
What is the pressure drop (in mmHg) in the blood as it passes through a capillary 1mm long and 2$\mu m$in radius if the speed of the blood through the center of the capillary is 0.66mm/s? (The viscosity of whole blood is $4 \times {10^{ - 3}}$poise).
Answer
132k+ views
Hint: Blood passes through a capillary only if there is a pressure difference between two points. Maximum velocity is at the center of the tube as near the periphery, due to the friction between liquid and walls of the tube, the flow of liquid slows down. The maximum velocity at the center of tube is determined by the formula ${v_{\max }} = \dfrac{{({P_1} - {P_2}){R^2}}}{{4\eta L}}$. With the help of this equation, the pressure difference can be found out. Now, to note the pressure in terms of mercury level, we equate the pressure difference with$\rho gh$. Here, h will give the pressure difference in terms of height of mercury column giving the same pressure as is taken as the density of mercury.
Complete step by step solution:
To find the pressure difference we use the formula
${v_{\max }} = \dfrac{{({P_1} - {P_2}){R^2}}}{{4\eta L}}$
Here, P1 and P2 are the pressure at two points at spacing L and radius R for liquid with coefficient of viscosity ŋ.
Substituting the values, we get
\[
{P_1} - {P_2} = \dfrac{{4\eta L{v_{\max }}}}{{{R^2}}} \\
\Rightarrow {P_1} - {P_2} = \dfrac{{4 \times 4 \times {{10}^{ - 3}} \times (.66 \times {{10}^{ - 3}}) \times {{10}^{ - 3}}}}{{{{(2 \times {{10}^{ - 3}})}^2}}} \\
\Rightarrow {P_1} - {P_2} = 2.64 \times {10^3} = \rho gh \\
\Rightarrow h = \dfrac{{2.64 \times {{10}^3}}}{{\rho g}} \\
\Rightarrow h = 19.5mm \\
\\
\\
\]
Therefore, the pressure difference can be said to be 19.5 mm of Hg.
The value of this answer signifies that if the amount of pressure 19.5mm of mercury would create is equal to the pressure difference across two points.
Note: This phenomenon occurs due to viscosity of the fluid. Viscosity is the property by the virtue of which, friction between the layers of the liquid and with the surfaces in contact. Here, the fluid close to the surface therefore has lesser velocity than the fluid at the center. Fluid always flows from a region of high pressure to a region of low pressure.
Complete step by step solution:
To find the pressure difference we use the formula
${v_{\max }} = \dfrac{{({P_1} - {P_2}){R^2}}}{{4\eta L}}$
Here, P1 and P2 are the pressure at two points at spacing L and radius R for liquid with coefficient of viscosity ŋ.
Substituting the values, we get
\[
{P_1} - {P_2} = \dfrac{{4\eta L{v_{\max }}}}{{{R^2}}} \\
\Rightarrow {P_1} - {P_2} = \dfrac{{4 \times 4 \times {{10}^{ - 3}} \times (.66 \times {{10}^{ - 3}}) \times {{10}^{ - 3}}}}{{{{(2 \times {{10}^{ - 3}})}^2}}} \\
\Rightarrow {P_1} - {P_2} = 2.64 \times {10^3} = \rho gh \\
\Rightarrow h = \dfrac{{2.64 \times {{10}^3}}}{{\rho g}} \\
\Rightarrow h = 19.5mm \\
\\
\\
\]
Therefore, the pressure difference can be said to be 19.5 mm of Hg.
The value of this answer signifies that if the amount of pressure 19.5mm of mercury would create is equal to the pressure difference across two points.
Note: This phenomenon occurs due to viscosity of the fluid. Viscosity is the property by the virtue of which, friction between the layers of the liquid and with the surfaces in contact. Here, the fluid close to the surface therefore has lesser velocity than the fluid at the center. Fluid always flows from a region of high pressure to a region of low pressure.
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