Euler’s equation and Bernoulli’s principle

Ask about two points in one flow

A liquid moves through a smoothly narrowing ideal flow region. At two points on its centreline, its speed is 2 m/s and 6 m/s. Both points are at the same height. The upstream pressure is 150 kPa absolute. If the flow pattern stays unchanged, can a moving fluid particle still accelerate? What pressure would match the faster downstream motion?

Pressure p is local normal force per area exerted by the fluid; its unit is the pascal, Pa = N/m². A velocity gives both direction and rate of motion at a location. Speed V is the magnitude of that velocity, measured in m/s. A streamline is a curve tangent to the local velocity at one instant. Our two points lie on the same streamline. In a steady flow, the field seen at a fixed position does not change with time.

For this original ideal example, assign one constant density ρ = 1000 kg/m³ and neglect viscous stresses. Pressure and gravity still act. The values describe a model, rather than a measured liquid or a viscous nozzle experiment. We will calculate on the centreline first, then give a complete smooth velocity and pressure field as an optional check.

For a first reading, follow the moving particle, the pressure-force balance, the three Bernoulli terms and the rotating-flow counterexample. Optional published refreshers are “Units, standards, and physical constants” and “Newton’s laws and free-body diagrams”. “Derivatives, Taylor series, and local approximation” and “Vectors and vector spaces” help with the later field derivation. The neighbouring “Conservation of mass, momentum, and energy” develops the wider balances; the local balance needed now is explained below.

Follow a particle through a steady field

Imagine standing beside the first point: the same local velocity is present whenever you look. A particle passing that point subsequently reaches locations with different velocities. Its acceleration depends on the velocities it encounters along its path. Steady describes the field at fixed positions; it does not mean that each moving particle keeps one velocity.

Write the velocity field as u(x,t). The material derivative D/Dt means the rate of change while following a fluid particle. Its acceleration is Du/Dt = ∂u/∂t + (u·∇)u. The first term is local time change at a fixed location. The second is change caused by travelling through a spatially varying field. The gradient symbol ∇ collects spatial derivatives; u·∇ weights them by the particle’s local velocity components.

Along our straight centreline, x runs from 0 to 1 m and points downstream. Prescribe V(x) = U + kx, with U = 2 m/s and k = 4 s⁻¹. In steady motion ∂V/∂t = 0, but a_x = V dV/dx = kV. The centreline acceleration rises from +8 m/s² to +24 m/s² as the particle travels downstream. At x = 0.5 m, V = 4 m/s and a_x = +16 m/s².

An acceleration can also change direction without changing speed. A particle moving around a circular streamline has inward centripetal acceleration. Euler’s equation includes both this directional change and the speeding-up motion just calculated. A flat time trace at one fixed location is therefore insufficient evidence for zero particle acceleration.

Balance pressure and body force

Apply Newton’s second law to a small moving parcel of an inviscid fluid. Pressure acts on all its faces. If pressure is slightly larger on the upstream face than on the downstream face, the difference supplies a downstream force. Per unit volume, the net pressure force is −∇p. This sign says that the pressure-force contribution points toward decreasing pressure.

Euler’s equation in an inertial frame is ρ[∂u/∂t + (u·∇)u] = −∇p + ρb. Here b is body force per unit mass, in m/s². For gravity with z upward, b = −g e_z and g = 9.81 m/s². Pressure is a contact force distributed over parcel faces; gravity acts throughout its mass. Inviscid means that viscous stresses are omitted from this momentum balance, rather than that pressure or acceleration is omitted.

On the horizontal centreline, gravity has no x component. Thus dp/dx = −ρa_x = −ρkV. The required gradient is −8000 Pa/m at x = 0 and −24000 Pa/m at x = 1 m. Pressure must fall along this accelerating centreline. Integrating gives p(x,0,0) = 150000 Pa − ρ[Ukx + k²x²/2]. It falls by 16000 Pa between the two endpoints.

Euler’s equation itself can describe unsteady flow and variable density. Such cases require the accompanying mass balance and an appropriate relation for the fluid state; Euler alone does not determine every unknown. At rest its gravity component reduces to dp/dz = −ρg. The steady constant-density Bernoulli relation below is a conditional consequence of Euler, rather than a replacement for every momentum or energy balance.

Integrate along a streamline

Let s measure distance along a steady streamline in the direction of motion. The acceleration component along it is V dV/ds. If gravity is the only body force, projecting Euler onto that direction gives V dV/ds = −(1/ρ)dp/ds − g dz/ds. This is a local force balance: a pressure change and an elevation change accompany a change of speed.

For the stated constant density, integrate along one streamline: B = p/ρ + V²/2 + gz is constant. This is Bernoulli’s principle in energy-per-mass form. Every term has units J/kg = m²/s². For two points on that streamline, p₁/ρ + V₁²/2 + gz₁ = p₂/ρ + V₂²/2 + gz₂.

Use this simple form for smooth steady inviscid flow with constant density and conservative body force. Gravity is conservative because its force per mass is −∇(gz). The comparisons here contain no pump, turbine, dissipative shock or significant viscous loss between their points. The flow may be rotational: constancy along a streamline does not, by itself, grant the same constant to a different streamline.

For our equal-height endpoints, p₂ = p₁ + ρ(V₁² − V₂²)/2. The change is 1000(4 − 36)/2 Pa = −16000 Pa, so p₂ = 134000 Pa = 134 kPa absolute. The integrated particle force balance and the Bernoulli calculation agree. Faster motion and lower pressure are related here because the two points satisfy the stated same-streamline, same-elevation conditions.

Read pressure, speed and height terms

The pressure term p/ρ describes pressure flow work per mass in this mechanical balance; V²/2 is kinetic energy per mass, and gz is gravitational potential energy per mass. Their sum B is not the entire thermodynamic energy of a fluid: internal energy is not included in this three-term expression. At the upstream point the terms are 150, 2 and 0 J/kg, giving B = 152 J/kg.

Multiplying by constant ρ gives p + q_d + ρgz = ρB, where q_d = ρV²/2 is called dynamic pressure. It has pressure units and represents a kinetic term. It is not an additional isotropic pressure to add to every fluid traction. Our downstream centreline has p = 134 kPa and q_d = 18 kPa; their equal-height sum is 152 kPa.

As a separate comparison, keep the same upstream state but prescribe a downstream point with V = 6 m/s at z = 1 m. The gravitational term becomes 9.81 J/kg, so its pressure term is 152 − 18 − 9.81 = 124.19 J/kg. The resulting absolute pressure is 124.19 kPa. This elevated point-state comparison obeys the same conditional Bernoulli relation; it is not asserted to belong to the horizontal planar field.

Dividing B by g gives head H = p/(ρg) + V²/(2g) + z, with all terms in metres. Pressure head, velocity head and elevation head can trade while H stays fixed in the same ideal comparison. Use a common pressure reference: changing every pressure by the same constant shifts B and H but not their differences. Absolute pressure is needed when checking a physical pressure limit such as phase change.

Compare static and stagnation pressure

Static pressure is the local p even when the fluid is moving. Stagnation pressure p₀ is the pressure reached when that local flow is brought smoothly to rest under a specified ideal process. For this steady constant-density inviscid comparison at the same elevation, p₀ = p + ρV²/2. At the faster centreline point, p₀ = 134 kPa + 18 kPa = 152 kPa. The fluid near a probe tip can stop locally while fluid elsewhere continues to flow.

A Pitot-static arrangement compares a flow-facing stagnation opening with a static-pressure connection. Under the ideal conditions, V = √[2(p₀ − p)/ρ]. For the separately supplied difference p₀ − p = 5000 Pa and ρ = 1000 kg/m³, this predicts V = √10 m/s ≈ 3.162278 m/s. The difference, density and units are essential; a static pressure alone does not supply a speed.

The same principle can estimate an ideal outlet speed. Compare a large reservoir’s free surface with a free jet at the same prescribed ambient pressure. If the free-surface speed is negligible and the head above the outlet is h, the quasi-steady inviscid prediction is V = √(2gh). For h = 0.2 m, it gives approximately 1.980909 m/s. This is an ideal speed, not a measured discharge, a loss coefficient or a prediction of the complete draining history.

At different elevations, p + ρV²/2 need not be constant even when B is: the gravity term also changes. A real probe needs appropriate alignment, pressure sampling and a valid fluid regime. Compressibility, a shock, dissipation or a blocked connection can invalidate the simple pressure-difference formula. A larger dynamic-pressure term is a description of this kinetic scale, rather than proof that every measured pressure must decrease.

Keep different streamline constants separate

Now prescribe a separate steady ideal fluid rotating like a solid body in a horizontal plane. Its velocity is u = (−Ωy, Ωx, 0), with Ω = 2 rad/s. At radius r its speed is V = Ωr, and positive-radius streamlines are circles. A particle has inward radial acceleration a_r = −Ω²r even though the field is steady.

Take pressure at the reference axis as 100000 Pa. The inward pressure force requires dp/dr = ρΩ²r, so at z = 0, p(r) = 100000 Pa + ρΩ²r²/2. At r = 0.5 m the speed is 1 m/s and pressure is 100.5 kPa. At r = 1 m they are 2 m/s and 102 kPa. Pressure and speed both rise in this radial comparison because we are crossing different circular streamlines, not following one accelerating streamline.

The Bernoulli value in that plane is B(r) = 100 J/kg + Ω²r². It is 101 J/kg on the r = 0.5 m circle and 104 J/kg on the r = 1 m circle. On each circle r stays fixed, so B is constant along that streamline. The axis r = 0 supplies the reference value; it is not a circular streamline. Spin-up, wall torque, viscosity and a free surface are not modelled in this example.

A sufficient stronger condition for a common spatial B is irrotational flow: ∇×u = 0, along with the same steady inviscid constant-density conservative-force assumptions in a connected smooth region. The symbol ∇×u is vorticity, a measure of local fluid rotation; the next planned subject “Vorticity and circulation” develops it. Rotational flow can have different streamline constants, as the example shows. Irrotationality is a sufficient condition, not a claim that every rotational flow must have different constants.

Check the density, time and loss regime

Constant density is an explicit model choice. If density varies but is a single-valued function of pressure, a barotropic fluid, the pressure term in a steady streamline integral becomes ∫ dp/ρ(p). It cannot generally be replaced by p/ρ. If density depends independently on temperature or composition, additional thermodynamic information is needed. The constant-density formulas here are not a high-speed gas-flow rule.

Unsteady motion retains ∂u/∂t, so the simple steady B need not stay constant along an instantaneous streamline. An optional stronger result is available when the smooth inviscid constant-density flow is irrotational and admits a velocity potential φ with u = ∇φ. Under gravity, ∂φ/∂t + |∇φ|²/2 + p/ρ + gz = C(t) in that connected region. The time-derivative term matters; this is not permission to apply the steady formula during arbitrary start-up.

Real viscous flow can convert mechanical energy to internal energy. A pressure drop in an equal-speed horizontal pipe, for example, can accompany friction rather than a kinetic increase. Mechanical head-loss or machine-work terms require an appropriate extended balance and data; they are not recovered by silently setting the ideal three-term sum constant. The neighbouring conservation lesson treats the full bookkeeping. Boundary layers, separation and turbulence require their own assessment, even when an outer region is approximated as inviscid.

Before calculating, name the compared points, their streamline relationship, elevation, pressure reference, density regime and time dependence. Decide whether viscous loss or a machine matters, and whether the flow remains smooth and single-phase. Bernoulli relates an established flow and pressure field; it does not determine an unknown flow geometry from a slogan. It also does not imply equal travel times for fluid passing different sides of an object.

Return to the complete ideal field

This optional field check explains how the centreline example fits a complete smooth model. Use u = (U + kx, −ky, 0) with the assigned U and k. Its divergence is k − k = 0, so the constant-density volume balance holds. The negative transverse component accompanies the increasing downstream component; simply writing u = (U + kx, 0, 0) would fail that incompressibility check.

A stationary streamline obeys dy/dx = u_y/u_x = −ky/(U + kx). Its solution is y(x) = y(0)U/(U + kx). Choose the two boundary streamlines with y(0) = ±0.05 m and span 0.03 m in the z direction. Their section area shrinks as the axial speed increases, keeping the volume flow 0.006 m³/s. The inviscid boundaries allow tangential slip and have no normal crossing; this is not a no-slip viscous wall solution.

Direct differentiation gives (u·∇)u = (k(U + kx), k²y, 0). The pressure field p = 150000 Pa − ρ[Ukx + k²(x² + y²)/2 + gz] supplies the axial and transverse pressure forces, while its z gradient balances gravity. Its centreline at y = z = 0 gives the earlier table. The transverse pressure variation is retained rather than assuming uniform pressure across every section.

Here ∇×u = 0 and a velocity potential is φ = Ux + k(x² − y²)/2. Substituting the complete speed and pressure into B cancels both the x and y dependence: B = 150 J/kg + 2 J/kg = 152 J/kg. This exact ideal field explains why a global constant is justified for this example. The separate rotating field does not share that irrotational condition. The derivations check the prescribed models; they do not establish a measured device’s losses or fluid regime.

Worked examples

Try it yourself

1. At x = 0.5 m in the prescribed centreline field V = 2 + 4x in SI units, what are the local time derivative, material acceleration and pressure gradient?

   Show the answer and reasoning

   ∂V/∂t = 0, a_x = +16 m/s² and dp/dx = −16000 Pa/m.

   The field is steady, but V = 4 m/s and dV/dx = 4 s⁻¹. A moving particle therefore has V dV/dx = 16 m/s². Horizontal Euler gives dp/dx = −ρa_x with ρ = 1000 kg/m³.

2. An ideal equal-elevation stagnation comparison in the assigned constant-density liquid gives p₀ − p = 5000 Pa. What speed follows, and which pressure difference did you use?

   Show the answer and reasoning

   V = √10 m/s ≈ 3.162278 m/s, using stagnation pressure minus local static pressure.

   Use V = √[2(p₀ − p)/ρ] with ρ = 1000 kg/m³. A static pressure alone would not determine the speed. The result assumes smooth steady inviscid deceleration to rest at the same elevation, with the stated density model.

3. At the separate elevated point with p = 124190 Pa, V = 6 m/s and z = 1 m, is p + ρV²/2 equal to 152000 Pa?

   Show the answer and reasoning

   No. It is 142190 Pa; the gravity term ρgz = 9810 Pa completes the Bernoulli pressure sum.

   The kinetic term is 18000 Pa. Add it to the local static pressure to get the local equal-elevation stagnation pressure. The same streamline’s Bernoulli invariant includes elevation as well, so stagnation pressure alone need not match between different heights.

4. In the supplied rotating field, find speed, static pressure and B at r = 0.75 m and z = 0. Does higher pressure there imply slower motion than at r = 0.5 m?

   Show the answer and reasoning

   V = 1.5 m/s, p = 101125 Pa and B = 102.25 J/kg. No: both pressure and speed are higher.

   V = Ωr and p − p_axis = ρΩ²r²/2 = 1125 Pa. The kinetic term is 1.125 J/kg, so B = 101.125 + 1.125 = 102.25 J/kg. The two radii identify different circular streamlines; the equal-height same-streamline shortcut does not compare them.

5. A large reservoir has a free surface 0.2 m above a free jet. Both are at the same ambient pressure and the surface speed is negligible. What quasi-steady ideal outlet speed follows?

   Show the answer and reasoning

   V = √(2 × 9.81 × 0.2) m/s ≈ 1.980909 m/s.

   Equal ambient pressure terms cancel; the negligible upstream kinetic term leaves gh = V²/2. This is a same-streamline inviscid estimate. Real losses, a finite surface speed or start-up would require further information, and no complete draining-time prediction follows.

6. A flow is inviscid and steady. One comparison crosses unrelated streamlines; another has density varying with temperature independently of pressure. Are those two facts enough to use one constant p/ρ + V²/2 + gz everywhere?

   Show the answer and reasoning

   No. The streamline relationship and the density model still need justification.

   Along-streamline Bernoulli need not have one constant across rotational streamlines. For variable density, p/ρ is not generally the integrated pressure term. A barotropic relation permits ∫ dp/ρ(p); independent temperature dependence needs additional thermodynamic information. Inviscid and steady alone do not settle either comparison.

What to carry forward

• A steady velocity field can accelerate a moving particle through spatial change or a change of direction.
• Euler balances material acceleration with pressure gradients and body force; neglecting viscosity does not remove pressure forces.
• The simple Bernoulli sum p/ρ + V²/2 + gz uses stated smooth steady inviscid constant-density conservative-force conditions and is constant along a streamline.
• A common constant across streamlines needs more justification; irrotationality is a sufficient stronger condition in a connected smooth region.
• Dynamic pressure is a kinetic scale. Ideal local stagnation pressure is p + ρV²/2 at the same elevation, while the full Bernoulli sum also includes gravity.
• Density variation, time dependence, viscous loss, machines and nonsmooth flow require the appropriate extended model rather than an unconditional faster-flow slogan.