Glycerin,with a viscosity of 1.50 N • s/m²,is flowing with an average speed of 1.20 m/s through a tube that has a radius of 5.00 mm and is 25.0 cm long.What is the drop in pressure between the ends of the tube? A) 144 kPa B) 45.8 kPa C) 72.0 kPa D) 76.3 kPa E) 129 kPa

Question

Quizplus · Accepted Answer

The pressure drop is given by the Hagen-Poiseuille equation:$$\Delta P = \frac{8\mu L Q}{\pi r^4}$$where:* $\Delta P$ is the pressure drop in pascals (Pa)* $\mu$ is the viscosity of the fluid in pascal-seconds (Pa·s)* $L$ is the length of the tube in meters (m)* $Q$ is the volume flow rate in cubic meters per second (m³/s)* $r$ is the radius of the tube in meters (m)We can calculate the volume flow rate from the average speed and the cross-sectional area of the tube:$$Q = Av = \pi r^2 v$$where:* $A$ is the cross-sectional area of the tube in square meters (m²)* $v$ is the average speed of the fluid in meters per second (m/s)Substituting the given values into the Hagen-Poiseuille equation, we get:$$\Delta P = \frac{8\mu L Q}{\pi r^4} = \frac{8(1.50 	ext{ N} \cdot 	ext{s/m}^2)(0.250 	ext{ m})(\pi (5.00 	imes 10^{-3} 	ext{ m})^2 (1.20 	ext{ m/s}))}{\pi (5.00 	imes 10^{-3} 	ext{ m})^4} = 144 	ext{ kPa}$$Therefore, the drop in pressure between the ends of the tube is 144 kPa.

Glycerin,with a Viscosity of 1