Question

If cholesterol buildup reduces the diameter of an artery by 20%, by what % will the blood flow rate be reduced, assuming the same pressure difference?

delta Q / Q = ? %

Express answer using two significant figures.

Answer 1

Let the initial diameter is DA
The final diameter, DB = (100%-20%)DA
So, the final radius radiused to
RB = (100%-20%)RA
= (80%)RA
____________________________________________________
The volume flow rate Q is given by
Q = pi*R^4(P2-P1)/8nL
The volume flow rate of the narrow pipe is
QA = pi*R_A^4(P2-P1)/8nL
The volume flow rate of the narrow pipe is
QB = pi*R_B^4(P2-P1)/8nL
Dividing the equation (2) with equation (1), we get
QB/QA = [R_B/RA_A]^4
= [(80%)R_A/R_A]^4
= [80/100]^4
= [0.8]^4
= 0.4096
= 40.96