This document contains 10 questions regarding hydrological losses and infiltration modeling. It provides solutions to problems involving calculating evaporation rates, evapotranspiration, and infiltration using equations like Horton's infiltration model. Key concepts covered include evaporation pan measurements, Penman method for evapotranspiration, and determining parameters for Horton's infiltration equation based on given cumulative infiltration depths over time.

1. 1
Course: Engineering Hydrology (CE 606) | Assignment#01
Program: Bachelor in Civil Engineering | Year/Part: III/I
Chapter 3: Hydrological Losses
A) Evaporation/Evapotranspiration
Q#01. Observation at a circular evaporation pan of 1.5m diameter for the period of 9 AM to 7
PM showed total precipitation (P) of 10 mm, quantity of water added to keep the water level in
the pan constant is 0.005 m3
, and leakage from the pan as 0.001 m3
. Estimate rate of evaporation
(mm/h).
Solution:
Diameter of pan (d) = 1.5m
Area of pan (Ap) =
𝜋
4
𝑥1.52
=1.768 m2
Quantity of water added (Q) = 0.005 m3
Volume of precipitation (P) = precipitation depth x pan’s area = 10 mm x 1.768 m2
= 10/1000 * 1.767
m3
= 0.01768 m3
Leakage (L) = 0.001 m3
.
Therefore, amount of evaporation (E) = Q+P-L = 0.005+0.01767-0.001 = 0.02168m3
Time duration (t) = 9 AM to 7 PM = 10h
Rate of evaporation (Er) =
𝐸
𝐴𝑝 𝑥 𝑡
=
0.02167 𝑚3
1.767 𝑚2 𝑥10 ℎ𝑟
=
0.02167
1.767 𝑥10
𝑥 1000 mm/m2
/hr = 1.23mm/hr.
Q#02. In a pan set-up adjacent to a reservoir, water depth at the beginning of a certain period
was 200mm. The pan received 50 mm of rainfall in that period and 20 mm of water was
removed from the pan to keep the water level within the specified range. Assuming that the
water depth at the end of the period as 180mm, pan coefficient of 0.7, and surface area (or water
spread) of the reservoir as 2 km2
, estimate the volume of water evaporated from the reservoir in
that period.
Solution:
Initial water level (I) = 200 mm
Precipitation (P) = 50 mm
Water taken out from the pan (T) = 20 mm
Final water level (O) = 180mm
Pan coefficient (K) = 0.7
Surface area of the reservoir (A) = 2 km2
= 2 *106
m2
.
Volume of water evaporated from the reservoir (Ev) = ?
Here,
Pan evaporation, Ep = (I+P-T) – O = (200 + 50 - 20) – 180 = 50 mm
Evaporation from the reservoir, E = K*Ep = 0.7*50 = 35 mm
Volume of water evaporated from the reservoir (Ev) = E * A = 35/1000 m * 2 * 106
m2
= 0.07 * 106
m3
= 0.07 Million Cubic Meters (MCM).

2. 2
Q#03. Calculate evaporation rate from an open water source, if the net radiation is 300 W/m2
,
air temperature is 300
C, and latent heat of evaporation is 2,430 kJ/kg. Assume value of zero for
sensible heat, ground heat flux, heat stored in water body and advected energy. The density of
water at 300
C = 996 kg/m3
.
Solution:
Net radiation (Hn) = 300 W/m2
Temperature (T) = 300
C
Density of water at 300
C (ρ) = 996 kg/m3
Latent heat of evaporation (L) = 2,430 kJ/kg
Evaporation (E) = ?
Sensible heat (Ha), Ground heat flux (Hg), Heat stored in body (Hs) and Advected energy (Hi) are zero.
Using energy budget method, Evaporation is estimated as;
𝐸 =
𝐻𝑛−𝐻𝑔−𝐻𝑠−𝐻𝑖
𝜌𝐿(1+𝛽)
; where, Hn is net radiation (= 300 W/m2), Hg is ground heat flux (= 0), Hs is heat
stored in the water body ( = 0), Hi is advected energy ( = 0), ρ is density of water ( = 996 kg/m3
), L is
latent heat of evaporation (= 2,430 kJ/kg), and β is Bowen ratio.
Bowen ratio (β) = Ha/He, where Ha is heat transfer by sensible heat ( =0)
Therefore, β = 0
𝐸 =
300 − 0 − 0 − 0
996𝑥2430 𝑥 1000 (1 + 0)
= 1.24x10-7
m/s
= 1.24x10-7
x1000x24x3600 mm/day = 10.7 mm/day
Q#04. A large reservoir with a surface area of 250 ha had the following average values of
parameters for a month (30 days): Water surface temperature = 250
C, Relative humidity =
65%, Wind velocity at 2m above ground = 10 km/h. For 250
C air temperature, the saturation
vapor pressure = 23.76 mm of Hg. Estimate the average daily evaporation from the reservoir
and the volume of water evaporated from the reservoir during the month. Use Meyer’s formula.
Solution:
Surface area of reservoir (A) = 250 ha = 250x104
m2
Relative humidity (RH) = 65%
Wind velocity at 1m above ground (u) = 10 km/h
For 250
C, the saturation vapor pressure (ew) = 23.76 mm of Hg
Wind speed at a height of 2 m (i.e., Z2 = 2 m) = 10 km/hr
Daily evaporation (E) = ?
Volume of water evaporated in a week (V) = ?
Here,
Actual vapor pressure (ea) = RH x ew = 0.65x23.76 = 15.44 mm of Hg
Wind velocity at h = 9m is

3. 3
𝑢9
𝑢2
= (
𝑍9
𝑍2
)
1/7
𝑢9
𝑢2
= (
9
2
)
1/7
𝑢9 = 12.40km/h
For a large reservoir (or water body), Meyer’s coefficient (KM) = 0.36
From Mayer’s formula
𝐸 = 𝐾𝑀 (𝑒𝑤 − 𝑒𝑎) (1 +
𝑢9
16
)
= 0.36 (23.76 − 15.44) (1 +
12.40
16
)
= 5.32 mm/day
Volume of water evaporated in a month (30 days) = 30𝑥
5.32
1000
𝑥250𝑥104
= 399,000 m3
Q#05. If a location (Latitude = 26.50
, Longitude = 84.50
) has following data in December,
estimate potential evapotranspiration (PET) in that month by Penman method: Mean monthly
temperature = 11.50
C, Mean relative humidity = 75%, Mean sunshine hours = 9h, Potential
sunshine hours = 11.6 h, Wind velocity at 2m height = 100 km/day, Albedo = 0.15, Upper
terrestrial solar radiation = 8 mm of water/day. Take values of saturated vapor pressure at
11.50
C as 10.4 mm of Hg, slope of saturated vapor pressure curve as 1.24 mm/0
C, Psychrometric
constant as 0.49mm/0
C, and Boltzman constant = 2.01x10-9
mm/day.
Solution:
Mean monthly temperature (Ta) = 11.50
C = 11.5+273 = 284.5K
Mean relative humidity (RH) = 75%
Mean sunshine hours (n) = 9h
Potential sunshine hours (N) = 11.6 h
Wind velocity at 2m height (u2) = 100 km/day
Albedo (r) = 0.15
Upper terrestrial solar radiation (Ha) = 8mm of water/day
Latitude (φ) = 26.5°
Saturated vapor pressure at 11.50
C (ew) = 10.4 mm of Hg
Slope of saturated vapor pressure curve (A) = 1.24mm/0
C
Psychrometric constant (γ) = 0.49mm/0
C
Boltzman constant (σ) = 2.01x10-9
mm/day
Potential evapotranspiration (PET) = ?
𝑎 = 0.29𝐶𝑜𝑠𝜑 = 0.29𝐶𝑜𝑠26.5 = 0.26
b = 0.52
Actual vapor pressure (ea) = RH x ew = 0.75x10.4 = 7.8 mm of Hg
Net radiation (Hn) is computed by
𝐻𝑛 = 𝐻𝑎(1 − 𝑟) (𝑎 + 𝑏
𝑛
𝑁
) − 𝜎𝑇𝑎
4(0.56 − 0.092√𝑒𝑎)(0.1 + 0.9
𝑛
𝑁
)

4. 4
= 8(1 − 0.15) (0.26 + 0.52
9
11.6
) − 2.01𝑥10−9
𝑥284.54(0.56 − 0.092√7.8) (0.1 + 0.9
9
11.6
)
= 1.323 mm/day
Ea is computed by
𝐸𝑎 = 0.35 (1 +
𝑢2
160
) (𝑒𝑤 − 𝑒𝑎)
= 0.35 (1 +
100
160
) (10.4 − 7.8)
=1.479 mm/day
According to Penman equation,
𝑃𝐸𝑇 =
𝐴𝐻𝑛 + 𝐸𝑎𝛾
𝐴 + 𝛾
𝑃𝐸𝑇 =
1.24𝑥1.326 + 1.479𝑥0.49
1.24 + 0.49
=1.367 mm/day
B) Infiltration – Horton’s equation
Q#06. The Horton’s infiltration equation for a basin is given by f = 10 + 30e-0.8t
where f is in
mm/hr and t is in hours. What are the values of f0, fc and k? If a storm occurs on the basin with
an intensity of more than 40mm/h, determine the depth of infiltration for the first 1 hour and
the average infiltration rate for the first 2 hours.
Solution:
Given equation
f = 10 + 30e-0.8t
Comparing with Horton’s equation f = fc + (f0-fc) e-Kt
fc = 10 mm/hr
f0-fc = 30 → f0 = 40 mm/hr
k = 0.8 /hr
If a storm or more than 40 mm/hr intensity occurs on the basin, the rainfall intensity becomes more
than f0, therefore, infiltration takes place at the capacity rate throughout the storm.
Cumulative depth of infiltration for the first 1 hour is given by
𝐹 = ∫ 𝑓. 𝑑𝑡 =
1
0
∫ (10 + 30𝑒−0.8𝑡)𝑑𝑡
1
0
= [10𝑡 − 37.5𝑒−0.8𝑡]0
1
= 30.65mm
Cumulative depth of infiltration for the first 2 hours is given by
𝐹 = ∫ 𝑓. 𝑑𝑡 =
2
0
∫ (10 + 30𝑒−0.8𝑡)𝑑𝑡
2
0
= [10𝑡 − 37.5𝑒−0.8𝑡]0
2
= 49.93mm
Average infiltration rate for first 2 hours = F/t = 49.93/2 = 24.96 mm/h

5. 5
Q#07. The cumulative depth of infiltration is given by F = 0.3 t0.45
, where F is in cm and t is in
minutes. Determine the equation for infiltration rate and the average infiltration rate.
Solution:
Cumulative depth of infiltration is given by F = 0.2 t0.55
Equation for infiltration rate (𝑓) =
𝑑𝐹
𝑑𝑡
=
𝑑(0.2𝑡0.55)
𝑑𝑡
= 0.11 t -0.45
cm/min
Average infiltration rate = F/t = 0.2 t0.55
/t = 0.2 t -0.45
cm/min
Q#08. A storm with a uniform intensity of 1.5 cm/hr for a period of 10 hours occurring over a
basin of area 600 km2
produced a runoff estimated to be 50 million m3
. Find the average
infiltration rate during the storm.
Solution:
Total rainfall = 1.5x10 = 15 cm
Total runoff depth =
𝑅𝑢𝑛𝑜𝑓𝑓 𝑣𝑜𝑙𝑢𝑚𝑒
𝐵𝑎𝑠𝑖𝑛 𝑎𝑟𝑒𝑎
=
50𝑥106
600𝑥106 = 0.083 m = 8.3 cm
Total Infiltration for 10 hrs = Rainfall – Runoff = 15.0 – 8.3 = 6.7 cm.
Infiltration rate = Infiltration/Duration = 6.7/10 = 0.67 cm/h
Q#09. If an initial infiltration rate for a catchment is 2.5 cm/hr and infiltrates a total volume of
5.5 cm until it attains a constant infiltration rate of 0.5 cm/hr after 10 hours, calculate the
Horton Constant (k).
Solution:
Initial infiltration (f0) = 2.5 cm/hr
Constant infiltration (fc) = 0.5 cm/hr
Total infiltration volume (F) = 5.5cm
Time (t) = 10 hrs
Horton constant (K) =?
Horton’s Formula for infiltration is given by
𝑓 = 𝑓
𝑐 + (𝑓0 − 𝑓
𝑐)𝑒−𝐾𝑡
Cumulative infiltration is given by
𝐹(𝑡) = ∫ 𝑓(𝑡)𝑑𝑡
𝑡
𝑜
= ∫[𝑓
𝑐 + (𝑓0 − 𝑓
𝑐)𝑒−𝑘𝑡]𝑑𝑡
𝑡
𝑜
= 𝑓
𝑐𝑡 + (𝑓0 − 𝑓
𝑐) |
𝑒−𝑘𝑡
−𝑘
|
0
𝑡
𝐹(𝑡) = 𝑓
𝑐𝑡 +
𝑓0 − 𝑓
𝑐
𝑘
(1 − 𝑒−𝑘𝑡)
For large t, the value of e-kt
becomes negligible. Hence above equation reduces to

6. 6
𝐹(𝑡) = 𝑓
𝑐𝑡 +
𝑓0 − 𝑓
𝑐
𝑘
𝑘 =
𝑓0 − 𝑓
𝑐
𝐹(𝑡) − 𝑓
𝑐𝑡
𝐾 =
2.5−0.5
5.5−0.5𝑥10
= 4/hr
Q#10. A 24hr storm occurred over a catchment of 1.8km2
and the total rainfall observed is
10cm. An infiltration capacity of 1cm/hr initially and finally 0.3cm/hr is obtained from Horton’s
curve with K = 0.5 hr-1
. An evaporation pan installed in the catchment indicated a decrease of
0.6cm in the water level (after allowing for rainfall) during 24 hours of its operation. Determine
the runoff from the catchment. Take pan coefficient = 0.7.
Solution:
Catchment area (A) = 1.8km2
= 1.8x106
m2
Rainfall (R) = 10cm
Initial rate of infiltration (f0) = 1 cm/hr
Finial rate of infiltration (fc) = 0.3 cm/hr
Horton coefficient (K) = 0.5 hr-1
Pan Evaporation (Ep) = 0.6cm
Pan coefficient (Cp) = 0.7
Runoff (Q) = ?
Total infiltration (F) during t = 24 hr is
𝐹 = 𝑓
𝑐𝑡 +
𝑓0 − 𝑓
𝑐
𝐾
(1 − 𝑒−𝐾𝑡)
= 0.3𝑥24 +
1 − 0.3
0.5
(1 − 𝑒−0.5𝑥24)
= 8.6cm
Evaporation (E) = Cp Ep = 0.7x0.6 = 0.42cm
Runoff (Q) = R-F-E = 10-8.6-0.42 = 0.98cm
Runoff volume = Qx A = (0.98/100) x1.8x106
= 17640m3
Q#11. The infiltration rates (f) observed during a test on a double ring infiltrometer are as given
below:
t(hrs) 0.0417 0.125 0.333 0.75 1.5 2.5 3.5 4.5 5.5
f (cm/h) 0.781 0.747 0.662 0.535 0.370 0.255 0.224 0.218 0.207
Determine the constants f0, fc and K of the Horton’s equation which fits the above data.
Solution:
Horton’s infiltration equation is given by
f = fc + (f0-fc) e-Kt
f -fc = (f0-fc) e-Kt

7. 7
Taking log on both sides
ln (f -fc ) = ln (f0 -fc ) –Kt
Let, y = ln (f -fc ), c = ln (f0 -fc ). Then above equation reduces to
y = -Kt + c (linear equation)
Taking, fc = 0.2cm/h
Determination of constants by least square method (We can do the same by plotting Graph as well!)
t (hrs) =x f (cm/h) y =ln(f-fc) xy x2
0.0417 0.781 -0.54 -0.02264 0.001739
0.125 0.747 -0.60 -0.07541 0.015625
0.333 0.662 -0.77 -0.25714 0.110889
0.75 0.535 -1.09 -0.82022 0.5625
1.5 0.37 -1.77 -2.65794 2.25
2.5 0.255 -2.90 -7.25106 6.25
3.5 0.224 -3.73 -13.054 12.25
4.5 0.218 -4.02 -18.0782 20.25
5.5 0.207 -4.96 -27.2901 30.25
Sum 18.7497 -20.39 -69.51 71.94
𝐾 = −
𝑁 ∑ 𝑥𝑦−∑ 𝑥 ∑ 𝑦
𝑁 ∑ 𝑥2−(∑ 𝑥)2 = −
9∗−69.51−18.7497𝑥−20.39
9𝑥71.94−(18.7497)2 = 0.822
𝑐 =
∑ 𝑦−(−𝐾) ∑ 𝑥
𝑁
=
−20.39+0.822𝑥18.7497
9
= -0.55346
c = ln (f0 -fc )
f0 -fc = exp (c) = exp (-0.55346) = 0.575
f0-0.2 = 0.575
f0 = 0.775 cm/h
C) Infiltration – Infiltration Indices
Q#12. The rainfall on 5 successive days on a catchment were 2, 6, 9, 5 and 3 cm. If the φ-index
for the storm is 3 cm/day, find the total surface runoff.
Solution:
Daily infiltration = φ Δt =3x1 = 3cm
𝑇𝑜𝑡𝑎𝑙 𝑟𝑢𝑛𝑜𝑓𝑓 = ∑(𝑅𝑎𝑖𝑛𝑓𝑎𝑙𝑙𝑖 − 𝐼𝑛𝑓𝑖𝑙𝑡𝑟𝑎𝑡𝑖𝑜𝑛𝑖)
= 0 + (6-3) + (9-3) + (5-3) + (3-3)
(for φ-index>rainfall, there is no runoff, i.e. runoff = 0)
= 11 cm.
Q#13. The mass curve of a rainfall of duration 100 min. is given below

8. 8
Time from start of rainfall (min) 0 20 40 60 80 100
Cumulative rainfall (cm) 0 0.5 1.2 2.6 3.3 3.5
If the catchment had an initial loss of 0.6 cm and a φ-index of 0.6 cm/hr, calculate the total
surface runoff from the catchment.
Solution:
Time from start of rainfall (min) 0 20 40 60 80 100
Cumulative rainfall (cm) 0 0.5 1.2 2.6 3.3 3.5
Incremental rainfall (cm) - 0.5 0.7 1.4 0.7 0.2
Since initial abstraction is 0.6 cm, there is no runoff in the first 20 minutes from rainfall of 0.5cm, and
for the second 20 minutes, the abstraction of 0.1cm is subtracted to compute runoff (i.e., 0.7 – 0.6).
With φ-index = 0.6 cm/hr, infiltration for 20 minute = φ Δt = 0.6x20/60 = 0.2 cm.
𝑇𝑜𝑡𝑎𝑙 𝑟𝑢𝑛𝑜𝑓𝑓 = ∑(𝑅𝑎𝑖𝑛𝑓𝑎𝑙𝑙𝑖 − 𝐼𝑛𝑓𝑖𝑙𝑡𝑟𝑎𝑡𝑖𝑜𝑛𝑖)
= 0 + (0.7-0.2) + (1.4-0.2)+ (0.7-0.2)
= 2.2 cm
Q#14. Precipitation falls on a 500 km2
drainage basin according to the following schedule:
Time (30 min. period) 1 2 3 4
Rainfall intensity (cm/hr) 8 4 12 10
Determine the total storm rainfall in cm. Determine also the φ-index for the basin if the net
storm runoff is 6 cm.
Solution:
Total rainfall = 8𝑥
30
60
+ 4𝑥
30
60
+ 12𝑥
30
60
+ 10𝑥
30
60
= 17 cm
Net runoff = 6 cm
Infiltration = 17-6 = 11cm
First trial: Take effective duration of rainfall (te) = 120 min (or 2 hr; i.e. entire duration)
φ-index = 11/2 = 5.5 cm/h
With this value of φ-index, the excess rainfall is
Excess rainfall = (8 − 5.5)𝑥
30
60
+ 0 + (12 − 5.5)𝑥
30
60
+ (10 − 5.5)𝑥
30
60
= 6.75 cm → this is higher
than net runoff of 6cm.
Second trial: φ-index = 6 cm/hr. [Why? → by reducing 1 duration (30 minutes) for ineffective
rainfall, te = 1.5 hrs. Then φ-index = (17 – 6 – 4*30/60)/1.5 = 6.0 cm/h.
With this value of φ-index,

9. 9
Excess rainfall = (8 − 6)𝑥
30
60
+ 0 + (12 − 6)𝑥
30
60
+ (10 − 6)𝑥
30
60
= 6.0 cm which is equal to the net
runoff of 6cm.
Hence φ-index = 6 cm/hr
Q#15. A catchment of 0.8 km2
receives a rainfall storm of 20 cm and produces surface runoff of
92,800 m3
. Estimate φ-index of the storm if temporal distribution of rainfall is as follows:
Duration (hr) 1 2 3 4 5 6 7 8
Incremental rainfall (cm) 0.8 1.8 3 4.6 3.6 3.2 2 1
Solution:
Total precipitation = 20cm
Runoff depth = 92,800 m3
/0.8 km2
= 11.6 cm.
Infiltration = 20-11.6 = 8.4cm
First trial: Take effective duration of rainfall (te) = 8 hours
φ-index = 8.4/8 = 1.05 cm/h
With this φ-index, rainfall excess is
Rainfall excess = 0 +(1.8-1.05)+ (3-1.05) +(4.6-1.05)+ (3.6-1.05)+(3.2-1.05)+(2-1.05)+0 = 11.9 cm
→ This value is higher than net runoff of 11.6cm. So increase φ-index.
Second trial: by reducing 2 hrs of ineffective rainfall, take te = 6 hrs. Then φ-index = (20-11.6-0.8-
1.0)/6 = 1.1 cm/h (because, incremental rainfall of two hrs (1st
& 8th
hrs) were also deducted to re-
calculate φ-index. → Take new φ-index = 1.1 cm/h
With this φ-index, rainfall excess is
Rainfall excess = 0 + (1.8-1.1) + (3-1.1) + (4.6-1.1) + (3.6-1.1) + (3.2-1.1) +(2-1.1) + 0 = 11.6 cm
which is equal to the net runoff.
Hence φ-index = 1.1 cm/h
Q#16. The following is the set of observed data for successive 15-minute period of 105 minutes’
storm in a catchment of 100 sq.km: 2, 2, 8, 7, 1.25, 1.25, 4 (cm/hr). If the value of φ-index is 3
cm/hr and initial loss (Ia) is 0.8 cm, estimate W-index.
Solution:
𝑇𝑜𝑡𝑎𝑙 𝑟𝑢𝑛𝑜𝑓𝑓 (𝑄) = ∑(𝑅𝑎𝑖𝑛𝑓𝑎𝑙𝑙 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦𝑖 − 𝜑 𝑖𝑛𝑑𝑒𝑥)∆𝑡
= 0 + 0 + (8 − 3)
15
60
+ (7 − 3)
15
60
+ 0 + 0 + (4 − 3)
15
60
= 2.5 cm
𝑇𝑜𝑡𝑎𝑙 𝑟𝑎𝑖𝑛𝑓𝑎𝑙𝑙 (𝑃) = 2𝑥
15
60
+ 2𝑥
15
60
+ 8𝑥
15
60
+ 7𝑥
15
60
+ 1.25𝑥
15
60
+ 1.25𝑥
15
60
+ 4𝑥
15
60
= 6.375 cm
Time duration for effective rainfall (te ) = 105/60 = 1.75 hr
𝑊 − 𝑖𝑛𝑑𝑒𝑥 =
𝑃−𝑄−𝐼𝑎
𝑡𝑒
=
6.375−2.5−0.8
1.75
= 1.76 cm/hr