PSV Discharge Piping and Backpressure 2

Perth1,

The exit loss is real and is going to happen, therefore it must be included. It accounts for converting the kinetic energy (velocity) of the stream inside the exit to the potential energy level (pressure) of the atmosphere outside the exit.

Good luck,
Latexman

Latexman,

Thanks for your reply. However if there is no tailpipe on the PSV and the valve is reliefing directly to atmosphere, then the backpressure is considered zero.
Hence by adding a tailpipe I'm not sure that adding the exit loss to the pressure loss from the pipe run and elbows is correct. The exit loss has just been transferred from the discharge flange of the PSV to the discharge of the tailpipe.

Due to a low PSV setting, the backpressure calculation is important and I wonder if anybody else has come up against this problem.

Saying the backpressure with a PSV open to atmosphere is 0.0 isn't correct in my opinion though people may just write that down on the data sheet. To get the flow from the outlet of the PSV to the atmosphere takes a positive dP which implies 'some' back pressure on the PSV.

Now, for most applications even with a conventional PSV that allows you up to 10% backpressure, it's very likely a moot point.

I either include the exit loss or if I have chocked flow through the outlet piping, I'll calculate the outlet pressure that gives me sonic flow and then calculate the pressure back at the PSV outlet.

Perth1,

There may be a basis for you to exclude the exit loss when piping is discharging to atmosphere if you are relieving a compressible fluid.

There used to be a very good website that covered compressible flow and had a good discussion about your question but I've had problems accessing it lately. Here's the actual link that discusses your question..

Just in case, here's cached version of it....

A similar discussion can be found in the "Guidelines for Pressure Relief and Effluent Handling Systems" which is an AICHE, CCPS publication. Here's an excerpt...

"The pressure change at the exit (change from station 2 to station 3) is not included in the calculations on this section for the following reasons:
- If the flow is choked at station 2, P3 cannot be computed from a value of
P2 (see para 2.10.2 for a discussion of the pressure discontinuity across a
choke point).
- If the flow is not choked and the pipe discharges to a large reservoir (or
to a same-size section of piping), P2 = P3 and no computations across the
exit are required (exit loss and velocity head recovery are equal).
In contrast to the above conditions, a pressure change at the exit must be determined if the pipe connects to a downstream piping run of a smaller size (reduction). The pressure change must also be determined if the pipe connects to a downstream piping run of larger size (expansion), if the flow is not choked at station 2."

I would suggest you get the AICHE publication and review the complete section on the referenced subject.

Typically, I have always accounted for an exit loss in the outlet piping when discharging to atmosphere and it is only recently that I have found the above references. For new designs, I'll continue to leave the exit loss in the calculation just as an additional allowance for piping runs that may not be well defined. For checking existing systems, certainly no harm to include the exit loss if pressure drop is not a problem. If it is a problem, I may have to think more about that since old habits die hard.

SME sect VIII has a non-mandatory appendix that includes explicit calculation methods for computing the relief valve backpressure.

Do they include the exit loss or not?

Good luck,
Latexman

You can include the exit loss, by your determination of the total fL/d loss for the piping , elbows, entrance , exit losses. The curves that are porvided for determining the pressure ratio are based on the Bectel Lios paper ( whose objective was to prevent blowback in the drippan) and the use of Fanno relationships for flow that is choked.

Just for clarification, ASME Section VIII does have a non-mandatory Appendix M which gives some general design criteria for relief valve piping but I believe it is the Power Piping Code, ASME B31.1 to which Davefitz is referring. ASME B31.1 has a non-mandatory Appendix II that goes into detail as to how to calculate the backpressure in the outlet piping and uses the fanno line approach to solve.

ASME B31.1 does include an example problem for a relief valve which discharges steam through a typical atmospheric vent type arrangement. The vent arrangement baically has the valve discharge connected to an elbow then a short section of pipe which is then inserted (not connected) into a larger, straight run pipe. For the example shown, it appears that sonic flow is attained at the ends of both sections of pipe diameters and it does not look like an exit loss is counted which would be consistent with references I've cited and what TD2K has mentioned about sonic flow.

Consdider incompressible flow or low Mach numbers
H=p/rho subscript ex exit
amb ambient

Hex + Vex^2/2 = Hamb +Vamb^2/2 + losses (1)

Losses are based on higher velocity = K*Vex^2/2
For sudden exp to ambient K=1

Eq (1) becomes
Hex + Vex^2/2 = Hamb +Vamb^2/2 + 1* Vex^2/2 (2)
or
Hex - Hamb = Vamb*2/2
Vamb approx =0 or much lower than V ex

Therefore back pressure = exit pressure.

K=1 may be derived from energy and momentum equations.

Equation (1) and equation (2) are the same equation, just restated. With one equation and one unknown (losses or 1*Vex^2/2), on the second pass through the equation you got a trivial solution, Vamb = 0, which was your assumption on the first pass. I don't see how this proves exit pressure = ambient pressure.

Good luck,
Latexman

Looking at both the web reference and the AICHE Guidelines I mentioned before, it appears that both are saying that an exit term should be included for incompressible flow which would include liquids or gas but gas only at low pressure drop/velocity. I'm not able at the moment but for what it is worth, I'll try to get some more info out of the AICHE Guidelines for reference and post it.

Latexman (Chemical)

Consdider incompressible flow or low Mach numbers into a large flow area called the ambient.
H=p/rho subscript ex exit
amb ambient
A= area
Hex + Vex^2/2 = Hamb +Vamb^2/2 + losses (1)

consveration of mass with rho approx const.
With
Aex*Vex =Aamb*Vamb
(1) becomes
Hex + Vex^2/2= Hamb +(Aex/Amb)^2*Vex^2 + K*Vex^2

Hex = Hamb +[(Aex/Amb)^2-1]*Vex^2 + K*Vex^2

With the area of the exit much smaller than the area of the ambient one obtains the result

Hex=Hamb + (K-1)*Vex^2

For discharge to the atmosphere, K=1
and Hex=Hamb

Don't know if this will help any but here's a link to some excerpts of the AICHE Guidelines for those who are interested.

I'm trying a little trick I learned from TD2K and this is the first time to try so I hope it works okay. The link is only good for 7 days.

It worked EGT01

Latexman, EGT01 and others.
Are there any disagreements/agreements or further comments on my post of Backpressure=Pambient=Pdischarge pressure?

Sailoday28,

I'm still trying to come to an understanding of what I've read in the references cited above and I may have misinterpreted the references to say that you should include an exit "fitting" for incompressible fluids or compressible fluids at low velocity.

From typical text book presentations, the exit loss is equal to the velocity head just before the expansion.

As I read from the other references cited above, when the pressure recovery equals the exit loss, then you can ignore the exit as a "fitting" in the piping system.

Looking at page 269 and 270 of the AICHE Guidelines, there is usually a pressure rise across an expansion but some designers follow the practice of taking no credit for a possible pressure increase on expansion.

In another section of the AICHE Guidelines, not included in the sections I posted, there was mention of accounting for the pressure rise across expansions in cases when the designer wanted to avoid overdesigning the outlet piping. This was specifically mentioned when an expansion fitting was used directly at the PSV outlet flange.

Sailoday28,

I accept you proved K = 1. I cannot see where exit pressure = ambient pressure is proven though. I do know from many textbooks on compressible flow they say exit pressure = ambient pressure if the flow is from subsonic to just sonic at the exit.

Good luck,
Latexman

EGT01 (Chemical)states
From typical text book presentations, the exit loss is equal to the velocity head just before the expansion."
From my previous postings, K*Vamb^2/2 = Vexit^2/2
I agree and with that statement

Hex=Hamb +Vamb^2/2 and the velocity in the much large area will be very small. Hence Hex=Hamb

Generall the losses in a reducer/expander are based on the higher velocity. Therefore losses to atmosphere (where area is large compared to exit)
are
K*Vexit^2/2=Vexit^2/2 and K=1



Sorry for harping on this.

We haven't heard from Perth1 so I'm not sure if any of this is any help for them but I appreciate the discussion.

I may be finally understanding how all this relates to the condition that when the pressure recovery equals the exit loss, the pressure at the end of the pipe is equal to that outside the pipe.

To take it from another approach, look at equation 18 in the linked pipe%20flow2.pdf,
h[sub] L[/sub] = (u[sub]1[/sub][sup]2[/sup] - u[sub]2[/sub][sup]2[/sup])/2g - (p[sub]2[/sub] - p[sub]1[/sub])/[ρ]g

and rearrange
(p[sub]2[/sub] - p[sub]1[/sub])/[ρ]g = (u[sub]1[/sub][sup]2[/sup] - u[sub]2[/sub][sup]2[/sup])/2g - h[sub] L[/sub]

If all there is between points 1 and 2 is a pipe exit, I believe the right side of the equation represents the pressure recovery and exit loss respectively.

then for h[sub] L[/sub] representing the head loss for the exit, and as defined otherwise
h[sub] L[/sub] = K * u[sub]1[/sub][sup]2[/sup]/2g
and for an exit loss K = 1
h[sub] L[/sub] = u[sub]1[/sub][sup]2[/sup]/2g

Then for (u[sub]1[/sub][sup]2[/sup] - u[sub]2[/sub][sup]2[/sup])/2g representing the pressure recovery and if we use the relationship
u[sub]2[/sub] = u[sub]1[/sub]*A[sub]1[/sub]/A[sub]2[/sub] and substitute

(p[sub]2[/sub] - p[sub]1[/sub])/[ρ]g = (u[sub]1[/sub][sup]2[/sup] - u[sub]1[/sub][sup]2[/sup]*A[sub]1[/sub][sup]2[/sup]/A[sub]2[/sub][sup]2[/sup])/2g - u[sub]1[/sub][sup]2[/sup]/2g
or
(p[sub]2[/sub] - p[sub]1[/sub])/[ρ]g = u[sub]1[/sub][sup]2[/sup]*(1 - A[sub]1[/sub][sup]2[/sup]/A[sub]2[/sub][sup]2[/sup])/2g - u[sub]1[/sub][sup]2[/sup]/2g

In the case that A[sub]2[/sub] is very large compared to A[sub]1[/sub], then A[sub]1[/sub][sup]2[/sup]/A[sub]2[/sub][sup]2[/sup] ~ 0 and
(p[sub]2[/sub] - p[sub]1[/sub])/[ρ]g = u[sub]1[/sub][sup]2[/sup]*(1 -0)/2g - u[sub]1[/sub][sup]2[/sup]/2g
or the pressure recovery equals the exit head loss and
(p[sub]2[/sub] - p[sub]1[/sub])/[ρ]g = 0 or
p[sub]2[/sub] = p[sub]1[/sub]

Then it would seem for any discharge to atmosphere, you can neglect the exit fitting loss if you choose to account for the pressure recovery. If you ignore the pressure recovery, there will be a difference in pressure between the end of the pipe and outside the pipe which will be equal to the head loss due to the exit fitting.

.....Right?