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Main points of this past exam are: Environmentally Aware River Engineering, Manning Coefficient, Flow Classification, Chezy Coefficient, Hydraulic Jump, Velocity of Flow, River Water Level Measurement, Data Discharges Water
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Answer five questions in total, a minimum of Examiners: Dr. J. Harrington two questions from each Section Dr. J.D. Murphy Use separate answer books for each section Prof. P. O’ Donoghue Programmable calculators are not permitted Mr. T. Corcoran All questions carry equal marks Attachments including figures and equations
Note: Attachment with equations
What would be the depth of water in the channel to pass twice this volume per second (Q 2 ) if both the slope and the value of n were unaltered? (4 marks)
(b) Water enters a rectangular channel of width 0.3m with a bed slope of 1 in 100 (Chezy Coefficient C = 56) with a depth of 0.15m. Determine the velocity of flow and the flow classification. Under what conditions will a hydraulic jump be formed downstream of this entry section to the channel? What will happen the specific energy if a hydraulic jump is formed? ( marks)
(b) An alluvial river with the following data discharges water into a downstream reservoir of 5 x 10 5 m^3 capacity. Width = 12m, Depth = 4m, Slope = 3 x 10 -4^ , Discharge = 20 m^3 /sec, d 50 = 0.1mm, Sediment Density = 2600 kg/m^3 , Sediment Porosity = 0.3, Suspended sediment concentration at 2m depth = 25 mg/l
Determine the total suspended sediment load (assuming a homogenous vertical concentration distribution), and the life expectancy of the reservoir. Comment on the accuracy of the assumption of a homogenous vertical concentration distribution. Comment on any other possible sediment transport mechanisms. (8 marks)
(c) A wide channel has a flow depth of 1.7m with a mean velocity of 2.5m/sec. Find the minimum size of bed material needed to obtain a stable bed assuming the sediment to have a density of 2650 kg/m^3. Take the channel Manning roughness coefficient to be 0.02. (5 marks)
(b) A river with a low flow of 7.5 m^3 /sec with a corresponding 1 mg/l BOD 5 level receives a discharge from a wastewater treatment plant serving an urban area with a population equivalent of 70,000, which complies with the Urban Wastewater Directive 25/35 standard. Determine the maximum downstream BOD 5 level. Comment on the assumptions made in your analysis. Do you consider the downstream water quality to be adequate? Please support your answer. (7 marks)
(c) Discuss water demand under the following headings: (i) Agricultural Use (2 marks) (ii) Distribution Losses (3 marks) (iii) Estimating Future Demand (3 marks)
(b) A water treatment plant built in Ireland in the 1970’s supplies a community with a population equivalent of 25,000 with water. It receives its supply from a nearby lake which now suffers eutrophication problems.
Estimate the demand form the community. Outline the likely treatment process within the plant, size the main elements for the treatment process and quantify the daily amount of any chemicals used. (8 marks)
The local authority is now investigating measures to improve the water quality of the supply. Suggest at least 2 possible changes to the treatment process. Support your answer with preliminary design calculations as appropriate. (5 marks)
Supporting data is presented overleaf.
Chezy Eqn.: v = C mS 0
Manning Eqn.: 01 /^2
2 / 3 S n
m v =
ku *
v
r
r r
s
Y y
Y y y
y C
C
−
= −
Shear Velocity: u * (^) = gmS 0
Stoke’s Law:
μ
ρ ρ 18
(^2) − s =^ p
gd v
Bed Shear Stress: τ 0 = ρgmS (^0)
Threshold Entrainment Function: τcr / [(ρs-ρ)gD] = 0.
Question 5 Hydrology (20 marks)
5.1 The internal dimensions of a culvert are 5 m wide by 2.5m deep. The culvert is laid at a slope of 1 in 300. What is the capacity of the culvert if it is open? What is the capacity of the culvert if it is closed? (8 marks)
5.2. The 20 year return period multiplier for Ireland is 1.54. Estimate the catchment area of river that the culverts outlined in 5.1 can cater for if the river has the following characteristics: 7 streams Index of soil type 0. SAAR 1200mm pa RSMD 2.48(SAAR) 0.5^ – 40 mm Lake area 15 Ha (12 marks)
The Five Variable Equation Q = C x Area0.87^ x STMFRQ 0.31^ x SOIL1.23^ x RSMD1.17^ (1 + LAKE) 0. Where, Q = flow m^3 /s Area = Catchment area km^2 STMFRQ = Stream Frequency, number of junctions /km^2 SOIL = index of soil type RSMD = 2.48(SAAR) 0.5^ – 40 mm LAKE = Fraction of catchment area covered in lake C = Regional Multiplier, 0.0183 for Ireland
Mannings Equation Q = (1/n) x A x R2/3^ x s 1/ Where, Q = flow m^3 /s n = 0.01 for smooth concrete A = cross section area of water P = wetted perimeter R = hydraulic radius = A/P s = gradient of energy or water surface line expressed as a decimal
Question 7 Urban Wastewater (20 marks)
With reference to table 7.1, figure 7.1 and formula 7.1, list the differences between conventional aeration and extended aeration. Outline under what circumstances each would be used.
Outline your understanding of biological nitrogen removal. Sketch and describe the Bardenpho process.
Outline your understanding of biological phosphorous removal. Sketch and describe the Phoredox process.
Process F/M KgBOD/d KgMLSS
BOD loading KgBOD/m 3 /d
Hydraulic retention time hours
MLSS
mg/L
BOD removal efficiency %
Sludge production KgDS/kgBOD removed Extended Aeration
0.05 – 0.15 0.25 – 0.3 20 – 30 2000 – 6000
95 0.
Conventional Aeration
0.2 – 0.5 0.5 – 1.5 5 – 14 2000 - 3000
90 0.
Table 7.1. Loading and operating parameters for selected wastewater treatment processes.
Figure 7.1. Relationship between temperature and solid retention time for production of nitrifying bacteria,
OD = 0.75 Q (BODi – BODe) + 2Va.MLSS Formula 7. Where OD is oxygen demand g/h 0.75 applies only for settled wastewater Q is flow through plant m^3 /h BODi and BODe are the influent and effluent BOD levels in mg/L Va is the volume of the basin in m^3 MLSS is the concentration of mixed liquor suspended solids in g/L