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 1.18
1.18 Consider a short pipe section with an internal diameter of 1000 mm and a celerity of 500 m/s. The flow velocity in the pipeline is 1 m/s, and the associated headloss is 1.54m. The head at the upstream end (left end) of the pipe section is 100m and the flow is from left to right.
a. A pressure wave of -50m approaches the pipe section from the left side. Compute the reflected and transmitted pressure waves using Wave Plan Method (WPM). The WPM models pipe frictional resistance as an equivalent “orifice,” which is to say a single point where the friction loss is applied, located at the midpoint of the pipe section (see Figure 1.26 and other figures and equations referenced below in” Surge Analysis and the Wave Plan Method “).
b. A pressure wave of 100m approaches from the right side. Compute the magnitude and sign of the reflected and transmitted pressure waves.
c. A pressure wave of -50m approaches from the left side and another pressure wave of 100m approaches from the right side. Compute the magnitude and sign of the reflected and transmitted pressure waves. Compute the pressure heads after the wave action on both the upstream and downstream sides of the pipe section.
Use Eq. 1.57 to compute the flowrate after the wave action at the orifice, and then use Eqs. 1.47 and 1.51 to compute the reflected and transmitted pressure waves.
Pipe diameter D = 1m, Area A = 0.7854 m 2
Velocity V = 1 m/s è Flowrate Q = VA = 0.7854 m 3 /s
Headloss (at V = 1 m/s) Δh = 1.54m
Pipe (orifice) resistance K = Δh/Q 2 = 2.4965 s 2 /m 5
Celerity c = 500 m/s
Pipe elastic factor F = c/gA = 500/(9.81*0.7854) = 64.89 s/m 2
H 1 = 100m, H 2 = 100 – 1.54 = 98.46m, Q X = Q = 0.7854 m 3 /s
Substituting all the known quantities in Eq. 1.57:
2.4965 Q Y 2 + 129.78 Q Y – 3.4692 = 0
Q Y = 0.0267 or – 52.01 m 3 /s
Use Eq. 1.47 and 1.51 to compute the reflected and transmitted pressure waves.
ΔH T = -50 – 64.89 (0.0267 – 0.7854) = 0.77m
ΔH T = 64.89 (0.0267 – 0.7854) = -49.23m
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