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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. |
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Answer» Correct Answer - `+29625J//m^(3),-30000J//m^(3)` From equation of continuity `A_(P)V_(P)=A_(Q)V_(Q)` `4xx10^(-3)xx1=8xx10^(-3)xxV_(Q)` `V_(Q)=(1)/(2)m//s` From `W_(g)=-DeltaU` `=-mg(h_(2)-h_(1))` work done by gravity per unit volume `=-rhog(h_(2)-h_(1))` `=-1000xx9.8xx(5-2)" "=-2.94xx10^(4)J//m^(3)" "=-29400J//m^(3)` change in kinetic energy per unit volume `=(1)/(2)p(V_(Q)^(2)-v _(P)^(2))` `Delta K=(1)/(2)xx1000xx[((1)/(2))^(2)-1]` `=-375 J` work done by all forces per unit voume `=DeltaK` work done by pressure perunit volume `+` work done by gravity per unit volume `=DeltaK` work done by pressure per unit volume `-29400=-375` work done by pressure per unit voume `=29025J//m^(3)` |
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