Chapter 2 – Why WPM?
Surge Analysis and the Wave Plan Method
Supplementary Material: Example Problems and Solutions
Chapter 2 – Problem 2.15
2.15 As described in Appendix B (the companion section to Chapter 2), the computational time expended in each time step (i.e., determining the pressure waves at all node elements including junctions, pumps, valves, tanks, etc.) of a pipe network is about the same for both WPM and MOC. Therefore, when comparing the computational efficiencies of WPM and MOC, it would be sufficient to compare the total number of pipe sections (associated with each method) where computations are performed.
Consider a water distribution network model comprising 1000 pipe elements, 800 junction nodes, and 25 other elements such as pumps, tanks, reservoirs, and control valves. All pipe elements are shorter than 200m, and the highest total frictional headloss in any pipe element is less than 20m. All pipe elements are metallic with an average celerity of 1000 m/s. The total length of pipe in the entire distribution network is 125 km. The desired computational time step is 0.01s.
a. Determine how many pipe sections (each of which represent one calculation summing the effect of all pressure waves present at each computational time step) are needed for MOC and WPM? [Hint: Use the Courant number to compute the pipe sections with MOC and the desired number of friction orifices with WPM].
b. Suppose WPM takes N FLOPs (floating point operations) for each friction orifice, first-order MOC takes 0.75N FLOPs for each pipe section, and second-order MOC takes 2N FLOPs for each pipe section. Compute the total number of FLOPs to perform unsteady flow simulation (surge analysis) for 100s with WPM, first-order MOC, and second-order MOC at the desired computational time step of 0.01s. [Answers: WPM 10N million FLOPs, first-order MOC 93.75N million FLOPs, and second-order MOC 250N million FLOPs].
c. In part b of the above question, if the time step is reduced to 0.005s, what are the associated number of FLOPs required by each of the three methods? [Answers: WPM 20N Mega FLOPs, first-order MOC 375N Mega FLOPs, and second-order MOC 1N Giga FLOPs].
d. In part c of the above question, if the time step is reduced to 0.001s, what are the associated number of FLOPs required by each of the three methods? [Answers: WPM 0.1N Giga FLOPs, first-order MOC 9.375N Giga FLOPs, and second-order MOC 25N Giga FLOPs].
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