Hydrologic Routing
Reading: Applied Hydrology Sections
8.1, 8.2, 8.4
Flow Routing
Q
• Procedure to
determine the flow
hydrograph at a
point on a
watershed from a
known hydrograph
upstream
• As the hydrograph
travels, it
– attenuates
– gets delayed
t
Q
t
Q
t
Q
t
2
Why route flows?
Q
t


Account for changes in flow hydrograph as a flood wave passes
downstream
This helps in


Accounting for storages
Studying the attenuation of flood peaks
3
Types of flow routing
• Lumped/hydrologic
– Flow is calculated as a function of time alone at a
particular location
– Governed by continuity equation and
flow/storage relationship
• Distributed/hydraulic
– Flow is calculated as a function of space and time
throughout the system
– Governed by continuity and momentum
equations
4
Hydrologic Routing
Discharge
Discharge
I (t )
Inflow
Transfer
Function
Q(t )
Outflow
I (t )  Inflow
Q(t )  Outflow
Upstream hydrograph
Downstream hydrograph
Input, output, and storage are related by continuity equation:
dS
 I (t )  Q(t ) Q and S are unknown
dt
Storage can be expressed as a function of I(t) or Q(t) or both
S  f (I ,
dI
dQ
,  , Q,
, )
dt
dt
For a linear reservoir, S=kQ
5
Lumped flow routing
• Three types
1. Level pool method (Modified Puls)
– Storage is nonlinear function of Q
2. Muskingum method
– Storage is linear function of I and Q
3. Series of reservoir models
– Storage is linear function of Q and its time
derivatives
6
S and Q relationships
7
Level pool routing
• Procedure for calculating outflow hydrograph
Q(t) from a reservoir with horizontal water
surface, given its inflow hydrograph I(t) and
storage-outflow relationship
8
Level pool methodology
Discharge
dS
 I (t )  Q(t )
dt
Inflow
I j 1
Outflow
S j 1
( j 1) t
( j 1) t
Sj
j t
jt
 dS 
Ij
Q j 1
Qj
S j 1  S j
t
jt
( j  1)t
Time
Storage
t
2 S j 1
t
 Idt 

 Qdt
I j 1  I j
2

 Q j 1  I j 1  I j 
Unknown
Need a function relating
S j 1
Sj
9
Time
Q j 1  Q j
2S
 Q, and Q
t
Storage-outflow function
2
2S j
Known
t
Qj
Level pool methodology
•
Given
– Inflow hydrograph
– Q and H relationship
•
Steps
1. Develop Q versus Q+ 2S/t relationship using
Q/H relationship
2S j 1
2S j
2. Compute Q+ 2S/t using t  Q j 1  I j 1  I j  t  Q j
3. Use the relationship developed in step 1 to get Q
10
Ex. 8.2.1
Given I(t)
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
Inflow
(cfs)
0
60
120
180
240
300
360
320
280
240
200
160
120
80
40
0
0
0
0
0
0
0
400
300
Inflow (cfs)
Time
(min)
Given Q/H
200
100
0
0
50
100
150
200
Tim e (m in)
Area of the reservoir = 1 acre, and outlet diameter = 5ft
11
250
Elevation H Discharge Q
(ft)
(cfs)
0
0
0.5
3
1
8
1.5
17
2
30
2.5
43
3
60
3.5
78
4
97
4.5
117
5
137
5.5
156
6
173
6.5
190
7
205
7.5
218
8
231
8.5
242
9
253
9.5
264
10
275
Ex. 8.2.1 Step 1
Develop Q versus Q+ 2S/t relationship using Q/H relationship
S  Area  Height  43560  0.5  21,780 ft 3
2S
2  21780
Q 
 3  75.6 cfs
t
10  60
300
250
Outflow Q (cfs)
Elevation H Discharge Q Storage S 2S/ t + Q
3
(ft )
(ft)
(cfs)
(cfs)
0
0
0
0
0.5
3
21780
75.6
1
8
43560
153.2
1.5
17
65340
234.8
2
30
87120
320.4
2.5
43
108900
406
3
60
130680
495.6
3.5
78
152460
586.2
4
97
174240
677.8
4.5
117
196020
770.4
5
137
217800
863
5.5
156
239580
954.6
6
173
261360
1044.2
6.5
190
283140
1133.8
7
205
304920
1221.4
7.5
218
326700
1307
8
231
348480
1392.6
8.5
242
370260
1476.2
9
253
392040
1559.8
9.5
264
413820
1643.4
10
275
435600
1727
200
150
100
50
0
0
500
1000
2S/ t + Q (cfs)
12
1500
2000
Step 2
Compute Q+ 2S/t using
2S j 1
t
 Q j 1  I j 1  I j 
2S j
t
Qj
At time interval =1 (j=1), I1 = 0, and therefore Q1 = 0 as the reservoir is empty
Write the continuity equation for the first time step, which can be
used to compute Q2
 2S 2

 2S

 Q2   I 2  I1    1  Q1 

 t

 t

 2S 2

 2S

 Q2   I 2  I1    1  Q1   0  60  60

 t

 t

13
Step 3
300
Use the relationship between 2S/t + Q versus Q to
compute Q
 2S 2

 Q2   60

 t

Outflow Q (cfs)
250
200
150
100
50
0
0
500
1000
1500
2000
2S/ t + Q (cfs)
Use the Table/graph created in Step 1 to compute Q
What is the value of Q if 2S/t + Q = 60 ?
Q 0
(3  0)
(60  0)  2.4 cfs
(76  0)
So Q2 is 2.4 cfs
Repeat steps 2 and 3 for j=2, 3, 4… to compute
Q3, Q4, Q5…..
14
Elevation H Discharge Q Storage S 2S/ t + Q
3
(ft )
(ft)
(cfs)
(cfs)
0
0
0
0
0.5
3
21780
75.6
1
8
43560
153.2
1.5
17
65340
234.8
2
30
87120
320.4
2.5
43
108900
406
3
60
130680
495.6
3.5
78
152460
586.2
4
97
174240
677.8
4.5
117
196020
770.4
5
137
217800
863
5.5
156
239580
954.6
6
173
261360
1044.2
6.5
190
283140
1133.8
7
205
304920
1221.4
7.5
218
326700
1307
8
231
348480
1392.6
8.5
242
370260
1476.2
9
253
392040
1559.8
9.5
264
413820
1643.4
10
275
435600
1727
Ex. 8.2.1 results
2S j
t
2S j 1
 Qj 
2S j
t
 Q j  2Q j
15
t
 Q j 1  I j 1  I j 
2S j
t
Qj
Ex. 8.2.1 results
12.0
Storage (acre-ft)
10.0
Outflow
hydrograph
8.0
6.0
4.0
2.0
0.0
0
20
40
60
80
100
120
140
160
180
200
220
Time (minutes)
400
350
Inflow
Discharge (cfs)
300
Peak outflow intersects with the receding
limb of the inflow hydrograph
250
200
150
Outflow
100
50
0
0
20
40
60
80
100
120
TIme (minutes)
16
140
160
180
200
220
Q/H relationships
http://www.wsi.nrcs.usda.gov/products/W2Q/H&H/Tools_Models/Sites.html
17 an NRCS Reservoir
Program for Routing Flow through
Hydrologic river routing (Muskingum
Method)
Wedge storage in reach
S Prism  KQ
S Wedge  KX ( I  Q)
Advancing
Flood
Wave
I>Q
K = travel time of peak through the reach
X = weight on inflow versus outflow (0 ≤ X ≤ 0.5)
X = 0  Reservoir, storage depends on outflow, no
wedge
X = 0.0 - 0.3  Natural stream
S  KQ  KX ( I  Q)
S  K [ XI  (1  X )Q]
I
Q
I Q
Q
Q
I
Q
Receding
Flood
Wave
Q>I
QI
I
I
Muskingum Method (Cont.)
S  K [ XI  (1  X )Q]
S j 1  S j  K{[ XI j 1  (1  X )Q j 1 ]  [ XI j  (1  X )Q j ]}
Recall:
S j 1  S j 
I j 1  I j
2
t 
Q j 1  Q j
2
Combine:
Q j 1  C1I j 1  C2 I j  C3Q j
t
t  2 KX
2 K (1  X )  t
t  2 KX
C2 
2 K (1  X )  t
2 K (1  X )  t
C3 
2 K (1  X )  t
C1 
If I(t), K and X are known, Q(t) can be calculated using above
19
equations
Muskingum - Example
• Given:
– Inflow hydrograph
– K = 2.3 hr, X = 0.15, t = 1 hour,
Initial Q = 85 cfs
• Find:
– Outflow hydrograph using
Muskingum routing method
t  2 KX
1  2 * 2.3 * 0.15

 0.0631
2 K (1  X )  t 2 * 2.3(1  0.15)  1
t  2 KX
1  2 * 2.3 * 0.15
C2 

 0.3442
2 K (1  X )  t 2 * 2.3(1  0.15)  1
2 K (1  X )  t 2 * 2.3 * (1  0.15)  1
C3 

 0.5927
2 K (1  X )  t
2 * 2.3(1  0.15)  1
C1 
Period
(hr)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Inflow
(cfs)
93
137
208
320
442
546
630
678
691
675
634
571
477
390
329
247
184
134
108
90
20
Muskingum – Example (Cont.)
Q j 1  C1I j 1  C2 I j  C3Q j
Period
(hr)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
C1 = 0.0631, C2 = 0.3442, C3 = 0.5927
800
700
Discharge (cfs)
600
500
400
300
200
100
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
Time (hr)
21
Inflow
(cfs)
93
137
208
320
442
546
630
678
691
675
634
571
477
390
329
247
184
134
108
90
C1Ij+1
0
9
13
20
28
34
40
43
44
43
40
36
30
25
21
16
12
8
7
6
C2Ij
0
32
47
72
110
152
188
217
233
238
232
218
197
164
134
113
85
63
46
37
C3Qj
0
50
54
68
95
138
192
249
301
343
369
380
376
357
324
284
245
202
162
128
Outflow
(cfs)
85
91
114
159
233
324
420
509
578
623
642
635
603
546
479
413
341
274
215
170