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Derive an expression for the kinetic energy, potential energy and pressure energy per unit mass of a liquid in a steady flow. |
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Answer» Solution :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 heighth 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` |
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