Question 59 of 192

A liquid of density $750kg{m}^{-3}$ flows smoothly through a horizontal pipe that tapers in cross-sectional area from ${A}_{1}=1.2\times {10}^{-2}{m}^{2}$ to ${A}_{2}=\dfrac{{A}_{1}}{2}$. The pressure difference between the wide and narrow sections of the pipe is $4500Pa$. The rate of flow of liquid is _____ $\times {10}^{-3}{m}^{3}{s}^{-1}$.

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Answer: 24

From the equation of continuity, ${v}_{2}=2{v}_{1}$.
Applying the Bernoulli's Theorem,
${P}_{1}+\dfrac{\rho {{v}_{1}}^{2}}{2}={P}_{2}+\dfrac{\rho 4{\left({v}_{1}\right)}^{2}}{2}$ $\Rightarrow {P}_{1}-{P}_{2}=\dfrac{3\rho {{v}_{1}}^{2}}{2}$
$\Rightarrow 4500=\dfrac{3\times 750\times {{v}_{1}}^{2}}{2}\Rightarrow {v}_{1}=2m{s}^{-1}$
Now volume flow rate,$Q=A{v}_{1}=1.2\times {10}^{-2}\times 2=24\times {10}^{-3}{m}^{3}{s}^{-1}$