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18.2 Distribution & Hydraulic Math Workshop

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California Water/Wastewater Operator Exam

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1 Chapter 1: California Water Regulatory Framework & Certification Ladders

1.1 State Water Board & Regional Boards Structure 1.2 Drinking Water Certification: T1-T5 & D1-D5 Requirements 1.3 Wastewater Certification: Grades I-V & OIT Requirements 1.4 Safe Drinking Water Standards, MCLs & Emerging Contaminants 1.5 California Waterworks Standards (Title 22, Chapter 16) 1.6 Public Water System Types, Permits & the Sanitary Survey

2 Chapter 2: California Water Rights, Title 22 & Environmental Policy

2.1 California Title 22 Recycled Water Standards 2.2 California Environmental Quality Act (CEQA) & Proposition 218 2.3 Water Rights & California Water Resources

3 Chapter 3: Drinking Water Source Quality & Coagulation/Flocculation

3.1 Source Water Quality & Watershed Management 3.2 Coagulation Chemistry & Flash Mixing 3.3 Flocculation & Jar Testing 3.4 Wells, Aquifers & Groundwater Hydraulics 3.5 Raw Water Storage, Clearwells & Reservoir Management

4 Chapter 4: Clarification & Granular Media Filtration

4.1 Sedimentation & Advanced Clarification 4.2 Granular Media Filtration Mechanics & Media Types 4.3 Filter Operation, Backwash Cycles & SWTR Compliance

5 Chapter 5: Disinfection, Oxidation & Advanced Water Treatment

5.1 Chlorination Chemistry & Breakpoint Curves 5.2 Disinfection Kinetics & CT Calculations 5.3 Alternative Disinfection: Ozone, UV & Chlorine Dioxide 5.4 Advanced Treatment: Membranes, AOP & PFAS Adsorption 5.5 Chloramination & Nitrification Control 5.6 Iron & Manganese Removal 5.7 Fluoridation Practice & Compliance 5.8 Corrosion Control Treatment & Water Stability 5.9 Taste & Odor Control and Algal Toxin Management

6 Chapter 6: Wastewater Preliminary & Primary Treatment

6.1 Preliminary Treatment: Screening & Grit Removal 6.2 Primary Clarification Operations 6.3 Primary Sludge Characteristics & Pumping 6.4 Stabilization Ponds, Lagoons & Small-Plant Processes

7 Chapter 7: Secondary Biological Treatment & Activated Sludge

7.1 Activated Sludge Process Fundamentals 7.2 Process Control Parameters: F/M, MCRT & SVI 7.3 Activated Sludge Microbiology & Troubleshooting 7.4 Fixed-Film & Hybrid Biological Processes

8 Chapter 8: Biological Nutrient Removal, Water Reuse & Solids Handling

8.1 Biological Nitrogen Removal: Nitrification & Denitrification 8.2 Biological Phosphorus Removal (EBPR) & Chemical Precipitation 8.3 Anaerobic & Aerobic Digestion 8.4 Sludge Thickening, Dewatering & Biosolids Disposal 8.5 Wastewater Disinfection: Chlorination, Dechlorination & UV

9 Chapter 9: Distribution Systems, Collection Networks & Hydraulics

9.1 Hydraulics, Pressure & Flow Dynamics 9.2 Centrifugal Pumps, Motors & VFDs 9.3 Distribution Network Infrastructure & Cross-Connection Control 9.4 Wastewater Collection Systems, SSMP & FOG Management

10 Chapter 10: Laboratory Analytics, Safety & Water Math

10.1 Water & Wastewater Laboratory Methods 10.2 Laboratory Quality Assurance & Regulatory Reporting 10.3 Chemical Safety, Hazardous Gases & Confined Space 10.4 Essential Water & Wastewater Math for Operators

11 Chapter 11: Federal & California Drinking Water Rules

11.1 The Revised Total Coliform Rule & Assessments 11.2 The Lead and Copper Rule and Service Line Replacement 11.3 Surface Water Treatment Rules & Log Removal Credit 11.4 Stage 1 & Stage 2 Disinfection Byproduct Rules 11.5 Secondary MCLs, Notification Levels & Response Levels 11.6 Public Notification, Consumer Confidence Reports & Recordkeeping

12 Chapter 12: Distribution System Design, Demand & Hydraulics

12.1 System Layout, Pressure Zones & Distribution Maps 12.2 Water Demand, Peaking Factors & Fire Flow 12.3 Distribution Storage: Types, Sizing & Turnover 12.4 Head Loss, C-Factor & Distribution Hydraulic Calculations 12.5 Water Hammer, Surge Control & Air Management

13 Chapter 13: Distribution Equipment Operation, Maintenance & Inspection

13.1 Valves: Types, Operation & Exercising Programs 13.2 Fire Hydrants: Types, Testing & Maintenance 13.3 Water Meters: Types, Accuracy & Meter Management 13.4 Instrumentation, In-Line Sensors, Telemetry & SCADA 13.5 Electrical Systems, Motors & Standby Generators 13.6 External Corrosion, Cathodic Protection & Asset Inspection

14 Chapter 14: Water Mains, Piping & Field Construction

14.1 Pipe Materials & Pipe Selection 14.2 Joints, Fittings, Thrust Restraint & Bedding 14.3 Main Installation, Tapping & Service Line Installation 14.4 Leak Detection, Water Loss Auditing & Main Repair 14.5 Main Cleaning & Maintenance: Flushing, Pigging & Lining 14.6 Excavation, Trench Safety & Utility Locating

15 Chapter 15: Disinfecting Distribution Facilities & Maintaining Water Quality

15.1 Water Main Disinfection (AWWA C651) 15.2 Disinfecting Storage Reservoirs (C652) & Wells (C654) 15.3 Maintaining Distribution Water Quality: Residual, Biofilm & Discolored Water 15.4 Cross-Connection Control Program Administration & Device Testing

16 Chapter 16: Plant Equipment, Chemical Feed & Maintenance

16.1 Chemical Feed Systems: Metering Pumps, Dry Feeders & Calibration 16.2 Gas Chlorination Equipment & Hypochlorite Systems 16.3 Blowers, Compressors & Aeration Equipment 16.4 Flow Measurement Devices: Meters, Weirs & Flumes 16.5 Preventive Maintenance, CMMS & Asset Management

17 Chapter 17: Utility Management, Administration & Emergency Response

17.1 Planning, Organizing, Directing & Controlling a Utility Work Group 17.2 Budgets, Water Rates & Financial Management 17.3 Operator Training, Staffing & Safety Program Development 17.4 Energy Management & Optimization 17.5 Emergency Response Planning, Risk & Resilience & Security 17.6 Water Conservation Planning & Drought Response

18 Chapter 18: Applied Operator Mathematics Workshop

18.1 The Exam Conversion Sheet, Units & Dimensional Analysis 18.2 Distribution & Hydraulic Math Workshop 18.3 Treatment & Wastewater Process Math Workshop

18.2 Distribution & Hydraulic Math Workshop

Key Takeaways Explain This Topic AI

Pressure and head convert by 1 psi equals 2.31 feet and 1 foot of water equals 0.433 psi.

Specific capacity is well yield in gpm divided by drawdown in feet, and drawdown is pumping water level minus static water level.

Thrust equals internal pressure times projected area, and thrust block bearing area equals thrust divided by allowable soil bearing capacity.

Hydrant discharge equals 29.83 times the outlet coefficient times the square of the outlet diameter in inches times the square root of the pitot pressure.

Water horsepower equals gpm times head in feet divided by 3,960, and wire-to-water efficiency is pump efficiency multiplied by motor efficiency.

Last updated: September 2026

Pressure, Head, and Elevation Head (ft) = Pressure (psi) × 2.31 Pressure (psi) = Head (ft) × 0.433 \text{Head (ft)} = \text{Pressure (psi)} \times 2.31 \qquad \text{Pressure (psi)} = \text{Head (ft)} \times 0.433 Head (ft) = Pressure (psi) × 2.31 Pressure (psi) = Head (ft) × 0.433

Problem 1. A hilltop tank has an overflow elevation of 655 ft and is currently 9 ft below overflow. A service connection sits at ground elevation 488 ft . What is the static pressure?

  • Water surface = 655 − 9 = 646 ft
  • Elevation difference = 646 − 488 = 158 ft
  • Pressure = 158 x 0.433 = 68.4 psi

Problem 2. A pressure gauge at a pump discharge reads 118 psi . To what elevation could that pump lift water above the gauge, ignoring friction?

Head = 118 x 2.31 = 272.6 ft

Problem 3. A service reads 22 psi during a fire flow. Is the system in compliance with 22 CCR 64602?

Yes , but with only 2 psi of margin above the 20 psi floor. Any additional flow, or a 5-foot drop in tank level (which removes 2.2 psi), puts the system in violation.

Well Hydraulics Drawdown = Pumping Water Level − Static Water Level \text{Drawdown} = \text{Pumping Water Level} - \text{Static Water Level} Drawdown = Pumping Water Level − Static Water Level Specific Capacity (gpm/ft) = Well Yield (gpm) Drawdown (ft) \text{Specific Capacity (gpm/ft)} = \frac{\text{Well Yield (gpm)}}{\text{Drawdown (ft)}} Specific Capacity (gpm/ft) = Drawdown (ft) Well Yield (gpm) ​

Problem 4. A well has a static level of 112 ft , a pumping level of 189 ft , and produces 1,050 gpm .

Drawdown = 189 − 112 = 77 ft Specific capacity = 1,050 / 77 = 13.6 gpm/ft

Problem 5. Three years later the same well produces 1,050 gpm at a pumping level of 214 ft , with the static level unchanged at 112 ft.

  • Drawdown = 214 − 112 = 102 ft
  • Specific capacity = 1,050 / 102 = 10.3 gpm/ft , a 24 percent loss

Static level unchanged plus specific capacity falling means screen or gravel pack fouling , not aquifer decline. Video log and rehabilitate.

Problem 6. How many gallons stand in a 12-inch casing with 265 ft of water?

  • Gal/ft = 0.0408 x 12² = 0.0408 x 144 = 5.88
  • Volume = 5.88 x 265 = 1,558 gallons
  • Storage Volume and Level

Problem 7. A cylindrical tank is 68 ft in diameter with a water depth of 31 ft .

Volume = 0.785 x 68² x 31 x 7.48 = 0.785 x 4,624 x 31 x 7.48 = 841,900 gallons

Problem 8. That tank drops 4.5 ft in 3 hours with the inlet closed. What is the outflow?

  • Volume per foot = 0.785 x 4,624 x 1 x 7.48 = 27,158 gal/ft
  • Volume drawn = 27,158 x 4.5 = 122,211 gal
  • Flow = 122,211 / 180 min = 679 gpm

This is the drop test , the standard field method for verifying a flow meter.

Problem 9. A system needs 2,500 gpm of fire flow for 3 hours , operating storage of 20 percent of a 4.2 MGD maximum day demand , and emergency storage of one average day at 2.4 MGD .

Fire = 2,500 x 180 = 450,000 gal Operating = 0.20 x 4,200,000 = 840,000 gal Emergency = 2,400,000 gal Total = 3,690,000 gallons

Hydrant Flow Testing Q = 29.83 × c × d 2 × p Q = 29.83 \times c \times d^{2} \times \sqrt{p} Q = 29.83 × c × d 2 × p ​

Problem 10. A 2.5-inch outlet with c = 0.90 shows a pitot reading of 20 psi .

Q = 29.83 x 0.90 x 6.25 x √20 = 29.83 x 0.90 x 6.25 x 4.472 = 750 gpm

Problem 11. The same test at a 4.5-inch steamer outlet, c = 0.90 , pitot 12 psi .

Q = 29.83 x 0.90 x 20.25 x √12 = 29.83 x 0.90 x 20.25 x 3.464 = 1,883 gpm

Thrust and Bearing Area T = P × A A circle = 0.785 × D ( in ) 2 T = P \times A \qquad A_{\text{circle}} = 0.785 \times D^{2}_{(\text{in})} T = P × A A circle ​ = 0.785 × D ( in ) 2 ​

  • Configuration Multiplier on P × A
  • Dead end, cap, plug, tee branch, closed valve 1.000
  • 90° bend 1.414
  • 45° bend 0.765
  • 22.5° bend 0.390
  • 11.25° bend 0.196

Problem 12. A 14-inch main tested at 200 psi with a 45° bend .

  • Area = 0.785 x 196 = 153.9 in²
  • Thrust = 0.765 x 200 x 153.9 = 23,547 lb

Problem 13. That thrust is resisted by a block bearing on soil with an allowable capacity of 1,500 lb/ft² .

Bearing area = 23,547 / 1,500 = 15.7 ft²

Problem 14. The same 14-inch main terminates in a cap at 200 psi.

Thrust = 1.000 x 200 x 153.9 = 30,780 lb - over 15 tons. Nobody stands in line with that cap during the test.

Flushing Velocity and Volume Q gpm = 2.448 × D ( in ) 2 × V ( ft/s ) Q_{\text{gpm}} = 2.448 \times D^{2}_{(\text{in})} \times V_{(\text{ft/s})} Q gpm ​ = 2.448 × D ( in ) 2 ​ × V ( ft/s ) ​

Problem 15. What flow achieves 6 ft/s in a 6-inch main?

Q = 2.448 x 36 x 6 = 529 gpm

Problem 16. What flow achieves 6 ft/s in a 12-inch main?

Q = 2.448 x 144 x 6 = 2,115 gpm - four times the 6-inch requirement, because area scales with the square of diameter. This is why UDF segments are sized to the available hydrant flow.

Problem 17. How long to flush three pipe volumes through 900 ft of 8-inch main at 5 ft/s?

  • Gal/ft (8-inch) = 0.0408 x 64 = 2.61
  • Volume = 2.61 x 900 = 2,350 gal; three volumes = 7,050 gal
  • Q = 2.448 x 64 x 5 = 783 gpm
  • Time = 7,050 / 783 = 9.0 minutes

C-Factor and Head Loss Ratios h f 2 h f 1 = ( C 1 C 2 ) 1.85 h f 2 h f 1 = ( Q 2 Q 1 ) 1.85 \frac{h_{f2}}{h_{f1}} = \left(\frac{C_1}{C_2}\right)^{1.85} \qquad \frac{h_{f2}}{h_{f1}} = \left(\frac{Q_2}{Q_1}\right)^{1.85} h f 1 ​ h f 2 ​ ​ = ( C 2 ​ C 1 ​ ​ ) 1.85 h f 1 ​ h f 2 ​ ​ = ( Q 1 ​ Q 2 ​ ​ ) 1.85

Problem 18. C-factor falls from 120 to 60 .

Ratio = (120/60)^1.85 = 2^1.85 = 3.61 times the head loss

Problem 19. Flow doubles in the same pipe.

Ratio = 2^1.85 = 3.61 times the head loss - the same exponent, which is why doubling flow and halving C-factor cost the same energy penalty.

Problem 20. Cleaning and lining restores C from 65 to 130 . What happens to friction energy?

Ratio = (130/65)^1.85 = 3.61 , so friction head loss falls to about 28 percent of its prior value.

Pumping Power and Cost WHP = Q gpm × H ft 3,960 BHP = WHP Pump Eff MHP = BHP Motor Eff \text{WHP} = \frac{Q_{\text{gpm}} \times H_{\text{ft}}}{3{,}960} \qquad \text{BHP} = \frac{\text{WHP}}{\text{Pump Eff}} \qquad \text{MHP} = \frac{\text{BHP}}{\text{Motor Eff}} WHP = 3 , 960 Q gpm ​ × H ft ​ ​ BHP = Pump Eff WHP ​ MHP = Motor Eff BHP ​ Cost ($) = HP × 0.746 × Hours × $ / kWh \text{Cost (\$)} = \text{HP} \times 0.746 \times \text{Hours} \times \$/\text{kWh} Cost ($) = HP × 0.746 × Hours × $/ kWh

Problem 21. A booster pump delivers 1,100 gpm at 240 ft , pump efficiency 75 percent , motor efficiency 91 percent .

WHP = (1,100 x 240) / 3,960 = 66.7 hp BHP = 66.7 / 0.75 = 88.9 hp MHP = 88.9 / 0.91 = 97.7 hp Wire-to-water = 0.75 x 0.91 = 68.3 percent

Problem 22. That pump runs 16 hours per day at $0.18/kWh . Annual cost?

  • kW = 97.7 x 0.746 = 72.9 kW
  • Annual kWh = 72.9 x 16 x 365 = 425,736 kWh
  • Cost = 425,736 x 0.18 = $76,633 per year

Problem 23. A VFD lets the same pump run at 85 percent speed for half its operating hours. What is the power at 85 percent speed?

Power ratio = 0.85³ = 0.614 , so 72.9 x 0.614 = 44.8 kW - a savings of 28.1 kW during those hours.