Chapter 10, Problem 57P

(a)

To determine

To Find: The blood flow through the aorta is laminar or turbulent.

Answer to Problem 57P

Blood flow is turbulent in first case.

Explanation of Solution

Given info:

Radius, $r=1.2\text{cm}=0.012\text{m}$

Average speed, $\overline{v}=40\text{cm/s}=0.4\text{m/s}$

The density of blood plasma, ${\rho }_{\text{blood}}=1050{\text{kg/m}}^{\text{3}}$

The viscosity coefficient, $\eta =4\times {10}^{-3}\text{Pa·s}$

Formula used:

The Reynolds number is,

  $R_e=\frac{2\overline{v}r\rho }{\eta }$

Where, $\overline{v}$ average speed

  $\rho$ Is the density of fluid

  $\eta$ Is the viscosity coefficient

  $r$ Is the radius

Calculation:

Substitute the values in the given equation, then

  $\begin{array}{l}{R}_e=\frac{2\overline{v}r\rho }{\eta }=\frac{2\times 0.4\times 0.012\times 1050}{4\times {10}^{-3}}=\left(\frac{9.840}{4}\right)\times {10}^3\\ =2520\end{array}$

This value is approximately equal to the given value of the Reynolds number. So, when the blood is flowing through the aorta then it is turbulent.

Conclusion:

Blood flow is turbulent in the first case.

(b)

To determine

To Find: The blood flow through the aorta is laminar or turbulent.

Answer to Problem 57P

Blood flow is laminar in this case.

Explanation of Solution

Given info:

Radius, $r=1.2cm=0.012m$

Average speed, $\overline{v}=40cm/s=0.4m/s$

The density of blood plasma, ${\rho }_{blood}=1050kg/m^3$

The viscosity coefficient, $\eta =4\times {10}^{-3}Pas$

Formula used:

The Reynolds number is,

  $R_e=\frac{2\overline{v}r\rho }{\eta }$

Where, $\overline{v}$ average speed

  $\rho$ Is the density of fluid

  $\eta$ Is the viscosity coefficient

  $r$ Is the radius

Calculation:

In this case the speed of the blood flow is double which means its velocity is $80cm/s$ then the Reynolds number is also doubled. So, its value is $5040$ which is far away from its value.

Conclusion:

So, in this case the blood flow is laminar.