Bernoulli's theorem
According Bernoulli's theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non- viscous fluid in a streamlined flow remains a constant Mathematically,
`(P)/(rho)+(1)/(2)v^(2)+gh=` constant
This is known as Bernoulli's equation.
Proof :
Let us consider a flow of liquid through a pipe `AB` as shown in Figure. Let `V` be the volume of the liquid when it enters `A` in a time `t` which is equal to the volume of the liquid of the liquid leaving `B` in the same time. Let `a_(A)`, `v_(A)` and `P_(A)` be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at `A` respectiviely.
Let the force exerted by the liquid at `A` is
`F_(A)=P_(A)a_(A)`
Distance travelled by the liquid in time `t` is
`d=v_(A)t`
Therefore, the work done is
`W=F_(A)d=P_(A)a_(A)v_(A)t`
But `a_(A)v_(A)t=a_(A)d=V`, volume of the liquid entering at `A`.
Thus, the work done is the pressure energy
(at`A`), `W=F_(A)d=P_(A)V`
Pressure energy per unit volume at
`A=("Pressure energy")/("volume")=(P_(A)V)/(V)=P_(A)`
Pressure energy per unit mass at
`A=("Pressure energy")/("mass")=(P_(A)V)/(m)=(P_(A))/((m)/(V))=(P_(A))/(rho)`
Since `m` is the mass of the liquid entering at `A` in a given time, therefore, pressure energy of the liquid at `A` is
`E_(PA)=P_(A)V=P_(A)Vxx((m)/(m))=m(P_(A))/(rho)`
Potential energy of the liquid at `A`,
`PE_(A)=mg h_(A)`,
Due to the flow of liquid, the kinetic energy of the liquid at `A`,
`KE_(A)=(1)/(2)m v_(A)^(2)`
Therefore, the total energy due to the flow of liquid at `A`, `E_(A)=E_(PA)+KE_(A)+PE_(A)`
`E_(A)=m(P_(A))/(rho)+(1)/(2)mv_(A)^(2)+mg h_(A)`
Similarly, let `a_(B)`, `v_(B)` and `P_(B)` be the area of cross section of the tube, velocity of the liquid , and pressure exerted by the liquid at `B`. Calculating the total energy at `E_(B)`, we get
`E_(B)=m(P_(B))/(rho)+(1)/(2)mv_(B)^(2)+mgh_(B)`
From the law of conservation of energy,
`E_(A)=E_(B)`
`m(P_(A))/(rho)+(1)/(2)mv_(A)^(2)+mgh_(A)=m(P_(B))/(rho)+(1)/(2)mv_(B)^(2)+mgh_(B)`
`(P_(A))/(rho)+(1)/(2)v_(A)^(2)+gh_(A)=(P_(B))/(rho)+(1)/(2)v_(B)^(2)+gh_(B)=` constant
Thus, the above equation can be written as
`(P)/(rho g)+(1)/(2)(v^(2))/(g)+h=` constant.
