Euler and Bernoulli equations with applications

Euler's equation, derivation and assumptions of Bernoulli's equation, heads and grade lines, the mechanical energy balance with pumps and losses, and Torricelli and siphon applications.

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Why it matters

Bernoulli's equation and its engineering extension, the mechanical energy balance, are used to size pumps, predict tank-draining times, design siphons and transfer lines, and interpret every venturi, orifice and pitot reading. Knowing exactly where it applies (and where friction, pumps or compressibility must be added) separates a correct design from one that starves a pump or under-delivers flow.

Key ideas

Euler's equation. Newton's second law applied to an inviscid fluid particle: mass × acceleration = pressure force + gravity force. In vector form ρ·DV/Dt = −∇p + ρg. Along a streamline (coordinate s, elevation z) it becomes ∂V/∂t + V·∂V/∂s + (1/ρ)·∂p/∂s + g·∂z/∂s = 0. Euler's equation is the Navier–Stokes equation with viscosity set to zero.

From Euler to Bernoulli. Integrate Euler's equation along a streamline with four assumptions:

  1. Steady flow (∂V/∂t = 0).
  2. Incompressible fluid (ρ constant).
  3. Inviscid (frictionless) flow.
  4. Along a single streamline (or anywhere in an irrotational flow). The result: p/ρ + V²/2 + gz = constant. No shaft work and no heat transfer are also implied.

Physical meaning. Each term is energy per unit mass (J/kg). Dividing by g gives heads in metres: pressure head p/ρg, velocity head V²/2g and elevation head z. Their sum is the total head. The hydraulic grade line (HGL) plots p/ρg + z; the energy grade line (EGL) lies a distance V²/2g above it. Where velocity rises, pressure falls (venturi throat); where the fluid rises, pressure falls (siphon crest).

Stagnation pressure. Bringing flow to rest isentropically at a stagnation point gives p₀ = p + ρV²/2. The dynamic pressure ρV²/2 is what a pitot tube measures.

Extended (engineering) Bernoulli / mechanical energy balance. Real flows have friction losses h_L and may include a pump (adds head h_p) or turbine (removes h_t). With the kinetic-energy correction factor α (α = 2 for laminar pipe flow, ≈ 1.05 for turbulent, usually taken as 1): p₁/ρg + α₁V₁²/2g + z₁ + h_p = p₂/ρg + α₂V₂²/2g + z₂ + h_t + h_L.

Applications.

  • Torricelli's theorem: efflux from a tank orifice, V = √(2gh). The real discharge needs a coefficient of discharge C_d.
  • Siphon: velocity set by the drop to the outlet; the crest is below atmospheric pressure, so the crest height is limited by the vapour pressure of the liquid (theoretically about 10 m for water, less in practice).
  • Venturi, orifice and pitot meters: see the flow-measurement topic.

Limits of validity. Bernoulli fails across pumps and turbines (shaft work), across shocks, in regions of strong viscous dissipation (long pipes, boundary layers, wakes), across streamlines in rotational flow, and for gases when the density changes by more than about 5 % (Mach number above about 0.3).

Formulas

  • ρ·DV/Dt = −∇p + ρ·g — Euler's equation; ρ (kg/m³), V (m/s), p (Pa), g (m/s²).
  • p/ρ + V²/2 + g·z = constant — Bernoulli per unit mass (J/kg); steady, incompressible, inviscid, along a streamline.
  • p/(ρg) + V²/(2g) + z = H — Bernoulli in heads (m); H total head.
  • p₀ = p + ρ·V²/2 — stagnation pressure (Pa).
  • V = √(2·g·h) — Torricelli efflux velocity (m/s); h head above the orifice (m).
  • p₁/ρg + α₁V₁²/2g + z₁ + h_p = p₂/ρg + α₂V₂²/2g + z₂ + h_L — mechanical energy balance with a pump; h_p pump head, h_L head loss (m).
  • P = ρ·g·Q·h_p/η — pump shaft power (W); Q (m³/s), η efficiency.

Worked examples

Example 1 (standard). Water drains from a large open tank through a small, well-rounded nozzle of 50 mm diameter located 4 m below the free surface. Neglecting losses, find the jet velocity and discharge.

  1. Apply Bernoulli from the free surface (p = p_atm, V ≈ 0, z = 4 m) to the jet (p = p_atm, z = 0): V = √(2·g·h).
  2. V = √(2 × 9.81 × 4) = 8.86 m/s.
  3. Q = A·V = (π/4)(0.05)² × 8.86 = 0.0174 m³/s. V ≈ 8.86 m/s, Q ≈ 0.0174 m³/s (17.4 L/s). A real orifice delivers less; multiply by C_d from your data book.

Example 2 (GATE level). A 50 mm siphon draws water from an open tank. The outlet is 3 m below the tank's free surface and the crest is 1.5 m above it. Neglect friction. Take p_atm = 101.325 kPa. Find (a) the velocity, (b) the discharge, (c) the absolute pressure at the crest.

  1. Bernoulli from free surface (1) to outlet (2), both at atmospheric pressure: V = √(2g × 3) = 7.67 m/s.
  2. Q = (π/4)(0.05)² × 7.67 = 0.0151 m³/s.
  3. Bernoulli from free surface (1) to crest (C), with z_C − z_1 = 1.5 m and the crest velocity equal to the pipe velocity: p_C = p_atm − ρ·g·(z_C − z_1) − ρ·V²/2.
  4. ρV²/2 = ρg × 3 m (from step 1), so p_C = p_atm − ρg(1.5 + 3) = 101 325 − 1000 × 9.81 × 4.5 = 101 325 − 44 145 = 57 180 Pa. (a) 7.67 m/s, (b) 0.0151 m³/s, (c) p_C ≈ 57.2 kPa absolute (44.1 kPa vacuum). Raising the crest, or lengthening the outlet leg, lowers p_C towards the vapour pressure, where the siphon breaks.

Example 3 (energy balance). A pump lifts 0.01 m³/s of water from an open sump to an open tank 20 m higher. Friction losses are 3 m and pump efficiency is 70 %. Velocities at both free surfaces are negligible. h_p = 20 + 3 = 23 m. P = 1000 × 9.81 × 0.01 × 23 / 0.7 = 3223 W. Shaft power ≈ 3.22 kW.

Common mistakes

  • Applying Bernoulli across a pump or between points with significant friction without adding h_p or h_L.
  • Mixing gauge and absolute pressures at the two points.
  • Forgetting the continuity step: V₂ comes from A₁V₁ = A₂V₂, not from assuming it.
  • Taking velocity at a large tank surface as anything other than approximately zero, or forgetting the velocity head at a jet outlet.
  • Using Bernoulli for gas flows with large pressure changes.
  • Sign of elevation: z is measured upward from one chosen datum used for both points.

For GATE CH

Expect numericals combining continuity and Bernoulli (reducers, nozzles, siphons, tank draining), mechanical energy balances to find pump head or power, and conceptual MCQs on the assumptions and on HGL/EGL. Draining-time problems (integrating Torricelli's law) appear frequently in chemical engineering papers; practise t = (A_tank/A_orifice)·√(2/g)·(√h₁ − √h₂).

Quick check

  1. List the four assumptions behind Bernoulli's equation.
  2. What is the velocity head of water moving at 4 m/s?
  3. Why is the crest of a siphon below atmospheric pressure?
  4. A pitot tube in water reads a stagnation-minus-static difference of 2 kPa. What is V?

Answers: 1. Steady, incompressible, inviscid, along a streamline. 2. 4²/(2 × 9.81) = 0.815 m. 3. The fluid there is both higher than the surface and moving, so pressure head must fall to supply elevation and velocity head. 4. V = √(2 × 2000/1000) = 2 m/s.

Interview questions

All Fluid Mechanics interview questions

Try answering each one aloud before you open it.

  1. 1.What is Euler's equation of motion in fluid mechanics?Concept

    Euler's equation of motion describes the relationship between the velocity field and the pressure field within a fluid flow. It is derived from Newton's second law and is expressed as: ρ(du/dt + u·∇u) = -∇p + ρg, where ρ is the fluid density, u is the velocity vector, p is the pressure, and g is the gravitational acceleration vector.

  2. 2.Explain Bernoulli's equation and its assumptions.Concept

    Bernoulli's equation is a principle of fluid dynamics that describes the conservation of energy in a flowing fluid. It states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant along a streamline. The equation is given by: p + 0.5ρv² + ρgh = constant. The assumptions include incompressible flow, steady flow, and no viscous effects.

  3. 3.How does Bernoulli's equation apply to the operation of a Venturi meter?Application

    A Venturi meter uses Bernoulli's principle to measure the flow rate of a fluid. As fluid flows through the constricted section of the Venturi meter, its velocity increases, causing a decrease in pressure. By measuring the pressure difference between the wider and narrower sections, the flow rate can be calculated using Bernoulli's equation.

  4. 4.Why is Bernoulli's equation not applicable to viscous flows?Application

    Bernoulli's equation assumes no viscous effects, meaning it is only applicable to ideal, inviscid flows. In viscous flows, energy is lost due to friction, which is not accounted for in Bernoulli's equation. Therefore, it cannot accurately describe the energy distribution in real, viscous fluid flows.

  5. 5.What happens to the pressure in a fluid as its velocity increases, according to Bernoulli's principle?Application

    According to Bernoulli's principle, as the velocity of a fluid increases, its pressure decreases. This is because the total mechanical energy of the fluid is conserved, so an increase in kinetic energy (due to higher velocity) must be balanced by a decrease in pressure energy.

  6. 6.Explain how Euler's equation is used to derive Bernoulli's equation.Concept

    Euler's equation of motion can be simplified under the assumptions of steady, incompressible, and inviscid flow along a streamline. By integrating Euler's equation along a streamline, we obtain Bernoulli's equation, which relates pressure, velocity, and elevation in the fluid.

  7. 7.In what scenarios would you prefer using Euler's equation over Bernoulli's equation?Application

    Both neglect viscosity, but Bernoulli is Euler's equation integrated under extra restrictions: steady, incompressible flow along a streamline. So you go back to Euler's equation when the flow is unsteady (for example the oscillation of liquid in a U-tube or the start-up of flow in a pipe, which needs the ∂V/∂t term), when density varies (compressible gas flow, integrated with an equation of state), or when you need pressure variation across streamlines, such as the radial pressure gradient ∂p/∂r = ρV²/r in curved flow. If viscous effects matter, neither applies and you need Navier–Stokes or an energy balance with losses.

  8. 8.Calculate the pressure difference between two points in a horizontal pipe where the fluid velocity changes from 3 m/s to 5 m/s. Assume the fluid density is 1000 kg/m³.Numerical

    Using Bernoulli's equation for a horizontal pipe, the pressure difference Δp can be calculated as: Δp = 0.5ρ(v2² - v1²). Substituting the given values: Δp = 0.5 * 1000 * (5² - 3²) = 0.5 * 1000 * (25 - 9) = 0.5 * 1000 * 16 = 8000 Pa.

  9. 9.What are the limitations of using Bernoulli's equation in real-world applications?Application

    Bernoulli's equation assumes no energy loss due to friction, no heat transfer, and no work done by or on the fluid, which are not true in real-world applications. It also assumes steady, incompressible, and inviscid flow, which limits its applicability to idealized scenarios. In practice, corrections for viscosity and other factors are often needed.

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