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State and Prove Bernoulli’s Theorem | Bernoulli’s Theorem | Bernoulli’s Theorem Derivation

[State and Prove Bernoulli’s Theorem]

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.,

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

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 = $\displaystyle \frac{1}{2}(\Delta m){{v}^{2}}$

Kinetic energy per unit volume = $\displaystyle \frac{1}{2}(\frac{\Delta m}{\Delta V}){{v}^{2}}=\frac{1}{2}\rho {{v}^{2}}$

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 = $\displaystyle \Delta mgh$

Potential energy per unit volume = $\displaystyle \frac{\Delta m}{\Delta V}gh=\rho gh$

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, $\displaystyle {{W}_{1}}={{F}_{1}}{{x}_{1}}\cos 0$

$\displaystyle \Rightarrow {{W}_{1}}={{P}_{1}}{{A}_{1}}{{v}_{1}}t$

Work done at input Y in time t is, $\displaystyle {{W}_{2}}={{F}_{2}}{{x}_{2}}\cos 180$

$\displaystyle \Rightarrow {{W}_{2}}=-{{P}_{2}}{{A}_{2}}{{v}_{2}}t$

Hence total work done, $\displaystyle W={{W}_{1}}+{{W}_{2}}={{P}_{1}}{{A}_{1}}{{v}_{1}}t-{{P}_{2}}{{A}_{2}}{{v}_{2}}t$

Further by equation of continuity, $\displaystyle {{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}=\frac{Volume\text{ }Of\text{ }Liquid\text{ }Flowing\text{ }In\text{ }Or\text{ }Out}{Time}$

$\displaystyle \Rightarrow {{A}_{1}}{{v}_{1}}t={{A}_{2}}{{v}_{2}}t=\frac{m}{\rho }=\Delta V$

So work done

$\displaystyle W={{W}_{1}}+{{W}_{2}}=({{P}_{1}}-{{P}_{2}})\frac{m}{\rho }=({{P}_{1}}-{{P}_{2}})\Delta V$

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.,

ΔU=Δmg (h₂–h₁)

(ii) In changing the kinetic energy from $\displaystyle \frac{1}{2}\Delta mv_{1}^{2}$ to $\displaystyle \frac{1}{2}\Delta mv_{2}^{2}$ i.e.,

$\displaystyle \Delta K=\frac{1}{2}\Delta m\left( v_{2}^{2}-v_{1}^{2} \right)$

Now as the liquid is non–viscous, by conservation of mechanical energy,

W = ΔU+ΔK

$\displaystyle ({{P}_{1}}-{{P}_{2}})\Delta V=\Delta mg({{h}_{2}}-{{h}_{1}})+\frac{1}{2}\Delta m(v_{2}^{2}-v_{1}^{2})$

$\displaystyle ({{P}_{1}}-{{P}_{2}})=\frac{\Delta m}{\Delta V}g({{h}_{2}}-{{h}_{1}})+\frac{1}{2}\frac{\Delta m}{\Delta V}(v_{2}^{2}-v_{1}^{2})$

$\displaystyle ({{P}_{1}}-{{P}_{2}})=\rho g({{h}_{2}}-{{h}_{1}})+\frac{1}{2}\rho (v_{2}^{2}-v_{1}^{2})$

Rearranging the terms

$\displaystyle {{P}_{1}}+\rho g{{h}_{1}}+\frac{1}{2}\rho v_{1}^{2}={{P}_{2}}+\rho g{{h}_{2}}+\frac{1}{2}\rho v_{2}^{2}$

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

This equation is the Bernoulli’s equation and represents conservation of mechanical energy in case of moving fluids.

State and Prove Bernoulli’s Theorem | Bernoulli’s theorem | Bernoulli’s Theorem Derivation

State and Prove Bernoulli’s Theorem | Bernoulli’s Theorem | Bernoulli’s Theorem Derivation

[State and Prove Bernoulli’s Theorem]

Proof of Bernoulli’s Theorem [State and Prove Bernoulli’s Theorem]

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