Chapter 11, Problem 1P

To determine

Critical transition distance.

Explanation of Solution

Given:

The free stream velocity of the oil $\left(U\right)$ is $4\text{ ft/s}$.

The temperature of the oil $\left(T\right)$ is $68°\text{F}$.

Calculation:

Consider the flow is steady and the kerosene is incompressible.

Refer Appendix A, “Physical Properties of Liquids at Standard Atmospheric Pressure 14.70 psi, and 68°F (FPS Units)”

The transition from a laminar boundary layer takes place at a critical Reynolds number ${\left({\mathrm{Re}}_x\right)}_{\text{cr}}$ of $5\times {10}^5$.

The density of the kerosene $\left({\rho }_{ke}\right)$ is $1.58{\text{ slug/ft}}^3$.

The kinematic viscosity $\left(v_{\text{ke}}\right)$ is $25.4\times {10}^{-6}{\text{ ft}}^{\text{2}}\text{/s}$.

Write the Expression for the Reynolds numbers ${\left({\mathrm{Re}}_x\right)}_{\text{cr}}$ in terms of specific weight.

  $\begin{array}{c}{\left({\text{Re}}_x\right)}_{\text{cr}}=\frac{U{x}_{\text{c}}}{v_{\text{ke}}}\\ 5\times {10}^5=\frac{4\text{ ft/s}\times x_{\text{cr}}}{25.4\times {10}^{-6}{\text{ ft}}^{\text{2}}\text{/s}}\\ x_{\text{cr}}=3.175\text{ ft}\end{array}$

Thus, the critical distance for transition is $\underset{\_}{\text{3}\text{.175 ft}}$.