SYNTHETIC UNIT HYDROGRAPH IS DERIVED FROM KNOWN GAUGED CATCHMENT AREA.

1. SYNTHETIC UNIT
HYDROGRAPH
Presented by
Uzma shaikh (MT17WRE011)

2. Introduction:
• In India, only a small number of streams are gauged (i.e., stream flows due to
single and multiple storms, are measured)
• There are many drainage basins or catchments for which no stream flow records
are available and unit hydrographs may be required for such basins In such cases,
hydrographs may be synthesized directly from other catchments, which are
hydrologically and meteorologically homogeneous, or indirectly from other
catchments through the application of empirical relationship.
• Methods for synthesizing hydrographs for ungauged areas have been developed
from time to time which are
i. SNYDER’S METHOD
ii. SCS DIMENSIONLESS UNIT HYDROGRAPH

3. 1.Snyder’s method:
• Snyder’s (1938),based on the study of large number of catchments ,developed
a set of empirical equations for synthetic unit hydrograph.
• The most important characteristics of a basin affecting a hydrograph
due to a storm is a basin lag or time lag.
Fig.1 Snyder’s parameter

4. tp = Ct(L Lca)0.3
where
tp= basin lag in hours
L =Basin length measured along the water course from the basin divide to the
basin length in Km
Lca=Distance along the main water course from the gauging station to a point opposite
to watershed centroid
Ct=A regional constant representing watershed slope and storage effects, ranged from
1.35 to 1.65
• Snyder’s adopted a standard duration tr hours of effective rainfall given by
tr=tp/5.5

5. • The peak discharge Q ps (cumec) of a unit hydrograph of standard duration tr hours
is given by
Qps=2.78 Cp A/ tp
where
A= Catchment area in Km2
CP=A regional constant range from 0.56 to 0.69 for snyder’s study areas
• If a non-standard rainfall duration tR h is adopted ,the modified basin lag is
given by
tp’= tp+ (tR - tr )/4
=21/22 tp + tR/4
• Thus the peak discharge for a non-standard ER of duration tR is
Qp=2.78 Cp A/ tp’

6. • Time base of a unit hydrograph is given by Snyder as
Tb=72+3 tp’ hours ,for large catchments
Tb=5(tp’ +tR/2) hours, for small catchments
• To assisst in the sketching of a unit hydrographs,the widths of a unit hydrographs
at 50 and 75% of the peak have been found .these widths (in time units) are
correlated to the peak discharge intensity and are given by
W50=5.87/q1.08
W75=W50/1.75

7. Where
W50 , W75 =are widths of unit hydrograph in h
at 50% and 75% peak discharge
q =QP/A in m3/s/Km2
since the coefficients Cp and Ct are vary from region to region ,in practical
applications it is advisable that the value of these coefficients are determined from
Known unit hydographs of a meteorologically homogeneous catchment and then used
in the basin under study.

8. • Example 1. Two catchments A and B are considered meteorologically similar. Their
catchment characteristics are given below.
Catchment A Catchment B
L = 30Km L = 45Km
Lca= 15Km Lca = 25Km
A = 250Km2 A = 400Km2
For catchment A , a 2-h unit hydrograph was developed and found to
have peak discharge of 50 cumecs . The time to peak from the beginning of the
rainfall excess in this unit hydrograph was 9 h .Using Snyder’s method develop a
unit hydrograph for catchment B

9. 2.SCS Dimensionless unit hydrograph;
• A typical dimensionless unit hydrograph is developed by the US Soil Conservation
Services (SCS)
Fig.2.1 Dimensionless SCS Unit Hydrograph

10. • The coordinates of the SCS dimensionless unit hydrograph is given in table .1 for
use in developing a synthetic unit hydrograph in place of Snyder’s equation
Table.1 Coordinates of SCS Dimensionless Unit Hydrograph

11. 2.1 SCS Triangular Unit Hydrograph;
• The value of Qp and T p may be estimated using a simplified model of a triangular
unit hydrograph (Fig 2.2) suggested by SCS . This triangular unit hydrograph has
the same percentage of volume on the rising side as the dimensionless unit
hydograph of Fig 2.1
Fig 2.2 SCS Triangular Unit Hydrograph

12. • In fig 2.2
Qp=peak discharge (cumecs)
tr =duration of effective rainfall
T p=time of rise=(tr/2)+tp
tp=lag time
Tb =lag time
• Time of recession = Tb - T p =1.67 Tp
thus,
Tb=2.67 Tp
T p= tr /2 +0.6 tc
tp~ 0.6tc
Q p=2.08 A/Tp

13. • Example 2.Develop a 30 minute SCS triangular unit hydrograph for a watershed of
area 550 ha and time of concentration of 50 minutes.

14. Thank you