A non-viscous liquid of constant density 1000 ~kg ~m^ -3 flows in… - Vidyadip

Question

A non-viscous liquid of constant density $1000 \mathrm{~kg} \mathrm{~m}^{-3}$ flows in a streamline motion along a tube of variable crosssection. The tube is a kept inclined in the vertical plane as shown in figure. The area of cross-section of the tube at two points $P$ and $Q$ at heights of 2 m and 5 m are respectively, $4 \times 10^{-3} \mathrm{~m}^2$ and $8 \times 10^{-3} \mathrm{~m}^2$. The velocity of the liquid at point $P$ is $1 \mathrm{~m} \mathrm{~s}^{-1}$. 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

Given, $\rho=100\text{kg/m}^3,\text{ v}_1=1\text{m/s,}$ $\text{a}_1=4\times10^{-3}\text{m}^2,$$\text{a}_2=8\times10^{-3}\text{m}^2,\text{ h}_1=2\text{m},$ $\text{ h}_2=5\text{m}$
Apply Bernoull's theorem,$\text{p}_1+\frac{1}{2}\rho\text{v}_1^2\rho\text{h}_1=\text{p}_2+\frac12\rho\text{v}^2_2+\text{g}\rho\text{h}_2$
$(\text{p}_1-\text{p}_2)=\frac12\rho(\text{v}^2_2-\text{v}^2_1)+\rho\text{g}(\text{h}_2-\text{h}_1)$
Where, $(p_1 - p_2)$ = Work done by pressure per unit volume, i.e. $\Big(\frac{\text{W}}{\text{Volume}}\Big)_\text{p}=\frac12\rho(\text{v}^2_2-\text{v}^2_1)+\rho\text{g}(\text{h}_2-\text{h}_1)\dots\text{(i})$

From equation of continuity,$\text{a}_1\text{v}_1=\text{a}_2\text{v}_2$
$\text{v}_2=\frac{\text{a}_1\text{v}_1}{\text{a}_2}$
$=\frac{4\times10^{-3}\times1}{8\times10^{-3}}=0.5\text{m/s}$
$\Big(\frac{\text{W}}{\text{Volume}}\Big)_\text{p}=\frac12\times1000[0.25-1]+1000\times10(5-2)$
$=-375+30,000$
$=29,625\text{J/m}^3$
Work done per unit volume by the gravitational force,$=\rho\text{g}(\text{h}_1-\text{h}_2)$
$=100\times10(2-5)$
$=-3\times10^4\text{J/m}^3$

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