Question

state and prove Bernaulius theorem​

Answer 1

Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy. ... Let the velocity, pressure and area of the fluid column be p1, v1 and A1 at Q and p2, v2 and A2 at R.

Answer 2

Question:

State and prove Bernoulli's theorem.

Solution:

Bernoulli's theorem states that sum of pressure energy, potential energy and kinetic energy per unit mass of an incompressible fluid in a streamlined flow remains constant.

Bernoulli's equation mathematically can be written as :-

$\quad \diamond \ \bf \dfrac{P}{\rho} + \dfrac{1}{2} v^2 + gh \ = \ constant$

Proof:

Let us consider a fluid with negligible viscosity is moving with laminar flow.

Let,

$\bf At\ Q\begin{cases}\bullet\ \sf Let\ the\ velocity\ of\ the\ fluid\ be\ 'V_1'.\\ \bullet\ \sf Let\ the\ pressure\ of\ the\ fluid\ be\ 'P_1'.\\ \bullet\ \sf Let\ the\ area\ of\ the\ fluid\ be\ 'A_1'. \end{cases}$

$\bf At\ R\begin{cases}\bullet\ \sf Let\ the\ velocity\ of\ the\ fluid\ be\ 'V_2'.\\\bullet\ \sf Let\ the\ pressure\ of\ the\ fluid\ be\ 'P_2'.\\ \bullet\ \sf Let\ the\ area\ of\ the\ fluid\ be\ 'A_2'. \end{cases}$

Now,

Volume bounded by Q and R move to S and T.

Where,

• $\sf QS = L_1$
• $\sf RT = L_2$

If the fluid is incompressible then,

★ Work done by pressure difference per unit volume = Gain in PE per unit volume + Gain in KE per unit volume.

Therefore,

• $\sf A_1L_1 = A_2L_2$

Now,

Work done, w = f × d

=> P × Volume

=> $\sf W_{net} = P_1 - P_2$

=> $\sf KE = \dfrac{1}{2}mv^2$

=> $\sf \dfrac{1}{2}v \rho v^2$

=> $\sf \dfrac{1}{2} \rho v^2$

$\therefore \sf KE \ gained \ = \ \dfrac{1}{2} \rho (v_2^2 - v_1^2)$

Therefore,

=> $\sf P + \dfrac{1}{2} \rho v^2 + pgh = Constant$

For Horizontal tube,

• $\sf h_1 = h_2$
• $\sf P + \dfrac{1}{2} \rho v^2 = Constant$

• This proves Bernoulli's theorem.

Hence Proved ✔︎