Tutorial 3 - Solutions
Tutorial 3 - Solutions
Tutorial 3 - Solutions
1)
a) Vapour pressure, e(T) = 0.6108 exp (17.27T / (T + 237.3 )) …………….(1)
= 1.78 KPa
b) Tdew = 22.3 0C
= 2.69 KPa
= 0.000662 0C-1
= 1.91 KPa
3) Latitude = 13.1oN
= 0.2286
From table, extraterrestrial radiation Ra = 38 and Maximum possible duration of
sunshine hours N = 12.6
= 7.1 hours/day
Rns = (1-α) Rs
= (1-0.23) * 20.21
, ,
Rnl = 0.34 0.14 1.35 0.35
Tavg = 22.1 oC
= 2.66 kPa
Rso = (as + bs ) * Ra
= (0.25+ 0.5)20.21
Rn = Rns – Rnl
= 15.562-6.01 = 9.552 MJ m-2 day -1
5) Tmax = 34.8oC
Tmin = 25.6oC
Tmean = (34.8 + 25.6 ) / 2
= 30.2 oC
Altitude, z = 920 m
= 0.665 *10-3 P
= 0.0604 0C-1
.
.
Δ= .
= 0.245 kPa oC-1
.
=
.
N = 12.3 hours
.
= = 0.69
.
Rs = ( as + bs ) Ra
Rs = ( as + bs) Ra
Rns = (1-α) Rs
= (1-0.23) * 22.6
, ,
Rnl = 0.34 0.14 1.35 0.35
Tmax,k = 34.8 oC
Tmin,k = 25.6 oC
Tavg = 30.2 oC
ea = 2.85 kPa
. . .
Rnl = 0.34 0.14√2.85 1.35 ∗ 0.792 0.35
Rn = Rns – Rnl
. ∆
ET0=
∆ .
= 5.71 mm/day
(6) Plot the hyetograph for the given problem. The solution is a trial and error method.
This gives the surface runoff = 1.6 1 3.6 1 5 1 2.8 1 2.2 1 ∗ 0.5 5.1 > 3.6 cm
This gives the surface runoff = 3.6 1.6 5 1.6 2.8 1.6 2.2 1.6 ∗ 0.5 3.6
Rainfall rate(cm/hr)
4
0
0‐0.5 0.5‐1 1‐1.5 1.5‐2 2‐2.5 2.5‐3
Time(hours)
14
Rainfall rate(cm/hr)
12
10
8
6
4
2
0
0‐0.5 0.5‐1 1‐1.5 1.5‐2 2‐2.5 2.5‐3 3‐3.5
Time(hours)
Therefore, surface runoff = 4.0 3.5 12 3.5 8.5 3.5 4.5 3.5 4.5 3.5 ∗ 0.5 8
(8)
Type 1 soil
5
Rainfall rate(cm/hr)
4
3
2
1
0
0‐10 10‐20 20‐30 30‐40 40‐50 50‐60
Time(minutes)
Using the trial and error method mentioned in problem (6), we get the ф-index = 1.5 cm/hr
10
This gives surface runoff = 2.0 1.5 4 1.5 3 1.5 2 1.5 ∗ 0.8333
60
Type 2 soil
5
Rainfall rate(cm/hr)
4
3
2
1
0
0‐10 10‐20 20‐30 30‐40 40‐50 50‐60
Time(minutes)
Using the trial and error method mentioned in problem (6), we get the ф-index = 0.6 cm/hr
10
This gives surface runoff = 2.0 0.6 4 0.6 3 0.6 2 0.6 1 0.6 ∗ 1.5
60
Type 2 soil
5
Rainfall rate(cm/hr)
4
3
2
1
0
0‐10 10‐20 20‐30 30‐40 40‐50 50‐60
Time(minutes)
Using the trial and error method mentioned in problem (6), we get the ф-index ≥ 4 cm/hr
This gives surface runoff = 0 cm (i.e. all the rain water gets infiltrated)
(9) Δθ = (1-se). θe
= 1.27 cm
Substituting F= 1.27 into the right hand side of the equation gives F= 1.79 cm, and after a
number of iterations, F converges to a constant value of 3.17 cm. The infiltration rate after one
hour is found from the equation,
ψ∆θ
f= K[ 1]
.
= 0.65[ 1] = 1.81 cm/h
.
ψ∆θ
Ponding time, tp =
From problem (10), tp = 0.17 h and Fp = 0.85 cm under rainfall intensity 5 cm/h. For t= 1 h, infiltration
depth is given by
ψ∆θ
F – Fp – ψ∆θ ln [ ] = K(t-tp)
ψ∆θ
.
F – 0.85 – 5.68 ln [ ] = 0.65(1-0.17) = 0.54
. .
ψ∆θ
f= K[ 1]
.
= 0.65[ 1] = 1.87 cm/h
.
(12) The precipitation and infiltration rates versus time are plotted as shown.
In the Horrton’s equatio
on, the Horton
n’s constant k=
k
= 8.25 ( . 2 min) = 8.25 x = 0..275 cm
. . /
Thus, k= = 12 hr-1
.
-kt
Thus Hortton’s equation
n is f = fc + ( )e 1 + (4.5-1.22) e-12t
= 1.2
.
f= 1.2 +
where f is in cm/hr an
nd t is in hr
Total rain
n , P= 68.75 units = 68.7
75 x = 2.29
9 cm
3.3 3.3
= 1.2x 0.5 . 0
.
= 0.6 + (1- = 0.88 cm