Supplementary Material 5.2

Chapter 5 – Surge Protection

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Surge Analysis and the Wave Plan Method

Supplementary Material: Example Problems and Solutions

Chapter 5 – Problem 5.2

5.2 For the pressure relief valve described in the previous example, what is the longest possible time required to fill the diaphragm chamber avoiding drip flow across the tapping points A and C? What should the minor loss coefficient of the needle valve be to limit the flowrate to the desired value? What is the associated resistance across the needle valve? What is the ratio of the open area of the needle valve when it is closed to the point of drip flow with respect to the fully open area of the valve?

Solution:

In household faucets, a drip flow of roughly 4000 drops amount to 1 liter of water with an approximate drip size of ¼ ml (https://water.usgs.gov/edu/activity-drip.html). The fastest possible drip flow is about 240 drips per minute (after that the flow is usually considered to be continuous).

240 drips per minute = 60 ml/min or 1 ml/s or 0.001 l/s or 0.000001 m3/s.

Therefore, the longest possible time required to fill the 1-liter volume of the diaphragm chamber (avoiding drip flow) is 1000 s.

Velocity associated with 0.000001 m3/s flowrate = 0.0796 m/s

50.98 = (0.07962) * [ (0.015 * 1 /(2 * 9.81 * 0.004)) + (m / (2 * 9.81))]

m = 8050.248 *  (2 * 9.81) = 1641.23

Resistance = K = headloss across the needle valve / Q2 = ∆H­n / Q2

K = (m v2/(2g)) / Q2 = 1641.23 * 0.07962 / (2 * 9.81 * 0.0000012) = 0.5297E+12

The area ratio for any valve is related to its resistance ratio by an inverse square root relationship as described in Section 4.1 of the book “Pressure Wave Analysis of Transient Flow in Pipe Distribution Systems” by Wood, Lingireddy and Boulos (2005).

Ao/Af = √(Kf/Ko), where Kf is the resistance of a fully open valve with an area of Af and Ko is the resistance of a partially closed valve of area Ao.

Ko = 0.53E+12

Kf = (mv2/(2g)) / Q2 = 3* 12.172 / (2 * 9.81 * 0.0001532) = 9.67E+08

Ao/Af = (9.67E+08 / 0.5297E+12)0.5 = 0.0427 Ao is roughly 4% of full open area.


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