wastewater -flow : when-not-to-discharge...

Wastewater-Flow :When-not-to-discharge CASE

A wastewater treatment plant (WWTP) is allowed to discharge its effluentInto a river, provided the flow in the river is more than 1 m3/s.After a long dry period the flow in the river on April 4 has decreased To only 5 m3/s. The depletion curve during the preceding week is given. The curve may be described with the equation Qt=Qo e–at

Compute the date on which the WWTP has to terminate the dischargeof the effluent into the river, assuming that the drought continues.of the effluent into the river, assuming that the drought continues.

Q (m3/s)10

5

28 29 30 31 1 2 3 4 5 6 7 8 9 April

7Q10Low-flow Analysis

1971 1,72 1976 4,23 1981 4,48 1986 5,39

1972 3,03 1977 4,11 1982 3,03 1987 3

The following 7-d low flows were compiled in the following example:

1972 3,03 1977 4,11 1982 3,03 1987 3

1973 2,76 1978 1,92 1983 2,84 1988 2,5

1974 1,65 1979 2,14 1984 3,66 1989 2,47

1975 2 1980 1,48 1985 1,87 1990 3,07

Use this data to determine the 7Q10?

FREQUENCY ANALYSIS

?The objective ?

FREQUENCY ANALYSIS

�The objective?To relate the magnitude of extreme events

to their frequency of occurrence through the use of probability distributions.the use of probability distributions.

The hydrologic data analyzed are assumed to be independent and identically distributed.

FREQUENCY ANALYSIS

�The results of flood flow frequency analysis can be used:� Design of dams, bridges, culverts, and flood

control structures� Determine the economic value of flood � Determine the economic value of flood

control projects

FREQUENCY ANALYSIS

�Hydrologic system are sometimes impacted by extreme events: severe storms, floods, and droughts.The magnitude of an extreme event is �The magnitude of an extreme event is inversely related to its frequency of occurrence, very severe events occurring less frequently than more moderate events.

Periode Ulang (Return Period)

Suppose that an extreme event is defined to have oc curred if a random Variable X is greater than or equal to some level x T.

The recurrence interval σ is the time between occurrences of X > xT

Example: Example: Table shows the record of annual maximum discharges of the Guadalupe River near Victoria, Texas, from 1935 to 1978, plotted from the data given. If xT = 50,000 cfs, it can be seen that the maximum disch arge exceeded this level nine times during the period of record, with recurrence Interval ranging from 1 year to 16 years.

The return period T of the event X > xT is the expected value of σ, E(σ), Its average value measured over a very large number of occurrences.


Year 1930 1940 1950 1960 1970

0 55,900 13,300 23,700 9,190

Annual maximum discharges of the Guadalupe river ne ar Victoria, Texas 1935-1978, in cfs

0123456789

38,500179,00017,20025,4004,940

55,90058,00056,0007,71012,30022,00017,90046,0006,97020,600

13,30012,30028,40011,6008,5604,9501,73025,30058,30010,100

23,70055,80010,8004,1005,72015,0009,79070,00044,30015,200

9,1909,74058,50033,10025,20030,20014,10054,50012,700


For the Guadalupe River data, there are 8 recurrenc e intervals coveringa total period of 41 years between the first and la st exceedences of 50,000 cfs, so the return period of a 50,000 cfs an nual maximum Discharge on the Guadalupe River is approximately σ’ = 41/8 = 5.1 years.

Thus, the return period of an event of a given magn itude may be defined Thus, the return period of an event of a given magn itude may be defined the average recurrence interval between events equalling or exceedingA specified magnitude.

Excee-dence year

1936 1940 1941 1942 1958 1961 1967 1972 1977 Average

Rucurre-nce interval (years)

4 1 1 16 3 6 5 5 5.1


T = 1/p; that is, the probability of occurrence of an event in any observation is the inverse of its return period:

P(X > xT) = 1/T

for example, the probability that the maximum discharge in the Guadalupe River will equal or exceed 50,000 cfs in any year is approximately p = 1/σ’ = 1/5.1 = 0.195


What is the probability that a T-year return period event will occur at least once in N years?

To calculate this, first consider the situation where no T-year event occurs in N years. This would require a sequence of N successive “failures”.

P(X< xT each year for N years) = (1-p)N

The complement of this situation:

P(X> xT at least once in N years) = 1- (1-p)N

= 1- (1-1/T)N


Estimate the probability that the annual maximum dischargeQ on the Guadalupe River will exceed 50,000 cfs at least once during the next three years.

From the discussion,

P(Q > 50,000 cfs in any year) ~ 0.195So, P(Q> 50,000 cfs at least once during the next three years)

= 1- (1-p)N = 1- (1-0.195)3

= 0.48

Selection of the Design Level

Empirical ApproachRisk Analysis

Hydroeconomic Analysis


Periode waktu rata-rata yang diharapkan terjadiantara dua kejadian yang berurutan. Bukanantara dua kejadian yang berurutan. Bukanberarti bahwa dua peristiwa banjir akan terjadisecara berurutan dengan waktu yang tetap.

Tr = 1/p

Return Period (Tr) vs Probabilitas (P)

Beberapa Probabilitas yang umum dijumpai:1. Probabilitas terjadinya banjir (F) adalah:

P(F)=1/TrP(F)=1/Tr2. Probabilitas untuk tidak terjadi banjir (F’) adal ah:

P(F’) = 1-1/Tr3. Probabilitas untuk tidak terjadi banjir selama n tahun

berturut-turut:P(F’)n= {1-(1/Tr)} n

4. Risiko (R) bahwa akan terjadi banjir (F) paling sedikit sekali dalam n tahun adalah:R=1-(1-1/Tr)n

STATISTIKA DALAM HIDROLOGI

Contoh Kasus 2:Periode ulang berapakah yang harus digunakan oleh seorang ahli drainase jalan raya apabila dia bersedia untuk menerima ahli drainase jalan raya apabila dia bersedia untuk menerima risiko terjadinya banjir sebesar 10% dalam kurun waktu 5 tahun?


Contoh Kasus 2:Periode ulang berapakah yang harus digunakan oleh seorang ahli drainase jalan raya apabila dia bersedia untuk menerima ahli drainase jalan raya apabila dia bersedia untuk menerima risiko terjadinya banjir sebesar 10% dalam kurun waktu 5 tahun?R = 1 – (1-1/Tr)n

0.10 = 1 – (1-1/Tr)5

Tr = 48,1 tahun


Contoh Kasus 2:A culvert has an expected life of 10 years. If the acceptable risk of at least one event exceeding the culvert capacity during the design life is 10 percent, what design return period should be used? R = 1 – (1-1/Tr) nR = 1 – (1-1/Tr) n


Contoh Kasus 2:A culvert has an expected life of 10 years. If the acceptable risk of at least one event exceeding the culvert capacity during the design life is 10 percent, what design return period should be used?

R = 1 – (1-1/Tr)n

0.10 = 1 – (1-1/Tr)10

Tr = 95 tahun


Contoh Kasus 2:A culvert has an expected life of 10 years. If the acceptable risk of at least one event exceeding the culvert capacity during the design life is 10 percent, what design return period should be used? What is the chance that a culvert designed for an event of this What is the chance that a culvert designed for an event of this return period will not have its capacity exceeded for 50 years?R = 1 – (1-1/Tr)n

0.10 = 1 – (1-1/Tr)10

Tr = 95 tahunR = 1(1-1/Tr)n


Contoh Kasus 2:A culvert has an expected life of 10 years. If the acceptable risk of at least one event exceeding the culvert capacity during the design life is 10 percent, what design return period should be used? What is the chance that a culvert designed for an event of this What is the chance that a culvert designed for an event of this return period will not have its capacity exceeded for 50 years?R = 1 – (1-1/Tr)n

0.10 = 1 – (1-1/Tr)10

Tr = 95 tahunR = 1(1-1/Tr)n

R = 1(1-1/95)50

= 0.41 �� so, the probability that the capacity will not be exceeded during this 50-year period is 1-0.41=0.59 or 59%

Type of structure Return Period (years)

Highway culverts

Low traffic 5-10

High 50-100

Urban drainage

General design criteria for water-control structures (Chow et al., 1996)

Storm sewers in small cities 2-25

Storm sewers in large cities 25-50

Levees

on farms 2-50

around cities 50-200

Dams with no likelihood of loss of life (low hazard)

small dams 50-100

intermediate dams 100+


The probability that the most extreme event of the past N years will be



The probability that the most extreme event of the past N years will be equaled or exceeded once during the next n years can be estimated as:

P(N,n) = (n)/(N+n)

If a drought lasting m years is the critical event of record over an N-yearPeriod, what is the probability P(N,m,n) that a worse drought will occur Within the next n years?

P(N,m,n) = (n-m+1)/(N+n-2m+2)




If the critical drought of record, as determined from 40 years of hydrologic data, lasted 5 years, what is the chance that a more severe drought will occur during the next 20 years?

P(N,m,n) = (n-m+1)/(N+n-2m+2)P(40,5,20) = (20-5+1)/(40+20-2(5)+2)

= 0.308


The annual expected damage cost for a structure designed for return period


Hydroeconomic AnalysisThe annual expected damage cost for a structure designed for return periodT is given by:

DT = Σ[(D(xi-1)+D(xi))/2] [P(x>xi-1) – P(x>xi)]

By adding DT to the annualized capital cost of the structure, the total cost can be found; the optimum design return period is the one having the minimumTotal cost.


For events of various return periods at a given location, the damage costs and the annualized capital costs of structures designed to control the events, are shown in columns 4 and 7, respectively. Determine the expected annual damages if no structure is provided, andCalculate the optimal design return period

Increment i Return period, T (years)

Annual exceedence probability

Damage ($) Incremental expected damage ($/year)

Damage risk cost ($/year)

Capital cost ($/year)

Total cost ($/year)

(1) (2) (3) (4) (5) (6) (7) (8)

1 ? 0 01 ? 0 0

1 2 20,000 3,000

2 5 60,000 14,000

3 10 140,000 23,000

4 15 177,000 25,000

5 20 213,000 27.000

6 25 250,000 29,000

7 50 300,000 40,000

8 100 400,000 60,000

9 200 500,000 80,000








Total cost ($/year)

(1) (2) (3) (4) (5) (6) (7) (8)

1 1.000 0 01 1.000 0 0

1 2 0.500 20,000 3,000

2 5 0.200 60,000 14,000

3 10 0.100 140,000 23,000

4 15 0.067 177,000 25,000

5 20 0.050 213,000 27.000

6 25 0.040 250,000 29,000

7 50 0.020 300,000 40,000

8 100 0.010 400,000 60,000

9 200 0.005 500,000 80,000








Total cost ($/year)

(1) (2) (3) (4) (5) (6) (7) (8)

1 1.000 0 01 1.000 0 0

1 2 0.500 20,000 ? 3,000

2 5 0.200 60,000 14,000

3 10 0.100 140,000 23,000

4 15 0.067 177,000 25,000

5 20 0.050 213,000 27.000

6 25 0.040 250,000 29,000

7 50 0.020 300,000 40,000

8 100 0.010 400,000 60,000

9 200 0.005 500,000 80,000

Annual expected damage

∆Di = [(D(xi-1)+D(xi))/2] [P(x>xi-1) – P(x>xi)]

∆D1 = [(0+20,000)/2] [1.0 – 0.5]= $5,000/year








Total cost ($/year)

(1) (2) (3) (4) (5) (6) (7) (8)

1 1.000 0 01 1.000 0 0

1 2 0.500 20,000 5,000 3,000

2 5 0.200 60,000 12,000 14,000

3 10 0.100 140,000 10,000 23,000

4 15 0.067 177,000 5,283 25,000

5 20 0.050 213,000 3,250 27.000

6 25 0.040 250,000 2,315 29,000

7 50 0.020 300,000 5,500 40,000

8 100 0.010 400,000 3,500 60,000

9 200 0.005 500,000 2,250 80,000

Annual expected damage $49,098








Total cost ($/year)

(1) (2) (3) (4) (5) (6) (7) (8)

1 1.000 0 49,098 01 1.000 0 49,098 0

1 2 0.500 20,000 5,000 44,098 3,000

2 5 0.200 60,000 12,000 32,098 14,000

3 10 0.100 140,000 10,000 22,098 23,000

4 15 0.067 177,000 5,283 16,815 25,000

5 20 0.050 213,000 3,250 13,565 27.000

6 25 0.040 250,000 2,315 11,250 29,000

7 50 0.020 300,000 5,500 5,750 40,000

8 100 0.010 400,000 3,500 2,250 60,000

9 200 0.005 500,000 2,250 0 80,000









Total cost ($/year)

(1) (2) (3) (4) (5) (6) (7) (8)

1 1.000 0 49,098 0 49,0981 1.000 0 49,098 0 49,098

1 2 0.500 20,000 5,000 44,098 3,000 47,098

2 5 0.200 60,000 12,000 32,098 14,000 46,098

3 10 0.100 140,000 10,000 22,098 23,000 45,098

4 15 0.067 177,000 5,283 16,815 25,000 41,815

5 20 0.050 213,000 3,250 13,565 27.000 40,565

6 25 0.040 250,000 2,315 11,250 29,000 40,250

7 50 0.020 300,000 5,500 5,750 40,000 45,750

8 100 0.010 400,000 3,500 2,250 60,000 62,250

9 200 0.005 500,000 2,250 0 80,000 80,000


Hydroeconomic

Analysis

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