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CELE Hydraulics & Fluid Mechanics — Hydrology and Water SupplyDetailed Explanation

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Detailed explanation of Hydrology and Water Supply for the CELE 2026. Full depth, full reasoning — exactly what you need when Professional Regulation Commission (PRC) — Board of Civil Engineering tests this chapter with applied or scenario-based questions in the CELE Hydraulics & Fluid Mechanics subtest.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Hydraulics & Fluid Mechanics section sits under a "Core" weighting, and Hydrology and Water Supply is the 10th chapter in the 10-chapter CELE Hydraulics & Fluid Mechanics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Hydraulics & Fluid Mechanics.

Hydrology and Water Supply - Detailed Explanation

Hydrology is the science that quantifies how precipitation becomes the runoff that civil engineers must safely convey, store, and distribute. For the PRC Civil Engineer Licensure Examination, this chapter is one of the highest-yield topics in Hydraulics and Fluid Mechanics. Examinees are expected to apply the Rational Method for peak runoff, compute runoff volumes, and solve basic water-demand and well-hydraulics problems — all in SI units. This explanation covers every concept systematically, from the hydrologic cycle down to water-supply demand calculations, with board-style worked problems, pitfall warnings, and exam strategy.

Concepts

The Hydrologic Cycle

The hydrologic cycle describes the continuous movement of water in the Earth–atmosphere system. For engineering purposes, the cycle is simplified into: Precipitation (P) → Interception and Infiltration → Surface Runoff (Q) → Streamflow → Evapotranspiration (ET) → back to Precipitation. The water balance for a catchment is: P = Q + ET + ΔS, where ΔS is change in storage (groundwater and soil moisture). Engineers are primarily concerned with the runoff fraction because that is what floods roads, fills reservoirs, and recharges wells. The key insight is that not all rain becomes runoff; the portion that does depends on land cover, soil type, antecedent moisture, and storm characteristics. In the Philippine context, typhoon-driven rainfall events (like those hitting Luzon and Visayas) produce very high intensities over short durations, making proper hydrologic design critical for drainage and flood control.

Examples

This illustrates that the runoff fraction (100/200 = 0.50) depends on losses. The effective runoff coefficient C ≈ 0.50 for this event.

Scenario

A catchment receives 200 mm of rainfall. Evapotranspiration accounts for 70 mm and soil storage increases by 30 mm. What is the surface runoff depth?

Solution

Using the water balance: P = Q + ET + ΔS → 200 = Q + 70 + 30 → Q = 100 mm. Runoff depth = 100 mm.

Applications

Sizing storm drains and culverts for road crossings (DPWH standard drainage design).
Flood routing and flood-plain delineation for urban planning under RA 9729 (Climate Change Act).
Reservoir yield analysis for water-supply projects.
Groundwater recharge estimation for well-field design.

Misconceptions

Misconception: All rainfall becomes runoff. Reality: Infiltration, evaporation, and storage absorb a significant fraction first.
Misconception: Evaporation and transpiration are separate engineering parameters in basic calculations. Reality: They are combined as evapotranspiration (ET) for catchment water-balance work.
Misconception: The hydrologic cycle is unaffected by land use. Reality: Urbanisation increases imperviousness, dramatically raising C and peak runoff.

Related Concepts

Runoff coefficient (C)
Rational Method
Infiltration and Green-Ampt model
Unit hydrograph theory

Common Exam Questions

Example

Which process returns water from the land surface to the atmosphere without runoff? Answer: Evapotranspiration.

Approach

Identify which component of the hydrologic cycle corresponds to a described process.

Question Type

Conceptual identification

Key Points To Remember

Water balance: P = Q + ET + ΔS; for design purposes often simplified to Q = CP (C = runoff coefficient).
Evapotranspiration (ET) = evaporation from surfaces + transpiration from plants; it reduces available runoff.
Infiltration fills soil pores before runoff begins; hence initial losses reduce peak runoff.
The hydrologic cycle is driven by solar energy (evaporation) and gravity (precipitation, runoff).
For exam purposes, focus on the runoff-generating part of the cycle, not atmospheric processes.

The Rational Method — Peak Runoff

The Rational Method is the standard technique for estimating the peak discharge from small catchments (typically < 80 ha for urban drainage). It is the most frequently tested hydrologic method in the PRC board exam. The governing formula in SI practical units is: Q = (C × i × A) / 360 where: • Q = peak discharge (m³/s) • C = dimensionless runoff coefficient (0.10 for heavily vegetated to 0.95 for impervious pavement) • i = rainfall intensity (mm/hr) for a storm duration equal to the time of concentration tc • A = catchment area (hectares, ha) • 360 = unit-conversion constant (= 1000 mm/m × 1 hr/3600 s × 1 m²/10000 m² per ha × 10000) Derivation of the constant 360: Q [m³/s] = C × (i/1000 m/mm) × (1/3600 hr/s) × (A × 10000 m²/ha) = C × i × A × (10000 / (1000 × 3600)) = C × i × A / 360 The method assumes: (1) the storm lasts at least as long as tc so the entire catchment contributes; (2) intensity is uniform over the catchment; (3) the return period of Q equals the return period of i; (4) the runoff coefficient C accounts for all losses. Time of Concentration (tc): The travel time for runoff from the hydraulically most remote point in the catchment to the outlet. Design intensity i is taken from the IDF curve at this duration. Longer tc → lower i → lower Q (for the same storm frequency). Runoff Coefficient (C) typical values: • Pavement / rooftops: 0.75–0.95 • Urban residential (lawns, some pavement): 0.25–0.50 • Agricultural / cultivated: 0.20–0.40 • Forests / heavy vegetation: 0.10–0.20 For composite catchments (multiple land uses), the weighted average C is: C_weighted = Σ(Ci × Ai) / A_total

Examples

This is the standard board-exam template. Always verify units before substituting: C is dimensionless, i is in mm/hr, A is in ha. The answer in m³/s is direct.

Scenario

A 20 ha urban catchment has C = 0.60 and design rainfall intensity i = 50 mm/hr. Compute the peak runoff.

Solution

Q = CiA/360 = (0.60 × 50 × 20) / 360 = 600/360 = 1.67 m³/s

Higher C (more impervious urban area) and higher intensity produce significantly larger peak flows, illustrating the sensitivity of Q to both parameters.

Scenario

A 35 ha urban catchment has C = 0.75 and design intensity i = 80 mm/hr. Find the peak discharge.

Solution

Q = CiA/360 = (0.75 × 80 × 35) / 360 = 2100/360 = 5.83 m³/s

Always compute C_weighted for composite catchments first. This is a commonly tested multi-step board problem.

Scenario

A mixed 10 ha catchment is 40% pavement (C₁ = 0.90) and 60% lawn (C₂ = 0.20). Design intensity = 60 mm/hr. Find Q.

Solution

A₁ = 0.40 × 10 = 4 ha; A₂ = 0.60 × 10 = 6 ha. C_weighted = (0.90×4 + 0.20×6) / 10 = (3.60 + 1.20) / 10 = 4.80/10 = 0.48 Q = (0.48 × 60 × 10) / 360 = 288/360 = 0.80 m³/s

Applications

Design of storm sewers, culverts, and roadside channels per DPWH Drainage Design Guidelines.
Sizing inlet grates and catch basins in urban subdivisions.
Flood control channel design for barangay-level drainage projects.
PRC board exam: Rational Method problems appear in almost every exam cycle.

Misconceptions

Using Q=CiA without the 360 divisor when i is in mm/hr and A in ha — this gives answers ~360× too large.
Using the design storm duration as tc without justification — tc must be the actual travel time from the catchment's farthest point.
Assuming C is constant regardless of storm intensity — in reality, C increases for longer-return-period storms, but for board exams use the given C.
Forgetting to convert km² to ha (1 km² = 100 ha) when the area is given in km².

Related Concepts

Time of concentration (tc)
Rainfall IDF curves
Runoff coefficient C
Runoff volume calculation

Common Exam Questions

Example

Given: C=0.65, i=75 mm/hr, A=0.50 km². Find Q. Solution: Convert A: 0.50 km² = 50 ha. Q = (0.65×75×50)/360 = 2437.5/360 = 6.77 m³/s.

Approach

Substitute given C, i, A directly into Q = CiA/360. Watch for unit traps (km² given instead of ha).

Question Type

Direct application

Example

40% pavement (C=0.9), 60% lawn (C=0.2), A=10 ha, i=60 mm/hr → C_w=0.48, Q=0.80 m³/s.

Approach

Compute area-weighted C first, then apply Q = CiA/360.

Question Type

Composite catchment

Example

Q=2.50 m³/s, C=0.70, A=25 ha. Find i. i = Q×360/(C×A) = 2.50×360/(0.70×25) = 900/17.5 = 51.4 mm/hr.

Approach

Given Q and two of the three parameters, solve for the unknown.

Question Type

Back-calculation

Key Points To Remember

Formula: Q = CiA/360 (i in mm/hr, A in ha, Q in m³/s). The constant 360 is non-negotiable for these units.
The design intensity i is read from IDF curves at duration = tc and the chosen return period.
C is dimensionless; higher imperviousness = higher C.
For composite catchments: C_weighted = Σ(Ci × Ai) / ΣAi.
Rational method is valid for small, urban catchments (< 80 ha). For larger catchments, use unit hydrograph or SCS-CN.
Pure formula Q = CiA (without dividing by 360) only applies when i is in m/s and A is in m² — never mix units on the board exam.

Runoff Volume

While the Rational Method gives peak discharge (m³/s), engineers also need the total volume of runoff from a storm event — essential for reservoir sizing, detention basin design, and water-harvesting projects. The runoff volume equation is: V_runoff = C × P × A where: • C = runoff coefficient (dimensionless) • P = storm rainfall depth (m or mm — keep consistent with A units) • A = catchment area (m² if P in metres, or consistent units) In practical SI form with P in mm and A in m²: V_runoff [m³] = C × (P/1000) × A Or with P in mm and A in km²: V_runoff [m³] = C × P × A × 1000 (because 1 km² = 10⁶ m² and 1 mm = 10⁻³ m → net factor = 10³) Physical meaning: V_runoff is the fraction C of the total precipitation volume (P × A) that becomes surface runoff. The remainder (1–C)×P×A is lost to infiltration and evaporation. Key distinction from Rational Method: • Rational Method → peak flow rate Q [m³/s] (intensity-based) • Runoff volume equation → total volume V [m³] (depth-based) Both C values are numerically the same in simple analyses.

Examples

Convert km² to m² and mm to m first. The answer 648,000 m³ (648 ML) represents the volume that must be safely routed or stored.

Scenario

A 120 mm storm falls on a 12 km² catchment with C = 0.45. Find the runoff volume in m³ and in megalitres (ML).

Solution

A = 12 km² = 12 × 10⁶ m²; P = 120 mm = 0.12 m V = C × P × A = 0.45 × 0.12 × 12×10⁶ = 648,000 m³ In ML: 648,000 m³ ÷ 1000 = 648 ML

Straightforward unit-consistent solution. Always convert P to metres when A is in m².

Scenario

An 80 mm storm on a 5 km² catchment (C = 0.40). Find runoff volume.

Solution

V = C × (P/1000) × A = 0.40 × 0.080 m × 5×10⁶ m² = 160,000 m³

Applications

Sizing rainwater harvesting tanks for residential and agricultural use.
Detention/retention basin design to attenuate peak flows in new subdivisions.
Reservoir active-storage volume estimation for municipal water supply.
Runoff-volume-based irrigation scheduling.

Misconceptions

Using intensity i (mm/hr) instead of rainfall depth P (mm) in the volume formula — the two are entirely different.
Forgetting to convert km² to m², giving answers 10⁶ times too small.
Confusing V_runoff with total precipitation volume (P×A); runoff volume is only the fraction C of that total.

Related Concepts

Rational Method peak discharge
Reservoir mass-balance analysis
SCS Curve Number method (for larger catchments)

Common Exam Questions

Example

P=100 mm, A=8 km², C=0.50. V = 0.50×0.10×8×10⁶ = 400,000 m³.

Approach

Convert units first (mm→m, km²→m²), then apply V = C×P×A.

Question Type

Direct volume computation

Example

V=648,000 m³ = 648 ML = 0.648 GL.

Approach

Express final answer in ML or GL as requested (1 ML = 1,000 m³; 1 GL = 10⁶ m³).

Question Type

Unit conversion of result

Key Points To Remember

V_runoff = C × P × A; strictly verify units before computing.
P is storm depth (mm or m), not intensity (mm/hr) — do not confuse with the Rational Method input.
1 mm of rain on 1 ha = 10 m³ of water (useful check: 1 ha = 10,000 m², 0.001 m × 10,000 = 10 m³).
For A in km² and P in mm: V [m³] = C × P × A × 1000.
Runoff volume is the basis for reservoir and detention pond sizing.

Rainfall Intensity-Duration-Frequency (IDF) Curves

IDF curves relate rainfall intensity (mm/hr) to storm duration (minutes or hours) and return period (years). They are the source of the design intensity i used in the Rational Method. Key relationships: 1. Intensity vs Duration: For a fixed return period, intensity DECREASES as duration increases. A 15-minute storm at 100 mm/hr has the same return period as a 1-hour storm at perhaps 50 mm/hr. 2. Intensity vs Return Period: For a fixed duration, intensity INCREASES with longer return period (rarer, more intense storms). 3. Design Storm Return Period: Selected based on the consequence of failure — minor drainage: 5–10 years; major drainage/culverts: 25–50 years; critical infrastructure: 100 years. For the board exam, IDF curve values are always given in the problem statement. You do not need to construct IDF curves — only use the given i at the specified return period and at duration = tc. The design procedure: 1. Determine tc for the catchment. 2. Select return period based on design standard. 3. Read i from the IDF curve at (tc, return period). 4. Apply Q = CiA/360.

Examples

The design intensity is always taken at duration equal to tc. Using a shorter duration would give higher (unconservative for volume, but used for peak) intensity; using a longer duration gives lower intensity and underestimates peak.

Scenario

From an IDF table, a 30-min storm in Manila at 25-year return period gives i = 90 mm/hr. The tc for a catchment is 30 min. What intensity should be used for design?

Solution

Design intensity i = 90 mm/hr (read directly at duration = tc = 30 min, T = 25 years).

Applications

Selection of design intensity for storm sewer design.
Flood frequency analysis for bridge design.
Stormwater management planning in LLDA-regulated watersheds.

Misconceptions

Using intensity at an arbitrary duration (e.g., 1 hour) instead of at tc — this underestimates peak Q when tc < 1 hr.
Confusing return period with frequency (T=25 yr does NOT mean the storm occurs exactly every 25 years; it means 4% annual probability).

Related Concepts

Time of concentration (tc)
Rational Method
Flood frequency analysis

Common Exam Questions

Example

tc=45 min, T=10 yr, IDF table gives i=65 mm/hr at (45 min, 10 yr) → use i=65 mm/hr in Q=CiA/360.

Approach

Read i from a given IDF table at the tc and return period specified, then use in Rational Method.

Question Type

Interpretation and application

Key Points To Remember

Intensity DECREASES with increasing storm duration (for fixed return period).
Intensity INCREASES with increasing return period (for fixed duration).
Design duration = tc (time of concentration), not an arbitrary duration.
Return period T = 1/p, where p = annual exceedance probability. T=50 yr → 2% annual chance.
IDF values are locality-specific; PAGASA provides IDF curves for Philippine stations.

Water Supply Fundamentals — Demand and Sizing

Water-supply engineering ensures that adequate clean water is available to meet the varying demands of a community. The PRC board exam tests three core calculations: average daily demand, peak demands, and basic reservoir or pipe sizing. 1. AVERAGE DAILY DEMAND Q_avg [L/day] = Population × Per-capita consumption [L/person/day] Typical per-capita rates in the Philippines: 150–200 L/person/day (NWRB standards). 2. PEAK DEMAND FACTORS Demand varies by time of day and season. Engineers must size systems for peak, not average: Max daily demand = 1.5 × Q_avg (or as specified by local code) Peak hourly demand = 2.0 to 3.0 × Q_avg (typical factor 2.5) Max hourly demand = (Peak factor) × Q_avg These multipliers account for the fact that demand is not constant — morning and evening peaks require higher capacity. 3. UNITS CONVERSION (critical for the exam) 1 m³/day = 1000 L/day Q [m³/s] = Q [L/day] / 86,400,000 Q [L/s] = Q [L/day] / 86,400 4. WATER SOURCES • Surface water (rivers, reservoirs): Requires treatment; seasonal variability. • Groundwater (wells): Relatively reliable; pumping governed by Darcy's Law and Thiem's equation for steady radial flow. 5. WELL HYDRAULICS (basic) For steady-state radial flow to a well in a confined aquifer (Thiem equation): Q = [2π × k × b × (h₂ - h₁)] / ln(r₂/r₁) where k = hydraulic conductivity, b = aquifer thickness, h = piezometric head, r = radial distance. For the board exam at this level, well problems usually ask for yield from given drawdown, using the above formula or a simplified version.

Examples

This three-part format is the most common board-exam demand problem. Note the conversion from m³/day to L/s in part (c): divide L/day by 86,400.

Scenario

A town of 10,000 people consumes 150 L/person/day. Find the (a) average daily demand, (b) maximum daily demand, and (c) peak hourly demand (factor = 2.5).

Solution

(a) Q_avg = 10,000 × 150 = 1,500,000 L/day = 1,500 m³/day (b) Max daily = 1.5 × 1,500 = 2,250 m³/day (c) Peak hourly = 2.5 × 1,500 = 3,750 m³/day = 3,750,000 L/day = 3,750,000/86,400 = 43.4 L/s

For large cities, the answer in m³/s becomes significant — useful for sizing transmission mains and treatment plants.

Scenario

A city of 250,000 people uses 200 L/c/day. Find average demand in m³/s and peak hourly demand (factor 2.5) in m³/s.

Solution

Q_avg = 250,000 × 200 = 50,000,000 L/day = 50,000 m³/day Q_avg [m³/s] = 50,000 / 86,400 = 0.579 m³/s Peak hourly = 2.5 × 0.579 = 1.447 m³/s

Applications

Sizing water transmission mains and service reservoirs for LGU water districts.
Water treatment plant capacity determination.
Fire-flow demand calculations (fire demand added to peak domestic demand).
Design of elevated storage tanks (typically sized for peak hour minus average inflow).

Misconceptions

Designing for average daily demand — always design conveyance and distribution for peak demand.
Forgetting that 1 m³/day ≠ 1 L/s (1 m³/day = 0.01157 L/s); the conversion factor 86,400 is essential.
Applying the max-day factor (1.5) when the problem asks for peak-hour demand — use the specified peak-hour factor (2.0–3.0).
Ignoring fire demand in total demand calculation for distribution system design.

Related Concepts

Population projection methods
Water distribution network (Hardy-Cross method)
Well hydraulics and Thiem equation
Service reservoir sizing

Common Exam Questions

Example

Pop=50,000, 180 L/c/day, peak factor=2.0. Q_avg=9,000,000 L/day; Peak=18,000,000 L/day=208.3 L/s.

Approach

Compute Q_avg = Pop × per-capita, apply peak factor, then convert to required units.

Question Type

Demand calculation with unit conversion

Example

P₀=20,000, r=2%/yr, n=20 yr → P=20,000×(1.02)²⁰=29,737. Q_avg=29,737×150=4,460,530 L/day.

Approach

Project future population (geometric growth: P=P₀(1+r)ⁿ), then compute demand.

Question Type

Population projection + demand

Key Points To Remember

Q_avg = Population × Per-capita rate (L/person/day).
Max day = 1.5 × Q_avg; Peak hour = 2.0–3.0 × Q_avg (use given factor).
Always convert the final answer to the unit requested (m³/day, L/s, m³/s).
Design distribution mains for peak hourly demand; design transmission mains for max daily demand.
Per-capita consumption in Philippine design practice: 150–200 L/person/day (NWRB).
1 m³ = 1,000 L; 1 day = 86,400 seconds — memorise these conversions.

Practice Problems

Standard application of the Rational Method. The 360 divisor converts units (i in mm/hr, A in ha) to give Q in m³/s. Always verify all three inputs are in correct units before computing.

Problem

PROBLEM 1 (Rational Method — Direct). A 35 ha urban catchment has a runoff coefficient C = 0.75 and a design rainfall intensity of i = 80 mm/hr. Determine the peak runoff discharge in m³/s.

Solution

Q = CiA / 360 Q = (0.75 × 80 × 35) / 360 Q = 2100 / 360 Q = 5.83 m³/s

Always compute area-weighted C before applying Q = CiA/360. This two-stage process is the most common board-exam Rational Method variant.

Problem

PROBLEM 2 (Composite Catchment). A 15 ha catchment consists of 40% pavement (C₁ = 0.90) and 60% lawn (C₂ = 0.20). Design intensity = 70 mm/hr. Find the peak discharge.

Solution

Step 1 — Compute sub-areas: A₁ (pavement) = 0.40 × 15 = 6.0 ha A₂ (lawn) = 0.60 × 15 = 9.0 ha Step 2 — Weighted C: C_w = (C₁A₁ + C₂A₂) / A_total C_w = (0.90×6.0 + 0.20×9.0) / 15 C_w = (5.40 + 1.80) / 15 = 7.20 / 15 = 0.48 Step 3 — Peak discharge: Q = C_w × i × A / 360 = (0.48 × 70 × 15) / 360 Q = 504 / 360 = 1.40 m³/s

Unit conversion is critical: km² must become m², and mm must become m. The two-step unit conversion before applying V=CPA is the most common source of errors on the board exam.

Problem

PROBLEM 3 (Runoff Volume). A storm of depth 120 mm falls on a 12 km² catchment with C = 0.45. Find the runoff volume in (a) m³ and (b) megalitres (ML).

Solution

Convert units: A = 12 km² = 12 × 10⁶ m² P = 120 mm = 0.120 m (a) V = C × P × A = 0.45 × 0.120 × 12×10⁶ V = 0.45 × 0.120 × 12,000,000 V = 648,000 m³ (b) 1 ML = 1,000 m³ V = 648,000 / 1,000 = 648 ML

Three-part demand problem — the most common format. Note that in (c), multiply the average L/day by the peak factor first, then divide by 86,400 to get L/s. Do not apply the peak factor to m³/day and then convert — be consistent with units throughout.

Problem

PROBLEM 4 (Water Demand). A municipality has a population of 80,000 people with a per-capita water consumption of 175 L/person/day. Determine: (a) average daily demand in m³/day, (b) maximum daily demand in m³/day, and (c) peak hourly demand in L/s (use factor = 2.5).

Solution

(a) Q_avg = 80,000 × 175 = 14,000,000 L/day = 14,000 m³/day (b) Max daily = 1.5 × 14,000 = 21,000 m³/day (c) Peak hourly demand (in L/day): Q_peak = 2.5 × 14,000,000 = 35,000,000 L/day Convert to L/s: Q = 35,000,000 / 86,400 = 405 L/s

Back-calculation problems test algebraic manipulation of the Rational Method formula. Since 0 < C = 0.56 < 1.0, the answer is physically reasonable. A value > 1.0 would indicate an error in the computation.

Problem

PROBLEM 5 (Back-calculation). The peak discharge from a 25 ha catchment is measured as Q = 3.50 m³/s during a storm with i = 90 mm/hr. Determine the effective runoff coefficient C.

Solution

From Q = CiA/360, solve for C: C = Q × 360 / (i × A) C = 3.50 × 360 / (90 × 25) C = 1260 / 2250 C = 0.56

Population projection before demand computation is a frequently tested multi-step problem. The geometric growth formula P=P₀(1+r)ⁿ is standard for board-exam purposes. Always round population to whole persons before computing demand.

Problem

PROBLEM 6 (Population Projection + Demand). A barangay currently has 5,000 residents growing at 3% per year. The per-capita consumption is 160 L/person/day. Find the maximum daily demand after 10 years (max-day factor = 1.5).

Solution

Step 1 — Future population (geometric growth): P = P₀ (1 + r)ⁿ = 5,000 × (1.03)¹⁰ (1.03)¹⁰ = 1.3439 P = 5,000 × 1.3439 = 6,720 persons Step 2 — Average daily demand: Q_avg = 6,720 × 160 = 1,075,200 L/day = 1,075.2 m³/day Step 3 — Maximum daily demand: Q_max = 1.5 × 1,075.2 = 1,612.8 m³/day ≈ 1,613 m³/day

Exam Preparation Tips

MEMORISE the Rational Method formula Q = CiA/360 and know WHY the divisor is 360 — it comes from unit conversion (mm/hr and ha to m³/s). Examiners sometimes give the formula without the 360 and change units to trap unwary examinees.
ALWAYS check units first. Write down Q [=] m³/s, C [=] dimensionless, i [=] mm/hr, A [=] ha before plugging numbers. This prevents the #1 mistake on the board exam.
For COMPOSITE CATCHMENTS, never average C values directly — always compute C_weighted = Σ(CᵢAᵢ)/ΣAᵢ.
For RUNOFF VOLUME problems, remember P is depth (mm), not intensity (mm/hr). Write V = C × P × A and convert P to metres or A to the unit consistent with P before computing.
DEMAND PROBLEMS: memorise the tier of multipliers — avg → max day (×1.5) → peak hour (×2.0 to 3.0). Never design a water distribution system for average demand.
UNIT CONVERSIONS to drill before exam day: 1 km² = 100 ha = 10⁶ m²; 1 m³ = 1,000 L; 1 day = 86,400 s; 1 ML = 1,000 m³.
For IDF-related questions, the design intensity is ALWAYS at duration = time of concentration tc, never at an arbitrary duration. If tc is not given explicitly, it must be calculated (Kirpich, FAA, or other given equation).
Practice BACK-CALCULATION: isolate C, i, or A from Q = CiA/360. This algebraic manipulation appears frequently in multiple-choice problems where Q and two of the three parameters are given.
DIMENSIONAL ANALYSIS is your safety net. After computing any answer, perform a rough magnitude check: For a 20-ha urban catchment with moderate rain, expect Q in the range 0.5–5 m³/s. Wildly different answers signal a unit error.
Review the HYDROLOGIC CYCLE conceptually — know which process reduces available runoff (infiltration, ET) and which increases it (urbanisation, deforestation). Conceptual questions appear alongside numerical ones.

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In summary

Hydrology and Water Supply is a high-yield chapter in the PRC Civil Engineer Licensure Examination under Hydraulics and Fluid Mechanics. The entire chapter can be distilled into four core competencies: (1) understanding the hydrologic cycle as the conceptual framework for all runoff calculations; (2) applying the Rational Method Q = CiA/360 correctly — using mm/hr, ha, and m³/s — including weighted-C for composite catchments and back-calculation problems; (3) computing runoff volumes using V = CPA with meticulous unit conversion between mm, m, ha, km², and m³; and (4) calculating average, maximum-daily, and peak-hourly water demand and converting results to any required unit. The most critical exam pitfall remains the 360 divisor — it exists specifically because intensity is given in mm/hr and area in hectares, not base SI units. Master this chapter by drilling the six practice problems above, checking your units at every step, and remembering that civil engineers design for peak demand and peak flow — never for average conditions. With disciplined formula application and systematic unit checking, this chapter is one of the most reliable sources of correct answers on the licensure examination.

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