Wetted Surface Area In Fire Psv Sizing
#1
Posted 26 October 2009  12:39 AM
I am sizing a PSV for fire case for which I need to calculate wetted surface area of vessel. I am confused what does it mean here? I have found equation for horizontal vessels but i can’t find any equation for vertical vessels. Can anybody help me? Thanks
#2
Posted 26 October 2009  09:33 AM
Hi,
I am sizing a PSV for fire case for which I need to calculate wetted surface area of vessel. I am confused what does it mean here? I have found equation for horizontal vessels but i can't find any equation for vertical vessels. Can anybody help me? Thanks
P. Engr,
Here is the equation for a vertical vessel with 2:1 ellipsoidal head
A_{t} = A_{ev} + A_{cv} = (1.2*Pi*D^{2}/4) + (Pi*D*H) Where
A_{t} = total wetted area of vertical vessel A_{ev} = wetted area of bottom ellipsoidal head
A_{cv} = wetted area of cylindrical portion
D = diameter of vessel
H = Depth (Height) of liquid in cylindrical portion Use consistent units. I am using this equation for our fire case relief valve sizing for vertical vessels.
This formula can be traced from: "Chapter 7, Page 452, Applied Process Design for Chemical & Petrochemical Plants" by E. Ludwig
Hope this helps.
Regards,
Ankur
#3
Posted 26 October 2009  10:46 PM
Thanks for taking interest. The equation you have provided seems very simple. Take a look to the equation I was using for horizontal vessels.
A=Awet, cyl=2LR x COS1[(RH)/R]
Awet, head= (2.178/2п) (2R) 2 COS1(RH/R)
Where,
Awet, cyl = wetted surface area of the cylindrical portion of vessel, m2
Awet, head = wetted surface area of one elliptical head, m2
L = tangenttotangent length of the cylindrical section of the vessel, m
R = vessel inside radius, m (also equal to D/2 where D is the vessel inside diameter)
H = maximum liquid depth, m
It’s a complex equation. Have you ever used this equation for horizontal vessels?
#4
Posted 26 October 2009  11:15 PM
Ankur,
Thanks for taking interest. The equation you have provided seems very simple. Take a look to the equation I was using for horizontal vessels.
A=Awet, cyl=2LR x COS1[(RH)/R]
Awet, head= (2.178/2п) (2R) 2 COS1(RH/R)
Where,
Awet, cyl = wetted surface area of the cylindrical portion of vessel, m2
Awet, head = wetted surface area of one elliptical head, m2
L = tangenttotangent length of the cylindrical section of the vessel, m
R = vessel inside radius, m (also equal to D/2 where D is the vessel inside diameter)
H = maximum liquid depth, m
It's a complex equation. Have you ever used this equation for horizontal vessels?
Hi,
I thought your question was for vertical vessels. For horizontal vessels with 2:1 ellipsoidal post I have prepared an excel sheet which is posted at the following link:
http://webwormcpt.bl...eareafor.html
This should solve your query.
Regards,
Ankur.
#5
Posted 26 October 2009  11:55 PM
I thought your question was for vertical vessels. For horizontal vessels with 2:1 ellipsoidal post I have prepared an excel sheet which is posted at the following link:
http://webwormcpt.bl...eareafor.html
This should solve your query.
Regards,
Ankur.
[/quote]
My question was definitely for vertical vessels but i was asking for simpler equation for horizontal vessels like your equation for vertical vessels.
#6
Posted 16 November 2009  04:13 PM
I tried to find the reference you made in your response earlier to page 452 of the book by Ludwig. I was unable to find those equations or anything in reference to that topic in that book. I am also working on calculating surface area for hemispherical heads and spherical tanks. Do you or anyone else know if the following is correct:
FOR Vertical Vessel: A(wet,hemispherical head) = 0.5*PI*D^2
FOR Horizontal Vessel: A(wet,hemispherical head) = 0.5*PI*H*D
FOR Spherical Vessel: A(wet) = PI*D*H
Thanks!
Abby
#7
Posted 18 November 2009  03:16 AM
Ankur,
I tried to find the reference you made in your response earlier to page 452 of the book by Ludwig. I was unable to find those equations or anything in reference to that topic in that book. I am also working on calculating surface area for hemispherical heads and spherical tanks. Do you or anyone else know if the following is correct:
FOR Vertical Vessel: A(wet,hemispherical head) = 0.5*PI*D^2
FOR Horizontal Vessel: A(wet,hemispherical head) = 0.5*PI*H*D
FOR Spherical Vessel: A(wet) = PI*D*H
Thanks!
Abby
Abby,
The wetted area & vessel section volume questions have been debated so many times on the forum that I now feel that some sort of attachment giving information on this matter has become an absolute necessity. The attachment is a relevant page from the "Pressure Vessel Design Manual" by D.R. Moss for volumes & surface areas of vessel sections.
Hopefully this should go a long way in resolving the frequent questions on volumes & surface areas.
Regards,
Ankur
Attached Files
#8
Posted 18 November 2009  08:51 AM
That was definitely helpful although I am working on surface areas of partially filled vessels (wetted area). Your other postings for a horizontal vessel with 2:1 Ellipsoidal Head was a great help. I have been searching for days and this task has been more difficult than I anticipated. I found an equation from an article for calculating the surface area for a hemispherical head of a partially filled vessel (A=PI*R*H), but I have unable to verify that from any other source.
Thanks.
Abby
#9
Posted 24 May 2012  11:37 PM
Please help me up. thanks in advace.
Saad
#10
Posted 25 May 2012  12:10 AM
_{my calculation : 13.04 m2}
_{Breizh}
Edited by breizh, 25 May 2012  12:19 AM.
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