Answer:
The flow rate is [tex]9.7\times 10^{- 11}\ m^{3}/s[/tex]
Solution:
As per the question:
Pressure drop, = 1.45 kPa = 1450 Pa
Radius of the artery, R = [tex]2.50\times 10^{- 5}\ m[/tex]
length of the artery, L = [tex]1.10\times 10^{- 3}\ m[/tex]
Temperature, T = [tex]37^{\circ}C[/tex]
Viscosity, [tex]\eta = 2.084\times 10^{- 3}\ Pa.s[/tex]
Now,
The flow rate is given by:
[tex]Q = \frac{\pi R^{4}P}{8\eta L}[/tex]
[tex]Q = \frac{\pi (2.50\times 10^{- 5})^{4}\times 1450}{8\times 2.084\times 10^{- 3}\times 1.10\times 10^{- 3}} = 9.7\times 10^{- 11}\ m^{3}/s[/tex]
