CELE Hydraulics & Fluid Mechanics — Orifices, Weirs, Tubes and NozzlesExam Answer Templates

Marks

1

Topic

Orifice Discharge Coefficients

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners want to see that you connect Cd to both physical phenomena (contraction and friction). Mentioning Cd = Cv × Cc immediately signals mastery.

Model Answer

The coefficient of discharge (Cd) is the ratio of the actual discharge to the theoretical discharge through an orifice. It accounts for both the contraction of the jet (Cc) and the reduction in velocity due to friction (Cv), such that Cd = Cv × Cc ≈ 0.61 for a sharp-edged orifice.

Question Type

very_short_answer

Answer Structure

• Line 1: State Cd as the ratio of actual to theoretical discharge and give the relation Cd = Cv × Cc [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'ratio of velocity' instead of 'ratio of discharge' — loses the mark entirely
• Omitting the relationship Cd = Cv × Cc — answer is incomplete
• Giving a wrong typical value (e.g., Cd = 1.0) — shows conceptual misunderstanding

Key Phrases To Include

• ratio of actual discharge to theoretical discharge
• Cd = Cv × Cc
• coefficient of velocity
• coefficient of contraction
• approximately 0.61

Marks

2

Topic

Weirs — Rectangular and Triangular

Difficulty

easy

Template Id

T2

Examiner Tip

Always link a structural difference to a hydraulic consequence. 'V-notch has a triangular shape' alone scores 0; connecting shape to the H^5/2 relationship and small-flow accuracy scores full marks.

Model Answer

1. Flow formula: Rectangular weir — Q = (2/3)Cd√(2g) L H^(3/2); discharge varies as H^(3/2). Triangular weir — Q = (8/15)Cd√(2g) tan(θ/2) H^(5/2); discharge varies as H^(5/2). 2. Range of application: The rectangular weir is suitable for large discharges because it has a wider opening. The V-notch weir is more accurate for small discharges because the H^(5/2) relationship gives greater sensitivity (larger change in H for a small change in Q) at low heads.

Question Type

very_short_answer

Answer Structure

• Point 1: State the different discharge formulas and H-exponents (H^3/2 vs H^5/2) [1 mark]
• Point 2: State the application range difference — rectangular for large flow, V-notch for small flow with justification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Swapping the exponents (writing H^5/2 for rectangular) — loses that mark
• Stating differences only in shape without linking to hydraulic performance — too superficial
• Writing 'more accurate' without explaining why (H^5/2 sensitivity) — incomplete

Key Phrases To Include

• H^(3/2) for rectangular
• H^(5/2) for triangular
• greater sensitivity at low head
• suitable for small discharges
• suitable for large discharges

Marks

2

Topic

Orifice Flow — Vena Contracta

Difficulty

easy

Template Id

T3

Examiner Tip

Mentioning Cc = Ac/A connects definition to formula and signals that you understand how vena contracta feeds into the discharge coefficient — examiners reward this integration.

Model Answer

The vena contracta is the section of minimum cross-sectional area of a jet issuing from an orifice, located a short distance downstream from the orifice plane. It occurs because fluid approaching the orifice from all directions has lateral momentum (inward velocity components) that cannot be instantly redirected axially. These curved streamlines continue to converge after passing through the orifice opening, causing the jet to contract further until all streamlines become parallel. The ratio of the vena contracta area (Ac) to the orifice area (A) is the coefficient of contraction, Cc = Ac/A ≈ 0.62 for a sharp-edged orifice.

Question Type

short_answer

Answer Structure

• Sentence 1: Define vena contracta — section of minimum jet area downstream of the orifice [1 mark]
• Sentence 2: Explain the cause — lateral/inward momentum of converging streamlines that persist beyond the orifice plane [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining vena contracta as 'inside the orifice' — incorrect location loses the definition mark
• Saying only 'streamlines curve' without explaining the momentum cause — too vague
• Confusing vena contracta with the orifice itself — fundamental error

Key Phrases To Include

• minimum cross-sectional area
• downstream of the orifice
• converging streamlines
• lateral momentum
• coefficient of contraction Cc = Ac/A

Marks

3

Topic

Orifice Discharge — Numerical

Difficulty

medium

Template Id

T4

Examiner Tip

Always show √(2gh) as a separate computed value before multiplying. This intermediate step makes it easy for the examiner to award the formula mark even if the final arithmetic has a minor error.

Model Answer

Given: Diameter of orifice, d = 75 mm = 0.075 m Head, h = 3.6 m Coefficient of discharge, Cd = 0.62 g = 9.81 m/s² Required: Actual discharge Q (L/s) Solution: Step 1 — Orifice area: A = π d² / 4 = π(0.075)² / 4 = 4.418 × 10⁻³ m² Step 2 — Apply orifice discharge formula: Q = Cd × A × √(2gh) Q = 0.62 × 4.418 × 10⁻³ × √(2 × 9.81 × 3.6) Q = 0.62 × 4.418 × 10⁻³ × √(70.632) Q = 0.62 × 4.418 × 10⁻³ × 8.404 Q = 2.739 × 10⁻² × 0.62 Wait — computing correctly: Q = 0.62 × 4.418 × 10⁻³ × 8.404 = 0.62 × 3.712 × 10⁻² = 2.301 × 10⁻² m³/s Step 3 — Convert to L/s: Q = 2.301 × 10⁻² × 1000 = 23.01 L/s Answer: Q ≈ 23.0 L/s

Question Type

numerical

Answer Structure

• Given block: List d, h, Cd, g clearly [no mark, but expected for organization]
• Step 1: Compute orifice area A = πd²/4 [1 mark]
• Step 2: Apply Q = Cd A √(2gh) with correct substitution [1 mark]
• Step 3: Correct arithmetic and unit conversion to L/s with boxed answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using d instead of d²/4 for area — incorrect area loses Step 1 mark
• Forgetting to multiply by Cd (using ideal formula) — loses 1 mark
• Leaving answer in m³/s when L/s is requested — loses final mark
• Using g = 10 m/s² without justification — may lose accuracy mark

Key Phrases To Include

• Q = Cd × A × √(2gh)
• A = πd²/4
• √(2 × 9.81 × h)
• convert m³/s to L/s by multiplying by 1000

Marks

3

Topic

Rectangular Weir — Numerical

Difficulty

medium

Template Id

T5

Examiner Tip

Write out the constant (2/3)√(2g) = (2/3)(4.429) = 2.953 as a single number to simplify computation. This also demonstrates formula mastery.

Model Answer

Given: Weir length, L = 2.5 m Head over crest, H = 0.35 m Cd = 0.62 g = 9.81 m/s² End contractions and velocity of approach neglected. Required: Q (m³/s) Solution: Step 1 — Write the rectangular weir formula: Q = (2/3) × Cd × √(2g) × L × H^(3/2) Step 2 — Compute √(2g): √(2 × 9.81) = √19.62 = 4.429 m^(1/2)/s Step 3 — Compute H^(3/2): H^(3/2) = (0.35)^(3/2) = (0.35)^1 × (0.35)^(1/2) = 0.35 × 0.5916 = 0.2071 m^(3/2) Step 4 — Substitute all values: Q = (2/3) × 0.62 × 4.429 × 2.5 × 0.2071 Q = 0.6667 × 0.62 × 4.429 × 2.5 × 0.2071 Q = 0.4133 × 4.429 × 0.5178 Q = 0.4133 × 2.294 Q = 0.9481 × ... Let me compute step by step: 0.6667 × 0.62 = 0.4133 0.4133 × 4.429 = 1.8305 1.8305 × 2.5 = 4.5763 4.5763 × 0.2071 = 0.9477 Answer: Q ≈ 0.948 m³/s

Question Type

numerical

Answer Structure

• Step 1: Write the correct rectangular weir formula [1 mark]
• Step 2–3: Compute √(2g) and H^(3/2) correctly [1 mark]
• Step 4: Correct substitution and final answer with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using H^(5/2) instead of H^(3/2) — classic swap with triangular weir formula
• Using Q = Cd L H^(3/2) without the (2/3)√(2g) factor — missing constant loses formula mark
• Not stating the assumption about end contractions and velocity of approach

Key Phrases To Include

• Q = (2/3) Cd √(2g) L H^(3/2)
• H^(3/2) not H^2
• √(2g) = 4.429
• neglect end contractions
• velocity of approach neglected

Marks

3

Topic

Triangular V-Notch Weir — Numerical

Difficulty

medium

Template Id

T6

Examiner Tip

For 90° V-notch, tan(θ/2) = 1.0 simplifies the formula. Explicitly state 'tan(45°) = 1.0' — it shows deliberate computation, not luck.

Model Answer

Given: Apex angle, θ = 90° → θ/2 = 45° → tan(45°) = 1.0 Head, H = 0.30 m Cd = 0.58 g = 9.81 m/s² Required: Q (m³/s) Solution: Step 1 — Write the triangular weir formula: Q = (8/15) × Cd × √(2g) × tan(θ/2) × H^(5/2) Step 2 — Compute constants: √(2g) = √19.62 = 4.429 m^(1/2)/s tan(45°) = 1.0 H^(5/2) = (0.30)^(5/2) = (0.30)^2 × (0.30)^(1/2) = 0.09 × 0.5477 = 0.04929 m^(5/2) Step 3 — Substitute: Q = (8/15) × 0.58 × 4.429 × 1.0 × 0.04929 = 0.5333 × 0.58 × 4.429 × 0.04929 = 0.3093 × 4.429 × 0.04929 = 1.3699 × 0.04929 = 0.06753 m³/s Answer: Q ≈ 0.0675 m³/s (= 67.5 L/s)

Question Type

numerical

Answer Structure

• Step 1: Identify θ = 90°, hence tan(θ/2) = tan(45°) = 1 — write the V-notch formula [1 mark]
• Step 2: Compute H^(5/2) and √(2g) correctly [1 mark]
• Step 3: Correct substitution and final numerical answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using H^(3/2) instead of H^(5/2) — swapping with rectangular weir
• Writing tan(90°) instead of tan(θ/2) = tan(45°) — common trigonometric error
• Using (2/3) instead of (8/15) as the weir constant

Key Phrases To Include

• Q = (8/15) Cd √(2g) tan(θ/2) H^(5/2)
• θ = 90°, tan(45°) = 1
• H^(5/2)
• (8/15) = 0.5333

Marks

2

Topic

Submerged Orifice

Difficulty

medium

Template Id

T7

Examiner Tip

Draw a quick 2-second sketch showing upstream surface at h₁, downstream at h₂, and label h_eff. The sketch replaces 50 words of explanation and guarantees the mark.

Model Answer

A submerged orifice (also called a drowned orifice) is one where the downstream water level is above the bottom of the orifice opening, so the jet does not flow freely into air but discharges into a submerged pool. For a free orifice, the driving head is h measured from the free surface to the orifice center. For a submerged orifice, the effective head is the difference in water surface elevations on the upstream and downstream sides: h_eff = h₁ − h₂ where h₁ = upstream head above the orifice datum and h₂ = downstream head above the same datum. The discharge formula remains the same in form: Q = Cd × A × √(2g × h_eff) = Cd × A × √(2g(h₁ − h₂)) The Cd value for a submerged orifice is generally similar to the free-flow value, but the driving head is reduced, resulting in lower discharge.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Define submerged orifice — downstream water above orifice, jet into submerged pool [1 mark]
• Sentence 3–5: State modified head h_eff = h₁ − h₂ and write the formula Q = Cd A √(2g h_eff) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using only upstream head instead of the difference — fundamental concept error
• Not writing the modified formula — answer is incomplete for 2 marks
• Confusing submerged orifice with a pipe flow scenario

Key Phrases To Include

• downstream water level above the orifice
• effective head h_eff = h₁ − h₂
• difference in water surface elevations
• Q = Cd A √(2g(h₁ − h₂))

Marks

5

Topic

Time to Empty a Tank

Difficulty

hard

Template Id

T8

Examiner Tip

This is a derivation question. Every step must be shown with a brief justification (one line per step). Examiners award marks at each logical step — if you write only the answer, you get 1 out of 5.

Model Answer

Given: Tank plan area = As (constant) Orifice area = Ao Discharge coefficient = Cd Initial head = h₁; Final head = h₂ (h₂ < h₁) g = 9.81 m/s² Derivation: Step 1 — Set up the continuity equation. At any instant, the head in the tank is h. The instantaneous discharge through the orifice is: Q = Cd × Ao × √(2gh) ... (outflow) Step 2 — Apply continuity (volume balance). In a time interval dt, the volume of water leaving the tank through the orifice equals the drop in water volume in the tank: Q × dt = −As × dh (negative because h decreases as water leaves) Substituting Q: Cd × Ao × √(2gh) × dt = −As × dh Step 3 — Rearrange and integrate. dt = −(As / (Cd × Ao × √(2g))) × dh / √h dt = −(As / (Cd × Ao × √(2g))) × h^(−1/2) dh Integrating from h₁ to h₂ (tank draining, so h goes from h₁ down to h₂): t = (As / (Cd × Ao × √(2g))) × ∫[h₂ to h₁] h^(−1/2) dh Note: The limits flip to remove the negative sign. ∫ h^(−1/2) dh = [2h^(1/2)] = 2√h t = (As / (Cd × Ao × √(2g))) × [2√h₁ − 2√h₂] Step 4 — Final formula: ┌─────────────────────────────────────────────────────┐ │ t = 2As(√h₁ − √h₂) / (Cd × Ao × √(2g)) │ └─────────────────────────────────────────────────────┘ Physical interpretation: • t is proportional to As/Ao — a larger tank or smaller orifice takes more time. • t depends on (√h₁ − √h₂), not (h₁ − h₂) — the draining rate slows as head decreases. • To empty completely: set h₂ = 0, giving t = 2As√h₁/(Cd Ao √(2g)).

Question Type

long_answer

Answer Structure

• Step 1: State the instantaneous orifice discharge formula Q = Cd Ao √(2gh) [1 mark]
• Step 2: Write the continuity/volume balance equation Q dt = −As dh [1 mark]
• Step 3: Separate variables and set up the integral correctly [1 mark]
• Step 4: Evaluate the integral ∫h^(−1/2) dh = 2√h and apply limits [1 mark]
• Step 5: State the final formula with boxed result AND physical interpretation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Missing the negative sign in Q dt = −As dh and not explaining why the limits flip — loses logic mark
• Incorrect integration: writing ∫h^(−1/2) dh = h^(1/2) (missing the factor of 2) — loses integral mark
• Not boxing or explicitly stating the final formula — loses the last mark
• Skipping the derivation and just writing the formula — scores 1 mark at most for a 5-mark derivation

Key Phrases To Include

• Q = Cd Ao √(2gh)
• Q dt = −As dh
• separate variables
• integrate h^(−1/2) dh = 2√h
• t = 2As(√h₁ − √h₂)/(Cd Ao √(2g))

Marks

5

Topic

Time to Empty a Tank — Numerical

Difficulty

hard

Template Id

T9

Examiner Tip

The most common error in this problem type is using (h₁ − h₂) instead of (√h₁ − √h₂). Memorize: the formula involves square roots of heads, not the heads themselves.

Model Answer

Given: Tank dimensions: 3 m × 4 m → As = 3 × 4 = 12 m² Orifice diameter: d = 100 mm = 0.10 m Cd = 0.60 Initial head: h₁ = 4.0 m Final head: h₂ = 1.0 m g = 9.81 m/s² Required: Time t (seconds) Solution: Step 1 — Orifice area: Ao = π d² / 4 = π(0.10)² / 4 = 7.854 × 10⁻³ m² Step 2 — Apply the draining time formula: t = 2 As (√h₁ − √h₂) / (Cd × Ao × √(2g)) Step 3 — Compute each component: √h₁ = √4.0 = 2.000 m^(1/2) √h₂ = √1.0 = 1.000 m^(1/2) √h₁ − √h₂ = 2.000 − 1.000 = 1.000 m^(1/2) √(2g) = √(2 × 9.81) = √19.62 = 4.429 m^(1/2)/s Cd × Ao × √(2g) = 0.60 × 7.854 × 10⁻³ × 4.429 = 0.60 × 0.03479 = 0.020874 m^(5/2)/s Step 4 — Compute numerator: 2 × As × (√h₁ − √h₂) = 2 × 12 × 1.000 = 24.0 m^(5/2) Step 5 — Final answer: t = 24.0 / 0.020874 = 1149.8 s ┌──────────────────────┐ │ t ≈ 1150 seconds │ │ (≈ 19.2 minutes) │ └──────────────────────┘

Question Type

numerical

Answer Structure

• Given block and orifice area calculation Ao = πd²/4 [1 mark]
• State the correct formula t = 2As(√h₁ − √h₂)/(Cd Ao √(2g)) [1 mark]
• Correct computation of √h₁ − √h₂ and √(2g) [1 mark]
• Correct denominator: Cd × Ao × √(2g) [1 mark]
• Correct final answer ~1150 s with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using h₁ − h₂ = 3.0 instead of √h₁ − √h₂ = 1.0 — most common error, loses 2 marks
• Using d instead of d²/4 for orifice area — loses area mark
• Forgetting Cd in the denominator — loses 1 mark
• Incorrect unit for final answer (e.g., minutes without conversion)

Key Phrases To Include

• As = 3 × 4 = 12 m²
• Ao = πd²/4
• t = 2As(√h₁ − √h₂)/(Cd Ao √(2g))
• √4.0 = 2.0, √1.0 = 1.0
• √(2g) = 4.429

Marks

3

Topic

Rectangular Weir — Design Problem

Difficulty

medium

Template Id

T10

Examiner Tip

For 'find L' or 'find H' problems, show the rearrangement explicitly. Examiners need to see you can manipulate the formula, not just substitute known values.

Model Answer

Given: Q = 1.8 m³/s H = 0.45 m Cd = 0.62 End contractions neglected; velocity of approach neglected g = 9.81 m/s² Required: Crest length L (m) Solution: Step 1 — Write the rectangular weir formula: Q = (2/3) × Cd × √(2g) × L × H^(3/2) Step 2 — Solve for L: L = Q / [(2/3) × Cd × √(2g) × H^(3/2)] Step 3 — Compute components: √(2g) = √19.62 = 4.429 m^(1/2)/s H^(3/2) = (0.45)^(3/2) = (0.45)(0.6708) = 0.3019 m^(3/2) (2/3) × 0.62 × 4.429 = 0.4133 × 4.429 = 1.8305 Step 4 — Substitute: L = 1.8 / (1.8305 × 0.3019) L = 1.8 / 0.5527 L = 3.256 m Answer: L ≈ 3.26 m

Question Type

numerical

Answer Structure

• Step 1: Write rectangular weir formula and rearrange for L [1 mark]
• Step 2: Compute H^(3/2) and the constant term (2/3)Cd√(2g) [1 mark]
• Step 3: Correct substitution and final answer L ≈ 3.26 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using H^2 or H^(5/2) — wrong exponent for rectangular weir
• Arithmetic error in H^(3/2): computing (0.45)³ instead of (0.45)^1.5
• Not showing the rearrangement step — loses the method mark

Key Phrases To Include

• Q = (2/3) Cd √(2g) L H^(3/2)
• solve for L
• H^(3/2) = (0.45)^1.5
• L = Q / [(2/3) Cd √(2g) H^(3/2)]

Marks

2

Topic

Rectangular Weir — End Contractions

Difficulty

medium

Template Id

T11

Examiner Tip

Memorize the coefficient 0.1 in the Francis formula. A common distractor in MCQ is 0.2 — that applies to velocity-of-approach corrections in some references, not end contractions.

Model Answer

End contractions occur when the sides of the weir opening are not flush with the channel walls, causing the flow to contract horizontally as it approaches the weir. This reduces the effective length of the nappe below the full crest length L, resulting in a lower discharge than computed by the standard formula. The Francis formula accounts for end contractions by reducing the effective crest length: L' = L − 0.1 n H where n = number of end contractions (n = 1 for one contracted end; n = 2 for both ends contracted), H = head over the crest, and L = actual crest length. This reduced length L' is then substituted into the standard rectangular weir discharge formula: Q = (2/3) Cd √(2g) L' H^(3/2)

Question Type

short_answer

Answer Structure

• Sentence 1–2: Explain end contractions — lateral contraction of flow, reduces effective length, lowers Q [1 mark]
• Sentence 3–4: State Francis formula L' = L − 0.1nH and explain n [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing L' = L − 0.2nH (wrong coefficient) — loses the formula mark
• Stating end contractions increase discharge — fundamentally wrong
• Failing to define n — incomplete answer

Key Phrases To Include

• end contractions
• effective length reduced
• L' = L − 0.1nH
• n = number of end contractions
• Francis formula

Marks

5

Topic

Orifice Coefficients — Back-Calculation

Difficulty

hard

Template Id

T12

Examiner Tip

This question tests whether you know Cd = Q_actual/Q_ideal and Cv = v_actual/v_ideal as independent ratios. Always compute the ideal (theoretical) value first, then apply the coefficient.

Model Answer

Given: Q = 35 L/s = 0.035 m³/s h = 2.5 m d = 90 mm = 0.090 m Cv = 0.97 g = 9.81 m/s² Required: (a) Cd, (b) actual velocity v at vena contracta Solution: Part (a) — Coefficient of discharge: Step 1 — Orifice area: A = πd²/4 = π(0.090)²/4 = 6.362 × 10⁻³ m² Step 2 — Theoretical (ideal) discharge: Q_th = A × √(2gh) = 6.362 × 10⁻³ × √(2 × 9.81 × 2.5) = 6.362 × 10⁻³ × √49.05 = 6.362 × 10⁻³ × 7.004 = 4.456 × 10⁻² m³/s Step 3 — Coefficient of discharge: Cd = Q_actual / Q_theoretical Cd = 0.035 / 0.04456 Cd = 0.785 Wait — let me recheck: √(2 × 9.81 × 2.5) = √49.05 = 7.004 m/s Q_th = 6.362 × 10⁻³ × 7.004 = 4.456 × 10⁻² m³/s = 44.56 L/s Cd = 35 / 44.56 = 0.785 Note: Cd ≈ 0.785 (higher than sharp-edged orifice 0.61 — suggesting this may be a well-rounded or nozzle-type orifice. The problem tests the method.) Part (b) — Actual velocity at vena contracta: Step 4 — Theoretical velocity: v_th = √(2gh) = √(2 × 9.81 × 2.5) = 7.004 m/s Step 5 — Actual velocity: v_actual = Cv × v_th = 0.97 × 7.004 = 6.794 m/s Answers: ┌─────────────────────────────┐ │ (a) Cd = 0.785 │ │ (b) v = 6.79 m/s │ └─────────────────────────────┘

Question Type

numerical

Answer Structure

• Part (a) Step 1: Compute orifice area A = πd²/4 [1 mark]
• Part (a) Step 2: Compute theoretical discharge Q_th = A √(2gh) [1 mark]
• Part (a) Step 3: Cd = Q_actual / Q_theoretical with correct answer [1 mark]
• Part (b) Step 4: Compute theoretical velocity v_th = √(2gh) [1 mark]
• Part (b) Step 5: Actual velocity v = Cv × v_th with correct answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using Q_actual for both and not computing Q_theoretical separately
• Confusing Cv with Cd — applying Cd to compute velocity
• Not converting Q from L/s to m³/s before computing

Key Phrases To Include

• Cd = Q_actual / Q_theoretical
• Q_th = A √(2gh)
• v_th = √(2gh)
• v_actual = Cv × v_th
• A = πd²/4

Marks

2

Topic

Mouthpiece vs. Orifice

Difficulty

medium

Template Id

T13

Examiner Tip

Examiners expect you to cite typical Cd values (0.62 for orifice, 0.82 for mouthpiece). Numerical recall separates engineering answers from general statements.

Model Answer

An orifice is a sharp-edged opening in a thin plate through which fluid discharges as a free jet, with the plate thickness not exceeding the orifice diameter. A mouthpiece (or short tube) is a short pipe or tube of length approximately 2 to 3 times its diameter attached to the orifice opening. The mouthpiece generally increases the coefficient of discharge compared to a sharp-edged orifice. For a standard (cylindrical, internal) mouthpiece, the jet re-expands to fill the full tube bore after an internal vena contracta, so the area at exit = tube area (Cc ≈ 1.0). While Cv is slightly lower due to friction, the net Cd = Cv × Cc is typically ≈ 0.82 — significantly higher than the sharp-edged orifice Cd ≈ 0.62. This makes mouthpieces preferable when higher discharge is needed.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Define orifice vs. mouthpiece by geometry (thin plate vs. short tube 2–3 diameters long) [1 mark]
• Sentence 3–4: State effect on Cd — mouthpiece Cd ≈ 0.82 > orifice Cd ≈ 0.62 due to Cc ≈ 1.0 at exit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating mouthpiece reduces Cd — opposite of the correct effect
• Confusing mouthpiece with a long pipe (orifice in a thick plate context)
• Not providing numerical values — answer lacks precision for engineering-level response

Key Phrases To Include

• sharp-edged opening in thin plate
• short tube 2–3 diameters long
• Cc ≈ 1.0 at mouthpiece exit
• Cd ≈ 0.82 for mouthpiece
• higher discharge coefficient

Marks

3

Topic

V-Notch Weir — Finding Head

Difficulty

hard

Template Id

T14

Examiner Tip

Reverse-solving V-notch problems requires the step H = (H^5/2)^(2/5). Write this step explicitly — it is easy to forget and is where marks are lost.

Model Answer

Given: Apex angle, θ = 60° → θ/2 = 30° → tan(30°) = 0.5774 Cd = 0.62 Q = 0.025 m³/s g = 9.81 m/s² Required: Head H (m) Solution: Step 1 — Write the V-notch weir formula: Q = (8/15) × Cd × √(2g) × tan(θ/2) × H^(5/2) Step 2 — Solve for H^(5/2): H^(5/2) = Q / [(8/15) × Cd × √(2g) × tan(θ/2)] Step 3 — Compute the denominator: (8/15) = 0.5333 √(2g) = 4.429 tan(30°) = 0.5774 Denominator = 0.5333 × 0.62 × 4.429 × 0.5774 = 0.3307 × 4.429 × 0.5774 = 1.4642 × 0.5774 = 0.8456 Step 4 — Find H^(5/2): H^(5/2) = 0.025 / 0.8456 = 0.02957 m^(5/2) Step 5 — Solve for H: H = (0.02957)^(2/5) = (0.02957)^0.4 Compute: ln(0.02957) = −3.521 0.4 × (−3.521) = −1.408 H = e^(−1.408) = 0.2447 m Answer: H ≈ 0.245 m

Question Type

numerical

Answer Structure

• Step 1: Write V-notch formula and identify tan(30°) = 0.5774 [1 mark]
• Step 2–3: Rearrange for H^(5/2) and compute denominator correctly [1 mark]
• Step 4–5: Correct H^(5/2) then solve H = (H^5/2)^(2/5) with correct answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using tan(60°) instead of tan(30°) — wrong angle halving
• Solving H^3/2 instead of H^5/2 (treating it like a rectangular weir)
• Not raising to the power 2/5 to recover H from H^5/2

Key Phrases To Include

• θ = 60°, tan(θ/2) = tan(30°) = 0.5774
• H^(5/2) = Q / [(8/15) Cd √(2g) tan(θ/2)]
• H = (H^5/2)^(2/5)
• raise to the power 2/5

Marks

2

Topic

Orifice Coefficients — Cv and Cc

Difficulty

easy

Template Id

T15

Examiner Tip

In 2-mark questions with two sub-parts, allocate exactly one mark per part. If you write only definitions without typical numerical values, you will likely lose the precision marks.

Model Answer

Coefficient of velocity (Cv): Cv accounts for the energy loss due to friction as fluid passes through the orifice. The actual jet velocity is less than the theoretical (Torricelli) velocity, so: v_actual = Cv × √(2gh) Typical value for a sharp-edged orifice: Cv ≈ 0.97–0.99. Coefficient of contraction (Cc): Cc accounts for the reduction in jet cross-sectional area at the vena contracta compared to the orifice area, due to the inward curvature of streamlines: Ac = Cc × A Typical value for a sharp-edged orifice: Cc ≈ 0.61–0.64. The combined effect is the coefficient of discharge: Cd = Cv × Cc ≈ 0.61.

Question Type

short_answer

Answer Structure

• Part 1: Define Cv — velocity ratio, accounts for friction, v = Cv √(2gh), Cv ≈ 0.97–0.99 [1 mark]
• Part 2: Define Cc — area ratio at vena contracta, Ac = Cc A, Cc ≈ 0.62, and state Cd = Cv × Cc [1 mark]

Scoring Breakdown

Common Mark Deductions

• Swapping definitions of Cv and Cc — fundamental error loses both marks
• Giving Cv ≈ 0.62 (which is Cd's value) — shows confusion of coefficients
• Not stating Cd = Cv × Cc — misses the synthesis required at this level

Key Phrases To Include

• Cv = v_actual / v_theoretical
• friction loss
• Cv ≈ 0.97–0.99
• Cc = Ac / A
• vena contracta
• Cc ≈ 0.62
• Cd = Cv × Cc