Syphones
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Syphones
• A siphon is a long bent used for carrying water from a reservoir at a higher level to another reservoir at a lower level when the two reservoirs are separated by a hill or high level ground.
• The point C is highest point of siphon is called as summit.
• The pressure at point C is less than atmospheric pressure as it lies above the free water surface in the tank A.
• The pressure at C can be reduced, theoretically, to-10.3 m of water but in actual practice is only -7.6 m water of water (or 10.3 – 7.6 = 2.7 m of water absolute).
• If the pressure becomes less than 2.7m of water absolute, the dissolved air and other gases would come out from water and get collected at the summit. It may obstruct the flow of water.
(b) To drain out water form a channel without any outlet.
(c) To take out the water from a tank which does not have any outlet.
• The flow through siphon then remains continuous till pressure in siphon pipe remains negative but less than separation pressur.
• At summit, we find minimum pressure. Velocity or discharge through siphon can be obtained by applying Bernoulli’s equation between A and B as shown in.
Now applying Bernoulli’s equation between section A and B.
PA/ γ + V2A/2g + ZA = PB/ γ + V2B / 2g + ZB + losses
But PA/ γ = PB/ γ = 0 (atmospheric) and VA = VB = 0
ZA – ZB = losses
H = Loss head + Head loss due + Head loss due
at entry to friction to exit
H = 0.5 V2/2g + f LV2/2gd + V2/2g
H = [1.5 + f L/d ] V2/2g
Where f = friction factor
L = length of siphon AB
d = diameter of the siphon
v = velocity of flow
Now, applying Bernoulli’s equation between A and C.
PA / γ + V2A/2g + ZA = Pc / γ + V2c/2g + Zc + Losses between AC
But PA/ γ = 0, VA = 0, Vc = V
ZA = Pc/ γ + V2/2g + Zc + (0.5V2/2g + f L1V2/2gd)
but ZA – Zc = -bc
Pc/ γ = [hc + (1.5 + fL1/d )V2/2g]
This equation is used to find out,
Pc/γ= pressure head at summit
hc = height of summit above the surface in the higher level
L1 = length of inlet leg.
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• The point C is highest point of siphon is called as summit.
• The pressure at point C is less than atmospheric pressure as it lies above the free water surface in the tank A.
• The pressure at C can be reduced, theoretically, to-10.3 m of water but in actual practice is only -7.6 m water of water (or 10.3 – 7.6 = 2.7 m of water absolute).
• If the pressure becomes less than 2.7m of water absolute, the dissolved air and other gases would come out from water and get collected at the summit. It may obstruct the flow of water.
Syphon is used in following cases:
(a) To take out water from one reservoir to another reservoir separated by a hill or ridge.(b) To drain out water form a channel without any outlet.
(c) To take out the water from a tank which does not have any outlet.
Working principle of siphon:
• Negative or vacuum pressure is created in the siphon, so that liquid gets pushed into it.• The flow through siphon then remains continuous till pressure in siphon pipe remains negative but less than separation pressur.
• At summit, we find minimum pressure. Velocity or discharge through siphon can be obtained by applying Bernoulli’s equation between A and B as shown in.
Now applying Bernoulli’s equation between section A and B.
PA/ γ + V2A/2g + ZA = PB/ γ + V2B / 2g + ZB + losses
But PA/ γ = PB/ γ = 0 (atmospheric) and VA = VB = 0
ZA – ZB = losses
H = Loss head + Head loss due + Head loss due
at entry to friction to exit
H = 0.5 V2/2g + f LV2/2gd + V2/2g
H = [1.5 + f L/d ] V2/2g
Where f = friction factor
L = length of siphon AB
d = diameter of the siphon
v = velocity of flow
Now, applying Bernoulli’s equation between A and C.
PA / γ + V2A/2g + ZA = Pc / γ + V2c/2g + Zc + Losses between AC
But PA/ γ = 0, VA = 0, Vc = V
ZA = Pc/ γ + V2/2g + Zc + (0.5V2/2g + f L1V2/2gd)
but ZA – Zc = -bc
Pc/ γ = [hc + (1.5 + fL1/d )V2/2g]
This equation is used to find out,
Pc/γ= pressure head at summit
hc = height of summit above the surface in the higher level
L1 = length of inlet leg.
For more help in Syphones click the button below to submit your homework assignment