Research Article

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

Table 8

Secant procedure. Option 8: two initial starting points λ0 and λ−1 required: starting point λ−1 is with fixed value λ−1 = 0.024069128765101 (i.e., x−1 = 6.445695939) as in Section 2.2.1, while starting point λ0 depends on input parameters: Equation (2), and direct calculation of λ: Equation (18).

Re = 5·106, ε/D = 2.5·10−5f(λi−1), Equation (5)f(λi), Equation (5)λ−1 = 0.024069128765101
λ0 = 0.009352225155363

Iteration 10.495092014−3.554956084−275.19702550.011151270814558
Iteration 2−0.4080719810.495092014−502.02394290.010338417191085
Iteration 3−0.029111936−0.408071981−466.20945510.010275973292109
Iteration 40.001836644−0.029111936−495.62215910.010279679026163
Iteration 5−0.0000078280.001836644−497.7344480.010279663299743
Iteration 6−0.000000002−0.000007828−497.6011214λ = 0.010279663295529
Control step0.000000000−0.000000002−497.60121790.010279663295529

Re = 3·104, ε/D = 9·10−3f(λi−1), Equation (5)f(λi), Equation (5)λ−1 = 0.024069128765101
λ0 = 0.036588313752304

Iteration 11.3917123940.143632267−99.693400790.038029053721052
Iteration 20.1436322670.041110009−71.159445850.038606770549177
Iteration 30.0411100090.001619232−68.356632510.038630458556837
Iteration 40.0016192320.000018909−67.55839020.038630738444645
Iteration 50.0000189090.000000009−67.52699052λ = 0.038630738574792
Control step0.0000000090.000000000−67.526622120.038630738574792