Supplementary Material 5.1

Chapter 5 – Surge Protection

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Surge Analysis and the Wave Plan Method

Supplementary Material: Example Problems and Solutions

Chapter 5 – Problem 5.1

5.1 Consider a hydraulic control valve shown in Figure 5.8. This valve is used as a pressure relief valve (i.e., surge protection device) on a pipeline system. Suppose the valve is in the fully open position at certain time = t (with port C connected to port B) when the pilot triggers the valve to start closing (port A now connects to port C). The volume of the diaphragm control chamber is 1 liter, the pipe connecting ports A and C is a 1m long 4 mm internal diameter copper tube with a Darcy-Weisbach friction factor of 0.015. The minor loss coefficient for the 4 mm diameter fully open needle valve on the copper tube is 3. If the pressure differential across ports A and C is 5 bar and this difference is assumed to remain constant throughout the closing cycle, compute the shortest time required to fill the diaphragm chamber effecting the valve closure.

Solution:

Volume of the diaphragm chamber ꓯ = 1 liter = 0.001 m3

Pressure differential across ports A and C, ∆HAC = 5 bar = 50.98m

Pressure differential across ports A and C must be equal to the sum of the frictional loss within the copper tube and the minor loss across the needle valve, neglecting the minor losses at the tapping points A and C.

∆HAC = (fLv2/(2gD)) + (mv2/(2g)), where f is friction factor, L is length of the copper tube, D is diameter of copper tube, and m is needle valve’s minor loss coefficient.

To fill the diaphragm chamber in the shortest time, the needle valve should be kept in the fully open position and therefore the fully open minor loss coefficient should be used for m.

50.98 = v2 [ (0.015 * 1 /(2 * 9.81 * 0.004)) + (3 / (2 * 9.81))] = v2

[0.1911 + 0.1529] = 0.344 v2

v = 12.17 m/s

v = 12.17 m/s implies the flowrate Q = 0.000153 m3/s

Time taken for filling the diaphragm chamber (at constant differential pressure across ports A and C) =ꓯ/Q = 0.001/0.000153 = 6.53 s


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