The volume flow rate in an artery supplying the brain is 4.7 x 10-6 m3/s. if the radius of the artery is 4.5 mm, determine the average blood speed. (b) find the average blood speed as a constriction in the artery if the constriction reduces the radius by a factor of 3. assume that the volume flow rate is the same as that in part (a).

a) the blood speed is 0.0738 m/s = 7.38 cm/s

b) when the artery contracts by a factor of 3 , the velocity speeds up to a factor of 9 , and thus is equal to 0.6642 m/s =66.42 cm/s

Explanation:

Assuming that an artery has a constant radius ( or its change is negligible in the region of study), it can be modelled as a cylinder of length L with constant area A

Therefore its cross sectional area A (area perpendicular to the flow) is

A= πR²

and its volume is

V = A * L

since the flow rate Q= dV/dt  , the velocity is v=dL/dt  and A is constant

Q= dV/dt = d (A*L) / dt = A dL/dt = A * v

Q= A * v = πR² * v

therefore

v = Q / (πR²)

replacing values

v = Q / (πR²) = 4.7 x 10⁻⁶ m³/s / [π(4.5mm)²]  *(10⁶ mm²/m²) =  0.0738 m/s = 7.3 cm/s

b) if the volume flow is the same, the velocity (v₂) when the radius reduces to 3 is

v₂ = Q / (πR₂²)

v₁ = Q / (πR₁²)

dividing the equations

v₂/v₁   = Q / (πR₂²) / [Q / (πR₁²)] = (R₁/R₂)²

v₂/v₁= (R₁/R₂)²

since R₂=R₁/3

v₂/v₁=3²=9

therefore

v₂= 9*v₁ =9* 0.0738 m/s = 0.6642 m/s

speeds up due to the area contraction , because it has to cover more length in the same time in order to achieve the same volumetric flow rate

Note

- we assumed that velocity is uniform in the cross sectional area (planar velocity profile in the area) , which is a good approximation if the fluid is in a turbulent motion , but the velocity changes with the distance from the artery walls if it flows in laminar motion

- we assumed steady state ( Q and v does not change with time)

where is the circuit diagram ?

d eras and periods