CELE Hydraulics & Fluid Mechanics — Hydrodynamics and Fluid MachineryExam Answer Templates

Marks

1

Topic

Force of a Jet on a Flat Plate

Difficulty

easy

Template Id

T1

Examiner Tip

For a 1-mark VSA, the examiner wants the formula only — do not waste time deriving it. Write it cleanly, define the symbols in one line, and move on.

Model Answer

The force of a jet striking a stationary flat plate normally is given by: F = ρQv = ρAv² where ρ = fluid density (kg/m³), Q = discharge (m³/s), v = jet velocity (m/s), and A = jet cross-sectional area (m²).

Question Type

very_short_answer

Answer Structure

• Line 1: Write the correct formula F = ρQv or F = ρAv² [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing F = ρQΔv without simplifying for a stationary plate (v_out = 0, so Δv = v)
• Using γ (specific weight) instead of ρ (density) in the momentum equation
• Omitting units or writing force in Pa instead of N

Key Phrases To Include

• F = ρQv
• ρAv²
• momentum
• stationary plate
• normal

Marks

2

Topic

Force on Stationary and Moving Curved Vanes

Difficulty

medium

Template Id

T2

Examiner Tip

Examiners specifically set 2-mark questions on this distinction because it is the most common error. Explicitly write the phrase 'relative velocity (v − u)' — that phrase alone signals that you understand the concept.

Model Answer

For a stationary curved vane: F = ρQv(1 − cosθ) where v = absolute jet velocity, Q = actual discharge. For a moving curved vane (vane speed = u): F = ρQ_rel · (v − u)(1 − cosθ) where (v − u) = relative velocity of jet with respect to vane, and Q_rel = A(v − u) is the relative discharge striking the vane. Key difference: A moving vane intercepts less flow (relative discharge), so both the relative velocity (v − u) and the effective discharge are reduced, resulting in a smaller force than for an identical stationary vane.

Question Type

short_answer

Answer Structure

• Line 1–2: Stationary vane formula with definition of v and Q [1 mark]
• Line 3–5: Moving vane formula with relative velocity (v − u) and Q_rel, plus explanation of the key difference [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using absolute velocity v instead of relative velocity (v − u) in the moving-vane formula
• Using the same Q for both cases without noting that moving vane intercepts less flow
• Confusing θ (turning angle) with the angle of the jet to the horizontal

Key Phrases To Include

• relative velocity
• (v − u)
• relative discharge
• Q_rel = A(v − u)
• stationary vane
• moving vane
• (1 − cosθ)

Marks

2

Topic

Force of a Jet on a Flat Plate

Difficulty

easy

Template Id

T3

Examiner Tip

Write 'Q = Av' as a separate numbered step — even for simple problems. It shows a systematic approach and protects your process mark if you make an arithmetic error.

Model Answer

Given: A = 0.005 m², v = 20 m/s, ρ = 1000 kg/m³ Step 1 — Discharge: Q = Av = 0.005 × 20 = 0.10 m³/s Step 2 — Force (momentum equation, stationary normal plate): F = ρQv = 1000 × 0.10 × 20 ∴ F = 2000 N = 2.0 kN

Question Type

numerical

Answer Structure

• Step 1: Compute Q = Av [½ mark — process]
• Step 2: Apply F = ρQv and substitute [½ mark — correct formula]
• Step 3: Correct numerical answer with unit (2000 N or 2.0 kN) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to compute Q first and plugging in A instead of Q in F = ρQv
• Using γ = 9810 N/m³ instead of ρ = 1000 kg/m³ in the momentum equation
• No unit in final answer

Key Phrases To Include

• Q = Av
• F = ρQv
• momentum equation
• 2000 N
• 2.0 kN

Marks

2

Topic

Pumps — NPSH and Cavitation

Difficulty

medium

Template Id

T4

Examiner Tip

For definition-plus-significance questions, always use the structure: 'X is defined as... Its significance is...' — this guarantees you address both marking points clearly.

Model Answer

Net Positive Suction Head (NPSH) is the difference between the absolute pressure head at the pump suction inlet and the vapor pressure head of the liquid at the operating temperature, measured in metres of fluid. Significance: If the available NPSH (NPSH_A) at the pump inlet falls below the required NPSH (NPSH_R) specified by the manufacturer, the local pressure drops to the vapor pressure of the liquid, causing cavitation — the formation and violent collapse of vapor bubbles. Cavitation causes noise, vibration, pitting of impeller surfaces, and a sudden drop in pump head and discharge, potentially destroying the pump. To prevent cavitation, the designer must ensure NPSH_A > NPSH_R at all operating conditions.

Question Type

short_answer

Answer Structure

• Sentence 1: Definition of NPSH (absolute suction pressure head minus vapor pressure head) [1 mark]
• Sentence 2–3: Consequence if NPSH_A < NPSH_R — cavitation, its effects, and the design rule NPSH_A > NPSH_R [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining NPSH as gauge pressure instead of absolute pressure
• Stating cavitation occurs without explaining the physical mechanism (pressure drops to vapor pressure)
• Omitting the design requirement NPSH_A > NPSH_R

Key Phrases To Include

• absolute pressure head
• vapor pressure head
• NPSH_A
• NPSH_R
• cavitation
• impeller pitting
• NPSH_A > NPSH_R

Marks

3

Topic

Pump Power and Efficiency

Difficulty

easy

Template Id

T5

Examiner Tip

Write the rule explicitly: 'For a pump, efficiency = output/input, so input = output/η.' This one-line statement earns a process mark even if your arithmetic is slightly off.

Model Answer

Given: Q = 0.05 m³/s, H = 30 m, η = 75% = 0.75 γ = 9.81 kN/m³ Step 1 — Water (output) power of the pump: P_water = γQH = 9.81 × 0.05 × 30 = 14.715 kW Step 2 — Input (shaft/motor) power: For a pump: efficiency = (output power)/(input power) ∴ P_input = P_water / η = 14.715 / 0.75 ∴ P_input = 19.62 kW The motor must supply at least 19.62 kW to drive the pump.

Question Type

numerical

Answer Structure

• Step 1: List all given data with units [½ mark]
• Step 2: Compute water power P_water = γQH (correct formula and substitution) [1 mark]
• Step 3: State efficiency relationship and compute P_input = P_water/η [1 mark]
• Step 4: Correct final answer with unit (19.62 kW) [½ mark]

Scoring Breakdown

Common Mark Deductions

• Multiplying by η instead of dividing (reversing pump and turbine efficiency rules)
• Using ρ = 1000 kg/m³ instead of γ = 9810 N/m³ (or 9.81 kN/m³) in the power formula
• Answering in watts without converting to kW when γ is used in kN/m³
• Omitting final unit or stating answer as 19.62 without 'kW'

Key Phrases To Include

• P_water = γQH
• P_input = γQH/η
• 14.715 kW
• 19.62 kW
• efficiency divides for pump

Marks

3

Topic

Turbine Power and Efficiency

Difficulty

easy

Template Id

T6

Examiner Tip

Contrast your turbine answer with the pump formula in a brief note: 'Turbine: P_out = η·γQH (multiply); Pump: P_in = γQH/η (divide).' Examiners appreciate when you show awareness of the distinction.

Model Answer

Given: H = 15 m, Q = 2 m³/s, η = 85% = 0.85 γ = 9.81 kN/m³ Step 1 — Available (gross) hydraulic power: P_hydraulic = γQH = 9.81 × 2 × 15 = 294.3 kW Step 2 — Shaft (output) power: For a turbine: P_output = η × P_hydraulic P_output = 0.85 × 294.3 ∴ P_output = 250.2 kW The turbine delivers 250.2 kW of shaft power.

Question Type

numerical

Answer Structure

• Step 1: List given data [½ mark]
• Step 2: Compute hydraulic power P = γQH [1 mark]
• Step 3: Apply P_output = η·γQH for turbine (multiply by η) [1 mark]
• Step 4: Correct final answer with unit (250.2 kW) [½ mark]

Scoring Breakdown

Common Mark Deductions

• Dividing by η instead of multiplying (confusing turbine with pump formula)
• Using H as gross head without noting the question specifies 'net head' (no deduction if net is already given, but must be acknowledged)
• Unit error: answering in watts (250,155 W) without converting to kW

Key Phrases To Include

• P_output = η·γQH
• 294.3 kW
• 250.2 kW
• net head
• turbine efficiency multiplies

Marks

3

Topic

Force on Curved Vane

Difficulty

medium

Template Id

T7

Examiner Tip

Always state explicitly: 'The turning angle θ = 120°, cos120° = −0.5, so (1 − cosθ) = 1 − (−0.5) = 1.5.' This step is the most common error source and writing it out completely earns you full process marks.

Model Answer

Given: d = 50 mm = 0.05 m, v = 25 m/s, θ = 120°, ρ = 1000 kg/m³ A = π/4 × (0.05)² = 1.9635 × 10⁻³ m² Step 1 — Discharge: Q = Av = 1.9635 × 10⁻³ × 25 = 0.04909 m³/s Step 2 — Force on vane (curved vane formula): F = ρQv(1 − cosθ) [θ = turning angle = 120°] cos120° = −0.5 F = 1000 × 0.04909 × 25 × (1 − (−0.5)) F = 1000 × 0.04909 × 25 × 1.5 F = 1227.2 × 1.5 ∴ F = 1840.8 N ≈ 1.84 kN The resultant force on the vane is approximately 1.84 kN.

Question Type

numerical

Answer Structure

• Step 1: Compute jet area A = π/4·d² [½ mark]
• Step 2: Compute discharge Q = Av [½ mark]
• Step 3: Apply F = ρQv(1 − cosθ) with θ = 120° and evaluate cos120° = −0.5 [1 mark]
• Step 4: Correct final answer ≈ 1.84 kN with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using θ = 60° (the supplement) instead of the actual turning angle θ = 120°
• Computing cos120° as +0.5 instead of −0.5 (sign error reducing force by half)
• Using jet diameter directly as radius in area calculation
• Not converting 50 mm to 0.05 m before squaring

Key Phrases To Include

• F = ρQv(1 − cosθ)
• cos120° = −0.5
• (1 − cosθ) = 1.5
• turning angle
• 1840 N
• 1.84 kN

Marks

3

Topic

Pump Affinity Laws

Difficulty

easy

Template Id

T8

Examiner Tip

Memorize using the mnemonic '1-2-3': Q is first power, H is second power, P is third power of N. Write all three in ratio form immediately — it takes 20 seconds and guarantees 3 marks.

Model Answer

The pump affinity laws describe how the performance of a geometrically similar centrifugal pump changes when its rotational speed N changes (at constant impeller diameter): 1. Discharge varies directly with speed: Q₂/Q₁ = N₂/N₁ (Q ∝ N) 2. Total head varies as the square of speed: H₂/H₁ = (N₂/N₁)² (H ∝ N²) 3. Power varies as the cube of speed: P₂/P₁ = (N₂/N₁)³ (P ∝ N³) Physical basis: Discharge follows impeller tip speed (∝ N); head is proportional to the square of tip speed (kinetic energy); power = γQH ∝ N · N² = N³.

Question Type

short_answer

Answer Structure

• Law 1: Q ∝ N with ratio form Q₂/Q₁ = N₂/N₁ [1 mark]
• Law 2: H ∝ N² with ratio form H₂/H₁ = (N₂/N₁)² [1 mark]
• Law 3: P ∝ N³ with ratio form P₂/P₁ = (N₂/N₁)³ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing exponents: writing H ∝ N³ and P ∝ N² (swapping the laws for head and power)
• Not writing the ratio form — examiners expect the usable equation, not just proportionality
• Applying laws to different-diameter impellers without noting the constant-diameter condition

Key Phrases To Include

• Q ∝ N
• H ∝ N²
• P ∝ N³
• affinity laws
• constant impeller diameter
• N₂/N₁

Marks

5

Topic

Pump Affinity Laws — Numerical Application

Difficulty

medium

Template Id

T9

Examiner Tip

Always compute the speed ratio r = N₂/N₁ as a single decimal first. Then: Q₂ = Q₁·r, H₂ = H₁·r², P₂ = P₁·r³. Boxing the ratio value prevents repetitive arithmetic errors across all three laws.

Model Answer

Given: N₁ = 1450 rpm, H₁ = 25 m, Q₁ = 0.03 m³/s N₂ = 1750 rpm Speed ratio: N₂/N₁ = 1750/1450 = 1.2069 Affinity Law 1 — New discharge: Q₂/Q₁ = N₂/N₁ Q₂ = Q₁ × (N₂/N₁) = 0.03 × 1.2069 Q₂ = 0.0362 m³/s Affinity Law 2 — New head: H₂/H₁ = (N₂/N₁)² H₂ = H₁ × (N₂/N₁)² = 25 × (1.2069)² H₂ = 25 × 1.4566 H₂ = 36.4 m Affinity Law 3 — Power ratio: P₂/P₁ = (N₂/N₁)³ = (1.2069)³ = 1.759 (Power increases by a factor of 1.76 — motor must be sized accordingly.) Summary: Q₂ = 0.0362 m³/s, H₂ = 36.4 m, P₂/P₁ = 1.76

Question Type

numerical

Answer Structure

• Step 1: Compute speed ratio N₂/N₁ = 1.2069 [½ mark]
• Step 2: Apply Law 1 to find Q₂ = 0.0362 m³/s [1 mark]
• Step 3: Apply Law 2 to find H₂ = 36.4 m [1½ marks]
• Step 4: Apply Law 3 to find P₂/P₁ = 1.76 [1½ marks]
• Step 5: Summary statement [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using N₁/N₂ (inverted ratio) leading to reduced values instead of increased ones
• Forgetting to square the ratio for H or cube it for P
• Not computing the speed ratio as a single number first — leads to arithmetic errors in squaring/cubing
• Omitting the power ratio or stating only the power ratio without computing H₂ and Q₂

Key Phrases To Include

• N₂/N₁ = 1.2069
• Q₂ = Q₁(N₂/N₁)
• H₂ = H₁(N₂/N₁)²
• P₂/P₁ = (N₂/N₁)³
• 36.4 m
• 1.76

Marks

5

Topic

Force on a Pipe Bend

Difficulty

hard

Template Id

T10

Examiner Tip

Draw the free-body diagram of the control volume FIRST. Label the four force contributions: two momentum fluxes (in and out) and two pressure forces (at each port). Missing any one of these four terms costs you marks.

Model Answer

Given: Q = 0.2 m³/s, D₁ = 0.30 m, D₂ = 0.20 m, p₁ = 200 kPa Bend turns flow from +x (inlet) to +y (outlet); 90° horizontal bend. ρ = 1000 kg/m³, γ = 9.81 kN/m³ Step 1 — Cross-sectional areas: A₁ = π/4 × (0.30)² = 0.07069 m² A₂ = π/4 × (0.20)² = 0.03142 m² Step 2 — Velocities: v₁ = Q/A₁ = 0.2/0.07069 = 2.829 m/s v₂ = Q/A₂ = 0.2/0.03142 = 6.366 m/s Step 3 — Find p₂ using Bernoulli (neglect elevation change): p₁/γ + v₁²/2g = p₂/γ + v₂²/2g p₂ = p₁ + ρ/2·(v₁² − v₂²) p₂ = 200,000 + 500·(2.829² − 6.366²) p₂ = 200,000 + 500·(8.003 − 40.525) p₂ = 200,000 − 16,261 = 183,739 Pa ≈ 183.7 kPa Step 4 — Momentum equation in x-direction: ΣFx = ρQ(v₂x − v₁x) Flow exits in +y direction, so v₂x = 0; flow enters in +x, so v₁x = v₁ = 2.829 m/s p₁A₁ − Rx = ρQ(0 − v₁) [Rx = reaction force component in x on fluid] Rx = p₁A₁ + ρQ·v₁ Rx = (200,000 × 0.07069) + (1000 × 0.2 × 2.829) Rx = 14,138 + 565.8 = 14,703.8 N = 14.70 kN Step 5 — Momentum equation in y-direction: Flow exits at +y with v₂y = v₂ = 6.366 m/s; v₁y = 0 −p₂A₂ − Ry = ρQ(v₂ − 0) [sign: p₂A₂ acts in −y on the control volume] Ry = −p₂A₂ − ρQ·v₂ |Ry| = (183,739 × 0.03142) + (1000 × 0.2 × 6.366) |Ry| = 5,771.7 + 1,273.2 = 7,044.9 N = 7.04 kN Step 6 — Resultant anchoring force: R = √(Rx² + Ry²) = √(14.70² + 7.04²) R = √(216.09 + 49.56) = √265.65 ∴ R = 16.30 kN Angle of resultant with x-axis: φ = arctan(7.04/14.70) = 25.6°

Question Type

numerical

Answer Structure

• Step 1–2: Compute areas and velocities [1 mark]
• Step 3: Apply Bernoulli to find p₂ [1 mark]
• Step 4: Apply momentum in x: include pressure force p₁A₁ AND momentum flux ρQv₁ [1 mark]
• Step 5: Apply momentum in y: include pressure force p₂A₂ AND momentum flux ρQv₂ [1 mark]
• Step 6: Combine Rx and Ry vectorially for resultant R = 16.30 kN [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting the pressure force terms p₁A₁ and p₂A₂ — this is the most common and costliest error (−2 marks)
• Not finding p₂ using Bernoulli — assuming p₂ = p₁ when areas differ
• Incorrect velocity direction assumptions for the 90° bend
• Not taking the vector sum — adding Rx + Ry algebraically instead of using Pythagoras

Key Phrases To Include

• control volume
• p₁A₁
• p₂A₂
• ρQ(v₂ − v₁)
• Bernoulli
• resultant
• √(Rx² + Ry²)
• anchoring force

Marks

3

Topic

Turbine — Discharge from Power

Difficulty

medium

Template Id

T11

Examiner Tip

When a problem gives P_output and asks for Q, the Pelton turbine is the tip-off for impulse type. The formula rearranges simply — write it, rearrange it, substitute. Three clear steps, three marks.

Model Answer

Given: P_output = 500 kW, H = 120 m, η = 0.88 γ = 9.81 kN/m³ Using the turbine power formula: P_output = η · γ · Q · H Solving for Q: Q = P_output / (η · γ · H) Q = 500 / (0.88 × 9.81 × 120) Q = 500 / 1035.936 ∴ Q = 0.4826 m³/s ≈ 0.483 m³/s The turbine requires a discharge of approximately 0.483 m³/s.

Question Type

numerical

Answer Structure

• Step 1: Write turbine power formula P = η·γ·Q·H [1 mark]
• Step 2: Rearrange for Q = P/(η·γ·H) and substitute [1 mark]
• Step 3: Correct answer Q ≈ 0.483 m³/s with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using P = γQH/η (pump formula) instead of P = η·γQH (turbine formula)
• Not converting γ to kN/m³ when P is given in kW — unit mismatch
• Arithmetic error in denominator: not multiplying η × γ × H correctly

Key Phrases To Include

• P_output = η·γ·Q·H
• Pelton turbine
• impulse turbine
• Q = P/(η·γ·H)
• 0.483 m³/s

Marks

2

Topic

Types of Turbines

Difficulty

medium

Template Id

T12

Examiner Tip

Memorize the one-line distinction: 'Impulse = jet in air; Reaction = runner submerged.' Then expand for marks. Philippine examples show local relevance and impress examiners.

Model Answer

Impulse Turbine (e.g., Pelton Wheel): All available head is converted to kinetic energy in nozzles before striking the runner. The runner operates in air (atmospheric pressure throughout); pressure does not change across the buckets. Used for high head, low discharge applications. Philippine example: Proposed or existing micro-hydro Pelton installations in the Cordillera highlands of Northern Luzon, where steep mountain streams provide heads exceeding 100 m. Reaction Turbine (e.g., Francis, Kaplan): Only part of the head converts to velocity in the guide vanes; the remainder converts to pressure energy within the runner itself. The runner is fully submerged; pressure drops across the runner. Used for low-to-medium head, high discharge applications. Philippine example: Francis turbines at the Pantabangan–Masiway hydroelectric complex (Pantabangan Dam, Nueva Ecija) and Angat Dam hydroelectric plant, operating under moderate heads with high flow.

Question Type

short_answer

Answer Structure

• Lines 1–2: Impulse turbine — mechanism (all head to KE, atmospheric pressure on runner) and head range [1 mark]
• Lines 3–4: Reaction turbine — mechanism (partial pressure conversion within runner, fully submerged) and head range, plus relevant Philippine example for each [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that impulse turbines use 'all the pressure' — they use kinetic energy from the converted pressure, not pressure directly
• Incorrectly classifying Francis turbine as impulse
• Not providing a Philippine-context example when asked

Key Phrases To Include

• impulse
• reaction
• Pelton
• Francis
• Kaplan
• atmospheric pressure
• high head
• submerged runner
• nozzle
• guide vanes

Marks

2

Topic

Pump Power and Efficiency

Difficulty

easy

Template Id

T13

Examiner Tip

A quick sanity check: motor power must always be GREATER than water power for a pump. If your answer is less than γQH = 35.3 kW, you have made the division/multiplication error.

Model Answer

Given: Q = 0.08 m³/s, H = 45 m, η = 0.70, γ = 9.81 kN/m³ Input (motor) power: P_input = γQH / η = (9.81 × 0.08 × 45) / 0.70 P_input = 35.316 / 0.70 ∴ P_input = 50.45 kW

Question Type

numerical

Answer Structure

• Step 1: Write formula P_input = γQH/η with values [1 mark]
• Step 2: Correct final answer 50.45 kW with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Multiplying instead of dividing by η (giving 24.72 kW — a sub-efficiency answer, which is physically impossible for pump input)
• Using ρ = 1000 kg/m³ and forgetting to divide by 1000 to convert N·m/s to kW

Key Phrases To Include

• P_input = γQH/η
• 35.316 kW
• 50.45 kW
• motor power

Marks

1

Topic

Turbine Classification — Specific Speed

Difficulty

medium

Template Id

T14

Examiner Tip

For 1-mark concept questions, write a single crisp sentence that contains both the definition and the application — examiners look for both in VSA scoring.

Model Answer

Specific speed (Ns) of a turbine is the rotational speed at which a geometrically similar turbine would produce unit power under unit head. It is used to classify turbine type: low Ns (10–50) → Pelton (impulse); medium Ns (50–300) → Francis; high Ns (300–900) → Kaplan/propeller (reaction).

Question Type

very_short_answer

Answer Structure

• One sentence: definition of specific speed [½ mark]
• One sentence: significance for turbine-type selection with approximate ranges [½ mark]

Scoring Breakdown

Common Mark Deductions

• Confusing turbine specific speed with pump specific speed (both exist but have different expressions)
• Stating Ns without noting its role in selecting turbine type

Key Phrases To Include

• specific speed
• unit power
• unit head
• turbine selection
• Pelton
• Francis
• Kaplan

Marks

5

Topic

Force and Power — Moving Flat Vane

Difficulty

hard

Template Id

T15

Examiner Tip

For any moving-vane problem, ALWAYS write 'v_rel = v − u' as the very first formula after listing given data. This single line earns a mark AND sets up every subsequent step correctly. If you use absolute velocity, every downstream answer is wrong.

Model Answer

Given: d = 30 mm = 0.03 m, v_jet = 15 m/s, u = 6 m/s (vane speed) ρ = 1000 kg/m³ A = π/4 × (0.03)² = 7.069 × 10⁻⁴ m² Step 1 — Relative velocity of jet with respect to vane: v_rel = v_jet − u = 15 − 6 = 9 m/s Step 2 — Relative discharge (flow striking the moving vanes): Q_rel = A · v_rel = 7.069 × 10⁻⁴ × 9 = 6.362 × 10⁻³ m³/s Step 3 — Force on vane (flat plate, normal impact): For a flat plate, the jet bounces back; relative velocity reverses direction. F = ρ · Q_rel · v_rel = 1000 × 6.362 × 10⁻³ × 9 F = 57.26 N Step 4 — Power developed by the moving vanes: P = F × u = 57.26 × 6 P = 343.6 W ≈ 344 W Note: For a moving flat vane, maximum power occurs at u = v/3 (one-third rule). Here u = 6 m/s = 0.4v — slightly past the maximum point. Summary: Force = 57.3 N; Power = 344 W

Question Type

numerical

Answer Structure

• Step 1: Compute jet area A [½ mark]
• Step 2: Compute relative velocity v_rel = v − u = 9 m/s [1 mark]
• Step 3: Compute relative discharge Q_rel = A·v_rel [1 mark]
• Step 4: Apply F = ρ·Q_rel·v_rel = 57.3 N [1 mark]
• Step 5: Compute P = F × u = 344 W [1½ marks]

Scoring Breakdown

Common Mark Deductions

• Using absolute velocity v = 15 m/s instead of relative velocity 9 m/s for a moving vane
• Using the absolute discharge Q = Av (based on absolute v) instead of relative discharge Q_rel = A·v_rel
• Computing power as P = γQH instead of P = F·u for a moving vane problem
• Not explicitly stating v_rel = v − u before substituting — loses the process mark

Key Phrases To Include

• relative velocity
• v_rel = v − u = 9 m/s
• relative discharge
• Q_rel = A·v_rel
• F = ρ·Q_rel·v_rel
• P = F·u
• 57.3 N
• 344 W