CELE Hydraulics & Fluid Mechanics — Fundamentals of Fluid FlowExam Answer Templates

Marks

1

Topic

Continuity / Discharge

Difficulty

easy

Template Id

T1

Examiner Tip

One-mark questions demand precision in one or two lines — no derivation needed, but the SI unit is non-negotiable for full credit.

Model Answer

Discharge Q is the volume of fluid passing through a cross-section per unit time. SI unit: m³/s.

Question Type

very_short_answer

Answer Structure

• Line 1: Concise definition linking volume, cross-section, and time [½ mark]
• Line 2: Correct SI unit stated explicitly [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'litres per second' without converting — must state m³/s for SI
• Omitting the word 'volume' and writing only 'flow per second'
• Confusing Q (volume flow) with mass flow rate ṁ (kg/s)

Key Phrases To Include

• volume of fluid
• per unit time
• cross-section
• m³/s

Marks

1

Topic

Continuity Equation

Difficulty

easy

Template Id

T2

Examiner Tip

Examiners want to see both the equation and the physical meaning (Q is constant). Two components in one mark — deliver both in two concise lines.

Model Answer

For steady, incompressible flow: Q = A₁v₁ = A₂v₂, where A is cross-sectional area (m²) and v is mean velocity (m/s). Discharge Q is constant along the flow path.

Question Type

very_short_answer

Answer Structure

• Line 1: State the equation in symbolic form [½ mark]
• Line 2: Confirm that Q is constant / conservation of mass statement [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing only Q = Av without the subscripts — misses the comparison between sections
• Not specifying the flow conditions (steady, incompressible)

Key Phrases To Include

• Q = A₁v₁ = A₂v₂
• steady
• incompressible
• constant along flow path
• conservation of mass

Marks

2

Topic

Continuity Equation

Difficulty

easy

Template Id

T3

Examiner Tip

The square on the diameter ratio is the key differentiator. Examiners mark this step explicitly — show it clearly.

Model Answer

Given: D₁ = 200 mm = 0.200 m, D₂ = 100 mm = 0.100 m, v₁ = 3 m/s Required: v₂ Governing equation (Continuity): v₂ = v₁(D₁/D₂)² Substitution: v₂ = 3 × (200/100)² v₂ = 3 × 4 v₂ = 12 m/s

Question Type

numerical

Answer Structure

• Line 1–2: Write Given and Required data [½ mark]
• Line 3: State the continuity formula v₂ = v₁(D₁/D₂)² [½ mark]
• Line 4: Substitute values correctly [½ mark]
• Line 5: State final answer with unit [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using the ratio D₁/D₂ instead of (D₁/D₂)² — forgetting to square the diameter ratio
• Leaving diameter in mm instead of converting to m (though ratio cancels, habit of non-conversion causes errors in Q)
• Omitting the unit 'm/s' on the final answer

Key Phrases To Include

• v₂ = v₁(D₁/D₂)²
• continuity
• velocity increases as area decreases
• 12 m/s

Marks

2

Topic

Energy Grade Line and Hydraulic Grade Line

Difficulty

medium

Template Id

T4

Examiner Tip

Board exams frequently ask this as a 2-mark conceptual question. The formula-based distinction plus the physical meaning (EGL is always higher by v²/2g) earns both marks.

Model Answer

The Energy Grade Line (EGL) represents the total head at each point along the flow: EGL = p/γ + v²/2g + z (m) The Hydraulic Grade Line (HGL) represents the piezometric head (pressure + elevation only): HGL = p/γ + z (m) Relationship: HGL = EGL − v²/2g The EGL always lies above the HGL by the velocity head v²/2g. Both lines slope downward in the direction of flow due to head losses.

Question Type

short_answer

Answer Structure

• Line 1–2: Define EGL with formula [½ mark]
• Line 3–4: Define HGL with formula [½ mark]
• Line 5: State the relationship HGL = EGL − v²/2g [½ mark]
• Line 6: Note the relative position and slope direction [½ mark]

Scoring Breakdown

Common Mark Deductions

• Defining HGL as 'pressure head only' — must include elevation head z
• Not stating the relationship between EGL and HGL
• Saying the EGL slopes upward — it only slopes upward when a pump is present

Key Phrases To Include

• total head
• piezometric head
• velocity head v²/2g
• EGL above HGL
• slope downward
• head loss

Marks

2

Topic

Energy Equation

Difficulty

medium

Template Id

T5

Examiner Tip

This equation is the backbone of every energy problem. Memorize the left-side vs right-side placement of machine heads: hA (pump) on left, hE and hL on right.

Model Answer

The general energy equation (extended Bernoulli) between sections 1 and 2: p₁/γ + v₁²/2g + z₁ + hA = p₂/γ + v₂²/2g + z₂ + hE + hL Where: p/γ = pressure head (m) v²/2g = velocity head (m) z = elevation head (m) hA = head added by pump (m) hE = head extracted by turbine (m) hL = head lost to friction and minor losses (m)

Question Type

very_short_answer

Answer Structure

• Line 1: Write the full symbolic equation correctly [1 mark]
• Line 2–7: Define each term with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Placing hA on the right side (same side as hL) — pump head must be on the upstream (input) side
• Omitting hE or hL, writing only the ideal Bernoulli equation
• Not defining each symbol — definition earns the second mark

Key Phrases To Include

• p/γ
• v²/2g
• z
• hA (pump head added)
• hE (turbine head extracted)
• hL (head loss)

Marks

3

Topic

Continuity and Bernoulli Equation

Difficulty

medium

Template Id

T6

Examiner Tip

State 'horizontal pipe ∴ z₁ = z₂, elevation terms cancel' explicitly. This shows the examiner you understand the simplification and earns the set-up mark.

Model Answer

Given: D₁ = 0.300 m, D₂ = 0.150 m p₁ = 250 kPa = 250,000 Pa, v₁ = 2 m/s z₁ = z₂ (horizontal pipe), γ = 9,810 N/m³ Required: Q and p₂ (a) Discharge: A₁ = π/4 × (0.300)² = 0.07069 m² Q = A₁v₁ = 0.07069 × 2 = 0.1414 m³/s (b) Velocity at section 2 (Continuity): v₂ = v₁(D₁/D₂)² = 2 × (300/150)² = 2 × 4 = 8.0 m/s Apply Bernoulli (no loss, no machine, horizontal → z terms cancel): p₁/γ + v₁²/2g = p₂/γ + v₂²/2g 250,000/9,810 + (2)²/(2×9.81) = p₂/9,810 + (8)²/(2×9.81) 25.484 + 0.204 = p₂/9,810 + 3.262 p₂/9,810 = 22.426 p₂ = 22.426 × 9,810 = 220,000 Pa ≈ 220.0 kPa

Question Type

numerical

Answer Structure

• Block 1: Given and Required section [½ mark]
• Block 2: Compute A₁ and Q correctly [½ mark]
• Block 3: Compute v₂ using continuity [½ mark]
• Block 4: Write Bernoulli equation with z terms cancelled [½ mark]
• Block 5: Substitute all values correctly [½ mark]
• Block 6: Solve for p₂ with correct unit [½ mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to square the diameter ratio when computing v₂
• Not cancelling elevation heads for horizontal flow — creates unnecessary algebra errors
• Using γ = 9.81 kN/m³ with pressure in Pa, giving wrong units in head calculation
• Reporting p₂ in Pa without converting to kPa — examiner expects consistent kPa

Key Phrases To Include

• A = π/4 × D²
• Q = Av
• v₂ = v₁(D₁/D₂)²
• Bernoulli
• z₁ = z₂ (horizontal)
• pressure head
• velocity head

Marks

3

Topic

Energy Equation with Pump and Head Loss

Difficulty

medium

Template Id

T7

Examiner Tip

For reservoir-to-reservoir problems, recognizing that p = 0 (gauge) and v ≈ 0 at both free surfaces simplifies the energy equation to a pure elevation + machine head + loss problem. State this explicitly.

Model Answer

Given: z₁ = 10 m (lower reservoir, free surface) z₂ = 35 m (upper reservoir, free surface) Q = 0.06 m³/s, hL = 5 m, η = 0.80 At both reservoir free surfaces: v ≈ 0, p = 0 (gauge) γ = 9.81 kN/m³ Required: hA and P_input Energy equation (section 1 = lower surface, section 2 = upper surface): p₁/γ + v₁²/2g + z₁ + hA = p₂/γ + v₂²/2g + z₂ + hL 0 + 0 + 10 + hA = 0 + 0 + 35 + 5 hA = 40 − 10 = 30 m Power input to pump: P_water = γQhA = 9.81 × 0.06 × 30 = 17.658 kW P_input = P_water / η = 17.658 / 0.80 = 22.07 kW ≈ 22.1 kW

Question Type

numerical

Answer Structure

• Block 1: Given with reservoir conditions (p = 0, v ≈ 0) stated [½ mark]
• Block 2: Write energy equation with hA on left, hL on right [½ mark]
• Block 3: Simplify and solve hA = 30 m [½ mark]
• Block 4: Compute P_water = γQhA = 17.66 kW [½ mark]
• Block 5: Apply efficiency P_input = P_water/η [½ mark]
• Block 6: Final answer P_input = 22.1 kW with unit [½ mark]

Scoring Breakdown

Common Mark Deductions

• Placing hA and hL on the same side of the equation — shows conceptual error
• Using P_input = ηγQH instead of P_input = γQH/η — the most common efficiency sign error
• Using γ = 9,810 N/m³ but reporting in kW — produces answer 1000× too large unless divided by 1000
• Not stating that p = 0 gauge and v ≈ 0 at reservoir surfaces

Key Phrases To Include

• free surface: p = 0, v ≈ 0
• hA on left side
• hL on right side
• P = γQH
• P_input = P_water/η
• kW

Marks

2

Topic

Momentum Equation

Difficulty

medium

Template Id

T8

Examiner Tip

The momentum equation is tested in the context of forces on hydraulic structures. Always mention pipe bends and nozzles as applications — these are the classic board exam scenarios.

Model Answer

The momentum equation for steady, uniform flow through a control volume states that the net external force on the fluid equals the rate of change of momentum: ΣF = ρQ(v₂ − v₁) [N] For 2-D flow, apply component-wise: ΣFx = ρQ(v₂x − v₁x) ΣFy = ρQ(v₂y − v₁y) Engineering applications: determining the force exerted on pipe bends, nozzles, reducing sections, and vanes/blades — any geometry that changes the magnitude or direction of flow velocity.

Question Type

short_answer

Answer Structure

• Line 1–2: State the equation ΣF = ρQ(v₂ − v₁) with definition of terms [1 mark]
• Line 3–4: State component-wise application and at least two engineering applications [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing F = ma without converting to ρQ(Δv) form — loses the hydraulic application mark
• Omitting the engineering application — the question explicitly asks for it
• Confusing the sign: it is (v₂ − v₁), outlet minus inlet

Key Phrases To Include

• ΣF = ρQ(v₂ − v₁)
• control volume
• rate of change of momentum
• pipe bend
• nozzle
• component-wise

Marks

3

Topic

Power of Flow and Turbine Efficiency

Difficulty

medium

Template Id

T9

Examiner Tip

The key distinction: for a pump, P_input = P_water/η (input is larger); for a turbine, P_output = η × P_water (output is smaller). One formula, opposite logic — boards test this repeatedly.

Model Answer

Given: Q = 4 m³/s, H_gross = 30 m, hL = 3 m, η = 0.85 γ = 9.81 kN/m³ Required: P_output Net head available at turbine: H_net = H_gross − hL = 30 − 3 = 27 m Water power (hydraulic power) to turbine: P_water = γQH_net = 9.81 × 4 × 27 = 1,059.48 kW Turbine output power: P_output = η × P_water = 0.85 × 1,059.48 P_output = 900.56 kW ≈ 900.6 kW

Question Type

numerical

Answer Structure

• Block 1: Given data, identify gross head vs net head [½ mark]
• Block 2: Compute H_net = H_gross − hL = 27 m [½ mark]
• Block 3: Write and evaluate P_water = γQH_net [½ mark]
• Block 4: Apply turbine efficiency P_output = η × P_water [½ mark]
• Block 5: Final answer 900.6 kW with unit [½ mark]
• Block 6: Correct use of kN/m³ for γ giving answer directly in kW [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using gross head instead of net head in the power formula
• For turbine: dividing by η instead of multiplying — the opposite of a pump
• Using γ = 9,810 N/m³ and not dividing by 1,000 to convert W to kW

Key Phrases To Include

• net head = gross head − losses
• P = γQH
• γ = 9.81 kN/m³
• P_output = η × P_water
• turbine multiplies by η

Marks

2

Topic

Total Head and Bernoulli Equation

Difficulty

easy

Template Id

T10

Examiner Tip

This conceptual question is a guaranteed board exam item. Memorize the exact formula H = p/γ + v²/2g + z and the phrase 'energy per unit weight in metres' verbatim.

Model Answer

In fluid mechanics, 'head' is energy per unit weight of fluid, expressed in metres (m) of fluid column. The three components of total head are: 1. Pressure head: p/γ — energy per unit weight due to fluid pressure (m) 2. Velocity head: v²/2g — kinetic energy per unit weight due to fluid motion (m) 3. Elevation head (datum head): z — potential energy per unit weight due to position above datum (m) Total head H = p/γ + v²/2g + z (m)

Question Type

short_answer

Answer Structure

• Line 1: Define 'head' as energy per unit weight in metres [½ mark]
• Line 2–4: List and define all three components with formulas [1 mark]
• Line 5: Write total head equation [½ mark]

Scoring Breakdown

Common Mark Deductions

• Defining head as 'pressure' only — missing velocity and elevation components
• Writing v/2g instead of v²/2g — missing the square on velocity
• Not including units (m) — head has physical dimensions

Key Phrases To Include

• energy per unit weight
• metres of fluid
• p/γ pressure head
• v²/2g velocity head
• z elevation head
• H = p/γ + v²/2g + z

Marks

3

Topic

Bernoulli Equation — Tank Discharge

Difficulty

medium

Template Id

T11

Examiner Tip

This is the classic tank-discharge / Torricelli problem. The result v₂ = √(2gH) should be recognizable. State 'v₁ ≈ 0 for large tank' and 'p = 0 gauge at free surface and exit' — these justify the simplification.

Model Answer

Given: Point 1 = tank free surface: p₁ = 0 (open to atmosphere, gauge), v₁ ≈ 0, z₁ = 4.5 m Point 2 = nozzle exit: p₂ = 0 (gauge, discharges to atmosphere), z₂ = 0 (datum) D₂ = 100 mm = 0.100 m, hL = 0 (no losses) Required: v₂ and Q (a) Apply ideal Bernoulli (points 1 → 2): p₁/γ + v₁²/2g + z₁ = p₂/γ + v₂²/2g + z₂ 0 + 0 + 4.5 = 0 + v₂²/(2 × 9.81) + 0 v₂² = 4.5 × 2 × 9.81 = 88.29 v₂ = √88.29 = 9.40 m/s (b) Discharge: A₂ = π/4 × (0.100)² = 7.854 × 10⁻³ m² Q = A₂v₂ = 7.854 × 10⁻³ × 9.40 Q = 0.07383 m³/s ≈ 0.0738 m³/s

Question Type

numerical

Answer Structure

• Block 1: Identify boundary conditions at both points (p = 0 gauge, v₁ ≈ 0) [½ mark]
• Block 2: Set up Bernoulli equation with correct simplifications [½ mark]
• Block 3: Solve for v₂ = 9.40 m/s (Torricelli's theorem form) [1 mark]
• Block 4: Compute A₂ correctly [½ mark]
• Block 5: Q = A₂v₂ = 0.0738 m³/s with unit [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using z₁ = 0 and z₂ = −4.5 m instead of setting z₂ = 0 at the nozzle — arithmetic is equivalent but sign errors are common
• Not specifying that p = 0 gauge at both free surface and exit — leaving examiner to infer
• Forgetting to square the radius when computing A₂ (using r instead of r²)

Key Phrases To Include

• p₁ = 0 gauge (open surface)
• v₁ ≈ 0 (large tank)
• z₂ = 0 (datum at nozzle)
• v₂ = √(2gH)
• Torricelli
• Q = A₂v₂

Marks

5

Topic

Energy Equation with Head Loss

Difficulty

hard

Template Id

T12

Examiner Tip

Five-mark problems are scored step-by-step. Even with an arithmetic error in v₁, you can still earn 4/5 if all subsequent steps are logically correct. Show every intermediate line — partial credit is awarded for method.

Model Answer

Given: D₁ = 0.200 m, D₂ = 0.100 m Q = 0.05 m³/s p₁ = 180 kPa, z₁ = z₂ = 2 m (horizontal) hL = 0.8 m, γ = 9.81 kN/m³ Required: p₂ Step 1 — Velocities: A₁ = π/4 × (0.200)² = 0.031416 m² v₁ = Q/A₁ = 0.05/0.031416 = 1.592 m/s A₂ = π/4 × (0.100)² = 7.854 × 10⁻³ m² v₂ = Q/A₂ = 0.05/7.854 × 10⁻³ = 6.366 m/s (Check: v₂ = v₁(D₁/D₂)² = 1.592 × 4 = 6.366 m/s ✓) Step 2 — Velocity heads: v₁²/2g = (1.592)²/(2 × 9.81) = 2.534/19.62 = 0.1292 m v₂²/2g = (6.366)²/(2 × 9.81) = 40.52/19.62 = 2.0653 m Step 3 — Energy equation (no machine, z₁ = z₂): p₁/γ + v₁²/2g = p₂/γ + v₂²/2g + hL 180/9.81 + 0.1292 = p₂/9.81 + 2.0653 + 0.8 18.349 + 0.1292 = p₂/9.81 + 2.8653 18.478 = p₂/9.81 + 2.8653 p₂/9.81 = 15.613 p₂ = 15.613 × 9.81 = 153.15 kPa ≈ 153.2 kPa

Question Type

numerical

Answer Structure

• Block 1: List all given data with unit conversions [½ mark]
• Block 2: Compute A₁ and v₁ correctly [½ mark]
• Block 3: Compute A₂ and v₂ correctly (and verify with ratio) [1 mark]
• Block 4: Compute velocity heads v₁²/2g and v₂²/2g [½ mark]
• Block 5: Write energy equation with z₁ = z₂ simplification and hL on correct side [1 mark]
• Block 6: Substitute all terms correctly [½ mark]
• Block 7: Solve for p₂ = 153.2 kPa with unit [½ mark]
• Block 8: Logical check — p₂ < p₁ (pressure drops) ✓ [½ mark]

Scoring Breakdown

Common Mark Deductions

• Computing velocity using D instead of A (writing v = Q/D — wrong formula)
• Putting hL on the wrong (inlet) side of the equation, which gives p₂ > p₁ — physically unreasonable
• Not computing velocity heads and using only pressure heads — missing the v²/2g terms loses 2 marks
• Final answer in Pa instead of kPa without explicit conversion
• Not performing a logical check — examiners reward students who verify the sign/direction of pressure change

Key Phrases To Include

• v = Q/A
• A = π/4 × D²
• velocity head = v²/2g
• hL on outlet side
• z₁ = z₂ cancelled
• p₂ < p₁ (pressure drops through constriction)

Marks

2

Topic

Head Loss

Difficulty

medium

Template Id

T13

Examiner Tip

The phrase 'irreversible conversion of mechanical energy to thermal energy (heat)' is the thermodynamically precise definition — use it to distinguish yourself from students who write only 'friction losses'.

Model Answer

Head loss hL represents the irreversible conversion of mechanical energy (useful head) into heat due to fluid friction and flow disturbances. It is expressed in metres of fluid and appears as a positive term on the downstream (outlet) side of the energy equation. Causes of head loss: • Major (friction) losses: viscous friction along pipe walls (computed by Darcy-Weisbach or Hazen-Williams) • Minor losses: at fittings, valves, bends, and sudden contractions/expansions Effect on EGL: The EGL drops continuously in the direction of flow by an amount equal to hL per unit length. A steep EGL slope indicates high friction losses; a sudden drop indicates a minor loss at a fitting.

Question Type

short_answer

Answer Structure

• Line 1–2: Define hL as irreversible energy conversion, units in metres [½ mark]
• Line 3–4: List two categories of causes (major and minor losses) [½ mark]
• Line 5–6: Describe how hL manifests as a downward slope/drop in the EGL [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining head loss as 'pressure loss' only — it is total energy loss (includes velocity and elevation components)
• Saying the HGL drops but not mentioning the EGL
• Not distinguishing between major and minor losses

Key Phrases To Include

• irreversible
• mechanical energy to heat
• major losses (friction)
• minor losses (fittings)
• EGL drops in flow direction
• Darcy-Weisbach

Marks

5

Topic

Pump System — Energy Equation and Power

Difficulty

hard

Template Id

T14

Examiner Tip

Always show hA = (z₂ − z₁) + hL = 17 + 4 = 21 m to make the physics visible. Examiners want to see that you understand pump head must overcome both the static head and the losses.

Model Answer

Given: z₁ = 5 m (suction reservoir free surface) z₂ = 22 m (discharge reservoir free surface) Q = 0.03 m³/s, hL = 4 m, η = 0.75 At both free surfaces: p = 0 (gauge), v ≈ 0 γ = 9.81 kN/m³ Required: hA, P_water, P_input (a) Pump Head hA — Energy equation (1 → 2): 0 + 0 + 5 + hA = 0 + 0 + 22 + 0 + 4 hA = 26 − 5 = 21 m (b) Water (hydraulic) power delivered to fluid: P_water = γQhA P_water = 9.81 × 0.03 × 21 P_water = 6.179 kW (c) Input power (power drawn from motor): P_input = P_water / η P_input = 6.179 / 0.75 P_input = 8.239 kW ≈ 8.24 kW Sanity check: P_input > P_water (pump input always exceeds output) ✓

Question Type

numerical

Answer Structure

• Block 1: Given data with boundary conditions at both reservoirs [½ mark]
• Block 2: Energy equation written with hA on left, z terms and hL on right [1 mark]
• Block 3: Solve hA = 21 m [½ mark]
• Block 4: P_water = γQhA = 6.179 kW [1 mark]
• Block 5: P_input = P_water/η = 8.24 kW [1 mark]
• Block 6: Sanity check stated [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using hA = z₂ − z₁ = 17 m (forgetting to add head loss) — hA must overcome both static head and friction
• Multiplying by η instead of dividing for pump input power
• Computing γ in N/m³ and Q in m³/s giving P in watts without converting to kW

Key Phrases To Include

• p = 0 and v ≈ 0 at free surfaces
• hA on inlet side
• hL on outlet side
• hA = static head + losses
• P_water = γQhA
• P_input = P_water/η
• P_input > P_water for pump

Marks

2

Topic

Power of Flow — Pumps and Turbines

Difficulty

easy

Template Id

T15

Examiner Tip

The γ vs ρ distinction is a board exam trap. γ = ρg = 9,810 N/m³. Power uses γ (energy per volume × flow rate = power). Memorize: turbine multiply by η, pump divide by η.

Model Answer

Power of a flowing stream: P = γQH (watts if γ in N/m³; kilowatts if γ in kN/m³) Where: γ = specific weight of fluid (9,810 N/m³ or 9.81 kN/m³ for water) Q = discharge (m³/s) H = net head across the device (m) For a turbine (extracts energy from fluid): The fluid delivers P_water = γQH to the turbine. Only a fraction η is converted to useful shaft output: P_output = η × γQH Note: For a pump (adds energy to fluid): P_input = γQH / η (input is always greater than useful output)

Question Type

short_answer

Answer Structure

• Line 1–2: State P = γQH with all symbols defined [1 mark]
• Line 3–4: Derive turbine output P_output = ηγQH with physical reasoning [½ mark]
• Line 5: Contrast with pump formula P_input = γQH/η [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing P = ρQH instead of P = γQH — confusing density ρ with specific weight γ
• Not specifying the unit of γ and hence the unit of P
• Applying the turbine formula to a pump or vice versa

Key Phrases To Include

• P = γQH
• γ = 9,810 N/m³
• Q in m³/s
• H in m
• turbine: P_out = ηγQH
• pump: P_in = γQH/η