Chapter 10, Problem 87GP

To determine

To find: The viscosity of the blood.

Answer to Problem 87GP

The Viscosity of the Blood is $\text{4}\text{.2732}\times {\text{10}}^{-3}\text{ Pa·s}$ .

Explanation of Solution

Given:

Inside diameter of needle $D=0.4\times {10}^{-3}\text{m}$

Therefore radius $r=0.2\times {10}^{-3}\text{m}$

Rate of flow, $\text{F}=6.833\times {10}^{-8}{\text{m}}^{\text{3}}\text{/s}$ .

Length of needle, $\text{L= 0}\text{.038m}$ .

Density of blood, $\rho =1060{\text{kg/m}}^{\text{3}}$ .

Formula used:

Pressure difference is given by , $\Delta \text{P=}\rho \text{gh}$ .

Where h is difference in height,

  $\rho$ is the density of the blood,

  $g$ is the acceleration due to gravity.

Using Poiseuille equation, Viscosity is given by

  $\eta \text{=}\frac{\text{π}\times \Delta \text{P}\times {\text{r}}^4}{8\text{LF}}$ .

Here, $r$ is the radius of the needle.

  $F$ is the rate of flow,

  $L$ is the length of needle,

  $\Delta P$ is the pressure difference.

Calculation:

Substituting the given values

  $\begin{array}{l}\Delta \text{P=1060}\times 10\times 1.66\\ \Delta \text{P=17660 Pa }\end{array}$

Therefore,

  $\begin{array}{l}\eta \text{=}\frac{\text{π}\times 17660\times {(0.2\times {10}^{-3})}^4}{8\times 0.38\times 6.833}\\ \eta \text{= 4}\text{.2732}\times {\text{10}}^{-3}\text{ Pa·s }\end{array}$

Conclusion:

Thus, The Viscosity of Blood is $\text{4}\text{.2732}\times {\text{10}}^{-3}\text{ Pa·s}$ .