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.

 

[dcasto]

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.

 

[BigInch]

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

 

[quark]

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.

 

[BigInch]

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

 

[KernOily]

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