Question 57 of 192

Glycerin of density $1.25\times {10}^{3}kg{m}^{–3}$ is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends are $10{cm}^{2}$ and $5{cm}^{2}$ and pressure drop across its length is $3N{m}^{–2}$. The rate of flow of glycerine through the pipe is $x\times {10}^{–5}{m}^{3}{s}^{–1}$. The value of $x$ is _____.

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Answer: 4

It is given that
${A}_{1}=10{cm}^{2};{A}_{2}=5{cm}^{2}$
$∆P=3N{m}^{-2}$
By continuity equation,
${A}_{1}{v}_{1}={A}_{2}{v}_{2} \Rightarrow {v}_{1}=\dfrac{{A}_{2}{v}_{2}}{{A}_{1}}...(i)$
By Bernoulli's equation,
$\dfrac{1}{2}\rho {{v}_{2}}^{2}-\dfrac{1}{2}\rho {{v}_{1}}^{2}=∆P...(ii)$
Using equation (i) in (ii)
$∆P=\dfrac{\rho }{2}({{v}^{2}}_{2}-{\left(\dfrac{{A}_{2}{v}_{2}}{{A}_{1}}\right)}^{2}) \Rightarrow ∆P=\dfrac{\rho {{v}^{2}}_{2}}{2}\left(1-{\left(\dfrac{{A}_{2}}{{A}_{1}}\right)}^{2}\right) \Rightarrow 3=\dfrac{1.25\times {10}^{3}\times {{v}^{2}}_{2}}{2}\left(1-\dfrac{1}{4}\right) \Rightarrow {{v}^{2}}_{2}=\dfrac{6\times 4}{1.25\times {10}^{3}\times 3} \Rightarrow {v}_{2}=8\times {10}^{-2}m{s}^{-1}$
So, the discharge rate is,
${A}_{2}{v}_{2}=x\times {10}^{-5} \Rightarrow x=\dfrac{5\times {10}^{-4}\times 8\times {10}^{-2}}{{10}^{-5}}=4$