Surface Water Hydrology Assignment 2: Analysis and Calculations

This document presents a comprehensive solution to a surface water hydrology assignment, addressing several key problems. The solution begins with calculations for different return periods, followed by hydrograph analysis. The hydrograph analysis includes the inflow and outflow hydrographs, along with the magnitude of peak attenuation and peak lag time. The assignment continues with a listing of bulk entitlement records and a calculation of the total available water in a water supply system. Finally, the assignment delves into the distribution of water entitlements among different stakeholders, including system operators, water corporations, and environmental agencies, showing the share commanded by each entitlement holder and the percentage of total water supplied. References to relevant literature are also included.

Surface Water Hydrology 1
SURFACE WATER HYDROLOGY
ENCIV7090
ASSIGNMENT 2
[Student number]
[Student name]
© May, 2018

Surface Water Hydrology 2
PROBLEM 1 (8 MARKS)
SOLUTION
X 2=Mean of Q+ K 2 S . D
Data ¿ Spread Sheet
Mean standard deviation Ӯ =0.735462863
Mean of log Q=1.838833904
Coefficient of skewness Cs=−1.567731336
Mean discharge Q=152.9531343 ML/d
Standard deviation of Q=150.3055677 ML
K T =Z +(Z 2 – 1) K + 1
3 (Z 3 – 6 Z )∗( K 2) – (Z 2 – 1)∗K 3+Z∗K 4+ 1
3 K 5( Singh ,1998) .
¿ K=C S / 6=−0.261289
Z=W− 2.515517+0.802853 W +0.01032W 2
1+1.432788 W +0.189269 W 2 +0.001308 W3
W ={ln( 1
0.52 )}0.5
Where P= 1
Return period T

Surface Water Hydrology 3
For 2 year return period;
P=( 1
2 )=0.5
W ={ln ( 1
0.52 )}0.5=1.17741
Substituting the value of W ∈Z ;
Z=W − 2.515517+0.802853 W +0.01032W 2
1+1.432788 W +0.189269 W2 +0.001308 W3 =0.00000365653
K T =Z +(Z 2 – 1) K + 1
3 (Z 3 – 6 Z )∗( K 2) – (Z 2 – 1)∗K 3+Z∗K 4+ 1
3 K 5
K=−0.261289
K 2=0.243
X 2=Mean of Q+ K 2 S . D
X 2=152.9531343+ 0.243(150.3055677)
X 2=189.48 Ml/day
For 5 year return period
TP=( 1
5 )=0.2
W ={ln( 1
0.22 ) }0.5=1.794123
Substituting the value of W ∈Z ;
Z=W − 2.515517+0.802853 W +0.01032W 2
1+1.432788 W +0.189269 W2 +0.001308 W 3 =0.841463
K T =Z +(Z 2 – 1) K + 1
3 (Z 3 – 6 Z )∗( K 2) – (Z 2 – 1)∗K 3+Z∗K 4+ 1
3 K 5
K=−0.261289
K 5=0.80687
X 5=Mean of Q+ K 5 S . D

Surface Water Hydrology 4
X 5=152.9531343+ 0.80687(150.3055677)
X 5=274.23 Ml /day
For a 10 year return period
TP=( 1
10 )=0.1
W ={ln( 1
0.12 ) }0.5=2.145966
Substituting the value of W ∈Z ;
Z=W − 2.515517+0.802853 W +0.01032W 2
1+1.432788 W +0.189269 W 2 +0.001308 W3 =1.2817362
K T =Z +(Z 2 – 1) K + 1
3 (Z 3 – 6 Z )∗( K 2) – (Z 2 – 1)∗K 3+Z∗K 4+ 1
3 K 5
K=−0.261289
K 10=1.26272
X 10=Mean of Q+ K 10 S . D
X 10=152.9531343+1.26272(150.3055677)
X 10=303.81 Ml/day
For a 25 year return period
Therefore ; P=( 1
25 )=0.04
W ={ln( 1
0.042 ) }0.5=2.537
Substituting the value of W ∈Z ;
Z=W − 2.515517+0.802853 W +0.01032W 2
1+1.432788 W +0.189269 W 2 +0.001308 W 3 =1.7510853
K T =Z +(Z 2 – 1) K + 1
3 (Z 3 – 6 Z )∗( K 2) – (Z 2 – 1)∗K 3+Z∗K 4+ 1
3 K 5
K=−0.261289

Surface Water Hydrology 5
K 25=1.12257
X 25=Meanof Q+ K 25 S . D
X 25=152.9531343+1.122568(150.3055677)
X 25=321.68 Ml /day
For a 50 year return period
P=( 1
50 )=0.02
W ={ln( 1
0.022 ) }0.5=2.79715
Substituting the value of W ∈Z ;
Z=W − 2.515517+0.802853 W +0.01032W 2
1+1.432788 W +0.189269 W 2 +0.001308 W3 =2.0541982
K T =Z +(Z 2 – 1) K + 1
3 (Z 3 – 6 Z )∗( K 2) – (Z 2 – 1)∗K 3+Z∗K 4+ 1
3 K 5
K=−0.261289
K 50=1.362747
X 50=Mean of Q+ K 50 S . D
X 50=152.9531343+1.362747 (150.3055677)
X 50=357.8 Ml /day
For 100 year return period
Therefore ; P=( 1
100 )=0.01
W ={ln( 1
0.012 ) }0.5=3.035
Substituting the value of W ∈Z ;
Z=W − 2.515517+0.802853 W +0.01032W 2
1+1.432788 W +0.189269 W 2 +0.001308 W 3 =2.32679567
K T =Z +(Z 2 – 1) K + 1
3 (Z 3 – 6 Z )∗( K 2) – (Z 2 – 1)∗K 3+Z∗K 4+ 1
3 K 5

Surface Water Hydrology 6
K=−0.261289
K 100=1.23162
X 100=Mean of Q+ K 100 S . D
X 100=152.9531343+1.362747 (150.3055677)
X 100=338.07 Ml /day

Surface Water Hydrology 7
PROBLEM 2 (7 MARKS)
SOLUTION
Column 4 is calculated using the formula; 2 S
Δt + Q (cfs) (Zoppou, 1999).
A ten minute routing interval Δt is used. (600 seconds)
Column 1
Head H (Ft)
Column 2
Discharge (Q)
Column 3
Storage S (Ft3)
Column 4
2 S
Δt + Q (cfs)
0 0 0 .00
0.5 3 43500 148.20
1.0 8 87,120 298.40
1.5 17 130,680 452.60
2.0 30 174.240 610.80
2.5 43 217,800 769.00
3.0 60 261.360 931.20
3.5 78 304,920 1094.40
4.0 97 348,480 1258.60
4.5 117 392,040 1423.80
5.0 137 435,600 1589.00
Table 2.1Elevation – Discharge-Storage data
 Find attached an MS Excel spreadsheet which re-creates the
results.

Surface Water Hydrology 8
 Graph which compares in the inflow and outflow hydrographs
0
0.805
3.21
10.2
21.24
28.34
30.285
28.62
24.445
18.51
14.055
10.965
8.55
0
10
20
30
40
50
60
70
inflows against outflows
inflows
Outflow (cfs)
Inflows I (cfs)
Figure 2.1 Graph which compares in the inflow and outflow
hydrographs
The magnitude of peak attenuation and peak lag time are crucial
indicators in hydrograph data interpretation. (Mitchell, 1954)
Magnitude of peak attenuation –
 Peak Inflow- 60 cfs
 Peak outflow – 21.21 cfs
The Peak lag time lies between 180- 240 minutes

Surface Water Hydrology 9
PROBLEM 3 (1 MARK)
List all of the bulk entitlement records for Central Highlands Water.
 Lal – Central Highlands
 Landsborough –Navarre
 Redbank
 Skipton
 Upper West Moorabool
System
 Yarrowee – White Swan
System
 Amphitheatre
 Avoca
 Ballan
 Beaufort
 Blackwood and Barry’s Reef
 Bullarook System – Central
Highlands Water
 Lexton
 Loddon System – Part
Maryborough – Central
Highlands Water
PROBLEM 4 (1 MARK)
Calculate the total available water (in mega liters) for use within a water
supply system
SOLUTION
→Available water= (143,500 – 7,500 + 52,000 – 15,500 – 24,000+ 17,000 –
27,500) ML
→Available water = 138,500 ML

Surface Water Hydrology 10
PROBLEM 5 (3 MARKS)
SOLUTION
Entitlemen
t
Water
Share A
(ML)
Water
Share B
(ML)
Water
Share C
(ML)
Water
Share D
(ML)
Water
Share E
(ML)
Water
Share F
(ML)
Total
(202200
-
174000)
= 28200
(174000
– 86400)
=87600
(86400-
56400)=
30000
(56400 -
45900)=
10500
(45900 -
0)=
45900
0
System
Operator
(150000
– 75000)
= 75000
(75000 –
56800)=
18200
(56800 –
40200) =
16600
(40200 –
32000 )=
8200
(32000 -
20000) =
12000
20000
Water
Corporatio
n A
(82000 -
82000) =
0
(82000 -
40700)=
41300
(40700 -
28700)=
12000
(28700 -
24500) =
4200
(24500 -
0)=
24500
0
Water
Corporatio
n B
(6500 –
2500)=
4000
(2500 –
1500) =
1000
(1500 –
1500) =
0
(1500 -
1400) =
100
(1400 –
0) =
1400
0
Water
Corporatio
n C
(8700 -
4500 ) =
4200
(4500 -
1700 )=
2800
(1700 -
1200 ) =
500
(1200 –
1000) =
200
(1000 –
0) =
1000
0
Environme
nt A
(40000 -
40000) =
0
(40000 -
35000 )
= 5000
(35000 -
25000) =
10,000
(25000 -
19000) =
6000
(19000 –
0)
= 19000
0
Environme
nt B
(65000 -
45000) =
20000
(45000–
7500) =
37500
7500 0 0 0
Table 5.1; Finding the share commanded by each entitlement holder in ML

Surface Water Hydrology 11
Entitlement
Water
Share
A
Water
Share
B
Water
Share
C
Water
Share
D
Water
Share
E
Water
Share
F
Total 14.0% 43.3% 14.7
% 5.3% 22.7% 0%
System Operator 150000 75000 5680
0 40200 32000 20000
Water
Corporation A 0.0% 20.4% 5.9% 2.1% 12.1% 0%
Water
Corporation B 2.0% 0.5% 0.0% 0.1%s 0.7% 0%
Water
Corporation C 2.1% 1.4% 0.2% 0.1% 0.5% 0%
Environment A 0.0% 2.5% 4.9% 3.0% 9.4% 0%
Environment B 9.9% 18.5% 3.7% 0% 0% 0%
Table 5.2; Share commanded by each entitlement holder as a percentage of
the total.
 Using the value of available water; water is distributed as follows
Entitlement
Water
Share A
(ML)
Water
Share B
(ML)
Water
Share C
(ML)
Water
Share D
(ML)
Water
Share E
(ML)
Water
Share
F(ML)
Total 138,500 119,100 59,139.5 38,780 31,439.5 0 %
System
Operator 150000 75000 56800 40200 32000 20000
Water
Corporation
A
0 28,254 8,171.5 2908.5 16,758.5 0
Water
Corporation
B
2770 692.5 0 138.5 969.5 0
Water
Corporation
C
2908.5 1939 277 138.5 692.5 0
Environment
A 0 3462.5 23,960.5 4155 13,019 0
Environment
B 13,711.5 25,622.5 5,124.5 0 0 0

Surface Water Hydrology 12
 Water share B has the greatest reliability entitlement. This is so since
the share supplies across all establishments and also commands the
largest percentage of the total water supplied, (Tisdell,1992)