An arterial constriction causes the radius of an artery to decrease from its nor

Question

An arterial constriction causes the radius of an artery to decrease from its normal value of 4 mm to 2.5 mm. The average speed of the blood in the artery is Vo in the unrestricted segment, and V in the restricted segment. The flow rate through this artery is 0.753 liters/min.

(i) Calculate the volume flow rate through the artery in SI units (m3 /s). Calculate Vo.
(ii) Calculate V, the average blood flow speed in constricted portion.
(iii) Use Bernoulli's equation to calculate the difference in pressure between the two ends of the segment of artery represented above.

Explanation / Answer

let,

radius r1=4mm

radius r2=2.5mm

avg speed of the blood is, v1=vo

avg speed of the blood is, v2=V

flow rate=0.753 L/min

i)

volume flow rate,

A*V=0.753*(10^-3/60) m^3/sec

==> A*V=1.255*10^-5 m^3/sec

and

A1*V1=1.255*10^-5

(pi*r1^2)*Vo=1.255*10^-5

(pi*(4*10^-3)^2)*Vo=1.255*10^-5

===> Vo=0.25 m^3

ii)

again,

A2*V2=1.255*10^-5

(pi*r2^2)*V=1.255*10^-5

(pi*(2.5*10^-3)^2)*V=1.255*10^-5

===> V=0.64 m^3

iii)

from the Bernoulli's equation,

P+(1/2*rho*v^2)+(rho*g*h)=constant

=====>

P1+(1/2*rho*v1^2)+rho*g*h1=P2+(1/2*rho*v2^2)+rho*g*h2

===>

P1+(1/2*rho*v1^2)=P2+(1/2*rho*v2^2)

P2-P1=1/2*rho_blood*(v2^2-v1^2)

=1/2*1060*(0.64^2-0.25^2)

=183.96 Pascal

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