A non-viscous liquid of constant density `1000kg//m^3` flows in a streamline motion along a tube of variable cross section. The tube is kept inclined in the vertical plane as shown in Figure. The area of cross section of the tube two point P and Q at heights of 2 metres and 5 metres are respectively `4xx10^-3m^2` and `8xx10^-3m^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.

Explanation

Continuity equation between `P` and `Q`
`a_1 v_1 = a_2 v_2`
`4 xx 10^-3 xx 1 = 8 xx 10^-3 v_2 rArr v_2 = 0.5 m//s`
Applying Bernoulli's theorem between `P` and `Q`
`P_1 + (1)/(2) P v_1^2 + rho gh_1 = P_2 + (1)/(2) rho v_2^2 + rho gh_2`
`P_1 -P_2 = (1)/(2) rho(v_2^2 - v_1^2)+ rho g(h_2 - h_1)`
=`(1)/(2) xx 1000(0.5^2 - 1^2)+ 1000 xx 10(5 -2)`
=`-375 + 30,000 = 29,625 N//m^2`
W.D. by pressure per unit volume `= P_1 - P_2 = 29,625 J//m^3`
W.D. by gravity `= rho g(h_1 - h_2) = 1000 xx 10 (2 - 5)`
=`- 30,000 J//m^3`.
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