This document provides solutions to hydrology problems and explanations of hydrological concepts. It includes: 1) Calculations of water balance components like precipitation, evaporation, runoff, and storage change for different systems. 2) Analysis of groundwater extraction impacts on evaporation and runoff. 3) Determination of seepage rates from polder water balances accounting for uncertainties in input data. 4) Explanations of concepts like the water balance of the root zone, open water vs grassland evaporation, and Thiessen polygons for areal rainfall estimation. 5) Calculations involving rainfall intensity-duration curves, extreme value analysis using the Gumbel distribution, and pan evap

1. 1
ANSWERS TO PROBLEMS ON HYDROLOGY
From the lecture notes Hydrology, LN0262/09/1
P.J.M. de Laat
1a Q = 0.5 m3
/s = 0.5x86400x365/(800x106
) = 0.0197 m/a = 19.7 mm/a
Water balance P – E – Q = ΔS/Δt
ΔS/Δt ≈ 0 because data refer to long-time averages, hence
E = P – Q = 200 – 19.7 = 180.3 mm/a
1b P – Erest – Eirr – Qnew = 0 (ΔS/Δt ≈ 0 because water balance components refer to long-time
averages)
Express all water balance components in volumes (m3
/a):
0.2 x 800x106
- 790x106
x0.1803 - Eirrx10x106
- 0.175x86400x365 = 0
Eirr = 1.2 m/a = 1200 mm/a
2a
Largest amount stored end of March
Smallest amount stored end of September
Difference = 220-(-225) = 445 mm = 0.445 m
Volume difference (given area = 120 km2
): 0.445 x 120x106
= 53.4x106
m3
2b Arid climate has P < 300 mm/a and humid tropical climate shows high evaporation in wet
period, hence this is a humid temperate climate (low evaporation is winter season).
3a Water balance root zone (see figure 1.2):
F + CR – ET –R ≈ 0, where infiltration F, capillary rise CR, evapotranspiration ET, recharge
groundwater system R
400 + CR – 340 -100 = 0, hence CR = 40 mm/a
3b Water balance groundwater system:
R – CR – Qb ≈ 0, where base flow is Qb
P E Q ΔS ΣΔS
Jan 250 5 150 95 95
Feb 205 25 110 70 165
Mar 165 30 80 55 220
Apr 50 50 5 -5 215
May 5 80 0 -75 140
Jun 0 100 0 -100 40
Jul 0 150 0 -150 -110
Aug 5 70 0 -65 -175
Sep 10 60 0 -50 -225
Oct 55 20 10 25 -200
Nov 65 10 15 40 -160
Dec 190 5 120 65 -95

2. 2
100 – 40 - Qb = 0, hence Qb = 60 mm/a
3c Interception water balance:
P – Ei – Ps ≈ 0, where interception evaporation Ei and precipitation reaching surface Ps
Ps = F + Qs, where surface flow is Qs
Q = Qb + Qs = 1.1x86400x365/(100x106
) = 0.347 m/a = 347 mm/a
Qs = 347 – 60 = 287 mm/a
Ps = 400 + 287 = 687 mm/a
Ei = 800 – 687 = 113 mm/a
Alternative solution:
P – Ei – ET – Q ≈ 0
800 – Ei – 340 – 347 = 0, hence Ei = 113 mm/a
3d E = ET + Ei = 340 + 113 = 453 mm/a
3e Water balance groundwater system:
R – CR – Qb – Qe ≈ 0, where the extracted groundwater is Qe
Qe = 0.16 m3
/s = 0.16x86400x365/(100x106
) = 0.05 m/a = 50 mm/a
100 – 0 – Qb -50 = 0, Qb = 50 mm/a
Total runoff Q = Qs + Qb = 287 + 50 = 337 mm/a, hence decreased by 10 mm/a
Water balance root zone:
F + CR – R – ET ≈ 0
400 + 0 – 100 – ET = 0, so ET = 300 mm/a, hence decreased by 40 mm/a
E = ET + Ei = 300 + 113 = 413 mm/a
As a result of groundwater extraction evaporation and runoff decreased!
4a Water balance polder (neglecting change in storage, because long-time average values):
P – Eo – Eg – Qout + Qin + S = 0, where open water evaporation Eo, evapotranspiration grass Eg,
water pumped out Qout, water let in Qin and seepage S
Total area A = Ao + Ag = 10x106
m2
, where open water area Ao = 2x106
m2
and grass area Ag
= 8 x 106
m2
Components of water balance are written in volumes:
0.8x10x106
- 0.6x2x106
- 0.75x0.6x8x106
- 5x106
+ 0.7x106
+ S = 0, hence
S = 1.1x106
m3
/a = 0.11 m/a = 110 mm/a
4b Inaccuracy in rainfall measurement at least a few percent (> 20 mm)
Error estimated evaporation possibly > 10% (> 60 mm)
So error seepage computed from these water balance components could be larger than 10 mm
4c Lower water level causes lower evapotranspiration of grass,
hence new Eg* = 0.9x0.75x600 = 405 mm/a
Lower water level causes the seepage to increase,
hence new S* = 1.1x110 =121 mm/a
Set up water balance in volumes:
0.8x10x106
– 0.6x2x106
– 0.405x8x106
– Qout* + 0.7x106
+ 0.121x10x106
= 0
Qout* = 5.47x106
m3
/a = 547 mm/a

3. 3
0
40
80
120
160
1 2 3 4 5
Precipitation(mm/day)
Days
Hyetograph
0
100
200
300
400
500
600
700
800
900
1000
0 200 400 600 800 1000 1200
A - B
B- C
A - C
Year A B C Sum A Sum B Sum C
0 0 0
1971 90 100 100 90 100 100
1972 60 100 80 150 200 180
1973 70 80 70 220 280 250
1974 80 120 100 300 400 350
1975 50 50 50 350 450 400
1976 50 50 50 400 500 450
1977 100 100 100 500 600 550
1978 50 100 80 550 700 630
1979 120 200 170 670 900 800
1980 60 100 100 730 1000 900
Weight (-) Rainfall (mm)
weighted
rainfall (mm)
A 8x8/2/120 = 0.27 75 20
B (8x8/2+8x2)/120 = 0.4 40 16
C 4x10/120 = 0.33 30 10
Thiessen mean areal rainfall = 46
5
6
Station A is not reliable, because it does not give a linear relation with B, nor with C.
7 Total area: 10x12 = 120 units
8a There is no linear increase of the mass curve between
3 and 5 hours since the rainfall rate is not constant.
8b Lowering the raingauge results in a larger catch. The
mass curve will show higher values, e.g. an increase
of 10 %.
Day Rain (%) Rain (mm/d)
1 10 24
2 10 24
3 10 24
4 60 144
5 10 24
Total 100 240

4. 4
Duration n (d) R depth (mm) Intensity (mm/d)
1 50 50
3 90 30
5 110 22
10 130 13
Year Rank Depth (mm) p=m/(N+1) T=1/p LogT
1986 1 110 0.09 11.0 1.041
1991 2 99 0.18 5.5 0.740
1992 3 93 0.27 3.7 0.564
1987 4 89 0.36 2.8 0.439
1990 5 86 0.45 2.2 0.342
1984 6 82 0.55 1.8 0.263
1989 7 80 0.64 1.6 0.196
1985 8 78 0.73 1.4 0.138
1993 9 76 0.82 1.2 0.087
1988 10 74 0.91 1.1 0.041
60
70
80
90
100
110
120
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
Dailyrainfalldepth(mm)
Logreturn period
9a
9b
10

5. 5
0
10
20
30
40
50
60
70
80
Jan Feb Mrt Apr Mei Jun Jul Aug Sep Okt Nov Dec
Precipitation Evaporation
LogT for T = 20 years equals 1.3. Interpolation in the above figure yields an extreme value for
the daily rainfall depth of about 119 mm/d.
11a For P(X < 100) = q = 0.90 It follows y = -ln(-ln q) = 2.25
For P(X < 130) = q = 0.98 It follows y = -ln(-ln q) = 3.90
Application of y = a(X-b) with the above data gives
a = 0.055 and b = 59.1,
hence the Gumbel equation reads: y = 0.055(X - 59.1)
For T = 100 years, p = 1/T = 0.01, hence q = 1-p = 0.99, which yields y = 4.6
4.6 = 0.055(X – 59.1) gives X (T = 100) = 142.7 mm
11b Gumbel distribution should be based on annual extremes
At least 20 years of data
Time series should be stationary
Data should be independent
Data are from the same population
12a
Only during April, May and June the rainfall is less than the potential evapotranspiration. This
is the “dry” period. Reduction occurs at the end of the dry period, hence in the month June.
12b Total potential evapotranspiration equals 630 mm, the actual evapotranspiration is given as 610,
hence there is a reduction of 20 mm, which occurs in June, so the actual evapotranspiration in
that month is 50 mm.
12c P – Q – E = ΔS/Δt = 0 or Q = P – E = 660 – 610 = 50 mm/a
13a Eo is generally (slightly) larger than Epot and Epot > Eact, moreover the actual evapotranspiration
Eact is in a natural catchment always less than the precipitation P, This results in the following
P = 100 mm/a
Eo = 3000 mm/a
Epot = 2500 mm/a
Eact = 40 mm/a
13b With an annual rainfall of less than 300 mm this is an arid climate.

6. 6
14a Curve A is the deep lake, it heats up slowly in spring, the shallow lake (curve B) heats up faster
and reaches peak evaporation earlier in the year. Curve C is pan evaporation, which is generally
higher that the evaporation of a nearby lake (see lecture notes for arguments).
14b Pan coefficients : fdeep = 18.7/23.4 = 0.8 and fshallow = 19.3/23.4 = 0.825
15a Dividing radiation given in J/d/m2
by the latent heat (L = 2.45x106
J/kg) yields the radiation in
terms of evaporation (mm/d) as follows:
Rs (august 15) = 20482000/(2.45x106
) = 8.36 mm/d
Rs (august 16) = 16758000/(2.45x106
) = 6.84 mm/d
The extraterrestrial radiation RA is read from table 3.3 for latitude 15o
N as (440+429)/2 = 434.5
W/m2
or expressed in terms of mm/d the value for RA = (434.5 x 86400)/(2.45x106
) = 15.3
mm/d
The equation Rs = RA(a + b n/N) could be written for both dates as:
8.36 = 15.3 (a + b 6.3/12.6) = 15.3 (a + 0.5b)
6.84 = 15.3 (a + b 4.2/12.6) = 15.3 (a + b/3)
The coefficients from these equations can now be solved as follows: a = 0.25 and b = 0.60
15b
Sunshine duration n 10.4 hr
Air temperature T 31.0 o
C
Relative Humidity RH 0.384
Wind speed U 2.0 m/s
Saturation vapour pressure es Eq. 3.11
0.6108*exp((17.27*T)/
(237.3+T))
4.49 kPa
Slope curve s Eq. 3.12 4098*4.49/(237.3+31)2 0.256
Dew point vapour pressure ed Eq. 2.1 4.49*0.384 1.724
Extraterrestrial Radiation RA Table 3.3 (452+423)/2 437.5 W/m2
Day length N Table 3.4 13.0 hr
Short wave radiation Rs Table 3.5 (0.25+0.6*10.4/13)*437.5 319.4 W/m2
Net long wave radiation RnL Eq. 3.4
5.6745*10-8
*(273+31)4
*
(0.34-0.139*SQRT(1.724))
*(0.1+0.9*10.4/13)
62.6 W/m2
Net radiation RN Eq. 3.5 (1 - 0.06)*319.4-62.6 237.6 W/m2
Aerodynamic resistance ra Eq. 3.13 245/(0.54*2+0.5) 155.1 s/m
Penman Eo Eq. 3.10
86400/(2.45*106
)*
(0.256*237.6+1004.6*
1.2047*(4.49-1.724)/155.1)/
(0.256+0.067)
9.0 mm/d
Radiation method ETMakkink Eq. 3.18
86400*0.8*0.256*319.4/
(0.256+0.067)/(2.45*106
)
7.14 mm/d

7. 7
Time Time step Vol added Vol per dt fp fp
minutes hours cm3 cm3 cm3/hr cm/hr
0 0 0
1 0.017 60 60 3600 6.00
3 0.033 162 102 3060 5.10
5 0.033 252 90 2700 4.50
10 0.083 427 175 2100 3.50
20 0.167 657 230 1380 2.30
40 0.333 837 180 540 0.90
60 0.333 957 120 360 0.60
90 0.500 1122 165 330 0.55
120 0.500 1272 150 300 0.50
0.00
1.00
2.00
3.00
4.00
5.00
6.00
7.00
0 20 40 60 80 100 120 140
fpincm/hr
time in minutes
12
13
14
15
16
17
18
19
0 2 4 6 8 10
fpincm/hr
time inhours
Few hours later
16a,b
16c fc = 0.5 cm/hr
17a fo = 20 mm/hr, fc = 4 mm/hr
For t =1, fp = 12: Eq. 4.2: 12 = 4 + 16 exp(-60 k), so k = 0.01155 min-1
17b Rainfall intensity i = 20 mm/hr, which is larger than infiltration capacity, so infiltration equal to
fp from t = 0 to t = ⅓ hr.
Hence for t = 0 fp = 20, for t = = ⅓ hr fp = 16, so average fp = 18 mm/hr during ⅓ hr, which
means ⅓ * 18 = 6 mm
Rainfall intensity i = 12 mm/hr, which is smaller than infiltration capacity, so infiltration equal
to rainfall intensity, hence ⅓ * 12 = 4 mm
18 First quarter of an hour: fp = 10 mm and rainfall = 10 mm, but only 8 mm reaches the ground,
because 2 mm is intercepted by the vegetation, hence 8 mm will infiltrate.
Second quarter: 10 mm of rain reaches the ground but infiltration capacity fp = 8 mm, hence 8
mm will infiltrate.
Total infiltration over 30 minutes = 8 + 8 = 16 mm
19a Surface runoff is 0.05 l/s = 0.05 * 10-3
m3
/s = 3600*0.05 * 10-3
m3
/s = 0.18 m3
/hr
Runoff from plot of 25 m2
= 0.18/25 = 0.0072 m/hr = 7.2 mm/hr
Rainfall intensity is 20 mm/hr, hence fc = 20 - 7.2 = 12.8 mm/hr
19b The initial infiltration capacity of a wet soil will be
smaller than a dry soil.

8. 8
0
10
20
30
40
50
1 2 3
P(mm/hour)
Hour

20a Available Moisture AM, Moisture Content Field Capacity MCFC and
Moisture Content Wilting point MCWP, Depth root zone Dr
Equation: AM = Dr (MCFC - MCWP)
MCFC MCWP AM (cm)
Sand 0.14 0.03 6.6
Loam 0.28 0.07 12.6
Clay 0.42 0.28 8.4
20b AM(Sand) = 66 mm, readily available 50% = 66/2 = 33 mm
Irrigation interval is 33/4 = 8 days.
Similar for loam: 63/4 = 15 days and Clay: 42/4 = 10 days
21 With infiltration F, percolation D and storage S, the water balance of the root zone reads:
F - D = ∆S = 500*(0.28 - 0.12) = 80 mm
D = F - ∆S = 110 - 80 = 30 mm
22a,b Apply Eq. 4.11: Q6 = Q0 exp(-6/K) = 1.8 = 10 exp(-6/K), hence K = 3.5 weeks. Use the equation
to compute Q at the end of each week (see table below).
Water released from the dam site Qout = 5 m3
/s = 5*86400*7/20,000 = 151.2 mm/week
Water balance reservoir: Q + P - E - Qout = ∆S/∆t
Week Q Av Q Q P E Qout ∆S/∆t Level
0 10.0 m3
/s mm/w mm/w mm/w mm/w mm/w 20.00
1 7.51 8.75 264.6 0 40 151.2 73.4 20.07
2 5.64 6.58 199.0 0 40 151.2 7.8 20.08
3 4.24 4.94 149.0 0 45 151.2 -42.2 20.04
4 3.18 3.71 112.2 0 45 151.2 -74.0 19.96
5 2.39 2.78 84.1 0 45 151.2 -112.1 19.84
6 1.80 2.10 63.5 23 45 151.2 -109.7 19.73
23a The total runoff from the catchment is the area under curve A: (15+60+65+30+12.5+2.5)*3600
= 666,000 m3
or 666000/33300000*1000 = 20 mm
Hence the runoff coefficient is 20/100 = 0.2 or 20%
23b Total losses are 100 - 20 = 80 mm. A first estimate of the loss
rate is 80/3 = 27 mm/hr.
This is more than the rainfall in the last hour (20 mm). The
losses in the first two hours are therefore 80 - 20 = 60 mm,
hence a constant loss rate of 60/2 = 30 mm/hr

9. 9
Day Q (mm/d) Log Q
1 1.72 0.24
2 1.5 0.18
3 1.31 0.12
4 1.14 0.06
5 1 0.00
6 6 0.78
7 10 1.00
8 6 0.78
9 4 0.60
10 3 0.48
11 2.62 0.42
12 2.29 0.36
13 2 0.30
14 1.75 0.24
0.00
0.20
0.40
0.60
0.80
1.00
1.20
0 5 10 15
LogQ
Day
0
2
4
6
8
10
12
0 5 10 15
Q(mm/d)
Day
Direct
Runoff
Base flow
rea Q (mm)
1 2.5
2 7
3 7
4 4
5 2.5
6 -5
noff (mm)= 18 0
2
4
6
8
10
12
0 5 10 15
Q(mm/d)
Day
Direct
Runoff
Base flow
1
2 3
4
5
6
Area Q (mm)
1 2.5
2 7
3 7
4 4
5 2.5
6 -5
Direct runoff (mm)= 18
23c Volume of flood hydrograph in point B = (0+10+40+45+20+5)*3600 = 432000 m3
.
Total infiltration between A and B = 666000-432000 = 234000 m3
over an area of 50*20000 =
1000000 m2
, which means an average depth of 234 mm. It takes 5 to 6 hours for the flood wave
to pass, hence the average infiltration rate is 234/5.5 = 42.5 mm/hr
24a
From a plot of Log Q against
time it is shown that the depletion
curve starts on day number 10.
24b
24c Depletion curve is Eq. 4.11:
1.0 = 1.72 exp(4/K), hence K = 7.4 days
QB = 3.0 mm/d (given)
Eq. 4.11 : QA= 1.0 exp(-5/7.4) = 0.5 mm/d

10. 10
Total runoff under depletion curve
0
K
tt
0
t
kQtdeQ=CT
0
0





Grey area is base flow produced by rainstorm
of 50 mm=
CT (B) – CT(A) + triangle under base flow
separation line = 7.4*3-7.4*0.5+5*(3-0.5) =
18.5 + 6.25 = 24.75 mm
25 From graph: Rainfall depth for tc = 3 hr equals 33 mm,
so intensity i = 11 mm/hr = 0.011/3600 m/s
Eq. 6.3: Qp = 0.4 * 0.011/3600 * 6 * 106
= 7.3 m3
/s