Bernoulli's Theorem, Steady and Unsteady

Derivation

Bernoulli's Theorem, Steady and Unsteady

D-320 Home PU-307 Threads energy · force Depends on Euler's Equations as the Inviscid Limit

Statement

Starting from Euler's equation for an inviscid fluid in a conservative body-force field, we derive two distinct Bernoulli relations. For steady flow the quantity \( \tfrac{1}{2}v^2 + \int \frac{dp}{\rho} + \Phi \) is constant along each streamline (and along each vortex line), even when the flow is rotational. For unsteady irrotational flow, writing \( \vec v = \nabla\varphi \), the quantity \( \dfrac{\partial \varphi}{\partial t} + \tfrac{1}{2}v^2 + \int \frac{dp}{\rho} + \Phi \) is spatially uniform, equal to a single function \( f(t) \) over the entire connected fluid domain.

Why it matters

Bernoulli's theorem is the energy integral of the momentum equation for an ideal fluid. It converts a coupled vector PDE into a single algebraic constraint linking pressure, speed and height, and underlies pitot tubes, venturi meters, aerofoil lift estimates, cavitation criteria, efflux from tanks and the water-hammer transient.

The steady and unsteady forms answer genuinely different questions. The steady form is a local conservation law valid along a streamline of an otherwise arbitrary (possibly rotational) steady flow; the unsteady irrotational form is a global statement whose strength comes from potential flow, and it is the tool that handles accelerating columns, sloshing, and impulsively started motion.

Assumptions

Inviscid fluid (no viscous stress).With viscosity the Navier–Stokes stress \( \mu\nabla^2\vec v \) adds a non-conservative term; mechanical energy is dissipated and the Bernoulli sum decreases downstream by the head loss.

Conservative body force, \( \vec g = -\nabla\Phi \).If the body force is not the gradient of a potential (e.g. a general non-inertial or magnetic force without a potential), it cannot be absorbed into the integrand and no scalar constant emerges.

Barotropic pressure–density relation, \( \rho = \rho(p) \).Only then is \( \frac{1}{\rho}\nabla p = \nabla\!\int\frac{dp}{\rho} \) a gradient; for a general baroclinic fluid \( \nabla p \) and \( \nabla\rho \) are misaligned, the pressure term is not exact, and vorticity is generated (Bjerknes).

Steady form: \( \partial_t\vec v = 0 \).If dropped, the local-acceleration term survives and the streamline integral picks up \( \int \partial_t\vec v\cdot d\vec\ell \), which is generally non-zero and path-dependent.

Unsteady form: irrotational flow, \( \vec\omega=\nabla\times\vec v=0 \), on a simply connected domain.If vorticity is present the term \( \vec v\times\vec\omega \) is not a gradient, so the sum is not uniform; on a multiply connected domain \( \varphi \) may be multivalued and circulation must be tracked separately.

Derivation

\[ \frac{\partial \vec v}{\partial t} + (\vec v\cdot\nabla)\vec v = -\frac{1}{\rho}\nabla p - \nabla\Phi \]

Euler's equation for an inviscid fluid, with the body force written as \( \vec g=-\nabla\Phi \) (prior result: euler-equations-inviscid-limit). A

\[ (\vec v\cdot\nabla)\vec v = \nabla\!\left(\tfrac{1}{2}v^2\right) - \vec v\times(\nabla\times\vec v) = \nabla\!\left(\tfrac{1}{2}v^2\right) - \vec v\times\vec\omega \]

Lamb (convective) vector identity, defining vorticity \( \vec\omega=\nabla\times\vec v \). This is an exact rewriting of the advection term. B

\[ \frac{1}{\rho}\nabla p = \nabla\!\left(\int\frac{dp}{\rho}\right) \equiv \nabla P \]

Barotropy \( \rho=\rho(p) \) makes the pressure term an exact gradient; define the pressure function \( P=\int dp/\rho \). For incompressible flow \( P=p/\rho \). B

\[ \frac{\partial \vec v}{\partial t} - \vec v\times\vec\omega = -\nabla\!\left(\tfrac{1}{2}v^2 + P + \Phi\right) \]

Substitute steps 2–3 into step 1 and collect the three gradients on the right. Define the Bernoulli head \( B \equiv \tfrac{1}{2}v^2 + P + \Phi \). B

\[ \text{Steady: } \quad \vec v\times\vec\omega = \nabla B \]

Set \( \partial_t\vec v=0 \). This is Crocco's form: the gradient of the Bernoulli head equals \( \vec v\times\vec\omega \), a vector orthogonal to both \( \vec v \) and \( \vec\omega \). C

\[ \vec v\cdot\nabla B = \vec v\cdot(\vec v\times\vec\omega) = 0, \qquad \vec\omega\cdot\nabla B = \vec\omega\cdot(\vec v\times\vec\omega)=0 \]

Dot step 5 with \( \vec v \) and with \( \vec\omega \); the triple products vanish because \( \vec v\times\vec\omega \) is perpendicular to each factor. So \( B \) has zero directional derivative along both streamlines and vortex lines. C

\[ \boxed{\ \tfrac{1}{2}v^2 + \int\frac{dp}{\rho} + \Phi = B\ \text{(const. on a streamline)}\ } \]

Integrating step 6 along a streamline: \( B \) is constant on that line. Different streamlines may carry different \( B \) unless the flow is irrotational or the inflow is uniform. B

\[ \text{Irrotational: } \vec\omega=0 \Rightarrow \vec v=\nabla\varphi,\qquad \frac{\partial\vec v}{\partial t}=\nabla\frac{\partial\varphi}{\partial t} \]

Zero vorticity on a simply connected domain guarantees a single-valued velocity potential \( \varphi \). Time and space derivatives commute, so the local-acceleration term is itself a gradient. C

\[ \nabla\!\left(\frac{\partial\varphi}{\partial t} + \tfrac{1}{2}v^2 + P + \Phi\right)=0 \]

Put \( \vec\omega=0 \) in step 4 and move \( \partial_t\vec v=\nabla\partial_t\varphi \) inside the gradient. The entire left side is the gradient of a single scalar, which must therefore be spatially uniform. C

\[ \boxed{\ \frac{\partial\varphi}{\partial t} + \tfrac{1}{2}v^2 + \int\frac{dp}{\rho} + \Phi = f(t)\ \text{(everywhere)}\ } \]

A field with zero gradient is a function of time alone, \( f(t) \). This is the unsteady (Bernoulli–Lagrange) integral; \( f(t) \) is fixed by a boundary condition and can be absorbed into \( \varphi \). B

Result

\[ \underbrace{\tfrac{1}{2}v^2 + \frac{p}{\rho} + gz = B}_{\text{steady, along a streamline}} \qquad\quad \underbrace{\frac{\partial\varphi}{\partial t} + \tfrac{1}{2}v^2 + \frac{p}{\rho} + gz = f(t)}_{\text{unsteady, irrotational, everywhere}} \]

Reading. Each term is energy per unit mass. The steady relation says that for an ideal fluid the sum of kinetic energy \( \tfrac12 v^2 \), flow work / pressure energy \( p/\rho \), and gravitational potential \( gz \) is conserved as a fluid parcel travels along its streamline: where it speeds up, pressure or height must fall. The unsteady form adds the local term \( \partial\varphi/\partial t \), the reversible work done to accelerate the whole irrotational field, and upgrades "along a streamline" to "at every point simultaneously".

Units check. In SI, \( v^2 \sim \mathrm{m^2\,s^{-2}} \); \( p/\rho \sim (\mathrm{Pa})/(\mathrm{kg\,m^{-3}}) = (\mathrm{N\,m^{-2}})(\mathrm{m^3\,kg^{-1}}) = \mathrm{J\,kg^{-1}} = \mathrm{m^2\,s^{-2}} \); \( gz \sim \mathrm{m\,s^{-2}\cdot m}=\mathrm{m^2\,s^{-2}} \); \( \partial\varphi/\partial t \sim (\mathrm{m^2\,s^{-1}})/\mathrm{s}=\mathrm{m^2\,s^{-2}} \). All four are specific energies \( \mathrm{J\,kg^{-1}} \). Multiplying through by \( \rho \) gives the pressure form (Pa); dividing by \( g \) gives the head form (m).

Limiting cases

Incompressible: \( \rho \) constant, \( \int dp/\rho \to p/\rho \), recovering the textbook \( \tfrac12 v^2 + p/\rho + gz \).
Hydrostatics: \( \vec v=0 \) gives \( p/\rho + gz=\text{const} \), i.e. \( p=p_0-\rho g z \).
Irrotational and steady: \( f(t)=B \) is the same constant on every streamline, so the sum is uniform throughout the field, not merely along lines.
Compressible adiabatic gas: \( \int dp/\rho = \frac{\gamma}{\gamma-1}\frac{p}{\rho} = \frac{c^2}{\gamma-1} \), giving \( \tfrac12 v^2 + \frac{c^2}{\gamma-1}=\text{const} \) along a streamline (stagnation-enthalpy conservation).
Rotating frame (solid-body \( \Omega \)): absorb the centrifugal potential \( -\tfrac12\Omega^2 r^2 \) into \( \Phi \); the Coriolis force does no work and drops out of the streamline integral.

Breaks when

Viscous / turbulent flow. Boundary layers, separation and turbulence dissipate mechanical energy; \( B \) falls downstream by the head loss \( h_f \). A pipe pressure drop attributed to "Bernoulli" alone is wrong — the sum is not conserved.
Across a shock or hydraulic jump. Entropy jumps discontinuously; the streamline sum (and stagnation pressure) drops even though mass, momentum and total enthalpy are conserved. Bernoulli holds up to, but not across, the discontinuity.
Baroclinic flow. When \( \nabla p\times\nabla\rho\neq0 \) (e.g. an ocean front, a heated room) the pressure term is not an exact gradient, vorticity is continuously generated, and no single Bernoulli constant exists.
Energy addition/extraction. A pump, fan or turbine on the streamline injects or removes work; the bare relation must be augmented with a shaft-work term.

Failure modes

Cross-streamline application. Comparing two points on different streamlines of a rotational flow using the steady form; legal only if the flow is irrotational or both points share a streamline.
Dropping \( \partial\varphi/\partial t \). Using the steady form on an accelerating column (starting siphon, sloshing tank, water hammer); the local-acceleration term is often the dominant one.
The "faster air, lower pressure" fallacy. Invoking Bernoulli where the two air parcels are not on a common streamline and no equal-transit-time constraint holds — Bernoulli does not by itself explain aerofoil lift.
Using \( p/\rho \) for a gas at high Mach number. Ignoring compressibility and writing \( p/\rho \) instead of \( \int dp/\rho \); the incompressible form errs by \( O(M^2) \).
Sign of \( gz \). Measuring \( z \) downward or mixing gauge and absolute pressure inconsistently between the two stations.
Forgetting the jet contraction. In tank efflux, using the orifice area rather than the vena-contracta area, overstating the flow rate.

Discussion

The two relations spring from the same equation but exploit different geometry. The steady form uses that \( \vec v\times\vec\omega \) is perpendicular to \( \vec v \): projecting Euler's equation onto the streamline annihilates the vorticity term regardless of its size, so the result survives in rotational flow but only along a line. The unsteady form instead kills the vorticity term by setting \( \vec\omega=0 \) outright, which lets the local-acceleration term become a gradient too, so the constant is global but the price is irrotationality. Crocco's theorem \( \nabla B=\vec v\times\vec\omega + T\nabla s \) (for a non-barotropic gas) unifies the picture: gradients of Bernoulli head across streamlines are exactly the signature of vorticity and entropy variation.

Physically Bernoulli is the first integral, i.e. the energy law, of ideal-fluid dynamics — the fluid analogue of \( \tfrac12 mv^2 + mgz=\text{const} \) for a frictionless bead, with \( p/\rho \) playing the role of the reversible flow work each parcel does against its neighbours. Because pressure work is internal to the fluid, no external agent is needed for it to shuttle energy between kinetic and pressure forms; a venturi is just this trade executed along a contracting tube.

The unsteady integral is the hydrodynamic shadow of a variational principle. For irrotational incompressible flow the pressure field is \( p/\rho = -\big(\partial_t\varphi + \tfrac12|\nabla\varphi|^2 + gz\big) + f(t) \), which is precisely (minus) the Lagrangian density evaluated on the potential; the momentum equation is its Euler–Lagrange equation. This viewpoint makes the free surface a genuine dynamical boundary — the two conditions of water-wave theory are Bernoulli (dynamic) plus kinematic — and is the natural setting for potential-flow wave dispersion and added-mass forces on accelerating bodies.

Common misconceptions. (i) Bernoulli is not conservation of energy for the whole fluid — it is the specific-energy integral along a streamline for an ideal fluid, and it is silent about heat and viscous losses. (ii) "Higher speed always means lower pressure" is only true along one streamline; between streamlines the pressure is set by the curvature (centripetal) balance, not Bernoulli. (iii) The relation does not require incompressibility — the barotropic form handles gases; incompressibility merely simplifies \( \int dp/\rho \) to \( p/\rho \).

Worked examples

Torricelli efflux from a large tank. Water, open surface at height \( h=2.0\,\mathrm{m} \) above a small side orifice; both surface and jet at atmospheric pressure. Find the jet speed.

\[ \tfrac12 v_1^2 + \frac{p_1}{\rho} + g z_1 = \tfrac12 v_2^2 + \frac{p_2}{\rho} + g z_2 \]

Steady incompressible Bernoulli along a surface-to-orifice streamline. A

\[ p_1=p_2=p_{\text{atm}},\quad v_1\approx0\ (\text{wide tank}),\quad z_1-z_2=h \;\Rightarrow\; \tfrac12 v_2^2 = g h \]

Equal pressures cancel; the surface descends negligibly slowly. A

\[ v_2=\sqrt{2gh}=\sqrt{2(9.81)(2.0)}\ \mathrm{m\,s^{-1}} \]

Solve symbolically, then insert numbers. A

\[ v_2 = 6.3\ \mathrm{m\,s^{-1}} \]

Reading. The efflux speed equals that of a body freely fallen through \( h \). A real orifice delivers \( \approx 0.62\,v_2 \) at the vena contracta because of contraction and slight viscous loss.

Impulsively started siphon (unsteady). A siphon tube of uniform bore, total length \( L=3.0\,\mathrm{m} \), is filled and the lower end (a height \( H=1.2\,\mathrm{m} \) below the reservoir surface) is suddenly opened. Find the initial acceleration of the water column at the instant \( v=0 \).

\[ \frac{\partial\varphi}{\partial t} + \tfrac12 v^2 + \frac{p}{\rho} + gz = f(t) \]

Unsteady Bernoulli along the tube; the column is (essentially) irrotational plug flow. B

\[ \int_1^2 \frac{\partial\vec v}{\partial t}\cdot d\vec\ell = \frac{dv}{dt}\,L,\qquad \Big[\tfrac12 v^2+\tfrac{p}{\rho}+gz\Big]_1^2 = -gH \ \ (v=0,\ p_1=p_2=p_{\text{atm}}) \]

With uniform bore the speed is the same everywhere, so \( \partial_t v \) integrates to \( \dot v\,L \); at \( v=0 \) only gravity and the unsteady term remain. B

\[ \frac{dv}{dt}\,L - gH = 0 \;\Rightarrow\; \frac{dv}{dt}=\frac{gH}{L}=\frac{(9.81)(1.2)}{3.0}\ \mathrm{m\,s^{-2}} \]

Rearrange for the acceleration, then substitute. B

\[ \dot v = 3.9\ \mathrm{m\,s^{-2}} \]

Reading. The column starts from rest with acceleration \( gH/L \) and asymptotes to the steady Torricelli speed \( \sqrt{2gH}=4.9\ \mathrm{m\,s^{-1}} \). The steady form alone could never give the transient — the \( \partial\varphi/\partial t \) term is the whole story at \( t=0 \).

Problems

(A) Pitot–static tube. An aircraft pitot tube reads a stagnation-to-static pressure difference \( \Delta p = 3.5\ \mathrm{kPa} \) in air of density \( \rho=1.2\ \mathrm{kg\,m^{-3}} \). Find the airspeed (incompressible model).
Solution
Along a streamline to the stagnation point, \( \tfrac12\rho v^2 = \Delta p \), so \( v=\sqrt{2\Delta p/\rho}=\sqrt{2(3500)/1.2}=\sqrt{5833}=76\ \mathrm{m\,s^{-1}} \) (about 275 km/h). Compressibility corrections are \( O(M^2)\approx5\% \) here.
(A) Venturi meter. Water (\( \rho=1000\ \mathrm{kg\,m^{-3}} \)) flows through a horizontal venturi with inlet area \( A_1=50\ \mathrm{cm^2} \) and throat \( A_2=20\ \mathrm{cm^2} \). The measured pressure drop is \( p_1-p_2=8.0\ \mathrm{kPa} \). Find the volume flow rate \( Q \).
Solution
Continuity: \( v_1=Q/A_1,\ v_2=Q/A_2 \). Horizontal Bernoulli: \( p_1-p_2=\tfrac12\rho(v_2^2-v_1^2)=\tfrac12\rho Q^2(1/A_2^2-1/A_1^2) \). With \( A_1=5.0\times10^{-3}\,\mathrm{m^2},\ A_2=2.0\times10^{-3}\,\mathrm{m^2} \): \( 1/A_2^2-1/A_1^2=2.5\times10^5-4.0\times10^4=2.1\times10^5\ \mathrm{m^{-4}} \). Then \( Q^2=2(8000)/[1000\cdot2.1\times10^5]=7.62\times10^{-5} \), so \( Q=8.7\times10^{-3}\ \mathrm{m^3\,s^{-1}}\approx8.7\ \mathrm{L\,s^{-1}} \).
(B) Compressible stagnation. Air (\( \gamma=1.4 \)) flows at \( v=200\ \mathrm{m\,s^{-1}} \), static temperature \( T=250\ \mathrm{K} \), \( c_p=1005\ \mathrm{J\,kg^{-1}K^{-1}} \). Using the compressible streamline Bernoulli (stagnation enthalpy), find the stagnation temperature \( T_0 \).
Solution
For an adiabatic gas \( \int dp/\rho=\frac{\gamma}{\gamma-1}p/\rho=c_pT \), so Bernoulli becomes \( c_pT+\tfrac12 v^2=c_pT_0 \). Thus \( T_0=T+v^2/(2c_p)=250+(200)^2/(2\cdot1005)=250+19.9=270\ \mathrm{K} \). Equivalently \( T_0/T=1+\tfrac{\gamma-1}{2}M^2 \) with \( M=v/\sqrt{\gamma R T}\approx0.63 \).
(B) Draining tank, quasi-steady. A cylindrical tank of cross-section \( A=0.30\ \mathrm{m^2} \) drains through a bottom orifice of area \( a=1.0\ \mathrm{cm^2} \). When the water depth is \( h=1.5\ \mathrm{m} \), find (i) the efflux speed and (ii) the rate of fall of the surface. State why the steady form is admissible.
Solution
(i) Torricelli: \( v=\sqrt{2gh}=\sqrt{2(9.81)(1.5)}=5.4\ \mathrm{m\,s^{-1}} \). (ii) Continuity \( A\,|\dot h|=a v \Rightarrow |\dot h|=(a/A)v=(1.0\times10^{-4}/0.30)(5.4)=1.8\times10^{-3}\ \mathrm{m\,s^{-1}} \). The quasi-steady (steady-form) approximation is valid because \( a/A\approx3\times10^{-4}\ll1 \): the unsteady term \( \sim L\dot v \) is smaller than \( \tfrac12 v^2 \) by \( O(a/A) \).
(C) Oscillating U-tube (unsteady Bernoulli). A liquid column of total length \( L=0.80\ \mathrm{m} \) in a uniform U-tube is displaced and released. Using the unsteady Bernoulli integral applied along the column, derive the equation of motion and find the oscillation period.
Solution
Let \( \xi \) be the displacement of the liquid surfaces from equilibrium; the two free surfaces are at atmospheric pressure and differ in height by \( 2\xi \). The unsteady integral along the column of uniform bore gives \( \int\partial_t\vec v\cdot d\vec\ell + [\tfrac12 v^2 + p/\rho + gz]_1^2 = 0 \). The convective end terms cancel (equal speeds, equal pressures); the local term is \( \ddot\xi\,L \) and the gravity term is \( g(2\xi) \). Hence \( L\ddot\xi + 2g\xi = 0 \), simple harmonic with \( \omega=\sqrt{2g/L} \). Period \( T=2\pi\sqrt{L/2g}=2\pi\sqrt{0.80/(2\cdot9.81)}=2\pi(0.202)=1.27\ \mathrm{s} \). The restoring "stiffness" is gravity; the inertia is the whole column — a result the steady form cannot produce.