7 replies to this topic

[CuriousGeorge]

I know this should be obvious. And, engineers say it everyday as if it goes without question. But, I am missing something basic here! When venting to atmosphere, engineers say "well, your pressure at the end of the line HAS to be atmospheric". Well, I don't see the "why" and "how" this is so obvious.

If I relieve at 100psig over a foot of pipe open atmosphere, and there's only 10 psi drop (let's assume) across the pipe, then how in the world will I ever reach 0 psig? Won't I be venting into the atmosphere at 90psig (in this example)?

From where I sit, I'd say something must physically happen (i.e., a physical restriction in the pipeline) to get me down to 0psig; yet, what is happening that makes it so obvious to others? Something so subtle must be eluding me. Please explain.

[djack77494]

George,
The simplest way to visualize this - work backwards. Starting with atmospheric pressure, back up into the tail pipe a short distance. What's the pressure? It's darn near atmospheric because the only resistance is the short peice of pipe, and that's not much resistance. How can it be otherwise?

OK, now that we "know" the answer, let's see how this looks if we instead work forwards. I assume you're talking about a relief valve (or PSV) that might have an appreciable set point. Let's call it 100 psig (or 7 barg if you prefer). During a discharge your fluid flows from containment at 100 psig thru your PSV, then a tail pipe, then out to atmosphere. (Or else into a nearly atmospheric flare header.) Between the containment and atmosphere, you have 100 psig - 0 psig = 100 psid of differential pressure. Alternately, you have 100 psi (7 bar) of hydraulic losses. The bulk of these losses typically occurs at the PSV. What else if there to take a large delta P? So, the pressure at the upstream connection to the PSV is nearly 100 psig, and the pressure at the downstream connection to the PSV is nearly atmospheric. (I'm not considering the possibility of sonic flow in the tail pipe. That would complicate things by negating our assumption of small dP in the tail pipe.)

[CuriousGeorge]

djack, I appreciate your help in trying to explain, so please bear with me.

Let's assume water relief to keep it very simple protecting a positive displacement pump of 100gpm:
1) I agree 100psig must be the differential pressure, i.e. dP = 100psig because I know the start and end point pressures.
2) In contrast, let's say I have 100-ft of straight pipe (tweaking my example a bit) of 3-in, Sch 40 std wt steel pipe gives me a loss of 2.37 ft (or ~1psi) per Cameron. This says the piping alone will not bring you down to 0psig.

Am I fixing too many parameters, or something?

[JoeWong]

CuriousGeorge,

Hope these posts will give you some clue.

• A refresh to Process Engineer on few phenomenons in restriction orifice
• Discussion on ISENTROPIC and ISENTHALPIC process via Relief Valve

[CuriousGeorge]

CuriousGeorge,

Hope these posts will give you some clue.

Thanks, Joewong. I guess the short answer is because it is the nature of the beast for a vena contracta.

[Qalander (Chem)]

CuriousGeorge,

Hope these posts will give you some clue.

Thanks, Joewong. I guess the short answer is because it is the nature of the beast for a vena contracta.

Dear CuriousGeorge Hello, Somehow your 'Nick' here reminds me of my original 'childhood' nature of being exra-ordinarily curious. Thus prompted to pop in this thread with a slight general conceptual explanation.
It is true that Liquids/Gases are Called Fluids.
It is also true that, while venting into atmosphere their i.e.liquids and Gases behavier greatly varies.
It is also true that your explanation using water does not cover all fluids.
It is also essentially true that to affect flow slight positive(i.e above -atmospheric) has be physically existant in close vicinity of vent pipe final end.
Now mathematical co-relationships can be worked out accurately to provide/prove the possible numerical values by your goodself to reply to your own question.
I expect that, this logical rationale may be accepable to you for resolution.
Best regards
Qalander

[djack77494]

djack, I appreciate your help in trying to explain, so please bear with me.

Let's assume water relief to keep it very simple protecting a positive displacement pump of 100gpm:
1) I agree 100psig must be the differential pressure, i.e. dP = 100psig because I know the start and end point pressures.
2) In contrast, let's say I have 100-ft of straight pipe (tweaking my example a bit) of 3-in, Sch 40 std wt steel pipe gives me a loss of 2.37 ft (or ~1psi) per Cameron. This says the piping alone will not bring you down to 0psig.

Am I fixing too many parameters, or something?

Having a pump capacity of 100 gpm has no relationship whatsoever with a differential pressure of 100 psid. Now you have 100 ft of pipe making too many 100's; hard to differentiate; your example is confusing and even nonsensical. The pressure at the discharge port of a PD pump will be anywhere from 0 psig to the mechanical limit of the pump and its driver. It will build to whatever pressure is needed to overcome discharge restrictions, up to the point where it stops working or breaks. If you removed all your discharge piping, the pressure would be atmospheric. It doesn't matter what the pump's nameplate says - that's just a maximum possible pressure. Add 100 ft of piping, and the pressure at the discharge port of the pump will be ~1 psig. Instead have 5000 ft of pipe and you're up to 50 psig. You get the idea. Nothing says that the pump discharge pressure WILL BE 100 psig. Hope that helps, and if you'd like to explore these concepts further, please strive to be precise and accurate. Those are becoming increasingly rare but remain absolutely essential in technical fields.

[CuriousGeorge]

[/quote]...Now mathematical co-relationships can be worked out accurately to provide/prove the possible numerical values by your goodself to reply to your own question. I expect that, this logical rationale may be accetpable to you for resolution.[/quote]

Thanks for assuring me I could derive my own mathematical proof and answer my own question. But, I am simply trying to conceptualize what the theory says is happening. If you prefer dealing in theory rather than the hands-on practical, to each his/her own.