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

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.

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

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).

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

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.

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

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)

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

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).

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

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

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

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

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

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

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

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.

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

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

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

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

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)

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

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

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

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)

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.

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

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.

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

Do

• Write the exact formula (e.g., R = A/P) on the first line.
• Identify each symbol in the formula briefly.
• Use engineering shorthand where accepted (e.g., Fr for Froude number).
• State SI units for the answer if it is a formula for a quantity.

Don't

• Do not write lengthy explanations or derivations — wastes time.
• Do not omit the formula and rely on words alone — likely to lose the mark.
• Do not confuse similar-looking formulas (e.g., R = A/P versus D_h = 4R).

Do

• Use the Given / Required / Solution structure even for 2-mark numericals.
• State the formula explicitly before substituting numbers.
• Include the unit in the final answer line.
• For Froude number questions, always state the flow classification.

Don't

• Do not skip intermediate steps — partial credit is awarded per step.
• Do not round excessively during intermediate calculations.
• Do not omit units in the final answer.

Do

• Number your steps (Step 1, Step 2, Step 3).
• Box or underline your intermediate results (A, P, R) so the examiner can check them.
• Write the answer as a clear boxed final statement.
• Verify the result if time permits (e.g., check Fr = 1 for critical depth).

Don't

• Do not skip the hydraulic radius calculation — it is always an intermediate mark.
• Do not mix up Manning's SI and US customary forms.
• Do not omit the classification step for Froude number or jump problems.

Do

• Allocate one step per mark — 5-mark problems typically have exactly 5 computable steps.
• State the governing design condition (e.g., most efficient section) at the very start.
• Show all intermediate computations (R^(2/3), S^(1/2)) as separate sub-results.
• Write a summary box at the end listing all required answers.
• Verify using an alternative formula where one exists (e.g., head-loss formula for hydraulic jump).

Don't

• Do not rush past the area and perimeter formulas — they carry 2 of the 5 marks.
• Do not skip the Given/Required section — it helps you organise and shows the examiner your understanding of the problem.
• Do not leave blanks — a partially correct step still earns partial credit.
• Do not use the US customary Manning constant 1.49 in Philippine board exams.