A medical technician is trying to determine what percentage of a patient’s arter

Question

A medical technician is trying to determine what percentage of a patient’s artery is blocked by plaque. To do this, he measures the blood pressure just before the region of blockage and finds that it is 1.20×10^4 Pa, whereas in the region of blockage it is 1.18×10^4 Pa . Furthermore, the technician knows that blood flowing through the normal artery just before the point of blockage is traveling at 30.0 cm/s , and the density of this patient’s blood is 1050 kg/m^3 .

What percentage of the cross-sectional area of the patient’s artery is blocked by the plaque?

Explanation / Answer

Given Initial pressure , P1 =1.20×10^4 Pa Final pressure , P2 =1.18×10^4 Pa Speed of the blood before the point of blockage is 30 cm /s Density of patient's blood is , = 1050 kg /m3    The height h1 = h2 According to Bernoulli's equation, P1 + 1/2 v12 = P2 + 1/2 v22 v2 = (2 (P1- P2) / ) +v12      = (2(1.20×10^4 Pa - 1.18×10^4 Pa) /1050 kg /m3 ) + (0.3 m/s)2 v2 = 0.68 m/s ------------------------------------------------------------------------- From equation of continuity A1v1 = A2 v2 A2 / A1 = 0.3 m/s / 0.68 m/s     A2 / A1 = 0.4411 = 44.11 % Percentage of cross sectional area blocked is 100% - 44.11% = 55.89 /% ˜ 56% v2 = (2 (P1- P2) / ) +v12      = (2(1.20×10^4 Pa - 1.18×10^4 Pa) /1050 kg /m3 ) + (0.3 m/s)2 v2 = 0.68 m/s ------------------------------------------------------------------------- From equation of continuity A1v1 = A2 v2 A2 / A1 = 0.3 m/s / 0.68 m/s     A2 / A1 = 0.4411 = 44.11 % Percentage of cross sectional area blocked is 100% - 44.11% = 55.89 /% ˜ 56% v2 = 0.68 m/s ------------------------------------------------------------------------- From equation of continuity A1v1 = A2 v2 A2 / A1 = 0.3 m/s / 0.68 m/s     A2 / A1 = 0.4411 = 44.11 % Percentage of cross sectional area blocked is 100% - 44.11% = 55.89 /% ˜ 56%

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