Question

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 the figure, The area of cross-section of the tube at two points P and Q at heights of $2m$and $5m$ are respectively $4\times {10}^{-3}{m}^2$ and $8\times {10}^{-3}{m}^2$. 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 1

Given: $A_1=4\times {10}^{-3}$ m${}^2$, $A_2=8\times {10}^{-3}$ m${}^2$
$h_1=2$ m, $h_2=5$ m
$v_1=1$ m/s
and $\rho ={10}^3$ kg/m${}^3$
From continuity equation, we have
$A_1{v}_1=A_2{v}_2$
or $v_2=\left(\frac{A_1}{A_2}\right)v_1$

or $v_2=\left(\frac{4\times {10}^{-3}}{8\times {10}^{-3}}\right)$ (m/s)

$v_2=\frac{1}{2}$ m/s

Applying Bernoulli's equation at sections 1 and 2
$p_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$
or
$P_1-P_2=\rho g(h_2-h_1)+\frac{1}{2}\rho (v_2^2-v_1^2)$ (i)

(i) Work done per unit volume by the pressure as the fluid flows from P to Q.
$W_1=p_1-p_2$

$=\rho g(h_2-h_1)+\frac{1}{2}\rho (v_2^2-v_1^2)$ [From Eq. (i)]

$=\{\left({10}^3\right)\left(9.8\right)(5-2)+\frac{1}{2}{\left(10\right)}^3(\frac{1}{4}-1)\}$ J/m${}^3$

$=[29400-375]$ J/m${}^3$=$29025$ J/m${}^3$

(ii) Work done per unit volume by the gravity as the fluid flows from P to Q.

$W_2=\rho g(h_2-h_1)=\left\{\left({10}^3\right)\left(9.8\right)(5-2)\right\}$ J/m${}^3$

or $W_2=29400$ J/m${}^3$