Question

Suppose an obstruction in an artery reduces its radius by 16%, but the volume flow rate of
the blood in the artery is remains the same. By what factor has the pressure drop across the
length of this artery increased?

Answer 1

Q = flow rate = v A
= v * pi * r^2
so
V1 R1^2 = V2 R2^2

V2 = V1 (R1/R2)^2
but we know R2 = (1.-16) R1 = .84 R1
so
V2 = 1.42 V1

then Bernoulii
p + 1/2 rho v^2 = constant
P1 +.5 rho V1^2 = P2 + .5 rho V2^2
P1- P2 = .5 rho (V2^2-V1^2)
so
(P1-P2)/P1 = (V2^2 -V1^1)/V1^2
= (1.4^2 V1^2 - V1^2)/V1^2
=1.4^2-1 = .96
or P2 = 1.96 P1

Answer 2

Thanks a lot.