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## 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.

• 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).

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.

Cite This Work

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