I do have five Question Engineering _Water system Assignment Question to completed due on 26/7/2015.
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SEV215: Assignment one Access Date: August 3, 2015 Due Date: August 28, 2015
1. Determine the gross water supply reservoir volume for the following information:
City population: 200,000
Water consumption rate: 568 Lpcd
Catchment area: 103 Km2
Rainfall rate: 110cm/yr
Annual soil evaporation: 50cm
Minimum net reservoir volume: 50% of annual yield or half of a year water supply whichever is greater.
If during the droughts 25% of water required is supplied from a deep well field, how many deep wells with 30 liter/second discharge are required if each well supplies water for 12 hours per day. (4Marks)
2. Estimate the capacity of the different components of the following water supply scheme, (Fig. Q2) Conduit I, LLP, TP, HLP, Conduit II, Service Reservoir and Conduit III, if the future population is estimated as 30,000 persons with 568 Lpcd and fire fighting capacity of 3875 L/min for 2 hours. If 30% of supplied water is retained within the consumption areas ( as garden water,…) determine daily average, daily maximum and daily minimum of wastewater using both tabulated coefficients and Gifft’s formula for peaking factor. ( 5Marks)
N.B. Conduit II is taking water directly to the Service Reservoir.
Fig: Q2
3. The upper part of a catchment in Mt Helen, Ballarat, Victoria shown as triangular ABC in Fig. Q3,has undergone extensive urbanization. Pre-development runoff coefficient for the whole catchment, ABDEC, is estimated 0.15 while post-development runoff coefficient for the upper part, ABC, has changed to 0.6. Runoff coefficient for the lower part of the catchment, BDEC, remains the same as that of pre-development. (Areas of ABC and BDEC are 125,000 and 250,000 m2 respectively).
i) Estimate time of concentration in minutes if velocity of flood flow in ABDEC catchment longest floodway thalweg reaches maximum of 1m/s.
ii) Calculate flood peaks for 5 year return period at the outlet, G, for both pre and post development conditions.
iii) Calculate flood peak for a culvert to be installed as the minor system of the artillery road shown on the map.
iv) What would be approximate diameter of the culvert if its full flow velocity is assumed 3m/s.
( 4 Marks)
Fig: Q3
4. In the water supply system shown in Fig. Q4, the pumping station (P) delivers the water from the water treatment plant (WTP) to City ABCDE. The water head delivered by the pumps is 28 m. The water level in the pumping station wet well is at 600.00 m. The value of C for all pipes is l00.
The following elevations are known:
A 570.00 m
B 575.00 m
C 573.00 m
Fig: Q4
(a) Find the required diameter of the force main PA, (dPA), if the pressure at point A is not allowed to drop below 370 kPa
(b) Determine the required diameter of pipe BC, (dBC), if the maximum allowed head loss in the network ABC is 0.003 m/m
(c) Calculate the actual residual pressures at points A and C if dPA= 500 mm
(5 Marks)
5. The water supply system shown in Fig. Q5 is designed to serve a small town ABDCA. The flow from the reservoir is delivered to the town by gravity through the 400 mm main. All pipes have a Hazen-Williams C of 100.
(i) Assuming an initial flow of 50L/s (clockwise direction) in BC, determine the flowrate in each pipe. Iterate your solution up to a correction of 0.5 L/s.
(ii) What is the flow in the mains from the reservoir to ABCD?
( 7 Marks)
Fig: Q5
Water supply pipes and reservoirs
Table of Contents
Answer 1. 2
Answer 2. 2
Answer3. 4
Answer 4. 6
Answer 5. 9
Answer 1.
Effective area: 0.25 x 103 km2 =
Reservoir storage volume on the Annual yield = (110-50) x 10-2 x 0.25 x 103 x 106 x 0.5
= 772500 m3
Volume of reservoir storage = 200,000 x 568 x 0.5 x 365 10-3
= 207320 m3
The, greater value is 772500 m3.
Answer 2.
Capacity of the water supply = per capita demand x population
= 568 x 30000
= 17040000 Lt
Average per capita demand (daily) = Quantity required in the 12 Months/ (365 x Population)
= 17040000/ (365 x 30000)
= 0.4664
Daily demand = 1.5 x average hourly demand
= 1.5 x 1937.5
= 2906.25
Answer3.
i)
The Kirpich Method
The Kirpich equation is:
Equation 4-15.
Where:
- tch = the time of concentration,
- K = units conversion coefficient, K = 0.0078 for the traditional unit and K = 0.0195 for the SI units
- L = channel of the flow length, as dictated by K
- S = dimensionless channel slope
If the low slope condition exists, then the user should consideredas an adjusted slope in calculating the concentration time.
Concentration
= 0.0195* 10.770 * 0.15-0.385
= -0.35349775
ii)
Flood peak flow
| year | Flood peak (cfs) |
| 1 | 22500 |
| 2 | 45000 |
| 3 | 67500 |
| 4 | 90000 |
| 5 | 112500 |
iii)
Flood peak flow= 0.15 × 0.6 (125000 + 250000)
= 135000 cfs
iv)
Diameter2 = 4 * flow rate / π * velocity
= 4 * 135000 / π * 3
= 540000
22/7 *3
= 540000 / 9.42
Diameter2 = 57324.84
Diameter = 239.42
Answer 4
a)
Pressure drop = P1
Pipe friction coefficient = λ
Length of pipe = L
Pipe diameter = D
Elevation = e
Flow velocity = w
Pressure drop = λ * L * e * w2
D 2
370 = λ * 3000 * 570 * (55)2
D 2
740 D = 5172750000 λ
Diameter = 6990202.70 λ
b)
h = head loss
L = length of the pipe
D = diameter
U = flow velocity (average)
G= acceleration due to gravity
F = dimensionless friction factor
H = f * L * u2
D 2g
0.18 =28* 800 * (40)2
D 2g
0.18 D = 35840000 / g
Diameter = 199111111.11 / g
C)
Q = related capacity
F = Total flow
H = static pressure
P = residual pressure
Q = F * (H/P) 0.54
600 = 500 (600/P) 0.54
1.2 = (600/P) 0.54
P=(600/2)0.54psi
Answer 5
i)
Fluidvelocity = 50
Diameter of pipe = 20 m
Cross section area = π/4 * (20)2
= 314.28 m 2
Flow rate (BC) = 50 * 1257.14 m2
= 15714.28 m3/s
Fluid velocity = 50
Diameter of pipe = 10 m
Area of cross section = π/4 * (10)2
= 78.57 m2
Flow rate (AB) = 50 * 78.57
= 3928.57 m3/s
Fluid velocity = 50
Diameter of pipe = 80 m
Cross section area= π/4 * (30)2
= 707.14 m2
Flow rate (CB) = 50* 707.14
= 35357 m3/s
ii)
Fluid velocity = 100
Diameter of pipe = 60 m
Cross section area= π/4 * (60)2
= 2828.57 m 2
Flow rate (ABCD) = 100 * 2828.57 m2
= 282857 m3/s
