Chapter 8.1, Problem 1P

To determine

Whether the air flow through the trachea of human body is laminar or turbulent?

Explanation of Solution

Writ the formula for Reynolds number.

  $\begin{array}{c}\text{Re}=\frac{\rho VD}{\mu }\because \left(\frac{\rho }{\mu }=\frac{1}{\nu }\right)\\ =\frac{VD}{\nu }\end{array}$        (I)

Here, the density is $\rho$, velocity is $V$, the characteristic dimension is $D$ in meter $\left(\text{m}\right)$, the dynamic viscosity is $\mu$, and the kinematic viscosity $\nu$.

When the Reynolds number is less than 2000, the flow is said as laminar flow. When it is greater than 4000, the flow is said as turbulent flow. When it is between 2000 and 4000, the flow is in transitional flow.

Consider that the air velocity through the human trachea is ranges from $1\text{m/s}$ to $2\text{m/s}$. The diameter of the adult human trachea is ranges from $2\text{cm}$ to $3\text{cm}$.

The average velocity is $1.5\text{m/s}$.

The diameter is $2.5\text{cm}$ or $0.025\text{m}$.

Refer Table B.4, “Physical Properties of Air at Standard Atmospheric Pressure (SI Units)”.

The kinematic viscosity $\left(\nu \right)$ of air corresponding to $15°\text{C}$ is $1.47\times {10}^{-5}{\text{m}}^2\text{/s}$.

Substitute $1.5\text{m/s}$ for $V$, $0.025\text{m}$ for $D$, and $1.47\times {10}^{-5}{\text{m}}^2\text{/s}$ for $\nu$ in Equation (I).

  $\begin{array}{c}\text{Re}=\frac{\left(1.5\text{m/s}\right)\left(0.025\text{m}\right)}{1.47\times {10}^{-5}{\text{m}}^2\text{/s}}\\ =2551.0204\\ \simeq 2551\end{array}$

The calculated Reynolds number lies in the range of 2000 to 4000. Thus, the flow of air through the trachea of human body is $\underset{\_}{\text{transitional flow}}$.