Question

The blood speed in a normal segment of a horizontal artery is 0.16 m/s. An abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to one-eighth the normal cross-sectional area. What is the difference in blood pressures between the normal and constricted segments of the artery?

Answer 1

Answer:

The difference in blood  pressures is $P_1 -P_2 = 854.8 \ Pa$

Explanation:

From the question we are told that

  The blood speed in a normal segment is  $v_1 = 0.16 \ m/s$

   The area  of the normal segment is  $A_1$

   The area of the abnormal segment is  $A_2 = \frac{A_1}{8}$

   

Generally from continuity equation we have that

     $v_1 A_1 = v_2 A_2$

=>    $v_1 A_1 = v_2 * \frac{A_1}{8}$

=>    $0.16 = v_2 * \frac{1}{8}$

=>    $v_2 = 1.28 \ m/s$

Generally from Bernoulli's equation the difference in pressure is mathematically represented as

    $P_1 -P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2 )$

Here  $\rho$ is the density of blood with value  $\rho = 1060 \ kg /m^3$

   $P_1 -P_2 = \frac{1}{2} (1060) ( 1.28 ^2 - 0.16 ^2 )$

=> $P_1 -P_2 = 854.8 \ Pa$