hydrology-extra-numerical.pdffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

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1. A semicircle of diameter of 40km with an equilateral triangle of side of 40km
below its diameter is a close approximation to a river basin. The position co-
ordinates of 5 rain gauge stations A, B, C, D and E located within the basin with
respect to the co-ordinate axes system whose X axis and origin coincident with
diameter and center of the circle are (10, 10), (-10, 10), (-10, -10), (10, -10) and
(0, 0) km respectively. If the rainfall recorded at these rain gauges are 80, 95, 76,

82, 107mm respectively, determine the average depth of rainfall using Thiessen
polygon method.

Solution:

Diameter of semicircle = 40km

Co-ordinates of stations with E as origin

A(10, 10), B(-10, 10), C{-10, -10), D(10,-10), E (0, 0)

2. 80mm, Pa = 95mm, Pc = 76mm, Po = 82mm, Pe = 107mm
Sides of equilateral triangle = 40km

The Thiessen polygon which bound each station is shown by solid lines.

NQ= PM = QM = V10? + 10? =10/2m

40? = 207 = 34.64m

Area of Thiessen polygon for station A (As) = Area of quarter circle-Area of
triangle PEN

164.16 m?

=4qx20? 1

=4nx20? - }x10x10
= An =264.16 m?
Area of triangle SER-Area of triangle

Area of Thiessen polygon for station B (A,

Area of Thiessen polygon for station C (Ac)
MEQ

x20x34.64 — 3 x10x10 = 296.4 m°

Area of Thiessen polygon for station D (Ap) = Ac =296.4 m?
Area of Thiessen polygon for station E (AE) = 10V2 x10V2 = 200 m?
Total area of basin (A) = 264.16+264.16+296.4+296.4+200 = 1321.12 m?

Average rainfall (Ps) = ?

DPA:
Fa A
Py Ap + Po Ag + PcAc + PpAn + PeAg
= |
80x264.16 + 95x264.16 + 76x296.4 + 82x296.4 + 107x200
ee 1321.12

= 86.64 mm

3. The shape of a catchment is in the form of a pentagon ABCDE. There are 4
raingauge stations P, Q, R and $ inside the catchment. The position co-ordinates
à km are: A(0, 0), B(50, 75), C(100, 70), D(150, 0), E(75, -50), P(50, 25), Q(100,
25), R(100, -25) and S(50, -25). if rainfalls recorded at P, Q, R and S are 88, 102,
2 and 116 mm respectively, determine the mean rainfall by Thiessen Polygon
nethod.

Solution:
CO = 75-70 = Skm
BO _ BM
co MN
100 _ 50
OA
MN = 2.5km

‘Computing area of polygon bounding each station
Ap = Area of MABE + Area BNKE

Ap = $x50x75 + (25x75 — 1x2.5x25) = 3718.75 km?
Ag = Area of ACDF + Area NCFK

Ag = $x50x70 + (25x75 -1x2.5x25 — 2.5125) = 3531.25 km?

Ag = Area of ADKE = $x75x50 = 1875 km?

4. As = Area of AAKE = 5275x50

Total area (A) = 11000 km?
— LPiAi

Payg = 2°
88x3718.75+102x3531.25+112x1875+116x1875
11000

1.36mm

5. Three points on rating curve of a stream gauging station obtained from observed
data have the following co-ordinates: (2 m’/s, 10.65m), (4 m’/s, 10.85m) and (8
m/s, 11.25m). Determine the equation of rating curve and compute the
discharge in the stream corresponding to a stage of 1.5m.

Solution:

Alm’/s) G(m)
2 10.65
4 10.85
8 11.25

Form of rating equation: Q = C,(G — a)

Here, Q2 =/Q1Q3

cd e 104
Finding B, C, algebraically,
CCE
a GG =a
4 _C,(10.85 - 10.45)
2 C,(10.65 — 10.45)P
2=(24
pel
Q = CG, a)
= C,(10.65 — 10.45)!
CG, =10

The equation of rating curve is
Q =10(G - 10.45)!

ForG=11.5m

Q = 10(11.5 — 10.45)! = 10.5 m°/s

6. The ordinates of Shr unit hydrograph are given below:

Time (hr) | o | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 | 55
UHordiate
3 o | 20 | 60 |150 |120 | 90 | 66 | 50 | 32 | 20 | 10 | O
(m= /s)

If a storm of 1.2cm effective rainfall with 5hr duration occurs in the catchment,
calculate the resulting hydrograph of flow. Assume base flow to be uniform at

10m*/s.
Solution:

Rainfall excess (Re)

Base flow (Q;)

For a single storm
Direct runoff (Qyr)
Total discharge (Q)

= 1.2cm
= 10 m?/s

Computation of hydrograph of flow

= UH ordinate x Re
= Qar + Qh

Time (hr) o {5 |10]15 | 20 | 25 |30 |35]40 | 45|s0]55
UH ordinate (m3/s) | o | 20 | 60 | 150 | 120] 90 | 66 |so|32 | 20] 10] 0
Qdr (m3/s) o | 24| 72 | 180 | 144 | 108 | 79.2 | 60 | 38.4 | 24] 12 | 0
BF (m3/s) 10 |10|10|10 |10 |10 [10 |10]10 | 10] 10] 10
Q (m3/s) 10 | 34 | 82 | 190 | 154 | 118 | 89.2 | 70 | 48.4 | 34 | 22 | 10

7. The 6hr unit hydrograph of a catchment is in the form of a triangle with the peak
of 100 m°/s occurring at 24hr from the start. The base is 72hr.
a) Determine the area of the catchment represented by this unit hydrograph.

b) Calculate the flood hydrograph due to a storm of rainfall excess of 2cm during
the first 6hr and 4cm during the second 6hr interval. The base flow can be

assumed to be 25m°/s constant throughout.

8. Solution:

a) Peak discharge (Qe) =100 m/s
Base =72hr = 72x36005
Catchment area (A) =?

Area of triangular UH = Volume of runoff (Va)

3x72x3600x100 =Vs

Va = 12960000m*

For UH, runoff depth (ra) = dem

Catchment area (A) = Va/ra = 12960000/1x10” = 1296000000m? = 1296 km?

b) Finding ordinates at 6hr interval (using similar triangles)

Time (h). UH (m?/s)
o (0722)100 0

6 (6/2100 25
2 (12724)100 50
18 (18/24)100 75
24 100
30 (42/48)100 875
36 (36/48)100 75
az (80/48)100 625
48 (24/28)100 50
54 (28/48)100 375
60 (12/48)100 25
66 (6/48)100 125
72 (0/48)100 0

Hx Rea (lagged by Ghr)

DRH = DRH;+DRH,

Q=DRH (0)+Q,

9. Time (h) | UH (m?/s) | DRH, | DRH: | DRH(O) | Qu (m/s) | Q(m?/s)
0 0 0 0 25 25
6 25 so lo 50 25 75
12 50 100 |100 | 200 25 | 225
18 75 |150 |200 | 350 25 | 375
24 |100 200 |300 | soo 25 525
30 87.5 175 | 400 |575 25 600
136 75 150 |350 | 500 25 525
42 62.5 125 |300 | 425 25 450
[48 50 100 |250 | 350 25 375
54 37.5 75 |200 |275 25 300
— 1

A |25 50 |150 | 200 25 225
66 12.5 [2s 100 | 125 25 150
72 0 0 50 |50 25 75
78 I To 0 25 25

10. Assignment

The following table gives the ordinates of a DRH resulting from two successive 3-
hour durations of rainfall excess value of 2 cm and 4 cm respectively.

| t (hr) ol 3 [6 9 [12 [15 | 18 | 21 [24/27/30]

| DRH
| (m/s)

O| 120 | 480 | 660 | 460 | 260 | 160 | 100 | 50 | 20| O

a. Derive the ordinates of 3-hr UH.

3
b. Using the B-hr UH derived in a, calculate the flood hydrograph for 2 successive
storms of 6.5 cm and 10.5 cm of 3 hours duration and -index of 0.2cm/hr.
Assume base flow = 20 m°/s