Question

State barnoles theorem. Prove that the total energy cause by a flowering ideal liquid is conserved

Answer 1

Bernoulli's theorem

When an incompressible and non-viscous liquid (or gas) flows in stream-lined motion from one place to another. then at every point of its path the total energy per unit volume (pressure energy + kinetic energy + potential eneryy) is constant. That is,

$\: \: \: \: \sf P + \dfrac{1}{2} \rho {v}^{2} + \rho \: gh = constant$

Proof -

Suppose an incompressible and non-viscous liquid is flowing in stream-lined motion through a tube XY of non-uniform cross-section. Let A1 and A2 be the areas of cross-section of the tube at the ends Yand Y respectively which are at heights h1 and h2 from the surface of the carth. Let P1 be the pressure and v1 the velocity of flow of the liquid at X; and P2 the respective quantities at Y. Since, the area A2 is smaller than A1 ne velocity v1 is greater than v2 (principle of continuity).

The liquid which enters at X travels a distance v1 in 1 second. On this liquid is acting a pressure force P1 x A1 (pressure x area).

∴ Work done per second on the liquid entering the tube at X is

$\sf \: force \times distance = P1 \times A1 \times v1$

Similarty, work done per second against the force P2×A2 by the iquid leaving the tube at Y is

$\sf P2 \times A2 \times v2$

∴ net work done on the liquid = (P1 × A1 × v1) - (P2 × A2 × v2).

But A1 v1 and A2 v2 are respectively the volumes of the liquid entering a X and living at Y per second. These volumes must be equal and so

$\sf \: A1 \times v1 = A2 \times v2 = \dfrac{m}{ \rho}$

where m is the mass of the liquid entering at X or leaving at Y in 1 second, and ρ is the density of the liquid. Substituting this in eq. (1), we have

net work done on the liquid = (P1 - P2)$\sf\dfrac{m}{\rho}$

The kinetic energy of the liquid entering at X in 1 second is $\dfrac{1}{2} m {v _{1} }^{2}$ and that of the liquid leaving at Y in 1 second is $\dfrac{1}{2} m {v _{2} }^{2}$

∴ increase in kinetic energy of the liquid =

$\sf \: \dfrac{1}{2} m( {v _{2} }^{2} - {v _{1} }^{2} )$

The potential energy of the liquid at X is mgh1 and that at Y is mgh2.

∴ decrease in potential energy of the liquid = mg(h1-h2).

Thus, net increase in the energy of the liquid =

$\sf \: \dfrac{1}{2} m( {v_{2} }^{2} - {v _{1} }^{2} ) - mg( h_{1} - h_{2})$

This increase in energy is due to the net work done on the liquid :

∴ net work done = net increase in energy

Hence by eq. (ii) and eq. (iii) we can write

${ \implies\sf \:( P_{1} - P _{2}) \dfrac{m}{ \rho} = \dfrac{1}{2} m( {v _{2} }^{2} - {v _{2} }^{2} ) \: - \: mg( h_{1} - h_{2})}$

${ \implies\sf \:( P_{1} - P _{2}) = \dfrac{1}{2} \rho( {v _{2} }^{2} - {v _{2} }^{2} ) \: - \: \rho \: g( h_{1} - h_{2})}$

${ \implies\sf \: P _{1} + \dfrac{1}{2} \rho {v_{1}}^{2} + \rho \: gh_{1} \: = \sf \: P _{2} + \dfrac{1}{2} \rho {v_{2}}^{2} + \rho \: gh_{2}}$

Hence

${ \boxed{\sf \dag\: P + \dfrac{1}{2} \rho {v_{}}^{2} + \rho \: gh_{} = constant \: \dag}}$