Answer:
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)
