Question

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

Answer 1

$A _{2}=\frac{ A _{1}}{2} $
$ P _{1}- P _{2}=4500\, Pa $
$ P _{1}+\frac{1}{2} \rho V _{1}^{2}+\rho gh = P _{2}+\frac{1}{2} \rho V _{2}^{2}+\rho gh$
$P _{1}- P _{2}=\frac{1}{2} \rho\left( V _{2}^{2}- V _{1}^{2}\right) ....$(1)
And $A _{1} V _{1}= A _{2} V _{2}$
$\Rightarrow V_{2}=2 V_{1} ....$(2)
$4500=\frac{1}{2} \times 750 \times 3 V _{1}^{2}$
$V _{1}=2\, m / s$
Volume flow rate $= A _{1} V _{1}=24 \times 10^{-3} m ^{3} s ^{-1}$