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Maximum height over which water can be siphoned

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Homework Statement


A student siphons water over a 8.5 m high wall at sea level. She then climbs to the summit of a mountain (elevation 4390 m , atmospheric pressure =58.5 kPa) and attempts the same experiment. Comment on her prospects for success.

Homework Equations


Bernoulli's equation: ##P/\rho + v^2/2 + zg = C##

##\rho## -density of water
##P## - pressure
##v## -velocity
##z## -elevation
##g## -acceleration due to gravity
##C## - constant

The Attempt at a Solution


I guess in order to find out whether the experiment works we need to find the maximum height over which water can be siphoned, this height depends on the pressure inside the pipe at the top most point ( my guess) and whether it is below or above the vapor pressure of water, if it is below the vapor pressure cavitation will occur and stop the flow right ?

Consider one end of the pipe as point 1 and the other end as point 2 ,now point 3 is in the middle part of the pipe that will be lifted to the maximum height.

Applying Bernoulli's equation between point 2 and 3

##P_2/\rho +v_2^2/2+ z_2g =P_3/\rho + v_3^2/2 +z_3g ##

##v_2 = v_3 ## since the pipe is of constant cross-section
##z_2 =0 ## as it is at sea level
##P_2=## atmospheric pressure at sea level
##P_3## is the pressure in the pipe at top most point = 58.5kPa correct ???
##z_3## is the maximum height to which the middle part of the pipe is elevated (=4390m correct ??)
assuming density to be constant , do i really need this ? should i just check whether the pressure at point 3 is above or below the vapour pressure of water and decide whether the experiment will work ?

I need to find the vapour pressure of water which according to wikipedia depends on temperature
https://en.wikipedia.org/wiki/Vapour_pressure_of_water
##P\text{ (mmHg)} = \exp\left(20.438 - \frac{5044}{T\text{ (K)}}\right) ##
Taking temperature as 260.77K based on altitude
http://www.engineeringtoolbox.com/air-altitude-temperature-d_461.html
, i get P =399.125 Pa as vapour pressure is that correct ?
does this mean water will not cavitate ? as the pressure at the altitude of 4390 m is 58.5kPa (> 0.399 kPa )
So the student's experiment will work ?
 
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  • #2
SteamKing
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Homework Statement


A student siphons water over a 8.5 m high wall at sea level. She then climbs to the summit of a mountain (elevation 4390 m , atmospheric pressure =58.5 kPa) and attempts the same experiment. Comment on her prospects for success.

Homework Equations


Bernoulli's equation: ##P/\textrho + v^2/2 + zg = C##

##\textrho## -density of water
##P## - pressure
##v## -velocity
##z## -elevation
##g## -acceleration due to gravity
##C## - constant

The Attempt at a Solution


I guess in order to find out whether the experiment works we need to find the maximum height over which water can be siphoned, this height depends on the pressure inside the pipe at the top most point ( my guess) and whether it is below or above the vapor pressure of water, if it is below the vapor pressure cavitation will occur and stop the flow right ?

Consider one end of the pipe as point 1 and the other end as point 2 ,now point 3 is in the middle part of the pipe that will be lifted to the maximum height.

Applying Bernoulli's equation between point 2 and 3

##P_2/\textrho +v_2^2/2+ z_2g =P_3/\textrho + v_3^2/2 +z_3g ##

##v_2 = v_3 ## since the pipe is of constant cross-section
##z_2 =0 ## as it is at sea level
##P_2=## atmospheric pressure at sea level
##P_3## is the pressure in the pipe at top most point = 58.5kPa correct ???
##z_3## is the maximum height to which the middle part of the pipe is elevated (=4390m correct ??)
assuming density to be constant , do i really need this ? should i just check whether the pressure at point 3 is above or below the vapour pressure of water and decide whether the experiment will work ?

I need to find the vapour pressure of water which according to wikipedia depends on temperature
https://en.wikipedia.org/wiki/Vapour_pressure_of_water
##P\text{ (mmHg)} = \exp\left(20.438 - \frac{5044}{T\text{ (K)}}\right) ##
Taking temperature as 260.77K based on altitude
http://www.engineeringtoolbox.com/air-altitude-temperature-d_461.html
, i get P =399.125 Pa as vapour pressure is that correct ?
does this mean water will not cavitate ? as the pressure at the altitude of 4390 m is 58.5kPa (> 0.399 kPa )
So the student's experiment will work ?
While bubbles in the top of the siphon will break suction, the maximum height over which the siphon can work is determined by the local ambient atmospheric pressure.

https://en.wikipedia.org/wiki/Siphon

At sea level, atmospheric pressure can support a column of fresh water approx. 10 m high, which is why the student had an initial success going over a wall 8.5 m tall. Can the student expect the same effect at an altitude of 4390 m?

BTW, what happens to liquid water when the temperature is 260 K?
 
  • #3
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BTW, what happens to liquid water when the temperature is 260 K?
oops!! i overlooked that
 
  • #4
rcgldr
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I'm wondering if the siphon actually stops, or if the siphon is effectively siphoning water vapor, which would still work, but at a much lower mass flow. It seems that presence of water vapor would mean that the pressure is never zero within the siphon, so there's still a pressure differential to drive the water vapor.
 
  • #5
rude man
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Homework Statement


A student siphons water over a 8.5 m high wall at sea level. She then climbs to the summit of a mountain (elevation 4390 m , atmospheric pressure =58.5 kPa) and attempts the same experiment. Comment on her prospects for success.

Homework Equations


Bernoulli's equation: ##P/\textrho + v^2/2 + zg = C##

##\textrho## -density of water
##P## - pressure
##v## -velocity
##z## -elevation
##g## -acceleration due to gravity
##C## - constant

The Attempt at a Solution


I guess in order to find out whether the experiment works we need to find the maximum height over which water can be siphoned, this height depends on the pressure inside the pipe at the top most point ( my guess) and whether it is below or above the vapor pressure of water, if it is below the vapor pressure cavitation will occur and stop the flow right ?
No, don't think it has anything to do with vapor pressure. We are ignoring water vapor.
Suggestion: think atmospheric pressure instead, which is mostly N2 and O2 in origin. Looking at your Bernoulli equation, what term is irrelevant and how do the other two terms combine? Thinks violation of Bernoulli = no siphoning!




[/QUOTE]
 
  • #6
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78
Looking at your Bernoulli equation, what term is irrelevant and how do the other two terms combine? Thinks violation of Bernoulli = no siphoning!
Hmm... the velocity terms get cancelled and since point 2 is considered to be at sea level ,##z_2 =0## so we get ##P_2/\textrho = P_3/\textrho + z_3g ## where ##P_2## is atmospheric pressure ,so ##P_3 = 58.5 kPa ## ? when ##z_3=4390m ## ? at what pressure will the water avoid freezing 260.77K temperature ? if this pressure is reached inside the pipe (at ##P_3##) then the water will not freeze and siphoning will continue right ?

n6LXj.gif


according to the above diagram water will freeze at 58.5 kPa at 260.77K so the experiment will not work ? freezing can stop siphoning i guess.
 
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  • #7
rude man
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according to the above diagram water will freeze at 58.5 kPa at 260.77K so the experiment will not work ? freezing can stop siphoning i guess.
You're still bogged down on worrying over state (p-V-T) diagrams. There is nothing in your problem statement that says that the water is frozen at the higher elevation. It might be a nice sunny summer day up there! So assume water, not ice, everywhere.

So, as I said, think atmospheric pressure, not vapor pressure.
Write the Bernoulli equation at the surface of the vessel from which siphoning is to take place. Write it again at the top of the siphon tube. This allows you to determine the pressure at the top of the tube. Assuming the same siphon height of 8.5m (between the surface and the top), what value of p do you get at the top when the surface is at sea level (approx. 100 kPa) and again when the surface is at 58.5 kPa?
(Hint: velocity v at the top must be >0).


-
 
  • #8
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At surface
##P = P_(atm) ##
##z=0##
##v=0## assuming a large vessel

At 8.5 m
## P=P_t ## pressure at the top
## z=8.5m ##
## v=v_t ## velocity at the top
So ,
## P_(atm)/\textrho = P_t/\textrho + v_t^2/2 + 8.5g ##


at the elevation of 4390m

## 58.5kPa/\textrho = P/\textrho +v^2/2 + 8.5g ##

there are two unknowns in both the equations !!
 
  • #9
rude man
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At surface
##P = P_(atm) ##
##z=0##
##v=0## assuming a large vessel

At 8.5 m
## P=P_t ## pressure at the top
## z=8.5m ##
## v=v_t ## velocity at the top
So ,
## P_(atm)/\textrho = P_t/\textrho + v_t^2/2 + 8.5g ##
at the elevation of 4390m
## 58.5kPa/\textrho = P/\textrho +v^2/2 + 8.5g ##
there are two unknowns in both the equations !!
This is all OK.
In your last equation above, what can you say about (pt/ρ + v2/2)?
Can you have negative pressure?
Compare with patm = 100kPa, notice what's different?

(BTW I misspoke last time when I said you can compute pressure at the top. You can only compute the sum of the pressure and the kinetic energy.)
 
  • #10
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We get the sum of pressure and kinetic energy as negative so that's the reason it won't work ?
 
  • #11
rude man
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We get the sum of pressure and kinetic energy as negative so that's the reason it won't work ?
Right! The kinetic energy term is always positive, so p would have to be negative.
Did you run the equation for sea level atmospheric pressure (~100kPa) & note the difference from the high-altitude case?
 
  • #12
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Right! The kinetic energy term is always positive, so p would have to be negative.
Did you run the equation for sea level atmospheric pressure (~100kPa) & note the difference from the high-altitude case?
Yea , i got a positive number so no need for pressure to be negative hence it is possible.

Thanks
 

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