Calculation of Discharges using LogPearson Type 3 Probability Distribution
Using the same data set you developed for problem 6 of assignment or problem 4 of assignment 1B (ENCIV4090), fit a LogPearson type 3 probability distribution and calculate the discharges which equates to the 2 year, 5 year, 10 year, 25 year, 50 year and 100 year return period.
As part of your work, you must show:
 All of your calculations and workings (4 marks);
 A graph which compares the actual measured data and your fitted probability distribution function (2 marks); and
 A discussion of how the measured data compares to the fitted PDF including comments on its goodness of fit and any other aspects you may think relevant (2 marks).
Make sure all units are correctly shown and that your graphs are appropriately scaled and titled.
X_{T} = Mean of Q+ K_{T}S_{.}D
Mean standard deviation ? = 0.827100252
Mean of log Q = 1.106109132
Coefficient of skewness Cs = 1.701861093
Mean discharge Q = 31.4258
Standard deviation of Q = 31.76046793
K_{T} = Z + (Z^{2 }– 1) K + (Z^{3} – 6 Z)*(K^{2)} – (Z^{2 }– 1)* K^{3 }+ Z*K^{4 }+ K^{5 }
And K = C _{S }/ 6=  0.28364 (BoBee, 1975)
Z = W 
W= {ln ()}^{ 0.5}
Where P =
Calculations for 2 year return period;
P = ( ) = 0.5
W= {ln ()}^{ 0.5}= 1.17741
Substituting the value of W in Z;
Z = W  = 0.00000365653
K_{T} = Z + (Z^{2 }– 1) K + (Z^{3} – 6 Z)*(K^{2)} – (Z^{2 }– 1)* K^{3 }+ Z*K^{4 }+ K^{5}
K = 0.28364
K_{2 }= 0.2602
X_{2} = Mean of Q+ K_{2}S_{.}D
X_{2 = }31.4258+ 0.2602 (31.76046793)
X_{2} = 39.69 Ml/day
Calculations for 5 year return period
TP = ( ) = 0.2
W= {ln ()}^{ 0.5}= 1.794123
Substituting the value of W in Z;
Z = W  = 0.841463
K_{T} = Z + (Z^{2 }– 1) K + (Z^{3} – 6 Z)*(K^{2}) – (Z^{2 }– 1)* K^{3 }+ Z*K^{4 }+ K^{5}
K = 0.28364
K_{5 }= 0.803
X_{5} = Mean of Q+ K_{5}S_{.}D
X_{5 }=_{ }31.4258+ 0.803 (31.76046793)
X_{5} = 56.93 Ml/day
For a 10 year return period
TP = ( ) = 0.1
W= {ln ()}^{ 0.5}= 2.145966
Substituting the value of W in Z;
Z = W  = 1.2817362
K_{T} = Z + (Z^{2 }– 1) K + (Z^{3} – 6 Z)*(K^{2}) – (Z^{2 }– 1)* K^{3 }+ Z*K^{4 }+ K^{5}
K = 0.28364
K_{10 }=0.97199
X_{10} = Mean of Q+ K_{10}S_{.}D
X_{10 = }31.4258+ 0.803 (31.76046793)
X_{10} = 62.30 Ml/day
For a 25 year return period
Therefore; P = ( ) = 0.04
W= {ln ()}^{ 0.5}= 2.537
Substituting the value of W in Z;
Z = W  = 1.7510853
K_{T} = Z + (Z^{2 }– 1) K + (Z^{3} – 6 Z)*(K^{2}) – (Z^{2 }– 1)* K^{3 }+ Z*K^{4 }+ K^{5}
K = 0.28364
K_{25}= 1.08511
X_{25}= Mean of Q+ K_{25} S_{.}D
X_{25 = }31.4258+ 1.154 (31.76046793)
X_{25} = 65.89 Ml/day
P = ( ) = 0.02
W= {ln ()}^{ 0.5}= 2.79715
Substituting the value of W in Z;
Z = W  = 2.0541982
K_{T} = Z + (Z^{2 }– 1) K + (Z^{3} – 6 Z)*(K^{2}) – (Z^{2 }– 1)* K^{3 }+ Z*K^{4 }+ K^{5}
K = 0.28364
K_{50 }=1.129
X_{50} = Mean of Q+ K_{50}S_{.}D
X_{50 = }31.4258+ 1.129 (31.76046793)
X_{50} = 67.28 Ml/day
For 100 year return period
Therefore; P = ( ) = 0.01
W= {ln ()}^{ 0.5}= 3.035
Substituting the value of W in Z;
Units Used and Graphs
Z = W  = 2.32679567
K_{T} = Z + (Z^{2 }– 1) K + (Z^{3} – 6 Z)*(K^{2}) – (Z^{2 }– 1)* K^{3 }+ Z*K^{4 }+ K^{5}
K = 0.28364
K_{100}= 2.1695
X_{100}= Mean of Q+ K_{100}S_{.}D
X_{100 = }31.4258+ 1.154 (31.76046793)
X_{100} = 68.078 Ml/day
 A graph which compares the actual measured data and your fitted probability distribution function (2 marks);
 A discussion of how the measured data compares to the fitted PDF including comments on its goodness of fit and any other aspects you may think relevant .
The fitted type (iii) log Pearson probability distribution function provides sensible estimates for maximum discharge over the various years.
Y = 0.4343ln(x)  3E15
R² = 1
Create an MS Excel spreadsheet which shows the computational steps to solve the following problem:
 Consider a 2acre storm water detention basin with vertical walls.
 Assume a triangular inflow hydrograph which increases linearly from zero to a peak of 60 cfs at 60 min and then decreases linearly to a zero discharge at 180 min.
 Route the inflow hydrograph through the detention basin using the headdischarge relationship for the 5ft diameter pipe spillway shown in Table 9.2.1 (see below) noting that the pipe is located at the bottom of the basin.
 Assuming the basin is initially empty, use the level pool routing procedure with a 10min time interval to determine the maximum depth in the detention basin.
Note that this problem has been taken from Chapter 9 of the text Mays (2001) Water Resources Engineering. John Wiley & Sons. You may like to refer to this text to assist you with this problem.
The results for this problem are shown in Table 9.2.2 below.
 Your task is to create an MS Excel spreadsheet which recreates these results (4 marks).
 In addition, create a graph which compares in the inflow and outflow hydrographs and provide a discussion on the magnitude of peak attenuation and also peak lag time (3 marks).
Ensure you spreadsheet is properly labeled so that others can read, interpret and understand your results.
Column 4 is calculated using the formula; + Q (cfs) (Zoppou, 1999).
A ten minute routing interval is used. (600 seconds)
Column 1 Head H (Ft) 
Column 2 Discharge (Q) 
Column 3 Storage S (Ft^{3}) 
Column 4 + 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 – DischargeStorage data
For the first time interval, S1 = Q1 = 0 because the reservoir is empty at t=0
Then 2s/ (Δt – Q1) = 0
The value of the storageoutflow function towards the end of the time span is;
_{ = }(I_{1 }+ I_{2}) + _{ }= (0 + 10) = 10
Value of Q_{2 }is determined using linear interpolation;
Q_{2 }= 0+ (10  0) =  0.2 cfs
With Q _{2} = 0.2, then 2S_{2}/ (Δt – Q_{2})
The next iteration is;
_{ = }_{ } 2Q_{2} = 10 – 2(0.2) = 9.6
 Find attached an MS Excel spreadsheet which recreates the results.
 Graph which compares in the inflow and outflow hydrographs
Graph which compares in the inflow and outflow hydrographs
Calculating maximum depth
One performs an interpolation between storage and head values function.
The maximum outflow of 30.285 cfs brings a maximum depth of approximately 2ft
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 – 285 cfs
The Peak lag time lies between 180 240 minutes
Problem 3
Go the Victorian Water Register at https://waterregister.vic.gov.au/waterentitlements/bulkentitlements and list all of the bulk entitlement records for Central Highlands Water.
SOLUTION
Bulk entitlements for Central Highlands Water.
 Amphitheatre
 Avoca
 Ballan
 Beaufort
 Blackwood and Barry’s Reef
 Bullarook System – Central Highlands Water
 Lal Lal – Central Highlands
 Landsborough –Navarre
 Lexton
 Loddon System – Part Maryborough – Central Highlands Water
 Redbank
 Skipton
 Upper West Moorabool System
 Yarrowee – White Swan System
Calculate the total available water (in mega liters) for use within a water supply system given the following information:
 Current volume in storage = 143,500 ML
 Total volume of dead storage = 7,500 ML
 Forecast inflows to end of water year = 52,000 ML
 Total volume of water already released = 15,500 ML
 Total volume of water to be reserved for next year = 24,000 ML
 Total volume of water carried over from previous year = 17,000 ML
 Total volume of water forecast to be lost by end of water year = 27,500 ML
SOLUTION
Available water= (143,500  7,500 + 52,000 +15,500  24,000 17,000  27,500) ML
Available water = 135,000 ML
Problem 5
Using the available water volume calculated in Problem 4, calculate what the allocation of this water will be for each entitlement holder, using the water sharing table shown below. Use linear interpolation techniques to find values between columns if needed. Provide your answer as both a volume (ML) and a percentage (%) of each entitlement. Note that Water Share A represents the full entitlement held by each entitlement holder, and so your allocation should be a fraction of this total amount. Describe who you think has the highest reliability entitlement.
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 
202200 
174000 
86400 
56400 
45900 
0 

System Operator 
150000 
75000 
56800 
40200 
32000 
20000 

Water Corporation A 
82000 
82000 
40700 
28700 
24500 
0 

Water Corporation B 
6500 
2500 
1500 
1500 
1400 
0 

Water Corporation C 
8700 
4500 
1700 
1200 
1000 
0 

Environment A 
40000 
40000 
35000 
25000 
19000 
0 

Environment B 
65000 
45000 
7500 
0 
0 
0 
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 
(202200  174000) = 28200 
(174000 – 86400) =87600 
(8640056400)= 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 Corporation A 
(82000 82000) = 0 
(82000 40700)= 41300 
(40700 28700)= 12000 
(28700 24500) = 4200 
(24500 0)= 24500 
0 

Water Corporation B 
(6500 – 2500)= 4000 
(2500 – 1500) = 1000 
(1500 – 1500) = 0 
(1500 1400) = 100 
(1400 – 0) = 1400 
0 

Water Corporation C 
(8700  4500 ) = 4200 
(4500  1700 )= 2800 
(1700  1200 ) = 500 
(1200 – 1000) = 200 
(1000 – 0) = 1000 
0 

Environment A 
(40000  40000) = 0 
(40000 35000 ) = 5000 
(35000  25000) = 10,000 
(25000  19000) = 6000 
(19000 – 0) = 19000 
0 

Environment B 
(65000  45000) = 20000 
(45000–7500) = 37500 
7500 
0 
0 
0 
Table 5.1; Finding the share commanded by each entitlement holder in ML
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 
56800 
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 
135,000 
116,00 
57645 
37,800 
30,645 
0 % 
System Operator 
150000 
75000 
56800 
40200 
32000 
20000 
Water Corporation A 
0 
27540 
7965 
2713.5 
16335 
0 
Water Corporation B 
2700 
675 
0 
135 
945 
0 
Water Corporation C 
2835 
1890 
270 
135 
675 
0 
Environment A 
0 
3375 
6615 
4050 
12690 
0 
Environment B 
13,365 
24,975 
4995 
0 
0 
0 
 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
References
Cunnane, C. (1989). “Statistical Distributions for Flood Frequency Analysis”. Operational Hydrology Report no. 33, World Meteorological Organization.
Das, S., (2010). “Estimation of Flood Estimation Techniques in the Irish Context”. PhD thesis, Department of Engineering Hydrology, National University of Ireland Galway
Milanovic, P. (2004). Water resources engineering in karst. CRC press.
Mays, L. W. (2010). Water resources engineering. John Wiley & Sons.
https://waterregister.vic.gov.au/weatherentitlements/bulkentitlements retrieved on 13th, May, 2018. At 1943 hours
Singh, V.P (1998). LogPearson type III distribution. In Hydrology (pp. 252274). Springer, Dordrecht.
Mitchell, W.D., 1954, Floods in Illinois: Magnitude and Frequency: Prepared in cooperation with the U.S. Geological Survey and Division of Waterways, Department of Public Works and Buildings, 386 p.
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