1. Question # 01
A Reservoir operator has to be release water from reservoir for being picked up at a distance
of 50 Km for downstream users. The average width of the stream for the anticipator discharge of
40km, The mean daily class a tank evaporation for this season is 0.5cm.
Estimate the mean daily evaporation from the stream in Acre-feet per day .If the pan co-efficient is
0.7. Assuming 15% loss due to seepage , find discharge at the head of canal if the required discharge
at the tail is to be 50 m3/s.
SOLUTION :
Discharge at tail Qtail = 50m3/s
Discharge at head Qhead =?
Surface area of stream = A =LxW
as L = 50 km= 50 x 1000= 50000 m
W = 40 m
A = 50000 x 40 = 2000000
Seepage Lossees = 15 %
Evaporation Losses :
PAN Evaporation E = 0.5 cm
PAN Co-efficient Kp = 0.7
LAKE Evaporation El =?
El = Kp.E
El = 0.7 x 0.5
El = 0.35
= 3.5 x10-3m Or 0.0035m
So Evaporation losses from stream per day
E stream = 0.0035 x 2000000
= 7000m3/day
= _ 7000_
24 x 3600
= 0.081m3/sec
As 1 Acer-ft = 43560ft
1 meter = 3.282
As = 7000 m3
day
= 7000 x (3.28)3
= 247013 ft3 / day
As 1 Acer-ft = 43560 ft
So = 247013
43560
= 5.6 Acer-ft
day

2. Question # 02
An Engineer-In-Charge of Reservoir operation has to release water from reservoir to
provide irrigation supplies at a discharge of 40 km for downstream Users. The average width
of the stream for the anticipates discharge is 25 km. The daily mean Class A pan evaporation
for this season is 5mm per day.
Estimate the daily evaporation losses from the stream in hector-meter-per day.
Solution:
Stream Length = 40 km
= 40 x 1000 = 40000 m
Stream width (W) = 25m
Surface Area of stream = 40000 x 25 = 1000000m2
Assuming Kp = 0.7
Pan Evaporation =E = 5mm/day
El = Kp E
= 0.7 x 5 = 3.5mnm or 0.0035m
E Stream = 0.0035 x 1000000m 3
= 3500m3
1 Hector = 10000m2
E Stream = 3500_
10000 = 0.35 hector-meter

3. Question # 03
A small catchment area 150 hectare received a rainfall of 10.5cm in 90 minutes, Due
to a storm draining the catchment was dry before the storm and experienced a runoff
lasting for 10 hours with an average discharge Value of 2m3/sec. The stream was given dry
after the runoff event.
(a) What is the amount of water in acre-feet which was not available to runoff alone to
combined effect of infiltration, evaporation and transpiration.
(b) What is the ratio of total and direct runoff to precipitation.
Solution :
Catchment Area “A” = 150 hectr as 1hectr = 10000m2
= 1500000m2
Precipitation “P” = 10.5cm = 0.105m
Total Runoff Volume = A x P
= 1500000 x .0.015
= 1575000m3
Time Duration of Rain = 90 min = 90 x 60 = 5400 Sec
Total Runoff Discharge Q Rain = 157500
5400
= 29.17m3/Sec
Q Runoff = 2m3/sec
Runoff time = 10 hrs
= 10 x 3600 = 36000 sec
Total Runoff Volume = 2 x 36000 = 72000m3
(a) For Water Lost :
Water Lost = Q Rain – Q Runoff
= 157500 – 72000
Water Lost = 85500 m3
as 1m3 = 35.315 ft3
1acre-f t = 43560 ft3
So,
Water Lost = 85500 x 35.315
43560
Water Lost = 69.3 Acer-Ft
(b) Ratio Between Total Runoff to Direct Runoff :
Direct Runoff = 72000m3
Total Runoff = 157500m3
Ratio = 72000_ = 0.457
157500
Percentage = 72000 x100
157500
= 45.7%

4. Question # 04
For Data given Example 4.4 Find total infiltration during the storm period Using
Horton’s Equation assuming fo = 1.5cm/h and fc = 0.5cm/h.
Solution :
Total Infiltration “F” = ?
Rain = 50mm
Time Duration = 2 HRS
Initial Infiltration Rate fo = 1.5cm/h
Constant Infiltration Rate fc = 0.5cm/h.
AS
Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet
K
= [1.5 – 0.5] (1 –e -1x2) + 0.5 x 2
1
= [1] (0.864) + 1
1
= 1.864 cm
Question # 05
A storm width 10cm precipitation produced a direct runoff as 5.8cm given the time

5. distribution of the storm in Table 4.3
Estimate the ɸ Index.
Solution :
Table 4.3 Time Distribution of the Storm.
Hour(Time) 1 2 3 4 5 6 7 8
International Rainfall (cm) 0.4 0.9 1.5 2.3 1.8 1.6 1.0 0.5
Calculation for Φ Index
Rainfall (cm) “f” values (cm)
0.45 .50 0.6
0.4 - - -
0.9 0.45 0.4 0.3
1.5 1.05 1.0 0.9
2.3 1.85 1.8 2
1.8 1.35 1.3 1.2
1.6 1.15 1.10 1
1.0 0.55 0.5 0.4
0.5 0.05 -
Total 6.45 6.10 5.8

6. Question # 07
For 3-Hours duration 225mm of total Rainfall we observed over 3200 km 2 catchment
Area. The infiltration capacity curve for this area can be given by Horton’s Equation
(Equation 4.6 & 4.7) in wich fo = 10mm/h and fc = 0.5mm/h. Evaporation and other losses
during the storm period were observed to be 50mm.
Find excess Rainfall over the Catchment
(a) Estimate Direct Runoff Volume in m3 & Hectare –m from excess Rainfall
(b) Total Run off in Hector-m
Solution :
Time Duration “t” = 3Hours
Rainfall “P” =225mm
Area “A” = 3200km2
Initial Infiltration Rate fo = 10mm/h
Constant Infiltration Rate fc = 0.5mm/h
K = 1h-1
Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet
K
= [10 – 0.5] (1 –e -1x3) + 0.5 x 3
1
= [9.5] (0.950) + 1.5
= 10.5 mm
Excess Rain Fall = Rain – F – Losses
= 225 – 10.5 – 50
= 164.5
Direct Runoff = Rainfall Excess x Area of Catchment
= 164.5 x 3200 (1000)2
1000
= 526400000m3
= 526400000
10000

7. = 52640 Hector – meter
Total Runoff = [Rain – Losses other than infiltration ] x A
= [225-50] x 3200 (1000)2 m3
1000
= 0.175 x 3200 (1000)2
= 560000000 m3
As 1 Hectr – meter = 560000000 hector – meter
10000
= 56000 hector – meter
Question # 08
An infiltration Capacity Curve prepared for a Catchment indicates an initial capacity
of 2.5cm/h and attains a constant value of 0.5 cm/h after 10 hour of Rainfall. With the
Horton’s constant K = 6 day-1
Solution :
Time Duration “t” = 10 Hours
Initial Infiltration Rate fo = 2.5 mm/h
Constant Infiltration Rate fc = 0.5 mm/h
K = 6 days-1 Or 6/24 = 0.25h-1
As Horton’s Equation
Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet
K
= [2.5 – 0.5] (1 –e -0.25x10) + 0.5 x 10
0.25
= [8] (0.918) + 5
= 7.34 mm
= 12.34 cm

8. Question # 09
In a project related to rainfall – runoff stadiesw f – curve was plotted to established
an equation of the form of Horton’s Equation. If F = 8.50 sq Units on the graph with each sq
represented 1 cm/h on the Vertical and 2 minute on the abscissa and f o = 4.5cm/h fc =
1.2cm/h
Determine the Horton’s Equation and Calculate “f” for t = 10 minutes
Solution :
Time Duration “t” = 10 Min
Initial Infiltration Rate fo = 4.5 mm/h
Constant Infiltration Rate fc = 1.2 mm/h
F = 8.50 Units
= 8.50 (1cm x 2 h)
h 60
As Horton’s Equation
Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet Let K = 1
K
= [4.5 – 0.5] (1 –e -1x.166) + 1.2 x 0.166
1
= [3.3] (0.152) + 2
= 5.3 mm
Question # 10
In a storm, total rain fall is 2.29cm and the total infiltration loss is 0.88cm,

9. Calculate the rain fall excess. Neglect evaporation during the period.
Solution :
Rainfall “P” = 2.29cm
Infiltration “F” = 0.88 cm
Rainfall excess =?
Rainfall excess =P–F
= 2.29-0.88 = 1.41cm
Question # 11
Determine the runoff from a catchment of area 2.3k over wich 7.5cm of Rainfall
occurred during 1 day Storm.
An infiltration of 0.6 cm/h and attained a constant Value of 0.15cm/h after 12 hours of
Rainfall with Horton’s Constant Value of K=3k-1. A class A-pom installed in the catchment
indicates a degree of 2.5 cm in water level on that day. All other losses were found to be
negligible.
Solution :
Catchment Area “A” = 2.3km2
Rainfall “P” = 7.5cm
Time Duration “t” = 1 day 24hours
Infiltration Losses :
Time Duration “t” = 12 Hours
Initial Infiltration Rate fo = 0.6 mm/h
Constant Infiltration Rate fc = 0.15 mm/h
K = 3 h-1
As Horton’s Equation :
Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet Let K = 1
K
= [0.6 – 0.15] (1 –e -3x12) + 0.15 x 24
1
= [0.15] (1) + (3.6)
= 3.75 mm
Evaporation Losses :
E = 2.5cm
Kp = 0.7
El =?
El = Kp x E
= 0.7 x 2.5
= 1.75cm
Rainfall Excess :

10. = P – F – El
= 7.5 -3.75 – 1.75
= 2.00 cm
Direct Runoff : = Rainfall Excess x Area of Catchment
= _2_ x 2.3 (1000)2
100
= 46000 m3
= 46000
10000
= 4.6 Hector – meter