Supplementary Material 1.28

Chapter 1 – Wave Plan Method

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Surge Analysis and the Wave Plan Method

Supplementary Material: Example Problems and Solutions

Chapter 1 – Problem 1.28

1.28 Consider a two-pipe junction. If a pressure wave of 100m approaches the junction from Pipe-1, compute the magnitude and sign of the reflected and transmitted pressure waves for the following conditions.

a. For pipes of the same material and thickness:

  1. If both pipes have identical diameters [Answer: 0 and 100m]
  2. If the diameter of Pipe-2 is 2 times the diameter of Pipe-1
  3. If the diameter of Pipe-2 is half that of Pipe-1
  4. If the diameter of Pipe-2 is 5 times that of Pipe-1
  5. If the diameter of Pipe-2 is 0.01 times that of Pipe-1

b. Compute the magnitude and sign of the reflected and transmitted pressure waves for all    the above cases if the Pipe-2 material is HDPE with a celerity of 400 m/s and the Pipe-1 is material ductile iron with a celerity of 1200 m/s.

Solution:

a.2.  If the diameter of Pipe-2 is 2 times the diameter of Pipe-1

Let A be the area of Pipe-1. As the diameter of Pipe-2 is 2 times the diameter of Pipe-1, the area of Pipe-2 is 4A. Both pipes have the same celerity values.

F1 = c/(gA), F2= c/(4gA), T = [(2gA)/c] / [(gA/c) + (4gA)/c)] = 2/5

R = T – 1 = -3/5 = -0.6

Reflected pressure wave: R∆H = -60m.

Transmitted pressure wave: T∆H = 40m.

a.3.  If the diameter of Pipe-2 is half the diameter of Pipe-1

Let 4A be area of Pipe-1, so the area of Pipe-2 would be A.

F1 = c/(4gA), F2= c/(gA), T = [(8gA)/c] / [(4gA/c) + (gA)/c)] = 8/5

R = T – 1 = 3/5 = +0.6,  R∆H = +60m, T∆H = 160m

a.4.  If the diameter of Pipe-2 is 5 times the diameter of Pipe-1

Let A be area of Pipe-1, so the area of Pipe-2 would be 25A.

F1 = c/(gA), F2= c/(25gA),

T = [(2gA)/c] / [(gA/c) + (25gA)/c)] = 2/26 = 1/13 = 0.076

R = T – 1 = -0.923

R∆H = -92.3m, T∆H = 7.6m.

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