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

Table 9

Secant procedure. Option 9: two initial starting points x₀ and x₋₁ required: starting point λ₋₁ is with fixed value x₋₁ = 6.445695939 (i.e., λ₋₁ = 0.024069128765101) as in Section 2.2.1, while starting point λ₀ depends on input parameters: Equation (2), and indirect calculation of λ through the transmission factor x: Equation (19).

Re = 5·10^6, ε/D = 2.5·10^−5	f(xi−1), Equation (8)	f(xi), Equation (8)		x−1 = 6.445695939	λ−1 = 0.024069128765101
				x0 = 10.34052343	λ0 = 0.009352225155363
Iteration 1	−3.554956084	0.495092014	1.039853012	9.864406125318800	0.010276804896656
Iteration 2	0.495092014	0.001422639	1.03686501	9.863034066961850	0.010279664332547
Iteration 3	0.001422639	−0.000000516	1.037241104	9.863034564456330	0.010279663295528
Iteration 4	−0.000000516	0.000000000	1.0372422	x = 9.863034564455800	λ = 0.010279663295529
Control step	0.000000000	0.000000000	1.037162162	9.863034564455800	0.010279663295529
Re = 3·10^4, ε/D = 9·10^−3	f(xi−1), Equation (8)	f(xi), Equation (8)		x−1 = 6.445695939	λ−1 = 0.024069128765101
				x0 = 5.227918429	λ0 = 0.036588313752304
Iteration 1	1.391712394	0.143632267	1.024883541	5.087773465040530	0.038631757665255
Iteration 2	0.143632267	−0.000068814	1.025374564	5.087840576494990	0.038630738523123
Iteration 3	−0.000068814	0.000000003	1.025426553	x = 5.087840573092420	λ = 0.038630738574792
Control step	0.000000003	0.000000000	1.025426591	5.087840573092420	0.038630738574792