18.2 Distribution & Hydraulic Math Workshop

Pressure, Head, and Elevation

$\text{Head (ft)} = \text{Pressure (psi)} \times 2.31 \qquad \text{Pressure (psi)} = \text{Head (ft)} \times 0.433$

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.

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Well Hydraulics

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

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

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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

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Hydrant Flow Testing

$Q = 29.83 \times c \times d^{2} \times \sqrt{p}$

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

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Thrust and Bearing Area

$T = P \times A \qquad A_{\text{circle}} = 0.785 \times D^{2}_{(\text{in})}$

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.

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Flushing Velocity and Volume

$Q_{\text{gpm}} = 2.448 \times D^{2}_{(\text{in})} \times V_{(\text{ft/s})}$

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

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C-Factor and Head Loss Ratios

$\frac{h_{f2}}{h_{f1}} = \left(\frac{C_1}{C_2}\right)^{1.85} \qquad \frac{h_{f2}}{h_{f1}} = \left(\frac{Q_2}{Q_1}\right)^{1.85}$

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.

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Pumping Power and Cost

$\text{WHP} = \frac{Q_{\text{gpm}} \times H_{\text{ft}}}{3{,}960} \qquad \text{BHP} = \frac{\text{WHP}}{\text{Pump Eff}} \qquad \text{MHP} = \frac{\text{BHP}}{\text{Motor Eff}}$

$\text{Cost (\$)} = \text{HP} \times 0.746 \times \text{Hours} \times \$/\text{kWh}$

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.