1. The property of fluid in motion by virtue of which it opposes relative motion between different layers and it opposes motion of another body through that is known as viscosity.

  2. The velocity of different layers of flowing liquid is different because of viscosity.

  3. Difference in velocity between laminar plates, that are unit distance from each other, is called velocity gradient (dv/dx).

  4. Ideal fluids have their viscosity and compressibility both zero.

Newton’s Formula

  1. The viscous force between two layers of liquids in motion is:

    F= -\eta A (\dfrac{dv}{dx})

    where, \dfrac{dv}{dx} = \dfrac{(v_1-v_2)}{(x_1-x_2)} = \dfrac{Δv}{Δx} is velocity gradient and A is the area of laminar plate.

    \eta = \dfrac{-F}{(A(\frac{dv}{dx})}

    Hence, the coefficient of viscosity (\eta) is defined as the viscous force per unit area to produce a velocity gradient of unity.

  2. S.I unit of ƞ is Pascal second i.e poiseuille.

    1 poiseuille = 1 pascal sec = 1Ns/m^2 = 10 Poise.

    1 decalpoise = 10 poise.

  3. The dimension of ƞ is [ML^{-1} T^{-1}]

Reynold’s Equation

  1. The maximum velocity of liquid flowing through a tube upto which the flow is streamline is called critical velocity.

  2. Reynold’s number indicates the nature of flow (i.e., streamline or turbulent).

  3. Mathematically, k = \dfrac{\rho v r}{\eta}

    where, \rho = density of liquid, r = radius of tube, \eta = coefficient of viscosity, v = critical velocity, K = Reynold’s number

    K is dimensionless pure member

  4. Beyond critical velocity (v), the flow will be turbulent.

    (i) If K < 2000, the flow is laminar.

    (ii) K > 3000, the flow will be turbulent

    (iii) 2000 < K < 3000, the flow will be unstable.

Equation of Continuity

  1. For an ideal liquid flowing through a non-uniform tube, under streamlined condition, the mass flowing per sec is same at each cross-section.

    i.e., \dot{m} = Av = Constant=

    or, A_1v_1= A_2v_2

    This is the equation of continuity.

  2. Clearly, the greater the cross-section, the velocity of flowing liquid will be smaller and vice-versa. Due to this reason,

    i. Deepwater appears to be still.

    ii. Falling stream water becomes narrower.

    iii. Fine jet is ejected by making cross-section narrower.

  3. Equation of continuity is based on conservation of mass.

  4. Equation of continuity also refers that volume of liquid flowing per second through any cross- section is constant.

  5. Equation of continuity is applied only in case of ideal liquid (non-viscous and incompressible).

  6. For laminar flow through non-uniform cylindrical tube,

    v \propto \dfrac{1}{A} \propto \dfrac{1}{r^2}

Bernoulli’s Theorem and Energy of Fluid in Motion

  1. A flowing liquid has three types of energies

    a. Pressure energy = pressure × volume = PV

    b. Kinetic energy = \dfrac{1}{2} mv^2

    c. Potential energy = mgh

  2. According to Bernoulli’s theorem, the total energy of an liquid under streamline flow remains constant i.e., PV + 1/2 mv^2+ mgh = constant ……….. (i)

    Conventionally, these energies are expressed per unit volume. So, this equation may be written as, P + 1/2 ρv^2+ ρgh = constant ………… (ii) (ρ=density)

    or, \dfrac{P}{ρg} + \dfrac{v^2}{2g}+ h = constant …………. (iii)


    \dfrac{P}{ρg}=Pressure head

    \dfrac{v^2}{2g}=Velocity head

    h = Gravitational head

  3. In case of horizontal flow ( i.e., ρgh=constant ), P + 1/2 ρv^2= Constant

  4. Bernoulli’s theorem is based on Principle of Conservation of mechanical energy.

  5. Pressure decreases liquid flows from broader to narrower portion of pipe.

  6. Applications of Bernoulli’s Principle

    1. Blowing of roofs by storms

    2. Attraction between two closely parallel moving boats in opposite direction.

    3. Repulsion between two closely parallel moving boats in opposite direction.

    4. Magnus effect in spinning ball

    5. Action of aspirator, carburetor, paint gun, scent spray, insect sprayer.

Torricelli’s Theorem and Velocity of Efflux

  1. It states that the velocity of efflux through an orifice at depth from liquid’s surface is the same as that of a freely falling body from height i.e.,

    v = \sqrt{2gh}=v \propto \sqrt{h}

  2. Torricelli’s theorem can be derived from Bernoulli’s principle.

  3. The velocity of efflux is independent of nature of liquid, quantity of liquid in the container and area of cross section of hole.

  4. Time taken for the liquid stream to reach the base of the tank (ground) is,

    T= \sqrt{\dfrac{2 (H-h)}{g}} \propto \sqrt{(H-h )} \propto \sqrt{x}

    Where (H-h) = x is distance of hole from base of tank.

  5. Range of liquid stream at base level is R = \sqrt{(2 h (H-h))}

  6. The range is maximum if h= H/2, The maximum range Rmax = H

  7. If a hole of cross-sectional area A_0 is made at the bottom of the tank of cross-sectional area A, then

    (i) Time taken to lower the height from H_1 to H_2 is,

    (ii) Time taken to empty is, 

    t = \dfrac{A}{A_0} \sqrt{\dfrac{2h}{g}}

  8. When a vertical tank is filled with a liquid up to height H and a hole at the bottom is opened then the time taken to lower the level by each H/n from the top are in the ratio,


    When a vertical tank is filled with liquid and several holes are made in vertical wall simultaneously then

    i. Velocity of efflux increases downward but ranges first increase becomes maximum for hole at centre and then decrease.

    ii. Range for stream escaping from two holes at equal distance from bottom and top are equal.

Stroke’s Law and Terminal Velocity

  1. When a body falls in a viscous medium, its velocity increases firstly and finally it attains a constant value called the terminal velocity.

  2. The viscous force on a spherical body of radius r and moving with velocity v is given by,

    F= 6πƞrv=F \propto v

    Where ƞ = coefficient of viscosity

  3. When a body attains terminal velocity (v) then it effective weight downward equals to viscous force upward. i.e.,


    Where, ρ=density of body and σ=density of viscous medium

    or, V(ρ-σ)g=6πƞrv_1

    or, \dfrac{4}{3} πr^2 (ρ-σ)g=6πƞrv_1

    Therefore, v_1 = \dfrac{2 (\rho- \sigma) gr^2}{9 \eta} \propto r^2 \propto \dfrac{1}{ƞ} (ρ-σ)