Viscous flow between parallel plates and in pipes

Why it matters

Laminar viscous flow is what happens in engine-oil galleries, hydraulic servo clearances, journal and thrust bearings, capillary tubes and viscometers. These are among the few flows with exact solutions of the Navier–Stokes equations, so they give clean closed-form answers for velocity profile, flow rate, wall shear and pressure drop — and GATE uses them often.

Key ideas

• Laminar vs turbulent. Flow in a circular pipe stays laminar when Re = ρVD/μ is below about 2000 (often quoted as 2300). Between parallel plates, laminar flow persists to a gap-based Reynolds number of roughly 1000 or more. Above these values, disturbances grow into turbulence (next topic).
• Fully developed flow. After an entrance length, the velocity profile stops changing along the flow. For laminar pipe flow L_e ≈ 0.05·Re·D. The results below apply only to the fully developed region.
• Force balance. In steady, fully developed flow the fluid does not accelerate, so the pressure force on an element balances the viscous shear force. This gives a linear shear-stress distribution: zero on the pipe axis (or plate mid-plane) and maximum at the wall. Combined with τ = μ·du/dy, it gives a parabolic velocity profile.
• Hagen–Poiseuille (circular pipe).
  • u(r) is parabolic; u_max on the axis = 2 × mean velocity.
  • Pressure drop is proportional to the first power of velocity and inversely proportional to D². For a given flow rate, Δp ∝ 1/D⁴, so halving the diameter raises Δp sixteen-fold.
  • In Darcy form the friction factor is f = 64/Re, independent of roughness.
  • The kinetic-energy correction factor α = 2 and momentum factor β = 4/3.
• Flow between fixed parallel plates (plane Poiseuille). The profile is parabolic across the gap h; u_max = 1.5 × mean velocity, and the flow rate per unit width varies as h³ — so leakage through a clearance is very sensitive to gap size.
• Couette flow. One plate moves at U and the other is fixed, with no pressure gradient. The profile is linear and the shear stress is uniform, τ = μU/h. This is the thin-film model used for bearings and viscometers in the fluid-properties topic.
• Combined Couette–Poiseuille flow. Both effects superpose because the governing equation is linear. An adverse pressure gradient (pressure rising in the plate direction) can produce back-flow near the fixed plate. This is the basis of hydrodynamic lubrication: the wedge builds pressure.
• Sign conventions. dp/dx is negative for flow driven in +x by pressure. Write formulas with (−dp/dx) to keep quantities positive.
• Viscometers. A capillary-tube viscometer uses Hagen–Poiseuille (μ from Q and Δp). Rotating-cylinder and cone-plate instruments use Couette flow; falling-sphere viscometers use Stokes' law (boundary-layer topic).

Formulas

Re = ρ·V·D/μ V mean velocity (m/s), D pipe diameter (m). Laminar if Re < about 2000.

u(r) = (−dp/dx)·(R² − r²)/(4μ) Pipe velocity profile; R pipe radius (m), r radial position (m), dp/dx pressure gradient (Pa/m).

u_max = 2·V V = (−dp/dx)·D²/(32μ)

Δp = 32·μ·V·L/D² and Q = π·Δp·D⁴/(128·μ·L) Hagen–Poiseuille; Δp pressure drop over length L (Pa), Q flow rate (m³/s).

f = 64/Re h_f = f·L·V²/(2g·D) = Δp/(ρg) Darcy friction factor (laminar pipe) and head loss (m).

τ_w = Δp·D/(4L) and τ(r) = (−dp/dx)·r/2 Wall and local shear stress (Pa); linear in r.

u(y) = (−dp/dx)·(h·y − y²)/(2μ) q = (−dp/dx)·h³/(12μ) u_max = 1.5·V Fixed parallel plates a distance h apart; y from one plate (m); q flow rate per unit width (m²/s). Wall shear τ_w = (−dp/dx)·h/2.

u(y) = U·y/h + (−dp/dx)·(h·y − y²)/(2μ) q = U·h/2 + (−dp/dx)·h³/(12μ) Couette–Poiseuille flow; lower plate fixed, upper plate moving at U (m/s).

P = Δp·Q Pumping power (W) to overcome the pressure drop.

Worked examples

Example 1 (standard) — oil in a pipe. Given: oil (ρ = 900 kg/m³, μ = 0.1 Pa·s) flows at Q = 2 L/s through a 50 mm pipe, 100 m long. Find the regime, pressure drop, head loss, wall shear stress, centre-line velocity and pumping power.

1. A = π × 0.05²/4 = 1.9635 × 10⁻³ m²; V = Q/A = 0.002/1.9635 × 10⁻³ = 1.0186 m/s.
2. Re = 900 × 1.0186 × 0.05/0.1 = 458 → laminar.
3. Δp = 32μVL/D² = 32 × 0.1 × 1.0186 × 100/0.0025 = 130 380 Pa.
4. Head loss: h_f = Δp/(ρg) = 130 380/(900 × 9.81) = 14.77 m. Check: f = 64/458.4 = 0.1396; h_f = 0.1396 × (100/0.05) × 1.0186²/(2 × 9.81) = 14.77 m.
5. τ_w = Δp·D/(4L) = 130 380 × 0.05/400 = 16.3 Pa.
6. u_max = 2V = 2.04 m/s.
7. P = Δp·Q = 130 380 × 0.002 = 260.8 W. Answer: laminar (Re ≈ 458), Δp ≈ 130.4 kPa, h_f ≈ 14.8 m of oil, τ_w ≈ 16.3 Pa, power ≈ 261 W.

Example 2 (GATE level) — leakage through a clearance and a moving wall. Given: (a) oil (μ = 0.2 Pa·s, ρ = 900 kg/m³) leaks between two fixed parallel plates with a gap of h = 8 mm under a pressure gradient of 1.5 kPa/m. Find the leakage per metre width, the mean and maximum velocities and the wall shear. (b) For a gap h = 5 mm filled with oil of μ = 0.1 Pa·s, with the upper plate moving at U = 1 m/s, find the pressure gradient that makes the net flow zero.

1. (a) q = (−dp/dx)·h³/(12μ) = 1500 × 0.008³/(12 × 0.2) = 1500 × 5.12 × 10⁻⁷/2.4 = 3.2 × 10⁻⁴ m²/s.
2. Mean velocity V = q/h = 3.2 × 10⁻⁴/0.008 = 0.04 m/s; u_max = 1.5V = 0.06 m/s.
3. Wall shear τ_w = (−dp/dx)·h/2 = 1500 × 0.004 = 6 Pa.
4. Check regime: Re = ρVh/μ = 900 × 0.04 × 0.008/0.2 = 1.44 → laminar.
5. (b) Zero net flow: U·h/2 + (−dp/dx)·h³/(12μ) = 0 → dp/dx = 6μU/h².
6. dp/dx = 6 × 0.1 × 1/0.005² = 24 000 Pa/m (positive — pressure rises in the direction of plate motion). Answer: (a) q = 3.2 × 10⁻⁴ m³/s per metre, V = 0.04 m/s, u_max = 0.06 m/s, τ_w = 6 Pa; (b) dp/dx = +24 kPa/m.

Common mistakes

• Using u_max = 2V for parallel plates (it is 1.5V) or 1.5V for a pipe (it is 2V).
• Using radius instead of diameter in 32μVL/D², or D² instead of D⁴ when working with Q.
• Applying f = 64/Re to turbulent flow, or applying Hagen–Poiseuille in the entrance region.
• Losing the sign of dp/dx — a favourable gradient is negative.
• Using the full gap where the half-gap appears (wall shear is (−dp/dx)·h/2 for plates, Δp·R/(2L) = Δp·D/(4L) for pipes).
• Forgetting to check Re before assuming laminar flow.

For GATE ME

Expect: pressure drop, flow rate or power in laminar pipe flow; the ratio of maximum to mean velocity; the position where local velocity equals the mean (r = R/√2 ≈ 0.707R in a pipe); shear stress at a given radius; leakage between plates; Couette flow with a pressure gradient (zero net flow or zero wall shear); and how Δp changes when the diameter or viscosity changes at fixed Q. Practise deriving the profiles once — then the formulas stick.

Quick check

1. What is f for laminar pipe flow at Re = 1600?
2. If the diameter of a laminar pipe is halved at the same Q, by what factor does Δp change?
3. What is u_max/V for flow between fixed parallel plates?
4. Where is the shear stress zero in fully developed pipe flow?
5. At what radius is the local velocity equal to the mean velocity in laminar pipe flow? Answers: 1. 64/1600 = 0.04. 2. It increases 16 times. 3. 1.5. 4. On the pipe axis. 5. r = R/√2 ≈ 0.707R.