Flow over a cylinder with and without circulation

Why it matters

The circular cylinder is the one body whose potential flow can be written down exactly, and it teaches the two central lessons of inviscid aerodynamics: without circulation there is no lift and no drag (d'Alembert's paradox), and with circulation there is lift L′ = ρ∞V∞Γ. Through conformal mapping, the lifting cylinder becomes the Joukowski airfoil, and the same circulation idea underpins all of airfoil theory.

Key ideas

Non-lifting cylinder = uniform flow + doublet. Superposing a uniform stream V∞ and a doublet of strength κ gives a stream function that is zero on the circle r = R with R² = κ/(2πV∞). That circle is a streamline, so it can be replaced by a solid cylinder. The flow outside is the potential flow past the cylinder.

• Surface velocity is purely tangential: V_θ = −2V∞ sin θ (θ measured from the downstream x-axis, anticlockwise). Stagnation points at θ = 0° and 180°; maximum speed 2V∞ at θ = 90° and 270°.
• Surface pressure coefficient C_p = 1 − 4 sin²θ: +1 at the stagnation points, −3 at the shoulders, and zero at θ = 30°, 150°, 210° and 330°.
• The distribution is symmetric top-to-bottom (no lift) and front-to-back (no drag). Zero drag on a body in steady inviscid flow is d'Alembert's paradox; real cylinders have large pressure drag because the boundary layer separates on the rear half and the rear pressure never recovers.

Lifting cylinder = uniform flow + doublet + vortex. Adding a point vortex of strength Γ at the centre keeps the circle a streamline (the vortex streamlines are circles) but makes the flow asymmetric. With Γ positive clockwise (Anderson's convention):

• Surface velocity V_θ = −2V∞ sin θ − Γ/(2πR) — the speed rises on top and falls underneath.
• Stagnation points move down to sin θ_s = −Γ/(4πV∞R), symmetric about the vertical axis.
  • Γ < 4πV∞R: two stagnation points on the lower surface.
  • Γ = 4πV∞R: they merge at the bottom (θ = 270°).
  • Γ > 4πV∞R: a single stagnation point leaves the surface and lies in the flow below the cylinder, with a closed recirculating region round the body.
• Integrating the surface pressure gives L′ = ρ∞V∞Γ (per unit span, perpendicular to V∞) and D′ = 0. This is the Kutta–Joukowski theorem in its simplest setting. The lift is independent of R.
• Circulation is not fixed by potential theory: any Γ gives a valid flow. For the cylinder it must be supplied externally (rotation); for an airfoil the sharp trailing edge fixes it (Kutta condition).

Magnus effect. A spinning cylinder or ball drags fluid round with it through viscosity, producing a real circulation and a side force — the curved path of a spinning cricket or tennis ball, and Flettner rotor ships. The real circulation is well below the "no-slip" estimate 2πR·ωR because of separation.

Assumptions: steady, 2-D, inviscid, incompressible, irrotational (except the vortex at the centre, which is inside the body).

Formulas

• ψ = V∞·r·sin θ·(1 − R²/r²) — non-lifting cylinder; R in m, R² = κ/(2πV∞).
• V_r = V∞(1 − R²/r²) cos θ, V_θ = −V∞(1 + R²/r²) sin θ — non-lifting field.
• V_θ(R) = −2V∞ sin θ, C_p = 1 − 4 sin²θ — surface, non-lifting.
• ψ = V∞·r·sin θ·(1 − R²/r²) + (Γ/2π)·ln(r/R) — lifting cylinder, Γ clockwise positive (m²/s).
• V_θ(R) = −2V∞ sin θ − Γ/(2πR) — surface, lifting.
• sin θ_s = −Γ/(4πV∞R) — stagnation points, valid for Γ ≤ 4πV∞R.
• C_p = 1 − [2 sin θ + Γ/(2πRV∞)]² — surface, lifting.
• L′ = ρ∞·V∞·Γ (N/m), D′ = 0.
• c_l = L′/(q∞·2R) = Γ/(R·V∞) — lift coefficient based on diameter.

Worked examples

Example 1 (standard). Air (ρ = 1.225 kg/m³) at V∞ = 20 m/s flows past a non-rotating cylinder of radius 0.5 m. Find C_p at θ = 30°, 60° and 90°, the maximum surface speed and p − p∞ at the top.

1. C_p = 1 − 4 sin²θ. θ = 30°: 1 − 4(0.25) = 0. θ = 60°: 1 − 4(0.75) = −2. θ = 90°: 1 − 4 = −3.
2. V_max = 2V∞ = 40 m/s at θ = 90°.
3. q∞ = 0.5 × 1.225 × 20² = 245 Pa; at the top p − p∞ = C_p·q∞ = −3 × 245 = −735 Pa.

Answer: C_p = 0, −2, −3; V_max = 40 m/s; p − p∞ = −735 Pa at the top.

Example 2 (GATE level). The same cylinder (R = 0.5 m, V∞ = 20 m/s, ρ = 1.225 kg/m³) now carries a clockwise circulation Γ = 20π m²/s. Find the stagnation points, the lift per unit span, c_l and C_p at the top and bottom.

1. sin θ_s = −Γ/(4πV∞R) = −20π/(4π × 20 × 0.5) = −0.5 ⇒ θ_s = 210° and 330° (30° below the horizontal on each side).
2. L′ = ρ∞V∞Γ = 1.225 × 20 × 20π = 1539 N/m.
3. c_l = Γ/(RV∞) = 20π/(0.5 × 20) = 6.28. Check: q∞·2R·c_l = 245 × 1 × 6.283 = 1539 N/m ✓.
4. Top (θ = 90°): V_θ = −2(20)(1) − 20π/(2π × 0.5) = −40 − 20 = −60 m/s; C_p = 1 − (60/20)² = −8.
5. Bottom (θ = 270°): V_θ = +40 − 20 = 20 m/s; C_p = 1 − 1 = 0.

Answer: stagnation at θ = 210° and 330°; L′ ≈ 1.54 kN/m; c_l ≈ 6.28; C_p = −8 (top), 0 (bottom).

Common mistakes

• Expecting potential theory to predict cylinder drag. It gives zero drag; real drag comes from boundary-layer separation.
• Forgetting the factor 2 in the surface speed (2V∞ sin θ) and hence getting C_p,min = 0 instead of −3.
• Getting the stagnation-point side wrong: for lift upward the flow is faster on top, and the stagnation points move to the lower surface.
• Using sin θ_s = −Γ/(4πV∞R) when Γ > 4πV∞R — there is then no surface stagnation point.
• Basing c_l on R instead of the diameter without saying so.
• Thinking lift depends on cylinder size at fixed Γ; L′ = ρV∞Γ does not contain R.

For GATE AE

Expect: C_p or velocity at a given surface angle; locating stagnation points for a given Γ; the Γ needed to make the stagnation points coincide; lift per unit span from Kutta–Joukowski; points where C_p = 0; and MCQs on d'Alembert's paradox and the Magnus effect. Write down the sign convention for Γ and θ before substituting.

Quick check

1. What is C_p at the shoulders (θ = 90°) of a non-lifting cylinder?
2. At what surface angles is C_p = 0 on a non-lifting cylinder?
3. What circulation makes the two stagnation points meet, for V∞ = 10 m/s and R = 0.2 m?
4. Why does potential flow predict zero drag on a cylinder?

Answers: 1. −3. 2. θ = 30°, 150°, 210°, 330°. 3. Γ = 4πV∞R = 8π ≈ 25.1 m²/s. 4. The pressure distribution is symmetric front-to-back because there is no viscosity and hence no separation (d'Alembert's paradox).