Supplementary Material, Appendix B – B.7

APPENDIX B – MOC and its Struggles to Compete with WPM


Surge Analysis and the Wave Plan Method

Supplementary Material: Example Problems and Solutions

Appendix B – Problem 7

B.7 A typical space-time grid for MOC calculations (Section B.2) is shown in the following figure.

Consider a 500 mm internal diameter pipeline (with a celerity = 1000 m/s) with a steady state flowrate of 0.5 m3/s and a constant Darcy-Weisbach friction factor of 0.02. Suppose the MOC grid size is Δx = 10m and Δt = 0.01s.

If the head and flowrate on the on the left side of grid point C at time t are 50m and 0.5 m3/s, and on the right side are 101.6692m and 0.4 m3/s, compute the head and flow value at grid point C at time t+Δt using both first and second-order MOC.


                Pipe diameter D = 0.5m, and the pipe cross sectional area A = 0.19635 m2

Pipe resistance K = ΔH/Q2 = [f Δx Q2/(2gDA2)] / Q2

     = (0.02*10)/(2*9.80665*0.5* 0.196352 ) = 0.52899

                                Pipe elastic factor F = c/gA = 1000 / (9.80665 * 0.19635) = 519.337

MOC first-order solution:

Use Equations B.15 and B.16 to compute the head and flowrate at the space-time grid point (C, t+Δt).

            Eq. B15:

HC,t+∆t    = 0.5[ (50+101.6692)+(0.5 – 0.4)519.337–0.52899(0.52–0.42)]


Eq. B16:

QC,t+∆t    = (0.5/519.337)[ (50-101.6692)+(0.5+0.4) 519.337-0.52899(0.52+0.42) ]

                = 0.400046 m3/s

MOC second-order solution:

Use Equations B.22 and B.23 to compute the head and flowrate at the space-time grid point (C, t+Δt).

HC,t+∆t    =101.7902m    

QC,t+∆t    = 0.40007 m3/s

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