Bernoulli theorem: According to bernoulli the sum of pressure energy kinetic energy and ponteniel energey per unit mass of an incompressible, non viscous fluid in a streamlined flow remains a constatnt, Mathmatically, `(P)/(rho)+(1)/(2)v_(2)+gh+Constant`
Proof: Let us consinder a flow of liquid through a pipe AB. 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 leaving B in the same time . Lrt `a_(A),v_(A)` and `P_(2)` br the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respetivily.
Let the force exerted by the liquid at a is `F_(2)=P_(A) a_(A)`
Distance travlled by the liquid in the yime t is `d=v_(A)t`
Therefore the work done is `W=F_(2)d=P_(A)a_(A)v_(A)t`
But `a_(A)v_(A)t=a_(A)d=V`, Volime 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)`
Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
`EP_(A)=P_(A)V=xx(m/m)=m(P_(A))/(rho)`
Due to the flow of liquid, the kinectic energy of the liquid at A,
`KE_(A)=(1)/(2)mv_(A)^(2)`
Therefore the total energey due to the flow of liquid at A,
`E_(A)=EP_(A)+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 `F_(B)`, we get
`E_(B)=m(P_(B))/(rho)+(1)/(2)mv_(B)^(2)+mg h_(B)` `m(P_(A))/(rho)+(1)/(2)mV_(A)^(2)+mgh_(A)=m(P_(B))/(rho)+(1)/(2)mv_(B)^(2)+mgh_(B)`
`m(P_(A))/(rho)+(1)/(2)mV_(A)^(2)+mgh_(A)=m(P_(B))/(rho)+(1)/(2)v_(B)^(2)+gh_(B)=`Constant
Thus the above equation can be written as `(P)/(rhog)+(1)/(2)(v^(2))/(g)+h= `Constant
