METHODS OF DETERMINING IUH

16.2 METHODS OF DETERMINING IUH

1. By the S-Curve Hydrograph In Fig. 16.4, S t is the S-curve ordinate at any time t (due to t r -hr UG) and S′ t is the ordinate at

time t of the S-curve lagged by t′ r -hr, then the t′ r -hr UGO at time t can be expressed as

( UHO) = I dS t t = Slope of

S - Curve

Discharge

Lag

Time (hr)

Fig. 16.4 IUH as S-curve derivative

As t′ r progressively diminishes, i.e., t′ r → 0, Eq. (16.6) reduces to the form (as can be seen from Fig. 16.4)

i.e., the ordinate of the IUH at any time t is simply given by the slope of the S-curve at time t; in other words, the S-curve is an integral curve of the IUH. Since the S-curve derived from the observed rainfall-runoff data can not be too exact, the IUH derived from the S-curve is only approximate. The IUH, reflects all the catchment characteristics such as length, shape, slope, etc., independent of the duration of rainfall, thereby eliminating one variable in hydrograph analysis. Hence, it is useful for theoretical investigations on the rainfall-runoff relationships of drainage basins. The determination of the IUH is analytically more tedious than that of UG but it can be simplified by using electronic computers.

The t′ r -UGO can be obtained by dividing the IUH into t′ r -hr time intervals, the average of the ordinates at the beginning and end of each interval being plotted at the end of the interval (Fig. 16.5).

I UH derived from S-curve

Discharge

B+C 2 B A+B

Time (hr)

Fig. 16.5 t r ′-hr UG derived from IUH

2. By Using a Convolution Integral By the principle of superposition in the linear-unit-hydrograph theory, when a net rainfall of

function i(t) of duration t 0 is applied, each infinitesimal element of P net will produce a DRO, i.e., Q (t) given by

Q (t) = ut ( − τ ).() i τ d τ

the upper limit t′ given by

′ = t, when t ≤ t t 0 ′=t t 0 , when t ≥ t 0

Eq. (16.8) is called the convolution integral (or Duhamel integral) in which u (t – τ) is a kernel function , i(τ) is the input function.

The shape of the IUH in Fig. 16.6 resembles a single peaked hydrograph. If the rainfall and runoff in the convolution integral are measured in the same units, the ordinates of the

IUH have the dimension

time

The properties of the IUH are given by

0 ≤ u(t) ≤ a positive peak value, for t > 0

u(t)dt = 1.0 and

u(t) dt = t

where t i = lag time of IUH = time interval between the centroid of P net and that of direct runoff. See next chapter 17 on IUH.

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time t

time t

Run-off hydrograph

t ¢ Q(t) = u (t – ).i( ) d t t t

Q(t)

t = t, when t ¢ £ t o t = t , when t > t ¢ o o

c. O

time t

Fig. 16.6 Convolution of i( τ) and IUH

3. By Conceptual Models Various conceptual models have been proposed to develop the IUH. They may be of physical

analogy or mathematical simulation composed of linear reservoirs, linear channels, or time- area diagrams.

(a) Linear Reservoirs

A mathematical simulation of a drainage basin consisting of a series of linear reservoirs as proposed by J.E. Nash (1957) is discussed below:

A linear reservoir is a fictitious reservoir in which the storage is directly proportional to outflow, i.e., S = KO.

From the principle of continuity

From the condition O = 0 when t = 0, and that S = KO,

...(16.11) when t = ∞, O = I, i.e., the outflow approaches an equilibrium condition and equals inflow. If

O = I(1 – e –t/k )

the inflow terminates at time t 0 since outflow began, a similar derivation gives the outflow at

time t in terms of out flow O 0 at t 0 , as

− ( t − O t =O e 0 )/ K

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For an instantaneous inflow, which fills the reservoir storage S in time t 0 = 0, and since S

S = KO, O 0 = , hence from Eq. (16.12) K

e –t/K

For a unit input or S = 1, the IUH of the linear reservoir is given by

u (t) =

e –t/K

This is represented by the hydrograph for the outflow from the first reservoir as shown in Fig. 16.7.

Hydrographs O 3 (O = Q ) i i

Linear storage

Fig. 16.7 Routing through linear reservoirs (Nash’s Model)

(b) Simulation of Linear Channels

A linear channel is a fictitious channel in which the time T required to translate a discharge Q of any magnitude through a given channel reach of length x is constant. Hence, when an inflow hydrograph is routed through the channel, its shape will not change. At a given section, the relation between the water area A and the discharge Q is linear (assuming velocity to be constant), i.e.,

A = CQ

where C = f(T) called ‘translation coefficient’ which is constant at a given section. If a segment of inflow of duration ∆t and volume S is routed through a linear channel,

Fig. 16.8, the outflow is given by

O = S δ(t, ∆t)

where

δ(t, ∆t) =

∆t

for 0 ≤ τ ≤ ∆ t and t = τ + T; it is zero otherwise, where τ is the time measured from the beginning of the segment. Eq. (16.16) is a ‘pulse function’. When ∆t → 0, this equation becomes an ‘im- pulse function’ δ(t), known as a ‘Dirac-delta function’, which represents the IUH for the linear channel.

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Inflow hydrograph

Outflow hydrograph

f(t) =

time t

Fig. 16.8 Routing through linear channel

4. Routing Time–Area Curve of Basins The principles of flood-routing can be used to derive unit hydrographs for a catchment where

there are no complete records of rainfall-runoff. The catchment may be divided into a series of sub-areas, each contributing inflow into

drainage channels (which have storage) due to a flash storm. The IUH can be divided into two parts—the first representing inflow of the rain, and second, the gradual withdrawal from the catchment storage, the dividing line being the inflection point on the recession limb, Fig. 16.9.

Short rain t t i i

Inflection point

, (cumec) Inflow of rain

Withdrawal

Discharge

from storage

Base flow Time (hr)

Fig. 16.9 Hydrograph from short rain (IUH)

Assuming that the catchment discharge (O) and the storage (S) are directly propor- tional

...(16.17) where K = storage coefficient. From the principle of continuity, if I = inflow resulting from the instantaneous rain

S = KO

(I – O) ∆t = ∆S

2 2 2 –S 1 ...(16.19) and

and hence, O 2 =C 0 I 2 +C 1 I 1 +C 2 O 1 ...(16.20)

where C 0 =

...(16.20a) K + 05 . t

1 K + 05 . t

2 K + 05 . t

which are same as Eqs. (9.12 a, b, c) with Muskingum approach, putting x = 0. and when a sub-area distribution or time-area graph is used and I 1 =I 2 , hence O 2 = C′I + C 2 O 1 ...(16.21)

From Eq. (16.18),

Using the condition O = 0, when t = 0, the equation can be solved as

...(16.22) Since the inflow ceases at the inflection point at time t i , the outflow at time t (in terms

O = I(1 – exp (– t/K ))

of the outflow O ti at t i ) is given by

i I HG K KJ

t =O ti exp −

Storage coefficient K can be determined from an observed hydrograph by noting two values of O, unit time apart at the point of inflection (Fig. 16.10).

A = K (O – O ) 1 2

Inflexion point O=O 1 t i O=O 2 t i+1 , cumec

– t–t i

O=O

Discharge Direct

runoff

O 1 time (t)

Fig. 16.10 Determination of storage coefficient t − t

O =O and O =O exp F − i 1 I ti 2 ti HG K KJ

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z t i HG K KJ HG K KJ 0 HG K QP KJ

the shaded area A =

...(16.23) Another observation that is to be made is the catchment lag (t i ), i.e., the maximum

∴ A = K(O 1 –O 2 )

travel time through the catchment. This may be taken as the time from the mass centre of the causative rain (flash storm or short rainburst to minimise error) to the inflection point on the recession limb.

The catchment is subdivided into isochrone such that the rain falling in any sub-area has the same time of travel to the outflow point O, (Fig. 16.11). The time-area graph (I) now has instantaneous unit rain applied to it and is routed through to obtain the outflow (O), Eq. (16.21). This outflow represents the IUH for the catchment and may be converted, if required to a t r -hr unit hydrograph.

Catchment boundary

2 Isochrones

A = 1040 km t = 9 hr i K = 8 hr

Fig. 16.11 Catchment divided into isochrones

This method is simple and the design rain can be applied directly to the time-area graph, with areal variation and with any desired intensity.

An estimate of K can also be had from data on the recession limbs of the basin hydrographs.

Example 16.1

A catchment of area 1040 km 2 is divided into 9-hourly divisions by isochrones (lines of equal travel time) in Fig. 16.11. From the observation of a hydrograph due to a short rain on the catchment, t i = 9 hr and K = 8 hr. Derive: (a) the IUH for the catchment. (b) a 3-hr UG.

Solution (i) It will be assumed that the catchment is divided into sub-areas such that all surface runoff from each of these areas will arrive during a 1-hr period at the gauging point. The areas are measured by planimetering each of the hourly areas as:

1 2 3 4 5 6 7 8 9 Area (km 2) :

Hour:

80 35 (ii) The time-area graph (in full lines) and the distribution graph of runoff (in dotted

lines) are drawn as shown in Fig. 16.12. The dotted lines depict the non-uniform areal distri- bution of rain.

0 1 2 3 4 5 6 7 8 9 Time (hr)

Fig. 16.12 Time-area graph for catchment Table 16.1 IUH by routing and derivation of 3-hr UG (Example 16.1)

Time Time-area

O 2 = IUH 3-hr UGO (hr)

previous (cumec)

= col (3) (cumec)

Area (km 2 )

× col (2) (cumec)

+ col (4) (cumec)

101.00 123.00 Plot col (1) vs. col (5) to get the IUH, and col (1) vs. col (6) to get the 3-hr UGO, as shown

in Fig. 16.13.

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(iii) From Eq. (16. 21), O 2 = C′I + C 2 O 1

Check: C′ + C =1 K + 0.5 t 8 + 0.5 × 1 5 2 Hence, the routing equation becomes

O 2 = 0.1177 I + 0.882 O 1

O 2 vs. time gives the required synthetic IUH from which the 3-hr UGO are obtained as computed in Table 16.1. The conversion constant for Col (3) is computed as

1-cm rain on 1 km 2 10 × − in 1 hr = 10 = 2.78 m 3 /s

The 3-hr UGO is obtained by averaging the pair of IUH ordinates at 3-hr intervals and writing at the end of the intervals.

3-hr UG (by averaging IUHO)

Time (hr) Fig. 16.13 IUH derived and 3-hr UG (Example 16.1)