CLARK’S METHOD
17.7 CLARK’S METHOD
In the Clark’s approach, the ordinates of TAD are converted to volume rate of runoff in cumec
catchment, as
cumec or m 3 /s
100 t × 60 × 60 t where t = Computation time interval, hr of TAD, i.e., isochrone interval or routing period
t c = t × N, N = No. of inter-isochrone areas or sub areas, A r km 2 Catchment are A = ΣA r
In Example 16.1, t c = 1 hr × 9 # = 9 hr ~ − t i ,N=9
Note: 8 Isochrones are drawn to yield 9 zones (N = 9) or Subareas A r .
The inflow (I) from the sub area A r calculated as above, i.e., the resulting translation hydrograph is then routed through a linear reservoir to simulate the storage effects of the basin; Clark’s method utilises Muskingum method of routing through a linear reservoir, i.e., x = 0 in Eq. (9.9); S = KQ, Q = O.
The general equations for the linear reservoir is (See Eq. (16.21))
IUHO = O 2 = Q = C′I + C 2 O 1 ,O 1 =Q 1
i.e.,
Q 2 = C′I + C 2 Q 1 ,
for IUH derivation
INSTANTANEOUS UNIT HYDROGRAPH (IUH)
The resulting IUHO (IUH ordinates) are averaged at t r intervals to produce a t r -hr UG, ANNACIVIL Ex. 16.1.
The routing coefficients are obtained from (see Eqs. 16.20, 16.20a, 16.21)
C ′=
...(17.14) k + /2 t
∴ C′ + C =1
u t + u tt −
Note: u t of t r -hr UG = u
r I of IUH =
of IUH, i.e., by averaging IUHO, assuming linear.
HG 2
2 KJ
Example 17.3 The recession ordinates of the flood hydrograph (FHO) for the Lakhwar dam site across river Yamuna are given below. Determine the value of K.
Time (hr): 30 36 42 48 54 60 66 72 78 FHO (cumec):
90 45 30 20 Solution Eq. (5.1) can be expressed in an alternative form of the exponential decay as
t Q t =Q 0 e –t/k , when K = ln ( Q 0 / Q t )
‘Q vs. t’ is plotted on the semi-log paper (Fig. 17.1). K is the slope of the recession-flood- hydrograph plot.
BLOG 100
6 Semi-log paper 500:5
cycle cycle
D ln 1000
Ordinates 150:1.5 raph
Time t (hr)
Fig. 17.1 Recession flood hydrograph
HYDROLOGY
= 31 hr – 59 hr, from the plot ANNACIVIL − 31 59 28 K = =− =− . 12 15 , say 12 hr
∆t = t
–t cumec
Example 17.4 The isochronal map of Lakhwar damsite catchment, Fig. 17.2 (a) has areas between successive 3 hr isochrones as 32, 67, 90, 116, 135, 237, 586 and 687 km 2 . Taking k = 12 hr (as determined in Ex. 17.3), derive the IUH of the basin by Clark’s approach and hence a 3-hr UG .
# hr Basin
hr
end
7 21 t = 24 hr c A r ll boundary
12 hr-4
R.Y R.Y am am una 9 una Hanuman
t=t hr-3 Ñ c
area
A 7 8 A Ganga
= 3 hr 6 hr-2
Lakhw Bhadri Gad Bhadri Gad
c BLOG
A 3 hr-Isochrones, 7 #
Inter-isochrone
subareas: A -A 1 8
R.Yamuna
A 3 A 6 A= S 1 r
A = 1950 km
A 5 t = 24 hr, t = t = 3 hr Ñ c c A 3 A A 4 Basin outlet
9h 12 15 18 3 hr A 1 ´ 2 8 # = 24 hr Mussorie
Time (hr) (a) Isochronal map of Lakhwar
DS Catchment. (b) Time-area diagram (TAD)
Fig. 17.2 Isochrones and TAD for Lakhwar Dam Site
Solution Note
A = ΣA r = 1950 km 2
t c = t × N = 3 × 8 = 24 hr, K = 12 hr
No. of isochrones = N – 1 = 8 – 1 = 7# 24 t
Computation interval t = ∆t c between successive isochrones = 3 hr = = c 8 N
Clark’s approach Eq. (17.13):
= 0.7778 k + /2 t
C ′+C 2 = 0.2222 + 0.7778 = 1
∴ O.K.
INSTANTANEOUS UNIT HYDROGRAPH (IUH)
ANNACIVIL Table 17.3 Computation of IUH by Clark’s approach and hence 3-hr UG.
From the sub areas A r , Eq. (17.11): I = 2.78
= 0.9267 A
Clark’s: Q 2 =C ′I + C 2 Q 1 , C 2 Q 1 = 0.7778 Q 1 Q 2 = IUHO
C ′I = 0.2222 × 0.9267 A r = 0.203 A r
1 2 3 4 5 6 Time
3-hr (hr)
A r 2 , (km )
C ′I
C 2 Q 1 IUHO
UGO TAD)
(from
= 0.203 A r
= 0.7778 Q 1 Q 2 (cumec)
=C ′I + C 2 Q 1 (cumec)
previous
(by averaging) 0 0 0 0 0
BLOG 139.5 159.2
294.5 peak of 245.7
230 r IUH 262.2 (peak of UG)
27 Σ A = 1950 km 2 0 230
Plot Col. (5) vs. col (1) to get IUH, and Col (6) vs. col. (1) to get 3-hr UG. Note that the two peaks are staggered by 3 hr; i.e., IUH is more skewed.
QUESTIONS
1 Define IUH and state its important properties illustrated in a neat sketch. What are the advan- tages of IUH over a UG of finite duration? 2 Determine the values of n and k, and hence derive an IUH for the drainage basin in Example 5.2.
State its peak and time to peak. 3 For the IUH obtained in (2) above, derive a 6-hr UG and compare with that obtained in
Example 5.2. 4 Derive an IUH by Clark’s approach and hence a 2-hr UG for a catchment of 140 km 2 ,t c = 18 hr,
K = 12 hr. The catchment was divided into 9 zones by drawing 8 isochrones. The areas planimetered between the successive 2-hr isochrones are 8 (near outlet) 12, 25, 35, 22, 16, 10, 8 and 4 (remote
end) in km 2 .
HYDROLOGY 5 (a) Explain the concept of Clark’s IUH.
ANNACIVIL IUH for the basin with K = 5 hr
(b) The shape of a catchment can be approximated to a square with diagonal of 60 km. The main channel is nearly along a diagonal; the isochrones may be assumed perpendicular to the main channel. The travel speed along the main channel may be taken as 5 km/hr. Derive the
Hint: t c = 12 hr. Divide the diagonal into 6 parts to obtain successive 2-hr isochrones. 6 The IUH of a basin can be approximated to a triangle of base 36 hr and peak of 30 cumec at 9 hr from the start. Derive a 3-hr UG for this basin.