CELE Hydraulics & Fluid Mechanics — Properties of FluidsExam Answer Templates

Marks

1

Topic

Density, Specific Weight, Specific Gravity

Difficulty

easy

Template Id

T1

Examiner Tip

Even in a 1-mark question, one correct keyword (mass per unit volume) plus the unit earns full marks. Keep it to one sentence maximum.

Model Answer

Density (ρ) is the mass per unit volume of a substance. SI unit: kg/m³. For water, ρ = 1000 kg/m³.

Question Type

very_short_answer

Answer Structure

• State definition (mass per unit volume) and symbol ρ [0.5 mark]
• State SI unit (kg/m³) and standard value for water [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'weight per unit volume' instead of mass — that is specific weight, not density
• Omitting the SI unit entirely
• Using N/m³ (which is the unit of specific weight, not density)

Key Phrases To Include

• mass per unit volume
• kg/m³
• ρ
• 1000 kg/m³ for water

Marks

2

Topic

Viscosity

Difficulty

easy

Template Id

T2

Examiner Tip

This is a classic 2-mark differentiation. Give one complete sentence per term — no more needed. Formula + unit per term = full marks.

Model Answer

Dynamic viscosity (μ) is a fluid's absolute resistance to shear deformation and is defined by Newton's law: τ = μ(dv/dy). Its SI unit is Pa·s (or N·s/m²). Kinematic viscosity (ν) is the ratio of dynamic viscosity to density: ν = μ/ρ. Its SI unit is m²/s. Kinematic viscosity accounts for the fluid's inertia, making it useful in flow analysis.

Question Type

short_answer

Answer Structure

• Line 1: Define dynamic viscosity μ with formula τ = μ(dv/dy) and unit Pa·s [1 mark]
• Line 2: Define kinematic viscosity ν = μ/ρ with unit m²/s [1 mark]

Scoring Breakdown

Common Mark Deductions

• Swapping units — writing m²/s for dynamic viscosity or Pa·s for kinematic
• Omitting the relationship ν = μ/ρ
• Not mentioning the Newton's law formula for dynamic viscosity

Key Phrases To Include

• τ = μ(dv/dy)
• Pa·s
• ν = μ/ρ
• m²/s
• resistance to shear

Marks

2

Topic

Density, Specific Weight, Specific Gravity

Difficulty

easy

Template Id

T3

Examiner Tip

Two-part numericals: earn one mark per sub-part. Show the formula before substituting — never go directly to the number.

Model Answer

Given: ρ = 850 kg/m³, g = 9.81 m/s² (a) Specific Weight: γ = ρg = 850 × 9.81 = 8338.5 N/m³ = 8.34 kN/m³ (b) Specific Gravity: s = ρ/ρ_water = 850/1000 = 0.85 (dimensionless)

Question Type

numerical

Answer Structure

• State given data clearly [0 marks, but required for structure]
• Part (a): Write γ = ρg, substitute, compute with unit N/m³ or kN/m³ [1 mark]
• Part (b): Write s = ρ/ρ_water, substitute, state result is dimensionless [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to convert N/m³ to kN/m³ when the question expects cleaner units
• Writing specific gravity with units (it is dimensionless — always state this)
• Using g = 9.8 or 10 m/s² without justification — use 9.81 m/s²

Key Phrases To Include

• γ = ρg
• 8338.5 N/m³
• 8.34 kN/m³
• s = ρ/ρ_water
• 0.85
• dimensionless

Marks

2

Topic

Viscosity

Difficulty

easy

Template Id

T4

Examiner Tip

When asked to 'identify each term,' the examiner wants the physical meaning AND the unit for each symbol — both are required for full marks.

Model Answer

Newton's Law of Viscosity states that the shear stress in a fluid is directly proportional to the velocity gradient (rate of angular deformation): τ = μ (dv/dy) Where: • τ = shear stress (Pa or N/m²) • μ = dynamic (absolute) viscosity (Pa·s) • dv/dy = velocity gradient perpendicular to flow (s⁻¹) Fluids that obey this law are called Newtonian fluids (e.g., water, air, light oils).

Question Type

short_answer

Answer Structure

• State the law in words (proportionality of shear stress to velocity gradient) [0.5 mark]
• Write the formula τ = μ(dv/dy) [0.5 mark]
• Identify all three terms with correct units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing dv/dt (time derivative) instead of dv/dy (spatial gradient)
• Omitting units for any of the three terms
• Not identifying what type of fluid obeys this law

Key Phrases To Include

• τ = μ(dv/dy)
• shear stress
• velocity gradient
• dynamic viscosity
• Pa·s
• Newtonian fluid

Marks

3

Topic

Viscosity

Difficulty

medium

Template Id

T5

Examiner Tip

The phrase 'linear velocity profile' or 'linear profile assumed' must appear in your solution — it justifies treating dv/dy as a simple ratio (V/h) rather than a calculus expression.

Model Answer

Given: • Oil film thickness: dy = 1.5 mm = 0.0015 m • Plate velocity: dv = 3 m/s (velocity at top; bottom is stationary, so dv = 3 m/s) • Dynamic viscosity: μ = 0.02 Pa·s Required: Shear stress τ Solution: Applying Newton's Law of Viscosity: τ = μ (dv/dy) Velocity gradient: dv/dy = 3/0.0015 = 2000 s⁻¹ Shear stress: τ = 0.02 × 2000 = 40 Pa ∴ The shear stress in the oil film is τ = 40 Pa.

Question Type

numerical

Answer Structure

• List all given data with unit conversions (mm to m) [0.5 mark]
• State the formula τ = μ(dv/dy) [0.5 mark]
• Compute velocity gradient dv/dy correctly in s⁻¹ [1 mark]
• Substitute and compute τ with correct unit Pa [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not converting mm to m before dividing — most common arithmetic error
• Forgetting to state that the profile is linear (which justifies using simple ratio dv/dy = V/h)
• Omitting the unit s⁻¹ for velocity gradient

Key Phrases To Include

• Newton's Law of Viscosity
• τ = μ(dv/dy)
• linear velocity profile
• dv/dy = 2000 s⁻¹
• τ = 40 Pa

Marks

3

Topic

Density, Specific Weight, Specific Gravity

Difficulty

medium

Template Id

T6

Examiner Tip

When asked to 'identify the likely liquid,' one logical inference based on typical specific gravity ranges earns the mark. State the typical range (e.g., s ≈ 0.80–0.85 for light oils) to show reasoning.

Model Answer

Definition: Specific gravity (s) is the dimensionless ratio of a fluid's density (or specific weight) to that of water at standard conditions. s = ρ_fluid/ρ_water = γ_fluid/γ_water Given: γ_liquid = 7848 N/m³ Reference: γ_water = 9810 N/m³ Solution: s = γ_liquid/γ_water = 7848/9810 = 0.80 Alternatively: ρ_liquid = γ/g = 7848/9.81 = 800 kg/m³ s = 800/1000 = 0.80 ∴ Specific gravity s = 0.80. This value is consistent with a light petroleum oil or gasoline (typical s ≈ 0.80–0.82).

Question Type

numerical

Answer Structure

• Define specific gravity correctly as a dimensionless ratio [0.5 mark]
• Write the formula s = γ_fluid/γ_water [0.5 mark]
• Substitute values and compute s = 0.80 [1 mark]
• Identify the likely liquid based on the computed s value [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not stating 'dimensionless' — specific gravity always has no unit
• Using γ_water = 9.81 (missing N/m³ scale factor) — use 9810 N/m³, not 9.81 kN/m³, if units are in N/m³
• Failing to identify the probable substance when the question asks for it

Key Phrases To Include

• dimensionless
• s = γ_fluid/γ_water
• s = ρ_fluid/ρ_water
• 0.80
• γ_water = 9810 N/m³

Marks

5

Topic

Surface Tension and Capillarity

Difficulty

hard

Template Id

T7

Examiner Tip

In 5-mark derivation-plus-calculation questions, the derivation earns roughly 40% of the marks. Never skip it even if you have memorized the formula. Examiners reward the force balance step explicitly.

Model Answer

CAPILLARY RISE DERIVATION: Consider a vertical capillary tube of diameter d (radius r = d/2) inserted in a liquid. The surface tension σ acts along the meniscus at contact angle θ with the tube wall. Upward force due to surface tension (along circumference): F_up = σ × (πd) × cos θ Downward force due to weight of liquid column: F_down = γ × (πd²/4) × h At equilibrium (F_up = F_down): σ(πd)cosθ = γ(πd²/4)h Solving for h: h = 4σcosθ / (γd) ← Capillary Rise Formula ───────────────────────────── CALCULATION: Given: • d = 1 mm = 0.001 m • σ = 0.0728 N/m • θ = 0° → cosθ = 1.0 (water wets glass completely) • γ = 9810 N/m³ Required: Capillary rise h Substituting: h = 4(0.0728)(cos 0°) / (9810 × 0.001) h = 4(0.0728)(1) / (9.81) h = 0.2912 / 9.81 h = 0.02969 m ∴ h ≈ 29.7 mm Physical note: For water on clean glass, θ ≈ 0° (complete wetting), producing a concave meniscus and an upward rise. For mercury on glass (θ > 90°), the formula yields a negative h, indicating capillary depression.

Question Type

long_answer

Answer Structure

• Draw or describe the capillary tube setup — forces acting [0.5 mark]
• Write upward surface tension force: F_up = σ(πd)cosθ [1 mark]
• Write downward weight force: F_down = γ(πd²/4)h [1 mark]
• Equate forces and derive h = 4σcosθ/(γd) [1 mark]
• Correctly substitute given values with unit conversion (mm to m) [0.5 mark]
• Correct final answer h ≈ 29.7 mm with physical interpretation of θ = 0° [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using radius r instead of diameter d in the final formula without consistent derivation
• Not converting d from mm to m (single most common arithmetic error in this problem)
• Skipping the derivation and writing the formula directly — loses the 2 derivation marks
• Not interpreting the physical meaning of θ = 0°
• Wrong answer due to using γ = 9.81 instead of 9810 N/m³

Key Phrases To Include

• surface tension force
• F_up = σ(πd)cosθ
• weight of liquid column
• equilibrium
• h = 4σcosθ/(γd)
• 29.7 mm
• contact angle
• wetting

Marks

3

Topic

Compressibility and Vapor Pressure

Difficulty

medium

Template Id

T8

Examiner Tip

Examiners want to see 'vapor pressure' and 'local pressure drops' in the same sentence. The NPSH connection for pumps is a bonus but strongly expected at licensure review level.

Model Answer

Cavitation is the sudden formation of vapor bubbles within a flowing liquid when the local pressure drops to (or below) the vapor pressure of the liquid at the prevailing temperature. Condition for occurrence: Cavitation occurs when: P_local ≤ P_vapor (at the given temperature) This can happen at high-velocity zones (e.g., pump impeller eye, pipe constrictions) where increased velocity corresponds to reduced pressure (Bernoulli effect). Engineering application: Cavitation is a critical concern in centrifugal pump design. When suction-side pressure falls below vapor pressure, vapor bubbles form and subsequently collapse violently near solid surfaces, causing pitting, noise, vibration, and loss of pump efficiency. This is why Net Positive Suction Head (NPSH) is specified for every pump.

Question Type

short_answer

Answer Structure

• Define cavitation correctly (bubble formation at vapor pressure) [1 mark]
• State the pressure condition: P_local ≤ P_vapor [1 mark]
• Give a relevant engineering example with brief technical consequence [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining cavitation as 'boiling' without mentioning pressure reduction — partial credit only
• Not stating the pressure condition quantitatively or symbolically
• Vague application (just writing 'pipelines') without specifying the mechanism

Key Phrases To Include

• vapor pressure
• local pressure
• P_local ≤ P_vapor
• vapor bubbles
• cavitation
• NPSH
• pump impeller

Marks

3

Topic

Compressibility and Vapor Pressure

Difficulty

medium

Template Id

T9

Examiner Tip

The phrase 'nearly incompressible' must appear in your answer for a full conclusion mark. A large E_B means the fluid resists volume change — state this explicitly.

Model Answer

Definition: The bulk modulus of elasticity E_B is a measure of a liquid's resistance to compression. It is defined as: E_B = -dp / (dV/V) = dp / (dρ/ρ) Where: • dp = change in pressure (Pa) • dV/V = volumetric strain (dimensionless) • The negative sign reflects that volume decreases with increasing pressure Calculation: Given: • dp = 3 MPa = 3 × 10⁶ Pa • E_B = 2.2 GPa = 2.2 × 10⁹ Pa Required: Fractional change in volume (dV/V) From the definition: dV/V = -dp / E_B dV/V = -(3 × 10⁶) / (2.2 × 10⁹) dV/V = -1.364 × 10⁻³ ∴ |dV/V| = 1.36 × 10⁻³ or approximately 0.136% This very small value confirms that water is nearly incompressible.

Question Type

numerical

Answer Structure

• State definition of E_B with formula and explain the negative sign [1 mark]
• Rearrange to dV/V = -dp/E_B and substitute given values (in consistent Pa units) [1 mark]
• Compute correct answer and comment on near-incompressibility [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting the negative sign in the definition — this is physically important
• Unit mismatch: mixing MPa and GPa without converting to common units (Pa)
• Not commenting on incompressibility — the examiner expects this interpretation

Key Phrases To Include

• E_B = -dp/(dV/V)
• bulk modulus
• 2.2 GPa
• fractional change in volume
• 1.36 × 10⁻³
• nearly incompressible

Marks

5

Topic

Density, Specific Weight, Specific Gravity

Difficulty

medium

Template Id

T10

Examiner Tip

Three-part numerical questions award one mark per sub-part (plus a presentation mark in 5-mark questions). Structure matters — label parts, show the formula, substitute, answer. Do not merge all three computations into one paragraph.

Model Answer

Given: • Specific gravity: s = 0.88 • Dynamic viscosity: μ = 0.072 Pa·s • ρ_water = 1000 kg/m³, γ_water = 9810 N/m³ ───────────────────────────── (a) Density: ρ = s × ρ_water = 0.88 × 1000 ρ = 880 kg/m³ ───────────────────────────── (b) Specific Weight: γ = ρg = 880 × 9.81 γ = 8632.8 N/m³ ≈ 8.63 kN/m³ Alternatively: γ = s × γ_water = 0.88 × 9810 = 8632.8 N/m³ ✓ ───────────────────────────── (c) Kinematic Viscosity: ν = μ/ρ = 0.072/880 ν = 8.18 × 10⁻⁵ m²/s (To express in cSt: 1 m²/s = 10⁶ cSt → ν = 81.8 cSt) ───────────────────────────── Summary: • ρ = 880 kg/m³ • γ = 8632.8 N/m³ (8.63 kN/m³) • ν = 8.18 × 10⁻⁵ m²/s

Question Type

numerical

Answer Structure

• Clearly list all given data [0 marks, but required for presentation]
• Part (a): ρ = s × ρ_water = 880 kg/m³ with formula [1 mark]
• Part (b): γ = ρg (or s × γ_water) = 8632.8 N/m³ [1.5 marks]
• Part (c): ν = μ/ρ with correct substitution and unit m²/s [1.5 marks]
• Clean summary box with all three answers labeled [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not labeling each sub-part (a), (b), (c) — examiner cannot award partial credit to unlabeled work
• Using γ_water = 9.81 instead of 9810 N/m³ when working in N/m³
• Writing the unit of ν as m/s² (acceleration) instead of m²/s

Key Phrases To Include

• ρ = s × ρ_water
• 880 kg/m³
• γ = ρg
• 8632.8 N/m³
• ν = μ/ρ
• 8.18 × 10⁻⁵ m²/s

Marks

2

Topic

Surface Tension and Capillarity

Difficulty

easy

Template Id

T11

Examiner Tip

For 'give examples' questions, one mark is reserved for the examples, not the definition. Ensure each example is a distinct phenomenon — both being capillarity earns only 0.5.

Model Answer

Surface tension (σ) is the force per unit length (or energy per unit area) acting along a liquid surface, arising from the net inward cohesive forces experienced by surface molecules. SI unit: N/m. Two practical effects: 1. Capillary action — water rises in fine soil pores and concrete capillaries, which is significant in seepage analysis and reinforced concrete durability (water ingress). 2. Formation of droplets and bubbles — liquid tends to minimize surface area, forming spherical droplets. In small-bore pipes used in HVAC or chemical dosing systems, surface tension affects drip formation.

Question Type

short_answer

Answer Structure

• Define surface tension with unit N/m [1 mark]
• State two distinct and valid practical effects, each in one sentence [1 mark total, 0.5 per effect]

Scoring Breakdown

Common Mark Deductions

• Defining surface tension as 'N/m²' (which is pressure/stress) — the unit is N/m
• Giving vague effects like 'water forms bubbles' without engineering context
• Listing effects that are actually capillarity, not surface tension per se, without distinction

Key Phrases To Include

• force per unit length
• N/m
• cohesive forces
• capillary action
• surface energy

Marks

3

Topic

Viscosity

Difficulty

medium

Template Id

T12

Examiner Tip

This is a two-step problem: first compute stress, then multiply by area for force. Examiners award the second mark only if F = τ × A is explicitly shown — do not just write the final number.

Model Answer

Given: • Plate dimensions: 200 mm × 200 mm → Area A = 0.2 × 0.2 = 0.04 m² • Plate velocity: V = 0.8 m/s • Oil film thickness: h = 0.4 mm = 0.0004 m • Dynamic viscosity: μ = 0.05 Pa·s • Linear velocity profile assumed Required: (a) τ, (b) Force F (a) Shear Stress: τ = μ (dv/dy) = μ (V/h) τ = 0.05 × (0.8/0.0004) τ = 0.05 × 2000 τ = 100 Pa (b) Force Required: F = τ × A = 100 × 0.04 F = 4.0 N ∴ Shear stress τ = 100 Pa; Force F = 4.0 N

Question Type

numerical

Answer Structure

• Convert units (mm to m) and compute area in m² [0.5 mark]
• Apply τ = μ(V/h) correctly and get τ = 100 Pa [1 mark]
• Apply F = τ × A and get F = 4.0 N [1 mark]
• State final answers with correct units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Not converting plate dimensions from mm to m when computing area
• Forgetting to convert oil film thickness from mm to m
• Computing F in kN or incorrect unit
• Not using F = τ × A (confusing force with stress)

Key Phrases To Include

• τ = μ(V/h)
• linear velocity profile
• A = 0.04 m²
• τ = 100 Pa
• F = τA = 4.0 N

Marks

2

Topic

Surface Tension and Capillarity

Difficulty

medium

Template Id

T13

Examiner Tip

This is a conceptual question. The key words are 'wetting' (θ < 90°) for water and 'non-wetting' (θ > 90°) for mercury, linked to the sign of cosθ in the formula.

Model Answer

The direction of capillary action depends on the contact angle θ between the liquid and the tube wall: • Water on clean glass: θ ≈ 0° (strongly wetting). Cohesive forces between water molecules are weaker than adhesive forces between water and glass. The meniscus is concave upward, and surface tension pulls the liquid column upward → capillary RISE. • Mercury on glass: θ ≈ 140° (non-wetting). Cohesive forces among mercury atoms are much stronger than adhesion to glass. The meniscus is convex, and the net surface tension force acts downward → capillary DEPRESSION. In both cases the formula h = 4σcosθ/(γd) applies: cosθ > 0 for water (h > 0, rise); cosθ < 0 for mercury (h < 0, depression).

Question Type

short_answer

Answer Structure

• Explain the role of contact angle θ and adhesion vs. cohesion [1 mark]
• Apply the formula sign convention (cosθ positive for water, negative for mercury) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying mercury is 'heavier' as the reason — density is irrelevant here; the mechanism is contact angle
• Not mentioning the formula or the sign of cosθ
• Confusing adhesion (liquid-solid) with cohesion (liquid-liquid)

Key Phrases To Include

• contact angle θ
• wetting
• non-wetting
• adhesion
• cohesion
• cosθ
• concave meniscus
• convex meniscus

Marks

1

Topic

Density, Specific Weight, Specific Gravity

Difficulty

easy

Template Id

T14

Examiner Tip

1-mark questions require only the formula and a single computation. More than 3 lines of work wastes time. Formula + substitution + answer = done.

Model Answer

γ = ρg For seawater: γ = 1025 × 9.81 = 10,055.25 N/m³ ≈ 10.06 kN/m³

Question Type

very_short_answer

Answer Structure

• Write γ = ρg [0.5 mark]
• Correct substitution and answer with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Reporting answer in N/m³ only and not converting when examiner expects kN/m³
• Using g = 10 m/s² without instruction to do so

Key Phrases To Include

• γ = ρg
• 10,055 N/m³
• kN/m³

Marks

5

Topic

Viscosity

Difficulty

hard

Template Id

T15

Examiner Tip

This is a classic 5-mark essay-type question. The split is: 2 marks liquids + 2 marks gases + 1 mark engineering context. Without the physical explanation (cohesion for liquids, molecular collision for gases), you earn at most half the available marks.

Model Answer

EFFECT OF TEMPERATURE ON VISCOSITY I. LIQUIDS (e.g., water, oil): Viscosity DECREASES as temperature INCREASES. Physical explanation: In liquids, viscosity is primarily due to intermolecular cohesive forces (attractive forces between neighboring molecules). As temperature rises, the kinetic energy of molecules increases, overcoming intermolecular attraction. The resistance to relative motion between layers decreases → μ decreases. Example: Engine oil at 20°C has μ ≈ 0.1 Pa·s; at 80°C it drops significantly, which is why engines run more easily when warm. II. GASES (e.g., air): Viscosity INCREASES as temperature INCREASES. Physical explanation: In gases, molecules are far apart and intermolecular forces are negligible. Viscosity arises from momentum exchange between faster and slower molecular layers. As temperature increases, molecular velocity and frequency of cross-layer collisions increase → momentum exchange increases → μ increases. III. EFFECT ON KINEMATIC VISCOSITY ν = μ/ρ: For liquids, both μ and ρ decrease with temperature, but μ decreases faster, so ν also decreases with temperature. IV. ENGINEERING SIGNIFICANCE: Viscosity variation with temperature is critical in: • Hydraulic system design (oil viscosity drop at high temperature reduces lubrication) • Pump selection (viscous fluids at startup vs. operating temperature) • Pipe flow analysis (Reynolds number Re = ρVD/μ changes with temperature) Summary: • Liquids: μ ↓ as T ↑ (cohesive forces weaken) • Gases: μ ↑ as T ↑ (molecular momentum exchange increases)

Question Type

long_answer

Answer Structure

• State the trend for liquids (μ decreases with T) and physical reason (cohesion) [1.5 marks]
• State the trend for gases (μ increases with T) and physical reason (molecular momentum exchange) [1.5 marks]
• Discuss kinematic viscosity ν = μ/ρ behavior [0.5 mark]
• Give at least one engineering application or example [0.5 mark]
• Summary table or conclusion statement [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that all fluids decrease in viscosity with temperature — ignores gases (major conceptual error)
• No physical reason given — just stating the trend earns only 1 out of 2 for each fluid type
• Confusing density change with viscosity change
• No engineering significance mentioned in a 5-mark question

Key Phrases To Include

• cohesive forces
• intermolecular forces
• momentum exchange
• μ decreases with temperature (liquids)
• μ increases with temperature (gases)
• kinematic viscosity ν = μ/ρ