Calculation of Discharges using Log-Pearson 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 Log-Pearson 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.
XT = Mean of Q+ KTS.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
KT = Z + (Z2 – 1) K + (Z3 – 6 Z)*(K2) – (Z2 – 1)* K3 + Z*K4 + K5
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
KT = Z + (Z2 – 1) K + (Z3 – 6 Z)*(K2) – (Z2 – 1)* K3 + Z*K4 + K5
K = -0.28364
K2 = 0.2602
X2 = Mean of Q+ K2S.D
X2 = 31.4258+ 0.2602 (31.76046793)
X2 = 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
KT = Z + (Z2 – 1) K + (Z3 – 6 Z)*(K2) – (Z2 – 1)* K3 + Z*K4 + K5
K = -0.28364
K5 = 0.803
X5 = Mean of Q+ K5S.D
X5 = 31.4258+ 0.803 (31.76046793)
X5 = 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
KT = Z + (Z2 – 1) K + (Z3 – 6 Z)*(K2) – (Z2 – 1)* K3 + Z*K4 + K5
K = -0.28364
K10 =0.97199
X10 = Mean of Q+ K10S.D
X10 = 31.4258+ 0.803 (31.76046793)
X10 = 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
KT = Z + (Z2 – 1) K + (Z3 – 6 Z)*(K2) – (Z2 – 1)* K3 + Z*K4 + K5
K = -0.28364
K25= 1.08511
X25= Mean of Q+ K25 S.D
X25 = 31.4258+ 1.154 (31.76046793)
X25 = 65.89 Ml/day
P = ( ) = 0.02
W= {ln ()} 0.5= 2.79715
Substituting the value of W in Z;
Z = W - = 2.0541982
KT = Z + (Z2 – 1) K + (Z3 – 6 Z)*(K2) – (Z2 – 1)* K3 + Z*K4 + K5
K = -0.28364
K50 =1.129
X50 = Mean of Q+ K50S.D
X50 = 31.4258+ 1.129 (31.76046793)
X50 = 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
KT = Z + (Z2 – 1) K + (Z3 – 6 Z)*(K2) – (Z2 – 1)* K3 + Z*K4 + K5
K = -0.28364
K100= 2.1695
X100= Mean of Q+ K100S.D
X100 = 31.4258+ 1.154 (31.76046793)
X100 = 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) - 3E-15
R² = 1
Create an MS Excel spreadsheet which shows the computational steps to solve the following problem:
- Consider a 2-acre 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 head-discharge relationship for the 5-ft 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 10-min 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 re-creates 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 (Ft3) |
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 – Discharge-Storage 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 storage-outflow function towards the end of the time span is;
= (I1 + I2) + = (0 + 10) = 10
Value of Q2 is determined using linear interpolation;
Q2 = 0+ (10 - 0) = - 0.2 cfs
With Q 2 = 0.2, then 2S2/ (Δt – Q2)
The next iteration is;
= - 2Q2 = 10 – 2(0.2) = 9.6
- Find attached an MS Excel spreadsheet which re-creates 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/water-entitlements/bulk-entitlements 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 |
(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 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/weather-entitlements/bulk-entitlements retrieved on 13th, May, 2018. At 1943 hours
Singh, V.P (1998). Log-Pearson type III distribution. In Hydrology (pp. 252-274). 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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