CELE Hydraulics & Fluid Mechanics — Flow in PipesRevision Notes

Formulas

Exam Tips

• Board problems will almost always give you one of: (a) the regime and ask for f, or (b) v, D, ν and ask for Re and regime. Memorize the three Re boundaries: 2000 and 4000.
• If the problem gives only the flow rate Q, compute v = Q/A = Q/(πD²/4) first, then find Re.
• For water at 20°C, Re = v(m/s) × D(mm) × 1000. This mental shortcut works because ν = 10⁻⁶ m²/s.
• Laminar flow in a board exam usually signals a viscous oil problem, not water.

Key Points

• The Reynolds number (Re) is a dimensionless ratio of inertial forces to viscous forces; it determines whether flow is laminar, transitional, or turbulent.
• Re = vD/ν = ρvD/μ, where v = mean velocity (m/s), D = pipe diameter (m), ν = kinematic viscosity (m²/s), μ = dynamic viscosity (Pa·s), ρ = fluid density (kg/m³).
• Laminar flow: Re < 2000 — fluid moves in smooth, parallel layers; velocity profile is parabolic.
• Transitional flow: 2000 < Re < 4000 — unstable, neither fully laminar nor fully turbulent; avoid this zone in design.
• Turbulent flow: Re > 4000 — chaotic fluid motion; velocity profile is flatter than laminar; most engineering pipe flows fall here.
• For water at 20°C: ν ≈ 1×10⁻⁶ m²/s (1 cSt). For oils, ν is much larger (e.g., 4×10⁻⁵ m²/s for light oil), making laminar flow more likely.
• The friction factor f is a strong function of Re and the relative roughness ε/D, read from the Moody chart for turbulent flow.
• In laminar flow, f = 64/Re (exact, derived analytically from Hagen-Poiseuille).

Definitions

Section Title

1. Flow Regimes and the Reynolds Number

Common Mistakes

• Forgetting to convert D from mm to m before computing Re — a diameter of 200 mm must be entered as 0.2 m.
• Using f = 64/Re for turbulent flow — this formula applies ONLY in the laminar regime.
• Using ν = 1×10⁻⁶ m²/s for fluids other than water at ~20°C without checking the problem statement.
• Confusing kinematic viscosity ν (m²/s) with dynamic viscosity μ (Pa·s) in the Re formula.

Formulas

Exam Tips

• If the board problem gives f directly, go straight to hf = f(L/D)(v²/2g). Compute v²/2g once and reuse it for minor losses.
• When Q is given instead of v: v = 4Q/(πD²). Substitute immediately.
• For Manning problems, remember the SI form: v = (1/n)R^(2/3)S^(1/2) with R = D/4 for full pipes.
• Hazen-Williams C values to memorize: C = 150 (PVC, new), 120–130 (new cast iron), 100 (average concrete), 70–80 (old corroded pipe).
• The question 'Find the diameter for a given Q and hf' is a common board trap — set up Darcy-Weisbach with v = 4Q/(πD²) and solve iteratively or use the given options.

Key Points

• The Darcy-Weisbach equation is the most general and physically rigorous pipe-friction formula; it is valid for laminar and turbulent flow, any fluid, any pipe material.
• hf = f(L/D)(v²/2g), where hf is the friction head loss in metres of fluid.
• The friction factor f depends on Re and relative roughness ε/D; for laminar flow f = 64/Re (no roughness dependence).
• Relative roughness ε/D: ε is the absolute pipe roughness (mm). Typical values: cast iron ε = 0.26 mm; commercial steel ε = 0.046 mm; PVC (smooth) ε ≈ 0.0015 mm.
• For fully turbulent rough flow, f depends only on ε/D (Moody chart, right-hand asymptote); f is independent of Re.
• The Swamee-Jain explicit approximation to the Colebrook-White equation (commonly used when Moody chart is not available): f = 0.25 / [log(ε/(3.7D) + 5.74/Re⁰·⁹)]²
• Since hf ∝ v², doubling the velocity quadruples the head loss — a critical scaling relationship.
• Manning's equation (SI): v = (1/n)R^(2/3)S^(1/2), with R = D/4 for a full circular pipe and S = hf/L. Commonly used in open channels but also applicable to full pipes.
• Hazen-Williams equation (water only): v = 0.849 C R^0.63 S^0.54, valid for water at ~15°C, v < 3 m/s. C ranges from 70 (old cast iron) to 150 (PVC, new).

Definitions

Section Title

2. Major (Friction) Head Loss — Darcy-Weisbach Equation

Common Mistakes

• Using R = D/2 (radius) instead of R = D/4 (hydraulic radius) in Manning and Hazen-Williams equations — the hydraulic radius of a full circular pipe is D/4.
• Confusing Darcy f (used in Darcy-Weisbach) with Fanning friction factor Cf = f/4 — board problems use Darcy unless stated otherwise.
• Applying Hazen-Williams to fluids other than water or to water at temperatures far from 15°C.
• Forgetting that hf ∝ v² — a 10% increase in velocity increases friction loss by 21%, not 10%.
• Leaving diameter in millimetres in the Darcy-Weisbach formula — D must be in metres.

Formulas

Exam Tips

• Compute v²/2g once at the start of a problem; all minor losses (and friction loss) are multiples of this value.
• If v2 is not given in an expansion problem, use continuity: A1v1 = A2v2, so v2 = v1(D1/D2)².
• Memorize K = 0.5 (sharp entrance), K = 1.0 (exit), K = 0.9 (90° elbow). Valves: gate (full open) = 0.2, globe (full open) = 10.0.
• A board problem asking for 'total head loss' means hf + Σhm; don't report only the friction loss.

Key Points

• Minor losses occur at pipe fittings, valves, bends, entrances, exits, and sudden changes in section. They are called 'minor' but can be significant in short pipe systems.
• General formula: hm = K(v²/2g), where K is the loss coefficient for each fitting.
• Common K values (approximate): sharp entrance = 0.5, re-entrant (projecting) entrance = 0.8–1.0, exit (pipe to reservoir) = 1.0, 90° elbow = 0.9, 45° elbow = 0.4, gate valve (fully open) = 0.2, globe valve (fully open) = 10.0.
• Sudden expansion loss (Borda-Carnot): he = (v1 – v2)²/2g, where v1 is the upstream velocity and v2 is the downstream velocity. This is an exact result from momentum and continuity.
• Sudden contraction: use hm = Kc(v2²/2g) with Kc ≈ 0.5 for sharp-edged contraction (v2 is the velocity in the smaller pipe).
• In long pipelines (L/D > 1000), minor losses are often less than 5% of friction losses and may be neglected in a first analysis.
• In short systems (pump discharge, building plumbing), minor losses can dominate and must be included.
• The equivalent length concept: Le = KD/f converts a fitting's K to an equivalent pipe length that produces the same loss.

Definitions

Section Title

3. Minor (Local) Head Losses

Common Mistakes

• Using the wrong velocity in hm = K(v²/2g) — for entrance losses use the pipe velocity (downstream of entrance); for exit losses use the pipe velocity (upstream of exit); for expansions use v2 (downstream).
• Forgetting exit loss K = 1.0 — all kinetic energy at the pipe exit is lost to the receiving reservoir.
• Applying he = (v1–v2)²/2g to a gradual expansion — this formula is for sudden (abrupt) expansion only.
• Omitting minor losses in short piping systems such as pump suction/discharge lines or building plumbing — they can be larger than friction losses here.

Formulas

Exam Tips

• For two parallel pipes, write the head-loss equality hf1 = hf2 and express everything in terms of Q1 and Q2 = Q – Q1. Solve for Q1.
• In series problems, find v in each pipe from Q = Av; then compute hf for each segment; sum them.
• For the three-reservoir problem: if HJ is between the two intermediate reservoir levels, flow direction is: highest reservoir → junction → two lower reservoirs.
• Hardy Cross: you may get a one-loop network on the board exam — know how to compute Σhf and Σ(hf/Q) for the loop and apply ΔQ.

Key Points

• Series pipe system: pipes of different diameters, roughness, or length connected end-to-end. The same discharge Q flows through every pipe. Total head loss = sum of individual losses: hL = h1 + h2 + ... + hn.
• Parallel pipe system: two or more pipes connecting the same two junction nodes. The head loss is identical across every branch: hA-B = h1 = h2 = ... Total flow = sum of branch flows: Q = Q1 + Q2 + ... + Qn.
• For two parallel pipes: set hf1 = hf2, substitute Darcy-Weisbach for each, and solve simultaneously with Q1 + Q2 = Q.
• The equivalent pipe concept: replace a series or parallel system with a single pipe of diameter De and length Le that carries the same Q for the same total head loss — useful for simplifying networks.
• For pipes in series: 1/(De^(5/2)) ~ sum of individual terms (exact form depends on friction factor assumptions).
• For pipes in parallel: combine branch conductances.
• Branch pipe problems (three-reservoir problem): an additional pipe connects from a junction J at piezometric head HJ to three reservoirs at known elevations. Set up energy equations from each reservoir to J; the unknown is HJ — solved iteratively or by trial.
• Hardy Cross method: for closed pipe networks with multiple loops, assign an initial Q to each pipe, compute the head-loss discrepancy in each loop, apply a correction ΔQ = -Σ(hf) / [2Σ(hf/Q)] per loop, and iterate until convergence.

Definitions

Section Title

4. Pipes in Series and Parallel

Common Mistakes

• Swapping series and parallel rules: in series hL adds; in parallel Q adds. Mix-up is the most common board error.
• In parallel pipe problems, using total Q = Q1 + Q2 but forgetting the constraint h1 = h2 — both conditions must be satisfied simultaneously.
• Applying the Hardy Cross correction with wrong sign convention — clockwise head losses are positive; counterclockwise are negative.
• Ignoring continuity at junctions — flow in must equal flow out at every node.

Formulas

Exam Tips

• When a board problem asks for pump power, always check whether efficiency is given; if not, assume η = 1.0 (100%, theoretical).
• Always label datum, known pressures (free surfaces = 0 gauge), and known velocities (large reservoirs: v ≈ 0) before writing the energy equation.
• In a simple pipe-between-two-reservoirs problem with no pump: hp = 0 and the energy equation reduces to Δz = hL (the elevation difference drives the flow against friction).
• EGL sketch: draw it first as a sloping line; steep slopes indicate high velocity (large pipe losses); flat sections mean small loss.

Key Points

• The total head loss in a pipeline system is: hL = hf + Σhm (friction loss + sum of all minor losses).
• Energy equation between two points (Bernoulli with losses): z1 + p1/γ + v1²/2g + hp = z2 + p2/γ + v2²/2g + hL, where hp = pump head added.
• Pipe sizing: given Q and allowable hf, solve for D. Requires trial-and-error or the use of design charts because f also depends on D (via Re and ε/D).
• Pump head requirement: hp = hL + (z2 – z1) + (p2 – p1)/γ + (v2² – v1²)/2g (all terms on the energy equation that work against the pump).
• Power consumed by pump: P = γQhp / η, where η = pump efficiency (decimal) and γ = 9810 N/m³ for water.
• Water hammer: a rapid valve closure creates a pressure surge Δp = ρcΔv (Joukowsky equation), where c = wave speed ≈ 1200 m/s for water in steel pipe. Design implication: slow valve closure reduces surge.
• Hydraulic Grade Line (HGL) = piezometric head = z + p/γ. Energy Grade Line (EGL) = HGL + v²/2g. EGL always slopes downward in the direction of flow (energy is always lost). A pump causes a sudden rise in EGL.
• For board exam problems: first identify the system type (series, parallel, branching), write the energy equation, identify unknowns, and solve systematically.

Definitions

Section Title

5. Practical Design Considerations and Board Exam Integration

Common Mistakes

• Forgetting to include pump head hp in the energy equation when a pump is present in the system.
• Using efficiency as a percentage (e.g., 80) instead of a decimal (0.80) in the power formula — gives answer 100× too large.
• Treating the HGL and EGL as the same line — HGL = EGL – v²/2g; they only coincide when velocity is zero.
• Not checking whether the HGL drops below the pipe elevation — this indicates negative gauge pressure and potential cavitation.