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Pipe Line With Negative Pressure (Siphon Phenomena) : Follow The Natural Contour of The Land

1) When a pipeline rises above its hydraulic gradient line, it creates a negative pressure region known as a siphon. 2) Negative pressure exists wherever the pipe is above the hydraulic gradient line, with the highest negative pressure (vacuum pressure) occurring at the summit point. 3) For water to continue flowing through the siphon, the pressure at the summit must remain above the vapor pressure of water, usually -7.6 meters of water head. Otherwise cavitation will occur and stop the flow.

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1K views8 pages

Pipe Line With Negative Pressure (Siphon Phenomena) : Follow The Natural Contour of The Land

1) When a pipeline rises above its hydraulic gradient line, it creates a negative pressure region known as a siphon. 2) Negative pressure exists wherever the pipe is above the hydraulic gradient line, with the highest negative pressure (vacuum pressure) occurring at the summit point. 3) For water to continue flowing through the siphon, the pressure at the summit must remain above the vapor pressure of water, usually -7.6 meters of water head. Otherwise cavitation will occur and stop the flow.

Uploaded by

charbel
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Pipe line with negative Pressure

(siphon phenomena)
S
Introduction:
• Long pipelines laid to transport
water from one reservoir to another
over a large distance usually follow
the natural contour of the land.

• A section of the pipeline may be raised to an elevation that is above the local
hydraulic gradient line (HGL). If this happens, then a siphon phenomena may occur
in the negative pressure region as shown above.
Characteristics of the syphon phenomena:
• Point “S” is known as the summit.
• All Points above the HGL have pressure less than atmospheric  therefore, they
have negative value (vacuum pressure).
• If the absolute pressure is used, then the atmospheric absolute pressure = 10.33 m.
(1 atm  10.33 m head)
• To avoid syphoning, it is important to maintain the pressure at all points above
H.G.L. greater than the water vapor pressure 
Pabove HGL > Pv
Siphoning
When a pipeline rises above its hydraulic line, it is termed “Siphon” such as the
situation below. e
+ Z = HGL

b
a
c
Notes:
• HGL is the line that connects water surfaces A and B

• If pipe is long  Minor losses can be neglected

• The dynamic pressure head at any section of the pipe = vertical distance between
HGL and the pipe centerline. Thus:
➢ Pe is the lowest dynamic pressure.

➢ Practice and calculation show that If Pe becomes < - h, then water start to
vaporize (natural cavitation). This causes the flow to stop 

Thus, the elevation of any section of the pipeline above HGL should be < h
Analysis S
ZA - ZS

P
ZA - ZB
Q

𝐿 𝑉2
𝑍𝐴 − 𝑍𝐵 = ℎ𝐿 = 𝑓 > 0 (below HGL)
𝐷 2𝑔

Apply Bernoulli between point A and S:

𝑉𝐴2 𝑃𝐴 𝑉𝑠2 𝑃 𝑉𝑆2 𝑃𝑆


+ + 𝑍𝐴 = + 𝑠 + 𝑍𝑠 + ℎ𝐿  𝑍𝐴 − 𝑍𝑠 = + + ℎ𝐿
2𝑔 𝛾 2𝑔 𝛾 2𝑔 𝛾

Above HGL  -ve value Below the atm. pressure  vacuum pressure
(Must be -ve value)
Therefore:
• Negative pressure exists in the pipelines wherever the pipe line is raised above the
hydraulic gradient line (between P & Q)
• The negative pressure at the summit point can reach theoretically -10.3 m water
head (gauge pressure) and zero (absolute pressure). However, since water contains
dissolved gasses, water will vaporize before reaching -10.3 m water head.
• Generally, this pressure reaches -7.6 m water head (gauge pressure) and 2.7 m
(absolute pressure).
Example:
A siphon pipe, between two reservoirs, has a diameter of 20 cm and length 500 m as
shown. The difference between reservoir levels is 20m. The length of the pipe from A to
S is LAS = 100m. Calculate the flow in the system and the pressure head at summit. Take
f = 0.02
Solution: D = 0.20 m, ZA – ZB = 20 m, L = 500 m, LAS = 100 m, f = 0.02
Since the pipe is long  minor losses are ignored
𝐿 𝑉2
𝑍𝐴 − 𝑍𝐵 = 𝑓  solve for V = 2.8 m/sec  𝑉ሶ = VA = 0.08796 m3/sec
𝐷 2𝑔

Apply Bernoulli between point A and S:

𝑉𝐴2 𝑃𝐴 𝑉𝑠2 𝑃 𝑉𝑠2 𝑃𝑠


+ + 𝑍𝐴 = + 𝑠 + 𝑍𝑠 + ℎ𝐿  𝑍𝐴 − 𝑍𝑠 = + + ℎ𝐿 
2𝑔 𝛾 2𝑔 𝛾 2𝑔 𝛾

2.82 𝑃 100 2.82


0–3= + 𝑠 + 0.02
2𝑔 𝛾 0.2 28

𝑃𝑠
= - 7.396 m of water gage < - 7. 6 m  water will flow through S
𝛾

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