Gravity Flow Question: I'm MIssing Something Here... 1

[KernOily]

Hi guys. I am working a system that drains one tank by gravity through a pipeline to another tank at lower elevation. My simulator gives me an answer I don't trust. I need some insight and counsel into what is actually happening.

The upper end of the pipeline is 30' higher than the lower end. The line consist of 827' of 10" s40 and 225' of 18" std. The tank at the upper end of the line has 24' of fluid in it and the lower tank has 2' of fluid above the outlet of the pipe. Both tanks are atmospheric, fluid T is 160 F, fluid is mostly water.

So to determine the rate in the pipe, I iterated flowrate until I used up the head difference. That rate is 3735 gpm. Problem is, the pressure in the line goes negative about 2/3 the way down the line because of the friction loss. So I start with 24' of tank head at the top going into the line, then at about L=850' the pressure in the line drops to -2.5 psig or so, then it recovers to 0.9 psig at the end (at the bottom) because that is the head of the fluid above the pipe exit in the bottom tank. I attached the pressure profile. It has a sharp discontinuity where the pressure goes negative. I've never seen this before.

I attached a sketch of the system and the pressure profile.

Is this line going into slack line? I say no, because -2.5 psig is still much higher than Pvap at 160 F.

I am missing something fundamental. It's right in front of my face but I don't see it. Thanks in advance for any insight/help/rude comments/reality slaps-in-the-face.

 

[Zapster]

Total hydraulic head is composed of pressure head, velocity head and elevation head. Is your plot total hydraulic head? If it is not total head, there is a logical explanation.

 

[quark]

I couldn't get your question.

If you say you iterated flowrates (and you got near correct answer) to maintain the frictional loss equal to the available head then where is the question of negative pressure existing in the system?

The pressure, ideally, should be 53 feet at the pipe exit from top tank and 2 feet at the entry into the bottom tank. The difference in static head is 51 feet and this is offset by friction.

However, the flowrate obtained by equating the static head to the frictional loss is not valid during the entire draining process. As the water level in the top tank reduces, the available head for flow decreases and thus flow decreases. The average flowrate will be half the flowrate you calculated.

PS: Check whether you are caluclating the pressure drop upto 1050 (instead of 827 feet with 10" and 225 feet with 18") feet with 10" pipe size. This is the only calculation error that gives you negative pressure.

 

[trashcanman]

The actual difference in head = (30-2)+23=51' (from the sketch)

Using Hazen-Williams equation,
hf=0.002083*((100/c)^1.85)*Q^1.85/(D^4.865)(per ft. of length)

Assuming C=120 and no fitting losses, then the 51' of head is used up to provide flow of Q which is the same for both lengths of pipe.

.002083*((100/120)^1.85)*((Q^1.85)/(10.02^4.865))*827+
.002083*((100/120)^1.85)*((Q^1.85)/(17.25^4.865))*225 = 51

solving for Q, Q = 3011 gpm

This assumes the head difference remains constant.

 

[BigInch]

Flowrate = 3920 gpm
Velocity Head in 10" = 3.63 ft
Velocity head in 18" = 0.43 ft

Plot these points,

Tank outlet Total Head (Ht)=53 ft

10" inlet Ht=53 ft,
Velocity Head (Hv)= 3.63 ft
Static Head (Hs)= 53 - 3.63 = 49.37 ft

10" head loss = 50.08 ft

10" outlet Ht = 2.92', Hv = 3.63', Hs = 2.92-3.63= -0.70 ft

10"/18" joint
(pressure head increases by 3.2 ft due to Bernoulli effect)

18" inlet Ht = 2.92', Hv = 0.43', Hs = 2.92-0.43= 2.50 ft
(some round off err there)

18" head loss = 0.92 ft

18" outlet Ht = 2.01, Hv = 0.43', Hs = 2.01-0.43 = 1.58 ft

Lower Tank Inlet
Hs = 1.58 ft
Hv = 0.43 ft
Total Head = 1.58 + 0.43 = 2.01 ft
Ht = 2.01 feet

Those were "gage heads", so add 33 ft to all to see absolute heads. Check your vapor pressures according to absolutes and its easier to see where you really are.

http://virtualpipeline.spaces.msn.com

"We can't solve problems by using the same kind of thinking we used when we created them." -Albert Einstein

 

[BigInch]

At least that's what I get using my 2 pipes, head loss between 2 tanks spreadsheet based on the Churchill eq and 0.25 wt for both pipes

http://virtualpipeline.spaces.msn.com

"We can't solve problems by using the same kind of thinking we used when we created them." -Albert Einstein

 

[dcasto]

I get similar results as the OP, 3700 gpm and a 9.6 psia at the change from 10" to 18". Its all reasonable.

What may happen in real life is a vapor formation at the low pressure point and the friction drop would increase and the flowrate would drop to compensate.

 

[BigInch]

dcasto, did you include velocity head?

in the 10" at 15 fps v^2/2/g = 3.5 ft and
in the 18" at 5 fps its v^2/2/g = 0.4 ft
change in velocity head = 3.1 ft
3.1 ft *.43 psi/ft = only a 1.34 psia change in pressure at the 10-18" junction.

http://virtualpipeline.spaces.msn.com

"We can't solve problems by using the same kind of thinking we used when we created them." -Albert Einstein

 

[BigInch]

P.S. No vapor formation, as absolute pressure at the 10-18 joint is 16 psia. Water would have to be over 212F or the elevation of this system would have to be at 5000 feet with 60F water, or some similar reduction of atmospheric pressure would have to had occured.

If anything some entrained air might form larger bubbles which reverse flow back up the top of the pipe to the high tank as the water flows slightly faster below or around any traveling bubbles.

http://virtualpipeline.spaces.msn.com

"We can't solve problems by using the same kind of thinking we used when we created them." -Albert Einstein

 

[BigInch]

sorry, 10,000 feet elevation.

http://virtualpipeline.spaces.msn.com

"We can't solve problems by using the same kind of thinking we used when we created them." -Albert Einstein

 

[BigInch]

DISTANCE FT 0 0.1 827 827.1 1052 1052.1
T-HEAD 53 53 2.92 2.92 2.01 2.01
P-HEAD FT 53 49.37 -0.70 2.50 1.58 2.00
PSIG 22.96 21.39 -0.31 1.08 0.69 0.87
PSIA 37.96 36.39 14.69 16.08 15.69 15.87

OK, I see 14.69 psia is the minimum, still around 212 F to boil.

http://virtualpipeline.spaces.msn.com

"We can't solve problems by using the same kind of thinking we used when we created them." -Albert Einstein

 

[katmar]

I agree that the pressure reaches a minimum at the point of the 10"/18" reducer, and my numbers are very close to what your simulator gives. Basically, I have done the same analysis as BigInch.

The numbers are not too important - we are looking at the principle here - but for the sake of example the numbers I used are

Density = 61 lb/ft³
Viscosity = 0.4 cP
10" pipe ID = 10.02"
18" pipe ID = 17.25"
Pipe roughness = 0.00197" (=0.05 mm)

Taking the exit velocity from BigInch's calcs gives me a velocity head loss into tank 2 of about 0.4 ft. This makes the overall head available 50.6 ft (i.e. 23+30-2-0.4). Using this head I get a flowrate (when tank 1 is full) of 3645 USGPM. As I said, the numbers are not too important but all further calcs are based on the flow being 3645 USGPM.

As BigInch has pointed out, the pressure increases as the liquid flows through the reducer because the velocity decreases and the velocity head is converted to static pressure. However, BigInch neglected the friction losses through the reducer. The velocity head reduces from 3.34 ft to 0.38 ft in going through the reducer, giving an available increase of 2.96 ft. But the friction loss due to what is basically a sudden expansion (assuming a standard pipe reducer) is 1.47 ft. The actual pressure increase across the reducer is therefore 2.96 - 1.47 = 1.49 ft.

Now we have to compare this with the pressure drop due to friction across the 225 ft of 18" std pipe. By my calcs this is only 0.78 ft. Now we can work back from the entrance to tank 2. The static head at the entrance to tank 2 is 2 ft. Adding the friction head in the 18" pipe to this would make the static head at the start of the 18" pipe 2.78 ft. However, if we assume that the slope of the whole 1052 ft (=827+225) of pipe is uniform then the reducer is 6.4 ft above the entrance to tank 2. This makes the static head at the start of the 18" pipe 2.78 - 6.4 = -3.62 ft.

The pressure increase previously calculated across the reducer must be subtracted from this to find the pressure at the end of the 10" pipe section. This makes the static head at the end of the 10" pipe (-3.62-1.49) -5.11 ft. This is -2.2 psi. We could argue over the actual friction loss in the reducer, and the pipe roughness, and the liquid physical properties and so on, but I believe my numbers are close enough to the simulator output to give you some confidence in your calcs.

Hope this helps
Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[katmar]

Further to my earlier post - although I agree with the minimum pressure point found, I disagree with the shape of the curve. The first section from tank 1 to the reducer would be as shown in the sketch, but there would be a sudden increase at the reducer and then a very gentle slope up to tank 2.

Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[BigInch]

I didn't neglect it, my fitting is freictionless.

http://virtualpipeline.spaces.msn.com

"We can't solve problems by using the same kind of thinking we used when we created them." -Albert Einstein

 

[BigInch]

Now we have to compare this with the pressure drop due to friction across the 225 ft of 18" std pipe. By my calcs this is only 0.78 ft. Now we can work back from the entrance to tank 2. The static head at the entrance to tank 2 in Tank 2 is 2 ft - vel head 0.4 = 1.6 ft at the exit of the 18". Adding the friction head in the 18" pipe to this would make the static head at the start of the 18" pipe 2.78 ft 1.6 + 0.78 - 0.4 = 2.38 ft. However, if we assume that the slope of the whole 1052 ft (=827+225) of pipe is uniform then the reducer is 6.4 ft above the entrance to tank 2. This makes the static head at the start of the 18" pipe 2.78 - 6.4 = -3.62 ft. 2.38 - 6.4 = -4.02 ft

http://virtualpipeline.spaces.msn.com

"We can't solve problems by using the same kind of thinking we used when we created them." -Albert Einstein

 

[BigInch]

1.6 + 0.78 = 2.38 ft.

http://virtualpipeline.spaces.msn.com

"We can't solve problems by using the same kind of thinking we used when we created them." -Albert Einstein

 

[BigInch]

Harvey, what's the head loss in your 10"

http://virtualpipeline.spaces.msn.com

"We can't solve problems by using the same kind of thinking we used when we created them." -Albert Einstein

 

[katmar]

Hi BigInch, I don't think that you can subtract the velocity head to find the static head at the entrance to tank 2. I believe the velocity component would be lost in turbulence in the tank and would not be recoverable. IMHO if you put a pressure gauge at the nozzle you would read 2 ft of head. But this is splitting hairs I suppose.

My pressure drop for the 10" section is 51.5 ft at a flow of 3645 GPM. This is made up of 1.7 ft for the flush entrance from tank to pipe, 46.5 ft for the 827 ft of straight pipe (including the frictionless elbows!) and 3.3 ft for the generation of the velocity head. The static height from the base of tank 1 to the reducer is 23.6 ft (assuming constant slope) giving a static head of 46.6 ft (i.e. 23 + 23.6). The static head under flow conditions at the end of the 10" line would therefore be (46.6-51.5) -4.9 ft. This compares with the -5.1 ft I got previously when calculating from the tank 2 end. The difference will be rounding errors I suppose.

Harvey


Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[BigInch]

OK, I prefer the (theoretical) change of velocity head to static head, so I don't have to assume there is the turbulent loss in tank 2 (A guy that uses frictionless joints doesn't like to consider tank turbulence). There is also an entrance coefficient there, in addition to the exit coefficient at tank 1, each of which I neglected and fjor which you considered them both, if I assume that exit coefficient makes up some of that "tank turbulence".

Where'd you get the frictionless elbows? I don't have any of those yet... just reducers. Trade?

http://virtualpipeline.spaces.msn.com

"We can't solve problems by using the same kind of thinking we used when we created them." -Albert Einstein

 

[KernOily]

Awesome replies guys - thanks for taking the time to look at my little problem and for getting back to me.

I have messed with this some more and I keep getting the same answer, more or less, within a few percent, as y'all did. Hand calc, using Darcy eqn - more or less same answer. Spreadsheet calc, also using Darcy - same answer. I did not add the 18x10 reducer in the simulator but I will do that. I also did not add any entrance or exit losses, which I need to do. Simulator of course accounts for all velocity head effects, which I ignored in my kwikie hand calc.

The shape of the pressure profile plotted by the simulator is controlled by the pipe length segmentation in the simulator. If I could divide the pipe to an infinite number of segments I would get a nice smooth curve, as Harvey suggests. The discontinuity at the point of low pressure in my original curve was messing with my head until I read Harvey's post on this and realized it is a segmentation thing.

I was concerned about slack line flow but I'm not near Pvap so that's not an issue, thank goodness.

This system actually is part of a classic three-reservoir problem. There is a third tank, at an elevation between these two, connected near the 18x10 reducer. I'm now on to that little problem, to determine whether the middle tank is a source or a sink.

Thanks again for the help. BigInch/katmar - Just want to say I appreciate your posts here on ET. You add a lot to the quality of the community here. Informative, accurate, cogent, helpful - a high signal-to-noise ratio, as it were.

BTW I have a pipeyard full of frictionless fittings here in Bakersfield - yours for the asking. One of my clients took them out of a system and replaced them with 'frictional' fittings because the frictionless kept making his pump run off the curve