EXAMPLE • A patient lying horizontally on a hospital bed is found to have an

enlargement of his blood vessel where the walls have weakened. The

cross-sectional area of the enlargement is 2.0A where A is the cross-

section of the normal aorta. The normal speed of blood through the

person's aorta is 0.40 m/s, and the density of blood is 1,060 kgm-3

Calculate

. a) the speed of blood in the enlargement.

• b) how much higher the pressure is in the enlargement.

Explanation:

a ) The volume of blood flowing per second throughout the vessel is constant .

a₁ v₁ = a₂ v₂

a₁ and a₂ are cross sectional area at two places of vessel and v₁ and v₂ are velocity of blood at these places .

2A x v₁ = A x .40

v₁ = .20 m /s

b )

Let normal pressure be P₁ when cross sectional area is 2A and at cross sectional area A , pressure is P₂

Applying Bernoulli's theorem

P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²

P₁ - P₂ = 1/2  ρ(v₂² - v₁² )

= .5 x 1060 ( .4² - .2² )

= 63.6 Pa .