Question 2.9: In a major human artery with an internal diameter of 5 mm, t...

The average velocity is given by the volumetric flow rate divided by the area for flow. Thus, upstream of the bifurcation, where the vessel diameter is 0.5 cm,

u_{up} = \frac{q}{A} = \frac{5  cm^{3} . s^{-1} }{(\pi /4)(0.5^{2}  cm^{2})} =25.5  cm.s^{-1}

Downstream of the bifurcation, the volumetric flow rate in each vessel is 2.5  cm^{3}·s^{−1}, and the vessel diameter is 0.3 cm. Thus,

u_{down} = \frac{2.5  m^{3} . s^{-1} }{(\pi /4)(0.3^{2}  cm^{2})} =35.4  cm.s^{-1}

The mass flow rate in the upstream vessel is given by the volumetric flow rate times the density:

\dot{m} _{UP} = 5  cm^{3}.s^{-1}\times 1.06  g⋅ cm^{-3} = 5.30  g.s^{-1}

Similarly, for each downstream vessel:

\dot{m} _{down} = 2.5  cm^{3}.s^{-3}\times 1.06  g⋅ cm^{-3} =2.65  g.s^{-1}

which is of course half the upstream value.