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CELE Hydraulics & Fluid MechanicsHydrology and Water SupplyRevision Notes

Quick revision notes for Hydrology and Water Supply — the one-page refresher for CELE aspirants. Every item on this page has appeared in recent CELE Hydraulics & Fluid Mechanics papers, so revising these is the shortest path to a confident performance in Professional Regulation Commission (PRC) — Board of Civil Engineering's CELE 2026.

Exam context

The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Hydraulics & Fluid Mechanics subtest is marked as "Core" in the official pattern, and Hydrology and Water Supply appears in position 10th of 10 in the CELE Hydraulics & Fluid Mechanics review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.

Hydrology and Water Supply - Revision Notes

Hydrology quantifies the movement and distribution of water on Earth, while water-supply engineering ensures adequate, safe water delivery to communities. For the PRC Civil Engineer Licensure Examination, this chapter focuses on the rational method for peak runoff, runoff volume calculations, rainfall intensity-duration-frequency (IDF) relationships, and water demand estimation. These topics appear regularly in the board exam under Hydraulics & Fluid Mechanics. Mastery of unit conversions, the 360-factor, and demand multipliers is critical to scoring well.

Sections

Formulas

Example

A storm delivers P = 120 mm. If F = 40 mm, E = 10 mm, and ΔS = 0, then R = 120 − 40 − 10 = 70 mm depth of runoff.

Formula

P = R + F + E + ΔS

Variables

P = precipitation (mm); R = runoff (mm); F = infiltration (mm); E = evapotranspiration (mm); ΔS = change in storage (mm)

Application

Mass-balance check over a catchment for a given storm period.

Exam Tips

  • Board problems on the hydrologic cycle are mostly conceptual or qualitative; know the sequence and the engineering terms.
  • tc is not directly computed on most board problems — it is usually given. Focus on using it correctly to find i from IDF data.
  • If the problem gives return period and duration, select i from the given IDF table or equation; do not invent a value.

Key Points

  • The hydrologic cycle describes the continuous movement of water through the environment: precipitation → interception/infiltration → surface runoff → streamflow → evaporation/transpiration → back to precipitation.
  • Engineering hydrology focuses on the runoff fraction — the portion of precipitation that flows over the surface and must be managed by drainage or storage structures.
  • Losses in the cycle include infiltration (absorbed by soil), interception (held by vegetation), evaporation (liquid to vapor from surfaces), and transpiration (vapor release from plants). The sum is called evapotranspiration (ET).
  • The water balance equation is: Runoff = Precipitation − Infiltration − Evapotranspiration − Surface storage.
  • Rainfall duration, intensity, and return period determine the design storm for any drainage or flood-control structure.
  • The time of concentration (tc) is the travel time from the most remote point of the catchment to the outlet; it governs the design rainfall duration in the rational method.

Definitions

Term

Catchment (Watershed)

Definition

The land area that drains to a common outlet point, bounded by topographic divides (ridges).

Importance

Area A in the rational method must be the contributing catchment area, measured in hectares for SI board problems.

Term

Time of Concentration (tc)

Definition

The time required for water to travel from the hydraulically most remote point of the catchment to the design outlet.

Importance

tc sets the design storm duration; intensity i is read from the IDF curve at duration = tc. Longer tc → lower i → may reduce Q.

Term

Evapotranspiration (ET)

Definition

Combined water loss from evaporation (from soil and water surfaces) and transpiration (from vegetation).

Importance

ET is the largest non-runoff loss in tropical catchments; reducing ET (e.g., paved areas) increases runoff C.

Term

Return Period (T)

Definition

The average interval in years between occurrences of a storm of given magnitude; probability of exceedance in any year = 1/T.

Importance

Design return periods: minor drainage 2–10 yr; major drainage 25–100 yr. Higher T → higher design intensity i.

Section Title

The Hydrologic Cycle

Common Mistakes

  • Confusing the time of concentration with the storm duration — they are set equal only for the rational method design condition.
  • Forgetting that intensity DECREASES as storm duration increases; using a short-duration intensity for a large catchment with long tc overestimates Q dangerously.
  • Treating the hydrologic cycle as only precipitation and runoff; ignoring infiltration and ET leads to over-design.

Formulas

Example

Given: C = 0.65, i = 75 mm/hr, A = 30 ha. Q = (0.65 × 75 × 30) / 360 = 1462.5 / 360 = 4.06 m³/s.

Formula

Q = (C · i · A) / 360

Variables

Q = peak discharge (m³/s); C = runoff coefficient (dimensionless); i = rainfall intensity (mm/hr) at duration = tc; A = catchment area (hectares, ha)

Application

Design of storm drains, culverts, channels, and small detention basins for peak inflow.

Example

40% pavement (C = 0.90, A1 = 8 ha) and 60% lawn (C = 0.20, A2 = 12 ha): C = (0.90×8 + 0.20×12) / (8+12) = (7.2 + 2.4) / 20 = 9.6 / 20 = 0.48.

Formula

C_composite = Σ(Ci · Ai) / Σ(Ai)

Variables

Ci = runoff coefficient of sub-area i; Ai = area of sub-area i (ha or consistent units)

Application

Mixed land-use catchments with different surface types.

Example

i = 50 mm/hr = 50/(1000×3600) = 1.389×10⁻⁵ m/s; A = 20 ha = 200 000 m²; C = 0.6. Q = 0.6 × 1.389×10⁻⁵ × 200 000 = 1.667 m³/s. Same as the 360-factor result.

Formula

Q = C · i · A [base SI: i in m/s, A in m²]

Variables

Q in m³/s; i in m/s; A in m²

Application

Dimensional check or when given intensity in m/s directly — rarely used in board problems but useful for unit verification.

Exam Tips

  • Memorize Q = CiA/360 exactly — this is the most-tested formula in board hydrology problems.
  • When given C as separate sub-areas, always compute composite C before applying the rational method.
  • Double-check units: i in mm/hr? A in ha? Then divide by 360. Any other combination requires unit conversion first.
  • The 360 factor derivation: 1 mm/hr × 1 ha = (1/1000 m/hr) × (10 000 m²) = 10 m³/hr = 10/3600 m³/s ≈ 1/360 m³/s. Hence Q[m³/s] = CiA/360.
  • Board exam C values are always given or standard tabulated values — you will not derive C from first principles.

Key Points

  • The rational method gives the PEAK discharge Q from a catchment for a storm whose duration equals tc.
  • The fundamental formula in pure consistent units is Q = C·i·A, where i must be in m/s and A in m² to get Q in m³/s.
  • The board-exam practical SI form uses intensity in mm/hr and area in hectares, requiring the 360 conversion factor: Q = C·i·A / 360.
  • The runoff coefficient C is dimensionless (0 < C ≤ 1); it accounts for imperviousness, slope, and land cover.
  • For mixed catchments, the weighted (composite) C is: C_composite = Σ(Ci·Ai) / ΣAi.
  • The rational method is valid for small catchments (typically < 80 ha or < 200 ha in some references); for larger areas, use hydrograph methods.
  • The method assumes: uniform rainfall over the catchment, storm duration ≥ tc, and steady-state runoff at the outlet.
  • Typical C values: pavement/roofs 0.85–0.95; lawns/parks 0.10–0.35; agricultural land 0.20–0.40; urban mixed 0.40–0.70.

Definitions

Term

Runoff Coefficient (C)

Definition

The fraction of rainfall that becomes surface runoff, accounting for infiltration, evaporation, depression storage, and surface characteristics.

Importance

C is the most judgment-dependent value in the rational method; boards typically give it or ask you to compute a composite C.

Term

Design Intensity (i)

Definition

Rainfall intensity (mm/hr) corresponding to the design return period and a duration equal to tc, read from IDF curves.

Importance

Always use i at duration = tc, not the peak intensity of the IDF curve. Using a shorter duration overestimates i and Q.

Term

Peak Discharge (Q)

Definition

The maximum instantaneous flow rate at the outlet when the entire catchment is contributing, i.e., at time = tc.

Importance

Q is the design flow for sizing drainage structures — channels, pipes, culverts, inlets.

Section Title

The Rational Method — Peak Runoff

Common Mistakes

  • CRITICAL: Forgetting the 360 factor when i is in mm/hr and A is in ha — omitting it gives a result 360× too large.
  • Using the wrong area units: if A is given in km², convert to ha first (1 km² = 100 ha) before applying the formula.
  • Applying the rational method to large catchments (>500 ha) where the steady-state assumption breaks down; board problems will stay within the valid range.
  • Selecting rainfall intensity at a duration shorter than tc; the correct design condition is duration = tc.
  • Forgetting to compute composite C for mixed land-use catchments — using only one C value introduces significant error.

Formulas

Example

Storm P = 90 mm = 0.090 m on A = 8 km² = 8 × 10⁶ m² with C = 0.50. V = 0.50 × 0.090 × 8 × 10⁶ = 360 000 m³ = 360 ML.

Formula

V_runoff = C · P · A

Variables

V_runoff = total runoff volume (m³); C = runoff coefficient (dimensionless); P = total rainfall depth (m); A = catchment area (m²)

Application

Sizing retention/detention ponds, reservoir yield analysis, and flood volume estimation.

Example

C = 0.40, P = 80 mm. Runoff depth = 0.40 × 80 = 32 mm over the catchment.

Formula

Runoff depth = C · P

Variables

Runoff depth in same units as P (mm or m); C = runoff coefficient; P = rainfall depth

Application

Express runoff as a depth over the catchment — useful for comparing catchments of different sizes.

Exam Tips

  • For volume problems, the key steps are: convert P to metres, convert A to m², then multiply C × P × A.
  • Board problems on volume often ask for the answer in ML — always check the requested unit.
  • If P is given as intensity × duration (e.g., 25 mm/hr for 3 hr), compute P = 75 mm before using V = CPA.
  • Runoff volume and peak discharge use the same C but different storm parameters (P for volume, i for peak Q).

Key Points

  • While the rational method gives PEAK FLOW (m³/s), runoff volume is the total volume of water (m³) generated by a storm.
  • Runoff volume formula: V_runoff = C · P · A, where P is rainfall depth (m) and A is catchment area (m²).
  • This formula treats C as a volume ratio: fraction of total rainfall depth that becomes runoff volume.
  • Runoff depth = C × P (in mm or m), which equals the equivalent depth of runoff uniformly spread over the catchment.
  • Volume can be expressed in m³, ML (megalitres, 1 ML = 1000 m³), or Mm³.
  • Volume calculations are used for reservoir sizing, detention pond design, and flood routing.
  • The SCS Curve Number (CN) method is an alternative volume method, but the PRC board primarily tests the C·P·A approach.

Definitions

Term

Rainfall Depth (P)

Definition

Total accumulated rainfall during a storm event, expressed in mm (equivalent depth over the catchment area).

Importance

P is total storm rainfall — not intensity. Intensity i (mm/hr) × duration (hr) = P (mm) if intensity is uniform.

Term

Runoff Volume (V)

Definition

Total volume of stormwater generated from a catchment during a storm event.

Importance

Used for storage design; distinct from peak flow Q used for conveyance design.

Section Title

Runoff Volume

Common Mistakes

  • Using P in mm and A in m² directly — always convert P to metres before multiplying: P(m) = P(mm)/1000.
  • Confusing runoff depth (C·P, units of mm) with runoff volume (C·P·A, units of m³).
  • Using area in km² without converting to m² (1 km² = 1 × 10⁶ m²) — a factor of 10⁶ error.
  • Reporting volume in m³ when the problem asks for ML or vice versa without conversion (1 ML = 1000 m³).

Formulas

Example

Given i = 3000 / (tc + 20), tc = 30 min. i = 3000 / (30 + 20) = 3000 / 50 = 60 mm/hr. Use this i in Q = CiA/360.

Formula

i = a / (t_c + b)

Variables

i = design intensity (mm/hr); a, b = regression constants for the locality; t_c = time of concentration (minutes)

Application

Board problems that provide the IDF equation; substitute t_c to find i, then use in rational method.

Exam Tips

  • Board problems will provide either a table, a graph, or an equation for IDF — you will not need to recall specific intensity values.
  • Practice substituting tc into IDF equations quickly; this is a straightforward algebra step.
  • If tc and return period are given, select i directly from a provided table — no interpolation needed for most board problems.

Key Points

  • IDF curves relate rainfall intensity (mm/hr) to storm duration (hr or min) for various return periods (years).
  • Key relationship: intensity DECREASES as duration INCREASES for the same return period.
  • Key relationship: intensity INCREASES as return period INCREASES for the same duration.
  • Design return periods by structure type: residential drainage 2–10 yr; arterial roads 10–25 yr; major infrastructure 50–100 yr.
  • For the rational method, always select the intensity at duration = tc from the appropriate return-period curve.
  • IDF equations are often given in board problems in the form: i = a / (t + b)^n or i = kT^m / t^n, where t is duration in minutes and T is return period in years.
  • PAGASA provides IDF data for Philippine stations — local IDF data may differ significantly from generic values.

Definitions

Term

Return Period (Recurrence Interval, T)

Definition

Average number of years between events of equal or greater magnitude; probability of exceedance per year = 1/T.

Importance

Determines the design storm severity; selecting too low a T risks flooding, too high increases construction cost unnecessarily.

Term

IDF Curve

Definition

Graphical or tabular relationship between rainfall intensity (mm/hr), storm duration, and return period for a specific location.

Importance

Source of design intensity i in the rational method; Philippine IDF data are from PAGASA records.

Section Title

Rainfall Intensity-Duration-Frequency (IDF)

Common Mistakes

  • Using the peak intensity from the IDF curve regardless of tc — always pick i at duration = tc.
  • Reading IDF intensity at a different return period than specified in the design criteria.
  • Confusing intensity (mm/hr) with depth (mm) — intensity is a rate, depth is the total accumulated volume per unit area.

Formulas

Example

N = 50 000 persons, q = 200 L/person/day. ADD = 50 000 × 200 = 10 000 000 L/day = 10 000 m³/day.

Formula

ADD = N × q

Variables

ADD = average daily demand (L/day or m³/day); N = population (persons); q = per-capita consumption (L/person/day)

Application

Baseline demand for water supply system design.

Example

ADD = 10 000 m³/day. MDD = 1.5 × 10 000 = 15 000 m³/day.

Formula

MDD = 1.5 × ADD

Variables

MDD = maximum daily demand (same units as ADD); 1.5 = maximum day factor

Application

Design of water mains, transmission lines, and treatment plant capacity.

Example

ADD = 10 000 m³/day, f_ph = 2.5. PHD = 2.5 × 10 000 = 25 000 m³/day = 25 000/86 400 = 0.289 m³/s.

Formula

PHD = f_ph × ADD

Variables

PHD = peak hourly demand; f_ph = peak hour factor (2.0–3.0, default 2.5 unless given); ADD = average daily demand

Application

Design of service reservoirs and distribution system pipes that must meet instantaneous peak demand.

Example

ADD = 1 500 m³/day. ADD = 1 500 / 86 400 = 0.01736 m³/s = 17.36 L/s.

Formula

ADD (m³/s) = ADD (m³/day) / 86 400

Variables

86 400 = seconds per day

Application

Convert daily demand to flow rate for hydraulic calculations.

Exam Tips

  • Memorize: ADD = N × q; MDD = 1.5 × ADD; PHD = 2.5 × ADD (unless the problem specifies a different factor).
  • For unit conversions: 1 m³/day = 1/86 400 m³/s ≈ 1.157 × 10⁻⁵ m³/s; 1 L/s = 86.4 m³/day.
  • Board problems on water demand are straightforward multiplication — set up unit conversions carefully and you will not lose marks.
  • If the problem asks for the design flow of a pipe serving a community, use MDD or PHD, not ADD.
  • Fire demand problems add a fixed fire flow (L/s) to MDD — check if the problem specifies fire demand consideration.

Key Points

  • Water demand is classified as: average daily demand (ADD), maximum daily demand (MDD), and peak hourly demand (PHD).
  • ADD = population × per-capita consumption (L/person/day). Typical Philippine per-capita: 150–250 L/person/day for urban areas.
  • MDD ≈ 1.5 × ADD (maximum day factor = 1.5).
  • PHD ≈ 2.0 to 3.0 × ADD (peak hour factor = 2–3, commonly 2.5 for board problems unless stated).
  • Design pipelines and pumps for MDD; design storage reservoirs for PHD or fire demand.
  • Water sources: surface water (rivers, lakes, reservoirs) and groundwater (wells, springs).
  • Well yield: the sustainable pumping rate without excessive drawdown, governed by Darcy's Law and Dupuit-Thiem well equations.
  • Reservoir sizing: mass-balance (Rippl diagram) — accumulate (inflow − demand) over time to find required storage volume.
  • Per-capita demand increases with economic development; Philippine water utilities (governed by LWUA and local ordinances) target 150–250 LCD for service levels.
  • Fire demand is an additional design consideration (not per capita) — typically added to MDD for distribution system design.

Definitions

Term

Average Daily Demand (ADD)

Definition

The mean volume of water consumed per day, equal to annual consumption divided by 365.

Importance

Baseline design parameter; all peak factors multiply ADD to give peak demands.

Term

Maximum Daily Demand (MDD)

Definition

The highest daily consumption observed (or designed for) in a year; typically 1.5× ADD.

Importance

Governs capacity of water treatment plants and transmission mains.

Term

Peak Hourly Demand (PHD)

Definition

The highest hourly demand in a day; typically 2–3× ADD expressed as an equivalent daily rate.

Importance

Governs sizing of distribution pipes and service reservoirs to prevent pressure deficiency during morning/evening peaks.

Term

Per-Capita Consumption (q)

Definition

Volume of water used per person per day (L/person/day or LCD), covering domestic, commercial, industrial, and leakage.

Importance

Primary input for demand calculation; varies by climate, income, metering, and pressure.

Term

Well Yield

Definition

The rate at which a well can supply water without excessive drawdown or aquifer depletion; expressed in L/s or m³/day.

Importance

Limits groundwater source capacity; must exceed peak demand for the supply zone.

Section Title

Water Supply Fundamentals — Demand and Sources

Common Mistakes

  • Using per-capita consumption in L/person/day without converting to m³/person/day when computing demand in m³: 1 L = 0.001 m³.
  • Forgetting to multiply ADD by the peak factor — designing for ADD alone will cause shortfalls during peak demand.
  • Confusing MDD (1.5× ADD for daily variation) with PHD (2–3× ADD for hourly variation); PHD is always larger.
  • Not converting daily demand to m³/s when comparing with pipeline capacity: divide m³/day by 86 400.
  • Assuming per-capita = domestic only; in Philippine water supply, per capita includes domestic + commercial + unaccounted-for water (UFW).

Formulas

Example

SOLUTION: Q = (0.75 × 80 × 35) / 360 = 2100 / 360 = 5.83 m³/s. Answer: 5.83 m³/s.

Formula

Q = (C · i · A) / 360

Variables

Problem 1 — Peak discharge for urban catchment

Application

35 ha urban catchment, C = 0.75, i = 80 mm/hr.

Example

SOLUTION: P = 0.120 m; A = 12 × 10⁶ m². V = 0.45 × 0.120 × 12 × 10⁶ = 648 000 m³ = 648 ML. Answer: 648 000 m³ (648 ML).

Formula

V = C · P · A

Variables

Problem 2 — Runoff volume from large catchment

Application

P = 120 mm on A = 12 km², C = 0.45.

Example

SOLUTION: ADD = 250 000 × 200 = 50 000 000 L/day = 50 000 m³/day. PHD = 2.5 × 50 000 = 125 000 m³/day = 125 000 / 86 400 = 1.447 m³/s. Answer: 125 000 m³/day or 1.447 m³/s.

Formula

PHD = f_ph × N × q

Variables

Problem 3 — Peak hourly demand

Application

Population 250 000, q = 200 L/c/day, peak hour factor = 2.5.

Example

SOLUTION: A_pavement = 0.40 × 20 = 8 ha; A_lawn = 0.60 × 20 = 12 ha. C = (0.90×8 + 0.20×12) / 20 = (7.2 + 2.4) / 20 = 9.6 / 20 = 0.48. Answer: C = 0.48.

Formula

C_composite = Σ(Ci·Ai) / ΣAi

Variables

Problem 4 — Composite runoff coefficient

Application

40% pavement (C = 0.90) and 60% lawn (C = 0.20). Total area = 20 ha.

Exam Tips

  • For Problem 4 shortcut: C_composite = (0.90)(0.40) + (0.20)(0.60) = 0.36 + 0.12 = 0.48 — using area fractions directly gives the same answer and is faster.
  • Always re-read the problem to check if the answer is needed in m³/s, m³/day, L/s, or another unit — unit mismatch is the top source of errors.
  • Write the formula first, then substitute — this habit prevents formula errors under exam pressure.

Key Points

  • Practice all three formula types: Q (peak runoff), V (runoff volume), and water demand.
  • Always write out given data, identify the formula, substitute with units, and check the answer for reasonableness.
  • Typical board problem structure: one unknown, all other values given, single formula needed.

Section Title

Worked Board-Style Problems

Common Mistakes

  • In composite C problems, using total area = 100% (fraction) instead of actual hectares — both give the same result only if computed correctly, but using fractions directly: C = Σ(Ci × fi) where fi is fraction, is an equivalent shortcut.
  • In Problem 3, forgetting to divide by 86 400 to get m³/s when the question asks for flow rate rather than daily volume.

Connections

  • Rational method peak discharge Q feeds directly into open-channel hydraulics (Manning's equation) to size drainage channels and culverts — a cross-chapter connection tested in combined board problems.
  • Runoff volume calculations connect to reservoir routing and mass-balance problems in hydrology — the same V = CPA framework applies to flood storage design.
  • Water demand (m³/s) connects to pipe flow (Hazen-Williams, Darcy-Weisbach) for distribution system design — demand is the design discharge for water supply pipes.
  • The IDF concept (intensity-duration-frequency) connects to statistics and probability — return period T is the inverse of annual exceedance probability, a topic in engineering statistics.
  • Runoff coefficient C conceptually links to soil mechanics (infiltration capacity, permeability of soil layers) and land use planning (imperviousness of urban surfaces).
  • Well hydraulics (Thiem equation) connects to groundwater flow and Darcy's Law from the fluid mechanics portion of the board exam.
  • The hydrologic cycle (evapotranspiration) connects to environmental engineering and water resources management, tested in the CE board under Sanitary Engineering.
  • Mass balance (P = R + F + E + ΔS) is an application of the continuity equation — the fundamental conservation law taught in fluid mechanics.

Exam Strategy

For the PRC CE board exam on Hydrology and Water Supply: (1) Prioritize the rational method formula Q = CiA/360 — it appears in nearly every board set; write it from memory and verify units before substituting. (2) For runoff volume, always convert P to metres and A to m² as the very first step to avoid unit errors. (3) For water demand, set up the problem as ADD = N×q, then multiply by the appropriate factor (1.5 for MDD, 2.0–3.0 for PHD); convert to m³/s if flow rate is required by dividing by 86 400. (4) For composite C problems, use the shortcut C = Σ(Ci × fi) where fi are area fractions — it saves 30 seconds per problem. (5) If an IDF equation is given, substitute tc immediately to get i, then proceed with the rational method — these are two-step problems. (6) Avoid overthinking: hydrology board problems are formula-direct with one unknown; identify what is asked, pick the right formula, convert units, and compute. (7) Allocate no more than 3–4 minutes per hydrology problem — they are among the more straightforward numerical problems on the board.

Quick Review Questions

A 20 ha catchment has C = 0.60 and design intensity i = 50 mm/hr. What is the peak discharge?

Apply Q = CiA/360 = (0.60 × 50 × 20) / 360 = 600 / 360 = 1.667 m³/s. The 360 factor is mandatory when i is in mm/hr and A is in hectares.

A storm deposits 80 mm of rain on a 5 km² catchment with C = 0.40. What is the total runoff volume?

Convert: P = 0.080 m; A = 5 × 10⁶ m². V = 0.40 × 0.080 × 5 000 000 = 160 000 m³. Check: 160 000 m³ = 160 ML.

A town of 10 000 people consumes 150 L/person/day. What is the maximum daily demand?

ADD = 10 000 × 150 = 1 500 000 L/day = 1500 m³/day. MDD = 1.5 × 1500 = 2250 m³/day.

What is the composite runoff coefficient of a catchment that is 40% pavement (C = 0.90) and 60% lawn (C = 0.20)?

C = (0.90 × 0.40) + (0.20 × 0.60) = 0.36 + 0.12 = 0.48. Using area fractions directly (since proportions sum to 1) is the fastest approach.

Why is it incorrect to use Q = C·i·A (without the 360 factor) when i is in mm/hr and A is in ha?

Dimensional check: 1 mm/hr × 1 ha = (1/1000 m/hr) × 10 000 m² = 10 m³/hr = 10/3600 m³/s = 1/360 m³/s. Hence Q[m³/s] = C·i[mm/hr]·A[ha]/360.

A water utility serves 250 000 people at 200 L/c/day. What is the peak hourly demand in m³/s using a peak hour factor of 2.5?

ADD = 250 000 × 200 = 50 × 10⁶ L/day = 50 000 m³/day. PHD = 2.5 × 50 000 = 125 000 m³/day. Converting: 125 000 / 86 400 = 1.447 m³/s.

What does the time of concentration (tc) govern in the rational method?

The rational method assumes the entire catchment contributes simultaneously, which occurs when storm duration = tc. Using a shorter duration gives a higher (non-representative) intensity; using a longer duration underestimates i for the design condition.

An IDF equation gives i = 4000/(tc + 25) mm/hr. For tc = 35 min, find i and then the peak discharge from a 15 ha catchment with C = 0.70.

Step 1: i = 4000/(35 + 25) = 4000/60 = 66.67 mm/hr. Step 2: Q = (0.70 × 66.67 × 15)/360 = 700/360 = 1.944 m³/s.

What are the three key water demand levels and their standard multipliers relative to ADD?

These multipliers reflect daily and hourly variations in consumption. MDD governs pipeline and treatment plant capacity; PHD governs distribution system and service reservoir sizing.

A 100 mm storm falls on a catchment of 8 km². The runoff coefficient is 0.55. Express the runoff volume in megalitres (ML).

P = 0.100 m; A = 8 × 10⁶ m². V = 0.55 × 0.100 × 8 × 10⁶ = 440 000 m³. Converting: 440 000 m³ ÷ 1000 = 440 ML (since 1 ML = 1000 m³).

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