Understanding the movement of a fluid in a vertical pipe

[whburling]

The actual problem I am trying to understand is the suction requirement to move #2 home heating fuel from the outdoor tank to the pump of the burner located in my cellar.

But first i just want to understand the movement of fluid in a vertical pipe without worrying about viscosity or temperature and assuming that I must suck the fluid in a vertical pipe to move it out the top.

Fundamentally, my vision of the problem is that i am limited by the suction i can create above the vertical pipe
that would lift the fluid's weight. In an ideal world (not worrying about the complexity of #2 fuel oil), I am limited by 14.7 psi suction. I am assuming I can not create any better suction than 14.7.

so now i am trying to figure out how i might calculate the weight of the fluid in the pipe that must be lifted. as I imagined this pipe, I began to see that I had to put the pipe in some sort of bathtub. If i filled the bathtub and put the tube into the bathtub so the vast majoirity of the tube is filled. I am guessing this case is the easiest. I only need to lift the fluid the height of the tube sticking above the bathtub fluid surface.

if I put the tube into the bathtub so the vast majority of the tube is outside of the bathtub (hence empty), I am guessing I will be lifting the most fluid weight (the weight of the fuild filling the empty part of the tube above the bathtub.

now given the above images, I am now having trouble thinking. I am guessing if i can compute the positive pressure necessary to push the fluid to the top of the pipe. that would also be the suction pressure necessary to suck the fluid from teh bathtub up the tube to the top.

i am not sure how to do this. I know P = F/A. I am assuming F = fluid density * volume of the empty pipe above the bathtub. A equals the cross sectional area of the pipe. Volume = A * Length of empty pipe.
Areas cancel out So I am left with P = Density * Length of empty Pipe.

the above P must be less than 14.7 psi otherwise the fluid will not reach the top of the pipe.

Is the above thinking correct? Is there a way of thinking about this problem which makes it more clear for my mind?

*******************
to move beyond the above ideal problem to #2 fuel oil, I must consider that at some vacuum the fuel oil comes apart and hence the light molecules collect in the inlet side of the pump (as gas) and hence stop the pumping action. so my limit is not 14.7 psi suction but some magic suction (probably around 10" Hg).

I also must consider that the system is dynamic and hence must consider pressure drop of fuel oil through the elbows and long runs of the copper tubing. this entails knowing viscosity of #2 fuel oil for a given temperature and the fact that the oil tends to generate parafin.at cold temperatures.

but until i understand a simple case, the above complexity is not of interest to me.

 

[Baluncore]

Welcome to PF.
You need to provide a diagram.
Is it a siphon? https://en.wikipedia.org/wiki/Siphon

 

[some bloke]

What you are essentially doing is using your pressure in the pipe to overcome the force due to gravity, the weight of the fluid.

You will have to consider the siphon effect - due to the incompressibility of fluids, if the exit of the pipe is lower than the entrance, then the fluid will flow through the pipe, even if it goes over an obstacle. This is how people steal petrol out of cars!

As such, you need to look at the entire pipe system, and not just the vertical section. If the vertical section is to climb a wall, but you subsequently drop down again (you said your boiler is in the cellar) then the only height which is of importance is the difference in height between the entrance and exit of the pipe, provided that the pipe has no airlocks or leaks.

It is also worth considering that the fluid level is where you take your entrance to be - so if you have a bath of fluid which is 0.5m deep, and the pipe is at the bottom, then if it sits without any external force, It will already be 0.5m up the pipe, so this is the starting point. if the bath will be replenished, use this; if not, then use the worst case scenario for your calculations - an almost-depleted bath. The difference between this entrance and your exit point is the Height, H.

Now onto the pressure difference you plan to use.

It's a lot easier for me to convert to SI units, so 14.7Psi = 101352.9 Pascal (Newtons of force per m²).

Force of fluid in pipe is H x A x p x g (Height (m) x cross sectional Area of pipe (m³) x Density of fluid (Kg per m³)x acceleration due to gravity (9.81))
Force of pressure in your pipe is P x A (Pressure (101352.9) x cross sectional Area of pipe (m³))

For the fluid to flow, the force due to pressure must exceed the force due to the fluid, so

H x A x p x g < P x A

Area cancels out, leaving

H x p x g < P

As the density of the fluid and acceleration due to gravity is a constant, H is proportional to P, as we would expect - more pressure = more height.

if you plug your numbers in, then you should get a good approximation of how high your pressure can push the fluid.

Assuming it was distilled water (p=1000)

H x 1000 x 10 < P
10,000H < 101352.9
ergo H < 10.13529

so it should suck water up by about 10m.

Make sure that someone else confirms my maths before you base your whole system off of it though!

Hope this helps!

 

[some bloke]

Hi all,

I answered a thread in the General Engineering forum, but once I re-read my conclusion, it seemed wrong. I checked and double checked the workings and it looks like I've done the right thing, but it still appears to be out by a factor of 10 to what I would expect, as the pressure involved is about 1 bar, and the result was that it would pump water up pipe by 10m; I would anticipate it pumping it up by 1m.

My (long winded) response:

What you are essentially doing is using your pressure in the pipe to overcome the force due to gravity, the weight of the fluid.

You will have to consider the siphon effect - due to the incompressibility of fluids, if the exit of the pipe is lower than the entrance, then the fluid will flow through the pipe, even if it goes over an obstacle. This is how people steal petrol out of cars!

As such, you need to look at the entire pipe system, and not just the vertical section. If the vertical section is to climb a wall, but you subsequently drop down again (you said your boiler is in the cellar) then the only height which is of importance is the difference in height between the entrance and exit of the pipe, provided that the pipe has no airlocks or leaks.

It is also worth considering that the fluid level is where you take your entrance to be - so if you have a bath of fluid which is 0.5m deep, and the pipe is at the bottom, then if it sits without any external force, It will already be 0.5m up the pipe, so this is the starting point. if the bath will be replenished, use this; if not, then use the worst case scenario for your calculations - an almost-depleted bath. The difference between this entrance and your exit point is the Height, H.

Now onto the pressure difference you plan to use.

It's a lot easier for me to convert to SI units, so 14.7Psi = 101352.9 Pascal (Newtons of force per m²).

Force of fluid in pipe is H x A x p x g (Height (m) x cross sectional Area of pipe (m³) x Density of fluid (Kg per m³)x acceleration due to gravity (9.81))
Force of pressure in your pipe is P x A (Pressure (101352.9) x cross sectional Area of pipe (m³))

For the fluid to flow, the force due to pressure must exceed the force due to the fluid, so

H x A x p x g < P x A

Area cancels out, leaving

H x p x g < P

As the density of the fluid and acceleration due to gravity is a constant, H is proportional to P, as we would expect - more pressure = more height.

if you plug your numbers in, then you should get a good approximation of how high your pressure can push the fluid.

Assuming it was distilled water (p=1000)

H x 1000 x 10 < P
10,000H < 101352.9
ergo H < 10.13529

so it should suck water up by about 10m.

Make sure that someone else confirms my maths before you base your whole system off of it though!

Hope this helps!

Did I get this right? I would hate to have given misinformation!

 

[jim hardy]

Is the above thinking correct? Is there a way of thinking about this problem which makes it more clear for my mind?

your thinking is right . What messes people up is using inconsistent units.

If you use density in pounds per cubic foot and height in feet you'll get pressure in pounds per square foot.

Might be easier to ratio it to water.
You know that pressure of one atmosphere, 14.7 psi, will push water up 32 feet.
It'll lift a lighter fluid like kerosene farther,
by the ratio $\frac{specific ~gravity~ of~ water}{specific ~gravity~ of ~kerosene}$

give that a try ? How many feet per psi do you come up with ?
Just be aware you can only raise vacuum to the point your fluid boils, (1 atmosphere) minus (fuel's vapor pressure).old jim

 

[whburling]

your thinking is right . What messes people up is using inconsistent units.

If you use density in pounds per cubic foot and height in feet you'll get pressure in pounds per square foot.

Might be easier to ratio it to water.
You know that pressure of one atmosphere, 14.7 psi, will push water up 32 feet.
It'll lift a lighter fluid like kerosene farther,
by the ratio $\frac{specific ~gravity~ of~ water}{specific ~gravity~ of ~kerosene}$

give that a try ? How many feet per psi do you come up with ?
Just be aware you can only raise vacuum to the point your fluid boils, (1 atmosphere) minus (fuel's vapor pressure).old jim

 

[whburling]

Hi guys!

Thank you for replying.

I need to add information to this problem in order to convey my concern.

Picture the oil tank above ground
And the boiler burner (which has the heating oil pump attached) is in the cellar.

Imagine NO oil is in the fuel line connecting the outside above ground oil tank to the pump in the cellar. Imagine the fuel line end is sitting in the tank 2” above the tank bottom.
The oil tank has enough oil to fill the tank 3” above the bottom (thus almost empty).

If i trace the fuel line elevation from tank to pump, the highest fuel line relative elevation is 3” above the top of the fuel tank.

I am assuming the pump must first suck the 3” of oil in the tank to the highest fuel line elevation. Once that happens i am assuming the pump suction does not matter anymore as long as there is enough fuel in the tank to completely fill the fuel line from tank to pump. Siphon action will fill the line as long as the slightest vacuum is maintained by the pump.

So how do i calculate the suction (temporarily setting aside the vapor pressure issue)?

I am guessing that i need to calculate the pressure delta that can lift the weight of the fuel oil occupying the fuel line from 3” off the bottom of the tank to 3” above the tank.

The pressure in the fuel line at 3” above the tank bottom is atmospheric pressure (assuming) and the suction pressure is pressure below atmospheric.

If the above is correct, then your calculations should give me my answer. Correct?

 

[some bloke]

Hi guys!

Thank you for replying.

I need to add information to this problem in order to convey my concern.

Picture the oil tank above ground
And the boiler burner (which has the heating oil pump attached) is in the cellar.

Imagine NO oil is in the fuel line connecting the outside above ground oil tank to the pump in the cellar. Imagine the fuel line end is sitting in the tank 2” above the tank bottom.
The oil tank has enough oil to fill the tank 3” above the bottom (thus almost empty).

If i trace the fuel line elevation from tank to pump, the highest fuel line relative elevation is 3” above the top of the fuel tank.

I am assuming the pump must first suck the 3” of oil in the tank to the highest fuel line elevation. Once that happens i am assuming the pump suction does not matter anymore as long as there is enough fuel in the tank to completely fill the fuel line from tank to pump. Siphon action will fill the line as long as the slightest vacuum is maintained by the pump.

So how do i calculate the suction (temporarily setting aside the vapor pressure issue)?

I am guessing that i need to calculate the pressure delta that can lift the weight of the fuel oil occupying the fuel line from 3” off the bottom of the tank to 3” above the tank.

The pressure in the fuel line at 3” above the tank bottom is atmospheric pressure (assuming) and the suction pressure is pressure below atmospheric.

If the above is correct, then your calculations should give me my answer. Correct?

Yes, essentially once your fluid has enough pressure to reach the peak, it should not need any additional pressure to continue flowing, as each unit of fluid which is raised is offset by a unit of fluid which falls. However, I believe that this will only be entirely true if the pipes are vertical - if a pipe is on a gentle slope, then the fluid moving 1m along it will not be dropping by 1m, so will not suck up 1m of fluid on a vertical pipe.

If you position your vacuum pump on the "down" side, below the level of the fluid held in the tank, then if it successfully sucks the fluid to fill the pipe up to itself, if it is turned off and a valve opened further down to let the fluid out, the siphon effect should be sufficient to "suck" the fluid through the pipe from there on.

If you placed the pump on the peak of the pipe system (3" above the tank) then you will suck the fluid to the peak, but then the siphon effect will simply suck it back. You need to get the "front" of the fluid to below where it started for it to siphon through. It's the weight of the fluid which pulls it down - think of it like a set of scales - the heaviest side moves down, the lighter side moves up. you need more weight of fluid on one side than the other in order to achieve a siphon.

I doubt that there will be an issue with vaporising the fuel, but if there is, then you might consider a positive pressure in the tank to increase the pressure difference instead of increasing your vacuum. Alternatively, you might be able to use one-way valves to push & pull the liquid through in several locations to the point where the siphon takes hold.

 

[whburling]

Yes, essentially once your fluid has enough pressure to reach the peak, it should not need any additional pressure to continue flowing, as each unit of fluid which is raised is offset by a unit of fluid which falls. However, I believe that this will only be entirely true if the pipes are vertical - if a pipe is on a gentle slope, then the fluid moving 1m along it will not be dropping by 1m, so will not suck up 1m of fluid on a vertical pipe.

If you position your vacuum pump on the "down" side, below the level of the fluid held in the tank, then if it successfully sucks the fluid to fill the pipe up to itself, if it is turned off and a valve opened further down to let the fluid out, the siphon effect should be sufficient to "suck" the fluid through the pipe from there on.

If you placed the pump on the peak of the pipe system (3" above the tank) then you will suck the fluid to the peak, but then the siphon effect will simply suck it back. You need to get the "front" of the fluid to below where it started for it to siphon through. It's the weight of the fluid which pulls it down - think of it like a set of scales - the heaviest side moves down, the lighter side moves up. you need more weight of fluid on one side than the other in order to achieve a siphon.

I doubt that there will be an issue with vaporising the fuel, but if there is, then you might consider a positive pressure in the tank to increase the pressure difference instead of increasing your vacuum. Alternatively, you might be able to use one-way valves to push & pull the liquid through in several locations to the point where the siphon takes hold.

You guys are so wonderful.

You are patient. All of you. You give me ways to visualize so that the mathematics makes sense (the equations no longer need to be memorized, but make sense).

I can't underscore, enough, how grateful i am to all of you.

Thank you.
bil

 

[some bloke]

You guys are so wonderful.

You are patient. All of you. You give me ways to visualize so that the mathematics makes sense (the equations no longer need to be memorized, but make sense).

I can't underscore, enough, how grateful i am to all of you.

Thank you.
bil

You are most welcome, I'm glad I can help. I'm the same, if I can't picture how something works and why it does, all the equations in the world will do me no good. calculating exact values is one thing, but without context, it's as useful a 2+2=4 - it's correct, but means nothing!

 

[jim hardy]

(the equations no longer need to be memorized, but make sense).

I'm the same, if I can't picture how something works and why it does, all the equations in the world will do me no good.

yep.

old jim