Question

If the flow rate in an intravenous (IV) setup is 2.0 cm³/min, how long does it take to empty a 1.0-liter bottle of IV solution?


A glucose solution is being administered intraveneously with a flow rate of 3.0 cm³/min. If the glucose solution is replaced by blood plasma having a viscosity of 1.5 times that of glucose, what is the new flow rate?


The aorta is 1.0 cm in radius and the average speed of blood in the aorta is 30 cm/s in a normal resting adult. Calculate the blood flow rate in the aorta in cubic centimeters per second and liters per minute.


Calculate the flow rate of blood through a capillary in cubic centimeters per second, given that the capillary has a diameter of 4.0 x 10-4cm and the average speed of the blood in the capillary is 3.0 x 10-2cm/s.


Calculate the Reynolds number for flow in the needle considered in Example 12.8 to verify the assumption that the flow is laminar. Assume that the density of the saline solution is 1025 kg/m³.(v=1.70 m/s, daimeter =0.150×10-3m and n= 1.00×10-3Ns/m²)


Blood is pumped from the heart at a rate of 5.0 L/min into the aorta (of radius 1.0 cm). Determine the speed of blood through the aorta.


Blood is flowing through an artery of radius 2 mm at a rate of 40 cm/s.Determine the flow rate and the volume that passes through the artery in a period of 30 s.


The pressure drop along a length of artery is 100 Pa, the radius is 10mm, and the flow is laminar. The average speed of the blood is 15 mm/s.(a) What is the net force on the blood in this section of artery? (b) What is the power expended maintaining the flow?

Answer 1

1000 cm³ / 2.0 cm³/min = 500 minutes


Viscosity of blood plasma = 1.5 x 3.0 cm³/min = 4.5 cm³/min


Blood flow rate = π(1.0 cm)² x 30 cm/s = 30π cm³/s = 94.25 cm³/s = 0.09425 L/s = 5.65 L/min


Flow rate = π(4.0 x 10-4 cm / 2)² x 3.0 x 10-2 cm/s = 1.13 x 10-10 cm³/s


Reynolds number = (1025 kg/m³)(1.50 x 10-3 m)(1.70 m/s)/(1.00 x 10-3 Ns/m²) = 2.44


Speed of blood = (5.0 L/min x 1000 cm³/L) / (π(1.0 cm)²) = 159.15 cm/s


Flow rate = π(2 mm)² x 40 cm/s = 502.65 mm³/s = 0.50265 cm³/s
Volume = 0.50265 cm³/s x 30 s = 15.08 cm³


(a) Net force = 100 Pa x π(10 mm)² = 0.0314 N
(b) Power = force x speed = 0.0314 N x 15 mm/s = 0.471 W