WATER & WASTEWATER ENGINEERING 2 Water Management & Background: (Hydraulics & Hydrology, Groundwater & Distribution System) John Manuel B.Vergel BS-CE, BS CE, MS MS-CE CE
Hydraulics & Hydrology y Water Pressure:
P = γH γ where: P = Pressure (KPa) γ = unit i weight i h off water (9 (9.81 81 KN/ KN/m3) H = head (m)
Pressure diagram
Hydraulics & Hydrology y Pressure-Velocity Head-Relationship:
Continuityy Equation: q Q = VA where: h Q = Discharge Di h ((m3/s) /) V = Velocity (m/s) A = Cross sectional area of flow(m2)
Q=A1V1=A2V2
Hydraulics & Hydrology y Pressure-Velocity Head-Relationship:
Energy gy Head:
where: Z = elevation l (m) P/γ = Pressure Head (m) V2/2gg = Velocity Head (m)
Hydraulics & Hydrology y Pressure-Velocity l Head-Relationship: d l h Head Loss: (Darcy Weisbach Equation)
where:
H=hL = head loss (m) f = ffriction i ti ffactor t L = length of pipe (m) V = velocity of flow (m/s) D = diameter of pipe (m)
H
Relative Roughness off pipe materials & friction factors for turbulent pipe flow (D of pipes in feet)
Hydraulics & Hydrology y Pressure-Velocity Head-Relationship:
Minor Head Loss: hL = K(V2/2g) /2 )
where:
hL = head loss (m) V = velocity of flow, (m/s) k = loss coefficient
Units in ft
Hydraulics & Hydrology y Example 1: Calculate the head loss in a 0.60 m (24-in)
diameter, 1,525 m long, smooth walled concrete (ε=0.001) pipeline carrying a water flow of 0.28 m3/s. V = Q/A = 0.28/[π(0.60)2/4] = 0.99 m /s ¾ From the h diagram: di (d 24 ε=0.001) (d=24in, 0 001) f = 0.017 0 017 ¾
HL = 0.017 (1,525/0.60)(0.992/19.62) = 22.16 16 m
Hydraulics & Hydrology y Example 2: A pump discharge line consist of 60 m of 0.30 m (12
in) new cast-iron pipe, three 90° medium radius bends, two gate valves and one swing check valve. valve Compute the headloss through the line at a velocity of 1m/s.
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The total equivalent pipe length is: (60 3 28)+(3 27)+(2 17)+(1 135)= 447 ft = 136.22m (60x3.28)+(3x27)+(2x17)+(1x135)= 136 22 From the diagram: (d=12in, cast iron) f = 0.019 HL = 0.019 (136.22/0.30)(12/19.62) = 0.44 m
Hydraulics & Hydrology y Example 3: Calculuate the headloss in the pipeline based on
the figure below. Based on the ff: Z1=4.5m Z2=9.3m P1=280 Kpa P2=200KPa V1=1.2m/s V2=1.2m/s ¾Z1+(P1/γ)+(V12/19.62) = ( 2/9.80)+(V ) ( 22/19.62)) +hL Z2+(P 4.5+(280/9.80)+(1.202/19.62) = . ( / . ) ( . 2//19.62) . ) + hL 9.3+(200/9.80)+(1.20 4.5+28.6+0.07=9.3+20.4+0.07+hL hL = 3.4m
Hydraulics & Hydrology y Flow in Pipes under Pressure:
for Pipe Flow: ((Hazen Williams)) Q=0.281CD Q 0. 8 C 2.63S0.54
Q=0.278CD Q 0. 78C 2.63S0.54
(SI Units)
where: Q = quantity of flow flow, (gpm, (gpm cms) C = coefficient (see table) D = Diameter of pipe (in, m) S = hydraulic gradient (ft/ft,m/m) This equation relates the quantity of g a circular turbulent water fflow through pipe flowing full
(Metric Units)
Hydraulics & Hydrology y Flow in Pipes under Pressure:
for headloss: (Hazen Williams)
100 1.85 Q1.85 100 1.85 Q1.85 hL=0.002083 L ( ) hL=0.002131 L ( ) 4.8655 C D C D4.8655 (SI Units)
where: hL = headloss ((ft,m)) L = length of pipe (ft, m) C = coefficient,, ((see table)) Q = quantity of flow (gpm, cms) D = Diameter of pipe (in (in,m) m)
Note: substitute S = hL/L
(Metric Units)
Hydraulics & Hydrology y Flow in Pipes under Pressure:
Hazen Williams
Correction Factors to Determine Head Losses atValues off C other than C=100 Example: if C=100; hl=8.5ft/1000ft. Therefore if C=130; 0.62x8.5 = 5.3ft/1000ft Nomograph in English Units for C=100 ((15-20 yyr-old ductile iron)) Example: for Q=500gpm, 8in diam [headloss=8.5 ft/1000ft, V=3.2ft/sec]
Hydraulics & Hydrology y Flow in Pipes under Pressure:
Hazen Williams
Nomograph in Metric Units for C=100 ((15-20 yyr-old ductile iron)) Example: for Q=30l/s, 200mm diam [headloss=0.0080 m/m, V=0.95m/sec]
Hydraulics & Hydrology y Example l 44: G Groundwater d from f a wellll ffield ld is pumpedd through h ha
14-in diameter transmission main for 10,500 ft to the treatment plant. Calculate the headloss for a flow rate of 1,400 gpm assuming C=100 and assuming C=140
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100 1.85 Q1.85 hL = 0.002083 L ( ) C D4.8655 1.85 100 1400 1.85 =0.002083 x10,500 ( ) = 38.3 ft 100 144.8655 Using nomograph for C=100; Q=1400gpm;d=14-in: hL = 3.6ft/1000 ft. Therefore: hL = 3.6/1000 x 10,500 = 38 ft
Hydraulics & Hydrology y Example 4: Groundwater from a well field is pumped through a
14-in diameter transmission main for 10,500 ft to the treatment plant Calculate the headloss for a flow rate of 11,400 plant. 400 gpm assuming C=100 and assuming C=140.
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100 1.85 Q1.85 hL = 0.002083 L ( ) D4.8655 C 1.85 100 1400 1.85 =0.002083 x10,500 ( ) = 20.6 ft 4.8655 140 14
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Using correction factor for C=140; K=0.54 (see table) Therefore: hL = 3.6/1000 x 10,500 x 0.54 = 20 ft
Hydraulics & Hydrology y Example 5: If a 200mm water main (C=100) is carrying a
flow of 30L/s. What is the velocity of flow and head loss.
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Usingg nomograph: g p Q=30L/s; d=200mm; HL/L = 0.008m/m ; V=0.95 m/s
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Q=VA; V = 0.03/[π(0.20)2/4] = 0.955 m/s 2 63S0.54 0 54; S0.54 0 54=0.30/(0.278x100x0.20 2 63) Q=0 278CD2.63 Q=0.278CD =0 30/(0 278 100 0 202.63 S = 0.0744(1/0.54) S=HL/L = 0.00817m/m
Hydraulics & Hydrology y Example 5: If a 200mm water main (C=100) is carrying a
flow of 30L/s. What is the velocity of flow and head loss. 1.85 100 Q 1.85 ¾ hL = 0.002131 0 002131 L ( ) C D4.8655 1.85 100 0 0.030 030 1.85 hL/L =0.002131 ( ) 100 0.204.8655
= 0.00817m/m
Hydraulics & Hydrology y Example 6: An extremely simplified water supply system
consisting of a reservoir with lift pumps, elevated storage, piping and loaded center (withdrawal point) is shown in the figure. a) Based on the ff: data, sketch the hydraulic gradient for the system: ZB=30ft; ZC=40ft; ZA=0ft; PA=80psi; PB=30psi; PC=100ft; (water level tank)) b) For these conditions compute the flow available at point B from both supply pumps and elevated storage storage. Use C=100 C 100 and pipe sizes as shown in the diagram
Hydraulics & Hydrology y Example 6:
a) Hydraulic Head : @ A = 0ft + (80 psi x 2.31ft/psi) = 185 ft @ B = 30ft + (30psi x 2.31 ft/psi) = 99ft @ C = 40ft f + 100ft f = 140ft f b) hL between A&B = 185-99 = 86ft hL per 1000ft = 86/5 = 17.2ft hL = 17.2ft/1000ft g p usingg nomograph: (hL = 17.2ft/1000ft; 12in diameter) Q = 2,160 gpm from A hL between C&B = 140-99 = 41ft hL per 1000ft = 41/3 = 13.7ft hL = 13.7ft/1000ft 13 7f /1000f using nomograph: (hL = 13.7ft/1000ft; 10in diameter) Q = 1,180 gpm from C Total available Q@B = 2160+1180=3,340gpm
Hydraulics & Hydrology y Centrifugal Pump Characteristics: ¾ Centrifugal g pumps p p are used commonlyy for low and high g
service to lift and transport water, reciprocating positivesp ace e t. displacement. ¾ It is popular because of its simplicity, compactness, low cost and ability to operate under a wide variety of conditions. conditions
Hydraulics & Hydrology y Pump Head-Discharge Curve: ¾ as the valve is ggraduallyy opened, p , allowingg increasingg of flow
of water, the pump head decreases. ¾ The pump efficiency rises with increasing rate of discharge to an optimum value and the decreases.
Hydraulics & Hydrology y Pump Characteristics: ¾ For a ggiven impeller p diameter ((D)) operated p at different
speeds (N):
Where:
Q = Discharge (gpm,Lps) H = Head (ft,m) Pi = Power P iinput (h (hp,kW) kW)
Hydraulics & Hydrology y Pump Characteristics: ¾ For a p pumpp operating p g at same speed p ((N)) . A change g in
impeller diameter (D) affects discharge,, head and power put: input:
Where:
D = Impeller diameter (in (in,cm) cm)
Hydraulics & Hydrology y Power and Efficiency: ¾ Power input is the motor power applied to a pump. ¾ Power output is the work done power unit of time lifting the
g elevation. water to a higher ¾ Efficiency of pump can be defined as:
¾ Centrifugal pump efficiency usually in the range of (60 to
85%)
Hydraulics & Hydrology y Power and Efficiency: ¾ The equation q relatingg discharge g to motor power p input p is
English units Where:
P = Motor power input (hp) g (gpm) g Q = Discharge H = Head (ft) γ = unit weight of water = (8.34lb/gal) EP = Pumpp efficiencyy 550 = ft-lbs/sec/hp 60 = seconds/min
Metric units Where:
P = Motor power input (kW) Q = Discharge g (lps) H = Head (m) γ = unit weight of water . /) = ((0.0098KN/l) EP = Pump efficiency
Hydraulics & Hydrology y Example 7: The characteristics of a centrifugal pump
operating at two different speeds are listed in the ff: chart. Graph these curves and connect the best efficiency points p with a dashed line. Calculate the head-discharge g values (bep) for an operating speed of 1450 rpm and plot the curve. p g envelope p between 60 & Finallyy sketch the ppumpp operating 120 % of the “bep”
Hydraulics & Hydrology y Example 7:
Plotting Pl tti th the H Head-discharge d di h value for operating speed of 1450rpm: (sample computation for: 1500gpm;H=216ft @1750 rpm) Q2=Q1(N2/N1)= 1500(1450/1750) = 1,240gpm H2=H1(N2/N1)2= 216(1450/1750)2 = 148ft @1750rpm: best efficiency @86%; Q = 3300gpm. 60%(3 300) = 2,000 60%(3,300) 2 000 gpm 120%(3,300)=4,000gpm @1150rpm: best efficiency @84%; Q = 2200gpm. 60%(2,200) = 1,300 gpm ( , ) , gp 120%(2,200)=2,600gpm Then plot
Hydraulics & Hydrology y System Characteristics: ¾
Hydraulic grade line 1: Water is being pumped through outlet and elevated storage. Hydraulic grade line 2: Water is discharging through outlet 2 f from pump andd elevated l d storage
Curve 1: Pump flow entering elevated storage Curve 2: Pump flow discharging at outlet 2
Hydraulics & Hydrology y Constant-Speed Pumps: ¾
Point A: Entire discharge flows into elevated storage. Point B: if free discharge is allowed only from the piping system at outlet 2 Point C: New operating point to raise hydraulic gradeline and the system head discharge
Hydraulics & Hydrology y Constant-Speed Pumps: ¾ Parallel installation arrangement g ((1,2 , same & 3 is higher) g ) The point of intersection of the combined head g curve and discharge the system curve gives the combined rate of discharge of the pumps and the operating head of pumps Pumps p are operated p individuallyy or in combination to meet the water demand by discharging into a common header and outlet pipe
Hydraulics & Hydrology y Variable-Speed Pumps: (operating w/ 2 impeller speeds) a) Head discharge curves with efficiency values for a pump operating at two impeller p speeds b) System-demand head curve for a constant pressure discharge corrected for transducer operation. (Pump speed increases when the pump discharge pressure reduces as a result of increasing d demand d andd ddecreases c w/increaseing /i c i demand d d pressure) c) Superimposedd pump head h d discharge d h andd demand curves d) Curves of speed and efficiency versus demand
Hydraulics & Hydrology y Variable-Speed Pumps: ¾ A variable speed p drive must be pprevented from operating p ga
pump at extremely low speeds. ¾ When the demand is less than the minimum required discharge, the pump is protected from damage by recirculating water through the pump. pump ¾ The recommended minimum discharge rate is generally 25 to 35% of the pumping rate at the best operating efficiency. ¾ Variable speed p pumps p p can be used in combination with constant speed pumps.
Hydraulics & Hydrology y Variable-Speed Pumps: ¾ Variable speed p can also be operated p in pparallel in multiple p
pump installations (load sharing and staggered operations) ¾ Load sharing – all pumps run at the same speed and discharge at
equal rates ¾ Staggered Operation – one or more of the pumps runs at optimum efficiency (constant speed) while the speed of only one pump is varied to meet changing demand.
Hydraulics & Hydrology y Example 8: Draw system head-discharge curves for the two
operating conditions in the simplified water system diagramed in Figure. The highest system head-discharge pp boundary) y occurs when ppumpp discharge g enters curve (upper the elevated storage tank with no system withdrawal. The p system y head-discharge g curve occurs when lowest anticipated the pressure at the load center is 45 psi and flow is entering y from both pump p p discharge g and elevated storage. g the system On the same head-discharge diagram, draw the characteristics pump curve from figure at an operating speed of 1750 rpm. Label the range of pump operations
Hydraulics & Hydrology y Example 8:
Total pump head @elevated storage = static head + headloss in 8000 ft of 16-in for a given Q S l solution: Sample l ti (f (for 2,000gpm) 2 000 ) 8,000ft x 0.0037 ft/ft = 30ft Total pump head = 150+30 = 180ft Upper curve
Operation: Upper curve: H d =200ft Head 200f Q Q=2620gpm 2620 Lower Curve: Head =165ft Q=3770gpm
Total pump head @load center = static head + headloss in 5000 ft of 16-in for a given Q p solution: (for f 2,000gpm) gp Sample L Lower curve 5,000ft x 0.0037 ft/ft = 19ft Total pump head = 104+19 = 123ft
Hydraulics & Hydrology y Equivalent Pipes: ¾ An equivalent q pipe p p is an imaginary g y conduit that replaces p a
section of a real system such that the headlosses in the two systemss are syste a e identical e t ca for o tthee quantity qua t ty oof flow. o . ¾ Example E l 9: 9 D Determine i an equivalent i l pipeline i li 2000ft 2000f in i
length to replace the pipe system illustrated in the figure @ Line BC: Assume headloss of 10ft g p Usingg nomograph: (d=8in; hl=10ft/1000ft) Q=550gpm (d=6in; hl=10ft/800ft=(12.5ft/1000ft)) Q 290gpm Q<290gpm (Q=550+290=840gpm;hl=10ft/1000ft) d=9.4in
Hydraulics & Hydrology y Equivalent Pipes: ¾ Example p 9: Determine an equivalent q pipeline pp 2000ft in
length to replace the pipe system illustrated in the figure Consider the 3 pipes in series: Assume Q=500gpm: (using nomograph) A-B: hl=0.4x8.3=3.3ft B C hl=3 B-C: hl=3.8ft 8ft C-D: hl=0.6x2.7=1.6ft Total hl=8.7ft in 2000ft = (4.4ft/1000ft) Equivalent Pipe A-D: (required loss 4.4ft/1000ft at Q=500gpm Therefore: d=9.2in
Hydraulics & Hydrology y Computer Program: ¾ Epanet p - is a public-domain, water distribution system modeling software ft package k developed d l d by b the th United States Environmental Protection Agency’s (EPA) Water Supply and Water Resources Division. It performs extendedpperiod simulation of hydraulic y and water-quality behavior within pressurized pipe networks and is designed to be "aa research tool that improves our understanding of the movement and fate of drinkingwater constituents within distribution systems". EPANET first appeared in 1993.
Hydraulics & Hydrology y Computer Program: ¾ WaterCad - is an easy-to-use hydraulic and water quality modeling solution for water distribution systems. It features advanced interoperability, model building, optimization, and asset management tools. tools WaterCAD helps engineers and utilities analyze, design, andd optimize ti i water t distribution systems.
Hydraulics & Hydrology y Gravity Flow in Circular Pipes: ¾ Manning’s g Formula – used for uniform,, steady, y, open p channel
flow ((English g units)) Where:
((Metric units))
Q = Quantity of flow (cfs, cms) g roughness g based on material n = Coeffficient of Manning’s A = cross-sectional area of flow (ft2,m2) R = Hydraulic radius = (A/wetted P) (ft,m) S = slope of hydraulic gradient (ft/ft (ft/ft, m/m)
Common Sewer Pipe Materials: a))Vitrified f clayy and smooth concrete: n=0.011 to 0.015 b) Corrugated steel pipe: n = 0.021 to 0.026
nn=00.013 013 (common adopted n for swer design
Hydraulics & Hydrolo y Gravity Flow in Circular Pipes:
Nomograph N h for f Mannings Formula based on n=0.013 Example: D=8in; S=0.02ft/ft Therefore: Therefore Q=760gpm V=4.9ft/s
Hydraulics & Hydrolo y Gravity Flow in Circular Pipes: ((q,v,a)) – symbols b l for partial flow. (Q,V,A) – symbols for full flow p Example: Flow in pipe at a depth of 30% of pipe diameter q/Q = 0.20 a/A = 0.25 0 25 v/V = 0.78 Relative quantity, velocity, cross sectional area of flow in a circular pipe for any depth of flow
The greatest quantity of flow occurs at 93% of depth
Hydraulics & Hydrology • Example 10: If a 10-in sewer is placed on a slope of 0.010
what is the flowing full quantity and velocity for a)n=0.013 b)n=0.015 Based on nomograph: (S=0.010; d=10in) Q = 990 990gpm; V=4ft/s V 4f / b. Q&V are inversely proportional to n, therefore: Q (@n=0.015) = 0.013x990/0.015 = 860gpm V (@n=0 (@n=0.015) 015) = 0.013x4.0/0.015 0 013x4 0/0 015 = 3.5ft/s 3 5ft/s a.
Hydraulics & Hydrology • Example 11: The measured depth of flow in a 48-in storm
sewer on a grade of 0.00015ft/ft is 30-in. What is the calculated quantity and velocity of flow? ¾ Based on nomograph: Q=7700gpm and V=1.4ft/s.
Depthh off flow D fl to diameter di off pipe i ratio: i d/D d/D=30/48=0.62 30/48 0 62 Based on graph (any depth of flow): q/Q=0.72; v/V=1.08 Therefore: q on depth 30in = 0.72x7700 = 5500gpm v on depth 30in = 1.08x1.4 1 08x1 4 = 1.5 1 5 ft/sec
Hydraulics & Hydrology • Example 12: An 18-in sewer pipe, n= 0.013 is placed on a
slope of 0.025. At what depth of flow does the velocity of flow equal to 2ft/s? ¾ Based on nomograph: V=3ft/s.
Therefore: Th f v/V=2/3=0.67 /V 2/3 0 67 Based on graph (any depth of flow): d/D=0.23 d (depth for v=2ft/s) = 0.23x18=4.1in
Hydraulics & Hydrology • Example 13: What is the flowing full quantity and velocity of flow
for a 450-mm-diameter sewer, n=0.013 on a slope of 0 00142m/m? Determine the quantity of flow for a depth of flow 0.00142m/m? equal to 300mm.
¾ Based on nomograph: Q=0.18cms; V=1.14m/s ¾ A=π(0 A=π(0.45) 45)2/4=0 /4=0.159m 159 2
R=A/P=(π)d2/4/(π)d=d/4=0.45/4=0.112m
Q Q=(1/0.013)(0.159)(0.112)(2/3)(0.00142(1/2)= (1/0.013)(0.159)(0.112)(2/3)(0.00142(1/2) 0.18cms V=Q/A = 0.18/0.159=1.14m/s
d/D = (300/450)=0 (300/450)=0.667; 667; Therefore: q/Q=0 q/Q=0.78 78 Therefore: Q@d=300mm = 0.78x0.18 = 0.14cms
Hydraulics & Hydrology y Flow Measurements in Pipe: ¾ HouseHold water meter
For low rates off fflow
¾ Turbine Type yp water meter
For high rates of flow
Hydraulics & Hydrology y Flow Measurements in Pipe: ¾ Compund p water meter
For high and low rates of flow
¾ Venturi Flow meter
Differential pressure meters
Hydraulics & Hydrology y Flow Measurement in Open Channels: ¾ Parshall Flume
Hydraulics & Hydrology y Amount of Storm Runoff: ¾ Rational Formula
Q=CiA (english units) Q=0.278CiA (metric units) Where:
Q = maximum rate of runoff (cfs,cms) yp and C = Coefficient of runoff based on type character on surface (see table on the left) i = average rainfall intensity, for the period of maximum rainfall of a given frequency of occurrence having a duration equal to the time required for the entire drainage area to contribute ib flow fl (inches/hr,mm/hr) (i h /h /h ) A = drainage area (acres,sq.km)
Hydraulics & Hydrology y Amount of Storm Runoff: ¾ Intensityy Flow Duration Curves
5-yr storm frequency = residential areas 10-yr storm frequency = business section 15-yr storm frequency = high value districts where flooding would result in considerable pproperty p y damage g The duration of rainfall frequency depends on time of concentration (inlet time + time of fflow through g the ppipe) p ) Inlet time generally ranges from 5-20 min
Hydraulics & Hydrology • Example 15: Compute the diameter
of the outfall sewer required to drain the storm water from the watershed described in figure, which gives the l lengths h off llines, ddrainage areas, andd inlet times. Assume the ff: C=0.30, T 5 years, V=2ft/s. T=5 V 2ft/ ¾ Time of Manhole1-2
t=d/V=400/2=200sec=3.33min ¾ Time of Manhole2-3 t=d/V=600/2=300sec=5min
Hydraulics & Hydrology • Example 15: ¾ Time of concentration from remote
points off the h tree separate areas to Manhole 3: t=5+3.3+5=13.3min 5+3 3+5 13 3 i for f Area A 1 t=5+5=10 min for Area2 t=8min for Area3 ¾ i=4.4in/hr (from graph) (for D=13.3min&T=5yrs) ¾ Q5=CiA=0.30(4.4)(3+6+4.5) Q5=18ft3/s = 8080gpm •Usingg Manning’s g nomograph: g p (Q=8080gpm&V=2ft/s) d=42in & slope=0.0004ft/ft
Hydraulics & Hydrology y Flow in Stream and Rivers ¾ Calculatingg Frequency q y Curve
Weibull plotting position
Hydraulics & Hydrology y Flow in Stream and Rivers ¾ Calculatingg Frequency q y Curve
For T=10yrs: P=90% (lowflow) Q=22.6cfs
Groundwater y Groundwater Hydrology ¾ Originates g as infiltration from precipitation, p p , streamflow,,
lakes and reservoir ¾ The surface of saturated zone is called water table table, and its depth is described by the level of free water in an observation well extending into a saturated zone zone. ¾ Porosity (n): Typical yp values of porosity: p y Where:
Vv = Volume of voids V = total Volume
n=0.2 to 0.4 (sands & gravel, depending on the grains size, size distribution and degree of compaction) n=0.1 to 0.2 (sand & stone) n= 01 to 00.11 (shale & limestone) n=.01
Groundwater y Groundwater Hydrology
Groundwater y Groundwater Hydrology ¾ Permeabilityy is the abilityy of porous p medium to transmit
water. ¾ Coefficient of permeability, permeability K, K by Darcy’s Darcy s Law: v=Ki Where:
v = velocity of flow (ft/s,mm/s) K = coefficient of Permeability (ft/s,mm/s) i = hydraulic gradient (ft/ft, (ft/ft m/m)
Ranges off values R l off K: K= 10x10-5 ft/s = 0.003 mm/s (fine grane deposit) Up to K= 1 ft/s = 300 mm/s g (course gravel)
Groundwater y Groundwater Hydrology ¾ Unconfined Aquifer q Where: Q=wellll discharge d h (cfs,l/s) f l K=coefficient of permeability (ft/s,mm/s) ho=saturated thickness of q before aquifer pumping (ft.m) ro=radius of the cone of depression (ft,m) (ft m) hw=depth of water in well while pumping (ft,m) rw=radius di off wellll (f (ft,m))
Groundwater y Groundwater Hydrology ¾ Confined Aquifer q
Where: B B=thickness thickness of aquifer (ft,m). Values of ro and hw may be assumed or measured from observation well data
Groundwater y Groundwater Hydrology ¾ Pumping p g test in unconfined acquifer q (Permeability ( y test))
For unconfined acquifer
For confined acquifer
Groundwater y Example l 16 16: A wellll withh a ddiameter off 2f 2ft is constructedd in a confined f d
aquifer as illustrated in the figure. The sand aquifer has uniform p layer y with a depth p of 115 thickness of 50ft overlain byy an impermeable ft. A pumping test was conducted to determine the coefficient of permeability of the aquifer. The initial piezometric surface was 49ft below the gground surface datum of the test well and observation wells. After water was pumped at rate of 0.46 cfs for several days, water level in the wells stabilized with the ff: drawdowns: 21ft (test well) 12.1ft (observation well at a distance of 30ft) 7 9 ft (2nd observation well at a distance of 100ft) 7.9 From these test data calculate the permeability of the aquifer. Then using the K value, estimate the well discharge with the drawdown in the well l lowered d on the h top off the h confined fi d aquifer. if
Groundwater y Example 16: =0.46cfs 49ft 21ft 115ft
2ft
100ft 30ft 12.1ft
ho hw
h1
7.9ft
ho=115-49=66ft hw=66-21=45ft h1 66 12 1 53 9ft h1=66-12.1=53.9ft h2=66-7.9=58.1ft
h2
50ft
K=0.00042ft/sec
Groundwater y Example 16: =0.46cfs 49ft 21ft 115ft
2ft
100ft 30ft 12.1ft
ho hw
h1
7.9ft
h2
hho=66ft =66ft hw=0ft (drawdown in the well at the top of sand aquifer) rw=1ft Assume: rcone=700ft
50ft
Q=1.3cfs
Distribution System y Introduction I d i ¾ Objectives of a municipal water system: To T provide id safe, f potable bl water ffor ddomestic i use; To provide an adequate quantity of water sufficient pressure for
fire protection and industrial water for manufacturing. manufacturing
¾ Typical waterworks consists of a source, source treatment treatment, pumping and
distribution system. ¾ Sources for municipal p supplies pp are deepp wells, shallow wells, rivers, lakes, and reservoir. ¾ About 2/3 of the water for public supplies comes from the surface-water f t sources.
Distribution System y Water Quantity and Pressure Requirements
The amount of water required q byy a municipality p y depends p on population; climate; commercial and industrial water userss co use conservation; se at o ; water ate reuse euse for o landscape a scape irrigation gat o and economic conditions ¾ Typical, Typical average average-day day municipal per capita, capita based on location, climate, and mix of residential versus commercial andd industrial i d t i l connections ti vary bbetween t 380 to t 780 L/person for all demands included. ¾ Residential demands alone are 300 to 500 L/person ¾
Distribution System y Water Quantity and Pressure Requirements
The recommended water ppressure in a distribution system y is 65 to 75 psi (450 to 520 kPa), which is considered aadequate equate to co compensate pe sate for o local oca fluctuations uctuat o s in consumption. ¾ For a residential service connections, connections the minimum pressure in the water distribution main should be 40 psi (280 kPa) kP ) ¾ Maximum pressure in a system are allowed for 150 psi. (most pipes and fittings are designed for 150 psi only) ¾
Distribution System y Well Construction ¾ Water in the voids of
underground sand or gravel e s can ca bee tappe tapped for o beds municipal supply by using a drilled well. well
Typical Well Construction
Gravel packed water well in a sand aquifer equipped with a two-stage vertical turbine pump
Distribution System y Surface Water Intakes ¾ Intake structure is required q to withdraw water from a river,,
lake or reservoir. ¾ Typical intakes are towers and shoreline structures. structures ¾ Their primary function is to supply the highest quality of water from f the h source.
Distribution System y Surface Water Intakes ¾ Towers are for lakes and
reservoirs with fluctuating ate levels e e s or o variations a at o s water of water quality with depth ¾ Shore intakes located
adjacent to a river
Distribution System y Piping Networks ¾ Municipal water distribution system includes a network of mains
withh storage reservoirs, pumping stations, ffire hhydrants d andd service lines. ¾ Arterial A i l mains i or ffeeders: d are pipelines i li or llarger size i that h are connected to the transmission lines that supply the water distribution. distribution ¾ Parallel feeders are cross-connected at interval of one mile, with valving, for isolation in emergency cases. ¾ Distribution lines are connected to arterial loops to form gridiron y which services fire hydrants y and domestic and commercial system consumers.
Distribution System y Piping Pi i Networks N k ¾The gridiron system: Best
arrangement for distributing water because: All arterial and secondaryy mains are interconnected Dead-ends are eliminated Water is circulated When piping repairs are necessary, the h area removedd from service can be reduced to one block if valves are properly located.
Distribution System y Piping Networks ¾ Dead-end system: y it should be avoided in new
area and can be corrected in existing systems by proper keeping. p g Water stagnation at deadends ma developp tastes and odors, to prevent this, deadends, mayy require q frequent q flushing.
Distribution System y Piping Networks y Service Connections: A typical service installation consists of a pipe from the
distribution main to turnoff valve located near the pproperty p y line. A special tapping machine is used to insert the corporation stop while the main is in service under pressure. Water meters are installed inside or outside the property. The Th estimated ti t d rate t ffor a ttypical i l hhouse hhaving i ttwo bbathrooms, th ffullll laundry, kitchen, and one or two house bibs is 15 gpm. A pressure of 15 psi is usually adequate to operate any fixture fixture, with the excerption of lawn sprinkler.
Distribution System y Kinds of Pipe: ¾In water distribution system, y , the followingg pipes p p are used: Ductile iron Plastic Pl ti Concrete Steel Copper (for house constructions)
Distribution System y Kinds Ki d off Pipe: Pi ¾Pipe materials must have the ff: properties: Adequate Ad tensile il andd bbending di strengthh to withstand i h d externall
loads. High bursting strength to withstand internal water pressure Ability to resist impact loads during transportation, handling, installation Smooth non-corrosive interior surface for minimum resistance of water flow An exterior unaffected by aggressive soil and groundwater Material that can be provided with tight joints and easy to tap for making ki connection ti
Distribution System y Kinds Ki d off Pipe: Pi ¾Ductile Iron Pipes: Are A notedd ffor llong lif life, toughness, h imperviousness, i i andd ease off
tapping as well as withstand internal pressure and external loads. p is stronger, g , tougher, g , and more elastic than cast This kind of ppipes iron. Iron pipes are available in sizes between 2 and 48 inches in diameter. (50 to 1200mm).
The selection pipe thickness depends on: • • • •
IInternal t l pressure External load Allowance for corrosion Design factor for safety
Distribution System y Kinds of Pipe: ¾Plastic Pipes: p They are not subject to corrosion
or deterioration by chemical or biological activities. They Th are smooth, h minimizing i i ii friction losses in water flow. PVC (Polyvinyl chloride) is the plastic refered for water distribution piping because of its strength and resistance to pressure. Co Commonly o y manufactured a u actu e ssizes es aaree 4 to 12 inches c es ((100 00 to 300mm) with internal pressure of 100, 150, and 200psi.
Distribution System y Kinds of Pipe: ¾Concrete Pipes: p Three types of reinforced
concrete pipes are used for pressure conduits. They Th hhave the h advantage d off durability, water tightness, and low maintenance costs. Theyy are applicable pp in larger g size (16 to 144inches).
Distribution System y Kinds of Pipe: ¾Steel Pipes: p Used in transmission lines They Th are off hi highh strength, t th able bl
to yield without breaking, resisting i i shocks. h k Protection against corrosion is necessary.
Distribution System y Distribution Pumping and Storage: ¾Pumps: High-lift pumps deliver water from treatment to distribution system Booster pumps to deliver water to elevated points in distribution area Common pumps in high h h service: a. b b.
Vertical turbine pumps Horizontal split-case split case centrifugal pumps
Why? y 1. 2. 3.
They have good efficiency. They have the capability to deliver water at high discharge heads This type of pumps operates at a range of capacities from design flow to shutoff without excessive loss of discharge pressure or efficiency.
Distribution System y Distribution Pumping and Storage: ¾Distribution storage g can be pprovided by: y
Elevated tanks 2 Stand 2. St d pipes i 3. Underground basins 4. Covered reservoir 1.
Distribution System y Distribution Pumping and
Storage: 1. Elevated Tanks: Have the advantage of that the pressure is derived from holding water that is higher than the surrounding terrain Can be made from steel (50,000 to 3 mil gallons)
Distribution System y Distribution Di ib i P Pumping i and d 2.
Storage: Ground Level Stand Pipes p or Reservoir: Provided when gravity water pressure is not necessary or when booster ppumps p are used. Usually available in sizes up to 5 mil gal. St l standpipes Steel t d i are applicable li bl where h the height of the tank exceeds diameter (If the diameter is greater than the height then it is a reservoir). reservoir) Concrete reservoirs can be constructed above or below the ground. d
Distribution System y Distribution Pumping and Storage: 4. a. b. a. b. c.
Covered Reservoir: Storage Basins need to be covered to: Reduce e uce tthee possibility poss b ty oof po pollution ut o Reduce the possibility of deterioration of the interior surface Exposure to the atmosphere will result in: Ai b Airborne contamination t i ti Algal growth due to penetration of sunlight Freezing of water surface in cold climates
Distribution System y Distribution Pumping and Storage: ¾ The choice between elevated and ground storage depends on: Topography SSizee of o community co u ty Reliability of water supply Economic aspects ¾In general, elevated tanks are more economical and are
recommended for small water systems. Reservoir are booster pumping i ffacilities ili i are usually ll less l expensive i in i llarge system
Distribution System y Distribution Di ib i P Pumping i and d SStorage: ¾ Functions of Distribution Storage: • Permit continuous treatment of water • Permit continuous uniform pumping rates of water into the distribution system • Store water in advance of actual needs at one or more locations ¾ Advantages of Distribution Storage: • Water W t ddemands d are nearly l equall att source, ttreatment, t t ttransmission, i i andd
distribution • Flow pressure of the system is stabilized throughout the service area • Reserve supplies are available for emergency cases such as fire fighting. ¾ To T Determine D t i storage t needed, d d the th ff: ff mustt be b considered: id d • Volume used to meet variations in demand • Amount related to emergency g y reserves
Distribution System y Example 1: Calculate the
distribution storage needed for both equalizing demand and for fire reserve based on the ff: information (see table). Fire flow q are 6000gpm gp for a requirements duration of 6-hr for the high gp value district with 2000gpm from storage.
Peak water consumption Data on Day of Max water usage
Distribution System Plot off hourlyy water consumption p rates to det. The storage needed to equalize at constant pumping rate
y Example 1:
For 8-hrs pumping: 1860gpmx24hrs/8hrs=5,580 gpm
Distribution System y Example 1:
The storage required to provide the entire reserve is equal to the fl rate times dduration: flow 2000 gal/min x 60min/hr x 6hr = 720,000 gal The total storage capacity required for equalizing demand for a continuous 24-hr pumping rate plus fire protection is: 500,000 gal + 720,000 gal = 1.22 mil gal The total considering an 8-hr pumping period plus fire reserve: 2.11 mil gal + 720,000 gal = 2.83 mil gal
Distribution System y Example 2: Consider a water supply system serving a city with the ff:
demand characteristics: average daily d il ddemandd = 4 0 mgdd (2780 gpm); 4.0 ) maximum day = 6.0 mgd (4170 gpm); peak hour = 9 0 mgd (6250 gpm); 9.0 gpm) required fire flow=7.2 mgd (5000gpm) resulting in a maximum 5-hr rate of 13.2 mgd(9170gpm) maximum daily demand plus fire flow.
Assume the minimum pressure to be maintained in the main district is 50 psi (115 ft) except during fire flow and that the piping system is equivalent to a 24-in diameter main with a C=100. Consider the system without h storage andd withh storage bbeyondd the h lloadd center.
Distribution System y Example l 22: ¾ Effect of no storage:
(L=29,000 (L=29 000 ft ft, D=24in C=100 & by using nomograph) For ave. dailyy demand: (Q=4mgd;s=0.9ft/1000ft) 115+(0.9x29)=140ft For F max. daily d l rate: (Q=6mgd;s=1.9ft/1000ft) ( ) 115+(1.9x29)=170ft For peak hourly demand: (Q=9mgd;s=4.0ft/1000ft) 115+(4.0x29)=230ft For max. daily rate plus fire flow: (Q=13 (Q 13.2mgd;s 2mgd;s=88.2ft/1000ft) 2ft/1000ft) 115+(8.2x29)=350ft
Distribution System y Example 2: ¾ Storage beyond Load
C t Center: (L=29,000 ft, D=24in C=100 & by using nomograph) For ave. daily demand:
(Q (Q=4mgd;s=0.9ft/1000ft) g ; f f) 120+(0.9x29)=146ft For max. dailyy rate: (Q=6mgd;s=1.9ft/1000ft)
For max. daily rate plus fire flow:
120+(1.9x29)=176ft For peak hourly demand:
(Q=3mgd; s=0.5ft/1000ft); Q=9-3=6mgd; s=1.9ft/1000ft) 120-(0.5x10)=115ft
115+(1.9x29)=171ft
(Q=4.8mgd;s=1.2ft/1000ft) (Q g ; f f) 120-(1.2x10)=108ft (Q=13.2-4.8=8.4mgd; s=3.5ft/1000ft) 108+(3.5x29)=210ft
Distribution System y ¾ • • • •
Valves: V l Valves are used in: Treatment plants Pumping stations Piping system Storage reservoirs
¾ Valve Function: • Control magnitude and direction of flow ¾How do they function: • They Th have h movable bl parts that h extendd iinto the h pipeline i li ffor opening i andd
closing the interior passage.
Distribution System y Valves: ¾ Kinds of Valves: a.
Shutoff Valve
b.
Check Valves
Distribution System y Valves: ¾ Kinds of Valves:
Small Pressure-Reducing and Pilot Valves d. Automatic Control Valves c.
Distribution System y Valves: ¾ Kinds of Valves: e.
Pressure Reducing Valves
f.
Altitude Valves
Distribution System y Valves: ¾ Kinds of Valves: e.
Solenoid Pilot Valves
f.
Air-Release Valves
Distribution System y Design D i Layout L off Di Distribution ib i SSystem: ¾ Arrangement of water system is dictated by: source off water supply, l topography of the distribution area, variations i ti iin water t consumption. ti ¾Basic Principles of Design: An arterial pipe network for a small system is shown with a high
p g station and elevation tank. ((see Figure g 6-33a)) service ppumping
Wells distributed throughout g the pipe p p network pump p p water
directly into the system at several locations, allowing the installation of smaller diameter pipes. (Figure 6-33b)
Distribution System y Design Layout of Distribution System: ¾
Distribution System y Evaluation of Distribution System: ¾ Quantity: Q y The supply source plus storage facilities should be capable of
yielding enough water to meet both the current daily demands and the anticipated consumption 10 years hence. General Rule: The storage g capacity p y of an impounding p g
reservoir should be equal to at least 30 days of maximum daily demand 5 years into the future.
Distribution System y Evaluation of Distribution System: ¾ Intake Capacity: A surface-water intake must be large enough to deliver
p use and treatment pplans sufficient water to meet municipal needs during any day of peak demand. A water intake system must be reliable; it must be located,
p protected, , or duplicated p such that no interruption p of service to customers or to fire protection occurs by reason of floods or other weather conditions or for reasons of breakdown, equipment repair, or power failure.
Distribution System y Evaluation of Distribution System: ¾ Pumping p g Capacity: p y In a typical surface-water supply system, low lift pumps draw
water from the source and transport it to the treatment plant. plant A pumping system must have sufficient capacity to provide the
amount of water at pressures and flow rates needed to meet both daily and hourly peak demand with require fire flow
Distribution System y Evaluation E l i off Distribution Di ib i SSystem: ¾ Piping Network: Arterial A i l andd secondary d feeder f d mains i should h ld be b ddesigned i d to supply l
water service for 40 or more years after installation.
Actual useful service life of mains under normal conditions is 50
y to 100 years.
Submains should be at least 6 in. in diameter in residential
districts, the minimum size in important districts should be 8 in. in diameter with 12-in. intersecting mains.
Distribution lines are laid out in a gridiron pattern avoiding deal
ends by proper looping. looping
