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17 March, 14:58

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).

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  1. 17 March, 18:35
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    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)
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