According to Poiseuille\'s Law, the velocity of blood flowing in a blood vessel

Question

According to Poiseuille's Law, the velocity of blood flowing in a blood vessel of radius R is

v(r) = k(R2?r2), where r is the distance from the center of the vessel and k is a constant.

Let v(r) be the velocity of blood in an arterial capillary of radius R = 2 E-5m. Use

Poiseuille's Law with k = 4E6 (ms)?1 to determine the velocity at the center

of the capillary and the flow rate (use correct units).

(Use decimal notation. Give your answer to four decimal places.)


I found the velocity at the center=1.6E-3


find the flow rate = ? ? e-12 m^3/s

Explanation / Answer

v(r) = k(R2-r2)

At the centre of the capillary, r = 0.

Thus v(0) = kR2 = 4*106 * (2*10-5)2 = 1.6*10-3 m/s = 1.6 mm/s

Flowrate Q = [Integral (v dA)]r=0r=R

Q = [Integral (k(R2-r2) (2r dr)]r=0r=R

Q = 2k (R2r2/2 - r4/4)r=0r=R

Q = 2k (R4/4)

Q = kR4/2

Q = 3.14*(4*106) * (2*10-5)4/2

Q = 1*10-12 m3/s

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