Derive an expression for the kinetic energy, potential energy and pressure energy per unit mass of a liquid in a steady flow.

Explanation

When a liquid is in a steady flow can possess three kinds of energy. They are (i) Kinetic energy, (ii) Potential energy, and (iii) Pressure energy, respectively.
(i) Kinetic energy: The kinetic energy of a liquid of mass m moving with a velocity v is given by
`KE=(1)/(2)mv^(2)`
The kinetic energy per unit mass `=(KE)/(m)`
`=((1)/(2)mv^(2))/(m)=(1)/(2)v^(2)`
The kinetic energy per unit volume
`=(KE)/("volume")=((1)/(2)mv^(2))/(V)=(1)/(2)((m)/(v))v^(2)`
`=(1)/(2)rhov^(2)`
(ii) Potential energy: The potential energ of a liquid of mass m at a height h above the ground level is given by
`PE=mgh`
The potential energy per unit mass
`=(PE)/(m)=(mgh)/(m)=gh`
The potential energy per unit
`"volume"=(PE)/("volume")=(mgh)/(V)=((m)/(V))gh`
`=rhogh`
(iii) Pressure energy: The energy acquired by a fluid by applying pressure of the fluid. We know that
`"Pressure"=("Force")/("Area")`
`rArr" ""Force"="Pressure"xx"Area"`
`Fxxd=(PA)xxd=P(Axxd)`
`rArr" "Fxxd=W=PV="pressure energy"`
Therefore, pressure energy, `E_(p)=PV`
The pressure energy per unit mass `=(E_p)/(m)`
`=(PV)/(m)`
`=(P)/(m//V)`
`=(P)/(rho)`
Similarly, the potential energy per unit
`"Volume"=(E_p)/("volume")=(PV)/(V)=P`