Question

Blood takes about 1.45 s 1.45 s to pass through a 2.00 mm 2.00 mm long capillary. If the diameter of the capillary is 5.00 μm 5.00 μm and the pressure drop is 2.35 kPa , 2.35 kPa, calculate the viscosity of blood. Assume laminar flow.

Answer 1

Answer:

266.21 Pa.s

Explanation:

By Hagen–Poiseuille equation,

ΔP = 128 μ LQ/ πd2

Where;

Δp is the pressure difference between the two ends,

L is the length,

μ is the dynamic viscosity,

Q is the volumetric flow rate,

d = diameter

t = is the time

Δp = 2.35 KPa = 2.350Pa

L= 2.00 mm = 0.002m

μ =  ??? (unknown)

Q = V × A

   = (L/t) A

   = Lπ$d^{2}$/ 4t ( since A = $d^{2}$)

d= 5.00 μm = 5.00 × 10 -6

t = 1.45s  

         

∴ 2350 = (128 μ L/ π$d^{2}$ ) × (L π$d^{2}$ / 4t)

Q(volumetric flow rate)  = 32μ$L^{2}$  / $d^{2}$t  

 

From ΔP = 128 μ LQ/ π$d^{2}$

μ = ΔP π$d^{2}$ / LQ

= (2350 $d^{2}$ / t) / 32L2

=  [2350 × (5.00 × 10 -6 ) ( 1.45) / 32( 0.002)2]

= [(0.034075/ (32 × 0.000004)]

= 0.034075/ 0.000128

= 266.21 Pa.s