BIO Artery Blockage. A medical technician is trying to determine what percentage of a patient’s artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 1.20 X 104 Pa, while in the region of blockage it is 1.15 X 104 Pa. Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 30.0 cm/s, and the specific gravity of this patient’s blood is 1.06. What percentage of the cross-sectional area of the patient’s artery is blocked by the plaque?

Solution 40E This question can be solved by combining Bernoulli’s equation and continuity equation. Bernoulli’s principle states that the summation of pressure energy, kinetic energy per unit volume and potential energy per unit volume of a flowing fluid remains constant. 4 Given that, P 1 = 1.20 × 10 Pa v 1 30.0 cm/s = 0.3 m/s P = 1.15 × 10 Pa 2 3 3 Density of blood d blood= 1.06 × 1000 kg/m = 1060 kg/m We need to calculate v . 2 From Bernoulli’s equation, 1 2 1 2 P 1 + 2 blood× v 1 = P 2 + 2 blood× v 2 (Potential energy term is zero as the heart remains at the same height) 1.20 × 10 +4 1 × 1060 × (0.3) = 1.15 × 10 + 4 1 × 1060 × v 2 2 2 2 4 2 0.05 × 10 + 47.7 = 530 × v 2 v 2 1.02 m/s Now, applying continuity equation, we get A v1= 1 v 2 2 A /1 = 2 /v 2 1 A /A = 1.02/0.3 1 2 A /1 = 2.4 A /2 = 1.29 Therefore, 29% of the patient’s artery is open. So, the block percentage is = (100 29)% = 71% Therefore, 71% of the cross-sectional area of the patient’s artery is blocked by the plaque.