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VIII. Protection of treated water

VIII. Protection of treated water. A. Hydraulics & B. Distribution systems. Objectives:. Describe water distribution system configurations and basic system requirements. Describe types and uses of water pumps. Describe and apply some equations used to determine hydraulic flow of water.

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VIII. Protection of treated water

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  1. VIII. Protection of treated water A. Hydraulics & B. Distribution systems

  2. Objectives: • Describe water distribution system configurations and basic system requirements. • Describe types and uses of water pumps. • Describe and apply some equations used to determine hydraulic flow of water.

  3. Distribution systems • System configurations: • Grid systems deliver water from two directions, so it is the preferred system configuration. • Branching systems deliver water from one direction only, so they have terminals or dead ends. • System configurations are usually a combination of grid and branching.

  4. Basic system requirements • System pressure should be high enough to meet consumer and fire-fighting needs: • Commercial systems require >60 psi. • Residential systems require 40-50 psi. • Pipe size should limit flow rate to 4-6 fps. • A minimum pipe diameter of 6 - 8” is necessary for fire-fighting.

  5. Pipe networks • Most municipal systems are complex mazes of pumps, storage, and pipelines of various sizes. • Hydraulic calculations must take into account looping characteristics of systems. • Network design must take into account head losses to deliver required water.

  6. Types of water pumps: • Displacement - piston driven. • Centrifugal - impeller driven. • Jet pumps - impeller + injector. • Air lift - compressed air + water rises. • Hydraulic ram - use hydraulic characteristics of water.

  7. Hydraulics: • Hydraulics is “the study of fluids at rest and in motion” including water. • Water is transported long distances through aqueducts. • Aqueducts may be open channels or pipelines.

  8. Open channels: • Hydraulics of open channel flow are determined by atmospheric pressure. • Channels may be lined with concrete, synthetic fabrics, etc. • Flumes are small channels often lined with concrete, steel, or even timber.

  9. Pipelines: • Built where topography limits the use of channels. • Usually built of concrete, steel, cast iron, or plastic (PVC). • Valves, drains, manholes, and pumping stations are required for maintenance.

  10. Design considerations: • Location of aqueducts is based on engineering and economic considerations. • Sizing is based on hydraulic, economic, and construction considerations. • Strength is important to resist internal and external pressure and stresses.

  11. Hydraulic calculations: • The Hazen-Williams formula for hydraulic flow is used for pressurized conduits: • V = 1.3CR.63 S.54 where • V = velocity of flow in feet/second, • C = coefficient of friction, • R = hydraulic radius (flow area/wetted perimeter), and • S = slope of grade in feet/feet.

  12. Hydraulic calculations: • The Hazen-Williams formula is modified for circular conduit flowing full as: • Q = 0.28CD2.63 S.54 where • Q = quantity of flow (in MGD), and • D = diameter of pipe. • S = slope of grade in feet/feet.

  13. Hydraulic calculations: • The Manning equation is often used in open channel problems: • V = 1.5R.66 S.50/n where • V = flow velocity in feet/second, • n = coefficient of friction, • R = hydraulic radius in feet, • S = slope of grade in feet/feet.

  14. Problem 1: • Water is drained from a reservoir down a 45 degree slope through 20 year-old circular cast iron pipe (full flow) with a radius of 1 foot. • You need 150 MGD for irrigation. Is this pipe sufficient to supply the water you need?

  15. Solution: • Q = 0.28CD2.63 S.54 (Hazen-Williams equation for circular conduit flowing full). • C = 100 (20 year old cast iron pipe) • D = 2 feet (radius of 1 foot  2) • S = 1 (450= 1 foot drop in 1 foot) • Q = 0.28 (100) (22.63) (1.54) = 173 MGD • 173 MGD > 150 MGD required (OK).

  16. Problem 2: • Water is flowing through a 6 inch PVC pipe which is half full down a 10 slope. • Is the velocity of flow within an acceptable range?

  17. Solution: • V = 1.3CR.63 S.54 V (Hazen-Williams formula for pressurized conduit). • C = 150 (plastic PVC pipe) • R = flow area [0.5 (3.14) .252]/wetted perimeter [(3.14) .25]= 0.1/0.8 = .13 • S = 0.01 (10 = 1 foot drop in 90 feet) • V = 1.3(150)(.13.63)(0.0154) = 4.4 feet/sec. • 4 - 6 fps range goal (OK).

  18. Summary: • Water distribution systems are highly complex, and may be grid, branching, or a combination of both. • Types of pumps include displacement, centrifugal, jet, air lift, and hydraulic ram. • Formulas for hydraulic calculations include Hazen-Williams and Manning.

  19. VIII. Protection of Treated Water: C. Cross-connections and cross-connection control

  20. Objectives: • Define cross-connections, backflow, and cross-contamination and distinguish among them. • List, describe, and identify appropriate uses for types of cross-connection control devices.

  21. Cross-connections: • A cross-connection is any physical connection between a potable water system and a nonpotable water supply. • This includes any waste pipe, sewer, drain, or any source of polluted water connected to a potable water supply.

  22. Backflow: • Backflow (flow of water or waste opposite the normal direction through plumbing) may occur by two means: • Backpressure - pressure in the nonpotable system exceeds that in the potable system. • Backsiphonage - pressure in the potable system is less than the nonpotable system.

  23. Cross-contamination: • Cross-contamination is the actual contamination of potable water with wastes or wastewater. • Cross-connectionsfacilitatebackflow, which may result in cross-contamination.

  24. Conditions that contribute to backflow: • Connection of potable water outlets to wastewater (or other waste) sources. • Negative pressure (partial vacuum) formed in pipes due to water flow. • A garden hose in a pool, tub, or laundry sink is the most common cause of backflow. • Prevention of partial vacuum formation is one method of preventing backflow.

  25. Cross-connection control (continued) • Cross-contamination can be prevented by breaking the connection between wastes and potable water. • If cross-connections are unavoidable, or occur due to accidents, backflow prevention devices can reduce risk of contamination (but not eliminate it).

  26. Backflow prevention devices: • Air gaps - physical separation of potable and non-potable system by air space. • Vacuum breakers - used only on non-potable systems. Examples include: • Atmospheric type • Pressure type • Double check valve assemblies which may also contain: • Reduced pressure zone devices.

  27. Air gaps: • Air gaps are the most safe and efficient design for backflow prevention. • They should always be used on potable water systems. • Practices that defeat air gaps should be avoided.

  28. Vacuum breakers: • Atmospheric type allows venting of low pressure to the atomosphere to prevent backflow (use float devices). • Pressure type vacuum breakers also vent low pressure to the atmosphere, but through a spring-loaded pressure relief valve.

  29. Double check valves: • Two independently operating valves prevent backflow of water. • Reduced pressure zones between check valves allow potentially contaminated water to escape.

  30. Summary: • Cross-connections facilitate backflow, which may lead to cross-contamination of drinking water with wastewater. • Devices for prevention of backflow include air gaps, atmospheric and pressure vacuum breakers, and double check valve assemblies (with or without reduced pressure zones).

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