Answer to Question #111429 in Physics for John Guz

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

Let us use Bernoulli's equation and the fact that the flux of the blood is continuous.

The flux of is time $\Delta t$ at two points with different diameter is equal, hence $S_1 v_1 \Delta t = S_2 v_2 \Delta t$, or in terms of diameter, $\pi \left(\frac{d_1}{2}\right)^2 v_1 \Delta t = \pi \left(\frac{d_2}{2}\right)^2 v_2 \Delta t$, from where $v_2 = \left(\frac{d_1}{d_2}\right)^2 v_1$, where $v_1$ is the speed before increase in diameter.

According to Bernoulli's equation: $p_1 + \rho g h_1 + \frac{\rho v_1^2}{2} = p_2 + \rho g h_2 + \frac{\rho v_2^2}{2}$. Since the height in both cases is the same, terms $\rho g h_1, \rho g h_2$ cancel and therefore $\Delta p = p_2 - p_1 = \frac{\rho}{2}[v_1^2 - v_2^2] = \frac{\rho v_1^2}{2}\left[1 - \left(\frac{d_1}{d_2} \right)^4\right]$ .

New diameter is $d_2 = (1+\eta)d_1$, where $\eta = 0.32$ (corresponds to 32%).

Hence, finally $\Delta p = \frac{\rho v_1^2}{2}\left[1 - \frac{1}{(1+\eta)^4}\right] \approx 506.9 Pa$.