Supplementary Material 3.6 | KYPipe

Supplementary Material 3.6 | KYPipe

Chapter 3 – What Causes Surge?

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Surge Analysis and the Wave Plan Method

Supplementary Material: Example Problems and Solutions

Chapter 3 – Problem 3.6

3.6 Consider a pipeline (with cross-sectional area of 1 m 2 and celerity = 981 m/s) where a down surge event has resulted in a 0.001 m 3 vapor cavity (see Figure 3.12). There is a dead end (closed valve) on one side of the vapor cavity and a stagnant water column on the other side at time t. The upsurge that begins at time t starts moving the stagnant water column into the vapor cavity. Assume that the water column moves at a constant acceleration of 2.5 m/s 2 .

a. What is the theoretical maximum vapor slam pressure when the vapor column collapses? Use a computational time step of 0.02s.

b. What is the theoretical maximum vapor slam pressure when the water column collapses if the initial vapor column volume was 0.005 m3.

c. What is the theoretical maximum vapor slam pressure when the water column collapses if the initial vapor column volume was 0.1 m3. [Answer: 75m]

d. What is the theoretical maximum vapor slam pressure when the water column collapses if the initial vapor column volume was 1 m3. [Answer: 225m]

e. Plot time vs water column velocity and time vs vapor cavity volume.

f. Would the actual slam pressures be less than or equal to the calculated pressures in questions a, b, c, and d above? Explain why. Assuming the cavity occupies the shape of a rigid column (i.e., horizontal cylinder with vertical boundaries) along the pipeline, longitudinal axis, what is the length of pipe it would occupy?

g. Repeat this question with a constant water column acceleration of 0.1 m/s2 and again at 5 m/s2. Is it reasonable to assume a constant acceleration, and what factors might force a significant variation in acceleration?

(a) What is the theoretical maximum vapor slam pressure when the vapor column collapses? Use a computational time step of 0.02s.

Given a pipe cross sectional area A = 1 m 2 , pipe celerity c = 981 m/s, and computational time step ∆t = 0.02s

At time t: Velocity of water column U = 0.

Kinematic equation: V = U + a * ∆t where U is initial velocity, a is acceleration of the water column, and V is final velocity after time ∆t.

The water column moves at a constant acceleration a = 2.5 m/s 2

The velocity of the water column after first ∆t of 0.02s = 0 + 2.5 * 0.02 = 0.05 m/s

The volume of water entering the vapor cavity in the first ∆t = V * A * ∆t = 0.05 * 1* 0.02 = 0.001 m 3 .

The vapor cavity volume after first ∆t = Initial vapor cavity volume – volume of water that entered the vapor cavity in the first ∆t = 0.001 – 0.001 = 0. That is, the vapor cavity collapses after the first ∆t. The velocity of water column just before the collapse of vapor cavity = 0.05 m/s. The water column comes to rest instantaneously after the collapse of vapor cavity. The change in velocity of water column resulting from the collapse of vapor cavity ∆V = 0.05 m/s

Vapor slam pressure = (c/g) ∆V = (981 / 9.81) * 0.05 = 5 m

(b) What is the theoretical maximum vapor slam pressure when the water column collapses if the initial vapor column volume was 0.005m 3

Initial vapor cavity volume = 0.005 m 3

Water volume entering the vapor cavity in the first ∆t = 0.001 m 3

Vapor cavity volume after the first ∆t = 0.005 – 0.001 = 0.004 m 3

The water column continues to move into the vapor cavity at 2.5 m/s 2 acceleration

Velocity of the water column after second ∆t = 0.05 + 2.5 * 0.02 = 0.1 m/s

Volume of water entering the vapor cavity in the second ∆t = V * A * ∆t = 0.1 * 1* 0.02 = 0.002 m 3 .

Vapor cavity volume after the second ∆t = 0.004 – 0.002 = 0.002 m 3

As the vapor cavity volume is greater than 0, the water column continues to move into the vapor cavity at 2.5 m/s 2 acceleration, therefore the velocity of water column after the third ∆t = 0.1 + 2.5 * 0.02 = 0.15 m/s

Volume of water entering the vapor cavity in the third ∆t = V * A * ∆t = 0.15 * 1* 0.02 = 0.003 m 3 .

Vapor cavity volume after the third ∆t = 0.002 – 0.003 = -0.001 m 3

As the vapor cavity volume is less than 0, the vapor cavity collapses between the second- and third-time steps. The change in velocity of water column resulting from the collapse of the vapor cavity is between 0.1 and 0.15 m/s, The resulting vapor slam pressure would be between 10 and 15m.

(e) Plot time vs water column velocity and time vs vapor cavity volume.

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