The Churchill formula for the friction factor is $$f=8\left[\left(\frac{8}{\mathrm{Re}}\right)^{12}+\frac{1}{(A+B)^{15}}\right]^{1 / 12}$$ where $$\begin{array}{l}
A=\left\{-2.457 \ln \left[\left(\frac{7}{\mathrm{Re}}\right)^{0.9}+\frac{\varepsilon}{3.7 D}\right]\right\}^{16} \\
B=\left(\frac{37.530}{\mathrm{Re}}\right)^{16}
\end{array}$$
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.