NASH CONCEPTUAL MODEL
17.4 NASH CONCEPTUAL MODEL
Nash considered that the IUH can be obtained by routing the instantaneous inflow through a cascade of linear channels (n numbers) with equal storage coefficient (Fig. 16.7). The out flow from the first reservoir is considered as inflow into the second reservoir, and so on. The outflow from the n th reservoir yields the IUH given by
k F t n u 1 (t) = e I , Γn = (n – 1) !
kn Γ
HG k KJ
The value of the parameter n, which is a shape parameter, is a measure of the catch- ment channel storage, which defines the shape of the IUH. A lower value of n yields a higher peak of IUH because of less storage flow attenuating the peak flow; a higher value of n leads to
a lower peak of IUH signifying higher storage for attenuating peak flow. The parameter K (delay time, hr), which is a scale parameter, represents the dynamics of rainfall-runoff proc- ess in the catchment. A smaller K-value reflects a lower time to peak of the runoff hydrograph and a higher K-value reflects a long time to peak.
The two parameters n and k may be computed by making an analysis of the observed rainfall-runoff data on the catchment as follows:
...(17.5) The second moment of the IUH about the origin (t = 0): M 2 = n(n + 1) k 2 ...(17.6)
The first moment of the IUH about the origin (t = 0): M 1 = nk
INSTANTANEOUS UNIT HYDROGRAPH (IUH)
By routing through the cascade of n-reservoirs, it can be shown that
where MQ 1 , MQ 2 are the first and second moments of direct runoff about the origin. ANNACIVIL and MI 1 , MI 2 are the first and second moments of the effective rainfall (rainfall excess P net )
MQ 1 – MI 1 = nk
...(17.7) MQ 2 – MI 2 = n(n + 1)k 2 + 2nk MI 1 ...(17.8)
about the origin. The parameters n and k may be evaluated by solving these two equations as illustrated
in the Example 17.1. Making use of these two parameters an IUH can be derived for a catch- ment of area eq. A km 2 ; and also a t r -hr UG, as illustrated in Example 17.2; i.e., from the available rainfall-runoff data on a catchment of area A, and IUH and a t r -hr UG can be de- rived, for which the procedure involves the steps of both the Examples, 17.1 and 17.2.
Example 17.1 The effective rainfall due to a 4-hr storm in the successive hours are: 2.6, 2.5, 2.3 and 2.4 cm. The resulting DRO’s in the successive hours are: 3, 15, 26, 40, 50, 35, 25, 20, 15, 10,
7, 4, 3 and 1 cumec. Determine the values of n and k.
Solution
Step 1 Evaluate the first and second moments of P net about the origin, i.e., t = 0 (commence- ment of P net and DRO)
19.2 cm . hr
49.75 cm . hr
MI =
~ _ 5 hr 2
9.8 cm Step 2 Evaluate the first and second moments of Q i about the origin;
2 BLOG
Similarly, Q t 2
= ΣQ 1 t 2 ;
MQ
Table 17.1 To evaluate MQ 1 and MQ 2
Time
DRO
t (hr)
(given) = Q
(cumec-hr)
(cumec-hr 2 ) 0 0 0 0
1000 (Contd.)...
ANNACIVIL 15 0 0 0
Eq. (17.8): MQ
2 – MI 2 = n(n + 1)k + 2 nk MI 1
40.5 – 5 = nk 2 (n + 1) + 2 × 3.8 × 2
2 35.5 = n k 2 + nk 2 + 15.2 nk 2 = 35.5 – 15.2 – (3.8) 2 = 5.87
= 2.45 ~ _ 2 (whole number)
Example 17.2 Derive an IUH and a 2-hr UG (UGO at 2-hr intervals) for a catchment of 240 km 2 , having n = 3 and k = 5 hr .
Solution NASH Model, Eq. (17.4):
BLOG
u (t) =
kn Γ
HG k KJ
k = 5 hr, n = 3 and Γn = (n – 1) ! = (3 – 1) ! = 2 ! = 2 × 1 = 2
5 F t I e 1 e − t / 5 u t (t) = . . = F I
5 × 2 HG 5 KJ 10 HG 5 KJ
Table 17.2 IUHO and 2-hr UGO computation
Time
IUHO
t 2 − t(hr)
e 5 F t I u(t)
5 HG 5 KJ
u(t)
2-hr UGO
(cm /hr)
(5) × 2.78 × 240 (by averaging)
(Contd.)...
INSTANTANEOUS UNIT HYDROGRAPH (IUH)
ANNACIVIL 14 2.8 0.061 7.84 0.0480 33.2 34.0
34.9 peak of IUH 35.5 peak of UG
2.67 3.00 Nash, from his study on some gauged catchments in UK, established a correlation be-
tween the IUH parameters n and k, and the basin parameters like length of main stream (L, miles), slope of the basin (S, parts per 1000) and the area (A, sq. miles), as
Using the above relations, the IUH of any ungauged basin in a hydrometeorologically homogeneous region can be obtained.
Clark in 1945, was the first to use the concept of IUH. The clark method requires three param- BLOG