Question

A liquid of density $750kg{m}^{-3}$ 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 $4500Pa$. The rate of flow of liquid is ________$\times 10^{-3}{m}^3{s}^{-1}$.

Answer 1

Answer: 24

$A_2=\frac{A_1}{2}$
$P_1-P_2=4500Pa$
$P_1+\frac{1}{2}\rho V_1^2+\rho gh=P_2+\frac{1}{2}\rho V_2^2+\rho gh$
$P_1-P_2=\frac{1}{2}\rho \left(V_2^2-V_1^2\right)....$(1)
And $A_1{V}_1=A_2{V}_2$
$\Rightarrow V_2=2V_1....$(2)
$4500=\frac{1}{2}\times 750\times 3V_1^2$
$V_1=2m/s$
Volume flow rate $=A_1{V}_1=24\times 10^{-3}{m}^3{s}^{-1}$