Advanced Iterative Procedures for Solving the Implicit Colebrook Equation for Fluid Flow Friction

Table 2

Newton–Raphson procedure. Option 2: fixed initial starting point λ₀ = 0.024069128765100981 from Section 2.2.1, calculation of λ: Equation (7), and analytical derivative f′(λ): Equation (6).

Re = 5·10^6, ε/D = 2.5·10^−5	f(λ), Equation (5)	f′(λ), Equation (6)	λ0 = 0.024069128765101
Iteration 1	−3.554956084	−139.7424853	−0.001370207567104
Iteration 2	17.630891548	−10069.59089	0.000380696888310
Iteration 3	42.275315189	−68216.8306	0.001000416608714
Iteration 4	22.325487096	−16105.99979	0.002386576262278
Iteration 5	10.932300910	−4398.30144	0.004872149626988
Iteration 6	4.615550920	−1516.202309	0.007916302041016
Iteration 7	1.426053458	−734.846953	0.009856914916156
Iteration 8	0.217044469	−529.7853757	0.010266598684182
Iteration 9	0.006507144	−498.5470019	0.010279650902858
Iteration 10	0.000006167	−497.602585	0.010279663295518
Iteration 11	0.000000000	−497.6016898	λ = 0.010279663295529
Control step	0.000000000	−497.6016898	0.010279663295529
Re = 3·10^4, ε/D = 9·10^−3	f(λ), Equation (5)	f′(λ), Equation (6)	λ0 = 0.024069128765101
Iteration 1	1.391712394	−137.1740994	0.034214720386916
Iteration 2	0.326434508	−80.9945153	0.038245048943635
Iteration 3	0.026240732	−68.54940037	0.038627849256271
Iteration 4	0.000195117	−67.53419088	0.038630738412914
Iteration 5	0.000000011	−67.52662412	λ = 0.038630738574792
Control step	0.000000000	−67.5266237	0.038630738574792