Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations KootK on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Leakage class / self-taught female / please help :) thank you 2

Status
Not open for further replies.

Allets

Industrial
Dec 22, 2023
5
Hi Team ! Glad to meet you all and hope I won't disturb you a lot.
I don’t even have a technical education, but I'm in a situation where I was forced to understand the leak class and calculation method myself. I wasted my job vacation for this lol and really hope and beg you to help, thank you ! I believe I'll be able to help with something else too.
I found Standards ANSI/FCI 70-2 1976(R1982) and EN 60534 which regulate this and this is what I got:

Initial data:
Control 2-way valve (regulator), size 3'', flow action: open
P1 = 5 bar
P2 = 2.5 bar / 2.1 bar
Rated CV = 75
Flow rate = 79,68 m3/h (or 70,92 m3/h)
FL: 0,9
Sound level: 85dB (A)
Question: if the Leakage class II is allowable ?

Calculation in accordance with EN 60534 for water:
1. Check for flow restriction
ΔP = P1-P2 = 2.5 bar
ΔPchoked = FL2 x (P1 + 0.99)
ΔPchoked = 0.9² x (5 + 0.99) = 4.8519 bar
Check: If the pressure difference ΔP of the test is smaller than the flow restricted pressure difference ΔPchoked, ΔP is used for the calculation, otherwise use ΔPchoked
So, we use ΔP (2.5 bar)
2. Calculate rated capacity Q [m3/h] in accordance with EN 60534-2-1
Q = CV x √ΔPsizing
Q = 75 x 1.5811 = 118,5825 m3/h

3. Multiply the rated capacity Q by the leakage factor in accordance with EN 60534-4
Factor for Leakage class II is 0.005
118,5825 x 0.005 = 0,5929125

0,5929125 is exactly 0.5% !

And Maximum Leakage Allowable for Leakage class II is not more than 0.5%.
Is Leakage class II is enough for the valve ?

Thank you !
 
 https://files.engineering.com/getfile.aspx?folder=c9089808-2a28-4b5f-a5eb-81e25a2734e4&file=WP-Leakage-calculation-in-accordance-with-standards-2020.pdf
Replies continue below

Recommended for you

I think you are asking if the leakage flow is OK for Class II.

Allowed Leakage limits are summarised here for all classes.

You have calculated the valve's flow rate and the allowable Class II leakage rate based on that value as 0.005 x Flow = 118m3/h

What I do not see is the actual leakage rate of the valve, which apparently should be the result of a leak test. If you do not know the actual leakage flow rate, you cannot compare it to what is allowed, so we don't know if the valve passes the test or not.

If this is not to compare with valve test data, then I assume there is a maximum leak rate limit derived from a buyer's valve specification, or somewhere else, but I don't see that either. So again, we cannot determine if the Class II allowable leakage rate meets the specified rate without knowing what that spec says is the maximum leakage it permits. We also do not know if a Class II valve is what is required.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
 
Thank you for your help and detailed explanation. Unfortunately, the problem is that we don't have any detailed specification for this valve. Anyways, you response is highly helpful !
 
Welcome.

Yes. The only thing we know is that the limiting leakage for Class II is 0.005 x Flow.


Enjoy your vacation.



--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
 
Most designs don't rely on a control valve sealing, even if specified as tight shut off due to the wear on the control surface. If you want isolation use an isolation valve.

Any leakage is still a leakage so not really sure what effect a class 2 valve has on your system.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
Thank you for the tip LittleInch,
This is a good idea on an isolation valve. Our valve is just VPR-2 (GE tag)
I believe, this part of my so called calculation is total mess [bigsmile]

3. Multiply the rated capacity Q by the leakage factor in accordance with EN 60534-4
Factor for Leakage class II is 0.005
118,5825 x 0.005 = 0,5929125
0,5929125 is exactly 0.5% !

Because any amount multiplied by 0.005 will ultimately produce this result - 0.5%

I'm so silly.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor