CELE Hydraulics & Fluid Mechanics — Hydrology and Water SupplyExam Answer Templates

Marks

1

Topic

Rational Method — Runoff Coefficient

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners expect the word 'dimensionless' and at least one anchor value (e.g., C ≈ 0.9 for pavement). One clean sentence with those elements earns the full mark.

Model Answer

The runoff coefficient C is a dimensionless ratio representing the fraction of rainfall that becomes surface runoff. It ranges from approximately 0.10 for permeable surfaces such as lawns to 0.95 for impervious surfaces such as paved roads.

Question Type

very_short_answer

Answer Structure

• Sentence 1: State what C represents (dimensionless fraction of rainfall that becomes runoff) [0.5 mark]
• Sentence 2: Give the numerical range with at least one example surface [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing C as a percentage (e.g., 10% to 95%) without converting — use decimal form
• Omitting the word 'dimensionless' or implying C has units
• Giving only the range without any qualifying context

Key Phrases To Include

• dimensionless
• fraction of rainfall
• becomes runoff
• impervious
• 0.10 to 0.95

Marks

2

Topic

Rational Method — Formula and Units

Difficulty

easy

Template Id

T2

Examiner Tip

The 360 factor is the most-tested detail in this formula. Always explain that 360 arises from unit conversion (1 mm/hr × 1 ha = 1/360 m³/s). Examiners penalize omission of this factor.

Model Answer

The Rational Method peak discharge formula in SI practical units is: Q = (C · i · A) / 360 Where: Q = peak runoff discharge (m³/s) C = dimensionless runoff coefficient (0.10 – 0.95) i = design rainfall intensity (mm/hr), taken at the time of concentration tc A = catchment area (hectares, ha) 360 = unit conversion factor (converts mm·ha/hr to m³/s)

Question Type

short_answer

Answer Structure

• Line 1: Write the formula clearly with the 360 denominator [1 mark]
• Lines 2–6: Define every variable with symbol, full name, and unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing Q = CiA without the 360 denominator — this is dimensionally incorrect for mm/hr and ha
• Omitting the units for i or A
• Not mentioning that intensity i is evaluated at the time of concentration tc

Key Phrases To Include

• Q = CiA / 360
• mm/hr
• hectares
• m³/s
• time of concentration
• unit conversion factor

Marks

2

Topic

Rational Method — Numerical Application

Difficulty

easy

Template Id

T3

Examiner Tip

Board-exam numerical questions are commonly presented exactly like this solved example from the reference notes. Memorize the 'Given → Formula → Solution → Answer' block format; it communicates clarity and earns process marks even if your final arithmetic has a slip.

Model Answer

Given: C = 0.6 i = 50 mm/hr A = 20 ha Formula: Q = (C · i · A) / 360 Solution: Q = (0.6 × 50 × 20) / 360 Q = 600 / 360 Q = 1.67 m³/s

Question Type

numerical

Answer Structure

• Block 1 — Given: List all known values with symbols and units [0.5 mark]
• Block 2 — Formula: Write Q = CiA/360 [0.5 mark]
• Block 3 — Substitution and arithmetic [0.5 mark]
• Block 4 — Boxed final answer with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to divide by 360, giving Q = 600 m³/s
• Correct number but missing unit m³/s
• Rounding error if more decimal places are used without noting 5/3 = 1.667

Key Phrases To Include

• Q = CiA / 360
• = 600 / 360
• 1.67 m³/s

Marks

3

Topic

Runoff Volume Calculation

Difficulty

medium

Template Id

T4

Examiner Tip

Examiners know that unit conversion is where most students lose marks on runoff volume problems. Show each conversion on its own line — you earn a dedicated mark for it.

Model Answer

Given: P = 80 mm = 0.080 m (storm rainfall depth) A = 5 km² = 5 × 10⁶ m² (catchment area) C = 0.40 (runoff coefficient) Formula: V_runoff = C · P · A Unit Conversions: P: 80 mm × (1 m / 1000 mm) = 0.080 m A: 5 km² × (1000 m / 1 km)² = 5,000,000 m² Solution: V_runoff = 0.40 × 0.080 × 5,000,000 V_runoff = 0.40 × 400,000 V_runoff = 160,000 m³

Question Type

numerical

Answer Structure

• Block 1 — Given: State P, A, and C with units [0.5 mark]
• Block 2 — Formula: Write V = CPA [0.5 mark]
• Block 3 — Unit conversions: mm→m and km²→m² shown explicitly [1 mark]
• Block 4 — Substitution and arithmetic [0.5 mark]
• Block 5 — Boxed answer: 160,000 m³ [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using P = 80 mm (not converted to metres) in V = CPA, giving 32,000,000 — wrong by factor of 1000
• Using A = 5 km² directly without converting to m², giving V = 0.032 m³
• Omitting unit from final answer
• Confusing the volume formula V = CPA with the peak discharge formula Q = CiA/360

Key Phrases To Include

• V_runoff = C · P · A
• 0.080 m
• 5 × 10⁶ m²
• 160,000 m³

Marks

3

Topic

Water Supply Demand Analysis

Difficulty

medium

Template Id

T5

Examiner Tip

Examiners reward the phrase 'sized for peak demand' because it shows engineering judgment, not just memorized definitions. Always close a demand question with a practical design statement.

Model Answer

Average Daily Demand (ADD) is the total water consumed in one day divided by the number of days in the period, calculated as: ADD = Population × Per-capita daily consumption (L/person/day) Peak-Hour Demand (PHD) is the maximum demand experienced during any single hour of the day. It accounts for the surge in usage (e.g., morning bathing and cooking) and is always greater than the ADD. Typical demand factors used in Philippine practice: Maximum Day Demand (MDD) = 1.5 × ADD Peak Hour Demand (PHD) = 2.0 to 3.0 × ADD (commonly 2.5× in design) Significance: Distribution mains, pumps, and service reservoirs must be sized for Peak Hour Demand, not merely the average, to maintain adequate pressure at all times.

Question Type

short_answer

Answer Structure

• Paragraph 1: Define ADD with formula [1 mark]
• Paragraph 2: Define PHD and explain why it exceeds ADD [1 mark]
• Paragraph 3: State demand factors (MDD ≈ 1.5×, PHD ≈ 2–3×) and design significance [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the factors (e.g., stating MDD = 2.5× and PHD = 1.5×)
• Defining ADD correctly but not defining PHD
• Failing to explain the design implication (that infrastructure must be sized for peak demand)

Key Phrases To Include

• average daily demand
• population × per-capita consumption
• peak-hour demand
• maximum day demand
• 1.5 × ADD
• 2.0 to 3.0 × ADD
• sized for peak

Marks

3

Topic

Water Supply Demand Calculation

Difficulty

easy

Template Id

T6

Examiner Tip

Always show the demand factor multiplication step (MDD = 1.5 × ADD) on its own line. Examiners check for the factor, not just the answer.

Model Answer

Given: Population = 10,000 persons Per-capita consumption = 150 L/person/day (a) Average Daily Demand (ADD): ADD = Population × Per-capita consumption ADD = 10,000 × 150 ADD = 1,500,000 L/day = 1,500 m³/day (b) Maximum Daily Demand (MDD): MDD = 1.5 × ADD MDD = 1.5 × 1,500 MDD = 2,250 m³/day (c) Peak-Hour Demand (PHD): PHD = 2.5 × ADD PHD = 2.5 × 1,500 PHD = 3,750 m³/day (or 3,750,000 L/day)

Question Type

numerical

Answer Structure

• Given block: Population and per-capita rate [0 marks — but must be present for clarity]
• Part (a): Formula ADD = Pop × q, correct answer 1,500 m³/day [1 mark]
• Part (b): MDD = 1.5 × ADD, correct answer 2,250 m³/day [1 mark]
• Part (c): PHD = 2.5 × ADD, correct answer 3,750 m³/day [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not converting L/day to m³/day (1000 L = 1 m³) when the question asks for m³
• Applying the MDD factor to PHD and vice versa
• Omitting the demand factor (e.g., just stating PHD = 3750 without showing × 2.5)

Key Phrases To Include

• ADD = Population × Per-capita
• 1,500,000 L/day
• MDD = 1.5 × ADD
• PHD = 2.5 × ADD
• 2,250 m³/day
• 3,750 m³/day

Marks

2

Topic

Rational Method — Time of Concentration

Difficulty

medium

Template Id

T7

Examiner Tip

This is a classic conceptual question. Examiners want to see both sides of the argument: duration < tc means partial catchment contribution, and duration > tc means lower intensity. State both sides to secure both marks.

Model Answer

The Time of Concentration (tc) is the time required for runoff to travel from the hydraulically most remote point in the catchment to the design outlet. It represents the minimum storm duration at which the entire catchment area contributes to peak discharge at the outlet. The design storm duration must equal tc because: if the storm lasts less than tc, only part of the catchment contributes, producing a lower peak; if the storm lasts longer than tc, the intensity (from IDF curves) decreases and again produces a lower peak. Therefore, the critical (maximum) peak discharge occurs precisely when duration = tc.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Define tc as travel time from the most remote point to the outlet [1 mark]
• Sentence 3–4: Explain why duration = tc produces the maximum peak (entire catchment contributes at the highest intensity) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining tc as just 'travel time' without specifying from the most remote point
• Not explaining why a duration longer than tc gives a lower peak
• Confusing tc with the storm return period

Key Phrases To Include

• hydraulically most remote point
• entire catchment contributes
• critical storm duration
• IDF curve
• intensity decreases with longer duration
• maximum peak discharge

Marks

2

Topic

Rational Method — Composite Runoff Coefficient

Difficulty

medium

Template Id

T8

Examiner Tip

The area-weighted average is the only accepted method for composite C. Examiners penalize the simple arithmetic mean (0.55). Always show the formula before substituting.

Model Answer

Given: A1 = 40% of total area, C1 = 0.90 (paved road) A2 = 60% of total area, C2 = 0.20 (lawn) Formula (area-weighted average): Cc = (C1·A1 + C2·A2) / (A1 + A2) Using percentage areas directly (A1 = 40, A2 = 60, total = 100): Cc = (0.90 × 40 + 0.20 × 60) / 100 Cc = (36 + 12) / 100 Cc = 48 / 100 Cc = 0.48

Question Type

numerical

Answer Structure

• Block 1 — Given: List sub-areas and their C values [0.5 mark]
• Block 2 — Formula: Area-weighted average for Cc [0.5 mark]
• Block 3 — Substitution [0.5 mark]
• Block 4 — Final answer Cc = 0.48 [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Simple average: (0.90 + 0.20)/2 = 0.55 — ignores area proportions, wrong
• Attaching a unit to Cc (it is dimensionless)
• Using raw areas without converting to fractions or percentages consistently

Key Phrases To Include

• area-weighted average
• Cc = (C1·A1 + C2·A2) / (A1 + A2)
• 0.48
• dimensionless

Marks

2

Topic

Hydrologic Cycle

Difficulty

easy

Template Id

T9

Examiner Tip

Whenever a question mentions 'describe and identify,' you need both a process description AND a specific list. Give the cycle description first, then use numbered items for the two losses — this signals to the examiner that you have answered both parts.

Model Answer

The hydrologic cycle is the continuous movement of water through the Earth-atmosphere system via the processes: precipitation → interception/infiltration → surface runoff → streamflow → evaporation/transpiration → atmospheric moisture → precipitation. The two main pathways by which precipitation is lost before becoming surface runoff are: 1. Infiltration — water percolates into the soil and eventually reaches groundwater or is taken up by plant roots. 2. Evapotranspiration — water returns to the atmosphere through direct evaporation from water surfaces and soils, plus transpiration from vegetation.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Describe the hydrologic cycle as a continuous process naming at least 4 stages [1 mark]
• Items 1–2: Identify and briefly explain infiltration and evapotranspiration as the two main losses [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing only evaporation without including transpiration in the second loss pathway
• Describing the cycle without showing the sequence (random list of processes)
• Omitting groundwater as a component of infiltration

Key Phrases To Include

• continuous
• precipitation
• surface runoff
• evapotranspiration
• infiltration
• percolates into the soil
• returns to the atmosphere

Marks

3

Topic

Rational Method — Numerical (Two-Unit)

Difficulty

medium

Template Id

T10

Examiner Tip

When a question asks for two units, always compute in m³/s first, then convert. Examiners look for the explicit conversion factor 1 m³/s = 1000 L/s written on a separate line.

Model Answer

Given: C = 0.75 i = 80 mm/hr (at time of concentration tc) A = 35 ha Formula: Q = (C · i · A) / 360 Solution: Q = (0.75 × 80 × 35) / 360 Q = 2,100 / 360 Q = 5.833 m³/s Conversion to L/s: Q = 5.833 m³/s × 1,000 L/m³ Q = 5,833 L/s ∴ Q = 5.83 m³/s ≈ 5,833 L/s

Question Type

numerical

Answer Structure

• Given block [no dedicated mark but required for full marks]
• Formula Q = CiA/360 stated [1 mark]
• Correct substitution and arithmetic: 2100/360 = 5.83 m³/s [1 mark]
• Unit conversion to L/s: × 1000, giving 5,833 L/s [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not dividing by 360 (gives Q = 2100, dimensionally inconsistent)
• Multiplying by 1000 instead of just stating the conversion, getting 5,833,000
• Rounding to 5.8 without showing the intermediate 5.833

Key Phrases To Include

• Q = CiA / 360
• 2,100 / 360
• 5.83 m³/s
• 1 m³/s = 1000 L/s
• 5,833 L/s

Marks

3

Topic

IDF Curves and Design Rainfall

Difficulty

medium

Template Id

T11

Examiner Tip

Examiners award marks for the connection between IDF and tc. The phrase 'design duration equals the time of concentration' is a trigger phrase that earns the third mark on its own.

Model Answer

An Intensity-Duration-Frequency (IDF) curve is a graphical relationship that shows how rainfall intensity (mm/hr) varies with storm duration (minutes or hours) for a given return period (years) at a specific location. Key characteristics: 1. For a fixed return period, intensity decreases as storm duration increases. 2. For a fixed duration, intensity increases with longer return periods (rarer storms). Use in hydrologic design: Step 1: Select the appropriate return period for the structure (e.g., 10-year storm for minor drainage, 100-year for major structures). Step 2: Compute the time of concentration tc for the catchment. Step 3: Enter the IDF curve at duration = tc and the chosen return period to read off the design intensity i (mm/hr). Step 4: Substitute i into the Rational Method Q = CiA/360 to obtain the peak design discharge.

Question Type

short_answer

Answer Structure

• Paragraph 1: Define IDF curve — relationship between i, duration, and return period [1 mark]
• Paragraph 2: State two key characteristics (intensity decreases with duration; increases with return period) [1 mark]
• Paragraph 3: Describe the design procedure using IDF in 4 steps [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining IDF without mentioning return period (frequency component)
• Stating that intensity increases with duration (this is the reverse of the correct relationship)
• Describing IDF without connecting it to the Rational Method application

Key Phrases To Include

• intensity-duration-frequency
• return period
• intensity decreases with duration
• time of concentration
• design storm
• Rational Method

Marks

5

Topic

Water Supply Demand — Full Analysis

Difficulty

hard

Template Id

T12

Examiner Tip

The 5-mark question always expects a summary. Write a clean 'Summary' block at the end listing all answers in one place — examiners scan for this when marking. The 86,400 s/day conversion is a frequently forgotten but high-value step.

Model Answer

Given: Population P = 250,000 persons Per-capita consumption q = 200 L/person/day MDD factor = 1.5 PHD factor = 2.5 (a) Average Daily Demand (ADD): ADD = P × q ADD = 250,000 × 200 ADD = 50,000,000 L/day = 50,000 m³/day = 50 ML/day (1 ML = 10⁶ L) (b) Maximum Daily Demand (MDD): MDD = 1.5 × ADD MDD = 1.5 × 50,000 MDD = 75,000 m³/day (c) Peak-Hour Demand (PHD): PHD = 2.5 × ADD PHD = 2.5 × 50,000 PHD = 125,000 m³/day Conversion to L/s for pump sizing: PHD in L/s = 125,000 m³/day × 1,000 L/m³ ÷ 86,400 s/day PHD = 125,000,000 / 86,400 PHD = 1,447 L/s (pump capacity required) Summary: ADD = 50,000 m³/day = 50 ML/day MDD = 75,000 m³/day PHD = 125,000 m³/day ≡ 1,447 L/s

Question Type

numerical

Answer Structure

• Given block: all data listed clearly [0.5 mark]
• Part (a): ADD formula and computation; conversion to ML/day shown [1 mark]
• Part (b): MDD = 1.5 × ADD with result 75,000 m³/day [1 mark]
• Part (c): PHD = 2.5 × ADD with result 125,000 m³/day [1 mark]
• Pump capacity: PHD converted to L/s using ÷ 86,400, result 1,447 L/s [1 mark]
• Summary/conclusion statement [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Not converting L/day to m³/day (factor of 1000)
• Not knowing 1 day = 86,400 seconds for the L/s conversion
• Applying the PHD factor to MDD or vice versa
• Missing the ML/day conversion (1 ML = 10⁶ L = 1000 m³)

Key Phrases To Include

• ADD = P × q
• 50,000 m³/day
• 50 ML/day
• MDD = 1.5 × ADD
• PHD = 2.5 × ADD
• 86,400 s/day
• 1,447 L/s

Marks

3

Topic

Runoff Volume — Unit Conversion

Difficulty

medium

Template Id

T13

Examiner Tip

The ML conversion (1 ML = 1,000 m³ = 10⁶ L) is a common board-exam unit trap. Write it explicitly as a conversion factor on a separate line.

Model Answer

Given: A = 12 km² = 12 × 10⁶ m² (area) P = 120 mm = 0.120 m (rainfall depth) C = 0.45 (runoff coefficient) Formula: V_runoff = C · P · A Unit Conversions: P: 120 mm ÷ 1000 = 0.120 m A: 12 km² × (1000)² = 12,000,000 m² Solution: (a) Volume in m³: V = 0.45 × 0.120 × 12,000,000 V = 0.45 × 1,440,000 V = 648,000 m³ (b) Volume in ML: V = 648,000 m³ × (1 ML / 1,000 m³) V = 648 ML ∴ V_runoff = 648,000 m³ = 648 ML

Question Type

numerical

Answer Structure

• Given block with unit conversions [0.5 mark]
• Formula V = CPA stated [0.5 mark]
• Part (a) correct arithmetic: 648,000 m³ [1 mark]
• Part (b) conversion 1 ML = 1,000 m³, answer 648 ML [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using P = 120 instead of 0.120 m — off by factor of 1000
• Using 1 ML = 1,000,000 m³ (confusing ML with GL)
• Computing in hectares: 12 km² = 1,200 ha — acceptable if P is in mm and result checked dimensionally

Key Phrases To Include

• V = C · P · A
• 0.120 m
• 12 × 10⁶ m²
• 648,000 m³
• 1 ML = 1,000 m³
• 648 ML

Marks

3

Topic

Water Supply Sources

Difficulty

easy

Template Id

T14

Examiner Tip

For 'state and describe' questions, use a numbered list. Each item must have a name AND at least one functional characteristic. Pure naming without description earns only half the mark per item.

Model Answer

Three common municipal water supply sources are: 1. Surface Water (Rivers and Streams): Water is collected directly from rivers, streams, or run-of-river diversion works. Surface water is generally abundant but requires treatment for turbidity, biological contaminants, and seasonal variability. 2. Impounding Reservoirs: Dams are constructed to store large volumes of rainwater and runoff. Reservoirs regulate seasonal flow variability and provide storage buffers for drought periods. Water quality is generally better than direct river intake but requires treatment before distribution. 3. Groundwater (Wells and Springs): Water is extracted from aquifers via tube wells, dug wells, or natural springs. Groundwater is naturally filtered by the soil, is often available year-round, and is generally closer to being potable, though it may contain dissolved minerals requiring treatment.

Question Type

short_answer

Answer Structure

• Source 1: Surface water — name, brief description, water-quality note [1 mark]
• Source 2: Reservoirs — name, description of storage and flow regulation function [1 mark]
• Source 3: Groundwater — name, description of aquifer extraction, quality note [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing four sources but only describing two — answer must match the number asked
• Describing treatment methods instead of the sources themselves
• Giving rivers and reservoirs as two separate categories without noting that reservoirs store surface runoff (acceptable if both are described distinctly)

Key Phrases To Include

• surface water
• rivers and streams
• impounding reservoir
• flow regulation
• groundwater
• aquifer
• treatment

Marks

5

Topic

Composite Catchment — Full Analysis

Difficulty

hard

Template Id

T15

Examiner Tip

In 5-mark problems, show every intermediate product (C1·A1, C2·A2, C3·A3) on individual lines. Examiners award method marks for each correct product even if the final answer is wrong due to arithmetic.

Model Answer

Given: A1 = 15 ha, C1 = 0.55 (residential) A2 = 5 ha, C2 = 0.85 (commercial) A3 = 10 ha, C3 = 0.15 (park) Total A = 15 + 5 + 10 = 30 ha i = 70 mm/hr (a) Composite Runoff Coefficient Cc: Formula: Cc = (C1·A1 + C2·A2 + C3·A3) / (A1 + A2 + A3) Numerator: C1·A1 = 0.55 × 15 = 8.25 C2·A2 = 0.85 × 5 = 4.25 C3·A3 = 0.15 × 10 = 1.50 Sum = 8.25 + 4.25 + 1.50 = 14.00 Cc = 14.00 / 30 Cc = 0.467 (dimensionless) (b) Peak Runoff Discharge Q: Formula: Q = (Cc · i · A) / 360 Q = (0.467 × 70 × 30) / 360 Q = (0.467 × 2,100) / 360 Q = 980.7 / 360 Q = 2.72 m³/s Summary: Cc = 0.467 Q = 2.72 m³/s

Question Type

numerical

Answer Structure

• Given block with all three sub-areas and total area computed [0.5 mark]
• Part (a) formula for area-weighted Cc [1 mark]
• Part (a) computation of numerator terms and correct Cc = 0.467 [1 mark]
• Part (b) formula Q = CiA/360 with Cc substituted [1 mark]
• Part (b) correct arithmetic Q = 2.72 m³/s [1 mark]
• Summary block with both answers clearly stated [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using simple arithmetic mean of C values: (0.55+0.85+0.15)/3 = 0.517 — incorrect
• Forgetting to include all three sub-areas in the numerator
• Not computing total area as 30 ha before substituting into Q formula
• Dropping the unit on Q

Key Phrases To Include

• Cc = (ΣCiAi) / ΣAi
• area-weighted average
• numerator = 14.00
• Cc = 0.467
• Q = CiA / 360
• 2.72 m³/s