PSV Discharge Piping and Backpressure 2

[Mech85]

We have an existing conventional pressure safety valve which relieves vapour to atmosphere (0 kPa gauge). The PSV is being checked for use for a new operating scenario. The PSV has a tailpipe (discharge piping) consisting of elbows and straight pipe run.

In determining the pressure drop ( ie back pressure ) for the PSV, I am unsure whether the exit loss from the discharge piping ( ie (k* v^2)/g) should be included or excluded. Section 5.4.1.3 of API 520 does not seem to clarify the issue.

Can anyone comment?

 

[sailoday28]

EGT01 (Chemical)"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?"
I believe what you are stating is correct.

In calculating a system resistance or equivalent K or L/d

IF the final downstream component is the atmosphere (a large volume and flow area), then if one is using gage pressures P/rho and V^2/2 should be taken as zero.

IF instead, one desires to use the exit area of the last fitting upstream of the (large area) atmosphere, then use at the exit plane P/rho=0 and the correct exit velocity.
Either approach should give the same answers.
Regards

 

[Mech85]

Gee, this query certainly started some good discussions! Thanks to everyone for all the input.

EGT01, thanks for the link to the AICHE guidelines. It is quite useful. In discussions with colleagues here, there is a tendecy to agree with ignoring the exit loss as we are discharging to atmosphere and only account for the pipe and fittings friction loss. Your posts seem to concur with this.

 

[sailoday28]

My original post stated K=1 may be derived from momentum equation.

Reference "Elementary Mechanics of Fluids--By Hunter Rouse, John Wiley and Sons
---Pg 259

I have modified his equations
H=P/rho
A1= flow area inlet
A2 flow area outlet
V velocity


Q=A1*V1=A2*V2
MOMENTUM EQUATION H1*A2-H2*A2 =Q(V2-V1)=A1*V1(V2-V1)
H1-H2=A1/A2*V1^2( A1/A2 -1)
(H1-H2)/(V1^2/2)= 2(A1/A2)[ (A1/A2)-1)

substitue in energy equation and K=1 when A1/A2 approaches zero