Question

A section of blood vessel wall has become weakened resulting in an increase in the vessel wall diameter of 32%. For this question treat blood as an incompressible non viscous fluid of density 1050 kg m−3. If the vessel normally carries blood at a speed of 1.2 m s−1 then what is the magnitude of the increase in pressure in the bulge in the blood vessel? (to 2 s.f and in Pa)

Answer 1

Density of blood = $\rho$ = 1050 kg/m³

Initial diameter of the vessel wall = d₁

New diameter of the vessel wall = d₂

A section of blood vessel wall has become weakened resulting in an increase in the vessel wall diameter of 32%

d₂ = 1.32d₁

Initial cross-sectional area = A₁

A₁ = $\pi$d₁²/4

New cross-sectional area = A₂

A₂ = $\pi$d₂²/4

Initial speed of blood = V₁ = 1.2 m/s

New speed of blood = V₂

The flow rate of blood remains same

A₁V₁ = A₂V₂

($\pi$d₁²/4)V₁ = ($\pi$d₂²/4)V₂

d₁²V₁ = d₂²V₂

d₁²(1.2) = (1.32d₁)²V₂

V₂ = 0.689 m/s

Initial pressure of blood in the flow = P₁

New pressure of blood in the flow = P₂

Increase in the pressure = $\Delta$P = P₂ - P₁

By bernoulli's equation,

$P_{1} + \frac{\rho V_{1}^{2}}{2} = P_{2} + \frac{\rho V_{2}^{2}}{2}$

$P_{2} - P_{1} = \frac{\rho(V_{1}^{2} - V_{2}^{2})}{2}$

$\Delta P = \frac{(1050)((1.2)^{2} - (0.689)^{2})}{2}$

$\Delta$P = 506.77 Pa

Increase in magnitude of pressure in the bulge of the blood vessel = 506.77 Pa