Answer:
Viscosity of blood will be [tex]918.54\times 10^{-6}Pa-sec[/tex]
Explanation:
We have given time to pass the capillary = 1.65 sec
Length of capillary L = 2 mm = 0.002 m
Pressure drop [tex]\Delta P=2.85kPa=2.85\times 10^3Pa[/tex]
Diameter [tex]d=5\mu m=5\times 10^{-6}m[/tex]
So radius [tex]r=\frac{5\times 10^{-6}}{2}=2.5\times 10^{-6}m[/tex]
Velocity will be [tex]v=\frac{L}{t}=\frac{0.002}{1.65}=1.212\times 10^{-3}m/sec[/tex]
We know that viscosity is given by [tex]\eta =\frac{r^2\Delta P}{8Lv}=\frac{(2.5\times 10^{-6})^2\times 2.85\times 10^3}{8\times 0.002\times 1.212\times 10^{-3}}=918.54\times 10^{-6}Pa-sec[/tex]
Viscosity of blood will be [tex]918.54\times 10^{-6}Pa-sec[/tex]
The viscosity of the blood flowing in the given capillary is [tex]9.27 \times 10^{-4} \ Pa.s[/tex]
The given parameters;
The velocity of the blood is calculated as follows;
[tex]v = \frac{L}{t} = \frac{2\times 10^{-3} \ m}{1.65 \ s} = 0.0012 \ m/s[/tex]
The radius of the capillary is calculated as follows;
[tex]r = \frac{d}{2} \\\\r = \frac{5\times 10^{-6}}{2} = 2.5 \times 10^{-6} \ m[/tex]
Assuming laminar flow, the viscosity of the blood is calculated as;
[tex]\mu = \frac{r^2 \times \Delta P}{8lv} \\\\\mu = \frac{(2.5\times 10^{-6})^2 \times 2850}{8 \times 2\times 10^{-3} \times 0.0012} \\\\\mu= 9.27 \times 10^{-4} \ Pa.s[/tex]
Thus, the viscosity of the blood flowing in the given capillary is [tex]9.27 \times 10^{-4} \ Pa.s[/tex]
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