Bernoulli's Theorem and Viscosity
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Direct answer
Bernoulli's theorem is energy conservation written per unit volume for streamline flow of a non-viscous, incompressible fluid: P + (1/2) rho v^2 + rho g h = constant along a streamline, so where the fluid runs faster its pressure must drop. Torricelli's result, efflux speed v = sqrt(2 g h) from a side hole, is Bernoulli applied to a tank. Viscosity enters through Newton's law F = −eta A (dv/dx) with eta in pascal-second, Stokes' drag on a sphere F = 6 pi eta r v, and the terminal velocity v(t) = 2 r^2 (rho − sigma) g/(9 eta); Poiseuille's volumetric flow Q = pi P r^4/(8 eta l) shows the fourth-power tyranny of radius in capillary flow.
What you must remember
- Bernoulli equation: P + (1/2) rho v^2 + rho g h = constant along a streamline; each term is an energy density — pressure energy, kinetic, potential, all in joules per cubic metre.
- Applications map: venturimeter measures flow rate from the pressure drop at the throat; pitot tube measures aircraft speed; a spinning ball curves (Magnus effect) because spin changes the relative air speed on the two sides.
- Torricelli's law: v = sqrt(2 g h), same as free fall through height h; the range of the jet is x = 2 sqrt(h(H − h)), maximum when the hole is at the middle of the column.
- Newton's law of viscosity: F = −eta A dv/dx; eta of water about 10^-3 Pa s at room temperature; liquid viscosity falls with temperature, gas viscosity rises — a viva staple.
- Stokes law and terminal velocity: F = 6 pi eta r v; setting drag plus buoyancy against weight gives v(t) = 2 r^2 (rho − sigma) g/(9 eta); doubling the drop's radius quadruples terminal speed.
- Poiseuille equation: Q = pi P r^4/(8 eta l); radius is the dominant variable — a slight narrowing of an artery raises the pressure the heart must supply dramatically.
- Reynolds number: Re = rho v D/eta; flow turns turbulent beyond roughly 2000, and Re is dimensionless.
- Pattern note: Main examines Torricelli and terminal-velocity numericals; Advanced builds multi-level tank problems and viscous-flow reasoning with the energy equation amended for head loss.
One tank, two holes
A tank filled to height H has a hole at depth h below the surface. Torricelli gives the jet speed sqrt(2gh), and projectile motion from the hole to the floor gives a landing distance that works out to x = 2 sqrt(h(H − h)). Two different holes — one at h, one at H − h — land at the same spot, and the maximum range belongs to the hole punched exactly at mid-depth, h = H/2. These three sentences describe the majority of JEE Main questions on this chapter; the remainder are terminal-velocity substitutions where the only trap is the (rho − sigma) difference: a bubble rising through water has sigma above rho, so the numerator changes sign, not magnitude — the terminal velocity formula returns the same size with reversed direction.
For Poiseuille flow the exam favourite is the artery: when a plaque halves the radius, flow falls sixteen-fold at the same pressure difference, so the body must raise the pressure drop sixteen-fold to keep perfusion constant — the physics of hypertension in one calculation.
Where students slip
Bernoulli's theorem assumes laminar, incompressible, non-viscous flow, and each assumption has anchored an assertion–reason question; applying it to turbulent flow is invalid. The second error is direction of pressure change: faster flow means lower pressure, and candidates who reverse this cannot explain lift at all. Third, terminal velocity questions forget buoyancy; a sphere falling through a fluid of density sigma has effective weight (4/3) pi r^3 (rho − sigma) g, and omitting sigma inflates v(t) for dense fluids.
Frequently asked questions
What does each term of Bernoulli's equation represent?
Pressure energy per unit volume P, kinetic energy per unit volume (1/2)rho v^2 and potential energy per unit volume rho g h, all conserved together along a streamline of an ideal flow.
Why does a spinning ball curve in flight?
Rotation makes air speed relative to the surface differ on the two sides; by Bernoulli the pressure differs sideways, and the net sideways force (Magnus effect) bends the trajectory.
How is terminal velocity derived for a sphere in a viscous fluid?
Balance weight minus buoyancy against Stokes drag 6 pi eta r v to get v(t) = 2 r^2 (rho − sigma) g/(9 eta), growing as the square of the radius.
What happens to flow if the radius of a pipe is halved?
Poiseuille's Q is proportional to r^4, so the flow rate falls sixteen times at the same pressure difference; small constrictions are expensive.
Why does efflux speed match free-fall speed from the same height?
Bernoulli between the top surface and the orifice at depth h, with both at atmospheric pressure and negligible surface speed, gives v = sqrt(2 g h) exactly.