Question

Oxygen-rich blood bound for the body tissues leaves the heart through the aorta, the major blood vessel emerging from the heart. It branches out to the arteries, which in turn branch into smaller vessels called arterioles (little arteries). These then branch into tiny, thin-walled capillaries. Blood travels slowly as the red blood cells squeeze through tiny capillaries. The radius of the aorta is approximately 9.8 mm and the speed of the blood there is 30 cm/s. A typical capillary has a diameter of approximately 6 µm. The flow speed in these capillaries is approximately 0.05 cm/s.

a. Draw a diagram of the blood vessels and indicate on the diagram the comparative flow speeds in these different-sized blood vessels. Discuss the direction and speed expected for the flow and the continuity equation for this system.

b. Using the given data, calculate how many capillaries a typical person would have.

c. What is the relationship between the combined cross-sectional area of the capillaries and the cross-sectional area of the aorta?

Answer 1

Let
R1 = 9.8 mm = 0.0098 m
v1 = 30 cm/s = 0.3 m/s

d2 = 6 micro m = 6*10^-6 m
v2 = 0.05 cm/s = 5*10^-4 m/s

a)

b) Let N is the number of capallaries that a typical person would have

use continuty equation,

A1*v1 = A2*v2

(pi*R1^2)*v1 = N*(pi*d2^2/4)*v2

==> N = 4*v1/(d2^2*v2)

= 4*0.3/((6*10^-6)^2*5*10^-4)

= 6.67*10^13 <<<<<<<--------------Answer

c) combined cross sectional area of the capaillaries, A2 = N*(pi*d2^2/4)

= 6.67*10^13*(pi*(6*10^-6)^2/4)

= 1886 m^2

cross sectional area of the aorta, A1 = pi*R1^2

= pi*(9.8*10^-3)^2

= 3.02*10^-4 m^2

so, the combined cross-sectional area of the capillaries = (1886/(3.02*10^-4))* the cross-sectional area of the aorta

= 6.24*10^6*the cross-sectional area of the aorta <<<<<<--------Answer