UNIT HYDROGRAPH FROM COMPLEX STORMS
5.5 UNIT HYDROGRAPH FROM COMPLEX STORMS
Unit hydrographs from complex storms, involving varying intensities of rain can be obtained by considering the complex storm as successive unit storms of different intensities and the runoff hydrograph (due to complex storm) as the result of superposition of the successive storm hydrographs. The ordinates of each storm hydrograph are obtained as ‘the storm inten- sity times the corresponding ordinate of the unit hydrograph’ as shown in Fig. 5.14. The unit
hydrograph ordinates u 1 , u 2 , ... are thus obtained by writing a series of equations for each of the ordinates Q 1 , Q 2 , .... of the runoff hydrograph (due to complex storm) and successively solving them. In Fig. 5.14,
and so on. Thus, the t r – hour unit graph ordinates can be determined. Although the method is straight forward, errors will creep in due to the assumptions on the intensity and duration of rainfall and deduction of an assumed base flow; many trials are required to get a reasonable unit graph.
HYDROGRAPHS
Table 5.1 Derivation of the 3-hour unit hydrograph. Example 5.1
Date Time TRO 1 BFO 2 DRO 2 UGO 4 Time 5 (hr)
(5) ÷P net from begin-
(cumec)
(cumec) ning of surface run off (hr)
1 2 3 4 5 6 7 1-3-1970
48 97 3.65 18 2-3-1970
Σ DRO = 986 cumec
1 TRO —Total runoff ordinate = gauged discharge of stream 2 BFO—Base flow ordiante read from graph separation line shown in Fig. 5.10 (a). N = 0.83 A 0.2
= 0.89(40) 0.2 = 1.73 days = 1.73 × 24 = 41.4 hr from peak, which is seen not applicable here; hence an arbitrary separation line is sketched. 3 DRO —Direct runoff ordinate = TRO—B.F.O.
4 UGO —Unit hydrograph ordinate
DRO 3 Σ DRO t . 986(3 60 × × 60)m
; P net =
40 6 10 m net 2 A ×
= 0.266 m = 26.6 cm
5 Time from beginning of direct surface runoff is at 5 hr on 1-3-1970, which is reckoned 0 hr for unit hydrograph. The time base for unit hydrograph is 33 hours.
The 3-hour unit hydrogryph is plotted in Fig. 5.10 (b) to a different vertical scale and its peak is 9.23 cumec.
HYDROLOGY
i (cm/hr) all
due to 3-storms
y cm y cm
nets
Rainf intensity x cm x cm
z cm z cm
u 1 1 Observed hydrograph zu z
1 1 Q=xu 1 1
(cumec)
u y yu
Q=xu+yu 2 2 1
Q = x u + y u + z u , etc 3 3 2 1
Successively solve 2 2 for u , u , u , etc. 1 2 3
which give t – hr UGO
0 0 0 Time t (hr)
Fig. 5.14 Derivation of unit hydrograph from multi-period storms
Example 5.2 The stream flows due to three successive storms of 2.9, 4.9 and 3.9 cm of 6 hours duration each on a basin are given below. The area of the basin is 118.8 km 2 . Assuming a constant base flow of 20 cumec, derive a 6-hour unit hydrograph for the basin. An average storm loss of 0.15 cm /hr can be assumed.
Time (hr): 0 3 6 9 12 15 18 21 24 27 30 33 Flow (cumec):
144 84.5 45.5 29 20 Solution Let the 6-hour unit hydrograph ordinates be u 0 ,u 1 ,u 2 ,u 3 ,u 4 , ...., u 7 at 0, 3, 6, 12, ....,
21 hours, respectively. The direct runoff ordinates due to the three successive storms (of 6 hours duration each) are obtained by deducting the base of flow of 20 cumec from the stream- flows at the corresponding time intervals as shown in Table 5.2. The net storm rains are ob- tained by deducting the average storm loss as
0-6 hr: x = 2.9 – 0.15 × 6 = 2 cm 6-12 hr: y = 4.9 – 0.15 × 6 = 4 cm
12-18 hr: z = 3.9 – 0.15 × 6 = 3 cm
The equations can be easily arrived by entering in a tabular column and successively solving them. The 6-hr unit hydrograph ordinates are obtained in the last column; of course the ordinates are at 3-hr intervals since the streamflows are recorded at 3-hr intervals. The
C-9\N-HYDRO\HYD5-1.PM5 HYDROGRAPHS
Table 5.2 Derivation of a 6-hour unit hydrograph from a complex storm, Example 5.2.
DRO due to**
Equation
Time
Solution (hr)
UGO*
1st storm
2nd storm
3rd storm
Total DRO = TRO— BFO
6-hr UGO 133
24 4u 6 3u 4 4u 6 + 3u 4 = 84.5 – 20
Σ u = 110
∴ u 5 = 8.5 U check for
27 4u 7 3u 5 4u 7 + 3u 5 = 45.5 – 20
V| UGO derived
W| above
*Except the first column, all other columns are in cumec. **x = 2 cm, y = 4 cm, z = 3 cm
HYDROLOGY
last four equations in the table serve to check some of the UGO’s derived. Another check for the UGO’s derived is that the area under the UG should give a runoff volume equivalent to
1 cm, i.e.,
Σu t = 1 cm, in consistent units
A Σu = sum of the UGO’s = 110 cumec
= 0.01 m, or 1 cm
Hence, the UGO’s derived are correct and is plotted in Fig. 5.15 (b). The UGO’s can also be derived by the method of least squares (Snyder, 1955) by writing
the direct runoff ordinate (Q) as
...(5.5) where a = 0 theoretically, b i =u i , and x 1 ,x 2 ,x 3 ,... are the rainfall excesses (P net ) in successive
Q =a+b 1 x 1 +b 2 x 2 +b 3 x 3 + ... + b i x i
periods. In the least square technique, data from a number of flood events, for which values of Q i and x i are established, are used to develop a set of average values of b i (= u i ) and the unit hydrograph derived. The method is more elegant but laborious.
P net due to 3 storms
Intensity (cm/hr)
a. observed 3u 3 hydrograph due
to 18-hr storm (u’s = 6-hr UGO)
3 3u
(cumec) 120 u 2 2 140 Q
4u 4 (cumec) 100 u 1 1 3 u 120 3 Q
runoff 4u 4 4u 4 w
60 u 4 4 80 Surf
Constant base flow = 20 cumec
0 3 6 9 12 15 18 21 24 27 30 33 Time t (hr)
Fig. 5.15 (a) Derivation of 6-hr UG from 18-hr complex storm hydrograph (Example 5.2)
25 6-hr UG derived
(cumec)
(u's = UGO)
Q 20 u u 2 2 15 u u 3 3
Discharge 10 u u 1 1 u u 4 4 5 u u s s u u 6 6
0 0 3 6 9 12 15 18 21 Time t (hr)
Fig. 5.15 (b) 6-hr unit hydrograph (derived) (Example 5.2)
Matrix method for unit hydrograph derivation from complex storm
In matrix notation
PU =Q
L p 1 0 0 0 00 O L u 1 O
M p 2 p 1 0 0 00 P M u 2 P
M p 3 p 2 p 1 0 00 P M u P
PP
where P = M
P U= M 3 P
Q= M 3
M 0 p 3 p 2 p 1 00 P M u 4 P
M 0 0 p 3 p 2 00 P M 0 P
M N 0 0 0 p 3 00 P Q M N 0 P Q
By deducting base flow from the stream flow hydrograph Q is known, and with the known rainfall excess P, the unit hydrograph U can be computed as
U =P –1 Q
The inverse for the precipitation matrix (P –1 ) exists only if it is a square matrix with a non-vanishing determinant. However, by the use of the transpose of the precipitation matrix, (P T ), a square matrix is obtained as
P T PU = P T Q
and the unit hydrograph matrix is obtained as
U = (P T P ) –1 P T Q
Usually, the solution of the unit hydrograph matrix is performed on a digital computer (Newton, 1967). Most computer installations have packaged programmes to work with large systems and for determining the inverse and transpose of matrices.
Also, the number of streamflow hydrograph ordinates (n) is given by
n =j+i–1
where j = number of UGO
i = number of periods of rainfall
HYDROLOGY
In Example 5.2, n = 12 ordinates at 3-hr interval (DRO of the first and last being zero) ;
i = 6 periods of 3-hr interval (= 3 × 2), then
Thus, the unit hydrograph consists of 7 ordinates at 3-hr intervals (of course, u 0 = 0, and u 7 = 0).
As an assignment, the student is advised first to become well versed with the matrix algebra and then apply the matrix method to derive the 6-hr UGO at 3-hr interval from the streamflow data given in Example 5.2.
Average Unit Hydrograph. It is better if several unit graphs are derived for different isolated (single peaked) uniform intensity storms. If the durations of storms are different, the unit hydrographs may be altered to the same duration (sect .... or art .....). From several unit hydrographs for the same duration, so obtained, an average unit hydrograph can be sketched by computing the average of the peak flows and times to peak and sketching a median line, (Fig. 5.16), so that the area under the graph is equal to a runoff volume of 1 cm.
P net = 1 cm
Average peak given by:
Qp 1 3 Qp 2 t t+t+t
peak = 1 2 3
1 2 Median line sketched so that area under curve = 1 cm runoff, gives t -hr r
average unit hydrograph
(cumec) Q
1 , 2 & 3 : t -hr r UG 's obtained from different
duration storms
Discharge
t t 1 1 peak peak
Time base T of ave. UG Time t (hr)
Fig. 5.16 Average unit hydrograph
Alteration of Unit Hydrograph Duration. It becomes necessary, in computation of flood hydrographs, that the duration of the unit graph available should be altered to suit the duration of the design storm (to be used for obtaining the flood hydrograph). Two cases arise:
HYDROGRAPHS
Case (i) Changing a Short Duration Unit Hydrograph to Longer Duration. If the desired long duration of the unit graph is an even multiple of the short, say a 3-hour unit graph is given and a 6-hour unit graph is required. Assume two consecutive unit storms, producing a net rain of 1 cm each. Draw the two unit hydrographs, the second unit graph being lagged by
3 hours. Draw now the combined hydrograph by superposition. This combined hydrograph will now produce 2 cm in 6 hours. To obtain the 6-hour unit graph divide the ordinates of the combined hydrograph by 2, Fig. 5.17. It can be observed that this 6-hour unit graph derived has a longer time base by 3 hours than the 3-hour unit graph, because of a lower intensity storm for a longer time.
Example 5.3 The following are the ordinates of a 3-hour unit hydrgraph. Derive the ordinates of a 6-hour unit hydrograph and plot the same.
Time (hr)
3-hr UGO
Time (hr)
3-hr UGO
Table 5.3 Derivation of the 6-hour unit hydrograph. (Example 5.3)
6-hr Time
UGO (hr)
UGO
UGO
(cumec)
(logged)
(cumec)
(cumec)
(cumec) 1 2 3 4 5 0 0 0 0
HYDROLOGY
The 6-hour unit graph derived as above is shown in Fig. 5.17.
1 cm
1 cm, P
Intensity (cm/hr)
net
2 unit storms
16 Combined hydrograph (2 cm in 6 hr)
14 3-hr UG's lagged by 3 hr (cumec) 12
Q 10 6-hr UG (1 cm in 6 hr)
8 (comb. hyd. ord. ¸ 2)
Discharge
0 3 6 9 12 15 18 21 24 27 Time t (hr)
Fig. 5.17 Changing a short duration UG to a long multiple duration
Case (ii) Changing a Long Duration Unit Hydrograph to a Shorter Duration by S-curve Technique