Bernoulli's theorem proof

Answer 1

Proof of Bernoulli's Theorem

To prove Bernoulli's theorem, we make the following assumptions:

1. The liquid is incompressible.

2. The liquid is non-viscous.

3. The flow is steady and the velocity of the liquid is less than the critical velocity for the liquid.

Imagine an incompressible and non-viscous liquid to be flowing through a pipe of varying cross-sectional area as shown in Fig. The liquid enters the pipe with a normal velocity v11 and at a height h1 above the reference level (earth's surface). It leaves the pipe with a normal velocity v2 at the narrow end B of cross-sectional area a2 and at a height h2 above the earth's surface. at its wide end A of cross-sectional area a

If r is the density of the incompressible liquid, then in accordance with the equation of continuity, the mass m of the liquid crossing any section of the pipe is given by

a1 v1 = a2 v2 = m (say) or a1 v1 = a2 v2 = m/ ..... (ii)

Let P1 and P2 be the values of the pressure due to the liquid at the ends A and B respectively. If the liquid moves from the end A to B under the action of pressure difference P1 - P2, then in accordance with energy conservation principle, the work done by the pressure energy of the liquid must appear as the increase in potential and kinetic energies of the liquid.

The pressure energy exerts a force P1 a1 on the liquid at the end A. The liquid covers a distance v1 in one second at the end A and therefore

Work done per second on the liquid at the end A = 1 a1 v1

The liquid reaches the end B against pressure P2 i.e. against a force P2 a2. At the end B, the liquid covers a distance v2 in one second and therefore

Work done per second by the liquid at the end B = 2 a2 v2

Hence, net work done by the pressure energy in moving the liquid from the end A to B in one second = 1 a1 v1 - 2 a2 v2

Using the equation (ii), we have

Net work done by the pressure energy per second = 1 m/ - p2 m/ = (1 - 2) m/ ... (iii)

When the mass m of the liquid flows in one second from the end A to B, its height increases from h1 to h2.

Therefore, increase in potential energy of the liquid per second

= m g h2 - m g h1 = m g (h2 - h1) … (iv)

Further, when the mass m of the liquid flows in one second from the end A to B, its velocity increases from v1 to v2.

Therefore, increase in kinetic energy of the liquid per second = 1/2 mv22 - 1/2 mv12 = 1/2 m(v22 - v12) ... (v)

According to work-energy conservation principle,

Work done by the pressure energy per second = increase in potential energy per second + increase in kinetic energy per second

Therefore (P1 - P2) = m/? = mg (h2 - h1) + 1/2 m(v22 - v02) or P1/? - P2/? = gh2 - gh1 + 1/2 v22 - 1/2 v12 or P1/? + gh1 + 1/2 v12 = P2/? + gh2 + 1/2 v22 ..... (vi)

or P/? + gh + 1/2 v2 = constant

It proves the Bernoulli's theorem. This is the most convenient form of Bernoulli's equation. However, it can be expressed in some other forms as explained below:

Multiplying both sides of the equation (vi) by ?, we have

P1 + ?gh1 + 1/2 ?v12 = P2 + ?gh2 + 1/2 ?v22 ... (vii)

In this equation, each term has got dimension of pressure.

Again, dividing both sides of the equation (vi) by g, we have

P1/?g + h1 + v12/2g = P2/?g + h2 + v22/2g ..... (viii)

In this equation, each term has got dimensions of length. The terms P/?g, h and v2/2g are called pressure head, elevation (or gravitational) head and velocity head respectively.

When the liquid flows through a horizontal pipe (h1 = h2), then the equation (vi) becomes

P1/ + 1/2 v12 + P2/ + 1/2 v22