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)`.

Options

A. zero
B. ` 29400 J//m^3 `
C. `-29400 J//m^3 `
D. `29025 J//m^ 3 `

Answer

D

Explanation

Given ` A_1 = 4 xx 10 ^(-3) m ^ 2, A _ 2 = 8 xx 10 ^(-3) m^2, h_1 = 2 m , 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))) (1m//s) rArr v _ 2 = (1)/(2) m//s `
Applying Bernoulli's equation at sections 1 and 2
` p_1 + (1)/(2) rho v _ 1 ^ 2 + rho g h _ 1 = p_2 + (1)/(2) rho v_2 ^ 2 + rho gh _ 2 `
or ` p _ 1 - p _ 2 = rho g (h _ 2 - h _ 1 ) + (1)/(2) rho (v_2 ^ 2 - v _ 1 ^2 ) " "`...(i)
(a) Work done per unit volume by the pressure as the fluid flows from P to Q.
` W _ 1 = p_1 - p_2 = rho h ( h _ 2 - h _ 1 ) + (1)/(2) rho (v _ 2 ^2 - v _ 1 ^ 2 ) " " ` [From equation (i) ]
` = { (10 ) ^3 (9.8) (5-2) + (1)/(2) (10 ) ^3 ((1)/(4) - 1))} = [24900- 375 ] = 29025 J//m^3 `