Question

State and prove Bernoulli’s theorem.

Answer 1

Bernoulli’s Theorem :
According to Bernoulli’s theorem, the sum of the energies possessed by a flowing ideal liquid at a point is constant provided that the liquid is incompressible and non-viseous and flows in streamline.
Potential energy + Kinetic energy + Pressure energy = Constant

where C is a constant.
This relation is called Bernoulli’s theorem.
Dividing eqn. (1) by g, we get,

where C is another constant.
For horizontal flow, h remains same throughout.
So,

Thus, for horizontal motion, the sum of static and dynamic pressure is constant. If , P₁,υ₁ and P₂, υ₂ represent pressure and velocities at two points. Then,

Derivation of Bernoulli’s Theorem :
The energies possessed by a flowing liquid are mutually convertible. When one type of energy increases, the other type of energy decreases and vice-versa. Now, we will derive the Bernoulli’s theorem using the work-energy theorem.

Consider the flow of liquid. Let at any time, the liquid lies between two areas of flowing liquid A₁ and A₂. In time interval ∆t, the liquid displaces from A₁ by ∆x₁ = υ₁∆t and displaces from A₂ by ∆ x₂ = υ₂ ∆t. Here υ₁ and υ₂ are the velocities of the liquid at A₁ and A₂.
The work done on the liquid is P₁A₁∆x₁ by the force and P₂A₂∆x₂ against the force respectively.
Net work done,

\(\because A_1v_1 = A_2v_2\)

So, more the cross-sectional area, lesser is the velocity and vice-versa.

So, from Bernoulli's theorem