Question

An ideal fluid of density $800kg{m}^{-3}$, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $\frac{a}{2}$. The pressure difference between the wide and narrow sections of pipe is $4100Pa$. At wider section, the velocity of fluid is $\frac{\sqrt{x}}{6}m{s}^{-1}$ for $x=$ (Given $g=10m^{-2}$ )

Answer 1

Answer: 363

From continuity equation
$a{v}_1=\frac{a}{2}{v}_2$
$v_2=2v_1$
From Bernoulli's theorem,
$P_1+\rho g{h}_1+\frac{1}{2}\rho v_1^2=P_2+\rho g{h}_2+\frac{1}{2}\rho v_2^2$
$P_1-P_2=\rho \left[\left(\frac{v_2^2-v_1^2}{2}\right)+g\left(h_2-h_1\right)\right]$
$4100=800\left[\left(\frac{4v_1^2-v_1^2}{2}\right)+10\times (0-1)\right]$
$\frac{41}{8}+10=\frac{3v_1^2}{2}$
$\frac{121}{8}\times \frac{2}{3}=v_1^2$
$v_1=\sqrt{\frac{I21}{4\times 3}\times \frac{3}{3}}$
$v_1=\frac{\sqrt{363}}{6}m/s$
$X=363.$