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)`.

Answer

25

Explanation

Given : `A_(1) = 4 xx 10^(-3)m^(2), A_(2) = 8 xx 10^(-3)m^(2)`
`h_(1) = 2m, h_(2) = 5m`
`v_(1) = 1m//s` and `rho = 10^(3) kg//m^(3)`
From continuity equation, we have
`A_(1)v_(1) = A_(2)v_(2)` or `v_(2) = ((A_(1))/(A_(2))v_(1))`
or `v_(2) = ((4 xx 10^(-3))/(8 xx 10^(-3))) (1 m//s)`
`v_(2) = (1)/(2) m//s`
Applying Beronulli's equation at section `1` and `2`
`P_(1) + (1)/(2) rhov_(1)^(2) + rhogh_(1) = P_(2) + (1)/(2)rhov_(2)^(2) + rhogh_(2)`
or `P_(1) - P_(2) = rhog (h_(2) - h_(1)) + (1)/(2) rho (v_(2)^(2) - v_(1)^(2)).......(2)`

`(i)` Work done per unit volume by the pressure as the fluid flows form `P` to `Q`.
`W_(1) = P_(1) - P_(2)`
`= rhog (h_(2) - h_(1)) + (1)/(2)rho (v_(2)^(2) - v_(1)^(2))` [From Eq. `(1)`]
`= {(10^(3))(9.8)(5 - 2) + (1)/(2)(10^(3))((1)/(4)-1)}J//m^(2)`
`= [29400 - 375] J//m^(3)`
`= 29025 J//m^(3)`
`1161 alpha = 29025 J//m^(3)`
`alpha = 25`