Potential flow

Start with flow around a post

Imagine a small circular post of radius 25 mm in a horizontal flow. Far from the post, the fluid moves to the right at 0.400 m/s. A proposed ideal model bends the flow around the post without allowing fluid through it. Can one scalar function describe this turning velocity field? If it also predicts a pressure pattern, does that make it a complete prediction of the force on a real post?

We will answer these questions in stages: first differentiate a simple scalar, then compare flow across and around circles, then follow an ideal flow around the post. The pressure calculation comes only after its extra physical assumptions have been introduced. The diagrams and numerical tables are the first-reading path; the final boundary-value discussion is an optional return path.

Geometry, trigonometry, and coordinate systems and Derivatives, Taylor series, and local approximation are useful preparation. Essential gradient, curl and circulation meanings are defined below. For the final derivations, return to Gradient, divergence, curl, and vector calculus and Integrals and multivariable calculus; Partial differential equations and Green functions is optional further preparation.

Use a fixed inertial frame with x right, y up on the page and z out of the page along the post. This is a horizontal cross-section viewed from above: page-up is not vertical height. Write velocity as v = (u, v_y). The distance from the centre is r = √(x² + y²), and θ increases counterclockwise from +x. Calculus uses radians; the sample tables display degrees.

Differentiate one scalar

A velocity potential φ is a scalar function whose spatial gradient gives velocity. With our sign convention, v = ∇φ, so u = ∂φ/∂x and v_y = ∂φ/∂y. A gradient records how rapidly a scalar changes in each spatial direction. Other sign conventions exist; all formulas here use the positive gradient.

Start with φ = Ux and U = 0.400 m/s. Increasing x by 0.025 m increases φ by 0.010 m²/s. Their ratio is 0.400 m/s, and φ has no y dependence. Thus u = U and v_y = 0: a uniform rightward flow.

The potential has units m²/s because one spatial derivative divides by metres to produce m/s. It is a velocity potential, not gravitational potential energy. Its value does not give speed directly. Adding the same constant everywhere changes no spatial derivative and therefore changes no velocity.

This scalar description is a kinematic statement about the velocity field. It does not by itself establish density, pressure, viscosity, boundary conditions or the force on a body. Those require additional equations and physical assumptions.

Separate curl from volume balance

Irrotational means that the local curl of velocity is zero. For a planar field, ω_z = ∂v_y/∂x − ∂u/∂y. This checks local rotation through neighbouring velocities; curved streamlines do not by themselves imply nonzero curl.

If φ has continuous second spatial derivatives, differentiating v = ∇φ gives ω_z = ∂²φ/(∂x∂y) − ∂²φ/(∂y∂x) = 0. A smooth velocity potential therefore gives an irrotational field in its admitted domain. The converse needs a local or global domain condition, which we examine shortly.

Incompressibility is a different statement: a moving fluid element preserves its volume. Its velocity field satisfies ∇·v = ∂u/∂x + ∂v_y/∂y = 0. For a velocity potential, this becomes ∇²φ = ∂²φ/∂x² + ∂²φ/∂y² = 0, called Laplace’s equation. A function satisfying this equation is harmonic.

A potential can be irrotational without being incompressible. For example, φ = Ax² has zero curl but divergence 2A. To use Laplace’s equation, we must impose incompressibility as well. In a steady compressible flow, mass conservation instead requires ∇·(ρ∇φ) = 0, where density ρ may vary; it does not automatically reduce to Laplace.

Nor does the existence of a potential prove that viscosity vanishes. Potential flow identifies a velocity structure. Our later pressure model explicitly makes an inviscid approximation, meaning that viscous stresses are omitted from its momentum balance.

Compare radial flow and swirl

Consider two separately prescribed fields in the same annulus, 0.010 ≤ r ≤ 0.100 m. The inner disk and the region outside this annulus are unassigned. A radial elementary flow uses φ = K ln(r/r_ref), with K = 0.001 m²/s and r_ref = 0.020 m. The dimensionless ratio inside the logarithm fixes a harmless reference value.

In polar coordinates, v_r = ∂φ/∂r and v_θ = (1/r)∂φ/∂θ. The radial potential therefore gives v_r = K/r and v_θ = 0. Its speed decreases as the same outward flow is spread around a larger circumference. Positive K is called a line source; negative K is a line sink. These names describe the elementary exterior field, without specifying what supplies or removes fluid behind the excluded boundary.

The second field has a local potential φ_local = Cθ, with C = 0.001 m²/s. It gives v_r = 0 and v_θ = C/r, a counterclockwise annular swirl. Here a local potential means an expression on an angular patch where θ is chosen continuously. Its ability to extend around the whole annulus is a separate question.

The outward normal flux around a circle is Q′ = ∮ v·n ds, with n = +e_r. In this two-dimensional cross-section, Q′ is volume flow per unit axial length and has units m²/s. The radial field gives Q′ = (K/r)(2πr) = 2πK. Its counterclockwise circulation Γ = ∮ v·dr is zero because radial velocity is perpendicular to the circle’s tangent.

The swirl reverses these roles: Q′ = 0, but Γ = (C/r)(2πr) = 2πC. Although these two integrals have the same units, one measures normal crossing and the other tangential motion. In the table the two chosen coefficients give equal speed magnitudes; their directions and integral meanings remain different.

No incompressible matter is created within the annular fluid. On its complete boundary the outer normal is +e_r, while the inner normal points toward the hole, −e_r. The radial field’s outward outer-boundary flux is balanced by an equal negative outward inner-boundary flux. A circle alone is not the complete boundary of this annular domain.

Check a potential around a hole

For a single-valued smooth potential, the line integral from one point to another is the change in φ. Returning to the same point gives ∮∇φ·dr = 0. Thus every closed-loop circulation must vanish if one global single-valued velocity potential exists.

The radial potential K ln(r/r_ref) is single-valued throughout the annulus, and its circulation vanishes. A hole does not forbid a global potential; it means that local zero curl alone is insufficient to guarantee one.

The swirl has zero local curl everywhere in the assigned annulus, yet one full counterclockwise turn accumulates Δφ_local = C(2π) = 0.002π m²/s ≈ 0.006283 m²/s. The same physical point would acquire two different potential values. Therefore φ_local = Cθ cannot be one single-valued smooth potential on the entire annulus.

A branch cut lets us choose one continuous angular coordinate on a smaller domain. It does not create a physical wall or change the velocity field. Similarly, a full disk crossing the missing centre is not an admitted smooth surface on which to apply Stokes’ theorem to this annular field. No singular vorticity or physical core is assigned inside the hole.

For a sufficiently smooth curl-free field on a simply connected domain, where every closed loop can be continuously shrunk to a point while staying inside, a global potential exists. On a domain with holes, also check circulation around independent holes. This is why stating the domain is part of specifying a potential-flow model.

Fit an impermeable circular wall

Return to the circular post. Its solid interior r < a is excluded, with a = 0.025 m. A stationary impermeable wall requires v·n = 0: fluid cannot cross it. In potential notation this is the normal-derivative condition ∂φ/∂n = 0. It does not require the tangential velocity to vanish; the ideal wall permits slip.

A uniform potential alone would send flow through the circle. Add a harmonic doublet term, a source–sink limiting field whose singular point lies inside the excluded solid. Choose the exterior potential φ = U(r + a²/r)cosθ, with U = 0.400 m/s. Its first term is Ux; its second term is Ua²cosθ/r. The sum is harmonic in the fluid exterior, although neither the singular centre nor the solid interior is part of that fluid.

Differentiate before judging the boundary: v_r = U(1 − a²/r²)cosθ and v_θ = −U(1 + a²/r²)sinθ. At r = a, the radial factor is zero, so no fluid passes through the wall. Far away, the a²/r² correction vanishes and the Cartesian velocity approaches (U, 0).

At the wall, v_θ = −2U sinθ. The surface speed is zero at θ = 0° and 180°: these are stagnation points. At 90° and 270° its magnitude reaches 2U = 0.800 m/s. The opposite signs of v_θ at the upper and lower points are compatible with rightward Cartesian motion at both points, because the local e_θ direction also changes.

We additionally prescribe zero circulation. In an exterior domain with a hole, the normal wall condition and far-field velocity alone would still permit a decaying circulation field to be added. The globally single-valued potential used here selects the noncirculating case; a different circulation condition changes the model.

Follow exterior streamlines

A streamline is a curve tangent to the instantaneous velocity field. For a steady field it also follows the spatial path of a moving parcel while that parcel has nonzero velocity. It is not a contour of constant speed, and a streamline plot alone gives no transit time.

For this two-dimensional incompressible model, introduce a streamfunction ψ through u = ∂ψ/∂y and v_y = −∂ψ/∂x. It has units m²/s. The cylinder field has ψ = U(r − a²/r)sinθ. Curves of constant ψ are streamlines because v·∇ψ = 0.

The wall has ψ = 0. The exterior x-axis branches also have ψ = 0, but the same value does not make them one smooth moving-parcel path through the solid. They meet wall branches at stagnation points, where direction and the passage time cannot be read from an ordinary nonzero tangent. The diagram keeps the excluded circle intact.

Contours of φ are equipotentials. Where the velocity is nonzero, it points normal to these contours and tangent to streamlines; the two contour families are orthogonal. At a stagnation point the gradient is zero, so no unique flow direction follows from that rule.

The plotted cylinder streamlines use x/a and y/a on equal scales. Changing a changes the physical size, while the same dimensionless streamline pattern remains for this family. Changing U rescales speed and ψ, while levels of ψ/(Ua) keep the displayed streamline shapes unchanged.

Add a conditioned pressure model

To infer pressure, add a steady inviscid, incompressible model with uniform density ρ = 1000 kg/m³. The body-force potential is constant across our horizontal cross-section. The far-field absolute pressure is p_∞ = 110000 Pa. There is no free-surface, compressibility or cavitation model. These assumptions supplement the prescribed velocity; they do not follow from the existence of φ alone.

For this irrotational ideal flow, Bernoulli’s constant is shared throughout the connected fluid exterior: p + ρ|v|²/2 = p_∞ + ρU²/2. Therefore p − p_∞ = (ρ/2)(U² − |v|²). Spatially changing velocity can be accompanied by a pressure gradient even when local curl is zero.

The far-field dynamic pressure q_∞ = ρU²/2 is 80 Pa. Define the dimensionless pressure coefficient C_p = (p − p_∞)/q_∞. On the wall, |v| = 2U|sinθ|, giving C_p = 1 − 4 sin²θ. It ranges from +1 at the stagnation points to −3 at the two maximum-speed points.

At a stagnation point the absolute pressure is 110080 Pa. At θ = 90° or 270° it is 109760 Pa. A negative C_p means pressure below the far-field reference, not negative absolute pressure. The difference between these two absolute pressures is 320 Pa.

This equation does not mean that pressure contributions from different elementary flows can be added separately. Potentials and velocities add linearly; pressure involves the square of the total speed and includes cross terms. Always apply the pressure closure to the combined field.

Interpret force and model limits

Pressure acts inward on the solid surface. With e_r pointing from the solid into the fluid, pressure traction on the post is −p e_r. A small arc a dθ therefore contributes −p a(cosθ, sinθ)dθ to force per unit axial length. Integrate around the complete circle; one high-pressure point cannot determine the resultant.

The surface pressure has fore–aft and upper–lower symmetry. Opposite surface points have equal pressure and opposite radial normals, so their pressure forces cancel. Equivalently, integrating the pressure expression gives F_x′ = F_y′ = 0 N/m. The uniform far-field pressure also cancels around a closed circle. This is a force per axial length, not a force on an unspecified finite-length post.

The zero resultant belongs to this steady, unbounded, inviscid, noncirculating ideal cylinder case. It is a form of d’Alembert’s paradox: a useful ideal velocity and pressure field can omit the mechanisms that produce real resistance. It is not a zero-force law for every potential flow. Circulation can change a transverse force, and unsteady flow can give forces even without the steady pattern used here.

A real viscous fluid at a stationary solid wall generally satisfies no slip, so its wall velocity is zero rather than the ideal tangential value. Boundary layers can contain strong velocity gradients and vorticity. Separation and the resulting wake can alter surface pressure; viscous shear also contributes to force. Our ideal pressure integral includes none of these effects and does not predict a real drag coefficient.

Potential models remain useful for describing an outer region where an irrotational approximation is appropriate, constructing boundary-value solutions, and interpreting how geometry redistributes velocity and ideal pressure. Whether a particular real flow has such a region needs physical evidence. Thin boundary layers alone do not guarantee that separation and wake effects are negligible.

Nearby walls, a free surface, appreciable density variation, imposed circulation and time dependence require changed domains, boundary conditions or equations. The later Boundary layers, lift, and drag subject develops wall and wake effects; it is further coverage, not a prerequisite assumed in this calculation.

Return to the boundary-value problem

Optional return path: in an incompressible potential model, solve ∇²φ = 0 throughout the fluid domain and specify the appropriate boundary data. A prescribed potential is a Dirichlet condition; a prescribed normal velocity ∂φ/∂n is a Neumann condition. At a moving impermeable wall, the fluid’s normal velocity equals the wall’s normal velocity; tangential equality does not follow.

For a bounded source-free incompressible domain, the total prescribed outward normal flux over its complete boundary must be zero. This follows by integrating ∇·v = 0 over the fluid volume. With compatible Neumann data on a connected domain, the potential is determined up to an additive constant under the usual regularity assumptions. Our annular radial example satisfies this compatibility only when both its inner and outer boundaries are included.

Laplace’s equation and fixed linear kinematic boundary conditions allow superposition: add harmonic elementary potentials, then verify that the combined field satisfies the desired boundary values and is regular everywhere in the fluid. Singular sources, doublets and excluded solid interiors must remain outside that domain. A pressure or free-surface boundary condition involving speed squared need not be linear.

In a hole-containing exterior domain, circulation is additional global information. Requiring a single-valued φ makes every closed-loop circulation zero; admitting a local angle potential permits a different circulation class. These alternatives must not be silently interchanged when solving the same-looking normal-wall problem.

The scalar-potential and Laplace ideas also apply in three dimensions. This lesson’s single streamfunction and elementary line flows are specifically two-dimensional. For a mathematical return path, study the gradient and integral identities first, then the harmonic boundary-value problem; two-dimensional complex-variable methods offer another route after those physical conditions are clear.

Worked examples

Try it yourself

1. Replace φ = (0.400 m/s)x by φ = (0.400 m/s)x + 0.003 m²/s on the same patch. What is the new velocity?

   Show the answer and reasoning

   u = 0.400 m/s and v_y = 0; the velocity is unchanged.

   The added constant has zero derivative in both spatial directions. A potential reference value is not a speed or an extra acceleration.

2. On 0 ≤ x ≤ 0.020 m and 0 ≤ y ≤ 0.030 m, prescribe φ = Ax² with A = 3 s⁻¹. Find velocity at x = 0.010 m, planar curl and divergence. Is it a constant-density incompressible field?

   Show the answer and reasoning

   At the sample, v = (0.060, 0) m/s. Curl is zero, but divergence is 6 s⁻¹, so it is not a constant-density incompressible field.

   Differentiation gives u = 2Ax and v_y = 0. Both cross derivatives vanish, whereas ∂u/∂x + ∂v_y/∂y = 2A = 6 s⁻¹. The potential is not harmonic. This prescribed kinematic example demonstrates that irrotational and incompressible are different conditions; no pressure model has been supplied.

3. In the same annulus, set K = −0.001 m²/s in φ = K ln(r/0.020 m). At r = 0.040 m, find signed v_r, outward normal flux Q′ per axial length and counterclockwise circulation.

   Show the answer and reasoning

   v_r = −0.025 m/s, Q′ = −0.002π m²/s ≈ −0.006283 m²/s and Γ = 0.

   Negative radial velocity is inward, so the flux with the chosen outward normal is negative. Radial velocity has no tangential component, making circulation zero. The inward annular boundary-flow model does not claim that matter is destroyed inside the fluid.

4. Keep a = 0.025 m, ρ = 1000 kg/m³, p_∞ = 110000 Pa and the same ideal cylinder conditions, but set U = 0.800 m/s. At θ = 90°, find speed, C_p and p − p_∞. Compare with the original U.

   Show the answer and reasoning

   Speed is 1.600 m/s, C_p = −3 and p − p_∞ = −960 Pa, so absolute pressure is 109040 Pa. Speed doubles and the pressure difference becomes four times as large in magnitude.

   The surface speed is 2U. Its ratio to U remains 2, so C_p is unchanged. Dynamic pressure is now ρU²/2 = 320 Pa and the pressure difference is −3 × 320 Pa = −960 Pa. This quadratic ideal-pressure scaling does not supply a real viscous drag law.

5. Follow the local potential φ_local = Cθ, C = 0.001 m²/s, once counterclockwise around r = 0.040 m in the assigned annulus, continuing θ from 0 to 2π. Find circulation and accumulated local potential change. Can φ be single-valued globally there?

   Show the answer and reasoning

   Both are 0.002π m²/s ≈ 0.006283 m²/s. No single-valued global velocity potential exists for this swirl on the full annulus.

   Tangential velocity C/r times circumference 2πr gives Γ = 2πC. Continuation gives Δφ_local = C(2π − 0) = 2πC. Returning to the same point cannot change a single-valued scalar, so this nonzero result forbids a global potential. It does not contradict zero local curl on the annulus or introduce a physical wall at an angular branch cut.

What to carry forward

• A smooth velocity potential gives v = ∇φ and zero local curl; incompressibility supplies the additional Laplace condition.
• Potential values have units m²/s, and an additive constant changes no velocity.
• Normal volume flux per axial length and tangential circulation have the same units but measure different components.
• Global potentials require zero circulation around every loop; a hole-containing domain needs more than a local zero-curl check.
• A harmonic superposition must satisfy the actual fluid-domain and boundary conditions; ideal no penetration permits tangential slip.
• The noncirculating ideal cylinder has stagnation points, a doubled maximum surface speed and a conditional symmetric pressure pattern.
• Zero ideal pressure resultant in this assigned steady case does not predict real wall shear, separation, wake pressure or a viscous drag coefficient.