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Published byRolf Harper Modified over 9 years ago
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Pipe Network Analysis 12’’- 1500’ 10’’- 3500’ 8’’- 1000’ 6’’- 1000’ 12’’- 3000’ 3.34 cfs 1.11 cfs 4.45 cfs Junction 3 Junction 4 Junction 2 Junction 1 1 2 4 3 5 Figure 1: A Small Pipe Network
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12’’- 1500’ 10’’- 3500’ 8’’- 1000’ 6’’- 1000’ 12’’- 3000’ 3.34 cfs 1.11 cfs 4.45 cfs Junction 3 Junction 4 Junction 2 Junction 1 1 2 4 3 5 Figure 1: A Small Pipe Network F1F2F3F4F1F2F3F4
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10’’- 3500’ 12’’- 1500’ 8’’- 1000’ 6’’- 1000’ 12’’- 3000’ 3.34 cfs 1.11 cfs 4.45 cfs Junction 3 Junction 4 Junction 2 Junction 1 1 2 4 3 5 Figure 2: A Small Pipe Network Loops Loop 1 Loop 2
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Dimensional coefficients D (pipe diameter)L (pipe length)C K (dimensional constant) Feet 4.73 InchesFeet8.56 X 10 5 Meters 10.67 Some typical values of roughness coefficient C HW MaterialC HW PVC150 Very Smooth Pipe140 Cement-Lined Ductile Iron140 New Cast Iron or Welded Steel130 Wood, Concrete120 Clay or New Riveted Steel110 Old Cast Iron, Brick100 Badly Corroded Cast Iron80
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10’’- 3500’ 12’’- 1500’ 8’’- 1000’ 6’’- 1000’ 12’’- 3000’ 3.34 cfs 1.11 cfs 4.45 cfs Junction 3 Junction 4 Junction 2 Junction 1 1 2 4 3 5 Loop 1 Loop 2
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10’’- 3500’ 12’’- 1500’ 8’’- 1000’ 6’’- 1000’ 12’’- 3000’ 3.34 cfs 1.11 cfs 4.45 cfs Junction 3 Junction 4 Junction 2 Junction 1 1 2 4 3 5 Loop 1 Loop 2
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Root sought 0xnxn x y Newton’s Method 1.Guess a first approximation to a root of the equation 2.Use the first approximation to get a second, the second to get a third, and so on, using the formula
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F1F1 F2F2 F3F3 F4F4 F5F5
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First Iteration MathCAD
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Second Iteration MathCAD
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Third Iteration MathCAD
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Fourth Iteration MathCAD
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Fifth Iteration MathCAD
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Sixth Iteration MathCAD
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Seventh Iteration MathCAD
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