Advances in Civil Engineering

Research Article

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

Table 6

Third-order Householder’s procedure. Option 6: starting point x₀ depends on input parameters: Equation (2), indirect calculation of λ through the transmission factor x: Equation (15), and the symbolic derivatives f′(x), f″(x), and f‴(x): Equations (11), (14), and (16).


Re = 5·10⁶, ε/D = 2.5·10⁻⁵	f(x), Equation (8)	f′(x), Equation (11)	f″(x), Equation (14)	f‴(x), Equation (16)	x₀ = 10.34052343	λ₀ = 0.009352225155363

Iteration 1	0.495092014	1.036495031	−0.001533392	0.000128855	9.863034531578420	0.010279663364062
Iteration 2	−0.000000034	1.037242198	−0.001596821	0.000136933	x = 9.863034564455800	λ = 0.010279663295529
Control step	0.000000000	1.037242198	−0.001596821	0.000136933	9.863034564455800	0.010279663295529

Re = 3·10⁴, ε/D = 9·10⁻³	f(x), Equation (8)	f′(x), Equation (11)	f″(x), Equation (14)	f‴(x), Equation (16)	x₀ = 10.34052343	λ₀ = 0.009352225155363

Iteration 1	0.143632267	1.025322691	−0.000738253	0.000043046	5.087840573035260	0.038630738575660
Iteration 2	0.000000000	1.025426528	−0.000744320	0.000043578	x = 5.087840573092420	λ = 0.038630738574792
Control step	0.000000000	1.025426528	−0.000744320	0.000043578	5.087840573092420	0.038630738574792