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Synthetic Unit Hydrograph
Prepared by
Prof. S. G. Taji
Dept. of Civil Engineering
S.R.E.S’s Sanjivani College of Engineering,
Kopargaon
Synthetic Unit Hydrograph
 Synthetic hydrographs are derived by
 Relating hydrograph characteristics such as peak flow, base
time etc. with watershed characteristics such as area and time
of concentration.
 Using dimensionless unit hydrograph
 Based on watershed storage
Need for synthetic UH
 UH is applicable only for gauged watershed and for
the point on the stream where data are measured
 For other locations on the stream in the same
watershed or for nearby (ungauged) watersheds,
synthetic procedures are used.
Snyder’s Method
UH
Snyder’s method allows the computations of
(a) lag time (tL);
(b) UH duration (tr);
(c) UH peak discharge (qp);
(d) Hydrograph time width at 50% and 75% (W50, W75)
of peak flow
Snyder’s Method
UH
1. Lag time (tL): time from the center of rainfall –
excess to the UH peak
tL = C1Ct (LLc)0.3
where tL = Time [hrs]; C1 = 0.75 for SI unit; 1.0 for
English unit; Ct = Coefficient which is a function of
watershed slope and shape, 1.8~2.2 (for steeper
slope, Ct is smaller); L = length of the main channel
[mi, km]; Lc = length along the main channel to the
point nearest to the watershed centroid
Snyder’s Method
UH
2. UH Duration (tr):
tr = tL / 5.5
where tr and tL are in [hrs]. If the duration of UH is other than tr,
then the lag time needs to be adjusted as
tpL = tL + 0.25 (tR - tr)
where tLR = adjusted lag time; tR = desired UH duration.
3. UH Peak Discharge (qp):
or
where C2 = 2.75 for SI unit; 640 for English unit; Cp = coefficient
accounting for flood wave and storage condition, 0.4 ~ 0.8;
qp = specific discharge, [m3/s/km2] or [ft3/s/mi2]
To compute actual discharge, Qp = Aqp
p
t
p
C
2
C
p
q 
pR
t
p
C
2
C
p
q 
Snyder’s Method
UH
4. Time Base (tb):
Assuming triangular UH,
tb = C3 / qp
where tb – [hrs]; C3 = 5.56 for SI unit, 1290 for English unit.
5. UH Widths: or
where
CW, 75 = 1.22 for SI unit; 440 for English unit;
CW, 50 = 2.14 for SI unit; 770 for English unit;.
W50, W75 are in hours; Usually, 1/3 of the width is distributed
before UH peak and 2/3 after the peak
Remember to check that the volume of UH is close to 1 cm or 1
inch
1.08
p
q
w,75
C
75
W 
1.08
p
q
w,50
C
50
W 

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Synthetic unit hydrograph

  • 1. Synthetic Unit Hydrograph Prepared by Prof. S. G. Taji Dept. of Civil Engineering S.R.E.S’s Sanjivani College of Engineering, Kopargaon
  • 2. Synthetic Unit Hydrograph  Synthetic hydrographs are derived by  Relating hydrograph characteristics such as peak flow, base time etc. with watershed characteristics such as area and time of concentration.  Using dimensionless unit hydrograph  Based on watershed storage
  • 3. Need for synthetic UH  UH is applicable only for gauged watershed and for the point on the stream where data are measured  For other locations on the stream in the same watershed or for nearby (ungauged) watersheds, synthetic procedures are used.
  • 4. Snyder’s Method UH Snyder’s method allows the computations of (a) lag time (tL); (b) UH duration (tr); (c) UH peak discharge (qp); (d) Hydrograph time width at 50% and 75% (W50, W75) of peak flow
  • 5. Snyder’s Method UH 1. Lag time (tL): time from the center of rainfall – excess to the UH peak tL = C1Ct (LLc)0.3 where tL = Time [hrs]; C1 = 0.75 for SI unit; 1.0 for English unit; Ct = Coefficient which is a function of watershed slope and shape, 1.8~2.2 (for steeper slope, Ct is smaller); L = length of the main channel [mi, km]; Lc = length along the main channel to the point nearest to the watershed centroid
  • 6. Snyder’s Method UH 2. UH Duration (tr): tr = tL / 5.5 where tr and tL are in [hrs]. If the duration of UH is other than tr, then the lag time needs to be adjusted as tpL = tL + 0.25 (tR - tr) where tLR = adjusted lag time; tR = desired UH duration. 3. UH Peak Discharge (qp): or where C2 = 2.75 for SI unit; 640 for English unit; Cp = coefficient accounting for flood wave and storage condition, 0.4 ~ 0.8; qp = specific discharge, [m3/s/km2] or [ft3/s/mi2] To compute actual discharge, Qp = Aqp p t p C 2 C p q  pR t p C 2 C p q 
  • 7. Snyder’s Method UH 4. Time Base (tb): Assuming triangular UH, tb = C3 / qp where tb – [hrs]; C3 = 5.56 for SI unit, 1290 for English unit. 5. UH Widths: or where CW, 75 = 1.22 for SI unit; 440 for English unit; CW, 50 = 2.14 for SI unit; 770 for English unit;. W50, W75 are in hours; Usually, 1/3 of the width is distributed before UH peak and 2/3 after the peak Remember to check that the volume of UH is close to 1 cm or 1 inch 1.08 p q w,75 C 75 W  1.08 p q w,50 C 50 W 