A non-viscous liquid of constant density `1000 kg//m^(3)` flows in a streamline notion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross-section of the tube at two points `P` and `Q` at heights of 2m and 5m are respectively `4xx10^(3)m^(2)` and `8xx10^(-3)m^(2)`. The velocity of the liquid at point P is `1m//s`. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q. Take `g=9.8m//s^(2)`.

Explanation

From equation of continuity at points `P` and `Q` we have
`a_(1)v_(1)=a_(2)v_(2)`
`v_(2)=(a_(1))/(a_(2))v_(1)=(4xx10^(-3))/(8xx10^(-3))xx10m//s=0.5m//s`
From Bernoulli's theorem at points `P` and `Q`
`p_(1)+1/2rhov_(1)^(2)=rhogh_(1)=p_(2)+1/2rhov_(2)^(2)+rhogh_(2)`
Each term in the equation represents energy per unit volume.
Work done by pressuure per unit volume is
`p_(1)-p_(2)=1/2 rho(v_(2)^(2)-v_(1)^(2))+rhog(h_(2)-h_(1))`
`1/2xx(1000)(0.5^(2)-1^(2))+1000xx10(5/2)`
`=1/2xx(1000)xx0.75+1000xx10xx3`
`=-30.375xx10^(3)`
`=3.0375xx10^(4)J//m^(3)`
Work done by gravity `=rhogh_(1)-rhogh_(2)`
`=rhog(h_(1)-h_(2))`
`=1000xx10xx2.5`
`=-2.5xx10^(4)J/m^(3)`