Unit Hydrograph

Unit Hydrograph

The Unit Hydrograph Unit Hydrograph Derivation Unit Hydrograph Application Synthetic Unit Hydrograph

Unit Hydrograph

The unit hydrograph is the unit pulse response function of a linear hydrologic system.

First proposed by Sherman (1932), the unit hydrograph (originally named unit-graph) of a watershed is defined as a direct runoff hydrograph (DRH) resulting from 1 in (usually taken as 1 cm in SI units) of excess rainfall generated uniformly over the drainage area at a constant rate for an effective duration.

Sherman originally used the word “unit” to denote a unit of time. But since that time it has often been interpreted as a unit depth of excess rainfall.

Sherman classified runoff into surface runoff and groundwater runoff and defined the unit hydrograph for use only with surface runoff.

Definition

Unit Hydrograph

The unit hydrograph is a simple linear model that can be used to derive the hydrograph resulting from any amount of excess rainfall. The following basic assumptions are inherent in this model;

Assumptions

The excess rainfall has a constant intensity within the effective duration.

The excess rainfall is uniformly distributed throughout the whole drainage area.

The base time of the DRH (the duration of direct runoff) resulting from an excess rainfall of given duration is constant.

The ordinates of all DRH’s of a common base time are directly proportional to the total amount of direct runoff represented by each hydrograph.

For a given watershed, the hydrograph resulting from a given excess rainfall reflects the unchanging characteristics of the watershed.

Unit Hydrograph DerivationDiscrete Convolution Equation

n M

n m n m 1m 1

Q P U

When Qn = Direct runoff

Pm = Excess rainfall

Un-m+1 = Unit hydrograph

Suppose that there are M pulses of excess rainfall N pulses of direct runoff in the storm considered, then N equations can be written for Qn = 1, 2, …,N in terms of N-M+1unknown values of unit hydrograph.

Unit Hydrograph Derivation

The set of equations for discrete time convolution

n M

n m n m 1m 1

Q P U1 1 1Q PU

2 2 1 1 2Q P U PU

3 3 1 2 2 1 3Q P U P U PU

M M 1 M 1 2 1 MQ P U P U ..... PU

M 1 M 2 2 M 1 M 1Q 0 P U ..... P U PU

N 1 M N M M 1 N M 1Q 0 0 ..... 0 0 ..... P U P U

N 1 M 1 N M 1Q 0 0 ..... 0 0 ..... 0 P U

n = 1, 2,…,N

Discrete Convolution Equation

Example 1

Find the half-hour unit hydrograph using the excess rainfall hyetograph and direct runoff hydrograph given in the table.

Solution.

The ERH and DRH in table have M=3 and N=11 pulses respectively.

Hence, the number of pulses in the unit hydrograph is N-M+1=11-3+1=9.

Substituting the ordinates of the ERH and DRH into the equations in table yields a set of 11 simultaneous equations.

Time (1/2hr)

Excess Rainfall (in)

Direct Runoff (cfs)

1 1.06 428

2 1.93 1923

3 1.81 5297

4 9131

5 10625

6 7834

7 3921

8 1846

9 1402

10 830

11 313

Example 1

11

1

Q 428U 404 cfs/ in

P 1.06

2 2 11

1

Q P U 1,928 1.93x404U 1,079 cfs/ in

P 1.06

3 3 1 2 2

31

Q P U P U 5,297 1.81x404 1.93x1,079U 2,343 cfs/ in

P 1.06

and similarly for the remain ordinates

4

9,131 1.81x1,079 1.93x2,343U 2,506cfs/ in

1.06

510,625 1.81x2,343 1.93x2,506

U 1,460 cfs/ in1.06

6

7,834 1.81x2,506 1.93x1,460U 453 cfs/ in

1.06

Example 1

n 1 2 3 4 5 6 7 8 9

Un (cfs/in) 404 1,079

2,343

2,506

1,460

453 381 274 173

7

3,921 1.81x1,460 1.93x453U 381cfs/ in

1.06

81,846 1.81x453 1.93x381

U 274 cfs/ in1.06

9

1,402 1.81x381 1.93x274U 173 cfs/ in

1.06

Unit hydrograph

Unit Hydrograph Application

Once the unit hydrograph has been determined, it may be applied to direct runoff and streamflow hydrograph.

Procedures:

A rainfall hyetograph is selected.

The abstractions are estimated.

The excess rainfall is calculated.

The time interval used in defining the excess rainfall hyetograph ordinates must be the same as that for which the unit hydrograph was specified.

The discrete convolution equation may then be used to yield the direct runoff hydrograph.

By adding an estimated baseflow to the direct runoff hydrograph, the streamflow hydrograph is obtained.

Example 2

Calculate the streamflow hydrograph for a storm of 6 in excess rainfall, with 2 in the first half-hour, 3 in in the second half-hour and 1 in in the third half-hour. Use the half-hour unit hydrograph computed in example 1 and assume the baseflow is constant at 500 cfs throughput the flood. Check that the total depth of direct runoff is equal to the total excess precipitation. (Watershed are = 7.03 mi2)

1 1 1Q PU 2.00x404 808cfs

2 2 1 1 2Q P U PU 3.00x404 2.00x1,079 1,212 2,158 3,370cfs

3 3 1 2 2 1 3Q P U P U PU 1.00x404 3.00x1,079 2.00x2,343

404 3,237 4,686 8,327cfs

Solution.

Example 2

Time (1/2 hr)

Excess Precipitatio

n (in)

Unit Hydrograph Ordinates (cfs/in) Direct Runoff (cfs)

Streamflow (cfs)

1 2 3 4 5 6 7 8 9

404 1079

2343

2506

1460

453 381 274 173

N=123456789

1011

2.003.001.00

8081212

404

2158

3237

1079

4686

7029

2343

5012

7518

2506

2920

4380

1460

906135

9453

7621143

381

548822274

346519173

8083370832713120127817792358121441549793173

130838708827136201328182924081264420491293673

Calculation of the direct runoff hydrograph and streamflow hydrograph

Baseflow = 500 cfs

Example 2

The total direct runoff volume is

Nd n

n 1

3

7 3

V Q t

54,438x0.5 cfs.h

ft .h 3,600s54,438x0.5 x

s 1s

9.80x10 ft

The corresponding depth of direct runoff is found by dividing by the watershed area A=7.03 mi2=7.03x5280 ft2=1.96x108 ft2

7

dd 8

V 9.80x10r ft 0.50 ft 6.00 in

A 1.96x10

Synthetic Unit Hydrograph

Unit Hydrograph:

developed from rainfall and streamflow data on a watershed applies only for that watershed and for the point on the stream where the streamflow data were measured.

Synthetic Unit Hydrograph:

Synthetic unit hydrograph procedures are used to develop unit hydrographs for other locations on the stream in the same watershed or for nearby watersheds of a similar character.

Synthetic Unit Hydrograph

There are three types of synthetic unit hydrograph:

Those relating hydrograph characteristics (peak flow rate, base time, etc.) to watershed characteristics. (Snyder, 1938)

Those based on a dimensionless unit hydrograph. (Soil Conservation Service, 1972)

Those based on models of watershed storage. (Clark, 1943)

Synthetic Unit HydrographSnyder’s UH

p rt 5.5t

0.3p 1 t ct C C LL

Snyder defined a standard unit hydrograph as one whose rainfall duration tr is related to the basin lag tp by

tp = the basin lag (hr)

L = Length of the main stream from the outlet to the upstream (km)

Lc = The distance from the outlet to a point on the stream nearest the centroid of watershed area.

C1 = 0.75

Ct = Coefficient derived from gaged watersheds in the same region.

For a standard unit hydrograph Snyder found that

The basin lag is

Synthetic Unit HydrographSnyder’s UH

2 pp

p

C Cq

t

r R

p pRt t

t t4

The peak discharge per unit drainage area in m3/s.km2 of the standard unit hydrograph is

C2 = 2.75 and Cp = Coefficient derived from gaged watersheds in the same region.

If tpR=5.5tR tR=tr, tpR=tp, qpR=qp

ct, cp are computed from 0.3p 1 t ct C C LL

If tpR5.5tR

tpR=tp, qpR=qp

ct, cp are computed from 0.3p 1 t ct C C LL

Synthetic Unit HydrographSnyder’s UH

2 pp

p

C Cq

t

3b

pR

Ct

q

The relationship between qp and the peak discharge per unit drainage area qpR of the required unit hydrograph is

The base time tb in hours of the unit hydrograph can be determined using the fact that the area under the unit hydrograph is equivalent to a direct runoff of 1 cm. Assuming a triangular shape for the unit hydrograph, the base time may be estimated by

C3 = 5.56

Synthetic Unit HydrographSnyder’s UH

1.08w pRW C q

The width in hours of a unit hydrograph at a discharge equal to a certain percent of the peak discharge qpR is given by

Cw = 1.22 for the 75 percent width and 2.14 for the 50 percent width.

Synthetic Unit Hydrograph

Standard UH Required UH

Tp=5.5tr

Tp5.5tr

Snyder’s UH

Example 3

From the basin map of a given watershed, the following quantities are measured: L = 150 km, Lc = 75 km, and drainage area = 3,500 km2. From the unit hydrograph derived for the watershed, the following are determined: tr = 12 hr, tpR = 34 hr, and peak discharge = 157.5 m3/s.cm. Determine the coefficients Ct and Cp for the synthetic unit hydrograph of the watershed.

Solution From the given data, 5.5tR = 66 hr, which is quite different from .

r Rp pR

t tt t

4

rt 1234

4 **

Example 3

Solving * and ** simultaneously gives tp = 32.5 hr and tr = 5.9 hr. To calculate Ct, use

0.3p 1 t ct C C (LL )

0.3t32.5 0.75C (150 75)

tC 2.65

The peak discharge per unit area is

The coefficient is calculated by

3 . 2.

qpR 157.5/ 3500

0.045m / skm cm

p pR p pRq ,q andt t

2 ppR

pR

C Cq

t

p2.75C0.045

34.0pC 0.56

Example 4

0.23 0.25 0.18 1.57pT 3.1L S I

3 0.96 1.07p pQ 31.62 10 A T

3 0.95B pT 125.89 10 AQ

3 0.93 0.9250 pW 16.22 10 A Q

Compute the six-hour synthetic unit hydrograph of a watershed having a drainage area of 2,500 km2 with L = 100 km and Lc = 50 km. This watershed is a sub-drainage area of the watershed in example 3.

Ct = 2.64

Synthetic Unit HydrographSCS Dimensionless UH

The SCS-UH is a synthetic unit hydrograph in which the discharge is expressed by

the ratio of discharge discharge (q) to peak discharge (qp)

the time the ratio of the time (t) to the time of rise of the unithydrograph (Tp)

Given the peak discharge and lag time for the duration of excess rainfall, the UH can be estimated from the synthetic dimensionless hydrograph for the given basin

Synthetic Unit HydrographSCS Dimensionless UH

pp

CAq

T

The figure shows a dimensionless hydrograph, prepared from the unit hydrographs of variety of watersheds. The values of qp and Tp may be estimated using a simplified model of triangular unit hydrograph.

SCS suggests the time of recession may be approximated as 1.67Tp.

As the area under the unit hydrograph should be equal to a direct runoff of 1 cm

Tp = peak time, hr

qp = peak discharge, cms.m

c = 2.08

A = the drainage area, sq.km.

Synthetic Unit HydrographSCS Dimensionless UH

rp p

tT t

2

A study of unit hydrographs of many large and small rural watersheds indicates that the basin lag

Time to rise, Tp can be expressed in terms of lag time, tp and the duration of effective rainfall, tr

p cT 0.6t Tc = time of concentration of watersheds

Example 5

Construct a 10-minute SCS-UH for a basin of area 3.0 km2 and time of concentration 1.25 hr.

Solution The duration tr = 10 min = 0.166 hr

Lag time tp = 0.6Tc = 0.6x1.25 = 0.75 hr

Rise time Tp = tr/2 + tp = 0.166/2 + 0.75 = 0.833 hr

qp = 2.08x3.0/0.833 = 7.49 m3/s.cm

The dimensionless hydrograph in the figure may be converted to the required dimensions by multiplying the values on the horizontal axis by Tp and those on the vertical axis by qp. Alternatively, the triangular unit hydrograph can be drawn with tb = 2.67Tp = 2.22 hr. The depth of direct runoff is checked to equal 1 cm.

pp

CAq

T

Unit HydrographDifferent Rainfall Durations

When a UH of a given express rainfall duration is available, the UH of other durations can be derived.

If other durations are integral multiples of the given duration, the new UH can be easily computed by application of the principles of superposition and proportionality.

However, a general method of derivation applicable to UH of any required duration may be used on the basis of the principle of superposition. This is the S-Hydrograph Method.

Unit HydrographDifferent Rainfall Durations

The theoretical S-Hydrograph is that resulting from continuous excess rainfall at a constant rate of 1 cm/hr for an indefinite period.

This is a step response function of a watershed system

The curve assume a deformed S shape and its ordinates ultimately approach the rated of excess rainfall at a time of equilibrium.

This step response function, g(t) can be derived from the unit pulse response function, h(t) of the unit hydrograph.

Unit HydrographDifferent Rainfall Durations

1

h(t 2 t) g(t 2 t) g(t 3 t)t

1

h(t t) g(t) g(t t) g(t 2 t)t

1

h(t) g(t) g(t t)t

g(t) t h(t) h(t t) h(t 2 t) ....

The response at time t to a unit pulse of duration t beginning at time 0 is

The response at time t to a unit pulse beginning at time t is equal to h(t-t)

The response at time t to a third unit pulse beginning at time 2t is

Continuing this process indefinitely, summing the resulting equations, rearranging, yields the unit step response function

S-Hydrograph

Unit HydrographDifferent Rainfall Durations

g(t) g(t t )

1h(t) g(t) g(t t )

t

After S-Hydrograph is constructed, the UH of a given duration can be derived as follows:

Advance the position of the S-Hydrograph by a period equal to the desired duration t’ called as Offset S-Hydrograph

The difference between the ordinates of the original S-Hydrograph and the Offset S Hydrograph, divided by t’ gives the desired UH

S-Hydrograph

Offset S-Hydrograph

Unit Hydrograph of Duration t’

Example 6

Use the 0.5 hr unit hydrograph in example 1 to produce the S-hydrograph and the 1.5 hr unit hydrograph for this watershed.

Solution The 0.5 unit hydrograph is shown in column 2. The S-hydrograph is found using (with t = 0.5 hr)

For t = 0.5 hr : g(t) = t.h(t) = 0.5x404 = 202 cfs

For t = 1.0 hr : g(t) = t.[h(t)+h(t-0.5)]

= 0.5x[1,079+404]

= 742 cfs For t = 1.5 hr : g(t) = t.[h(t)+h(t-0.5)+h(t-1.0)]

= 0.5x[2,343+1,079+404] = 1,913 cfs

g(t) t h(t) h(t t) h(t 2 t) ....

Example 6

1Time

(hr)

20.5-h Unit

Hydrograph

(cfs/in)

3S-hydrograph

(cfs)

4Lagged

S-hydrograph

(cfs)

50.5-h unitHydrograp

h

(cfs/in)

0.51.01.52.02.53.03.54.04.55.05.56.0

4041079234325061460453381274173

000

202742

1913316638964123431344504537453745374537

000

202742

1913316638964123431344504537

135495

1275197621031473765369276149580

g(t t ) h(t)g(t)h(t)t

Calculation of a 1.5 hr unit hydrograph by S-hydrograph method