Concepts: viscosity, surface tension and exercise practice
Mechanical Properties of Fluids26 min readLesson 4 of 6
4.1 Viscosity and Newton's law of viscous flow
Honey pours slowly and water pours fast because their internal friction differs. Viscosity measures that internal friction. Consider a liquid layer between a fixed plate and a moving plate: the layer touching the fixed plate is at rest, the layer touching the moving plate moves with it, and the layers in between have intermediate speeds. A force is needed to keep the moving plate moving, and for steady flow this force is proportional to the area of the plate and to the velocity gradient across the film: F = eta A dv/dx. The constant eta is the coefficient of viscosity, with SI unit Pa s and the older unit poise, where 1 Pa s = 10 poise. Liquids become less viscous when heated because their molecules gain energy and slip past each other more easily; gases become more viscous when heated because faster molecules exchange momentum more often.
4.2 Poiseuille's formula and Stokes' law
For steady laminar flow of a liquid through a horizontal tube of length l and radius r, the volume flowing per second is V/t = pi P r^4/(8 eta l), where P is the pressure difference between the ends. The fourth power of the radius is why a small narrowing of an artery raises the pressure needed so sharply, and why a wide pipe carries much more water than a narrow one at the same pressure. For a small sphere of radius r moving slowly through a liquid, Stokes showed that the drag is 6 pi eta r v. A body dropped in a viscous liquid therefore reaches a terminal speed when the net downwards force on it is zero: weight minus buoyancy equals the drag.
4.3 Surface tension
A molecule inside a liquid is pulled equally in all directions by its neighbours, while a molecule at the surface has fewer neighbours above it, so it is pulled inward. The surface therefore has extra energy and behaves like a stretched membrane. Surface tension T is the force per unit length along the surface, and it also equals the surface energy per unit area. Water has a large surface tension because of hydrogen bonding, which is why small insects can stand on water and why raindrops are nearly spherical. Detergents lower the surface tension of water so that it wets cloth and spreads over greasy surfaces.
4.4 Excess pressure and angle of contact
The curved surface of a liquid drop or bubble has a higher pressure on the concave side. For a spherical drop of radius r the excess pressure is 2T/r; for a soap bubble, which has two surfaces, it is 4T/r. A small bubble therefore has a larger excess pressure than a large one, which is why two bubbles joined by a tube behave so surprisingly. The angle of contact decides whether a liquid spreads on a solid or balls up. Water on clean glass has a small contact angle and wets the surface, while mercury on glass has an obtuse angle and stands as a dome. A liquid that wets the wall has a concave meniscus, and one that does not wet the wall has a convex meniscus.
4.5 Capillary rise
When a narrow tube dips into a liquid that wets the glass, the liquid rises to a height h given by h = 2T cos(theta)/(rho g r_tube), where theta is the angle of contact and r_tube is the inner radius. The rise is larger for a narrower tube and for a liquid with larger surface tension. Mercury, which does not wet glass, is pushed down instead of rising, and the same formula then gives a negative height. The rise of sap in tall trees depends partly on narrow xylem tubes, but the main lift comes from transpiration pull, because a capillary alone can raise water only about a metre. A U-tube with one narrow arm and one wide arm shows the same effect, with the liquid higher on the narrow side.
Exam
For a capillary tube question, check whether the liquid wets the glass. If the contact angle is obtuse, the liquid is depressed, so the formula gives a negative height. Mercury in glass is the standard example of depression.
4.6 Exercise practice: NCERT questions worked in short
9.1 A tank contains water of density 1.03 x 10^3 kg m^-3 to a depth of 3 m. The excess pressure at the bottom is rho g h = 1.03 x 10^3 x 9.8 x 3 = 3.03 x 10^4 Pa, and the absolute pressure adds the atmospheric value 1.01 x 10^5 Pa. 9.5 A pressure of 1.0 x 10^3 Pa acts on a piston of area 0.01 m^2; the force is P A = 10 N, and a second piston of area 0.5 m^2 connected to the same fluid would feel 500 N. 9.20 A tank with a small hole 2 m below the free surface gives v = sqrt(2 x 9.8 x 2) = 6.26 m s^-1 for the efflux speed. 9.21 An air bubble of radius 1 mm in water has an excess pressure 2T/r = 2 x 0.072/10^-3 = 144 Pa above the surrounding water. The remaining exercise questions 9.2 to 9.4, 9.6 to 9.19 and 9.22 to 9.28 use these same formulas with different numbers.