CELE Hydraulics & Fluid Mechanics — Flow in Open ChannelsExam Answer Templates

Marks

1

Topic

Uniform Flow — Hydraulic Radius

Difficulty

easy

Template Id

T1

Examiner Tip

The single mark is awarded only if the ratio is correct. A verbal description without the formula does not earn the mark.

Model Answer

The hydraulic radius R is the ratio of the cross-sectional flow area A to the wetted perimeter P of the channel: R = A / P It characterises the hydraulic efficiency of a channel cross-section.

Question Type

very_short_answer

Answer Structure

• One sentence: define R as A/P — [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing P as the total perimeter instead of the wetted perimeter — loses the mark entirely.
• Confusing hydraulic radius with hydraulic diameter (D_h = 4R).
• Omitting the formula and writing only a vague description.

Key Phrases To Include

• hydraulic radius
• flow area
• wetted perimeter
• R = A/P

Marks

1

Topic

Uniform Flow — Manning's Equation

Difficulty

easy

Template Id

T2

Examiner Tip

The coefficient '1' in the SI form is often omitted in writing but must be understood. Examiners expect the SI form explicitly.

Model Answer

Manning's equation (SI): v = (1/n) R^(2/3) S^(1/2) where v = mean flow velocity (m/s), n = Manning roughness coefficient (dimensionless), R = hydraulic radius (m), S = channel bed slope (m/m, dimensionless).

Question Type

very_short_answer

Answer Structure

• Write the formula — [0.5 mark]
• Identify all four variables with units — [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using coefficient 1.49 (US customary form) instead of 1 (SI form) — incorrect for Philippine board exams.
• Writing S as percentage instead of a dimensionless ratio.
• Omitting the exponents 2/3 or 1/2.

Key Phrases To Include

• v = (1/n)R^(2/3)S^(1/2)
• Manning roughness coefficient
• hydraulic radius
• bed slope

Marks

2

Topic

Most Efficient Rectangular Section

Difficulty

easy

Template Id

T3

Examiner Tip

Always give the physical reason (minimum P → maximum R → maximum Q) for concept questions; this earns the second mark.

Model Answer

For the most efficient (best hydraulic) rectangular section, the channel width b equals twice the flow depth y: b = 2y This condition minimises the wetted perimeter P for a given flow area A, which maximises the hydraulic radius R = A/P and therefore maximises discharge Q for a fixed slope S and roughness n. Under this condition, R = y/2.

Question Type

short_answer

Answer Structure

• Line 1: State the condition b = 2y — [1 mark]
• Line 2: Explain that this minimises P (or maximises R), maximising Q — [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating only b = 2y without any justification — earns only 1 of 2 marks.
• Confusing 'most efficient' with 'maximum area' rather than minimum wetted perimeter.
• Incorrect statement such as b = y.

Key Phrases To Include

• b = 2y
• minimum wetted perimeter
• maximum hydraulic radius
• R = y/2

Marks

2

Topic

Specific Energy

Difficulty

easy

Template Id

T4

Examiner Tip

The key distinguishing phrase is 'referenced to the channel bottom as datum' — this differentiates specific energy from total hydraulic energy.

Model Answer

Specific energy E is the total energy per unit weight of flowing water measured from the channel bottom as datum: E = y + v²/(2g) where y = flow depth (m), v = mean velocity (m/s), and g = 9.81 m/s². The first term y is the potential (depth) component and the second term v²/(2g) is the velocity head. For a rectangular channel with unit discharge q = Q/b, this becomes: E = y + q²/(2g y²)

Question Type

short_answer

Answer Structure

• Line 1: Define E as energy per unit weight above the channel bottom — [1 mark]
• Line 2: Write the correct formula E = y + v²/(2g) with variable identification — [1 mark]

Scoring Breakdown

Common Mark Deductions

• Referencing datum to a different point (e.g., sea level) rather than the channel bottom.
• Omitting the velocity head term.
• Using total head H (which includes elevation z) instead of specific energy E.

Key Phrases To Include

• energy per unit weight
• channel bottom datum
• E = y + v²/(2g)
• depth component
• velocity head

Marks

2

Topic

Froude Number and Flow Classification

Difficulty

easy

Template Id

T5

Examiner Tip

Always write the classification statement explicitly — 'Fr = 0.54 < 1, therefore subcritical.' Without the classification, the second mark is at risk.

Model Answer

Given: b = 3 m, y = 1.2 m, v = 1.86 m/s, g = 9.81 m/s² Required: Froude number Fr and flow classification Solution: Fr = v / √(g·y) Fr = 1.86 / √(9.81 × 1.2) Fr = 1.86 / √11.772 Fr = 1.86 / 3.431 Fr = 0.542 Since Fr < 1, the flow is SUBCRITICAL (tranquil).

Question Type

numerical

Answer Structure

• Line 1–2: Write Given data and Required — [0 marks but required structure]
• Line 3: State formula Fr = v/√(gy) — [1 mark]
• Line 4–5: Substitute and compute Fr = 0.542 — [0.5 mark]
• Line 6: Classify as subcritical since Fr < 1 — [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using Fr = v/√(g) without the depth in the denominator.
• Classifying incorrectly (e.g., calling Fr = 0.54 supercritical).
• Forgetting to take the square root of g·y.

Key Phrases To Include

• Fr = v/√(gy)
• Fr < 1
• subcritical
• tranquil flow

Marks

3

Topic

Critical Flow — Rectangular Channel

Difficulty

medium

Template Id

T6

Examiner Tip

The formula E_min = (3/2)y_c is specific to rectangular channels — mention this qualifier to show depth of knowledge.

Model Answer

Given: b = 3 m, Q = 6 m³/s, g = 9.81 m/s² Required: Critical depth y_c and minimum specific energy E_min Solution: Step 1 — Unit discharge: q = Q/b = 6/3 = 2.0 m²/s (m³/s per metre width) Step 2 — Critical depth: y_c = (q²/g)^(1/3) y_c = (2.0²/9.81)^(1/3) y_c = (4.0/9.81)^(1/3) y_c = (0.4077)^(1/3) y_c = 0.742 m Step 3 — Minimum specific energy: E_min = (3/2) y_c = 1.5 × 0.742 = 1.113 m ≈ 1.11 m Answer: y_c = 0.742 m; E_min = 1.11 m

Question Type

numerical

Answer Structure

• Step 1: Compute q = Q/b — [1 mark]
• Step 2: Apply y_c = (q²/g)^(1/3) and solve — [1 mark]
• Step 3: Apply E_min = (3/2)y_c — [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using Q instead of q (unit discharge) directly in the formula without dividing by b.
• Applying the exponent as 1/2 instead of 1/3.
• Forgetting the E_min = 1.5 y_c relationship and computing E from scratch with errors.

Key Phrases To Include

• q = Q/b
• y_c = (q²/g)^(1/3)
• E_min = (3/2)y_c

Marks

3

Topic

Uniform Flow — Manning's Equation

Difficulty

medium

Template Id

T7

Examiner Tip

Show the intermediate value R^(2/3) and S^(1/2) separately before multiplying — this demonstrates systematic computation and earns partial credit even with a later arithmetic error.

Model Answer

Given: b = 3 m, y = 1.2 m, n = 0.013, S = 0.001 Required: Discharge Q (m³/s) Solution: Step 1 — Cross-sectional area and wetted perimeter: A = b × y = 3 × 1.2 = 3.60 m² P = b + 2y = 3 + 2(1.2) = 3 + 2.4 = 5.40 m Step 2 — Hydraulic radius: R = A/P = 3.60/5.40 = 0.6667 m Step 3 — Manning's velocity (SI form): v = (1/n) R^(2/3) S^(1/2) v = (1/0.013)(0.6667)^(2/3)(0.001)^(1/2) v = 76.923 × 0.7631 × 0.031623 v = 1.856 m/s ≈ 1.86 m/s Step 4 — Discharge: Q = Av = 3.60 × 1.856 = 6.68 m³/s Answer: Q ≈ 6.68 m³/s

Question Type

numerical

Answer Structure

• Step 1: Compute A and P correctly — [1 mark]
• Step 2: Compute R = A/P — [0.5 mark]
• Step 3: Apply Manning's equation, compute v — [1 mark]
• Step 4: Compute Q = Av with units — [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using P = 2b + 2y (full perimeter) instead of P = b + 2y (wetted perimeter for open channel).
• Computing R^(2/3) incorrectly — use logarithms or direct computation: 0.6667^(2/3) = e^(2/3 × ln 0.6667) = 0.763.
• Using 1.49 as the Manning constant (US customary) instead of 1 (SI).
• Forgetting to compute Q = Av after finding v.

Key Phrases To Include

• R = A/P
• v = (1/n)R^(2/3)S^(1/2)
• Q = Av
• wetted perimeter

Marks

3

Topic

Hydraulic Jump

Difficulty

medium

Template Id

T8

Examiner Tip

Always verify the jump condition (Fr₁ > 1) with a tick mark (✓) — examiners reward this check as it shows conceptual understanding.

Model Answer

Given: y₁ = 0.40 m, v₁ = 6.0 m/s, g = 9.81 m/s² Required: Sequent depth y₂ Solution: Step 1 — Froude number upstream: Fr₁ = v₁/√(g·y₁) Fr₁ = 6.0/√(9.81 × 0.40) Fr₁ = 6.0/√3.924 Fr₁ = 6.0/1.981 = 3.029 Since Fr₁ = 3.03 > 1, the flow is supercritical — a hydraulic jump can occur. ✓ Step 2 — Conjugate depth formula: y₂/y₁ = (1/2)[√(1 + 8·Fr₁²) − 1] y₂/0.40 = (1/2)[√(1 + 8 × 3.029²) − 1] y₂/0.40 = (1/2)[√(1 + 8 × 9.175) − 1] y₂/0.40 = (1/2)[√(1 + 73.40) − 1] y₂/0.40 = (1/2)[√74.40 − 1] y₂/0.40 = (1/2)[8.626 − 1] y₂/0.40 = (1/2)(7.626) = 3.813 Step 3: y₂ = 0.40 × 3.813 = 1.525 m ≈ 1.53 m Answer: y₂ ≈ 1.53 m

Question Type

numerical

Answer Structure

• Step 1: Compute Fr₁ and verify Fr₁ > 1 — [1 mark]
• Step 2: State and apply conjugate-depth formula correctly — [1 mark]
• Step 3: Compute y₂ with units — [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not verifying Fr₁ > 1 before applying the jump formula.
• Using Fr₁² as Fr₁ × 2 rather than Fr₁ squared.
• Applying the formula as y₂/y₁ = (1/2)(√(1 + 8Fr₁²) + 1) — wrong sign, plus instead of minus.

Key Phrases To Include

• Fr₁ = v₁/√(gy₁)
• Fr₁ > 1 supercritical
• y₂/y₁ = (1/2)[√(1 + 8Fr₁²) − 1]
• conjugate depth
• sequent depth

Marks

3

Topic

Hydraulic Jump

Difficulty

medium

Template Id

T9

Examiner Tip

Mention the practical application (stilling basin, energy dissipator) to show engineering context — examiners appreciate applied understanding.

Model Answer

A hydraulic jump is an abrupt, localised transition from supercritical flow (Fr > 1) to subcritical flow (Fr < 1) in an open channel. It occurs when a high-velocity, shallow flow (e.g., at the base of a spillway or sluice gate) encounters a downstream condition that forces it to decelerate. Physically, the transition is caused by the momentum imbalance between the high-velocity supercritical stream and the lower-velocity subcritical regime downstream. The jump is characterised by intense turbulence, surface rollers, and significant air entrainment. Specific energy is NOT conserved across the jump — a portion of the upstream kinetic energy is irreversibly converted into heat (internal energy) by turbulence. Therefore: E₂ < E₁ (energy loss ΔE = E₁ − E₂ > 0) This energy dissipation makes the hydraulic jump useful as a natural stilling basin to protect channel structures from erosion.

Question Type

short_answer

Answer Structure

• Part 1: Define the hydraulic jump (supercritical → subcritical transition) — [1 mark]
• Part 2: Explain the physical cause (momentum imbalance, turbulence) — [1 mark]
• Part 3: State that specific energy decreases across the jump (E₂ < E₁) — [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying energy is conserved across the jump — this is wrong; energy is dissipated.
• Describing the jump as occurring from subcritical to supercritical — the direction is reversed.
• Not mentioning the Froude number condition.

Key Phrases To Include

• supercritical to subcritical
• Fr > 1 upstream
• turbulence
• energy dissipation
• E₂ < E₁
• stilling basin

Marks

5

Topic

Uniform Flow — Trapezoidal Channel

Difficulty

hard

Template Id

T10

Examiner Tip

The most common error in trapezoidal channels is the wetted perimeter. Write the formula P = b + 2y√(1+z²) explicitly before substituting numbers to signal to the examiner you know the correct geometry.

Model Answer

Given: b = 4 m, side slope z = 1.5 (horizontal:vertical), y = 1.5 m n = 0.015, S = 0.0008 Required: Discharge Q (m³/s) Solution: Step 1 — Cross-sectional area (trapezoidal): A = (b + z·y)·y A = (4 + 1.5 × 1.5)(1.5) A = (4 + 2.25)(1.5) A = 6.25 × 1.5 A = 9.375 m² Step 2 — Wetted perimeter: Side length = y√(1 + z²) = 1.5√(1 + 1.5²) = 1.5√(1 + 2.25) = 1.5√3.25 Side length = 1.5 × 1.8028 = 2.704 m P = b + 2 × side length = 4 + 2(2.704) = 4 + 5.408 = 9.408 m Step 3 — Hydraulic radius: R = A/P = 9.375/9.408 = 0.9965 m ≈ 0.997 m Step 4 — Manning's velocity (SI): v = (1/n)·R^(2/3)·S^(1/2) R^(2/3) = (0.9965)^(2/3) ≈ 0.9977 S^(1/2) = (0.0008)^(1/2) = 0.02828 v = (1/0.015)(0.9977)(0.02828) v = 66.667 × 0.9977 × 0.02828 v = 1.882 m/s Step 5 — Discharge: Q = A·v = 9.375 × 1.882 Q = 17.64 m³/s ≈ 17.6 m³/s Answer: Q ≈ 17.6 m³/s

Question Type

numerical

Answer Structure

• Step 1: Correct trapezoidal area formula A = (b + zy)y — [1 mark]
• Step 2: Correct wetted perimeter with inclined side length = y√(1+z²) — [1 mark]
• Step 3: Hydraulic radius R = A/P — [1 mark]
• Step 4: Manning's velocity with correct exponents — [1 mark]
• Step 5: Q = Av with units — [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using A = by (rectangular formula) instead of the trapezoidal formula.
• Using P = b + 2y (vertical sides) instead of P = b + 2y√(1+z²) for inclined sides.
• Plugging in S = 0.08% as 0.08 rather than 0.0008.
• Not showing intermediate computations — difficult to award partial marks.

Key Phrases To Include

• A = (b + zy)y
• side length = y√(1+z²)
• R = A/P
• v = (1/n)R^(2/3)S^(1/2)
• Q = Av

Marks

5

Topic

Most Efficient Rectangular Section — Design

Difficulty

hard

Template Id

T11

Examiner Tip

This is a classic design problem. The first and most critical step is invoking b = 2y — write it prominently. Board examiners give 1 mark just for this correct starting condition.

Model Answer

Given: Q = 5.0 m³/s, n = 0.013, S = 0.001 Required: Width b and depth y for the most efficient rectangular section. Concept: For the most efficient rectangular section, b = 2y, and R = y/2. Solution: Step 1 — Express A and R in terms of y only: Since b = 2y: A = b·y = 2y·y = 2y² P = b + 2y = 2y + 2y = 4y R = A/P = 2y²/(4y) = y/2 Step 2 — Apply Manning's equation: Q = A·v = A·(1/n)·R^(2/3)·S^(1/2) 5.0 = 2y² × (1/0.013) × (y/2)^(2/3) × (0.001)^(1/2) Step 3 — Evaluate known constants: (1/n) = 1/0.013 = 76.923 S^(1/2) = (0.001)^(1/2) = 0.031623 (y/2)^(2/3) = y^(2/3) / 2^(2/3) = y^(2/3) / 1.5874 Step 4 — Combine: 5.0 = 2y² × 76.923 × [y^(2/3)/1.5874] × 0.031623 5.0 = 2y² × 76.923 × 0.031623 × y^(2/3) / 1.5874 5.0 = 2 × 76.923 × 0.031623 / 1.5874 × y^(8/3) 5.0 = 2 × 1.5320 × y^(8/3) 5.0 = 3.0640 × y^(8/3) Step 5 — Solve for y: y^(8/3) = 5.0/3.0640 = 1.6319 y = (1.6319)^(3/8) ln y = (3/8) × ln(1.6319) = 0.375 × 0.4894 = 0.1835 y = e^(0.1835) = 1.2016 m ≈ 1.20 m Step 6 — Width: b = 2y = 2 × 1.20 = 2.40 m Verification: A = 2(1.20)² = 2.88 m², R = 1.20/2 = 0.60 m v = (1/0.013)(0.60)^(2/3)(0.001)^(1/2) = 76.923 × 0.7114 × 0.031623 = 1.730 m/s Q = 2.88 × 1.730 = 4.98 ≈ 5.0 m³/s ✓ Answer: y ≈ 1.20 m, b = 2y ≈ 2.40 m

Question Type

numerical

Answer Structure

• State the condition b = 2y and R = y/2 — [1 mark]
• Express A and R solely in terms of y — [1 mark]
• Set up Manning's equation with Q = 5.0 m³/s — [1 mark]
• Solve for y algebraically — [1 mark]
• Find b = 2y and verify (optional but rewarded) — [1 mark]

Scoring Breakdown

Common Mark Deductions

• Setting up the equation without first applying the b = 2y condition — introduces two unknowns, making the equation unsolvable.
• Algebraic errors when combining the y^2 and y^(2/3) exponents — the correct combined exponent is y^(8/3).
• Not verifying the answer — missing the chance to catch arithmetic errors.

Key Phrases To Include

• b = 2y
• R = y/2
• most efficient rectangular section
• Q = A(1/n)R^(2/3)S^(1/2)
• y^(8/3)

Marks

5

Topic

Critical Flow — Specific Energy

Difficulty

medium

Template Id

T12

Examiner Tip

Always verify your critical depth result by checking Fr = 1. This double-check takes 15 seconds and can earn an extra mark while confirming your answer.

Model Answer

Given: Unit discharge q = 3.0 m²/s (m³/s per metre width), g = 9.81 m/s² Required: y_c, E_min, v_c Solution: (a) Critical depth: y_c = (q²/g)^(1/3) y_c = (3.0²/9.81)^(1/3) y_c = (9.0/9.81)^(1/3) y_c = (0.9174)^(1/3) y_c = 0.9713 m ≈ 0.971 m (b) Minimum specific energy: E_min = (3/2)·y_c E_min = 1.5 × 0.9713 E_min = 1.457 m ≈ 1.46 m Verification using E = y + v²/2g: v_c = q/y_c = 3.0/0.9713 = 3.088 m/s E = 0.9713 + (3.088)²/(2×9.81) = 0.9713 + 0.4856 = 1.457 m ✓ (c) Critical velocity: v_c = q/y_c = 3.0/0.9713 = 3.09 m/s Check: Fr_c = v_c/√(g·y_c) = 3.09/√(9.81×0.9713) = 3.09/3.09 = 1.00 ✓ (critical) Answer: (a) y_c = 0.971 m (b) E_min = 1.46 m (c) v_c = 3.09 m/s

Question Type

numerical

Answer Structure

• (a) Apply y_c = (q²/g)^(1/3) — [2 marks]
• (b) Apply E_min = (3/2)y_c — [1 mark]
• (c) Compute v_c = q/y_c — [1 mark]
• Froude number check Fr = 1 confirms critical condition — [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using Q instead of q (unit discharge per metre width) — dimensional error.
• Forgetting to verify Fr = 1 at the critical condition — missed mark.
• Computing E_min independently as y_c + v_c²/2g without using the E_min = 1.5y_c shortcut (still correct but error-prone).

Key Phrases To Include

• y_c = (q²/g)^(1/3)
• E_min = (3/2)y_c
• v_c = q/y_c
• Fr = 1 at critical flow

Marks

5

Topic

Hydraulic Jump — Energy Loss

Difficulty

hard

Template Id

T13

Examiner Tip

The alternative formula h_L = (y₂−y₁)³/(4y₁y₂) is the direct hydraulic jump head-loss formula — knowing and applying it as a check demonstrates mastery and impresses examiners.

Model Answer

Given: b = 4 m, y₁ = 0.5 m, Q = 10 m³/s, g = 9.81 m/s² Required: Fr₁, y₂, h_L Solution: Step 1 — Upstream velocity and Froude number: v₁ = Q/(b·y₁) = 10/(4 × 0.5) = 10/2 = 5.0 m/s Fr₁ = v₁/√(g·y₁) = 5.0/√(9.81 × 0.5) = 5.0/√4.905 = 5.0/2.214 = 2.259 Fr₁ = 2.26 > 1 → supercritical flow confirmed ✓ Step 2 — Conjugate depth: y₂/y₁ = (1/2)[√(1 + 8·Fr₁²) − 1] y₂/0.5 = (1/2)[√(1 + 8 × 2.259²) − 1] y₂/0.5 = (1/2)[√(1 + 8 × 5.103) − 1] y₂/0.5 = (1/2)[√(1 + 40.82) − 1] y₂/0.5 = (1/2)[√41.82 − 1] y₂/0.5 = (1/2)[6.467 − 1] y₂/0.5 = (1/2)(5.467) = 2.734 y₂ = 0.5 × 2.734 = 1.367 m ≈ 1.37 m Step 3 — Downstream velocity: v₂ = Q/(b·y₂) = 10/(4 × 1.367) = 10/5.468 = 1.829 m/s Step 4 — Head loss (energy loss across the jump): E₁ = y₁ + v₁²/(2g) = 0.5 + 5.0²/(2×9.81) = 0.5 + 1.274 = 1.774 m E₂ = y₂ + v₂²/(2g) = 1.367 + 1.829²/(2×9.81) = 1.367 + 0.170 = 1.537 m h_L = E₁ − E₂ = 1.774 − 1.537 = 0.237 m Alternatively: h_L = (y₂ − y₁)³ / (4·y₁·y₂) h_L = (1.367 − 0.5)³ / (4 × 0.5 × 1.367) = (0.867)³ / (2.734) h_L = 0.652 / 2.734 = 0.238 m ✓ Answer: (a) Fr₁ = 2.26 (supercritical) (b) y₂ = 1.37 m (c) h_L ≈ 0.237 m

Question Type

numerical

Answer Structure

• Step 1: Compute v₁ = Q/(by₁) and Fr₁ — [1 mark]
• Step 2: Apply conjugate-depth formula to find y₂ — [2 marks]
• Step 3: Compute v₂ — [0.5 mark]
• Step 4: Compute E₁, E₂, and h_L = E₁ − E₂ — [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Forgetting to compute v₁ = Q/(by₁) before computing Fr₁.
• Arithmetic errors in Fr₁² — squaring Fr₁ first before multiplying by 8.
• Not computing the energy loss and leaving the answer at y₂ only.

Key Phrases To Include

• v₁ = Q/(by₁)
• Fr₁ = v₁/√(gy₁)
• y₂/y₁ = (1/2)[√(1+8Fr₁²)−1]
• h_L = E₁ − E₂
• (y₂−y₁)³/(4y₁y₂)

Marks

2

Topic

Flow Classification — Froude Number

Difficulty

easy

Template Id

T14

Examiner Tip

The downstream-control concept for subcritical and upstream-control concept for supercritical is a higher-order understanding that distinguishes above-average answers.

Model Answer

Subcritical flow (Fr < 1): 1. Flow is deep and slow (tranquil); gravity forces dominate over inertial forces. 2. Disturbances propagate both upstream and downstream — downstream conditions control the flow. Supercritical flow (Fr > 1): 1. Flow is shallow and fast (rapid/shooting); inertial forces dominate over gravity forces. 2. Disturbances can only propagate downstream — upstream conditions are unaffected by changes downstream.

Question Type

short_answer

Answer Structure

• Two characteristics of subcritical flow (Fr < 1) — [1 mark]
• Two characteristics of supercritical flow (Fr > 1) — [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the descriptions (calling Fr > 1 slow/deep).
• Omitting the Froude number values — the answer must reference Fr.
• Listing only one characteristic per flow type.

Key Phrases To Include

• Fr < 1 subcritical
• Fr > 1 supercritical
• deep and slow
• shallow and fast
• downstream control
• upstream propagation

Marks

1

Topic

Most Efficient Trapezoidal Section

Difficulty

easy

Template Id

T15

Examiner Tip

The key numbers to memorise: 60° angle and R = y/2. These two facts earn the mark in 1-mark VSA questions.

Model Answer

The most efficient trapezoidal section is a half-hexagon: the three sides (bottom and two inclined sides) are all equal in length, the side slopes are at 60° from the horizontal (side slope z = 1/√3 ≈ 0.577H:1V), and the hydraulic radius equals half the depth: R = y/2.

Question Type

very_short_answer

Answer Structure

• One statement: half-hexagon, equal sides, 60° inclination, R = y/2 — [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating 45° or 30° instead of 60°.
• Confusing this with the rectangular best section (b = 2y).
• Omitting R = y/2.

Key Phrases To Include

• half-hexagon
• 60° from horizontal
• equal sides
• R = y/2