Glycerine flows steadily through a horizontal tube of length $1.5 ext{ m}$ and radius $1.0 ext{ cm

Question

Glycerine flows steadily through a horizontal tube of length $1.5	ext{ m}$ and radius $1.0	ext{ cm}$. The amount of glycerine collected per second at one end is $4.0 	imes 10^{-3}	ext{ kg/s}$. What is the pressure difference between the two ends? (Density = $1.3 	imes 10^3	ext{ kg/m}^3$, viscosity = $0.83	ext{ Pa s}$)

Professor PC Thomas Classes · Answer

Volume flow rate $Q = \frac{4.0 	imes 10^{-3}}{1.3 	imes 10^3} \approx 3.08 	imes 10^{-6}	ext{ m}^3	ext{/s}$. By Poiseuille's law, $Q = \frac{\pi \Delta P r^4}{8 \eta L} \implies \Delta P = \frac{8 \eta L Q}{\pi r^4} = \frac{8 	imes 0.83 	imes 1.5 	imes 3.08 	imes 10^{-6}}{3.14 	imes (1.0 	imes 10^{-2})^4} \approx 9.8 	imes 10^2	ext{ Pa}$.