Question

Compute the decrease in the blood pressure of the blood flowing through an artery the radius of which is constricted by a factor of 3. Assume that the average flow velocity in the unconstricted region is 50 cm/sec.

Answer 1

Begin with Bernoulli's equation which is ..

P1 + 1/2 p v^2 + pgh = P2 + 1/2 pv2 + pgh

pgh does not enter into into it. h is the same.

P1 - P2 = 1/2p(v2)^2 - 1/2 p (v1)^2

P1 - P2 = 1/2 p(v2)^2 [1 - ( (v1/v2)^2 ]

switch our attention to the principle of Continuity which says

A1v1 = A2V2

A2/A1 = v1/v2

Then......

(A2/A1)^2 = (v1/v2)^2

P2 - P1 = 1/2 p(v2)^2 [ ( (A2 / A1)^2 - 1 ]

Put values

P2 - P1 = 0.5 * 1025 kg/m^3 * (0.50 m/s)^2 [9 - 1]

Calculate..

P2 - P1 = 1025 pa ANSWER..