In streamline flow, the energy of a fluid particle remains constant.

Suppose that, in a tube of flow a fluid enters the cross-section A_{1} with velocity v_{1} and leave the cross-section A_{2} with a velocity v_{2}. Let the pressures at A_{1} and A_{2} be P_{1} and P_{2} respectively.

Consider an element ABCD of the incompressible fluid entering the tube. In time Δt, the fluid element ABB’A’ enters the tube. The expelled fluid emerges from the tube at the other end and is denoted by DCC’D’.

Δm = ρA_{1}v_{1} Δt = ρ A_{2}v_{2} Δt

Work done by fluid pressure at 1 = (P_{1}A_{1}) v_{1}Δt = P_{1} Δm/ρ

Work done by fluid pressure at 2 = − (P_{2}A_{2}) v_{2} Δt = −P_{2} Δm/ρ

Work done by gravity = -Δm.g. (h_{2} – h_{1})

Change in kinetic energy = $ \displaystyle \frac{1}{2}\Delta m [v_2^2 -v_1^2 ] $

Using work energy theorem, W = ΔK

$ \displaystyle P_1\frac{\Delta m}{\rho}-P_2\frac{\Delta m}{\rho} -\Delta m(h_2-h_1) = \frac{1}{2}\Delta m(v_2^2 – v_1^2)$

$ \displaystyle \frac{P_1}{\rho}+gh_1 + \frac{v_1^2}{2}= \frac{P_2}{\rho} + gh_2 + \frac{v_2^2}{2} $

$ \displaystyle P_1 + \rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 $

$ \displaystyle P + \rho gh + \frac{1}{2}\rho v^2 = Constant $

where, P = Pressure energy per unit volume = Pressure at a cross-section.

ρgh = potential energy per unit volume and

(1/2) ρv^{2} = kinetic energy per unit volume.

The above equation is known as Bernoulli’s equation.

Solved Example : Water flows through a tunnel from the reservoir of a dam towards the turbine installed in its power plant. The power plant is situated h m below the reservoir. If the ratio of the cross-sectional areas of the tunnel at the reservoir and power station end is η , find the speed of the water entering into the turbine.

Solution: Applying Bernoulli’s theorem at reservoir and power plant for the flowing water, we obtain,

$ \displaystyle P_0 + \rho gh_1 + \frac{1}{2}\rho v_1^2 = P_0 + \rho gh_2 + \frac{1}{2}\rho v_2^2 $

$ \displaystyle \rho gh_1 + \frac{1}{2}\rho v_1^2 = \rho gh_2 + \frac{1}{2}\rho v_2^2 $

$ \displaystyle v_2^2 = v_1^2 + 2 g(h_1 -h_2 ) $

Putting (h_{1} – h_{2}) = h, we obtain

$ \displaystyle v_2 = \sqrt{ v_1^2 + 2 g h} $ …….(1)

Equation of continuity yields A_{1}v_{1} = A_{2}v_{2} …….(2)

Eliminating v_{1} from above equations , we obtain

$ \displaystyle v_2^2 = (\frac{A_2}{A_1}v_2)^2 + 2 g h $

$ \displaystyle v_2^2(1-(\frac{A_2}{A_1})^2) = 2gh $

$ \displaystyle v_2 = \sqrt{\frac{2 g h}{1-(\frac{A_2}{A_1})^2}} $

$\large \frac{A_1}{A_2} = \eta $

$ \displaystyle v_2 = \sqrt{\frac{2 g h}{1-(\frac{1}{\eta})^2}} $

$\large v_2 = \eta \sqrt{\frac{2gh}{\eta^2 – 1}}$

Q: Why is the shape of a rotating liquid surface Paraboloidal ?

Ans: In a rotating liquid there is a pressure gradation from the axis to the wall . The pressure at a distance x from the axis is given by

$ \displaystyle p = p_0 + \frac{1}{2} \rho \omega^2 x^2 $

Consider a point P at a distance x from the axis and at a depth y from the liquid surface . Then

$\displaystyle p = p_0 + \rho g y $

Hence , $ \displaystyle p_0 + \rho g y = p_0 + \frac{1}{2} \rho \omega^2 x^2 $

$ \displaystyle y = \frac{\omega^2}{2 g} x^2 $

or , $ \displaystyle x^2 = ( \frac{2 g}{\omega^2} ) y $

On comparing this equation with the standard form of equation of a parabola x^{2} = 4 a y , we find that rotating liquid surface is a parabola about the vertical axis of rotation and the entire surface is therefore , Paraboloidal .