A large artery has a diameter of 7 mm and carries blood which flows with a peak

Question

A large artery has a diameter of 7 mm and carries blood which flows with a peak velocity of 0.15m/s. this vessel eventually feeds network of capillaries which together have area tea approximately400 times that of the large artery which feeds into them . In this model, the capillaries are identical to each other and have a diameter of 7.5 um. (ignore viscosity)

a. suppose that the diastolic blood pressure is 130 mmHg at the level of the heart and the blood velocity in the large artery at the heart is 0.15 m/s. What is the blood velocity in the artery at a point 1 m below the heart? (blood density = 1050 kg/m^3)

b. If the artery is severed at a point 1 m blow the heart , what is the maximum velocity of blood flow from the artery?

c. What is the blood velocity in capillary in the capillary net at this point (1m below the heart)

d. What is the blood pressure in the capillary net at this poin(in mmHg and with the assumption discussed above?)

e. If a capillary is served, what is the blood velocity leaving the wound?(again on the basis of this model)

Explanation / Answer

bernoulli principle states that, for incompressable flow,

P+0.5*pho*v^2+pho*g*h=constant

where P=pressure at a point

pho=density of the fluid

v=speed of the fluid at that point

g=9.8 m/s^2

h=height of the fluid as compared to a referrence point

part a:

let pressure, speed and height of fluid at point 1 are P1,v1 and h1 and those for point 2

are P2,v2 and h2.

(point 1 is at heart level and point 2 is at artery level)

let point 2 be of referrence point for height.

then h2=0

h1= 1 m

P2-P1=pho*g*h=1050*9.8*1=10290 N/m^2

v1=0.15 m/s

pho=1050 kg/m^3

using bernoulli's principle:

P1+0.5*pho*v1^2+pho*g*h1=P2+0.5*pho*v2^2+pho*g*h2

==>P1-P2+0.5*pho*v1^2+pho*g*h1-pho*g*h2=0.5*pho*v2^2

==>-10290+0.5*1050*0.15^2+1050*9.8*1-1050*9.8*0=0.5*1050*v2^2

==>v2=0.15 m/s

part b:

if artery is severed, the pressure at point 2 will be at 1 atm =101325 N/m^2

and P1 will be at 130 mm Hg above P2.

hence P1-P2=17331.9 N/m^2

hence

using bernoulli's principle:

P1+0.5*pho*v1^2+pho*g*h1=P2+0.5*pho*v2^2+pho*g*h2

==>P1-P2+0.5*pho*v1^2+pho*g*h1-pho*g*h2=0.5*pho*v2^2

==>17331.9+0.5*1050*0.15^2+1050*9.8*1-1050*9.8*0=0.5*1050*v2^2

==>v2=7.255 m/s

part c:

let area of artery=A1

let area of capillary=A2

given that A2=400*A1

then using the continuity equation:

A1*v1=A2*v2

==>v2=A1*v1/A2=v1/400=3.75*10^(-4) m/s

part d:

using bernoulli principle:

P1+0.5*pho*v1^2+pho*g*h1=P2+0.5*pho*v2^2+pho*g*h2

==>P1+0.5*1050*0.15^2+1050*9.8*1=P2+0.5*1050*(3.75*10^(-4))^2

==>P2=P1+10301.812 N/m^2=P1+77.2699 mm Hg

==>P2=207.2699 mm Hg

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