What is the pressure drop in mm Hg in the blood as it passes through a capillary 1 mm long and 2- m in radius if the speed of the blood through the centre of the capillary is 0-66 mm-s - The viscosity of whole blood is 4 - 10-3 poise-

See the expression of maximum velocity at centre of tube, v_max= ( p_1 p_2 )R^2/4η L ∴ nbsp; nbsp; ( P_1 P_2 )=4η Lv_max/R_4 nbsp; nbsp; ...

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Standard XII

Physics

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Question

What is the pressure drop (in mm Hg) in the blood as it passes through a capillary 1 mm long and $2 μ m$ in radius if the speed of the blood through the centre of the capillary is 0.66 mm/s ? (The viscosity of whole blood is $4 \times 10^{- 3}$ poise).

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Solution

See the expression of maximum velocity at centre of tube,
$v_{m a x} = \frac{(p_{1} - p_{2}) R^{2}}{4 η L}$
$∴$ $(P_{1} - P_{2}) = \frac{4 η L v_{m a x}}{R_{4}}$

$= \frac{(4 \times 4 \times 10^{- 3}) (10^{- 3}) (0.66 \times 10^{- 3})}{{(2 \times 10^{- 6})}^{2}}$
$= 2.64 \times 10^{3}$ N/m $^{2}$ = $h ρ g$
$h = \frac{2.64 \times 10^{3}}{ρ g}$
$= \frac{2.64 \times 10^{3}}{13.6 \times 10^{3} \times 9.81}$ m of Hg
$= 0.0195$ m of Hg
$= 19.5$ mm of Hg

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What is the pressure drop (in mm Hg) in the blood as it passes through a capillary 1 mm long and 2μm in radius if the speed of the blood through the centre of the capillary is 0.66 mm/s ? (The viscosity of whole blood is 4×10−3 poise).

See the expression of maximum velocity at centre of tube,vmax=(p1−p2)R24ηL∴ (P1−P2)=4ηLvmaxR4 =(4×4×10−3)(10−3)(0.66×10−3)(2×10−6)2 =2.64×103 N/m2=hρgh=2.64×103ρg =2.64×10313.6×103×9.81 m of Hg =0.0195 m of Hg =19.5 mm of Hg

What is the pressure drop (in mm Hg) in the blood as it passes through a capillary 1 mm long and $2 μ m$ in radius if the speed of the blood through the centre of the capillary is 0.66 mm/s ? (The viscosity of whole blood is $4 \times 10^{- 3}$ poise).