State and Prove Bernoulli’s Theorem | Bernoulli’s Theorem | Bernoulli’s Theorem Derivation
[State and Prove Bernoulli’s Theorem]
(State and Prove Bernoulli’s Theorem) According to Bernoulli’s Theorem , in case of steady flow of incompressible and non–viscous fluid through a tube of non–uniform cross–section, the sum of the pressure energy per unit volume, the potential energy per unit volume and the kinetic energy per unit volume is same at every point in the tube, i.e.,
Bernoulli’s equation is a mathematical expression of the law of mechanical energy conservation in fluid dynamics.
Bernoullis theorem is applied to the ideal fluids(SIIN Fluid).
Characteristics of ideal fluids are :-
- The fluid flow must be steady(Streamlined)
2. The fluid must be Incompressible
3. The fluid flow must be Irrotational &
4. The fluid must be Non–viscous
Before the proof of Bernoulli’s Theorem let us first discuss the three kinds of energies associated with an ideal fluid :-
- Kinetic Energy
If a liquid of mass (Δm) and volume (ΔV) is flowing with velocity (v) then it’s Kinetic Energy =
Kinetic energy per unit volume =
where ρ is the density of the fluid.
2. Potential Energy
If a liquid of mass (Δm) and volume (ΔV) is at height (h) from the surface of the earth then its Potential Energy =
Potential energy per unit volume =
3. Pressure Energy
The pressure energy is the energy of a fluid due to the applied pressure.
Let us consider a static fluid in an enclosed container having a circular hole on its bottom fitted with a piston of area “a”.
Due to the hydrostatic pressure let us assume that the piston is this place to by “L” as shown in figure below :-
Work done pressure, W = Force × displacement = Pressure × Area × displacement
⇒ W = PaL
This work done on the the liquid is equal to pressure energy, E = PaL
Pressure energy per unit volume = E/Volume = E/aL = PaL/aL = P
⇒ Pressure energy per unit volume = P (hydrostatic pressure of the liquid)
Proof of Bernoulli’s Theorem [State and Prove Bernoulli’s Theorem]
Consider an ideal liquid is flowing steadily through a tube of non–uniform area of cross–section as shown in figure.
If P1 and P2 are the pressures at the two ends of the tube respectively, work done in pushing the volume ΔV of incompressible liquid from point X to Y through the tube can be calculated as below :-
Work done at input X in time t is,
Work done at input Y in time t is,
Hence total work done,
Further by equation of continuity,
So work done
This work is used by the liquid in two ways :-
(i) In changing the potential energy of mass Δm (of volume ΔV) from Δmgh1 to Δmgh2 i.e.,
(ii) In changing the kinetic energy from to i.e.,
Now as the liquid is non–viscous, by conservation of mechanical energy,
W = ΔU+ΔK
Rearranging the terms
This equation is the Bernoulli’s equation and represents conservation of mechanical energy in case of moving fluids.