CELE Hydraulics & Fluid Mechanics — Flow in PipesExam Answer Templates

Marks

1

Topic

Reynolds Number & Flow Regimes

Difficulty

easy

Template Id

T1

Examiner Tip

A 1-mark definition question requires precision, not length. One complete sentence with the formula and the physical meaning is ideal.

Model Answer

Reynolds number (Re) is a dimensionless ratio of inertia forces to viscous forces in a fluid flow, defined as Re = vD/ν (or ρvD/μ), where v is the mean velocity, D is the pipe diameter, and ν is the kinematic viscosity. It is used to classify pipe flow as laminar (Re < 2 000), transitional (2 000 ≤ Re ≤ 4 000), or turbulent (Re > 4 000).

Question Type

very_short_answer

Answer Structure

• Line 1: State Re as a dimensionless ratio of inertia to viscous forces [½ mark]
• Line 2: Write the formula Re = vD/ν with variable definitions [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing the formula without defining the variables (–½ mark)
• Stating 'ratio of velocity to viscosity' without mentioning inertia vs. viscous forces
• Omitting units check — Re is dimensionless; stating incorrect units loses the mark

Key Phrases To Include

• dimensionless
• inertia forces
• viscous forces
• Re = vD/ν
• laminar
• turbulent

Marks

1

Topic

Reynolds Number & Flow Regimes

Difficulty

easy

Template Id

T2

Examiner Tip

The transitional zone (2 000–4 000) is frequently omitted by examinees. Always include all three regimes to guarantee the full mark.

Model Answer

Laminar: Re < 2 000; Transitional: 2 000 ≤ Re ≤ 4 000; Turbulent: Re > 4 000.

Question Type

very_short_answer

Answer Structure

• Line 1: List all three regimes with correct Re boundaries [1 mark]

Scoring Breakdown

Common Mark Deductions

• Missing the transitional regime (most common error)
• Stating Re > 2 000 as turbulent — ignores the transitional zone
• Reversing the boundaries

Key Phrases To Include

• laminar
• transitional
• turbulent
• Re < 2000
• Re > 4000

Marks

2

Topic

Reynolds Number & Flow Regimes

Difficulty

easy

Template Id

T3

Examiner Tip

Convert D to metres immediately upon reading the problem. Many examinees substitute D = 100 and get a nonsensical result. The unit check takes 5 seconds and saves a mark.

Model Answer

Given: v = 2 m/s, D = 0.10 m, ν = 1 × 10⁻⁶ m²/s Formula: Re = vD/ν Re = (2)(0.10) / (1 × 10⁻⁶) Re = 0.20 / (1 × 10⁻⁶) ∴ Re = 200 000 Since Re = 200 000 > 4 000, the flow is TURBULENT.

Question Type

numerical

Answer Structure

• Line 1: List given data with correct SI units [½ mark]
• Line 2: Write the formula Re = vD/ν [½ mark]
• Line 3: Substitute and compute Re = 200 000 [½ mark]
• Line 4: State regime classification with justification (Re > 4 000) [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using D in mm instead of m — gives Re = 200 (wrong by factor 1 000)
• Not converting ν to m²/s if given in cSt
• Stating turbulent without citing the Re > 4 000 criterion

Key Phrases To Include

• Re = vD/ν
• 200 000
• Re > 4 000
• turbulent

Marks

2

Topic

Major Losses — Darcy-Weisbach

Difficulty

easy

Template Id

T4

Examiner Tip

Always clarify that f in Darcy-Weisbach is the Darcy (Moody) friction factor = 4 × Fanning friction factor. This distinction appears in PRC board items that test formula literacy.

Model Answer

The Darcy-Weisbach equation is: hf = f (L/D)(v²/2g) where: • hf = friction head loss (m) • f = Darcy friction factor (dimensionless; from Moody chart or f = 64/Re for laminar flow) • L = pipe length (m) • D = pipe internal diameter (m) • v = mean flow velocity (m/s) • g = gravitational acceleration = 9.81 m/s² Note: For laminar flow (Re < 2 000), f = 64/Re exactly. For turbulent flow, f depends on Re and relative roughness ε/D (read from the Moody chart).

Question Type

short_answer

Answer Structure

• Line 1: Write the complete formula hf = f(L/D)(v²/2g) [1 mark]
• Line 2–7: Define all variables with correct SI units [½ mark]
• Line 8: State f = 64/Re for laminar flow and Moody chart for turbulent [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing Fanning friction factor formula (hf = 4f·L/D·v²/2g) instead of Darcy — common confusion in textbooks using different conventions
• Omitting the source of f (Moody chart or laminar formula)
• Writing v²/g instead of v²/2g

Key Phrases To Include

• Darcy-Weisbach
• hf = f(L/D)(v²/2g)
• Darcy friction factor
• Moody chart
• f = 64/Re
• relative roughness ε/D

Marks

3

Topic

Major Losses — Darcy-Weisbach

Difficulty

medium

Template Id

T5

Examiner Tip

Show every intermediate step on a separate line. PRC board checkers award partial marks for correct intermediate values even when the final answer is wrong due to an arithmetic slip.

Model Answer

Given: D = 200 mm = 0.20 m L = 100 m v = 3 m/s f = 0.02 g = 9.81 m/s² Formula (Darcy-Weisbach): hf = f(L/D)(v²/2g) Step 1 — Velocity head: v²/2g = (3)²/[2(9.81)] = 9/19.62 = 0.4587 m Step 2 — L/D ratio: L/D = 100/0.20 = 500 Step 3 — Head loss: hf = 0.02 × 500 × 0.4587 hf = 10 × 0.4587 ∴ hf = 4.59 m

Question Type

numerical

Answer Structure

• Block 1: State all given data with unit conversions (D in m) [½ mark]
• Block 2: Write Darcy-Weisbach formula [½ mark]
• Block 3: Compute velocity head v²/2g = 0.4587 m [½ mark]
• Block 4: Compute L/D = 500 [½ mark]
• Block 5: Substitute and compute hf = 4.59 m with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using D = 200 (mm) in the formula — gives hf 1 000× too small
• Computing L/D as D/L (inverting the ratio)
• Omitting the unit 'm' on the final answer
• Rounding v²/2g too early, causing cascade error

Key Phrases To Include

• hf = f(L/D)(v²/2g)
• v²/2g = 0.4587 m
• L/D = 500
• hf = 4.59 m

Marks

3

Topic

Major Losses — Laminar Flow & Friction Factor

Difficulty

medium

Template Id

T6

Examiner Tip

The laminar friction factor f = 64/Re is only valid when Re < 2 000. Always verify the regime before selecting the f formula. This two-step logic is what examiners test.

Model Answer

Given: ν = 4 × 10⁻⁵ m²/s v = 1.5 m/s D = 50 mm = 0.05 m (a) Reynolds number: Re = vD/ν = (1.5)(0.05)/(4 × 10⁻⁵) Re = 0.075/0.00004 ∴ Re = 1 875 (b) Flow regime: Re = 1 875 < 2 000 ∴ Flow is LAMINAR. (c) Friction factor (laminar flow formula): f = 64/Re = 64/1 875 ∴ f = 0.0341

Question Type

numerical

Answer Structure

• Block 1: List given data with D converted to m [½ mark]
• Block 2: Apply Re = vD/ν → Re = 1 875 [1 mark]
• Block 3: Classify as laminar (Re < 2 000) [½ mark]
• Block 4: Apply f = 64/Re → f = 0.0341 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Applying Moody chart or assuming turbulent f for laminar flow — wrong formula, zero marks for part (c)
• Not converting ν from cSt to m²/s (if given in cSt: 1 cSt = 10⁻⁶ m²/s)
• Computing f = Re/64 (inverting the formula)

Key Phrases To Include

• Re = vD/ν
• Re = 1 875
• Re < 2 000
• laminar
• f = 64/Re
• f = 0.0341

Marks

2

Topic

Minor Losses

Difficulty

easy

Template Id

T7

Examiner Tip

The word 'minor' does not mean 'small.' In short pipes with many fittings, minor losses can exceed major (friction) losses. Mentioning this earns bonus credit.

Model Answer

Minor losses (also called local or fitting losses) are energy losses caused by flow disturbances at pipe fittings, valves, bends, enlargements, contractions, and entrances/exits — i.e., at any point where the flow velocity or direction changes abruptly. General formula: hm = K (v²/2g) where: • hm = minor head loss (m) • K = loss coefficient (dimensionless; depends on fitting type and geometry) • v = mean flow velocity at the fitting (m/s) • g = 9.81 m/s² Typical K values: sharp-edged entrance ≈ 0.5; pipe exit ≈ 1.0; 90° elbow ≈ 0.9.

Question Type

short_answer

Answer Structure

• Line 1: Define minor losses with examples of fitting types [1 mark]
• Line 2: Write formula hm = K(v²/2g) with all variables defined [½ mark]
• Line 3: Cite at least two typical K values [½ mark]

Scoring Breakdown

Common Mark Deductions

• Confusing minor losses with major (friction) losses
• Writing hm = Kv/2g (missing the square on velocity)
• Not stating that K is dimensionless

Key Phrases To Include

• local losses
• fitting losses
• hm = K(v²/2g)
• loss coefficient K
• velocity head

Marks

3

Topic

Minor Losses

Difficulty

medium

Template Id

T8

Examiner Tip

Tabulate the K values before computing: list each fitting with its K, sum them, then apply the formula once. This avoids partial K-summation errors and is much faster under time pressure.

Model Answer

Given: D = 150 mm = 0.15 m Q = 0.03 m³/s K_entrance = 0.5, K_elbow = 0.9, K_valve = 4.0 g = 9.81 m/s² Step 1 — Cross-sectional area: A = π D²/4 = π(0.15)²/4 = 0.01767 m² Step 2 — Mean velocity: v = Q/A = 0.03/0.01767 = 1.698 m/s Step 3 — Velocity head: v²/2g = (1.698)²/[2(9.81)] = 2.883/19.62 = 0.1470 m Step 4 — Total K: ΣK = 0.5 + 0.9 + 4.0 = 5.4 Step 5 — Total minor loss: hm = ΣK × (v²/2g) = 5.4 × 0.1470 ∴ hm = 0.794 m

Question Type

numerical

Answer Structure

• Block 1: Convert D and compute area A = 0.01767 m² [½ mark]
• Block 2: Compute v = Q/A = 1.698 m/s [½ mark]
• Block 3: Compute velocity head = 0.147 m [½ mark]
• Block 4: Sum K values ΣK = 5.4 [½ mark]
• Block 5: hm = 5.4 × 0.147 = 0.794 m with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using D in mm in the area formula
• Forgetting to sum all K values (omitting one fitting)
• Computing v²/2g as v/2g (omitting the square)

Key Phrases To Include

• v = Q/A
• v²/2g
• ΣK = 5.4
• hm = ΣK(v²/2g)
• 0.794 m

Marks

2

Topic

Pipes in Series and Parallel

Difficulty

easy

Template Id

T9

Examiner Tip

The classic PRC board trap is reversing series/parallel rules. Use this memory trick: 'Series — Same Q flows through a single Straw (one path); Parallel — Same Pressure difference, flows split.'

Model Answer

Pipes in SERIES: • The same discharge Q flows through every pipe in the system. • The total head loss is the SUM of individual head losses: hL = h₁ + h₂ + h₃ + ... • Used when pipes are connected end-to-end along a single flow path. Pipes in PARALLEL: • The head loss across every parallel branch is EQUAL: h₁ = h₂ = h₃ = ... • The total discharge is the SUM of branch discharges: Q = Q₁ + Q₂ + Q₃ + ... • Used when two or more pipes share the same inlet and outlet nodes.

Question Type

short_answer

Answer Structure

• Lines 1–3: State Q-same and h-adds for series [1 mark]
• Lines 4–6: State h-same and Q-adds for parallel [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the rules — stating 'head loss same in series' or 'Q adds in series'
• Providing only one of the two conditions for each configuration
• Not writing the mathematical relationship (formula form)

Key Phrases To Include

• same Q
• head losses add
• same head loss
• discharges add
• Q = Q₁ + Q₂
• hL = h₁ + h₂

Marks

5

Topic

Pipes in Series

Difficulty

hard

Template Id

T10

Examiner Tip

In series pipe problems, always re-compute velocity separately for each pipe because velocity changes with diameter even though Q is constant. Many examinees use one velocity for all pipes and lose marks on every pipe calculation.

Model Answer

Given: Pipe A: DA = 0.20 m, LA = 500 m, fA = 0.02 Pipe B: DB = 0.15 m, LB = 300 m, fB = 0.025 Q = 0.05 m³/s (same through both — series) g = 9.81 m/s² Series condition: Q_A = Q_B = 0.05 m³/s --- PIPE A --- Step 1: Area_A = π(0.20)²/4 = 0.03142 m² Step 2: vA = Q/A_A = 0.05/0.03142 = 1.592 m/s Step 3: v_A²/2g = (1.592)²/[2(9.81)] = 2.534/19.62 = 0.1292 m Step 4: hfA = fA(LA/DA)(vA²/2g) = 0.02(500/0.20)(0.1292) = 0.02 × 2500 × 0.1292 = 6.46 m --- PIPE B --- Step 5: Area_B = π(0.15)²/4 = 0.01767 m² Step 6: vB = Q/A_B = 0.05/0.01767 = 2.829 m/s Step 7: v_B²/2g = (2.829)²/[2(9.81)] = 8.003/19.62 = 0.4079 m Step 8: hfB = fB(LB/DB)(vB²/2g) = 0.025(300/0.15)(0.4079) = 0.025 × 2000 × 0.4079 = 20.40 m --- TOTAL HEAD LOSS (Series) --- hL = hfA + hfB = 6.46 + 20.40 ∴ hL = 26.86 m

Question Type

numerical

Answer Structure

• Block 1: State given data and invoke series condition (Q same) [½ mark]
• Block 2: Compute area and velocity for Pipe A [½ mark]
• Block 3: Compute hfA using Darcy-Weisbach → 6.46 m [1 mark]
• Block 4: Compute area and velocity for Pipe B [½ mark]
• Block 5: Compute hfB using Darcy-Weisbach → 20.40 m [1 mark]
• Block 6: Apply series rule hL = hfA + hfB = 26.86 m [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using the same velocity for both pipes despite different diameters
• Not converting diameters to metres
• Applying parallel rule (setting head losses equal) to a series system
• Omitting the explicit statement of the series condition before solving

Key Phrases To Include

• series: Q same
• v = Q/A
• Darcy-Weisbach
• hfA = 6.46 m
• hfB = 20.40 m
• hL = hfA + hfB
• 26.86 m

Marks

2

Topic

Pipe Networks — Hardy Cross

Difficulty

medium

Template Id

T11

Examiner Tip

For 2-mark concept questions, always include: (1) what the method is, and (2) when/where it is used. Examiners want both theoretical definition and practical context.

Model Answer

The Hardy Cross method is an iterative numerical technique used to determine the distribution of flows and head losses in a pipe network with multiple loops (closed loops). It is applied when the network cannot be reduced to a simple series/parallel arrangement. Procedure summary: 1. Assume initial flow distribution satisfying continuity (ΣQ_in = ΣQ_out) at each node. 2. For each loop, compute the correction ΔQ = −(Σ hf) / (Σ 2hf/Q) per loop. 3. Apply ΔQ to update flows in each pipe and repeat until ΔQ ≈ 0 (convergence). Application: Water distribution networks in towns and cities (e.g., MWSS pipelines, barangay waterworks).

Question Type

short_answer

Answer Structure

• Lines 1–2: Define Hardy Cross as iterative method for multi-loop networks [1 mark]
• Lines 3–5: Outline correction formula and convergence criterion [½ mark]
• Line 6: Give practical application example [½ mark]

Scoring Breakdown

Common Mark Deductions

• Describing Hardy Cross as applicable to simple series/parallel systems
• Not mentioning the iterative or correction aspect
• Confusing Hardy Cross with Newton-Raphson or other numerical methods

Key Phrases To Include

• iterative
• looped network
• continuity
• flow correction ΔQ
• convergence
• head loss balance

Marks

3

Topic

Head Loss Equations — Comparison

Difficulty

medium

Template Id

T12

Examiner Tip

The PRC board often gives you a problem specifying which formula to use. Knowing the domain of each equation prevents you from applying the wrong one and ensures you select the correct coefficient (n vs C).

Model Answer

1. Darcy-Weisbach: hf = f(L/D)(v²/2g) • Universal — applies to any fluid, any flow regime (laminar or turbulent). • Requires friction factor f from the Moody chart (function of Re and ε/D). • Primary application: general fluid mechanics; design involving non-water fluids or precise Re-based analysis. 2. Manning's Equation: v = (1/n) R^(2/3) S^(1/2) [for full pipe: R = D/4] • Semi-empirical — originally for open-channel flow; extended to full pipes. • Uses roughness coefficient n (units: s/m^(1/3)). • Primary application: gravity-flow pipes, sewers, drainage channels in Philippine practice. 3. Hazen-Williams: v = 0.849 C R^(0.63) S^(0.54) • Empirical — developed specifically for turbulent water flow (not valid for other fluids). • Uses dimensioned roughness coefficient C (higher C = smoother pipe). • Primary application: water supply and distribution pipelines. Key distinction: Only Darcy-Weisbach is theoretically rigorous and fluid-independent. Manning and Hazen-Williams are empirical and restricted to their specific domains.

Question Type

short_answer

Answer Structure

• Block 1: Darcy-Weisbach — formula, universality, f from Moody [1 mark]
• Block 2: Manning — formula, n coefficient, open-channel origin, sewer application [1 mark]
• Block 3: Hazen-Williams — formula, water-only, C coefficient, waterworks application [½ mark]
• Block 4: Key distinction — theoretical vs empirical [½ mark]

Scoring Breakdown

Common Mark Deductions

• Mixing up which coefficient belongs to which equation (n with H-W, C with Manning)
• Stating Hazen-Williams applies to any fluid
• Not noting that Manning's R = D/4 for a full circular pipe

Key Phrases To Include

• Darcy-Weisbach
• Moody chart
• Manning n
• R = D/4
• Hazen-Williams C
• empirical
• water supply
• any fluid

Marks

3

Topic

Minor Losses — Sudden Expansion

Difficulty

hard

Template Id

T13

Examiner Tip

The sudden expansion formula is a perennial board topic. Memorize he = (v₁ − v₂)²/2g and know that it comes from combining momentum and energy equations. The derivation earns process marks even if you already know the result.

Model Answer

Head loss at a sudden expansion (Borda-Carnot equation): he = (v₁ − v₂)² / 2g Derivation outline (momentum + energy approach): 1. Continuity: Q = v₁A₁ = v₂A₂ → v₂ = v₁(A₁/A₂) 2. Momentum (control volume at expansion): p₁A₂ − p₂A₂ = ρQ(v₂ − v₁) [pressure at the expansion face acts on full area A₂] → (p₁ − p₂)/γ = (v₂² − v₁v₂)/g ... (i) 3. Energy equation (including loss he): p₁/γ + v₁²/2g = p₂/γ + v₂²/2g + he → (p₁ − p₂)/γ = he + (v₂² − v₁²)/2g ... (ii) 4. Equating (i) and (ii) and simplifying: he = (v₁² − 2v₁v₂ + v₂²)/2g ∴ he = (v₁ − v₂)² / 2g Note: v₁ > v₂ (flow decelerates at expansion), so he is always positive.

Question Type

short_answer

Answer Structure

• Line 1: State the Borda-Carnot formula he = (v₁−v₂)²/2g [1 mark]
• Block 2: Apply continuity and momentum equation [1 mark]
• Block 3: Apply energy equation and combine to derive formula [½ mark]
• Line last: Note physical consistency (v₁ > v₂, he > 0) [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using he = (v₁² − v₂²)/2g (wrong — this is kinetic energy change, not loss)
• Not noting that pressure acts on the full downstream area A₂ (critical momentum step)
• Confusing sudden expansion formula with contraction or exit loss

Key Phrases To Include

• Borda-Carnot
• he = (v₁ − v₂)²/2g
• momentum equation
• pressure on full area A₂
• continuity
• always positive

Marks

5

Topic

Total Head Loss — Major + Minor Combined

Difficulty

hard

Template Id

T14

Examiner Tip

In combined major + minor loss problems, compute the velocity head v²/2g once as a common factor, then multiply separately by f(L/D) for major and by ΣK for minor. This is faster and reduces errors.

Model Answer

Given: D = 300 mm = 0.30 m L = 250 m v = 2.5 m/s f = 0.018 K_entrance = 0.5, K_elbow = 0.9 g = 9.81 m/s² --- STEP 1: Velocity head --- v²/2g = (2.5)²/[2(9.81)] = 6.25/19.62 = 0.3186 m --- STEP 2: Major (friction) loss --- hf = f(L/D)(v²/2g) hf = 0.018 × (250/0.30) × 0.3186 hf = 0.018 × 833.3 × 0.3186 hf = 0.018 × 265.5 hf = 4.779 m --- STEP 3: Minor losses --- hm_entrance = K_entrance × (v²/2g) = 0.5 × 0.3186 = 0.1593 m hm_elbow = K_elbow × (v²/2g) = 0.9 × 0.3186 = 0.2867 m Total minor loss: hm = 0.1593 + 0.2867 = 0.4460 m --- STEP 4: Total head loss --- hL = hf + hm = 4.779 + 0.446 ∴ hL = 5.225 m (accept 5.22–5.23 m)

Question Type

numerical

Answer Structure

• Block 1: List given data with conversions [½ mark]
• Block 2: Compute velocity head = 0.3186 m [½ mark]
• Block 3: Compute major loss hf = 4.779 m [1½ marks]
• Block 4: Compute each minor loss and sum → hm = 0.446 m [1½ marks]
• Block 5: Total hL = hf + hm = 5.225 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing major loss only — missing minor losses entirely (–2 marks)
• Using D = 300 mm in the Darcy formula
• Computing L/D = 0.30/250 (inverted ratio)
• Not showing the velocity head computation as a separate step

Key Phrases To Include

• v²/2g = 0.3186 m
• hf = f(L/D)(v²/2g)
• hf = 4.779 m
• hm = K(v²/2g)
• hL = hf + hm
• 5.225 m

Marks

5

Topic

Pipes in Parallel

Difficulty

hard

Template Id

T15

Examiner Tip

Parallel pipe problems are most efficiently solved by expressing hf = RQ² where R = 8fL/(π²gD⁵). Setting R₁Q₁² = R₂Q₂² directly gives the Q ratio without trial-and-error. Always include the verification step — it is a recognized mark in PRC board rubrics.

Model Answer

Given: Pipe 1: D₁ = 0.25 m, L₁ = 400 m, f₁ = 0.02 Pipe 2: D₂ = 0.15 m, L₂ = 200 m, f₂ = 0.025 Q_total = 0.20 m³/s g = 9.81 m/s² Parallel conditions: (i) h_f1 = h_f2 [same head loss across each branch] (ii) Q₁ + Q₂ = 0.20 m³/s --- Express head loss in terms of Q --- For circular pipe: A = πD²/4, v = Q/A = 4Q/(πD²) v²/2g = 8Q²/(π²D⁴g) hf = f(L/D)(v²/2g) = f(L/D) × [8Q²/(π²D⁴g)] = 8fLQ²/(π²gD⁵) ... call this R × Q² Resistance coefficients: R₁ = 8f₁L₁/(π²gD₁⁵) = 8(0.02)(400)/[π²(9.81)(0.25)⁵] = 64/[π²(9.81)(9.766×10⁻⁴)] = 64/[9.870 × 9.81 × 9.766×10⁻⁴] = 64/0.09462 = 676.5 (m·s²/m⁶ → m/(m³/s)²) R₂ = 8f₂L₂/(π²gD₂⁵) = 8(0.025)(200)/[π²(9.81)(0.15)⁵] = 40/[9.870 × 9.81 × 7.594×10⁻⁵] = 40/0.007352 = 5 441.3 Parallel condition h_f1 = h_f2: R₁Q₁² = R₂Q₂² 676.5 Q₁² = 5441.3 Q₂² Q₁/Q₂ = √(5441.3/676.5) = √8.044 = 2.836 → Q₁ = 2.836 Q₂ Apply Q₁ + Q₂ = 0.20: 2.836 Q₂ + Q₂ = 0.20 3.836 Q₂ = 0.20 Q₂ = 0.05214 m³/s Q₁ = 0.20 − 0.05214 = 0.14786 m³/s ∴ Q₁ ≈ 0.148 m³/s ∴ Q₂ ≈ 0.052 m³/s Verification — head loss check: hf1 = 676.5 × (0.148)² = 676.5 × 0.02190 = 14.82 m hf2 = 5441.3 × (0.052)² = 5441.3 × 0.002704 = 14.71 m ✓ (≈ equal)

Question Type

numerical

Answer Structure

• Block 1: State parallel conditions (i) equal head loss, (ii) Q₁+Q₂ = Q_total [½ mark]
• Block 2: Express hf = R×Q² and compute resistance R₁ and R₂ [1½ marks]
• Block 3: Set R₁Q₁² = R₂Q₂², solve for Q₁/Q₂ ratio [1 mark]
• Block 4: Apply Q₁+Q₂ = 0.20 and solve for Q₁ and Q₂ [1 mark]
• Block 5: Verification by checking head losses are approximately equal [1 mark]

Scoring Breakdown

Common Mark Deductions

• Applying series rule (same Q) instead of parallel rule
• Not deriving hf as a function of Q (keeping v as the variable makes the system much harder to solve)
• Not performing the verification step — this earns a separate mark
• Arithmetic error in computing D⁵ (e.g., 0.25⁵ = 9.766 × 10⁻⁴, not 0.25)

Key Phrases To Include

• parallel: same head loss
• Q₁+Q₂=0.20
• hf = RQ²
• R₁Q₁² = R₂Q₂²
• Q₁/Q₂ = 2.836
• verification