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Water Demand Forecasting

Water-supply problems often estimate future population and convert demand into discharge or required pumping capacity.

$$P_t=P_0+kt\quad \text{arithmetic growth}$$$$P_t=P_0(1+r)^t\quad \text{geometric growth}$$$$Q=\frac{Pq}{86400}$$

Hydrologic Budget and Rainfall

Hydrology items include water balance, areal rainfall, rainfall intensity, and runoff volume.

$$P-Q-E-T-G=\Delta S$$$$\bar{P}=\frac{\sum P_iA_i}{\sum A_i}$$

Rational Method and Sewer Design

Storm sewer discharge is commonly estimated using the rational method. Keep units consistent with the coefficient form required by the problem.

$$Q=CiA$$

After finding design discharge, size the sewer using Manning's equation for the selected slope and roughness.

Population Served by a Reservoir

A reservoir supplies 0.12 m3/s. If consumption is 0.60 m3/person-day, estimate the population served.

$$V_{day}=0.12(86400)=10368\text{ m}^3/\text{day}$$$$P=\frac{10368}{0.60}=17280\text{ persons}$$

Answer: The reservoir can serve about 17,280 persons.

Problem: Population Forecast — Geometric Growth and Design Flow

A municipality had a population of 45,000 in 2005 and 63,000 in 2020. Using geometric (exponential) growth, forecast the population in 2045 and determine the required average daily supply pipe capacity in L/s if the per capita consumption is 180 L/person-day. Then compute the peak daily demand if the peak factor is 1.80.

Determine the geometric growth rate $r$ from 2005 to 2020 (15 years):

$$P_{2020} = P_{2005}(1+r)^{15}$$ $$63{,}000 = 45{,}000(1+r)^{15}$$ $$(1+r)^{15} = \frac{63{,}000}{45{,}000} = 1.4\overline{0}$$ $$1+r = (1.400)^{1/15} = e^{\ln(1.400)/15} = e^{0.3365/15} = e^{0.02244} = 1.02269$$ $$r = 0.02269 \quad (2.269\%\text{ per year})$$

Forecast for 2045 (40 years from 2005):

$$P_{2045} = 45{,}000(1.02269)^{40} = 45{,}000 \cdot e^{40(0.02244)} = 45{,}000 \cdot e^{0.8976}$$ $$e^{0.8976} = 2.453$$ $$P_{2045} = 45{,}000(2.453) = 110{,}385 \approx 110{,}400 \text{ persons}$$
$$Q_{avg} = \frac{P \times q}{86400} = \frac{110{,}400 \times 0.180}{86400} = \frac{19{,}872}{86400} = 0.2300 \text{ m}^3/\text{s} = 230 \text{ L/s}$$ $$Q_{peak} = 1.80 \times Q_{avg} = 1.80(230) = 414 \text{ L/s}$$

Answer: Forecast population 2045 ≈ 110,400. Average daily demand Q = 230 L/s. Peak daily demand = 414 L/s. Distribution mains must be sized for the peak demand, while treatment and source facilities are typically designed for average annual demand.

Problem: Rational Method — Storm Sewer Design Discharge

A storm drainage system serves a combined catchment area of 8.0 ha consisting of three sub-areas: residential rooftops (2.0 ha, C = 0.90), lawn/parks (3.5 ha, C = 0.35), and paved roads (2.5 ha, C = 0.85). The design rainfall intensity for the storm duration and return period is i = 75 mm/hr. Using the rational formula Q = CiA (with A in hectares and i in mm/hr, use appropriate conversion), find the design peak discharge in m³/s.

For a composite catchment, compute the area-weighted runoff coefficient:

$$C_{avg} = \frac{\sum C_i A_i}{\sum A_i} = \frac{0.90(2.0) + 0.35(3.5) + 0.85(2.5)}{2.0 + 3.5 + 2.5}$$ $$= \frac{1.80 + 1.225 + 2.125}{8.0} = \frac{5.150}{8.0} = 0.644$$

Apply the rational formula $Q = CiA$. Convert: $i = 75$ mm/hr $= 75/1000/3600$ m/s $= 2.083 \times 10^{-5}$ m/s. Area $A = 8.0$ ha $= 80{,}000$ m²:

$$Q = C_{avg} \cdot i \cdot A = 0.644(2.083 \times 10^{-5})(80{,}000)$$ $$= 0.644(1.6667) = 1.073 \text{ m}^3/\text{s}$$

Alternatively, using the shortcut formula $Q = CiA/360$ with $i$ in mm/hr and $A$ in hectares gives $Q$ in m³/s:

$$Q = \frac{CiA}{360} = \frac{0.644(75)(8.0)}{360} = \frac{386.4}{360} = 1.073 \text{ m}^3/\text{s}$$

Answer: Design peak discharge Q = 1.073 m³/s ≈ 1.07 m³/s. The weighted C of 0.644 reflects the dominant contribution of impervious surfaces (rooftops and roads), which produce high runoff. Green areas with C = 0.35 act as natural retention and reduce the overall coefficient slightly.

Problem: Reservoir Storage Volume — Mass Diagram

A water supply reservoir receives inflow and must meet a constant demand. Monthly inflow data (in million m³) for a dry year are: Jan=3.2, Feb=2.8, Mar=4.5, Apr=6.1, May=5.3, Jun=2.4, Jul=1.2, Aug=0.9, Sep=1.5, Oct=2.6, Nov=3.8, Dec=4.0. Total annual inflow = 38.3 million m³. If the demand is uniform at 38.3/12 = 3.192 million m³/month, determine the required storage capacity using the cumulative surplus/deficit method.

Monthly demand = 38.3/12 = 3.192 million m³/month. Compute monthly surplus (+) or deficit (−):

$$\begin{array}{lrrr} \text{Month} & \text{Inflow} & \text{Demand} & \text{Surplus(+)/Deficit(−)} \\ \text{Jan} & 3.2 & 3.192 & +0.008 \\ \text{Feb} & 2.8 & 3.192 & -0.392 \\ \text{Mar} & 4.5 & 3.192 & +1.308 \\ \text{Apr} & 6.1 & 3.192 & +2.908 \\ \text{May} & 5.3 & 3.192 & +2.108 \\ \text{Jun} & 2.4 & 3.192 & -0.792 \\ \text{Jul} & 1.2 & 3.192 & -1.992 \\ \text{Aug} & 0.9 & 3.192 & -2.292 \\ \text{Sep} & 1.5 & 3.192 & -1.692 \\ \text{Oct} & 2.6 & 3.192 & -0.592 \\ \text{Nov} & 3.8 & 3.192 & +0.608 \\ \text{Dec} & 4.0 & 3.192 & +0.808 \end{array}$$

The required storage is the maximum cumulative deficit (total drawdown from peak to trough). Track cumulative deficit from the point where reservoir is full (after the wet season surplus):

$$\text{Deficit period: Jun through Oct}$$ $$\text{Cumulative deficit} = 0.792 + 1.992 + 2.292 + 1.692 + 0.592 = 7.360 \text{ million m}^3$$

Answer: Required reservoir storage capacity = 7.36 million m³. This is the volume that must be stored during the high-inflow months (Mar–May) to sustain the constant demand through the low-inflow dry season (Jun–Oct). The mass curve (Rippl) method identifies this as the maximum cumulative net deficit during the dry season.

Problem: Sewer Pipe Design — Manning's Equation for Full Flow

A circular concrete sewer pipe (Manning's n = 0.013) must carry a peak design discharge of 0.85 m³/s flowing full. The available slope is S = 0.0006. Determine the required pipe diameter (to the nearest 50 mm standard size) and verify that the self-cleansing velocity criterion of V ≥ 0.60 m/s is met.

For a circular pipe flowing full: $A = \pi D^2/4$, $P = \pi D$, $R = D/4$.

$$Q = \frac{1}{n}AR^{2/3}S^{1/2} = \frac{1}{0.013}\cdot\frac{\pi D^2}{4}\cdot\left(\frac{D}{4}\right)^{2/3}(0.0006)^{1/2}$$ $$0.85 = 76.92 \cdot \frac{\pi D^2}{4} \cdot \frac{D^{2/3}}{4^{2/3}} \cdot 0.02449$$ $$4^{2/3} = (4)^{2/3} = 2.520$$ $$0.85 = 76.92(0.7854)(0.3969)(0.02449)\,D^{8/3}$$ $$0.85 = 76.92(0.07614)\,D^{8/3} = 5.857\,D^{8/3}$$ $$D^{8/3} = \frac{0.85}{5.857} = 0.14512$$ $$D = (0.14512)^{3/8} = (0.14512)^{0.375}$$

Compute: $\ln(0.14512) = -1.929$, $0.375(-1.929) = -0.7234$, $e^{-0.7234} = 0.4851$ m.

$$D_{required} = 0.485 \text{ m} \Rightarrow \text{Use standard } D = 500 \text{ mm}$$ $$A_{500} = \frac{\pi(0.50)^2}{4} = 0.1963 \text{ m}^2$$ $$V = \frac{Q}{A} = \frac{0.85}{0.1963} = 4.33 \text{ m/s} \gg 0.60 \text{ m/s} \checkmark$$

Answer: Required diameter ≈ 485 mm; use 500 mm standard sewer pipe. Full-pipe velocity = 4.33 m/s, which far exceeds the 0.60 m/s self-cleansing minimum. In practice, designers also check that velocity does not exceed about 3 m/s for non-reinforced concrete to prevent erosion — the 4.33 m/s here suggests a lined or reinforced concrete pipe should be specified.

Problem: Catchment Water Balance — Estimating Groundwater Recharge

A 500 km² catchment receives annual average rainfall of 1800 mm. Stream gauging at the catchment outlet records an average annual runoff of 720 million m³/year. Evapotranspiration estimated from the Penman method is 850 mm/year. No significant change in surface storage is observed (ΔS = 0). Compute: (a) total annual rainfall volume, (b) the annual evapotranspiration volume, (c) the groundwater recharge (deep percolation) from the water balance equation $P = Q + ET + G + \Delta S$.

$$A = 500 \text{ km}^2 = 500 \times 10^6 \text{ m}^2$$ $$P_{volume} = 1800 \text{ mm} \times 500 \times 10^6 \text{ m}^2 = 1.800 \text{ m} \times 500 \times 10^6 = 900 \times 10^6 \text{ m}^3/\text{year}$$ $$= 900 \text{ million m}^3/\text{year}$$ $$ET_{volume} = 0.850 \text{ m} \times 500 \times 10^6 \text{ m}^2 = 425 \times 10^6 \text{ m}^3/\text{year} = 425 \text{ million m}^3/\text{year}$$

Apply water balance: $P = Q + ET + G + \Delta S$ (with $\Delta S = 0$):

$$G = P - Q - ET = 900 - 720 - 425 = -245 \text{ million m}^3/\text{year}$$

A negative result means the stated values are inconsistent — outflow + ET > precipitation. This indicates either (1) groundwater is being mined (net outflow from storage), or (2) the ET estimate is too high, or (3) there is interbasin groundwater inflow. Recheck: if ET = 850 mm over the 500 km² catchment, ET volume = 425 million m³. But Q = 720 million m³ and P = 900 million m³, so $G = 900 - 720 - 425 = -245$ million m³.

Correcting the problem to ET = 550 mm (realistic for the given P and Q):

$$ET_{revised} = 0.550(500 \times 10^6) = 275 \text{ million m}^3/\text{year}$$ $$G = 900 - 720 - 275 = -95 \text{ million m}^3 \text{ (still negative)}$$

With Q = 480 million m³ instead (runoff coefficient = 480/900 = 0.533):

$$G = 900 - 480 - 275 = +145 \text{ million m}^3/\text{year (recharge)}$$

Answer (corrected scenario): P = 900 million m³/yr, ET = 275 million m³/yr (550 mm), Q = 480 million m³/yr. Groundwater recharge G = 145 million m³/yr (290 mm equivalent depth). The water balance equation $P = Q + ET + G$ is the fundamental tool in hydrology for partitioning precipitation into its components — errors in any one term propagate directly into the others.