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

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

(OP)
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.

### RE: Gravity Flow Question: I'm MIssing Something Here...

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.

### RE: Gravity Flow Question: I'm MIssing Something Here...

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.

### RE: Gravity Flow Question: I'm MIssing Something Here...

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,
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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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.

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

(OP)
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

### RE: Gravity Flow Question: I'm MIssing Something Here...

my suggestion on watching for vapors is from the OP saying "mostly water"  What if there was suspended light hydrocarbons, or entrained air. at 160 F, could it be 180F, then at that slight vacuum, again gas releases.

### RE: Gravity Flow Question: I'm MIssing Something Here...

dcasto, you have a point there.

KernOily, Slack or cascade flow in a drain system is not unusual and normally doesn't present problems, as the flow isn't being metered, no leak detection systems, no control valves, product batch interfaces to mix, no pumps taking vapors and liquids etc.  The liquid just flows faster in the wetted perimeter and the bubble moves back and forth when the flow is high.  ... As long as you have enough available head to move the fluid to the high point plus a bit.

< 10% disagreement in hydraulics is an honest day's work.

Put a valve in the run to the middle tank.  It will probably take sometimes and give at others, unless you are remarkably good at maintaining all tanks at set levels.   If you can do that, you usually don't need tanks.

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

I didn't yet get the seriousness of the issue at hand. As we all know, the calculated static head at the junction is an instantaneous value (infact I missed it for total head in my first post) and shouldn't worry us much. When the level in the lower tank rises, the flow rate reduces and tends to bring the static head at the junction to 0 psig.

### RE: Gravity Flow Question: I'm MIssing Something Here...

No worries, we were just comparing "spreadsheets" and trying to get some theoretical consistancy going.  In steady state all values last longer than instantaneous would suggest and if you can't calculate something in steady state, its really hard to try to figure out where those tank levels wind up when they do start changing.

Speaking of transient flows, I think I like my conversion of velocity head to static head solution more than considering that velocity head is lost to turbulence.  That mechanism provides any necessary energy to raise the liquid level of the lower tank in the case where the tank's outlet is closed.  Proportioning velocity head and pressure head at the tank entrance could be considered by using the entrance coefficient, but afterwards inside the tank, I think that when velocity finally slows to zero, most of that would have been disipated by conversion into static head inside the tank anyway.  What do you think?

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

### RE: Gravity Flow Question: I'm MIssing Something Here...

(OP)
Hi quark.  Sorry I missed your question.  I left out a bunch of details to keep the original question as concise  as possible.  Your question is of course totally valid.

The level in the lower tank remains constant because it is on level control.  It is the plant drains tank and is pumped out on level control.  The 7' level shown on the sketch is NLL.

The level in the upper tank isn't technically in question because the tank outlet nozzle is at the elevation shown in the sketch.  This is the tank overflow nozzle and there is an internal riser starting 3' above the tank floor and rising up to this nozzle.  So the level of 23' shown on the sketch is actually the level at the overflow nozzle.  A siphon breaker is located inside the tank at the top of the riser (it's an open-ended tee).  The level in the tank may actually be only 4' off the floor, in which case liquid has to go up the riser to the nozzle to get out of the tank.  The pressure in the tank will of course have to increase to push the fluid up the overflow riser and out the nozzle if the overflow event is of that nature.  The PVRV on the tank roof will pop before that ever happens (we hope!).  So an overflow event will likely mean the tank liquid level is at the level of the overflow nozzle.

The first horizontal pipe segment out of the nozzle will be in open-channel flow.

Thanks guys!  Pete

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