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A velocity potential satisfying uniform flow at infinity and no radial flow through $r=a$ isThereforeOn the cylinder, $u_r=0$ and $u_{\theta }=-2U\sin \theta$. The steady Bernoulli equation, compared with the uniform flow at infinity, givesHencethe pressure in potential flow around a circular cylinder.

The steady Bernoulli equation isAt the regular sonic point, part a givesDefineThe isentropic flow relation $c_s^2\propto {\rho }^{\gamma -1}$ and mass conservation between the stellar surface and the sonic point giveEquating the surface and sonic values of the Bernoulli function therefore yields the required relationAn outflow reaching infinity with positive terminal kinetic energy requires $E>0$, henceAt $\gamma =5/3$ the terminal state is marginal with $E=0$; larger $\gamma$ cannot support the stipulated wind to infinity.

Let $u=-v_r>0$ denote the inward speed. Steady spherical mass conservation and the radial momentum equation for an isothermal gas giveEliminating $d\log \rho /dr$ produces the Isothermal Bondi equationA smooth flow can cross Mach number one only where both sides vanish. Its Bondi sonic point is thereforeThe integrated Bernoulli equation which approaches rest and density ${\rho }_{\infty }$ at infinity isAt the sonic point this gives ${\rho }_s=e^{3/2}{\rho }_{\infty }$, and hence the unique regular transonic branch has

The steady Euler equations for an inviscid fluid areTaking the scalar product with $u$ and usinggivesThusThe Bernoulli function $H$ is constant along each streamline: fluid particles in steady flow exchange pressure, kinetic, and potential energy without changing their total mechanical energy density.

Let $y(t)$ be the downward displacement of the free surface from its initial level, and let $U$ be the speed in the tube. Conservation of volume givesThe free surface and the outlet are both at atmospheric pressure, and their vertical separation is $H+h_0-y$. Applying the Bernoulli equation between them, while retaining the small free-surface speed, givesThereforeThe surface reaches the upper tube end when $y=h_0$. Hence the draining time of a uniform tank through a siphon is

Define the indoor-to-outdoor reduced gravityBecause the whole opening lies above the interface, its indoor side contains upper-layer fluid of density ${\rho }_H$. Under hydrostatic pressure, the pressure difference varies linearly about a neutral pressure level. Equal opening geometry and equal discharge coefficients for inflow and outflow put that level at the vent midpoint. At vertical distance $y$ from it, the ideal Bernoulli equation gives speed $\sqrt{2g_V^{\prime }|y|}$.

Writing $Q_V$ for either one-way volumetric flow rate, integration over one half of the opening givesThe total unsigned exchange is $2Q_V$. The ideal sharp-edged inviscid model has $C_d=1$; an empirical $C_d<1$ represents contraction and losses.

The free surface is material, so applying the material derivative to $y-\eta (x,t)$ gives the exact kinematic boundary conditionFor an incompressible fluid with constant density, the Euler equations areSince the flow is irrotational, $\mathbf{u}=\nabla \varphi$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\nabla \varphi {|}^2/2)$. Integrating the resulting gradient equation gives a function of time on the right of Bernoulli equation. Adding a time-dependent constant to $\varphi$ removes that function, without changing velocity, and choosing the atmospheric-pressure reference givesThe pressure immediately above the surface is atmospheric; the liquid-side pressure includes the stated surface tension jump.

At first order in wave amplitude, the products ${\varphi }_x{\eta }_x$ and $|\nabla \varphi {|}^2$, the slope-squared correction to curvature, and displacement corrections in evaluating $\varphi$ at $y=\eta$ are all higher order. Hence, on $y=0$,For $\eta =A\cos (kx-\omega t)$, a harmonic velocity potential decaying as $y\to -\infty$ and satisfying the first condition isIndeed $\Delta \varphi =0$ and ${\varphi }_y{|}_0=A\omega \sin (kx-\omega t)={\eta }_t$. Substituting in the second condition yields the dispersion relation for a gravity-capillary wave:Taking the positive-frequency branch, the phase velocity and group velocity areSubtracting givesSince $\omega >0$,The threshold is also the wavenumber at the minimum phase speed of a gravity-capillary wave.

Orient $\partial B$ counterclockwise, and define positive circulation by that orientation. The outward normal from the body satisfies $nds=(dy,-dx)$. Pressure acts inward, soThe steady Bernoulli equation for this constant-density irrotational flow is $p+\rho |v{|}^2/2=C$. The constant has zero integral around a closed curve. With complex potential of a two-dimensional potential flow $w=\varphi +i\psi$, let $W=w^{\prime }=u-iv$. The no-penetration condition makes the boundary a streamline, so $dw=Wdz$ is real along it. Thus $Wdz=\overline{W}d\overline{z}$, and $|W{|}^{2}d\overline{z}=W^{2}dz$. Substitution proves Blasius force theorem:At infinity the Laurent expansion has the formThere is no source term because the impermeable body has zero net flux. The circulation term follows from $\oint Wdz=\Gamma$. The coefficient of $z^{-1}$ in $W^2$ is $\Gamma (U-iV)/(\pi i)$. Analyticity in the fluid lets the body contour be enlarged to a large counterclockwise circle, so the residue theorem gives $\oint W^{2}dz=2\Gamma (U-iV)$. HenceIts component along $(U,V)$ is zero: the drag is zero. The lift magnitude is $\boxed{\rho |\Gamma |\sqrt{U^2+V^2}}$. With the stated positive counterclockwise circulation, the signed lift along the left normal $(-V,U)/\sqrt{U^2+V^2}$ is $-\rho \Gamma \sqrt{U^2+V^2}$. Reversing the circulation convention reverses this signed formula. This is the Kutta–Joukowski theorem with its orientation made explicit.

The factor ${\eta }_{xx}/(1+{\eta }_x^2{)}^{3/2}$ is the signed curvature of the graph, so the pressure jump is surface tension times curvature, the planar Young–Laplace equation.

Take total velocity potentials ${\Phi }_{-}={\varphi }_{-}$ and ${\Phi }_{+}=Ux+{\varphi }_{+}$. Each disturbance potential satisfies the Laplace equation. The interface moves with the fluid: for $F=y-\eta (x,t)$, the material equation $F_t+v\cdot \nabla F=0$ gives ${\eta }_t+u{\eta }_x=v_y$. Keeping only first-order terms and evaluating at $y=0$ yieldsThe unsteady Bernoulli equation is ${\Phi }_t+|\nabla \Phi {|}^2/2+p/\rho =C(t)$, with any common gravitational contribution canceling across this equal-density interface. Subtracting the basic flow and linearizing gives $p_{-}^{\prime }=-\rho {\varphi }_{-,t}$ and $p_{+}^{\prime }=-\rho ({\varphi }_{+,t}+U{\varphi }_{+,x})$. The linearized pressure jump is $T{\eta }_{xx}$. HenceUse complex modes with real parts understood:The decaying vertical factors solve the Laplace equation in their respective half-planes. The kinematic conditions give $B_{-}=-i\omega A/k$ and $B_{+}=i(\omega -kU)A/k$. Substitution in the dynamic condition givesWhen the discriminant is nonnegative, the phase speeds are $U/2\pm \sqrt{Tk/(2\rho )-U^2/4}$. They are two dispersive capillary-wave branches advected by the mean velocity $U/2$.

When $\boxed{U^2>2Tk/\rho }$, the frequencies are complex conjugates and one mode grows exponentially, with growth rate $k\sqrt{U^2/4-Tk/(2\rho )}$. This is Kelvin-Helmholtz instability. Surface tension stabilizes sufficiently short waves, while for any nonzero shear, sufficiently long waves are unstable in this gravity-free equal-density model. A real cosine with real frequency describes the stable branches; the unstable case is represented by the real part of the corresponding growing complex mode.

For the intended ideal Venturi effect, take steady flow, incompressible flow, negligible viscous losses, and the same height at the two stations shown. The Bernoulli equation along a streamline givesUsing water mass density $\rho \simeq 1000\mathrm{kg}{\mathrm{m}}^{-3}$,Thus the outlet pressure is lower by $16\mathrm{kPa}$. The increased kinetic energy per unit volume comes from the drop in pressure.

The source's assumption of laminar flow alone does not imply negligible viscosity or losses. The numerical result uses the usual lossless Bernoulli equation approximation, with the quoted speeds interpreted as representative streamline speeds. A real tube with significant viscous losses requires additional data; in a uniform-velocity energy balance its pressure drop would include an extra positive loss term.

The Magnus effect is the transverse force produced when a spinning body moves relative to a fluid. Rotation changes the flow and circulation around the body; the resulting asymmetric pressure and wake deflect the fluid, with the opposite transverse force on the body. Examples include the curved flight of spinning balls and the lift of rotating-cylinder sails.

A useful quantitative model is steady two-dimensional incompressible flow past a cylinder of radius $a$, with fluid velocity $\mathbf{U}=U\hat{\mathbf{x}}$ far away. Let circulation $\Gamma$ be positive anticlockwise as viewed from $+z$. The potential-flow velocity at the cylinder surface isThe Bernoulli equation outside the boundary layer givesIntegrating the pressure force around the cylinder, per unit span, yieldsThus the Kutta–Joukowski theorem givesThe sign here refers to the fluid velocity relative to the cylinder. For a body moving through otherwise stationary fluid, that relative fluid velocity points opposite to the body's translational velocity.

Real viscosity and the no-slip boundary condition transmit the surface rotation to the fluid and help set the circulation and separation pattern. Pure inviscid flow with only an impermeability condition does not determine $\Gamma$ from the cylinder's angular speed. Therefore the circulation must not be assigned a universal spin-rate formula without additional assumptions. The cylinder result illustrates the lift mechanism; quantitative forces on a finite spinning ball also depend on its geometry, Reynolds number, rotation rate, surface roughness and wake.

Writing $z=r{e}^{i\theta }$ gives $\varphi =\Gamma \theta /(2\pi )$ and $\psi =-\Gamma \log r/(2\pi )$. Constant streamfunction means constant radius, so streamlines are circles. The tangential velocity is $v_{\theta }=\Gamma /(2\pi r)$, and its anticlockwise circulation is ${\int }_0^{2\pi }{v}_{\theta }rd\theta =\Gamma$.

Use an opposite vortex at $z=-1$ as the image of the vortex at one:On $z=iy$ the two distances are equal, so $\psi =0$ and the wall is a streamline. Its velocity follows from $w^{\prime }=-i\Gamma /[\pi (z^2-1)]$: at the wall $u=0$ and $v=-\Gamma /[\pi (1+y^2)]$.

The Bernoulli equation gives $p-p_{\infty }=-\rho {\Gamma }^2/[2{\pi }^2(1+y^2{)}^2]$. Relative to the ambient pressure force, the fluid's excess force on the wall is $-\int (p-p_{\infty })dy$ in the positive $x$ direction. Using the supplied integral,Thus the wall is attracted toward the fluid and vortex. Subtracting the uniform ambient pressure isolates the finite force due to the vortex.

For potential flow, each velocity potential satisfies Laplace's equation in its fluid. The exact kinematic boundary condition on $y=\eta$ is ${\eta }_t+{\varphi }_{j,x}{\eta }_x={\varphi }_{j,y}$. Keeping only first-order terms givesThe linearized Bernoulli equation is $p_j=p_0-{\rho }_j({\varphi }_{j,t}+g\eta )$ at the interface, so pressure continuity givesFor a decaying disturbance away from the interface, put $\psi =kx-\omega t$ and takeThey satisfy Laplace's equation and the kinematic boundary conditions for $\eta =A\cos \psi$. Substituting in the dynamic condition gives the dispersion relationIf the heavier fluid is on top, ${\omega }^2<0$. Instead of oscillation, a Fourier disturbance has an exponentially growing component with growth rate $\sqrt{gk({\rho }_2-{\rho }_1)/({\rho }_1+{\rho }_2)}$. This is the Rayleigh-Taylor instability in the absence of surface tension.

The velocity potential gives horizontal and vertical velocity components $({\varphi }_x,{\varphi }_z)$. The kinematic boundary condition at each fixed impermeable wall is zero normal fluid velocity, henceLet $\eta (x,t)$ be the surface elevation. To first order in its amplitude, the kinematic boundary condition for a free-surface graph gives ${\eta }_t={\varphi }_z$ at $z=0$. The dynamic boundary condition for an inviscid interface fixes the surface pressure to atmospheric pressure. The unsteady Bernoulli equation, after choosing the time-dependent gauge of $\varphi$, then gives ${\varphi }_t+g\eta =0$ at $z=0$: the quadratic velocity term is second order in amplitude. Differentiate this last equation in time and eliminate ${\eta }_t$ to obtain the linearized free-surface boundary conditions in combined form,

Uniform irrotational flow of speed $U$ past a circular cylinder of radius $a$, with zero circulation, has velocity potentialOn the cylinder, the tangential speed is $-2U\sin \theta$, so the Bernoulli equation gives

The pressure coefficient compares a local static pressure with a specified reference static pressure, using a fixed reference dynamic pressure:The reference mass density, speed, pressure and location must accompany a reported pressure coefficient. Keep the same $q_{\infty }$ throughout a surface distribution; replacing it by local $\rho U^2/2$ gives a different quantity and fails at a stagnation point. The definition itself requires neither Bernoulli's principle nor inviscid flow, so measured pressure coefficients remain useful when those idealizations fail.

Positive $C_p$ means $p>p_{\infty }$; zero means equality; negative $C_p$ means $p<p_{\infty }$. This sign does not specify the sign of absolute pressure. For stipulated $p_{\infty }=101325\mathrm{Pa}$ absolute and $q_{\infty }=240\mathrm{Pa}$, $C_p=-3$ gives $p=100605\mathrm{Pa}$ absolute: a $720\mathrm{Pa}$ deficit, not a negative absolute pressure. If $p_{\infty }$ equals atmospheric pressure, the numerator is gauge pressure; a wind tunnel's reference static pressure need not equal room atmospheric pressure. These are reference choices, not different physical pressures.

Under the same-height, constant-mass density $\rho ={\rho }_{\infty }$, steady flow and inviscid flow hypotheses of the Bernoulli equation, and with the same Bernoulli invariant as the reference flow,The first equation immediately gives the second after subtracting $p_{\infty }$ and dividing by $q_{\infty }$. Hence a lossless stagnation point has $C_p=1$, $U=U_{\infty }$ gives $C_p=0$, and $U=2U_{\infty }$ gives $C_p=-3$. For a real stationary wall satisfying the no-slip boundary condition, the velocity is zero right at the wall; inserting that wall velocity in an inviscid flow formula would wrongly predict $C_p=1$ everywhere. A thin boundary layer approximation instead relates the wall pressure to the outer-flow pressure and uses the outer velocity where the Bernoulli equation applies.

For potential flow around a circular cylinder of radius $a$ with zero circulation, the surface tangential velocity is $u_{\theta }=-2U_{\infty }\sin \theta$. Thus the ideal pressure coefficient isWith $\theta$ measured from downstream, the front is at $\pi$ and the rear at zero. Both ideal stagnation points have $C_p=1$, while the sides at $\pi /2$ and $3\pi /2$ have $C_p=-3$. This exact model is a useful baseline for an experiment; its complete rearward pressure recovery is not a prediction for a separated real fluid wake.

To obtain a force, integrate the local normal contributions with their directions. Let $\mathbf{n}$ be the unit normal vector from the body into the fluid and $\Sigma$ its closed wetted surface:The second equality removes a spatially constant reference pressure over the closed surface; it does not remove a nonuniform hydrostatic pressure or justify dropping the reference term on an isolated open patch. A negative $C_p$ contributes outwards in this reference-subtracted calculation, even though a positive absolute pressure still pushes inwards locally. Project ${\mathbf{F}}_p$ onto the chosen lift and drag directions. The full fluid force also includes ${\int }_{\Sigma }{\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}dA$ from the viscous stress tensor, including tangential shear stress; pressure tapping does not measure that contribution.

For the two-dimensional cylinder, force per unit span follows from $dA=ad\theta$ per unit span and $\mathbf{n}=(\cos \theta ,\sin \theta )$:Substituting $C_p=1-4{\sin }^2\theta$ gives $(0,0)$: opposite points have identical pressure coefficients and opposite normal vectors. Zero drag despite a strongly nonuniform pressure is the instructive limitation associated with D'Alembert's paradox. Real viscosity, boundary layers and flow separation can prevent rearward pressure recovery and produce pressure drag. Thus the measured curve can test the model and estimate a pressure contribution without proving that total drag vanishes.

A motivating measurement is Virginia Tech's Experiment 3 - Flow Past a Circular Cylinder. Its open-jet wind tunnel holds a rotatable $140\mathrm{mm}$ cylinder with end plates and 36 one-millimetre pressure tappings at ten-degree intervals. A reference Pitot-static tube supplies $p_{0\infty }$ and $p_{\infty }$ to one digital manometer; a second connects a surface tap to its positive port and reference static pressure to its negative port. Record the simultaneous readings:The last equality uses the low-speed stopping relation, not an additional speed measurement. The manual requires sealed tubing and zero checks with the tunnel off. It also provides a water-column manometer: read each column's initial level, and subtract the independently measured reference-to-atmosphere pressure from the tap-to-atmosphere readings. Record tap positions after rotation. Compare the two instruments and check spanwise uniformity; end plates do not guarantee two-dimensional flow. ^[ref]

The inferential distinction matters: the manometers directly measure pressure differences, the ratio estimates pressure coefficients, and a surface integral estimates the pressure part of the force. Measure room absolute pressure and temperature if converting the reference difference to speed through the ideal gas law $\rho =p_{\mathrm{abs}}/(RT)$; use thermodynamic temperature in that equation. For force reconstruction, cover the complete contour and use numerical integration with the tap geometry. Finite tap spacing, unknown end effects, probe disturbance and fluctuations limit that reconstruction; tubes and instruments also average or delay changing pressures. Retain the signed raw readings, repeated readings and reference readings so that these limits can be assessed. A small reference difference magnifies any zero error in the ratio, rather than producing a reliably large $C_p$.

Mechanical work transferred by a pressure acting through a displaced volume. In steady incompressible flow, the Bernoulli equation contains $p/\rho$ as flow work per unit mass; it is not an independently conserved internal energy of an isolated parcel.

For a steady rectangular-channel flow with discharge $Q=bhu$, the shallow-water specific energy relative to the bed is $E=h+Q^2/(2g{b}^2{h}^2)$. At fixed discharge and width, $E_h=1-F^2$, where the Froude number satisfies $F^2=Q^2/(g{b}^2{h}^3)$. The minimum is $E_c=3h_c/2$, with $h_c=(Q^2/(g{b}^2){)}^{1/3}$. Without dissipation, $E+H$ is the conserved Bernoulli equation head for bed elevation $H$.

For steady spherical isentropic flow with $p=K{\rho }^{3/2}$ around a point mass, put $j=r^2\rho v>0$ and $\mathcal{E}=v^2/2+2c^2-GM/r>0$. Eliminating the mass density gives $c^2=[j(3K/2{)}^2{]}^{2/5}{r}^{-4/5}{\mathcal{M}}^{-2/5}$. With $r_s=GM/(4\mathcal{E})$, $x=(r/r_s{)}^{1/5}$, $y={\mathcal{M}}^{2/5}$ and $\lambda =[j(3K/2{)}^2{]}^{2/5}/(2\mathcal{E}{r}_s^{4/5})$, the Bernoulli equation reduces to $\lambda F(y)=F(x)$, where $F(t)=t^4+4/t$. Since $F$ has its unique minimum at $t=1$, a regular sonic point requires $\lambda =1$ and $r=r_s$. One transonic branch is $y=x$, which has $v=\sqrt{2\mathcal{E}}$, $\rho \propto r^{-2}$ and $p\propto r^{-3}$. The decreasing branch has inner free fall and an outer nearly static reservoir, giving Bondi accretion for inward fluid flow. A nonradiative stationary shock wave keeps $j,\mathcal{E},r_s$ fixed but raises $K$, so $\lambda$ increases by the factor $(K_2/K_1{)}^{4/5}$. The inward transonic solution is the spherical $n=2$, $\gamma =3/2$ specialization of transonic accretion in a power-law tube.

The pressure obtained by bringing a flow to rest without losses; in compressible flow the usual definition specifies an isentropic process for stopping. For the constant-mass density, same-height Bernoulli equation discussed under dynamic pressure, $p_0=p+q$.

An ideal fluid issuing from a small hole a vertical depth $h$ below a free surface has speed

Here $h$ is the vertical distance from the free surface to the centre of the orifice, and $g$ is the positive magnitude of gravitational acceleration. The ideal prediction assumes incompressible flow, the same ambient pressure at the free surface and emerging jet, negligible velocity of the reservoir's free surface, and negligible viscous dissipation and surface tension effects. For a falling free surface, a quasi-steady approximation additionally assumes that the local outflow adjusts much faster than $h$ changes. It applies the Bernoulli equation at each instant, rather than claiming that the whole drainage is steady flow.^[ref]

A short proof uses a streamline connecting the reservoir's free surface at elevation $z_s$ to the emerging jet at elevation $z_o$. With constant mass density $\rho$, negligible surface speed and no mechanical energy losses, the Bernoulli equation givesEqual pressures cancel, leaving $v^2/2=gh$: a unit mass gains the same kinetic energy as a body in free fall through height $h$. This makes Torricelli's law a useful first estimate before the detailed velocity field is known. At one quarter of the original head, ideal jet speed is one half as large; it is not proportional to head.

Far from the accelerating outlet region, the reservoir has approximately hydrostatic pressure $p_{\mathrm{ambient}}+\rho gh$ at the outlet's elevation. The emerging jet instead has approximately ambient pressure when surface tension is negligible. Do not assign that bulk hydrostatic pressure to the emerging jet as well as adding the height difference: that would count the available mechanical energy twice. A sealed reservoir, a submerged outlet or significant surface tension changes the pressure conditions and requires a different balance.

Actual volumetric flow rate is commonly represented by a discharge coefficient:where $a$ is the geometric area of the orifice. This definition compares volumetric flow rates, not just speeds. If a nearly uniform jet at its vena contracta has area $a_j$ and speed $v_j$, define the two separate ratios $C_c=a_j/a$ and $C_v=v_j/\sqrt{2gh}$. Then $Q=a_j{v}_j$ gives $C_d=C_c{C}_v$. A contracted jet can consequently have a speed close to $\sqrt{2gh}$ while $C_d$ is appreciably below one. No particular $C_d$ is universal: calibrate it for the opening and operating range.