Questions: Problem 09.084 - Average speed of blood flow The average speed of blood in the aorta is 0.344 m / s, and the radius of the aorta is 1.00 cm . There are about 2.00 x 10^9 capillaries with an average radius of 6.50 μm. What is the approximate average speed of the blood flow in the capillaries? mm / s

Solution Steps

The cross-sectional area \( A \) of a circle is given by: \[ A = \pi r^2 \] where \( r \) is the radius.

For the aorta: \[ r_{\text{aorta}} = 1.00 \, \text{cm} = 0.0100 \, \text{m} \] \[ A_{\text{aorta}} = \pi (0.0100 \, \text{m})^2 = \pi (0.0001 \, \text{m}^2) = 0.0003142 \, \text{m}^2 \]

The radius of a capillary is: \[ r_{\text{capillary}} = 6.50 \, \mu\text{m} = 6.50 \times 10^{-6} \, \text{m} \]

The cross-sectional area of one capillary is: \[ A_{\text{capillary}} = \pi (6.50 \times 10^{-6} \, \text{m})^2 = \pi (4.225 \times 10^{-11} \, \text{m}^2) = 1.327 \times 10^{-10} \, \text{m}^2 \]

The total cross-sectional area of all capillaries is: \[ A_{\text{total capillaries}} = 2.00 \times 10^9 \times 1.327 \times 10^{-10} \, \text{m}^2 = 0.2654 \, \text{m}^2 \]

The volume flow rate \( Q \) must be the same in the aorta and the capillaries. The volume flow rate is given by: \[ Q = A \cdot v \] where \( v \) is the average speed.

For the aorta: \[ Q_{\text{aorta}} = A_{\text{aorta}} \cdot v_{\text{aorta}} = 0.0003142 \, \text{m}^2 \cdot 0.344 \, \text{m/s} = 0.0001081 \, \text{m}^3/\text{s} \]

For the capillaries: \[ Q_{\text{capillaries}} = A_{\text{total capillaries}} \cdot v_{\text{capillaries}} \]

Since \( Q_{\text{aorta}} = Q_{\text{capillaries}} \): \[ 0.0001081 \, \text{m}^3/\text{s} = 0.2654 \, \text{m}^2 \cdot v_{\text{capillaries}} \]

Solving for \( v_{\text{capillaries}} \): \[ v_{\text{capillaries}} = \frac{0.0001081 \, \text{m}^3/\text{s}}{0.2654 \, \text{m}^2} = 0.0004074 \, \text{m/s} \]

\[ v_{\text{capillaries}} = 0.0004074 \, \text{m/s} \times 1000 \, \text{mm/m} = 0.4074 \, \text{mm/s} \]

Final Answer