CELE Hydraulics & Fluid Mechanics — Fundamentals of Fluid FlowRevision Notes

Formulas

Exam Tips

• Memorize the shortcut v₂ = v₁(D₁/D₂)² — it saves 30–40 seconds per continuity problem.
• For three-pipe junctions, label flow directions clearly and write the junction equation before solving.
• Q is constant in a single pipe regardless of elevation changes, diameter changes, or bends — it only changes at branches or leaks.
• Double-check: if D₂ = D₁/2 then v₂ = 4v₁; if D₂ = D₁/√2 then v₂ = 2v₁ — useful mental checks.

Key Points

• For steady, incompressible flow, the volumetric flow rate Q is constant along any stream tube or pipe section.
• Q = A₁v₁ = A₂v₂, where A is the cross-sectional area (m²) and v is the mean velocity (m/s).
• For a circular pipe, A = πD²/4; substituting gives v₂ = v₁(D₁/D₂)².
• Velocity increases when area decreases (converging section) and decreases when area increases (diverging section).
• For branching pipes: Q_in = Q_out, i.e., Q₁ = Q₂ + Q₃ at a junction.
• Mass flow rate (kg/s): ṁ = ρAv = ρQ. For incompressible flow (constant ρ), continuity reduces to constant Q.

Definitions

Section Title

1. Conservation of Mass — Continuity Equation

Common Mistakes

• Forgetting to square the diameter ratio — using v₂ = v₁(D₁/D₂) instead of v₂ = v₁(D₁/D₂)².
• Using diameter instead of area when D is not doubled or halved — always derive A = πD²/4 first.
• At pipe junctions, not applying ΣQ_in = ΣQ_out — incorrectly equating only two of three pipe flows.
• Mixing units: diameter in mm but using it directly in the area formula without converting to metres.

Formulas

Exam Tips

• Always write the full extended Bernoulli first, then cancel zero or negligible terms (large reservoir → v ≈ 0; free jet → p = 0).
• For a large tank discharging through a pipe: p_surface = 0 (atm, gauge), v_surface ≈ 0, z_surface = known — three terms simplify immediately.
• EGL and HGL sketching is occasionally asked directly; remember: at a pump, EGL jumps up by h_A; at entry to a nozzle, HGL dips sharply.
• Check your answer using the EGL: if total head increases from 1 to 2 without a pump, you made a sign error.
• Board exams frequently combine continuity + Bernoulli in one problem — solve continuity first to get v₂, then apply energy equation.

Key Points

• The total mechanical energy per unit weight (total head, H_T) at any section equals the sum of pressure head (p/γ), velocity head (v²/2g), and elevation head (z).
• Extended form accounts for energy added by a pump (h_A), energy extracted by a turbine (h_E), and energy lost to friction and minor losses (h_L).
• The Ideal Bernoulli Equation is the special case where h_A = h_E = h_L = 0; total head is then constant.
• Energy Grade Line (EGL): graphical plot of total head H_T = p/γ + v²/2g + z along the pipe.
• Hydraulic Grade Line (HGL): plot of piezometric head = p/γ + z = EGL − v²/2g.
• EGL always slopes downward in the direction of flow (energy is lost); HGL can rise in a diverging section.
• When HGL drops below the pipe centerline, p/γ < 0 (sub-atmospheric pressure) — risk of cavitation.
• At a free discharge (jet to atmosphere), p = 0 (gauge) at the outlet section.

Definitions

Section Title

2. The Energy Equation (Extended Bernoulli)

Common Mistakes

• Placing h_A on the wrong side of the equation — pump head (h_A) is ADDED to the upstream side (or equivalently added to the left side); turbine head (h_E) is SUBTRACTED from the downstream total.
• Forgetting the velocity head term when pipe sizes change — this is the most common Bernoulli error on board exams.
• Mixing gauge and absolute pressures — always use gauge pressures consistently unless the problem involves cavitation (requires absolute pressure).
• Using γ = 9.81 kN/m³ with pressures in Pa (N/m²) — either use γ = 9810 N/m³ with Pa, or γ = 9.81 kN/m³ with kPa.
• Choosing the wrong datum for z — it can be any convenient horizontal plane but must be consistent for both points.

Formulas

Exam Tips

• Quick unit check: if γ = 9.81 kN/m³ (kN/m³) × Q (m³/s) × H (m) = kN·m/s = kW. Use this version for kW answers.
• Turbine problems almost always give efficiency and ask for output power — use P = η·γQH directly.
• Pump problems often ask for both pump head and input power — solve energy equation for h_A first, then apply P = γQh_A/η.
• If total efficiency is given (e.g., pump-motor set at 75%), apply it directly: P_input(motor) = γQh_A / 0.75.

Key Points

• Power (P) is the rate of energy transfer; for a fluid carrying flow rate Q across a head H, P = γQH.
• Units: γ in N/m³ (= 9810 N/m³ for water) × Q in m³/s × H in m = Watts (W); divide by 1000 for kW.
• For a pump: the power delivered to the fluid (water power or fluid power) is P_fluid = γQh_A.
• The input shaft power to the pump is greater: P_input = γQh_A / η_pump.
• For a turbine: the output shaft power is less than the fluid power: P_output = η_turbine × γQh_E.
• Overall efficiency of a pump-motor set: η_overall = η_motor × η_pump.
• 1 horsepower = 745.7 W (conversion occasionally needed if problem gives power in hp).

Definitions

Section Title

3. Power of a Flowing Stream

Common Mistakes

• Using γ = 9.81 instead of 9810 — off by factor of 1000; gives answer in kW when W is expected, or vice versa.
• Applying efficiency in the wrong direction — for pump, divide by η (input > fluid power); for turbine, multiply by η (output < fluid power).
• Ignoring efficiency when the problem specifies it — always re-read whether efficiency is given.
• Confusing net head H with gross head — net head already accounts for losses; if the problem gives separate h_L, compute h_A from the energy equation first.

Formulas

Exam Tips

• Most board exam momentum problems involve simple horizontal pipes or 90° bends — draw the FBD and assign positive x and y axes before writing equations.
• For a nozzle problem: if friction is neglected, use Bernoulli for velocities, then momentum for force.
• The momentum equation does not require frictionless conditions — this is its advantage over the energy equation for force calculations.
• ρQ = ṁ = mass flow rate; if the problem gives ṁ directly, ΣF = ṁ(v₂ − v₁) — no need to know Q and ρ separately.

Key Points

• Newton's second law applied to a control volume: ΣF = rate of change of momentum = ρQ(v₂ − v₁) for steady flow.
• Forces included: pressure forces, weight of fluid in CV, and reaction forces from pipe walls or vanes.
• Always apply component-wise: ΣF_x = ρQ(v₂ₓ − v₁ₓ) and ΣF_y = ρQ(v₂ᵧ − v₁ᵧ).
• The force on the fluid is what the momentum equation directly gives; the force on the pipe/vane is the reaction (Newton's 3rd law).
• For a pipe bend of angle θ: draw a free-body diagram of the fluid in the bend; include p₁A₁, p₂A₂, and the wall reaction R.
• For a flat stationary vane: jet deflected through 180° gives maximum force = 2ρQv.
• The momentum equation is independent of losses — it does not require frictionless flow.

Definitions

Section Title

4. Momentum Equation

Common Mistakes

• Forgetting pressure forces (p₁A₁ and p₂A₂) in the momentum equation for high-pressure systems — they are part of ΣF.
• Using mass (kg) instead of weight force — momentum equation uses ρQΔv, not γQΔv.
• Not drawing a free-body diagram — sign errors in force direction are the most common mistake in bend problems.
• Confusing force on fluid vs. force on pipe — the reaction force on the pipe is equal and opposite to the net force on the fluid.

Formulas

Exam Tips

• EGL and HGL sketching problems are common in the board exam — practise drawing these for: (a) pipe with reservoir, (b) pipe with pump, (c) pipe with nozzle.
• Key check: EGL must slope downward from section 1 to 2 (unless a pump is present); any upward slope without a pump means an error.
• At a nozzle exit to atmosphere: HGL = pipe centreline elevation, EGL = HGL + v²/2g (large velocity head makes EGL well above HGL).
• Sub-atmospheric pressure regions (HGL below pipe) are acceptable as long as absolute pressure stays above vapour pressure (~0.24 m abs for water at 20°C).

Key Points

• EGL and HGL are graphical tools that give an instant picture of how energy is distributed along a pipeline.
• EGL = p/γ + v²/2g + z (total head at each section); plots as a line above the pipe.
• HGL = p/γ + z (piezometric head); always lies v²/2g below the EGL.
• For a constant-diameter pipe with friction: EGL slopes linearly downward (constant friction slope S_f = h_L/L); HGL is parallel to EGL.
• At a sudden expansion (area increases): velocity decreases, velocity head drops, pressure head rises — HGL rises.
• At a sudden contraction or nozzle (area decreases): velocity head rises, pressure head falls — HGL drops sharply.
• Pump: EGL jumps upward by h_A at the pump location.
• Turbine: EGL drops abruptly by h_E at the turbine location.
• If HGL falls below pipe centerline: local pressure is sub-atmospheric (negative gauge pressure). If HGL falls below the vapour pressure head: cavitation occurs.
• For a free jet discharging to atmosphere: p = 0 at the outlet, so HGL coincides with the jet centreline at that point.

Definitions

Section Title

5. Energy Grade Line and Hydraulic Grade Line — Graphical Interpretation

Common Mistakes

• Drawing EGL below HGL — EGL is always at or above HGL (velocity head is always ≥ 0).
• Assuming EGL is always above the pipe centreline — it can be below the pipe if total head is less than pipe elevation (only possible with very high velocity head).
• Forgetting that at a free-surface reservoir, the EGL = HGL = water surface elevation (v ≈ 0).
• Not accounting for the abrupt EGL drop at minor losses (valves, bends) — in detailed problems, each loss shifts the EGL downward.