Chapter 8, Problem 13P

The pressure drop in a section of pipe can be calculated as

$\Delta p=f\frac{Lp{V}^2}{2D}$

where $\Delta p=$ the pressure drop $\left(\text{Pa}\right),f=$ the friction factor, $L=$ the length of pipe $\left[\text{m}\right],\rho =\text{ density }\left(\text{kg}/{\text{m}}^3\right),V=\text{ velocity }\left(\text{m}/\text{s}\right),\text{ and }D=\text{ diameter }\left(\text{m}\right)$ For turbulent flow, the Colebrook equation provides a means to calculate the friction factor,

$\frac{1}{\sqrt{f}}=-2.0\text{ log }\left(\frac{\epsilon }{3.7D}+\frac{2.51}{\text{Re }\sqrt{f}}\right)$

where $\epsilon =$ the roughness $\left(\text{m}\right),\text{ and Re}=$ the Reynolds number.

$\text{Re}=\frac{\rho VD}{\mu }$

where $\mu =$ dynamic viscosity $\left(\text{N}\cdot \text{s}/{\text{m}}^2\right)$.

(a) Determine $\Delta p$ for a 0.2-m-long horizontal stretch of smooth drawn tubing given $\rho =1.23\text{kg}/{\text{m}}^3,\mu =1.79\times {10}^{-5}\text{N}\cdot \text{s}/{\text{m}}^2,D=0.005\text{ m, }V=40\text{m}/\text{s, and }\epsilon =\text{0}\text{.0015}$ mm. Use a numerical method to determine the friction factor. Note that smooth pipes with $\text{Re}<{10}^5$, a good initial guess can be obtained using the Blasius formula, $f=0.316/{\text{Re}}^{0.25}$

(b) Repeat the computation but for a rougher commercial steel pipe $\left(\epsilon =0.045\text{mm}\right)$.