11. When inflow ceases
t h e working equation i s 02 = 0.6 01.
-
4. Do t h e routing.
Follow t h e s t e p s shown i n t h e remarks column of Table 17-14.
The computational work i n s t e p 4 can usually be done on most desk- calculators by using a system of making t h e two m u l t i p l i c a t i o n s and t h e
addition i n one machine operation.
The outflow hydrograph of Table 17-14 i s p l o t t e d on Figure 17-15. The
c i r c l e d points a r e the outflow discharges obtained i n t h e routing. D i s charges between t h e points a r e found by connecting t h e points with a
smooth curve. Sometimes t h e routing p o i n t s do not define t h e peak region
NEH Notice 4-102, A u g u s t 1972
3
17-55
L.
Table 17-14 Basic operations in the Convex routing method.
Time
(hrs)
Inflow, I
(cfs)
Remarks
Outflow, 0
(cfs
Given.
Oe = 0.6(0) + 0.4(0) = zero
O2 = 0.6(0) + 0.4(800) = 320
0, = 0.6(320)
+
O2 = 0.6(832)
+ 0.4(2400)
0.4(1600) = 832
= 1459
0, = 0.6(1459) + 0.4(3200) = 2155
02 = 0.6(2155)
+
0.4(4000) = 2893
02 = 0.6(2893) + 0.4(3520) = 3144
02 = 0.6(3144) + 0.4(3040) = 3102
0, = 0.6(3102) + 0.4(2560) = 2885
02 = 0.6(2885) + 0.4(2080) = 2563
O2 = 0.6(2563) +
02 = 0.6(2178) +
0, = 0.6(1755) +
0, = 0.6(1309) +
0.4(1600) = 2178
0.4(1120) = 1755
0.4(640) = 1309
0.4(160)= 849
0 2 = 0.6(849) = 509
I 1 = zero.
0, = 0.6(509) = 305
PI
tt
11
0, = 0.6(305) = 183
II
TI
11
0, = 0.6(183) = 110
etc
.
etc.
etc.
etc.
-1/ Outflow starts at At = 0.3 hrs.
-2/ Inflow ceases at 4.0 hrs.
NEH Notice 4-102, August 1972
" "
"
well enough; t h i s usually happens when t h e routing i n t e r v a l i s l a r g e .
I n such cases t h e peak i s estimated by use of a smooth curve o r t h e
routing i s repeated using smaller i n t e r v a l s [see Example 17-11 f o r use
of At*).
The recession curve o r t a i l of t h e outflow hydrograph continues t o i n f i n i t y , t h e discharges g e t t i n g smaller with every s t e p but'never becoming
zero. This i s a c h a r a c t e r i s t i c of most routing methods. I n p r a c t i c e
the recession curve i s e i t h e r a r b i t r a r i l y brought t o zero a t some convenient low discharge o r the routing i s stopped a t some low discharge a s
shown i n Figure 17-15.
The next example i s t y p i c a l of t h e routine used i n p r a c t i c e . Routing
parameters a r e obtained from reach data and l o c a l inflow i s added i n
t h e conventional manner. Local inflow i s t h e ( u s u a l l y ) small flow
from the contributing area between t h e head and foot of a reach. Local
inflow and t h e inflow i n t o t h e head of t h e reach together make up t h e
t o t a l flow from t h e drainage a r e a above t h e foot of t h e reach. The
l o c a l inflow i s generally given as a hydrograph made with reference t o
the foot of t h e routing reach. When it i s added t o t h e routed outflow
the sum i s t h e total. outflow hydrograph.
Example 17-8.--The inflow hydrograph i n Figure 17-16 i s t o be routed
tbrough a reach having a low-flow channel length of 14,900 f e e t and a
valley length of 12,400 f e e t . Stage-discharge and stage-end-area curves
f o r t h e reach a r e available (not i l l u s t r a t e d ) . A hydrograph of l o c a l
inflow i s given i n Figure 17-16. Obtain t h e t o t a l outflow hydrograph
f o r t h e reach.
1. Determine t h e discharge t o be used f o r g e t t i n g t h e v e l o c i t y V.
The inflow hydrograph has two peaks and it i s not r e a d i l y apparent which peak i s the dominant-one, t h e r e f o r e t h e r u l e expressed
by Equation 17-34 w i l l be used. The 3/4-discharge f o r t h e f i r s t
peak i s 3,750 c f s w i t h a duration of 2.63 hours; f o r t h e second,
2,680 c f s with a duration of 5.35 hours, Then Tq = 3750(2.63) =
9,850 cfs-hrs f o r t h e f i r s t peak and Tq = 2680(5.35) = 14,320 cfsh r s f o r t h e second, therefore t h e second discharge w i l l be used.
2. Determine t h e v e l o c i t y , V.
Enter t h e stage-discharge curve f o r t h e reach with t h e s e l e c t e d 3/4discharge from s t e p 1 a& f i n d t h e stage f o r t h a t flow. Then
e n t e r the stage-end-area curve with t h a t stage and get t h e enda r e a i n square f e e t . The v e l o c i t y i s t h e discharge divided by t h e
end area. For t h i s example V w i l l be taken as 3.0 fps.
3. Determine K.
The reach has two lengths, one f o r t h e low-flow channel, t h e o t h e r
f o r t h e valley. From an examination of t h e stage-discharge curve
and t h e inflow hydrograph it i s evident t h a t most of t h e f l o w w i l l
exceed t h e capacity of t h e low-flow channel, t h e r e f o r e use t h e
v a l l e y length. This i s given as 12,400 f e e t . By Equation 17-17,
using T t = K, t h e value of K = 12400/3600(3.0) = 1.15 hours by a
slide-rule computation.
NEH Notice 4-102, August 1972
ii
4. Determine C.
Enter Figure 17-13 with V = 3.0 fps and f i n d C = 0.65.
5. Compute A t .
Using r e s u l t s from s t e p s 3 and 4, and by Equation 17-27, A t = 0.65
(1.15) = 0.745 hours. Round t o 0.75 hours.
6. Prepare an operations t a b l e f o r t h e routing.
Suitable headings and arrangement a r e shown i n Table 17-15.
7.
Tabulate accumulated time a t i n t e r v a l s of A t and t h e discharges
f o r inflow and l o c a l inflow a t those times.
The times a r e given i n column 1 of Table 17-15, inflows i n column
2, and l o c a l inflows i n column 4. Inflows and l o c a l inflows a r e
taken from t h e given hydrographs, which a r e shown i n Figure 17-16.
8. Prepare t h e working equation.
From s t e p 4, C = 0.65 so t h a t (1- C ) = 0.35.
i s O2 = 0.35 O1 + 0.65 I1.
The working equation
9. Do t h e routing.
Follow t h e routine used i n Table 17-14 t o a e t t h e outflows f o r column 3 of Table 17-15.
10.
Get t h e t o t a l outflow hydrograph.
4, Table 17-15, t o t h e routed outflows of column 3 t o get t h e t o t a l outflows f o r Column 5. This
s t e p completes t h e example. The t o t a l outflow hydrograph i s shown
i n Figure 17-16.
Add t h e Local inflows of column
Note i n Figure 17-16 t h a t t h e routed outflow peaks are not much smalle r than the inflow peaks. The f i r s t routed outflow peak i s 93.0 percent of i t s respective inflow peak, and t h e second 97.7 percent of i t s
inflow peak. The reach has r e l a t i v e l y small storage when compared with
t h e inflow volumes; t h e f i r s t inflow peak has l e s s volume associated
with it than t h e second and it i s reduced more than t h e second.
The next example i l l u s t r a t e s a routine sometimes needed t o g e t t h e upstream hydrograph when t h e downstream one i s given. The uorking equat i o n f o r t h i s routine i s a rearranged form of Equation 17-21:
Example 17-9.--Obtain t h e inflow hydrograph of a reach from t h e t o t a l
outflow hydrograph by use of reverse routing. The t o t a l outflow hydrograph and l o c a l inflow a r e given i n Table 17-16.
1. Determine t h e routing c o e f f i c i e n t C and t h e routing; i n t e r v a l
At.
-
Follow t h e procedure of s t e p s 1 through 5 of Example 17-8.
example C = 0.44 and A t = 0.5 h r s
.
NEH Notice 4-102, August 1972
For t h i s
Table 17-15
Time
(hrs. )
Operations t a b l e f o r Example 17-8.
Inflow
(cfs)
Out flow
(cfs
Local
Inflow
(cfs)
Tot a1
Outflow
(cfs)
(2)
0
380
1400
3000
4450
5000
4600
3750
2800
2100
1600
1280
1150
1210
1480
21.00
21.75
27.50
23.25
24.00
24.75
25.50
26.25
27.00
27.75
28.50
29.25
30.00
etc.
1880
2360
2880
3250
3500
3580
3480
3240
2930
2600
2280
i.980
1730
1480
1280
1130
980
850
720
620
530
450
400
350
310
270
etc.
230
190
150
120
100
90
80
70
60
50
etc.
2105
1808
1548
1344
1165
1015
872
750
642
546
Lo
47 4
30
20
etc.
383
345
Outflow s t a r t s a t A t = 0.75 hrs.
NEH Notice 4-102, August 1972
etc.
2. Prepare t h e operations t a b l e f o r t h e routing.
Suitable headings and arrangements a r e shown i n Table 17-16.
Tabulate accumulated time a t i n t e r v a l s of A" and t h e discharges f o r t o t a l outflow and l o c a l inflow a t those times.
The times a r e given i n column 1 of Table 27-16, t o t a l outflows
i n column 2, and l o c a l inflows i n column 3 . The t o t a l outflow
(but not t h e l o c a l inflow) i s shown i n Figure 17-17.
3.
4.
Determine t h e outflows t o be routed uustream.
A value i n column 2, Table 17-16, minus t h e corresponding value
i n column 3 gives t h e outflow f o r column
outflows t o be routed upstream.
4, which contains t h e
5. Prepare t h e working equation.
C i s given i n s t e p 1 as 0.44.
By Equation 17-35, 11 = 2.27 O2
1.27 O1.
-
6. Do t h e routing.
The routine i s s l i g h t l y d i f f e r e n t from t h a t i n Table 17-14. Using
values from Table 17-16, the sequence i s : f o r outflow time 0.5
h r s , Il = 2.27(0) - 1.27(0) = 0 , which i s recorded f o r inflow
time zero; a t outflow time 1 . 0 h r s , 11 = 2.27(163) - 1.27(0) =
370, recorded f o r inflow time 0.5 h r s ; f o r outflow 1.5, I1 =
2.27(478)
1.27(163) = 878, recorded f o r inflow time 1.0 h r s ;
and so on. The work i s e a s i l y done by accumulative p o s i t i v e
and negative m u l t i p l i c a t i o n on a desk c a l c u l a t o r . The inflow
hydrograph t o time 7,5 hours i s p l o t t e d on Figure 17-17.
-
It w i l l sometimes happen i n reverse routing t h a t t h e working equation
gives negative values f o r t h e inflow. This occurs when t h e t o t a l
outflow hydrograph o r t h e l o c a l inflow i s i n e r r o r .
The next example shows t h e downstream routing of a mass curve of inflow.
The routine i s t h e same as t h a t f o r Example 17-7. The outflow hydrograph can be obtained from t h e mass outflow curve by a s e r i e s of simple
calculations; t h e s e outflows must be p l o t t e d a t midpoints of time increments, not a t end points.
Example 17-10.--Route t h e mass curve of inflow of Figure 17-18 by t h e
Convex method. There i s no l o c a l inflow.
1. Determine t h e routing coefficient C and t h e routing i n t e r v a l
At.
Follow t h e procedure
of s t e p s 1 through 5 of Example 17-8.
t h i s example C = 0.40 and A t = 0.3 h r s .
2. Prepare t h e operations t a b l e f o r t h e routing.
Suitable headings and arrangement are shown i n Table 17-17.
NM Nbtice 4-102, August 1972
For
Table 17-16
-
p
p
Time
(hrs)
-
Operations t a b l e f o r Example L7-9
-
-
Tot a 1
Outflow
(cfs)
--
--
Local
Inflow
(cfs)
Outflou
t o be
routed
(cfs)
Inflow
(cfs)
(1)
0
.5
1.0
155
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
etc.
0
120
310
680
1250
1850
2490
3030
3440
3700
3900
3940
3840
3500
3000
2485
1960
etc.
-1/ Outflow starts a t A t
O0-1 /
31
etc.
=
163
478
932
1525
2165
2708
3122
3420
3631
3714
3637
3344
2915
2424
1929
etc.
0.5 hours.
NEH Notice 4-102, A u g u s t 1972
0
370
878
1508
2278
2978
3398
3648
3793
3899
3819
3539
2972
2370
1800
1300
etc.
3. Tabulate accumulated time a t i n t e r v a l s of A t and t h e mass
inflows a t those times.
The times are given i n column 1 of Table 17-17 and mass inflows
i n column 2.
4.
Prepare t h e working, equation.
C i s given i n s t e p 1 as 0.40. By Equation 17-21, O2 = 0.6 O1 + 0.4
5. Do t h e routing.
Tne routine i s exactly t h e same as t h a t i n Table 17-14. For examp l e , a t inflow time 2.7 h r s , O2 i s computed using inflow and outflow f o r t h e previous time or O2 = 0.6(3707) + 0.4(5952) = 4605 cfshrs.
(Note: I f only t h e mass outflow i s needed t h e work stops with s t e p 5.
I f t h e outflow hydrograph i s a l s o needed, t h e following s t e p s are a l s o
taken. )
6. Compute increments of outflow.
These a r e the differences shown i n column 4, Table 17-17.
7. Compute average r a t e s of outflow.
Dividing the increment of outflow of column 4 , Table 17-17, by
t h e increment of time ( i n t h i s case, 0.3 h r s ) gives t h e average
r a t e of outflow f o r the time increment. For example, between 1.8
and 2.1 hours i n Table 17-17, t h e time increment i s 0.3 hrs and
t h e outflow increment i s 906 cfs-hrs; then t h e average r a t e i s 9061
0.3 = 3,020 c f s . The average r a t e s must be p l o t t e d as midpoints
between t h e two accumulated times involved; f o r t h i s case, 3020
c f s i s p l o t t e d a t a time of (1.8 + 2.1112 = 1.95 hours.
The mass inflow, mass outflow, and r a t e hydrograph a r e p l o t t e d i n
Figure 17-18.
The next example shows how t o route any hydrograph through any reach
length. Methods 1 and 2 are compared.
Example 17-11.--Route t h e inflow hydrograph of Figure 17-19 through
a reach 30,000 f e e t long. Assume no l o c a l inflow.
I
I
Determine desired routing time i n t e r v a l , A t x .
Following t h e r u l e expressed i n Equation 17-30, At* w i l l be 0.4 hrs.
1.
2. Determine routing c o e f f i c i e n t , C, and routing i n t e r v a l A t .
If a stage-discharge-velocity t a b l e f o r a t y p i c a l s e c t i o n i n
t h e reach i s used, t h e average v e l o c i t y V i s determined using
t h e method from page 17-54, and C i s computed using Equation 17-28.
I f a r a t i n g t a b l e i s not used t h e C or V must be assumed; i n t h i s
case, l e t C = 0.72. Rearranging Eq. 17-28 gives V = 1 . 7 ~ / ( 1 - C )=
NEH Notice 4-102, August 1972
I
Table 17-17
Time
(hrs. 1
Operations t a b l e f o r Example 17-10.
Mass
Inflow
(cfs-hrs)
Mass
Outflow
(cfs-hrs)
Increment of
Outflow
(cfs-hrs)
Outflow
Rate
(cfs)
- .
5.2
7992
7817
5.5
7992
7887
etc.
etc.
etc.
-1/ Outflow
70
etc.
starts a t A t = 0.3 hours
iTBl Notice 4-102, August 1972
233
etc.
( 1 . 7 ) ( .72)/0.28 = 4.37 fps. Combining Equations 17-17 and 17-27,
At
CL
= - or A t =
= (.72)(30000)1(3600)(4.37) =
K =
C
3600V
-
3600V
1.37 h r s . Use 1 . 4 h r s .
t h e e n t i r e reach.
A t i s a l s o t h e wave t r a v e l time through
3. Determine C*
Using Equation 17-31 with A t = 1 . 4 h r s , At* = 0.4 h r s , and C = 0.72
0.4+0.5(1.4)
1.1
1
(3.T)
( 1.j ( 1 . 4 )
C* = 1 (1-.72)
= 1-(.28) 0.524 =
= 1-(.28)
-
4. Prepare an operations t a b l e f o r t h e routing.
Suitable headings and arrangement a r e shown i n Table 17-18.
5. Tabulate accumulated time i n t e r v a l s of At* and t h e inflow
discharges f o r those times.
The times a r e given i n column 1 Table 17-18, t h e inflows
taken from Figure 17-19 i n column 2.
6. Prepare t h e working equation.
C*) O1 +
From s t e p 3 , C* = 0.49. Using Equation 17-21 O? = ( 1
C X I l o r O2 = 0.51 O1 + 0.49 I1. Solutions of t h r s equation can
e a s i l y be made by accumulative m u l t i p l i c a t i o n or a desk c a l c u l a t o r .
-
7. Do t h e routing.
Follow t h e routine o f . Table 17-14.
umn 3 of Table 17-18.
The outflows a r e shown i n col-
8. Determine t h e times f o r t h e outflow.
Outflow begins a t t h e end of t h e f i r s t A t (not A t * )
interval.
J i t h A t = 1 . 4 h r s , show t h i s time i n column 4 of Table 17-18 i n t h e
?ow where outflow begins. Get succeeding times by adding A t * i n t e r r a l s , 0.4 hours i n t h i s case, a s shown i n column 4. I n p l o t t i n g o r
~ t h e r w i s edisplaying t h e inflow and outflow hydrographs they a r e put
in t h e i r proper time o r d e r , using columns 1 and 4 , as shown i n f i g u r e
L7-19.
1. Determine desired routing time i n t e r v a l , At*.
Same as Method 1, At* = 0.4 hr.
2. Determine r o u t i n 4 coefficient C.
The routing c o e f f i c i e n t YC" f o r each subreach i s computed from
t h e outflow hydrograph of t h e preceding subreach a s done i n Step
2, Method 1. A constant C may be used f o r t h e e n t i r e reach but
t h e r e s u l t a n t hydrograph w i l l vary from one produced by recornputin@
C f o r each subreach. For s i m p l i c i t y i n t h i s example, a constant
C = 0.72 i s assumed. V = 4.37 f p s .
NM Notice 4-102, August 1972
Table
17-18
Operations t a b l e f o r Example 17-11 Method 1.
Time
Inflow
Time
(hrs)
Inflow
(cfs)
Outflow
(cfs)
(1)
(2)
(3)
.4
260
980
2100
3120
3500
0
.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6
4.0
4.4
4.8
5.2
5.6
6.0
6.4
6.8
7.2
7.6
8.0
8.4
8.8
9.2
9.6
10.0
etc.
0
3220
2630
1960
1470
1120
840
630
455
345
265
180
130
100
75
60
45
35
20
10
0
0
0-1/
127
545
1307
2195
2834
3023
2830
2404
1946
1541
1198
920
692
522
396
290
212
157
117
89
67
51
36
23
etc.
12
6
3
2
1
etc.
-1/ Outflow
starts a t A t = 1 . 4 hours
NEH Notice 4-102, August 1972
Outflow
(hrs)
(4)
1.4
1.8
2.2
2.6
3.0
3.4
3.8
4.2
4.6
5.0
5.4
5.8
6.2
6.6
7.0
7.4
7.8
8.2
8.6
9.0
9.4
9.8
10.2
10.6
11.0
11.4
11.8
12.2
12.6
13.0
etc.
3. Determine length of subreach L*.
This i s t h e length of reach required t o s a t i s f y t h e r e l a t i o n s h i p
of Equation 17-26 with C = 0.72 and A t * = 0.4 h r s . Combining Equa) e have L* = (at)(v)( ~ ~ o o ) =/ c
t i o n s 17-26 and 17-17 ( l e t K = T ~ w
(0.4)(4.37)(3600)/0.72 = 8740 f t ;
4.
Compare t h e t o t a l of subreach lengths, CL* with t h e given
reach length, L.
For ELX<L
- 80 t o s t e o -5
For EL*>L go t o s t e p 7
-
I n t h i s example CL*n=l = 8740
= 17480
CL*,=~ = 26220
XL*,4
= 34960
Therefore, t h e f i r s t t h r e e rout.ings a r e ma
t h e l a s t routing by going t o s t e p 7.
t o s t e p 5 and
5. Prepare working equation and do t h e routing.
Using Equation 17-21 and t h e routing c o e f f i c i e n t computed i n s t e p
The outflows f o r each
2, O2 = (1 C)O1 + C I 1 = 0.28 O1 + 0.7211.
subreach a r e shown i n Table 17-19.
-
6.
Go . t o s t e p 2 .
7. Determine t h e length of t h e reinaining subreach t o be routed.
Subtract t h e EL* of t h e 3 completed routings, i . e , , 26220 f t from
t h e t o t a l reach length t o get t h e remaining reach length t o be routed
L** = 30000 - 26220 = 3780 ft.
8. Determine t h e A t time i n t e r v a l f o r t h e remaining subreach.
The time i n t e r v a l used here i s t h e same as t h e wave t r a v e l time
through t h e remaining subreach. Combining Equations 17-17 and 17-27
as i n s t e p 2 Method 1 At** = CL** = (0.72)(3780)/(3600)(4.37) =
0.173 h r s .
3600~
-
9. Determine t h e modified routing c o e f f i c i e n t C*.
Usinn Eouation 17-31 with At** = 0.173. A t * = 0.4 and C = 0.72,
10. Prepare working equation.
Following Method 1 O2 = (1-C*)O1 + C*Il
.09 O1 + 0.91 11.
= (1- .91) O1 + 0.9111 =
NEH Notice 4-102, August 1972
11. Do t h e routinq.
The outflow f o r t h e f r a c t i o n a l r o u t i n g a r e shown i n column 6
Table 17-19.
12. Determine t h e time f o r the routing.
The hydrograph f o r each subreach routing i s s e t back one A t time
i n t e r v a l . I n t h i s example t h e f i r s t t h r e e routings a r e s e t back
0.4 h r s each and t h e l a s t ( f r a c t i o n a l ) routing i s s e t back 0.173
h r s (round t o 0.2 h r s ) . See column 7 Table 17-19 and Figure 17-19.
When C* and At* a r e used and l o c a l inflow i s t o be added, t h e
l o c a l inflow must be used i n i t s a c t u a l time p o s i t i o n regardless
of A t and At*. That i s , t h e l o c a l inflow i s not s h i f t e d back o r
f o r t h because i t i s not affected by t h e use of C* and A t * .
*
*
Effects of transmission l o s s e s on routed flows
A flood hydrograph i s a l t e r e d by transmission l o s s e s occurring during
passage of t h e flow through a reach. The amount of l o s s depends on t h e
percolation r a t e of t h e channel, t h e wetted perimeter of channel during
flow, and t h e duration of flow f o r a p a r t i c u l a r wetted perimeter (Chapt e r 1 9 ) . Transmission l o s s v a r i e s with t h e mount of flow i n t h e
channel which means t h a t t h e most accurate method of deducting t h e t r a n s mission l o s s from t h e routed flow w i l l be on an incremental flow b a s i s .
It i s seldom worthwhile t o handle it i n t h i s manner unless t h e t r a n s mission l o s s i s very l a r g e .
An acceptable p r a c t i c e f o r handling transmission l o s s e s i s t o r o u t e t h e
inflow hydrograph i n t h e usual manner and afterwards deduct a s u i t a b l e
quantity of flow from t h e outflow hydrograph (mainly from t h e r i s i n g
limb). If t h a t outflow i s t o be routed downstream again, t h e manner of
flow deduction w i l l not be c r i t i c a l , I n some cases it may be reasonable
t o assume t h a t l o c a l inflow w i l l be completely absorbed by transmission
l o s s e s , thus no l o c a l inflow i s added t o t h e unmodified outflow hydrograph. I n other cases l o c a l r a i q f a l l may completely s a t i s f y transmission
l o s s e s , requiring unmodified l o c a l inflow t o be added t o t h e unmodified
outflow hydrograph. The use of d e t a i l e d procedures outlined i n Chapter
1 9 , "Transmission Losses", may be necessary f o r more complex S i t u a t i o n s .
a system of channels
Routing thro*
The methods of channel routing given i n Examples 17-7 throuah
- 17-11 a r e
used f o r individual reaches 02 a, stream. 0 r k n a r i l y a routing progress e s from reach through reach u n t i l t h e s t a g e s , r a t e s , o r amounts of flow
a r e known f o r s e l e c t e d p o i n t s i n t h e e n t i r e stream system of a watershed.
The method of progression w i l l be i l l u s t r a t e d using a schematic diagram
o r " t r e e grapht' of a stream system. A t y p i c a l graph i s given i n Figure
17-20. It does not need t o be drawn t o s c a l e . The main purpose of t h e
graph i s t o show t h e reaches i n t h e i r proper r e l a t i o n s h i p t o each other,
but various kinds of d a t a can be w r i t t e n down a t t h e i r respective p o i n t s
of application t o make t h e graph a complete reference during t h e routing.
Routing through a stream system begins a t t h e head of t h e uppermost reach.
I f t h e r e i s more than one possible s t a r t i n g p l a c e , a s i n Figure 17-20,
t h e most convenient should be chosen.
NEH Notice 4-102, A u g u s t 1972
Operation t a b l e f o r Example 17-11 Method 2
Table 17-19
Time
Inflow
lhrs)
(1)
0
.4
.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6
4.0
4.4
4.8
5.2
5.6
6.0
6.4
6.8
7.2
7.6
8.0
8.4
8.8
9.2
9.6
10.0
etc.
Inflow
(cfs)
(2)
0
260
980
2100
3120
3500
3220
2630
1960
1470
1120
840
630
455
345
265
180
130
100
75
60
45
35
20
10
0
etc.
Outflow
Outflow
C~*=8740 ZL*=17480
(cfs)
(cfs)
(3)
0-1 1
187
758
1724
2729
32e4
3238
2800
2195
1673
1275
962
723
530
397
302
214
154
115
86
67
51
40
25
14
4
1
0
etc.
(4)
0
135
584
1405
2358
3025
3178
2906
2394
1875
1442
1096
827
613
457
345
251
181
133
99
76
58
45
31
19
8
3
1
0
etc.
Outflow
Outflow
~ ~ * = 2 6 2 2 0 CL*=30000
(cfs)
(cfs)
(5)
0
97
447
1137
2016
2743
3056
2948
2549
2064
1617
1242
944
706
527
396
292
212
155
115
87
66
51
36
24
13
6
2
1
0
etc.
Outflow
Time
(hrs )
(6)
0-21
88
415
1072
1931
2670
3021
2955
2586
2111
1661
1280
974
730
545
409
303
220
161
119
90
68
53
38
25
14
7
2
1
0
etc.
-1/ Outflow
from subreach 1, 2 , & 3 starts A t * = 0.4 hours a f t e r i n f l o w
starts i n t o each subreach.
-2/
Outflow from subreach 4 starts A t = 0.2 hours (rounded from 0.17
h o u r s ) a f t e r inflow starts i n t o subreach 4.
NEE Notice 4-102, August 1972
m e f i r s t major s t e p i n routing through a stream system i s t o develop
t h e routing parameters, C and A t , f o r each reach. Many times it i s
necessary t o use At* t o make t h e routing i n t e r v a l uniform through t h e
stream system; these parameters should be obtained before t h e routing
begins. The method of developing t h e parameters C, K , and A t i s given
i n s t e p s 1 through 5 of Example 17-8. The method of determining C*
and At* i s given i n s t e p s 1 through 3 of Example 17-11.
The second major s t e p i s t h e development of t h e inflow hydrographs
a t heads of uppermost reaches and of l o c a l inflow hydrographs f o r a l l
reaches. The methods of Chapter 16 a r e used.
The t h i r d major s t e p i s t h e routing. For routing any p a r t i c u l a r flood
on t h e stream system pictured i n Figure 17-23 a s u i t a b l e sequence i s as
follows :
1. Route t h e inflow hydrograph a t ( a ) through reach ( a , b )
.
2 . Add l o c a l inflow of reach ( a , b ) t o t h e routed outflow t o get
t h e t o t a l outflow hydrograph, which becomes t h e inflow hydrograph f o r
reach ( b , c ) . It should be noted here t h a t l o c a l inflow f o r a reach
i s usually added a t t h e foot of t h e reach. These may be circumstances,
however, i n which t h e l o c a l inflow should be added a t t h e beginning of
t h e reach. The proper sequence f o r adding l o c a l inflow can be determined
only by evaluating each reach.
3.
the
Route t h e t o t a l outflow from reach ( a , b ) through reach ( b , c ) .
4. Add l o c a l inflow of reach ( b , c ) t o t h e routed outflow t o get
t o t a l outflow hydrograph f o r t h a t t r i b u t a r y .
5.
Route t h e inflow hydrograph a t ( d ) through reach ( d , c )
6. Add l o c a l inflow of reach ( d , c ) t o t h e routed outflow t o get
t h e t o t a l outflow hydrograph a t point ( c ) .
steps
7.
Add t h e t o t a l outflow hydrographs from. reaches ( b , c ) and (d,c)
t o get t h e t o t a l outflow hydrograph a t point ( c ) .
8.
Route t h e t o t a l hydrograph a t p o i n t ( c ) through reach ( c , f ) .
4 and 6,
9 . Add l o c a l inflow of reach ( c , f ) t o t h e routed outflow t o get
t h e t o t a l outflow hydrograph a t point (f)
.
10.
Route t h e inflow hydrograph a t point ( e ) through reach ( e , f ) .
11. Add l o c a l inflow of reach ( e , f ) t o t h e routed outflow t o get
t h e t o t a l outflow hydrograph f o r t h a t t r i b u t a r y .
12.
Route t h e inflow hydrograph a t p o i n t ( g ) through reach ( g , f ) .
13. Add l o c a l inflow of reach ( g , f ) t o t h e routed outflow t o get
t h e t o t a l outflow f o r t h a t t r i b u t a r y a t p o i n t ( f ) .
NM
Notice 4-102, August
1972
,
Add t h e t o t a l outflow hydrographs from reaches ( c , f ) , ( e , f ) ,
and ( g , f ) , s t e p s 9, 11, and 1 3 t o get t h e t o t a l outflow hydrograph a t
point ( f ) .
14.
15.
Route -the t o t a l hydrograph a t point ( f ) through reach ( f , h ) .
16. Add l o c a l inflow of reach (f ,h) t o t h e routed outflow t o
get t h e t o t a l outflow hydrograph f o r reach ( f , h ) .
17.
Route t h e t o t a l hydrograph a t point ( h ) through reach ( h , i ) .
18. Add l o c a l .inflow of reach ( h , i ) t o the routed outflow t o get
t h e t o t a l outflow hydrograph f o r reach ( h , i ) .
19.
Route t h e inflow hydrograph a t point ( j) through reach
( 3 ,k)
.
20. Add l o c a l inflow of reach ( j , k ) t o t h e routed outflow t o get
t h e t o t a l outflow hydrograph f o r reach ( j ,LC).
21.
Route t h e hydrograph a t point (k) through reach ( k , i )
22. Add l o c a l inflow of reach ( k , i ) t o t h e routed outflow t o get
t h e t o t a l outflow hydrograph f o r t h i s t r i b u t a r y .
23. ~ d tdh e t o t a l outflow hydrographs from reaches ( h , i ) and ( k , i ) ,
s t e p s 18 snd 22, t o get the t o t a l outflow hydrograph f o r point ( i ) .
24.
Route t h e hydrograph a t point ( i ) through reach ( i ,I).
t o t h e routed outflow t o get
25. Add l o c a l inflow of reach (i,l?)
This completes t h e routing
t h e t o t a l outflow hydrograph a t p o i n t (l?).
f o r a p a r t i c u l a r flood on t h i s stream system.
When manual computations a r e used, an operations t a b l e with times, inflow
hydrographs and l o c a l inflows tabulated i n t h e i r proper sequence i s usef u l . Blank columns a r e l e f t f o r t h e routed outflows and t o t a l outflows,
which a r e t a b u l a t e d as routing progresses. Above t h e appropriate columns
t h e required data and routing parameters a r e tabulated so t h a t t h e t a b l e
becomes a complete reference f o r t h e routing. A sample operations t a b l e
f o r routing by Method 2 i s shown a s Table 17-20. After t h e inflow hydrograph and l o c a l inflows are t a b u l a t e d t h e sequence of t h e work i s as
follows :
Tabulate the reach numbers i n t h e order i n which t h e routing w i l l
progress; p e r f o m t h e routings as shown i n Example 17-11 and continue
i n t h i s manner through t h e stream system. Note t h e routed outflow a t
1.17 h r s which i s rounded t o 1.0 h r s . Theoretically, the outflow hydrograph should be i n t e r p o l a t e d on a multiple of A t t o properly position
t h e hydrograph i n r e l a t i o n t o time. The l i n e a r i n t e r p o l a t i o n equation i s :
9.I. = 9 i
+
-
( ~ i ~ qi)
+ ~x
At*
- At
At*
NEX Notice 4-102, August 1972
where: qi and qi+i a r e consecutive discharges, A t * i s t h e desired time
i n t e r v a l and A t 1 s t h e required time i n t e r v a l of the p a r t i a l routing.
When using Method 2, A t i s always l e s s than At*.
If t h e i n t e r p o l a t i o n s t e p i s omitted and t h e s t a r t i n g times rounded as
i n Table 17-20 it i s recognized an e r r o r i s introduced, t h e magnitude
of which depends on t h e r e l a t i v e values of A t and At*.
NEH Notice 4-102, August 1972
Table 17-20.
REACH
L(ft)
V(fps)
Portion of a t y p i c a l o p e r a t i o n s t a b l e f o r r o u t i n g through a stream system.
.63
Atx(hrs)l.OO
LX(ft)16541.
A t (hrs)
C*
.63
1.00
2759.
.986
Time
Inflow
Outflow
14
15
4000.
4.8
.74
1.00
4000.
.17
1.00
16
19300.
2.9
Outflow
Local
Total
37000.
3.2
.65
1.00
17486.
Outflow
Local
Total
.65
1.00
17486.
Outflow
2028.
.12
.998
Outflow
0
97
498
15
etc.
'1
450
etc.
735
etc.
739
etc.
5
etc.
744
etc.
Outflow begins a t 1 . 0 0 h r s . , rounded from 1.17 h r s .
Outflow begins a t 0.00 h r s . , rounded from 0.17 h r s .
Outflow begins a t 2.00 h r s . , rounded from 2.12 h r s .
794
etc.
35
etc.
779
etc.
1279
etc.
Outflow
Lacal
Total
80
etc.
2117
etc.
031
97
498
1371
2560
3887
5007
5588
1371
2560
3887
5007
5588
5525
4963
4173
3365
2640
5525
4963
4173
3365
2640
2034
etc.
2037
etc.
NEH Notice 4-10?,
August 1972
Figure 17-13.
Convex routing coefficient versus velocity.
NM Notice 4-102, August 1972
r
I
P
~fA t * >KO or A t > 6.0
then d i v i d a both by 2 to
t i c
Eaampl. : Ata=3.0
ond A t =8.2. Use Ata:l.s
andAf=+.t.
Th. A t and At* must
ba in identical units
of time (minutes or
hours or days. etc.)
SYMBOLS
K
C
I
At
At'
/'
/
/
-
C*
- raach t r a w l tima
- rou!inq co.fiici.nt
- complied routing intonol
- desired routinq iekrval
-
required m u l l n ~c a r t t i c i ~ n l
EXAMPLE
I.
C = 0.8, Kz3.0, and A1':l.o
Compute A t = C K = 0.8(3.0) = 2.4
2.
U n A t ond A t o f o r mova I.
Giwn
3. U n C in mova 2 ond rood
At
U. 8. mARl%ENT OP ADRlCUL'IVILE
REFERENCE
Equat~onby W. A. Styner
Nomoqraph by Glenn 0. Commons
-
D l m O l i m1Y)m
Figure 17-14.
:0.628
STANDARD D M . NO.
DATE 6 . 71
ES-1025 rev. sheet 1 of 2.
NEX Notice 4-102,August 1972
HYDROLOGY:
at*+At
NOMOGRAPH FOR SOLUTION OF c*=I-(I-C)
I
tEFERENCE
Equation by W.A. Stynar
Nomograph by Glenn G. Commons
gure
17-14.
U. S. DEP-
OF AOBICULTURE
SOIL CONSERVATION SERVICE
R I G M W M O DnmnON .RmROIDCY aa*llCn
ES-1025 rev. sheet 2 of 2.
NEH Notice 4-102, August 1972
I
1.5At
STANDARD DWG. NO.
2
3
4
TIME, HOURS
Figure 17-15.
Inflow and routed outflow hydrograph f o r Example 17-7
X!i Nctice 4-102, August
1972
0
5
10
15
TIME, HOURS
20
25
30 7
-4
-4
4000
TOTAL OUTFLOW --\
ROUTED INFLOW
3
51
w
0
ct
P.
0
m
(TOTAL OUTFLOW MINUS LOCAL
C
I
N
"
C
10
ct
I-'
\O
-4
N
0
I
2
3
4
TIME, HOURS
5
6
7
8
I
2
3
4
5
TIME, HOURS
Figure 17-18. Mass inflow, mass outflow and r a t e hydrograph f o r
Eiample 17-10.
NM Notice 4-102, August 1972
4
TIME, HOURS
Figure 17-20.
Typical Schematic diagram for routing through a
system of channels.
NEH Notice 4-102, August 1972
Unit-Hydrograph Routing Methods
Principles of the u n i t hydrograph theory a r e given i n Chapter 16. They
apply t o single-peaked hydrographs o r i g i n a t i n g from uniform runoff on
t h e contributing a r e a b u t t h e y can be extended t o apply t o more complex
runoff conditions. Despite t h e l i m i t a t i o n s of t h e theory it has f e a t u r e s
t h a t can be used i n determining peak r a t e s i n stream reaches not only
when the watershed i s i n a "present" unreservoired condition but a l s o
when it i s controlled by many r e s e r v o i r s . It i s t h e ease w i t h which
complex systems of c o n t r o l s t r u c t u r e s a r e evaluated t h a t has made t h e
unit-hydrograph type of routing a popular method f o r many years. If
s u i t a b l e d a t a a r e used t h e r e s u l t s a r e usually as good a s those obtained
by more d e t a i l e d methods of routing.
I n t h i s p a r t of t h e chapter t h e basic equations f o r unit-hydrograph
routing w i l l be given and discussed ana some of t h e i r uses explained by
means of examples. The unit-hydrograph method of r o u t i n g gives only
t h e peak r a t e s of runoff. The peak-producing hydrograph, i f it i s needed,
must be obtained i n some other way.
Basic Equations
A l l of the unit-hydrograph working equations are derived from t h e r e l a tionship f o r the peak r a t e of a u n i t hydrograph:
where
qp = peak r a t e i n c f s
K = a constant (not t h e routing parameter used i n the Convex
method)
A
= drainage a r e a contributing runoff; i n square miles
Q
= average depth of runoff, i n inches, from t h e contributing
area
Tp = time t o peak, i n hours
By l e t t i n g q
f o r a watershed i n one condition
P' K, A, Q, and T stand
and using prmed symbols q;,
A ' , & I , and T' f o r t h e same watershed
i n a condition being studied, then by use of &tion
17-37 it i s evident t h a t :
%,
A' Q' Tp
(Eq. 17-38)
which i s a t y p i c a l working equation of t h e u n i t hydrograph method. It
can be used, f o r example, i n determining t h e peak r a t e s a f t e r establishment
of land use and treatment measures on a watershed. I n such work t h e pres e n t peaks, areas, runoff amounts, and peak times a r e known and it i s
only a matter of f i n d i n g t h e change i n runoff by use of Chapter 10 methods.
The areas and peak times a r e assumed t o remain constant.
NM Notice 4-102, August 1972
d
i
When a floodwater retarding s t r u c t u r e , o r other s t r u c t u r e c o n t r o l l i n g a
p a r t of the watershed, i s being used i n t h e "future" condition then t h e
value of A' i s reduced. And i f t h e r e a r e r e l e a s e s from t h e s t r u c t u r e
then they must a l s o be taken i n t o account. For a project having struct u r e s controlling a t o t a l of A* square miles and having an average r e l e a s e
r a t e of q* csm, t h e peak r a t e equation becomes:
When using Equation 17-39 t o f i n d t h e reduced peak r a t e t h e major assumpt i o n i s t h a t t h e s t r u c t u r e s a r e about uniformly d i s t r i b u t e d over t h e
watershed. Another assumption i s t h a t all s t r u c t u r e s contribute t o q*,
but t h i s i s sometimes too conservative an assumption (see t h e section
t i t l e d "Use of Equation 17-43 on l a r g e watersheds").
When A' = A and T ' = T , which i s t h e usual case when evaluating land
use and treatment Peffects, Equation 17-38 becomes:
which i s one of t h e basic expressions of t h e u n i t hydrograph theory. I f
t h e same s i m p l i f i c a t i o n applies when evaluating s t r u c t u r e s then Equation
17-39 becomes:
Equation 17-41 can be f u r t h e r s i m p l i f i e d by using:
(Eq. 17-42)
where r i s t h e f r a c t i o n of drainage a r e a under control or t h e percent of
control divided by 100. Using Equation 17-42 i n Equation 17-41 gives:
Effects of storm duration and time of concentration
When the e f f e c t s of a change i n e i t h e r t h e storm duration o r t h e time of
concentration must be taken i n t o account, one way t o do it i s t o use t h e
following r e l a t i o n from Chapter 16:
where
Tp = time t o peak, i n hours
a
= a constant
D
= storm duration, i n hours, during which runoff i s
generated; it i s u s u a l l y - l e s s than t h e t o t a l storm duration.
NFX Notice 4-102, August 1972
b
= a constant
T, = time of concentration, i n hours
As shown i n Chapter 16, t h e constants a and b can be taken a s 0.5 and
0.6 respectively, f o r most problems, i n which case Equation 17-44
becomes :
Using Equation 17-45 i n equations 17-37. 17-38, and 17-39 produces working equations i n which e i t h e r t h e storm duration or t h e time of concent r a t i o n can be changed and t h e e f f e c t of t h e change determined. Such
equations a r e not often used because t h e main comparison i s usually
between present and f u t u r e conditions i n which only runoff amount and
drainage area w i l l change. I n s p e c i a l problems where storm duration
must be taken i n t o account t h e r e a r e other approaches t h a t a r e more
applicable ( s e e t h e s e c t i o n t i t l e d "Use of Equation 17-43 on l a r g e
watersheds").
Elimination of Tp
I n many physiographic a r e a s t h e r e i s a consistent r e l a t i o n between Tp
and A because t h e r e i s a t y p i c a l storm condition o r pattern. The r e l a t i o n s h i p i s usually expressed as:
where c i s a constant m u l t i p l i e r and d i s a constant exponent.
s t i t u t i n g C A ~f o r Tp i n Equation 17-37 gives:
q p = k A (1- d)
where k = K/c.
Sub-
(Eq. 17-47)
Letting (1 - d ) = h, Equation 17-47 becomes:
qp = k Ah Q
(Eq. 17-48)
which i s the working equation i n p r a c t i c e . The parameters k and h are
obtained from d a t a f o r a l a r g e storm over t h e watershed o r region being
studied. Values of q a t s e v e r a l l o c a t i o n s a r e obtained e i t h e r from
P by means of slope-area measurements (Chapter 14):
streamflow s t a t i o n s or
values of Q associated with each qp are obtained f r o m t h e s t a t i o n d a t a
o r by use of r a i n f a l l and watershed data and methods of chapter 10; and
drainage areas a t each l o c a t i o n a r e determined. A p l o t t i n g of q /Q
against A i s made on log paper and a l i n e of best f i t i s drawn t rough
t h e p l o t t i n g . The m u l t i p l i e r k i s the i n t e r c e p t of t h e l i n e where A = 1
square mile and t h e exponent h is t h e slope of t h e l i n e . See t h e s e c t i o n
t i t l e d "Use .of Equations 17-48, 17-50, and 17-52" f o r an a p p l i c a t i o n of
t h i s procedure.
%
After h i s known, t h e equivalent of Equation 17-38 i s :
NEH Notice 4-102, August 1972
i/
The k ' s cancel out i n making t h i s change.
In t h e "Concordant Flow" method of peak determination, Equation 17-48 i s
modified t o take i n t o account t h e e f f e c t s of control s t r u c t u r e s and t h e i r
r e l e a s e r a t e s , with t h e working equation being:
which i s t h e same as Equation 17-43 i n form but where qp i s now determined
from Equation 17-48.
Equations 17-39, 17-41, 17-43, 17-50, and 17-51 should be used only when
t h e storm runoff volume does not exceed t h e storage capacity of t h e s t r u c t u r e with t h e smallest capacity. If t h e runoff does exceed t h a t capacity
these equations must be modified further. Equation 17-50, f o r example,
becomes:
C
where Qs i s the average storage capacity of t h e s t r u c t u r e s . It i s shown
i n Example 17-20 how Equation 17-51 and similar equations can be used
even when t h e capacity v a r i e s from s t r u c t u r e t o s t r u c t u r e .
Working equations f o r s p e c i a l cases
Additional equations can be developed from those given i f a s p e c i a l problem a r i s e s i n watershed evaluation. For an example, suppose t h a t Equat i o n 17-43 i s t o be used f o r determining the e f f e c t s of a proposed system
of floodwater r e t a r d i n g s t r u c t u r e s i n a watershed, and t h a t the evaluation
reaches a r e so long t h a t t h e percent of area reservoired v a r i e s s i g n i f i cantly from the head t o t h e foot of the reach. To modify Equation 17-43
f o r t h i s case, l e t A* be t h e a r e a reservoired, A t h e t o t a l area, and r =
A*/A f o r t h e head of t h e reach; and l e t B* be the t o t a l a r e a reservoired
(including A*), B t h e t o t a l area (including A ) , and r" = BY/B f o r t h e
foot of t h e reach. For evaluations t o be made a t t h e foot of t h e reach,
Equation 17-43 then becomes:
After f i r s t computing ( 2 - r - r 1 ' ) / 2 = C ' and (A* + B*)/2 = C" f o r t h e
reach, t h e working equation becomes:
q;, =
9p
C'
+
q* C"
NEH Notice 4-102, August 1972
(Eq. 17-54)
where C' and C" are t h e computed coefficients.
requires i t s own s e t of c o e f f i c i e n t s .
Each evaluation reach
Examples
The problems i n the following examples range from t h e very simple t o
t h e complex, t h e l a t t e r being given t o show t h a t unit-hydrograph methods
have wide application. For some complex problems, however., it w i l l
generally be more e f f i c i e n t t o use t h e SCS electronic-computer evaluation
program.
-
This b a s i c expression of t h e u n i t hydrograph
Use of Equation 17-40.
theory has many uses. The major l i m i t a t i o n i n i t s use i s t h a t Q and
Q' must be about uniformly d i s t r i b u t e d over t h e watershed being studied.
The following i s a t y p i c a l but simple problem.
&ample 17-12.--A watershed has a peak discharge of 46,300 c f s from a
storm t h a t produced 2.54 inches of runoff. What would t h e peak r a t e
have been f o r a runoff of 1.68 inches?
1. Apply Equation 17-40.
For t h i s problem q = 46,300 c f s , Q = 2.54 inches, and Q' = 1.68
P
inches. By Equation
17-40 q; = 46300(1.68/2.54) = 30,604 c f s , rihich
i s roun'ded t o 30,600 c f s .
Use of Equation 17-43 - The major l i m i t a t i o n s i n t h e use of t h i s equation
a r e t h a t both t h e runoffs and t h e s t r u c t u r e s must be about uniformly dist r i b u t e d over t h e watershed and t h a t t h e stream t r a v e l times f o r the
1,
future" condition must be about t h e same as f o r t h e "present." The
following i s a t y p i c a l but simple problem.
d
Example 17-13.--A watershed of 183 square miles has a flood peak of 37,800
c f s . I f 42 square miles of t h i s watershed were controlled by floodwater
r e t a r d i n g s t r u c t u r e s having a n average r e l e a s e r a t e of 15 csm, what would
t h e reduced peak be?
1. Computer.
By Equation 17-42 r = 42/183 = 0.230 because A* = 42 and A = 183
square miles.
2. Apply Equation 17-43.
For t h i s problem, q, = 37,800 c f s , r = 0.230 from s t e -p 1,
. q-* = 15
csm, and A* = 42 s & r e miles. ~y Equation 17-43 q' = 37800(1
0.230) + 15(42) = 29,736 c f s , which i s rounded t o 25,700 c f s . This
i s t h e reduced peak.
-
Use of Equation 17-43 on l a r g e watersheds.- I f Equation 17-43 i s used f o r
evaluating t h e e f f e c t s of s t r u c t u r e s i n a l a r n-e watershed or r i v e r basin
t h e releases from s t r u c t u r e s far upstream may not add t o t h e peak r a t e s
i n t h e lower reaches of t h e main stem. And i f r e l e a s e s from c e r t a i n upstream s t r u c t u r e s do not a f f e c t peaks f a r downstream then those s t r u c t u r e s
a l s o a r e not reducing t h e peak r a t e s , t h e r e f o r e t h e i r drainage areas
should not be used i n t h e equation.
NM
Notice 4-102, August 1972
L/
L
I n problems of t h i s kind t h e approach t o be taken i s r e l a t i v e l y simple
though there a r e supplementary computations t o be made before t h e equat i o n i s used. The key s t e p i n t h e approach i s finding the Tp f o r an
evaluation flood and using only those areas and s t r u c t u r e s close enough
t o t h e sub-basin o u t l e t t o a f f e c t t h e peak r a t e of t h a t flood. How t h i s
i s done w i l l be i l l u s t r a t e d using the data and computations of Table
17-21. The data a r e f o r a sub-basin of 620 square miles, with a time of
concentration of 48 hours. Storm durations f o r t h e floods t o be evaluated w i l l vary from 1 t o over 72 hours, which means t h a t t h e sub-basin
Tp w i l l a l s o vary considerably.
Table 17-21 i s developed as follows:
Column 1 l i s t s t h e t r a v e l times on t h e sub-basin main stem from t h e
o u t l e t point t o selected points upstream, which a r e mainly junctions
with major t r i b u t a r i e s . The f i r s t entry i s f o r t h e o u t l e t p o i n t .
Column 2 gives t h e t o t a l drainage a r e a above each s e l e c t e d p o i n t .
Column 3 gives t h e increments of area.
L
Column 4 gives t h e accumulated a r e a s , going upstream. These a r e
the contributing areas when t h e flood's Tp i s within t h e l i m i t s shown i n
column 1. For example, when Tp i s between 3.5 and 9 . 1 hours, t h e cont r i b u t i n g drainage a r e a i s 74 square miles. Tp must be a t l e a s t 48 hours
before the e n t i r e watershed contributes t o t h e peak r a t e .
Column 5 shows t h e t o t a l areas controlled by s t r u c t u r e s .
Column 6 gives t h e increments of controlled area.
Column 7 gives t h e accumulated controlled a r e a s , going upstream.
Column 8 gives values of r , which a r e computed using e n t r i e s of
columns 7 and 4.
Column 9 gives values of (1
of column 8.
-
r ) , which a r e computed using e n t r i e s
Column 10 gives t h e t o t d l average r e l e a s e r a t e i n c f s f o r t h e cont r o l l e d areas of column 7. For t h i s t a b l e t h e average r e l e a s e r a t e q*
i s 7 csm. Therefore t h e q*(A*) entry f o r a p a r t i c u l a r row i s t h e column
7 a r e a of t h a t row multiplied by t h e average r a t e i n csm.
Ll
Only columns 1, 9, and 10 are used i n t h e remaining work. To determine
t h e e f f e c t of t h e s t r u c t u r e s t h e qp and T of t h e evaluation flood must
be known, t h e proper e n t r i e s taken from t g e t a b l e , and Equation 17-43
applied. For example, i f qp = 87,000 cfs and Tp = 24 hours f o r a p a r t i c u l a r flood, f i r s t e n t e r column 1 with Tp = 24 hours and f i n d t h e row t o
be used, i n t h i s case it i s between Tt values of 21.1 and 28.0 hours;
next s e l e c t ( 1 r ) = 0.459 from column 9 of t h a t row and q * ( A W ) = 1,491
c f s from column 10; f i n a l l y , use Equation 17-43 which gives q i = 87000
(0.459) + 1491 = 41,424 c f s , which i s rounded t o 41,400 c f s .
-
NM Notice 4-102, August 1972
Table
17-21
D a t a and working t a b l e f o r
Tt
(hrs)
A
(sq.mi.)
(1)
(2)
o
AA
AU
A*
(sq.mi.) (sq.mi.) ( s q . m i . )
(3)
(4)
(5)
620
612
3.5
606
9.1
546
15.3
456
21.1
376
28.0
226
31.0
125
42.0
27
'48.0
0
AA*
4:
(sq.mi.) ( s q . m i . )
(6)
(7)
a l a r g e watershed
(1 - r )
r
6
60
(cfs)
(8)
(9)
(lo)
8
0
1.000
0
14
74
0
0
3
3
.214
.786
21
24
27
.365
.635
189
43
70
.426
.574
490
56
126
517
.483
882
87
213
.541
.459
1491
'77
290
.586
.414
2030
48
338
5'70
.430
2366
21
359
.580
.420
2513
359
356
332
90
80
164
244
289
233
150
101
394
495
146
69
98
593
27
620
21
o
--
-1/ Using
q*(~*p'
359
8
2.0
use of Equation 17-43 on
an average rate of q* = 7 c s m .
'..-
If any o t h e r p o i n t i n t h e sub-basin i s a l s o t o be used f o r evaluation of
s t r u c t u r e e f f e c t s t h e n a s e p a r a t e t a b l e i s needed f o r t h a t p o i n t .
Use of Equations 17-48, 17-50, and 17-52.- When streamflow d a t a o r slopea r e a measurements and Q e s t i m a t e s a r e a v a i l a b l e f o r a watershed and i t s
v i c i n i t y , t h e information can be used t o c o n s t r u c t a graph of qp/Q and A
as shown i n Figure 17-21.
This i s t h e g r a p h i c a l form of Equation 17-48.
If a l i n e with an i n t e r c e p t of 484 c f s / i n . and slope of 0 . 4 can be reasonably w e l l f i t t e d t o t h e d a t a , as i n t h i s case, it means t h a t t h e hydrograph shapes of t h e s e watersheds c l o s e l y resemble t h e shape of t h e u n i t
hydrograph of Figure 16-1 ( s e e Chapter 1 6 ) . Usually t h e slope w i l l be
0.4 f o r o t h e r shapes of hydrographs ( t h e reason f o r t h i s i s discussed
i n Chapter 1 5 ) but t h e i n t e r c e p t w i l l vary. kor t h e l i n e of Figure 17-21,
Equation 17-48 can be w r i t t e n :
The following examples show some t y p i c a l uses of t h e graph o r i t s equation.
Example 17-14.--For a watershed i n t h e region t o which Figure 17-21
a p p l i e s , A = 234 square miles and Q = 3.15 inches f o r a storm event.
What i s q 7
P'
1.
Find qp/Q f o r t h e given A.
Enter t h e graph with A = 234 square miles and a t t h e l i n e of r e l a t i o n f i n d qp/Q = 4,290 c f s / i n .
2.
Compute q,.
Multiplying %/Q by Q g i v e s qp, t h e r e f o r e , qp = 3.15(4290) = 13,500
c f s by a s l i d e - r u l e computation.
If p a r t of a watershed i s c o n t r o l l e d by floodwater r e t a r d i n g s t r u c t u r e s
t h e graph can be used t o g e t h e r with equation 17-50, as follows:
Example 17-15.--A watershed o f 234 square miles has a system of floodwater r e t a r d i n g s t r u c t u r e s on it c o n t r o l l i n g a t o t a l of 103 square
miles. Each s t r u c t u r e has a s t o r a g e c a p a c i t y of 4.5 inches b e f o r e
discharge begins through t h e emergency spillway. Each s t r u c t u r e has an
average r e l e a s e r a t e of 1 5 csm. When t h e storm runoff Q i s 4 . 1 inches
what i s t h e pea& r a t e with ( a ) s t r u c t u r e s not i n p l a c e , and ( b ) s t r u c t u r e i n place?
1. Determine t h e f l o o d peak f o r t h e watershed with s t r u c t u r e s not
i n place.
Use t h e method of Example 17-14. Enter Figure 17-21 with A = 234
square miles and f i n d qp/Q = 4,290 c f s / i n . Multiplying t h a t r e s u l t
by Q = 4.1 inches gives qp = 4.1(4290) = i 7 , 6 0 0 c f s by a s l i d e r u l e computation. This discharge i s (k A Q ) i n Equation 17-50.
NEH Notice 4-102, A u g u s t 1972
-
2. Determine (1 r ) .
From Equation 17-42 r = A*/A = 1031234 = 0.440.
1 - 0.440 = 0.560.
3.
Then ( 1
- r) =
Determine t h e flood peak f o r t h e watershed with s t r u c t u r e s i n
place.
Use Equation 17-50 with t h e r e s u l t s of s t e p s 1 and 2 and t h e given
data f o r controlled a r e a and r e l e a s e r a t e : q$ = 17600(0.560) +
15(103) =. 11,410 c f s , using a s l i d e - r u l e f o r t h e m u l t i p l i c a t i o n s .
Round t h e discharge t o 11,400 c f s .
If t h e storm runoff exceeds t h e storage c a p a c i t i e s of t h e s t r u c t u r e s but
t h e capacities a r e t h e same f o r a l l s t r u c t u r e s then Equation 17-52 can
be applied as shown i n t h e following example.
Example.17-16.--For the same watershed and s t r u c t u r e s used i n Example
17-18 find t h e peak r a t e s without and with s t r u c t u r e s i n place when t h e
storm runoff i s 6.21 inches.
1. Determine t h e flood peak f o r t h e watershed with s t r u c t u r e s not
i n place.
Use t h e method of Example 17-14. h t e r Figure 17-21 with A = 234
square miles and f i n d q p / ~= 4,290 c f s / i n . This i s (k ~ h i)n Equat i o n 17-52. Multiplying t h a t r e s u l t by Q = 6.21 inches gives q =
6.21 (4290) = 26,700 c f s by a s l i d e - r u l e computation. This i s ?he
peak r a t e without s t r u c t u r e s i n place.
2. Determine r.
From Equation 17-42 r = AY/A = 1031234 = 0.440
3. Determine t h e flood peak f o r t h e watershed with s t r u c t u r e s i n
place.
Use Equation 17-52 with (k A ~ =
) 4,290 c f s / i n . from s t e p 1; Q =
6.21 inches, as given; r = 0.440, from s t e p 2; and Qs = 4.5 inches,
q* = 15 csrn, and A* = 103 square miles as given i n Example 17-17.
O.bkO(4.5)) + 15(103) = 18160 + 1540 =
Then q$ = 4290(6.21
19,700 cfs.
-
Note t h a t t h e e f f e c t of t h e r e l e a s e r a t e on reducing t h e storm runoff
amount i s not taken i n t o account i n t h i s example. This means t h a t t h e
peak of 19,700 c f s i s s l i g h t l y too l a r g e and t h a t t h i s approach gives a
conservatively high answer.
I f t h e storage c a p a c i t i e s of t h e s t r u c t u r e s vary then Equation 17-52 i s
used with ( Q - r Q s ) computed by a more d e t a i l e d method, as shown i n
t h e following example.
Example 17-17.--A watershed of 311 square miles has a system of floodwater retarding s t r u c t u r e s c o n t r o l l i n g a t o t a l of 187 square miles and
having average r e l e a s e r a t e s of 8 csm. Storage c a p a c i t i e s of t h e s t r u c t u r e s are shown i n column 3 of Table 17-22; t h e s e a r e t h e c a p a c i t i e s
before emergency spillway discharge begins. When t h e storm runoff i s
NEH Notice 4-102, August 1972
L
uniformly 7.5 inches over the watershed, what i s t h e peak r a t e of flow
with ( a ) no s t r u c t u r e s i n place and ( b ) s t r u c t u r e s i n place?
1. Determine the flood peak f o r t h e watershed with s t r u c t u r e s not
i n place.
Use t h e method of Example 17-14. Enter Figure 17-21 with A = 311
square miles and f i n d q p / 4 = 4,800 c f s / i n . This . i s ( k A ~ i) n Equat i o n 17-52. Multiplying t h a t r e s u l t by Q = 7.5 inches gives qp =
36,000 cfs by a slide-rule computation. This i s t h e peak r a t e without s t r u c t u r e s i n place.
2. Compute the equivalent of ( r QS) i n Equation 17-52.
The f a c t o r ( r Qs) can a l s o be expressed a s :
where Ax i s t h e drainage a r e a i n square miles of t h e x-th s t r u c t u r e
and Qsx i s t h e reservoir capacity i n inches f o r t h a t s t r u c t u r e .
In
Table 17-22 each drainage a r e a of column 2 i s multiplied by t h e respective storage of column 3 t o get t h e entry f o r column 4. But note
t h a t when t h e storage exceeds t h e storm runoff it i s t h e storm runoff
amount, i n t h i s case 7.5 inches, which i s used t o get t h e e n t r y f o r
column 4. Equation 17-56 i s solved f o r ( r Q s ) by dividing t h e sum
of column 4 by the t o t a l watershed area:
( r Qs)=
967.26
311
= 3.11 inches
(Note: Column 4 is not needed if the c a l c u l a t i o n s a r e made by
accumulative multiplication on a desk-calculator. )
3.
Determine the flood peak f o r t h e watershed with s t r u c t u r e s i n
place.
Use Equation 17-52 with (k A ~ =) 4,800 c f s / i n . from s t e p 1; Q = 7.5
inches, as given; ( r QS) = 3.11 inches as computed i n s t e p 2; and
q* = 8 csm and A* = 187 square miles, a s given. This gives: q i =
4800(7.5 - 3.11) + 8(187) = 21100 + 1495 = 22,595 c f s , which i s
rounded t o 22,600 c f s . This i s t h e peak r a t e with s t r u c t u r e s i n
place.
DISCUSSION. These examples are a sample of t h e many ways i n which t h e
unit-hydrograph methoa of routing can be used. Accuracy of t h e method
depends on what has been ignored, such as v a r i a b l e r e l e a s e r a t e , surcharge
s t o r a g e , and so on. In general, t h e method gives conservative results-t h a t i s , t h e e f f e c t s of s t r u c t u r e s , f o r example, a r e usually underestimated so t h a t t h e peak r a t e i s s l i g h t l y too high.
L
The examples a l s o show t h a t as t h e problem contains more d e t a i l s t h e procedure g e t s more complex. It i s e a s i l y possible t o make t h i s "short-cut"
method so complicated it becomes d i f f i c u l t t o get t h e solution. For t h i s
reason, and f o r reasons of accuracy, it i s b e t t e r t o use t h e SCS e l e e t r o n i c computer program f o r complex routing problems.
NHI Notice 4-102, August 1972
Table 17-22
Floodwater
retarding
structure
*
Area and storage data f o r Example 17-17
Gontribut ing
drainage
area
( s q . mi.)
Storage
(in.)
Ax
X
Qsx
( s q . mi. x i n . )
This i s (drainage a r e a ) x (storm runoff of 7.5 inches)
because t h e storage g r e a t e r than t h e runoff i s ineffect i v e and should not be used i n the computation.
NEB ~ o t ' i c e4-102, August 1972