Question

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 1.20×10⁴ Pa , while in the region of blockage it is 1.15×10⁴ Pa . Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 33.0 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?

Answer 1

Re-arranging the Bernoulli’s equation -

p1 / w + v1^2/2g = p2/w + v2^2/2g --------------------------------------(i)

Where –

w = specific weight = 1.06 x 10^4 n/m^3

v1 = 33.0 cm/s = 0.33 m/s

p1 = 1.20 x 10^4 Pa

p2 = 1.15 x 10^4 Pa

Put the above values in equation (i) -

(1.2 x 10^4)/ (1.06 x 10^4) + 0.33^2/(2 x 9.81) = (1.15 x 10^4) /(1.06x10^4)+v2^2/(2 x 9.81)

=> 1.132 + 0.0056 = 1.085 + v^2/19.62

=> v^2/19.62 = 0.0526

=> v^2 = 0.0526 x 19.62

=> v2 = 1.016 m/s

Now -

Q= A1v1 = A2v2

=> A2/A1 = v1/v2 = 0.33 / 1.016 = 0.325

Therefore, percentage of the cross-sectional area of the patients artery blocked

                                                    = 100 – 32.5 = 67.5 % (Answer)