The Churchill formula for the friction factor is [ f=8left[left(frac{8}{operatorname{Re}} ight)^{12}+frac{1}{(A+B)^{1.5}} ight]^{1 / 12} ] where [

The Churchill formula for the friction factor is

\[ f=8\left[\left(\frac{8}{\operatorname{Re}}\right)^{12}+\frac{1}{(A+B)^{1.5}}\right]^{1 / 12} \]

where

\[ \begin{aligned} & A=\left\{-2.457 \ln \left[\left(\frac{7}{\operatorname{Re}}\right)^{0.9}+\frac{\varepsilon}{3.7 D}\right]\right\}^{16} \\ & B=\left(\frac{37,530}{\operatorname{Re}}\right)^{16} \end{aligned} \]

Compare this equation for \(f\) for both the laminar and turbulent regimes for \(\varepsilon / D=0.00001,0.0001,0.001\), and 0.01 and Reynolds numbers of \(10,10^{2}, 10^{3}, 10^{4}, 10^{5}, 10^{6}\), and \(10^{7}\) with the Moody chart and decide whether it is an acceptable replacement for the Colebrook formula.