Question

Liquid mercury (of viscosity $1.55\times {10}^{-3}N-s/m^2$) flows through a horizontal pipe of internal radius $1.88cm$ and length $1.26m.$ The volume flux is $5.35\times {10}^{-2}L/min.$ Then

(Given critical value of Reynold's number is $2000$, density of mercury is $13600kg/m^3$)

• A. the flow is turbulent
• B. the flow is laminar
• C. the difference in pressure between the two ends of pipe is $0.0355Pa.$
• D. the difference in pressure between the two ends of pipe is zero

Answer 1

Answer: (B), (C)

The correct options are
B the difference in pressure between the two ends of pipe is $0.0355Pa.$
C the flow is laminar

Average speed of fluid in piple $=v=\frac{5.35\times {10}^{-2}}{\pi (1.88{)}^2}$

= 4.82 x ${10}^{-3}$ lit/min.cm${}^2$

$\Rightarrow v=0.8\times {10}^{-3}m/s$

($\because 1L=1000c{m}^3$, $1min=60s$)

Reynold's number R = $\frac{\rho Dv}{\eta }=\frac{\left(13600\right)\left(0.0376\right)(0.8\times {10}^{-3})}{(1.55\times {10}^{-3})}$

$=264<<$critical value

$\Rightarrow$ It is laminar.

By Poiseuille's law, $\Delta P=\frac{8\eta L}{\pi r^4}\frac{dV}{dt}(\because \rho \frac{dV}{dt}=\frac{dm}{dt}$ when $\rho$ is constant )

= $\frac{8(1.55\times {10}^3)\left(1.26\right)}{\pi (1.88\times {10}^2{)}^4}(8.92\times {10}^{-7})=0.0355Pa$