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Lecture 10
Design of Water Distribution
System
 The design requirements of water distribution system are
to satisfied:
The water need ,and
The minimum residual pressure at each point of the
water distribution system
Pressure in Water Distribution
System
 Pressure in distribution system varies with consumption.
 Min. Pressure at peak flow(not less than 150 kPa to avoid
infiltration, proper flow to other buildings)
 Max. pressure during low flows
 Residential areas (3 stories)-150-300kPa(15-30m)
 Residential areas (firefighting )-400kPa(40m)
 Commercial areas-500KPa(50m)
Velocities in Water Supply
System
 Velocities in water supply system<1m/s
 2m/s upper limit may be reached near fire flows
 Min velocity-0.25m/s (WASA)
Hazen-Williams equation for
pipe flow
Headloss in pipes(water supply network)
Empirical
Named after Allen Hazen and Gardner Stewart Williams.
 H= head loss(m)
 Q= flow rate(m3/sec)
 L= length of pipe(m)
 d= diameter(m)
 C= Hazen William’s coefficient
The Hazen–Williams equation is an empirical relationship which relates the
flow of water in a pipe with the physical properties of the pipe and the
pressure drop caused by friction.
Hazen-Williams Equation for
Pipe Flow
 Hazen William greatly depends upon Roughness of pipe.
 Basic Hazen William Eq is
 Where,
V= velocity ,m/s
C=Hazen William co-efficient
R=Hydraulic radius or hydraulic mean depth
R=A/P(A=Area, P= Wetted Perimeter)
S= Hydraulic gradient=HL / L
H-head loss in meters
Q=flow in cu meter per sec
D= diameter in m
L= length of pipe in meters
For safety factor C=100
Hazen-Williams equation for
pipe flow
Advantages
 Coefficient C is rough measure of relative roughness
 Effect of Reynolds number is included in formula
 Effect of roughness on velocity are given directly
Disadvantages
 It is only valid for water
 Does not differentiate completely between laminar and
turbulent flow
 It does not account for the temperature or viscosity of the
water.error in water pipes can not be applied to all
fluids in all conditions
Hazen William
 Problem 1: Calculate the diameter of pipe 1Km laid to
discharge a flow of 1000m3/day under a head loss of
10m(C=100)
 Problem 2: A 6-km-long, new cast-iron pipeline carries
320 l /s of water .The pipe diameter is 30 cm. Find the
head loss .
Design of Water Distribution
System
Design Criteria for Distribution System
Design Parameter Value
Design flow Peak flow/Max.daily
demand
Population Population per plot
Peaking Factor 2.25/1.5
Minimum Size 75mm
Minimum Residual Head 14m
Input head 20m
Pipe Material AC or PVC, GI, steel pipe
C 100 or 110
Hardy’s Cross Method
Basic Principle
1. Sum of inflows is equal to the sum of outflows at any
junction or any node.
∑inflows=∑outflows
2. Sum of head losses around an elementary loop must be
zero.
∑HL = 0
Hardy’s Cross Method
Procedure
1. Draw the layout of the system.
2. Assign area to each node.
3. Calculate the population for
each node
4. Calculate the average
consumption at each node.
5. Calculate design flow for each
node which should be equal to
peak hourly demand.
6. Measure the length of each pipe.
Hardy’s Cross Method
Procedure
7. Assume flow in each pipe.
8. Assume the diameter of each pipe.
9. Assume any internally consistant
distribution of flow. The sum of flows
entering any junction must equal to
sum of leaving flows
10. Compute the head loss in each pipe by
means of equation(Hazen William) or
diagram. Conventionally clockwise flows
are positive and produce positive head
loss. Anticlockwise flows are negative
and produce negative head loss
Hardy’s Cross Method
Procedure
11. With due attention to the sign ,compute the total head loss
around each loop.
12. Compute without the regard to sign for the same loop for
sum of H/Q
13. Apply the correction to the flow in each line
16. Pipe line common to two loops receive both correction with
due attention to sign.
7. Balance the flow by Hardy cross method.
8. Calculate the residual pressure at all points of the system
and check its adequency.
Hardy’s Cross Method
(Example)
20 m3/min 5 m3/min
10 m3/min
5 m3/min
1500m
445mm
1100m
305mm
1000m
405mm
800m
225mm
1.
Hardy’s Cross Method
(Example)
20 m3/min 5 m3/min
10 m3/min
5 m3/min
1500m
445mm
1100m
305mm
1000m
405mm
800m
225mm
2.
250mm
1100m
Hardy’s Cross Method
(Example)
3.
20
Solution
21
22
23
24
25
26
27

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Lec-10-Week (7)( Hydraulics of water Distribution System).pdf

  • 2. Design of Water Distribution System  The design requirements of water distribution system are to satisfied: The water need ,and The minimum residual pressure at each point of the water distribution system
  • 3. Pressure in Water Distribution System  Pressure in distribution system varies with consumption.  Min. Pressure at peak flow(not less than 150 kPa to avoid infiltration, proper flow to other buildings)  Max. pressure during low flows  Residential areas (3 stories)-150-300kPa(15-30m)  Residential areas (firefighting )-400kPa(40m)  Commercial areas-500KPa(50m)
  • 4. Velocities in Water Supply System  Velocities in water supply system<1m/s  2m/s upper limit may be reached near fire flows  Min velocity-0.25m/s (WASA)
  • 5. Hazen-Williams equation for pipe flow Headloss in pipes(water supply network) Empirical Named after Allen Hazen and Gardner Stewart Williams.  H= head loss(m)  Q= flow rate(m3/sec)  L= length of pipe(m)  d= diameter(m)  C= Hazen William’s coefficient The Hazen–Williams equation is an empirical relationship which relates the flow of water in a pipe with the physical properties of the pipe and the pressure drop caused by friction.
  • 6. Hazen-Williams Equation for Pipe Flow  Hazen William greatly depends upon Roughness of pipe.  Basic Hazen William Eq is  Where, V= velocity ,m/s C=Hazen William co-efficient R=Hydraulic radius or hydraulic mean depth R=A/P(A=Area, P= Wetted Perimeter) S= Hydraulic gradient=HL / L
  • 7. H-head loss in meters Q=flow in cu meter per sec D= diameter in m L= length of pipe in meters For safety factor C=100
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  • 9. Hazen-Williams equation for pipe flow Advantages  Coefficient C is rough measure of relative roughness  Effect of Reynolds number is included in formula  Effect of roughness on velocity are given directly Disadvantages  It is only valid for water  Does not differentiate completely between laminar and turbulent flow  It does not account for the temperature or viscosity of the water.error in water pipes can not be applied to all fluids in all conditions
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  • 11. Hazen William  Problem 1: Calculate the diameter of pipe 1Km laid to discharge a flow of 1000m3/day under a head loss of 10m(C=100)  Problem 2: A 6-km-long, new cast-iron pipeline carries 320 l /s of water .The pipe diameter is 30 cm. Find the head loss .
  • 12. Design of Water Distribution System Design Criteria for Distribution System Design Parameter Value Design flow Peak flow/Max.daily demand Population Population per plot Peaking Factor 2.25/1.5 Minimum Size 75mm Minimum Residual Head 14m Input head 20m Pipe Material AC or PVC, GI, steel pipe C 100 or 110
  • 13. Hardy’s Cross Method Basic Principle 1. Sum of inflows is equal to the sum of outflows at any junction or any node. ∑inflows=∑outflows 2. Sum of head losses around an elementary loop must be zero. ∑HL = 0
  • 14. Hardy’s Cross Method Procedure 1. Draw the layout of the system. 2. Assign area to each node. 3. Calculate the population for each node 4. Calculate the average consumption at each node. 5. Calculate design flow for each node which should be equal to peak hourly demand. 6. Measure the length of each pipe.
  • 15. Hardy’s Cross Method Procedure 7. Assume flow in each pipe. 8. Assume the diameter of each pipe. 9. Assume any internally consistant distribution of flow. The sum of flows entering any junction must equal to sum of leaving flows 10. Compute the head loss in each pipe by means of equation(Hazen William) or diagram. Conventionally clockwise flows are positive and produce positive head loss. Anticlockwise flows are negative and produce negative head loss
  • 16. Hardy’s Cross Method Procedure 11. With due attention to the sign ,compute the total head loss around each loop. 12. Compute without the regard to sign for the same loop for sum of H/Q 13. Apply the correction to the flow in each line 16. Pipe line common to two loops receive both correction with due attention to sign. 7. Balance the flow by Hardy cross method. 8. Calculate the residual pressure at all points of the system and check its adequency.
  • 17. Hardy’s Cross Method (Example) 20 m3/min 5 m3/min 10 m3/min 5 m3/min 1500m 445mm 1100m 305mm 1000m 405mm 800m 225mm 1.
  • 18. Hardy’s Cross Method (Example) 20 m3/min 5 m3/min 10 m3/min 5 m3/min 1500m 445mm 1100m 305mm 1000m 405mm 800m 225mm 2. 250mm 1100m
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