8.5 Water Supply: Demand, Storage, Pumping, Distribution, and Treatment

8.5 Water Supply: Demand, Storage, Pumping, Distribution, and Treatment

The 2022 HGE table of specifications gives Water Supply 8% of the subject, or 4 of its 50 items. That is as much as buoyancy and flotation, and more than hydrodynamics. The items are computational and draw directly on pipe flow (Section 8.3) and on pumps and the energy equation (Section 8.1).

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Population Forecasting

Example. A town of 40,000 grew 3% per year. Projected to a 20-year design period: $P = 40{,}000(1.03)^{20} = 72{,}244$. An arithmetic increase of 1,200 per year would give $40{,}000 + 20(1{,}200) = 64{,}000$.

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Water Demand and Peaking Factors

Average day demand (ADD):

$\text{ADD} = P \times q \quad \text{(liters per capita per day)}$

Here $q$ includes domestic use, commercial and institutional use, and an allowance for non-revenue water (leakage).

Smaller communities have higher peak factors, because their demand is less spread out over time.

Example. A population of 72,000 uses 150 L/capita/day.

• ADD $= 72{,}000 \times 150 = 10.8 \times 10^6\text{ L/day} = 10{,}800\text{ m}^3/\text{day} = 0.125\text{ m}^3/\text{s}$.
• With MDD = 1.3 × ADD, MDD $= 0.1625\text{ m}^3/\text{s}$.
• With PHD = 2.0 × ADD, PHD $= 0.250\text{ m}^3/\text{s}$.

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Storage

Elevated tanks and ground reservoirs hold three volumes:

1. Equalizing (operational) storage covers hourly demand above the steady supply rate. It is found from a mass diagram: plot cumulative supply and cumulative demand over a day. The required storage is the largest surplus plus the largest deficit, measured from the supply line. Rule-of-thumb values are often around 15% to 25% of MDD.
2. Fire reserve equals fire flow times fire duration.
3. Emergency reserve covers power or source outages.

Example. A pumping plant delivers a steady $500\text{ m}^3/\text{h}$ for 24 hours. Cumulative demand exceeds cumulative supply by at most $1{,}100\text{ m}^3$ at 8 PM, and falls below it by at most $900\text{ m}^3$ at 5 AM. Equalizing storage is $1{,}100 + 900 = 2{,}000\text{ m}^3$.

Pressure. An elevated tank's water surface sets the hydraulic grade line. Residual pressure at a service point is the HGL elevation minus ground elevation, less head losses. Distribution systems typically aim to keep enough residual pressure for upper-floor fixtures without excessive pressure that drives leakage.

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Pumping

$P_{\text{water}} = \gamma Q H, \qquad P_{\text{brake}} = \frac{\gamma Q H}{\eta_p}, \qquad P_{\text{motor input}} = \frac{\gamma Q H}{\eta_p \eta_m}$

Total dynamic head is $H = (\text{static lift}) + (\text{friction losses}) + (\text{minor losses}) + (\text{velocity head at discharge, if free})$.

Example. A pump delivers $0.08\text{ m}^3/\text{s}$ from a well with its water level at elevation 12 m to a tank surface at elevation 52 m. Pipe losses total 6 m, and the pump efficiency is 0.75.

• $H = 40 + 6 = 46\text{ m}$.
• Brake power $= 9.81(0.08)(46)/0.75 = 48.1\text{ kW}$.

Cavitation. The available net positive suction head must exceed the pump's required NPSH, so suction lifts are kept short.

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Distribution Systems

Layouts:

• Branching (tree) systems are simple, but they have dead ends with stagnant water and a single supply path.
• Grid or loop systems give two-way flow, better pressure and reliability, and are analyzed by the Hardy Cross method (Section 8.3).

Pipe sizing. Choose a diameter so the friction slope keeps pressures acceptable and velocities moderate, roughly 0.6 to 2 m/s. With Hazen-Williams:

$h_f = \frac{10.67\, L\, Q^{1.852}}{C^{1.852} D^{4.87}}$

Example. A $1{,}500\text{ m}$ main carries the peak-hour flow of $0.25\text{ m}^3/\text{s}$, and the allowable head loss is 12 m with $C = 130$. Solving for D:

$D^{4.87} = \frac{10.67(1{,}500)(0.25)^{1.852}}{130^{1.852}(12)} = \frac{16{,}005(0.07651)}{8{,}186(12)} = 0.01247$

So $D = 0.01247^{1/4.87} = 0.406\text{ m}$. Select the next standard size, 450 mm.

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Treatment Hydraulics

Drinking-water quality limits in the Philippines are set by the Department of Health's Philippine National Standards for Drinking Water.

Example. A rectangular settling tank treats $0.20\text{ m}^3/\text{s}$ at an overflow rate of $30\text{ m}^3/\text{m}^2/\text{day}$.

• Surface area: $A = 0.20(86{,}400)/30 = 576\text{ m}^2$.
• With a depth of 3.5 m, detention time is $576(3.5)/0.20 = 10{,}080\text{ s} = 2.8\text{ h}$.